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Sharp ill-posedness and well-posedness results for the KdV-Burgers equation: the real line case
Luc Molinet, Stéphane Vento
To cite this version:
Luc Molinet, Stéphane Vento. Sharp ill-posedness and well-posedness results for the KdV-Burgers
equation: the real line case. Annali della Scuola Normale Superiore di Pisa, Classe di Scienze, Scuola
Normale Superiore 2011, 10 (3), pp.531-560. �hal-00436652v2�
Sharp ill-posedness and well-posedness results for the KdV-Burgers equation: the real line case
Luc Molinet and St´ ephane Vento
Abstract. We complete the known results on the Cauchy problem in Sobolev spaces for the KdV-Burgers equation by proving that this equation is well-posed in H
−1( R ) with a solution-map that is analytic from H
−1( R ) to C([0, T ]; H
−1( R )) whereas it is ill-posed in H
s( R ), as soon as s < −1, in the sense that the flow-map u
07→ u(t) cannot be continuous from H
s( R ) to even D
′( R ) at any fixed t > 0 small enough. As far as we know, this is the first result of this type for a dispersive-dissipative equation. The framework we develop here should be useful to prove similar results for other dispersive- dissipative models.
1 Introduction and main results
The aim of this paper is to establish positive and negative optimal results on the local Cauchy problem in Sobolev spaces for the Korteweg-de Vries- Burgers (KdV-B) equation posed on the real line :
u
t+ u
xxx− u
xx+ uu
x= 0 (1.1) where u = u(t, x) is a real valued function.
This equation has been derived as an asymptotic model for the propagation of weakly nonlinear dispersive long waves in some physical contexts when dissipative effects occur (see [17]). It thus seems natural to compare the well-posedness results on the Cauchy problem for the KdV-B equation with the ones for the Korteweg-de-Vries (KdV) equation
u
t+ u
xxx+ uu
x= 0 (1.2)
that correspond to the case when dissipative effects are negligible and for the dissipative Burgers (dB) equation
u
t− u
xx+ uu
x= 0 (1.3)
that corresponds to the case when dissipative effect are dominant.
To make this comparison more transparent it is convenient to define different notions of well-posedness (and consequently ill-posedness) related to the smoothness of the flow-map (see in the same spirit [13], [8]).
Throughout this paper we shall say that a Cauchy problem is (locally) C
0-well-posed in some normed function space X if, for any initial data u
0∈ X, there exist a radius R > 0, a time T > 0 and a unique solu- tion u, belonging to some space-time function space continuously embedded in C([0, T ]; X), such that for any t ∈ [0, T ] the map u
07→ u(t) is continuous from the ball of X centered at u
0with radius R into X. If the map u
07→ u(t) is of class C
k, k ∈ N ∪ {∞}, (resp. analytic) we will say that the Cauchy is C
k-well-posed (resp. analytically well-posed). Finally a Cauchy problem will be said to be C
k-ill-posed, k ∈ N ∪ {∞}, if it is not C
k-well-posed.
For the KdV equation on the line the situation is as follows: it is an- alytically well-posed in H
−3/4( R ) (cf. [14] and [11] for the limit case) and C
3-ill-posed below this index
1(cf. [4]). On the other hand the results for the dissipative Burgers equation are much clear. Indeed this equation is known to be analytically well-posed in H
s( R ) for s ≥ −1/2 (cf [7] and [1] for the limit case) and C
0ill-posed in H
sfor s < −1/2 (cf. [7] ). At this stage it is interesting to notice that the critical Sobolev exponents obtained by scaling considerations are respectively −3/2 for the KdV equation and −1/2 for the dissipative Burgers equation. Hence for the KdV equation there is an im- portant gap between this critical exponent and the best exponent obtained for well-posedness.
Now, concerning the KdV-Burgers equation, Molinet and Ribaud [16]
proved that this equation is analytically well-posed in H
s( R ) as soon as s > −1. They also established that the index −1 is critical for the C
2- well-posedness. The surprising part of this result was that, according to the above results, the C
∞critical index s
∞c(KdV B) = −1 was lower that the one of the KdV equation s
∞c(KdV ) = −3/4 and also lower than the C
∞index s
∞c(dB) = −1/2 of the dissipative Burgers equation.
In this paper we want in some sense to complete this study by proving that the KdV-Burgers equation is analytically well-posed in H
−1( R ) and C
0-ill-posed in H
s( R ) for s < −1 in the sense that the flow-map defined on H
−1( R ) is not continuous for the topology inducted by H
s, s < −1, with values even in D
′( R ). It is worth emphasizing that the critical index s
0c= −1 is still far away from the critical index s
c= −3/2 given by the scaling symmetry of the KdV equation. We believe that this result strongly
1See also [5] where it is proven that the solution-map is even not uniformly continuous on bounded sets below this index
suggest that the KdV equation should also be C
0-ill-posed in H
s( R ) for s < −1.
To reach the critical Sobolev space H
−1( R ) we adapt the refinement of Bourgain’s spaces that appeared in [20] and [19] to the framework de- veloped in [16]. One of the main difficulty is related to the choice of the extension for negative times of the Duhamel operator (see the discussion in the beginning of Section 4). The approach we develop here to overcome this difficulty should be useful to prove optimal results for other dispersive- dissipative models. The ill-posedness result is due to a high to low frequency cascade phenomena that was first observed in [2] for a quadratic Schr¨ odinger equation..
At this stage it is worth noticing that, using the integrability theory, it was recently proved in [13] that the flow-map of KdV equation can be uniquely continuously extended in H
−1( T ). Therefore, on the torus, KdV is C
0-well-posed in H
−1if one takes as uniqueness class, the class of strong limit in C([0, T ]; H
−1( T )) of smooth solutions. In the present work we use in a crucial way the global Kato smoothing effect that does not hold on the torus. However, in a forthcoming paper ([15]) we will show how one can modify the approach developed here to prove that the same results hold on the torus, i.e. analytic well-posedness in H
−1( T ) and C
0-ill-posedness in H
s( T ) for s < −1. In view of the result of Kappeler and Topalov for KdV it thus appears that, at least on the torus, even if the dissipation part of the KdV-Burgers equation (it is important to notice that the dissipative term −u
xxis of lower order than the dispersive one u
xxx) allows to lower the C
∞critical index with respect to the KdV equation, it does not permit to improve the C
0critical index .
Our results can be summarized as follows:
Theorem 1.1. The Cauchy problem associated to (1.1) is locally analyti- cally well-posed in H
−1( R ). Moreover, at every point u
0∈ H
−1( R ) there exist T = T(u
0) > 0 and R = R(u
0) > 0 such that the solution-map u
07→ u is analytic from the ball centered at u
0with radius R of H
−1( R ) into C([0, T ]; H
−1( R )). Finally, the solution u can be extended for all posi- tive times and belongs to C( R
∗+; H
∞( R )).
Theorem 1.2. The Cauchy problem associated to (1.1) is ill-posed in H
s( R ) for s < −1 in the following sense: there exists T > 0 such that for any 0 <
t < T , the flow-map u
07→ u(t) constructed in Theorem 1.1 is discontinuous
at the origin from H
−1( R ) endowed with the topology inducted by H
s( R )
into D
′( R ).
Acknowlegements: L.M. was partially supported by the ANR project
”Equa-Disp”.
2 Ill-posedness
The ill-posedness result can be viewed as an application of a general result proved in [2]. Roughly speaking this general ill-posedness result requires the two following ingredients:
1. The equation is analytically well-posed until some index s
∞cwith a solution-map that is also analytic.
2. Below this index one iteration of the Picard scheme is not continuous.
The discontinuity should be driven by high frequency interactions that blow up in frequencies of order least or equal to one.
The first ingredient is given by Theorem 1.1 whereas the second one has been derived in [16] where the discontinuity of the second iteration of the Picard scheme in H
s( R ) and H
s( T ) for s < −1 is established.
However, due to the nature of the equation, our result is a little better than the one given by the general theory developed in [2]. Indeed, we will be able to prove the discontinuity of the flow-map u
07→ u(t) for any fixed t > 0 less than some T > 0 and not only of the solution-map u
07→ u. Therefore for sake of completeness we will prove the result with hand here.
Let us first recall the counter-example constructed in [16] that we renor- malize here in H
−1( R ). We define the sequence of initial data {φ
N}
N≥1by
φ ˆ
N= N
−1χ
IN(ξ) + χ
IN(−ξ)
, (2.1)
where I
N= [N, N + 2] and ˆ φ
Ndenotes the space Fourier transform of φ
N. Note that kφ
Nk
H−1(R)∼ 1 and φ
N→ 0 in H
s( R ) for s < −1. This sequence yields a counter-example to the continuity of the second iteration of the Picard Scheme in H
s( R ), s < −1, that is given by
A
2(t, h, h) = Z
t0
S(t − t
′)∂
x[S(t
′)h]
2dt
′where S is the semi-group associated to the linear part of (1.1) (see (3.2)).
Indeed, computing the space Fourier transform we get F
x(A
2(t, φ
N, φ
N))(ξ) =
Z
R
e
−tξ2e
itξ3φ ˆ
N(ξ
1) ˆ φ
N(ξ − ξ
1) (iξ)
Z
t0
e
−(ξ21+(ξ−ξ1)2−ξ2)t′e
i(ξ31+(ξ−ξ1)3−ξ3)t′dt
′dξ
1= (iξ) e
itξ3e
−tξ2Z
R
φ ˆ
N(ξ
1) ˆ φ
N(ξ − ξ
1)
e
−(ξ12+(ξ−ξ1)2−ξ2)te
i3ξξ1(ξ−ξ1)t− 1
−2ξ
1(ξ − ξ
1) + i3ξξ
1(ξ − ξ
1) dξ
1, so that
kA
2(t, φ
N, φ
N)k
2Hs≥ Z
1/2−1/2
(1 + |ξ|
2)
s|F
x(A
2(t, φ
N, φ
N))(ξ)|
2dξ
= N
4Z
1/2−1/2
(1 + |ξ|
2)
s|ξ|
2Z
Kξ
e
−(ξ21+(ξ−ξ1)2)te
i3ξξ1(ξ−ξ1)t− e
−ξ2t−2ξ
1(ξ − ξ
1) + i3ξξ
1(ξ − ξ
1) dξ
12
dξ ,
where
K
ξ= {ξ
1/ ξ − ξ
1∈ I
N, ξ
1∈ −I
N} ∪ {ξ
1/ ξ
1∈ I
N, ξ − ξ
1∈ −I
N} . Note that for any ξ ∈ [−1/2, 1/2], one has mes(K
ξ) ≥ 1 and
3ξξ
1(ξ − ξ
1) ∼ N
22ξ
1(ξ − ξ
1) ∼ N
2, ∀ξ
1∈ K
ξ. Therefore, fixing 0 < t < 1 we have
Re (e
−(ξ21+(ξ−ξ1)2)te
i3ξξ1(ξ−ξ1)t− e
−ξ2t) ≤ −e
−t/4+ e
−2(N+2)2t, which leads for N = N (t) > 0 large enough to
Z
Kξ
e
−(ξ21+(ξ−ξ1)2)te
i3ξξ1(ξ−ξ1)t− e
−ξ2t−2ξ
1(ξ − ξ
1) + i3ξξ
1(ξ − ξ
1) dξ
1≥ C e
−t/4N
2and thus
kA
2(t, φ
N, φ
N)k
2Hs≥ Ce
−t/4≥ C
0(2.2)
for some positive constant C
0> 0. Since φ
N→ 0 in H
s( R ), for s < −1, this ensures that, for any fixed t > 0, the map u
07→ A
2(t, u
0, u
0) is not continuous at the origin from H
s( R ) into D
′( R ).
Now, we will use that A
2(t, φ
N, φ
N) is of order at least one in H
s( R ) to prove that somehow A
2(t, εφ
N, εφ
N) is the main contribution to u(t, εφ
N) in H
s( R ) as soon as s < −1, ε > 0 is small and N is large enough. The discontinuity of u
07→ u(t) will then follow from the one of u
07→ A
2(t, u
0, u
0).
According to Theorem 1.1 there exist T > 0 and ε
0> 0 such that for any
|ε| ≤ ε
0, any khk
H−1(R)≤ 1 and 0 ≤ t ≤ T , u(t, εh) = εS(t)h +
+∞
X
k=2
ε
kA
k(t, h
k)
where h
k:= (h, . . . , h), h
k7→ A
k(t, h
k) is a k-linear continuous map from
H
−1( R )
kinto C([0, T ]; H
−1( R )) and the series converges absolutely in C([0, T ]; H
−1( R )).
In particular,
u(t, εφ
N) − ε
2A
2(t, φ
N, φ
N) = εS(t)φ
N+
+∞
X
k=3
ε
kA
k(t, φ
kN) . On the other hand, kS(t)φ
Nk
Hs(R)≤ kφ
Nk
Hs(R)∼ N
1+sand
X
∞ k=3ε
kA
k(t, φ
kN)
H−1
≤ ε ε
0 3X
∞ k=3ε
k0kA
k(t, φ
N)k
H−1≤ Cε
3. Hence, for s < −1,
sup
t∈[0,T]
u(t, εφ
N) − ε
2A
2(t, φ
N, φ
N)
Hs(R)
≤ Cε
3+ O(N
1+s) .
In view of (2.2) this ensures that, fixing 0 < t < 1 and taking ε small enough and N large enough, ε
2A
2(t, φ
N, φ
N) is a “good” approximation of u(t, εφ
N). In particular, taking ε ≤ C
0C
−1/4 we get
ku(t, εφ
N)k
Hs(R)≥ C
0ε
2/2 + O(N
1+s) .
Since u(t, 0) ≡ 0 and φ
N→ 0 in H
s( R ) for s < −1 this leads to the
discontinuity of the flow-map at the origin by letting N tend to infinity. It
is worth noticing that since φ
N⇀ 0 in H
−1( R ) we also get that u
07→ u(t, u
0)
is discontinuous from H
−1( R ) equipped with its weak topology with values
even in D
′( R ).
3 Resolution space
In this section we introduce a few notation and we define our functional framework.
For A, B > 0, A . B means that there exists c > 0 such that A ≤ cB.
When c is a small constant we use A ≪ B. We write A ∼ B to denote the statement that A . B . A. For u = u(t, x) ∈ S
′( R
2), we denote by u b (or F
xu) its Fourier transform in space, and u e (or F u) the space-time Fourier transform of u. We consider the usual Lebesgue spaces L
p, L
pxL
qtand abbreviate L
pxL
ptas L
p. Let us define the Japanese bracket hxi = (1+|x|
2)
1/2so that the standard non-homogeneous Sobolev spaces are endowed with the norm kfk
Hs= kh∇i
sf k
L2.
We also need a Littlewood-Paley analysis. Let η ∈ C
0∞( R ) be such that η ≥ 0, supp η ⊂ [−2, 2], η ≡ 1 on [−1, 1]. We define next ϕ(ξ) = η(ξ) −η(2ξ).
Any summations over capitalized variables such as N, L are presumed to be dyadic, i.e. these variables range over numbers of the form 2
ℓ, ℓ ∈ Z . We set ϕ
N(ξ) = ϕ(ξ/N ) and define the operator P
Nby F(P
Nu) = ϕ
Nb u. We introduce ψ
L(τ, ξ) = ϕ
L(τ − ξ
3) and for any u ∈ S
′( R
2),
F
x(P
Nu(t))(ξ) = ϕ
N(ξ)ˆ u(t, ξ), F(Q
Lu)(τ, ξ) = ψ
L(τ, ξ)˜ u(τ, ξ).
Roughly speaking, the operator P
Nlocalizes in the annulus {|ξ| ∼ N } whereas Q
Llocalizes in the region {|τ − ξ
3| ∼ L}.
Furthermore we define more general projection P
.N= P
N1.N
P
N1, Q
≫L= P
L1≫L
Q
L1etc.
Let e
−t∂xxxbe the propagator associated to the Airy equation and define the two parameters linear operator W by
F
x(W (t, t
′)φ)(ξ) = exp(itξ
3− |t
′|ξ
2) ˆ φ(ξ), t ∈ R . (3.1) The operator W : t 7→ W (t, t) is clearly an extension to R of the linear semi-group S(·) associated with (1.1) that is given by
F
x(S(t)φ)(ξ) = exp(itξ
3− tξ
2) ˆ φ(ξ), t ∈ R
+. (3.2) We will mainly work on the integral formulation of (1.1):
u(t) = S(t)u
0− 1 2
Z
t 0S(t − t
′)∂
xu
2(t
′)dt
′, t ∈ R
+. (3.3)
Actually, to prove the local existence result, we will apply a fixed point argu-
ment to the following extension of (3.3) (See Section 4 for some explanations
on this choice).
u(t) = η(t) h
W (t)u
0− 1 2 χ
R+(t)
Z
t 0W (t − t
′, t − t
′)∂
xu
2(t
′)dt
′− 1 2 χ
R−(t)
Z
t 0W (t − t
′, t + t
′)∂
xu
2(t
′)dt
′i
. (3.4) If u solves (3.4) then u is a solution of (3.3) on [0, T ], T < 1.
In [16], the authors performed the iteration process in the space X
s,bequipped with the norm
kuk
Xs,b= khi(τ − ξ
3) + ξ
2i
bhξi
suk e
L2which take advantage of the mixed dispersive-dissipative part of the equa- tion. In order to handle the endpoint index s = −1 without encountering logarithmic divergence, we will rather work in its Besov version X
s,b,q(with q = 1) defined as the weak closure of the test functions that are uniformly bounded by the norm
kuk
Xs,b,q= X
N
h X
L
hN i
sqhL + N
2i
bqkP
NQ
Luk
qL2 xti
2/q1/2.
This Besov refinement, which usually provides suitable controls for nonlinear terms, is not sufficient here to get the desired bound especially in the high- high regime, where the nonlinearity interacts two components of the solution u with the same high frequency. To handle these divergences, inspired by [19], we introduce, for b ∈ {
12, −
12}, the space Y
s,bendowed with the norm
kuk
Ys,b= X
N
[hN i
skF
−1[(i(τ − ξ
3) + ξ
2+ 1)
b+1/2ϕ
Ne u]k
L1tL2x
]
21/2,
so that
kuk
Y−1,12∼ X
N
[hN i
−1k(∂
t+ ∂
xxx− ∂
xx+ I)P
Nuk
L1tL2x
]
21/2.
Next we form the resolution space S
s= X
s,12,1+Y
s,12, and the ”nonlinear space” N
s= X
s,−12,1+ Y
s,−12in the usual way:
kuk
X+Y= inf{ku
1k
X+ ku
2k
Y: u
1∈ X, u
2∈ Y, u = u
1+ u
2}.
In the rest of this section, we study some basic properties of the function
space S
−1.
Lemma 3.1. For any φ ∈ L
2, X
L
[L
1/2kQ
L(e
−t∂xxxφ)k
L2]
21/2. kφk
L2. Proof. From Plancherel theorem, we have
X
L
[L
1/2kQ
L(e
−t∂xxxφ)k
L2]
21/2∼ k|τ − ξ
3|
1/2F(e
−t∂xxxφ)k
L2.
Moreover if we set η
T(t) = η(t/T ) for T > 0, then
F (η
T(t)e
−t∂xxxφ)(τ, ξ) = η c
T(τ − ξ
3) φ(ξ). b
Thus we obtain with the changes of variables τ − ξ
3→ τ
′and T τ
′→ σ that k|τ − ξ
3|
1/2F(η
T(t)e
−t∂xxxφ)k
L2. kφk
L2k|τ
′|
1/2T η(T τ b
′)k
L2τ′
. kφk
L2. Taking the limit T → ∞, this completes the proof.
Lemma 3.2. 1. For each dyadic N , we have k(∂
t+ ∂
xxx)P
Nuk
L1tL2x
. kP
Nuk
Y0,12
. (3.5) 2. For all u ∈ S
−1,
kuk
L2xt
. kuk
S−1. (3.6)
3. For all u ∈ S
0,
X
L
[L
1/2kQ
Luk
L2]
21/2. kuk
S0. (3.7) Proof. 1. From the definition of Y
0,12, the right-hand side of (3.5) can be
rewritten as kP
Nuk
Y0,12
= k(∂
t+ ∂
xxx− ∂
xx+ I )P
Nuk
L1tL2x
.
Thus, by the triangle inequality, we reduce to show (3.5) with ∂
t+∂
xxxreplaced by I − ∂
xx. Using Plancherel theorem as well as Young and H¨older inequalities, we get
k(I − ∂
xx)P
Nuk
L1tL2x
. F
t−1ξ
2+ 1
i(τ − ξ
3) + ξ
2+ 1 (i(τ − ξ
3) + ξ
2+ 1)ϕ
Nu e
L1tL2ξ
.
In the sequel, it will be convenient to write ϕ
Nfor ϕ
N/2+ ϕ
N+ ϕ
2N. With this slight abuse of notation, we obtain
k(I − ∂
xx)P
Nuk
L1tL2x.
F
t−1ϕ
N(ξ)(ξ
2+ 1) i(τ − ξ
3) + ξ
2+ 1
L1tL∞ξ
k(∂
t+ ∂
xxx− ∂
xx+ I )P
Nuk
L1tL2x
. On the other hand, a direct computation yields
F
t−1ϕ
N(ξ)(ξ
2+ 1) i(τ − ξ
3) + ξ
2+ 1
= Cϕ
N(ξ)(1 + ξ
2)e
−t(1+ξ2)χ
R+(t)
so that
F
t−1ϕ
N(ξ)(ξ
2+ 1) i(τ − ξ
3) + ξ
2+ 1
L1tL∞ξ
. khNi
2e
−thNi2χ
R+(t)k
L1t
. 1, and the claim follows.
2. We show that for any fixed dyadic N , we have
kP
Nuk
L2. kP
Nuk
S−1. (3.8) Estimate (3.6) then follows after square-summing. Observe that (3.8) follows immediately from the estimate hN i
−1hL + N
2i
1/2& 1 if the right-hand side is replaced by kP
Nuk
X−1,12,1
, so it suffices to prove (3.8) with kP
Nuk
Y−1,12
in the right-hand side. But applying again Young and H¨older’s inequalities, this is easily verified:
kP
Nuk
L2= F
t−11
i(τ − ξ
3) + ξ
2+ 1 (i(τ − ξ
3) + ξ
2+ 1)ϕ
Ne u
L2tξ
.
F
t−1ϕ
N(ξ) i(τ − ξ
3) + ξ
2+ 1
L2tL∞x
kP
Nuk
Y0,12
. ke
−thNi2χ
R+(t)k
L2t
kP
Nuk
Y0,12
. hNi
−1kP
Nuk
Y0,12
. kP
Nuk
Y−1,12
. 3. First it is clear from definitions that P
L
[L
1/2kQ
Luk
L2]
21/2. kuk
X0,12,1
. Setting now v = (∂
t+ ∂
xxx)u, we see that u can be rewritten as
u(t) = e
−t∂xxxu(0) + Z
t0
e
−(t−t′)∂xxxv(t
′)dt
′.
By virtue of Lemma 3.1, we have X
L
[L
1/2kQ
Le
−t∂xxxu(0)k
L2]
21/2. ku(0)k
L2. kuk
L∞t L2x.
Moreover, we get as previously kuk
L∞t L2x.
F
t−11
i(τ − ξ
3) + ξ
2+ 1
L∞tξ
kuk
Y0,12
. kuk
Y0,12
. (3.9) Thanks to estimate (3.5), it remains to show that
X
L
h L
1/2Q
LZ
t0
e
−(t−t′)∂xxxv(t
′)dt
′L2
i
21/2. kvk
L1tL2x. (3.10) In order to prove this, we split the integral R
t0
= R
t−∞
− R
0−∞
. By Lemma 3.1, the contribution with integrand on (−∞, 0) is bounded by
.
Z
0−∞
e
t′∂xxxv(t
′)dt
′L2x
. kvk
L1tL2x
. For the last term, we reduce by Minkowski to show that
X
L
[L
1/2kQ
L(χ
t>t′e
−(t−t′)∂xxxv(t
′))k
L2tx
]
21/2. kv(t
′)k
L2x. This can be proved by a time-restriction argument. Indeed, for any T > 0, we have
X
L
[L
1/2kQ
L(η
T(t)χ
t>t′e
−(t−t′)∂xxxv(t
′))k
L2]
21/2. k|τ |
1/2b v(t
′)F
t(η
T(t)χ
t>t′)(τ )k
L2. kv(t
′)k
L2k|τ |
1/2F
t(η(t)χ
tT >t′)k
L2. kv(t
′)k
L2.
We conclude by passing to the limit T → ∞.
Now we state a general and classical result which ensures that our resolu- tion space is well compatible with dispersive properties of the Airy equation.
Actually, it is a direct consequence of Lemma 4.1 in [19] together with the
fact that the resolution space S
0used by Tao to solve 4-KdV contains our
space S
0thanks to estimate (3.5)
Lemma 3.3 (Extension lemma). Let Z be a Banach space of functions on R × R with the property that
kg(t)u(t, x)k
Z. kgk
L∞tku(t, x)k
Zholds for any u ∈ Z and g ∈ L
∞t( R ). Let T be a spacial linear operator for which one has the estimate
kT (e
−t∂xxxP
Nφ)k
Z. kP
Nφk
L2for some dyadic N and for all φ. Then one has the embedding kT (P
Nu)k
Z. kP
Nuk
S0.
Combined with the unitary of the Airy group in L
2and the sharp Kato smoothing effect
k∂
xe
−t∂xxxφk
L∞xL2t. kφk
L2, ∀φ ∈ L
2, (3.11) we deduce the following result.
Corollary 3.1. For any u, we have
1kuk
L∞t H−1x
. kuk
S−1, (3.12) kP
Nuk
L∞x L2t
. N
−1kP
Nuk
S0, (3.13) provided the right-hand side is finite. In particular, S
−1֒ → L
∞tH
−1.
4 Linear estimates
In this section we prove linear estimates related to the operator W as well as to the extension of the Duhamel operator introduced in (3.4).
At this this stage let us give some explanations on our choice of this extension. Let us keep in mind that this extension has to be compatible with linear estimates in both norms X
s,1/2,1and Y
s,1/2. First, since X
s,1/2,1is a Besov in time space we are not allowed to simply multiply the Duhamel term by χ
R+(t). Second, in order to prove the desired linear estimate in Y
s,1/2the strategy is to use that the Duhamel term satisfies a forced KdV-Burgers equation. Unfortunately, it turns out that the extension introduced in [16],
1Note that (3.12) can also be deduced from estimate (3.9).
that makes the calculus simple, does not satisfy such PDE for negative time. The new extension that we introduce in this work has the properties to satisfy some forced PDE related to KdV-Burgers for negative times (see (4.14)) and to be compatible with linear estimates in X
s,1/2,1. However the proof is now a little more complicated even if it follows the same lines than the one of Propositions 2.3 in [16], see also Proposition 4.4, [12].
The following lemma is a dyadic version of Proposition 2.1 in [16].
Proposition 4.1. For all φ ∈ H
−1( R ), we have
kη(t)W (t)φk
S−1. kφk
H−1. (4.1) Proof. We bound the left-hand side in (4.1) by the X
−1,12,1-norm of η(t)W (t)φ.
After square-summing in N , we may reduce to prove X
L
hL + N
2i
1/2kP
NQ
L(η(t)W (t)φ)k
L2xt
. kP
Nφk
L2(4.2) for each dyadic N . Using Plancherel, we obtain
X
L
hL + N
2i
1/2kP
NQ
L(η(t)W (t)φ)k
L2xt
. X
L
hL + N
2i
1/2kϕ
N(ξ)ϕ
L(τ ) φ(ξ)F b
t(η(t)e
−|t|ξ2)(τ )k
L2τ ξ
. kP
Nφk
L2X
L
hL + N
2i
1/2kϕ
N(ξ)P
L(η(t)e
−|t|ξ2)k
L∞ξ L2t
. Hence it remains to show that
X
L
hL + N
2i
1/2kϕ
N(ξ)P
L(η(t)e
−|t|ξ2)k
L∞ξ L2t
. 1. (4.3) We split the summand into L ≤ hN i
2and L ≥ hN i
2. In the former case, we get by Bernstein
X
L≤hNi2
hL + N
2i
1/2kϕ
N(ξ)P
L(η(t)e
−|t|ξ2)k
L∞ξ L2t
. X
L≤hNi2
hN iL
1/2sup
|ξ|∼N
kη(t)e
−|t|ξ2k
L1t
Also, one can bound kη(t)e
−|t|ξ2k
L1either by kηk
L1or by ke
−|t|ξ2k
L1t
∼ |ξ|
−2. It follows that
X
L≤hNi2
hL + N
2i
1/2kϕ
N(ξ)P
L(η(t)e
−|t|ξ2)k
L∞ξ L2t
. hN i
2min(1, N
−2) . 1.
Now we deal with the case L ≥ hN i
2. A standard paraproduct rearrange- ment allows us to write
P
L(η(t)e
−|t|ξ2) = P
LX
M&L
(P
Mη(t)P
.Me
−|t|ξ2+ P
.Mη(t)P
Me
−|t|ξ2= P
L(I) + P
L(II).
Using the Schur’s test, the term P
L(I ) is directly bounded by X
L≥hNi2
hL + N
2i
1/2kϕ
NP
L(I)k
L∞ξ L2t
. X
L
L
1/2X
M&L
kϕ
NP
Mη(t)k
L∞ξ L2t
kϕ
NP
.Me
−|t|ξ2k
L∞ξt. X
M
M
1/2kP
Mηk
L2t
. 1.
Similarly for P
L(II), we have X
L≥hNi2
hL + N
2i
1/2kϕ
NP
L(II)k
L∞ξ L2t
. X
L
L
1/2X
M&L
kϕ
NP
.Mη(t)k
L∞ξtkϕ
NP
Me
−|t|ξ2k
L∞ξ L2t
. X
M
M
1/2kϕ
NP
Me
−|t|ξ2k
L∞ξ L2t
.
Moreover, it is not too hard to check that if |ξ| ∼ N , then kP
Me
−|t|ξ2k
L2t
.
kP
Me
−|t|N2k
L2t
, thus X
L≥hNi2
hL + N
2i
1/2kϕ
NP
L(II)k
L∞ξ L2t
. X
M
M
1/2kP
Me
−|t|N2k
L2t. 1, where we used the fact that the Besov space ˙ B
2,11/2has a scaling invariance and e
−|t|∈ B ˙
2,11/2.
Lemma 4.1. For w ∈ S ( R
2), consider k
ξdefined on R by k
ξ(t) = η(t)ϕ
N(ξ)
Z
R
e
itτe
(t−|t|)ξ2− e
−|t|ξ2iτ + ξ
2w(τ e )dτ.
Then, for all ξ ∈ R , it holds X
L
hL + N
2i
1/2kP
Lk
ξk
L2t. X
L
hL + N
2i
−1/2kϕ
L(τ )ϕ
N(ξ) wk e
L2τ.
Proof. Following [16], we rewrite k
ξas k
ξ(t) = η(t)e
(t−|t|)ξ2Z
|τ|≤1
e
itτ− 1
iτ + ξ
2g w
N(τ )dτ + η(t) Z
|τ|≤1
e
(t−|t|)ξ2− e
−|t|ξ2iτ + ξ
2w g
N(τ )dτ + η(t)e
(t−|t|)ξ2Z
|τ|≥1
e
itτiτ + ξ
2g w
N(τ )dτ − η(t) Z
|τ|≥1
e
−|t|ξ2iτ + ξ
2g w
N(τ )dτ
= I + II + III − IV
where w
Nis defined by F
x(w
N)(ξ) = ϕ
N(ξ)F
x(w)(ξ).
Contribution of IV . Clearly we have kP
L(IV )k
L2t
. kP
L(η(t)e
−|t|ξ2)k
L2t
Z
|τ|≥1
| g w
N(τ )|
hiτ + ξ
2i dτ.
On the other hand, by Cauchy-Schwarz in τ , Z
|τ|≥1
| g w
N(τ )|
hiτ + ξ
2i dτ . X
L
hL+N
2i
−1kϕ
Lg w
Nk
L1τ. X
L
hL+N
2i
−1/2kϕ
Lg w
Nk
L2τ,
which combined with (4.3) yields the desired bound.
Contribution of II. By Cauchy-Schwarz inequality, kP
L(II)k
L2t
. kP
L(η(t)(e
(t−|t|)ξ2− e
−|t|ξ|2))k
L2t
×
Z | g w
N(τ )|
2hiτ + ξ
2i dτ
1/2Z
|τ|≤1
hiτ + ξ
2i
|iτ + ξ
2|
2dτ
!
1/2. kP
L(η(t)(e
(t−|t|)ξ2− e
−|t|ξ|2))k
L2t
× N
−2hN i X
L
hL + N
2i
−1/2kϕ
Lg w
Nk
L2τ. (4.4) Hence we need to estimate
X
L
hL + N
2i
1/2kP
L(η(t)(e
(t−|t|)ξ2− e
−|t|ξ|2))k
L2t
. X
L
hL + N
2i
1/2(kP
L(η(t)e
(t−|t|)ξ2)k
L2t
+ kP
L(η(t)e
−|t|ξ2)k
L2t
).
The second term in the right-hand side is bounded by 1 thanks to esti-
mate(4.3). Denote θ(t) = η(t)e
(t−|t|)ξ2. It is not too hard to check that one
integration by parts yields | θ(τ ˆ )| .
|τ|1whereas two integrations by parts give us | θ(τ ˆ )| .
hξi|τ|22. We thus infer that
X
L
hL + N
2i
1/2kϕ
Lθk ˆ
L2τ. X
L≤1
hN iL
1/2kθk
L1t+ X
1≤L≤hNi2
hN i
L
1/2+ X
L≥hNi2
hLi
1/2hN i
2L
3/2. hN i. (4.5) This provides the result for N ≥ 1. In the case N ≤ 1, we use a Taylor expansion and obtain
kP
L(η(t)(e
(t−|t|)ξ2− 1 + 1 − e
−|t|ξ2)k
L2t. X
n≥1
|ξ|
2nn!
kP
L(|t|
nη(t))k
L2t
+ 2
nkP
L(t
nη(t)χ
R−(t))k
L2t
.
According to the Sobolev embedding H
1֒ → B
2,11/2as well as the estimate kχ
R−f k
H1. kf k
H1provided f (0) = 0, we deduce
X
L
hL + N
2i
1/2kP
L(η(t)(e
(t−|t|)ξ2− e
−|t|ξ2))k
L2t
. ξ
2X
n≥1
1
n! (k|t|
nη(t)k
B1/22,1
+ 2
nkt
nη(t)χ
R−(t)k
B1/2 2,1) . N
2X
n≥1
2
nn! k|t|
nη(t)k
Ht1. N
2. Gathering this and (4.4) we conclude that
X
L
hL + N
2i
1/2kP
L(II)k
L2t. X
L
hL + N
2i
−1/2kϕ
Lg w
Nk
L2τ.
Contribution of I. Since I can be rewritten as I = η(t)e
(t−|t|)ξ2Z
|τ|≤1
X
n≥1
(itτ )
nn!
g w
N(τ ) iτ + ξ
2dτ, we have
kP
L(I)k
L2t
. X
n≥1
1
n! kP
L(t
nθ(t))k
L2t
Z
|τ|≤1
|τ |
n|iτ + ξ
2| | g w
N(τ )|dτ.
Using Cauchy-Schwarz we get, for n ≥ 1, Z
|τ|≤1
|τ |
n|iτ + ξ
2| | w g
N(τ )dτ .
Z | g w
N(τ )|
2hiτ + ξ
2i dτ
1/2Z
|τ|≤1
|τ |
2hiτ + ξ
2i
|iτ + ξ
2|
2dτ
!
1/2. hN i
−1X
L
hL + N
2i
−1/2kϕ
Lg w
Nk
L2τ.
Thus we see that it suffices to show that (see above the contribution of II for the definition of θ)
X
L
hL + N
2i
1/2X
n≥1
1
n! kP
L(t
nθ(t))k
L2t. hN i.
But again we have |F
t(t
nθ(t))| . 2
nmin(
|τ|1,
hξiτ22) and arguing as in (4.5), we get
X
L
hL + N
2i
1/2X
n≥1
1
n! kP
L(t
nθ(t))k
L2t. X
n≥1
hN i 2
nn! . hN i.
Contribution of III. Setting ˆ g(τ ) =
gwiτN+ξ(τ)2χ
|τ|≥1, we have to prove X
L
hL + N
2i
1/2kP
L(θg)k
L2t
. X
L
hL + N
2i
1/2kP
Lgk
L2t
. (4.6) Using the paraproduct decomposition, we have
P
L(θg) = P
LX
M&L
(P
.MθP
∼Mg + P
∼MθP
.Mg)
= P
L(III
1) + P
L(III
2)
and we estimate the contributions of these two terms separately.
Contribution of III
1. The sum over L ≥ hN i
2is estimated in the following way:
X
L≥hNi2
hL + N
2i
1/2kP
L(III
1)k
L2t
. X
L≥hNi2
hLi
1/2X
M&L
kP
.Mθk
L∞tkP
Mgk
L2t
. X
M
hM i
1/2kP
Mgk
L2t
.
Now we deal with the case where L . hN i
2. If ˆ θ is localized in an annulus {|τ | ∼ M }, we get from Bernstein inequality that
X
L≤hNi2
hL + N
2i
1/2X
M&L
kP
L(P
MθP
Mg)k
L2t
. X
M
hN i X
L.M
L
1/2kP
MθP
Mgk
L1t
. X
M
hN iM
1/2kP
Mθk
L2t
kP
Mgk
L2. X
M
hN ikP
Mgk
L2t
, (4.7)
where we used the estimate kP
Mθk
L2t. k
ϕMτ(τ)k
L2τ. M
−1/2. If ˆ θ is localized in a ball {|τ | ≪ M }, then we must have M ∼ L and thus
X
L≤hNi2
hL + N
2i
1/2X
M∼L
kP
L(P
≪MθP
Mg)k
L2t. X
L
hN ikP
≪Lθk
L∞tkP
Lgk
L2t,
which is acceptable.
Contribution of III
2. Consider the case L ≥ hN i
2. Since | θ| ˆ .
hξiτ22, we have kP
L(P
MθP
.Mg)k
L2t
. kϕ
Mθk ˆ
L1τkP
.Mgk
L2t
. hN i
2M kgk
L2t
. It follows that
X
L≥hN2i
hL + N
2i
1/2kP
L(III
2)k
L2t
. X
M&hNi2
M
1/2hN i
2M kgk
L2t
. hN ikgk
L2t
. It remains to establish the bound in the case L ≤ hN i
2. We may assume that ˆ g is supported in a ball {|τ | ≪ M} since the other case has already been treated (cf. estimate (4.7)). Therefore, M ∼ L and
X
L≤hNi2
hL + N
2i
1/2X
M&L
kP
L(P
MθP
≪Mg)k
L2t
. X
L
hN ikP
LθP
≪Lgk
L2t
. X
L
hN ikP
Lθk
L2t
X
M≪L
kP
Mgk
L∞t. X
L
hN iL
−1/2X
M≪L
M
1/2kP
Mgk
L2t
. X
M
hN ikP
Mgk
L2t
.
The proof of Lemma 4.1 is complete.
Proposition 4.2. Let L : f → Lf denote the linear operator Lf (t, x) = η(t)
χ
R+(t) Z
t0
W (t − t
′, t − t
′)f (t
′)dt
′+χ
R−(t)
Z
t0
W (t − t
′, t + t
′)f (t
′)dt
′. (4.8)
If f ∈ N
−1, then
kLf k
S−1. kf k
N−1. (4.9) Proof. It suffices to show that
kLf k
X−1,12,1
. kf k
X−1,−12,−1
(4.10) and
kLfk
Y−1,12. kf k
Y−1,−12
. (4.11)
Taking the x-Fourier transform, we get Lf (t, x) = U (t)
"
χ
R+(t) η(t) Z
R
e
ixξZ
t0
e
−|t−t′|ξ2F
x(U (−t
′)f (t
′))(ξ) dt
′dξ + χ
R−(t) η(t)
Z
R
e
ixξZ
t0
e
−|t+t′|ξ2F
x(U (−t
′)f(t
′))(ξ) dt
′dξ
#
= U (t)
"
η(t) Z
R
e
ixξZ
t0
e
−|t|ξ2e
t′ξ2F
x(U (−t
′)f (t
′))(ξ) dt
′dξ
# . Setting w(t
′) = U (−t
′)f (t
′), and using the time Fourier transform, we infer that
Lf(t, x) = U (t)
"
η(t) Z
R2
e
ixξe
itτe
(t−|t|)ξ2− e
−|t|ξ2iτ + ξ
2w(τ, ξ)dτ dξ ˜
# . Estimate (4.10) follows then easily from Lemma 4.1.
Now we turn to estimate (4.11). After square summing, it suffices to prove that for any dyadic N ,
k(∂
t+ ∂
xxx− ∂
xx+ I )P
NLf k
L1tL2x
. kP
Nf k
L1tL2x
. (4.12) In view of the expression of L it suffices to prove (4.12) separately for χ
R+Lf and χ
R−Lf First, a straightforward calculation leads to
(∂
t+ ∂
xxx− ∂
xx+ I)(χ
R+Lf (t))
= η(t)χ
R+(t)f(t) + (η
′(t) + η(t))χ
R+(t) Z
t0