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Sharp ill-posedness and well-posedness results for the KdV-Burgers equation: the periodic 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 periodic case. Transactions of the American Mathematical Society, American Mathe-
matical Society, 2013, 365 (1), pp.123-141. �hal-00467657v2�
Sharp ill-posedness and well-posedness results for the KdV-Burgers equation: the periodic case
Luc Molinet and St´ ephane Vento
Abstract. We prove that the KdV-Burgers is globally well-posed in H −1 ( T ) with a solution-map that is analytic from H −1 ( T ) to C([0, T ]; H −1 ( T )) whereas it is ill-posed in H s ( T ), as soon as s < −1, in the sense that the flow-map u 0 7→ u(t) cannot be continuous from H s ( T ) to even D ′ ( T ) at any fixed t > 0 small enough. In view of the result of Kappeler and Topalov for KdV it thus appears that even if the dissipation part of the KdV-Burgers equation allows to lower the C ∞ critical index with respect to the KdV equation, it does not permit to improve the C 0 critical index .
1 Introduction and main results
The aim of this paper is to establish positive and negative optimal results on the Cauchy problem in Sobolev spaces for the Korteweg-de Vries-Burgers (KdV-B) equation posed on the one dimensional torus T = R /2π Z :
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 by Ott and Sudan [16] as an asymptotic model for the propagation of weakly nonlinear dispersive long waves in some physical contexts when dissipative effects occur.
In order to make our result more transparent, let us first introduce differ-
ent notions of well-posedness (and consequently ill-posedness) related to the
smoothness of the flow-map (see in the same spirit [12], [9]). 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 solution 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 0 7→ u(t) is continuous from the ball of X cen-
tered at u 0 with radius R into X. If the map u 0 7→ 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.
In [15], Molinet and Ribaud proved that this equation is analytically well-posed in H s ( T ) 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 the C ∞ critical index s ∞ c (KdV B ) = −1 was lower that the one of the KdV equation
u t + u xxx + uu x = 0
for which s ∞ c (KdV ) = −1/2 (cf. [13], [7]) and also lower than the C ∞ index s ∞ c (dB) = s 0 c (dB) = −1/2 (cf. [1], [8]) of the dissipative Burgers equation
u t − u xx + uu x = 0 .
On the other hand, using the integrability theory, it was recently proved in [12] 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 −1 if one takes as uniqueness class, the class of strong limit in C([0, T ]; H −1 ( T )) of smooth solutions.
In [14] the authors completed the result of [15] in the real line case 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 ). To reach the critical Sobolev space H −1 ( R ) they adapted the refinement of Bourgain’s spaces that appeared in [19] and [18]
to the framework developed in [15]. The proof of the main bilinear estimate used in a crucial way the Kato smoothing effect that does not hold on the torus. Our aim here is to give the new ingredients that enable to overcome this lack of smoothing effects. The main idea is to weaken the space regu- larity of the Bourgain’s spaces in a suitable space-time frequencies region.
Note that our resolution space will still be embedded in C([0, T ]; H −1 ( T )) and that, to get the L ∞ ([0, T ]; H −1 ( T ))-estimate in this region, we use an idea that appeared in [4]. Finally, once the well-posedness result is proved, the proof of the ill-posedness result follows exactly the same lines as in [14]. It is due to a high to low frequency cascade phenomena that was first observed in [2] for a quadratic Schr¨ odinger equation.
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 1 allows to lower the C ∞ critical index with respect to the KdV
equation, it does not permit to improve the C 0 critical 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 ( T ). Moreover, at every point u 0 ∈ H −1 ( T ) there exist T = T(u 0 ) > 0 and R = R(u 0 ) > 0 such that the solution-map u 0 7→ u is analytic from the ball centered at u 0 with radius R of H −1 ( T ) into C([0, T ]; H −1 ( T )). Finally, the solution u can be extended for all posi- tive times and belongs to C( R ∗ + ; H ∞ ( T )).
Now that we have established analytic well-posedness, proceeding ex- actly as in [14] by taking as sequence of initial data
φ N (x) = N cos(N x) , we get the following ill-posedness result.
Theorem 1.2. The Cauchy problem associated to (1.1) is ill-posed in H s ( T ) for s < −1 in the following sense: there exists T > 0 such that for any 0 <
t < T , the flow-map u 0 7→ u(t) constructed in Theorem 1.1 is discontinuous at the origin from H −1 ( T ) endowed with the topology inducted by H s ( T ) into D ′ ( T ).
Acknowledgements: L.M. was partially supported by the ANR project
”Equa-Disp”.
2 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 × T ), we denote by u b (or F x u) 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 p x L q t and abbreviate L p x L p t as L p . Let us define the Japanese bracket hxi = (1+|x| 2 ) 1/2 so that the standard non-homogeneous Sobolev spaces are endowed with the norm kfk H
s= kh∇i s f k L
2.
We use 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(k) = η(k) − η(2k).
1
It is important to notice that the dissipative term −u
xxis of lower order than the
dispersive one u
xxx.
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 ∈ N ∪ {−1}.
We set ϕ
12
≡ η and for N ≥ 1, ϕ N (k) = ϕ(k/N ) and define the operator P N by F (P N u) = ϕ N u. We introduce b ψ L (τ, k) = ϕ L (τ − k 3 ) and for any u ∈ S ′ ( R × T ),
F x (P N u(t))(k) = ϕ N (k)ˆ u(t, k), F(Q L u)(τ, k) = ψ L (τ, k)˜ u(τ, k).
Roughly speaking, the operator P 1/2 and Q 1/2 localize respectively in the ball {|k| . 1} and {|τ −k 3 | . 1} whereas for N ≥ 1, the operator P N localizes in the annulus {|k| ∼ N } and Q N localizes in the region {|τ − k 3 | ∼ N }.
Furthermore we define more general projection P .N = P
N
1.N P N
1, Q ≫L = P
L
1≫L Q L
1etc.
Let e −t∂
xxxbe the propagator associated to the Airy equation and define the two parameters linear operator W by
W (t, t ′ )φ = X
k∈ Z
exp(itk 3 − |t ′ |k 2 ) ˆ φ(k) e ikx , t ∈ R . (2.1) The operator W : t 7→ W (t, t) is clearly an extension on R of the linear semi-group S(·) associated with (1.1) that is given by
S(t)φ = X
k∈ Z
exp(itk 3 − tk 2 ) ˆ φ(k) e ikx , t ∈ R + . (2.2) We will mainly work on the integral formulation of (1.1):
u(t) = S(t)u 0 − 1 2
Z t
0
S(t − t ′ )∂ x u 2 (t ′ )dt ′ , t ∈ R + . (2.3) Actually, to prove the local existence result, we will follow the strategy of [14] and apply a fixed point argument to the following extension of (2.3):
u(t) = η(t) h
W (t)u 0 − 1 2 χ R
+(t)
Z t 0
W (t − t ′ , t − t ′ )∂ x u 2 (t ′ )dt ′
− 1 2 χ R
−(t)
Z t
0
W (t − t ′ , t + t ′ )∂ x u 2 (t ′ )dt ′ i . (2.4) It is clear that if u solves (2.4) then u is a solution of (2.3) on [0, T ], T < 1.
In [14], adapting some ideas of [19] and [18] to the framework devel- oped in [15], the authors performed the iteration process in the sum space X −1,
12,1 + Y −1,
12where
kuk X
s,b,1= X
N
h X
L
hN i s hL + N 2 i b kP N Q L uk L
2xt
i 2 1/2
.
and
kuk Y
s,b= X
N
[hN i s kF −1 [(i(τ − k 3 ) + k 2 + 1) b+1/2 ϕ N e u]k L
1tL
2x] 2 1/2
,
so that
kuk Y
−1,12∼ X
N
[hN i −1 k(∂ t + ∂ xxx − ∂ xx + I)P N uk L
1t
L
2x] 2 1/2
. As explained in the introduction, due to the lack of the Kato smoothing effect on the torus, we will be able to control none of the two above norms in the region ”σ-dominant”. The idea is then to weaken the required x - regularity on the X s,b component of our resolution space in this region. For ε > 0 small enough, we thus introduce the function space X ε s,b,q endowed with the norm
kuk X
s,b,1ε
= X
N
h X
L≤N
3hN i s hL + N 2 i b kP N Q L uk L
2xt
i 2 1/2
+ X
N
h X
L>N
3hN i s−ε hL + N 2 i b kP N Q L uk L
2xt
i 2 1/2
. (2.5)
However, X ε s,b,1 is not embedded anymore in L ∞ ( R ; H −1 ( T )). For this rea- son we will take its intersection with the function space L ^ ∞ t H −1 , that is a dyadic version of L ∞ ( R ; H −1 ( T )), equipped with the norm
kuk L
∞t^ H
−1= X
N
[hN i −1 kP N uk L
∞tL
2x] 2 1/2
.
Finally, we also need to define the space Z s,−
12equipped with the norm kuk Z
s,−12= X
N
[hN i s kϕ N (k)hi(τ − k 3 ) + k 2 i −1 uk e L
2k
L
1τ] 2 1/2
.
We are now in position to form our resolution space f S ε s = (X s,
1 2
,1
ε ∩ L ^ ∞ t H −1 )+
Y s,
12and the ”nonlinear space” N ε s = (X s,−
1 2
,1
ε ∩ Z s,−
12) + Y s,−
12where the nonlinear term ∂ x u 2 will take place. Actually we will estimate k∂ x u 2 k N
εsin term of kuk S
εswhere S ε s = X s,
1 2
,1
ε + Y s,
12. Obviously kuk S
sε≤ kuk f S
sε
and the
first of these norms has the advantage to only see the size of the modulus of
the space-time Fourier transform of the function. This will be useful when dealing with the dual form of the main bilinear estimate.
Note that we endow these sum spaces with the usual norms:
kuk X+Y = inf{ku 1 k X + ku 2 k 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 ε −1 .
Lemma 2.1. For any φ ∈ L 2 , X
L
[L 1/2 kQ L (e −t∂
xxxφ)k L
2] 2 1/2
. kφk L
2. Proof. From Plancherel theorem, we have
X
L
[L 1/2 kQ L (e −t∂
xxxφ)k L
2] 2 1/2
∼ k|τ − k 3 | 1/2 F (e −t∂
xxxφ)k L
2. Moreover if we set η T (t) = η(t/T ) for T > 0, then
F (η T (t)e −t∂
xxxφ)(τ, k) = η c T (τ − k 3 ) φ(k). b
Thus we obtain with the changes of variables τ − k 3 → τ ′ and T τ ′ → σ that k|τ − k 3 | 1/2 F(η T (t)e −t∂
xxxφ)k L
2. kφk L
2k|τ ′ | 1/2 T b η(T τ ′ )k L
2τ′
. kφk L
2. Taking the limit T → ∞, this completes the proof.
Lemma 2.2. 1. For any ε ≥ 0 and all u ∈ S g ε −1 , we have kuk L
∞tH
−1( T ) . kuk g
S
ε−1and X
N
kP N Q ≤N
3uk 2 L
∞ tH
−1( T )
1/2
. kuk S
−1ε
. (2.6) 2. For any 0 ≤ ε ≤ 1/2 and all u ∈ S ε −1 , we have
kuk L
2xt
. kuk S
−1ε
. (2.7)
3. For all u ∈ Y 0,1/2 , X
L
[L 1/2 kQ L uk L
2] 2 1/2
. kuk
Y
0,12. (2.8)
Proof. 1. First it is fairly obvious that ^
L ∞ t H x −1 ֒ → L ∞ t H x −1 and that, according to the definition of X −1,
1 2
,1
ε ,
X
N
kP N Q ≤N
3uk 2 L
∞t
H
−1( T )
1/2
. kuk
X
−1,1 2,1 ε
.
Second, for any dyadic N , kP N uk L
∞tH
−1=
F t −1 hki −1
i(τ − k 3 ) + k 2 + 1 (i(τ − k 3 ) + k 2 + 1)ϕ N u e
L
∞tL
2k. F t −1 hki −1 ϕ N (k) i(τ − k 3 ) + k 2 + 1
L
∞tL
∞kkP N uk
Y
0,12. khki −1 ϕ N (k)k L
∞kke −thNi
2χ R
+(t)k L
∞tkP N uk
Y
0,12. hN i −1 kP N uk
Y
0,12. kP N uk
Y
−1,12.
This completes the proof of (2.6) after square summing in N.
2. In the same way, for any dyadic N kP N uk L
2. X
L≤N
3kP N Q L uk L
2+ X
L>N
3kP N Q L uk L
2. X
L≤N
3kP N Q L uk L
2+ X
L>N
3L 1/2 N −3/2 kP N Q L uk L
2. kP N uk
X
−1,1 2,1 1/2
. (2.9)
On the other hand, applying Young and H¨older’s inequalities, we get kP N uk L
2=
F t −1 1
i(τ − k 3 ) + k 2 + 1 (i(τ − k 3 ) + k 2 + 1)ϕ N e u
L
2tk.
F t −1 ϕ N (k) i(τ − k 3 ) + k 2 + 1
L
2tL
∞kkP N uk
Y
0,12. ke −thNi
2χ R
+(t)k L
2tkP N uk
Y
0,12. hN i −1 kP N uk
Y
0,12. kP N uk
Y
−1,12. This proves (2.7) after square summing in N .
3. Setting v = (∂ t + ∂ xxx )u, we see that u can be rewritten as u(t) = e −t∂
xxxu(0) +
Z t
0
e −(t−t
′)∂
xxxv(t ′ )dt ′ .
By virtue of Lemma 2.1, we have X
L
[L 1/2 kQ L e −t∂
xxxu(0)k L
2] 2 1/2
. ku(0)k L
2. kuk L
∞tL
2x.
Moreover, we get as previously kuk L
∞tL
2x. F t −1 1
i(τ − k 3 ) + k 2 + 1
L
∞tkkuk
Y
0,12. kuk
Y
0,12. Now it remains to show that
X
L
h L 1/2 Q L Z t
0
e −(t−t
′)∂
xxxv(t ′ )dt ′
L
2i 2 1/2
. kvk L
1t
L
2x, (2.10) since the right-hand side is controlled by
kP N vk L
1tL
2x. k(I − ∂ xx )P N uk L
1tL
2x+ kP N uk
Y
0,12. F t −1 ϕ N (ξ)(ξ 2 + 1) i(τ − ξ 3 ) + ξ 2 + 1
L
1tL
∞ξ+ 1 kP N uk
Y
0,12. kP N uk
Y
0,12.
In order to prove (2.10), we split the integral R t 0 = R t
−∞ − R 0
−∞ . By Lemma 2.1, the contribution with integrand on (−∞, 0) is bounded by
. Z 0
−∞
e t
′∂
xxxv(t ′ )dt ′
L
2x. kvk L
1t
L
2x. For the last term, we reduce by Minkowski to show that
X
L
[L 1/2 kQ L (χ t>t
′e −(t−t
′)∂
xxxv(t ′ ))k L
2tx
] 2 1/2
. kv(t ′ )k L
2x. This can be proved by a time-restriction argument. Indeed, for any T > 0, we have
X
L
[L 1/2 kQ L (η T (t)χ t>t
′e −(t−t
′)∂
xxxv(t ′ ))k L
2] 2 1/2
. k|τ | 1/2 b v(t ′ )F t (η T (t)χ t>t
′)(τ )k L
2. kv(t ′ )k L
2k|τ | 1/2 F t (η(t)χ tT >t
′)k L
2. kv(t ′ )k L
2.
We conclude by passing to the limit T → ∞.
3 Linear estimates
It is straightforward to check that estimates on the linear operator W (t) and on the extension of the Duhamel term proven in [14] on R still hold on T . We thus will concentrate ourselves on the X s,
1 2
,1
ε ∩ L ^ ∞ t H −1 component.
Proposition 3.1. 1. For any ε ≥ 0 and all φ ∈ H −1 ( T ), we have kη(t)W (t)φk g
S
ε−1. kφk H
−1. (3.1) 2. Let L : f → Lf denote the linear operator
Lf (t, x) = η(t) χ R
+(t)
Z t 0
W (t − t ′ , t − t ′ )f (t ′ )dt ′ +χ R
−(t)
Z t 0
W (t − t ′ , t + t ′ )f (t ′ )dt ′
. (3.2) If f ∈ N ε −1 with ε ≥ 0, then
kLfk g
S
ε−1. kf k N
ε−1. (3.3)
Proof. The first assertion is a direct consequence of the corresponding es- timate in X s,
12,1 proven in [14] together with the continuous embedding X s,
12,1 ֒ → L ^ ∞ t H −1 ∩ X s,
1 2
,1
ε for ε ≥ 0.
To prove the second assertion, it clearly suffices to show the three fol- lowing inequalities
kLf k
Y
0,12. kf k
Y
0,−12, (3.4)
kLf k
X
0,1 2,1 ε
. kf k
X
0,−1 2,1 ε
+ kf k
Z
0,−12, (3.5) kLf k L
∞t^ H
−1. kf k
Z
0,−12. (3.6)
Estimate (3.4) has been proved in [14]. To prove (3.5) we first note that according to [14], it holds
kLf k
X
s,12,1. kf k
X
s,−12,1. (3.7) It is then not too hard to be convinced that (3.5) is a consequence of the following estimate:
X
N ≥4
h X
L≤N
3hL + N 2 i 1/2 kP N Q L (L(Q ≥2N
3f ))k L
2i 2 1/2
. kf k
Z
0,−12+ X
N≥4
h X N
3L=N
3/2
hL + N 2 i −1/2 kP N Q L f k L
2i 2 1/2
. (3.8)
To prove (3.6) and (3.8) we proceed as in [14]. Using the x-Fourier expansion and setting w(t) = U (−t)f (t) it is easy to derive that
Lf (t, x) = U (t)
"
η(t) X
k∈ Z
e ixk Z
R
e itτ e (t−|t|)k
2− e −|t|k
2iτ + k 2 w(τ, k)dτ ˜
# .
In particular by Plancherel and Minkowski, kLf k ^
L
∞tL
2x= kU (−t)Lf k ^
L
∞tL
2x≤ X
N
η(t)ϕ N (k) Z
R
e itτ e (t−|t|)k
2− e −|t|k
2iτ + k 2 w(τ, k)dτ ˜ 2
L
2kL
∞t1/2
.
Now for k ∈ Z fixed and v ∈ S ( R ) we set K k (v)(t) = η(t)ϕ N (k)
Z
R
e itτ e (t−|t|)k
2− e −|t|k
2iτ + k 2 v(τ )dτ.
We thus are reduced to show that for any k ∈ Z , kK k (v)k L
∞t. ϕ N (k)v(τ ) hiτ + k 2 i
L
1τ(3.9)
and X
L≤N
3hL + N 2 i 1/2 kP L (K k (Φ ≥2N
3v))k L
2t.
ϕ N (k)v(τ ) hiτ + k 2 i
L
1τ+
N
3X
L=N
3/2
ϕ N (k)hL + N 2 i −1/2 kϕ L vk L
2τ. (3.10) where we set Φ ≥2N
3:= P
L≥2N
3ϕ L . To prove (3.9) it suffices to notice that kK k (v)k L
∞t. kηk L
∞ϕ N (k)
Z
R
|v(τ )|
|iτ + k 2 | dτ
which gives the result for k 6= 0. In the case k = 0 we use a Taylor expansion to get
kK 0 (v)k L
∞t. ϕ N (k) h kηk L
∞Z
|τ|≥1
|v(τ )|
|τ | dτ + X
n≥1
1
n! kt n ηk L
∞Z
|τ|≤1
|τ | n |v(τ )|
|τ | dτ i . ϕ N (k)
Z
R
|v(τ )|
hτ i dτ X
n≥0
1
n! kt n ηk L
∞which is acceptable since kt n ηk L
∞. 2 n . To get (3.10) we first rewrite K k (Φ ≥2N
3v) as η(t)e (t−|t|)k
2ϕ N (k)
Z
R
e itτ
iτ + k 2 Φ ≥2N
3(τ )v(τ )dτ −η(t)ϕ N (k) Z
T
e −|t|k
2iτ + k 2 Φ ≥2N
3(τ )v(τ )dτ.
The contribution of the second term is easily controlled by the first term of the right-hand side of (3.10) since, according to [14],
X
L
hL + N 2 i 1/2 kϕ N (k)P L (η(t)e −|t|k
2)k L
2t
. 1.
To treat the contribution of the first one, we set θ(t) = η(t)e (t−|t|)k
2and rewrite this contribution as P
L≤N
3I L with I L := hL + N 2 i 1/2 ϕ N (k)ϕ L (τ )
θ(τ b ′ ) ⋆ [Φ ≥2N
3(τ ′ ) v(τ ′ ) iτ ′ + k 2 ]
(τ )
L
2τ. (3.11) For L ≤ N 3 /4, by support considerations we may replace θ(τ b ′ ) by χ
|τ
′|≥
N23θ(τ b ′ ) in (3.11). Since it is not too hard to check that two integrations by parts yield | θ(τ ˆ )| . hki |τ|
22, this ensures that
X
L≤N
3/4
I L . X
L≤N
3/4
hL + N 2 i 1/2 ϕ N (k)kχ |τ|≥
N3 2b θk L
2v hiτ + k 2 i
L
1. X
L≤N
3/4
hL + N 2 i 1/2 N −5/2 ϕ N (k) v hiτ + k 2 i
L
1. ϕ N (k) v hiτ + k 2 i
L
1.
Now, for L = N 3 (Note that the case L = N 3 /2 can be treated in exactly the same way), we use that ϕ L ≡ η(·/2L)ϕ L and that by the mean value theorem, |ϕ L (τ ) − ϕ L (τ ′ )| . L −1 |τ − τ ′ |. Substituting this in (3.11) we infer that
I N
3. hN 3 + N 2 i 1/2 ϕ N (k) Z
R
b θ(τ − τ ′ )ϕ N
3(τ ′ ) Φ ≥2N
3(τ ′ )v(τ ′ ) iτ ′ + k 2 dτ ′
L
2τ+ hN 3 + N 2 i 1/2 ϕ N (k) η(τ /4N 3 )N −3 Z
R | θ(τ b − τ ′ )||τ − τ ′ | |v(τ ′ )|
|iτ ′ + k 2 | dτ ′
L
2τ:= I N 1
3+ I N 2
3.
Applying Plancherel theorem, H¨older inequality in t and then Parseval the- orem, the first term can be easily estimated by
I N 1
3. hN 3 + N 2 i 1/2 ϕ N (k)kθk L
∞ϕ N
3(τ )v(τ ) iτ + k 2
L
2. hN 3 i −1/2 ϕ N (k)kϕ N
3(τ )v(τ )k L
2which is acceptable. Finally, note that k θk b L
∞≤ kθk L
1≤ kηk L
1. 1 and that integrating by parts one time, it is not too hard to check that | b θ(τ )| . hτi 1 . This ensures that the second term can be controlled by
I N
3. hN 3 i 1/2 ϕ N (k)
η(τ /4N 3 )N −3 Z
R
|τ − τ ′ | hτ − τ ′ i
|v(τ ′ )|
|iτ ′ + k 2 | dτ ′ L
2. N −3/2 ϕ N (k)kη(τ /4N 3 )k L
2τv(τ ) iτ + k 2
L
1τ. ϕ N (k) v(τ ) iτ + k 2
L
1τ.
4 Bilinear estimate
In this section we provide a proof of the following crucial bilinear estimate.
Proposition 4.1. Let 0 < ε < 1/12. Then for all u, v ∈ S ε −1 it holds k∂ x (uv)k N
ε−1. kuk S
ε−1kvk S
ε−1. (4.1) We will need the following sharp estimates proved in [17].
Lemma 4.1. Let u 1 and u 2 be two real valued L 2 functions defined on R × Z with the following support properties
(τ, k) ∈ supp u i ⇒ |k| ∼ N i , hτ − k 3 i ∼ L i , i = 1, 2.
Then the following estimates hold:
ku 1 ⋆ u 2 k L
2τL
2(|k|≥N ) . min(L 1 , L 2 ) 1/2 max(L 1 , L 2 ) 1/4 N 1/4 + 1
ku 1 k L
2ku 2 k L
2and if N 1 ≫ N 2 ,
ku 1 ⋆ u 2 k L
2. min(L 1 , L 2 ) 1/2 max(L 1 , L 2 ) 1/2 N 1 + 1
ku 1 k L
2ku 2 k L
2.
Proof of Proposition 4.1. First we remark that because of the L 2 k struc- ture of the spaces involved in our analysis we have the following localization property
kf k S
ε−1∼ X
N
kP N f k 2 S
−1 ε1/2
and kf k N
ε−1∼ X
N
kP N f k 2 N
−1 ε1/2
.
Performing a dyadic decomposition for u, v we thus obtain k∂ x (uv)k N
ε−1∼ X
N
X
N
1,N
2P N ∂ x (P N
1uP N
2v) 2
N
ε−11/2
. (4.2) We can now reduce the number of case to analyze by noting that the right- hand side vanishes unless one of the following cases holds:
• (high-low interaction) N ∼ N 2 and N 1 . N ,
•• (low-high interaction) N ∼ N 1 and N 2 . N ,
• (high-high interaction) N ≪ N 1 ∼ N 2 .
The former two cases are symmetric. In the first case, we can rewrite the right-hand side of (4.2) as
k∂ x (uv)k N
−1ε
∼ X
N
kP N ∂ x (P .N uP N v)k 2 N
−1 ε1/2
, and it suffices to prove the high-low estimate
kP N ∂ x (P .N uP N v)k N
−1ε
. kuk S
−1ε
kP N vk S
−1ε
(HL)
for any dyadic N . If we consider now the third case, we easily get k∂ x (uv)k N
ε−1. X
N
1kP ≪N
1∂ x (P N
1uP N
1v)k N
ε−1, and it suffices to prove for any N 1 the high-high estimate
kP ≪N
1∂ x (P N
1uP N
1v)k N
−1ε
. kP N
1uk S
−1ε
kP N
1vk S
−1ε
(HH)
since the claim follows then from Cauchy-Schwarz.
Finally, since the S ε −1 -norm only sees the size of the modulus of the space- time Fourier transform we can always assume that our functions have real- valued non negative space-time Fourier transform.
Before starting to estimate the different terms we recall the resonance rela- tion associated with the KdV equation that reads
(τ 1 −k 3 1 )+(τ 2 −k 2 3 )+(τ 3 −k 3 3 ) = 3k 1 k 2 k 3 whenever (τ 1 , k 1 )+(τ 2 , k 2 )+(τ 3 , k 3 ) = 0.
(4.3)
4.1 High-Low interactions We decompose the bilinear term as
P N ∂ x (P .N uP N v) = X
N
1.N
X
L,L
1,L
2P N Q L ∂ x (P N
1Q L
1uP N Q L
2v).
Note first that we can always assume that N 1 ≫ 1, since otherwise, by using Sobolev inequalities and (2.7), it holds
X
N
1.1
kP N ∂ x (P N
1uP N v)k
Y
−1,−12. X
N
1.1
hN i −1 N kP N (P N
1uP N v)k L
1t
L
2x. X
N
1.1
kP N
1uk L
2t
L
∞xkP N vk L
2. X
N
1.1
N 1 1/2 kP N
1uk L
2kP N vk L
2. kuk S
ε−1kP N vk S
ε−1(4.4) as soon as ε ≤ 1/2. We now separate different regions. It is worth noticing that (4.3) ensures that max(L, L 1 , L 2 ) & N 2 N 1 .
4.1.1 L & N 2 N 1
We set L ∼ 2 l N 2 N 1 . Taking advantage of the X −1,−
1 2
,1
ε ∩ Z −1,−
12part of N ε −1 as well as the continuous embedding X −1,−
12,1 ֒ → X −1,−
1 2
,1
ε ∩ Z −1,−
12, by using Lemma 4.1 we get
I 1 : = X
1≪N
1.N
X
l≥0 L
1,L
2kP N Q L ∂ x (P N
1Q L
1uP N Q L
2v)k
X
−1,−12,1. X
1≪N
1.N
X
l≥0 L
1,L
22 −l/2 N −1 N 1 −1/2 kP N Q L (P N
1Q L
1uP N Q L
2v)k L
2. X
1≪N
1.N
X
l≥0 L
1,L
22 −l/2 N −1 N 1 −1/2 (L 1 ∧ L 2 ) 1/2 (L 1 ∨ L 2 ) 1/4 N 1/4 + 1
× kP N
1Q L
1uk L
2kP N Q L
2vk L
2Noticing that for any 0 < α < 1 it holds (L 1 ∧L 2 ) 1/2 (L 1 ∨ L 2 ) 1/4
N 1/4 +1
. (L 1 ∨L 2 ) −
α4N −
3−24α(L 1 +N 1 2 ) 1/2 (L 2 +N 2 ) 1/2 ,
we deduce that I 1 . X
1≪N
1.N
X
l≥0 L
1,L
22 −l/2 (L 1 L 2 ) −α/8 N 1 N
5−2α8kP N
1Q L
1uk
X
−9−28α,12,1kP N Q L
2vk
X
−9−28α,12,1Taking α > 0 small enough this proves with (2.8) that I 1 . kuk S
−1ε
kP N vk S
−1ε
whenever ε < 1/8.
4.1.2 L 1 & N 2 N 1 and L ≪ N 2 N 1
We can set L 1 ∼ 2 l N 2 N 1 with l ≥ 0. By duality, it is equivalent to show that 1
I 2 . kuk S
ε−1kP N vk S
ε−1kP N wk
X
1,12,∞where
I 2 := X
1≪N
1.N
X
l≥0 L
2,L≪N
2N
1P N Q L w , ∂ x (P N
1Q 2
lN
2N
1uP N Q L
2v)
L
2= X
1≪N
1.N
X
l≥0 L
2,L≪N
2N
1P N
1Q ^ 2
lN
2N
1u , ∂ x P ^ N Q L w ⋆ P ^ N Q ˇ L
2v)
L
2and ˇ θ(τ, k) = θ(−τ, −k). According to Lemma 4.1 we get
I 2 . X
1≪N
1.N
X
l≥0 L
2,L.N
2N
12 −l/2 N −1 N 1 −1/2 (L 1/2 1 kP N
1Q 2
lN
2N
1uk L
2)
∂ x P ^ N Q L w ⋆ P ^ N Q ˇ L
2v L
2. X
1≪N
1.N
X
l≥0 L
2,L.N
2N
12 −l/2 N −1 N 1 −1/2 (L 1/2 1 kP N
1Q 2
lN
2N
1uk L
2)
×(L ∧ L 2 ) 1/2 (L ∨ L 2 ) 1/4 N 1 1/4 + 1
k∂ x P N Q L wk L
2kP N Q L
2vk L
2.
1
The space X
1,12,∞is endowed with the norm
kuk
X1,12,∞:= X
N
sup
L
h hNi
shL + N
2i
bkP
NQ
Luk
L2 xti
21/2Since for any ε > 0 we have the estimate (L ∧ L 2 ) 1/2 (L ∨ L 2 ) 1/4
N 1 1/4 + 1
. (L ∨ L 2 ) −ε/2 N −
3 4
+ε
1 (L + N 2 ) 1/2 (L 2 + N 2 ) 1/2 , it follows that
I 2 . X
1≪N
1.N
X
l≥0 L
2,L.N
2N
12 −l/2 (LL 2 ) −ε/4 N −
1 4
+3ε 1
×kP N
1Q 2
lN
2N
1uk
X
−1−ε,12,1kP N Q L wk
X
1,12,∞kP N Q L
2vk
X
−1−ε,12,1which is acceptable whenever ε < 1/12.
4.1.3 L 2 & N 2 N 1 and L ∨ L 1 ≪ N 2 N 1
In this region thanks to the resonance relation (4.3) one has L 2 ∼ N 2 N 1 . We proceed as in the preceding subsection. We get
I 3 := X
1≪N
1.N
X
L∨L
1≪N
2N
1P N Q L w , ∂ x (P N Q N
2N
1vP N
1Q L
1u)
L
2. X
1≪N
1.N
X
L∨L
1.N
2N
1N −1 N 1 −1/2 (L 1/2 2 kP N Q N
2N
1vk L
2)
× (L ∧ L 1 ) 1/2 (L ∨ L 1 ) 1/4 N 1/4 + 1
k∂ x P N Q L wk L
2kP N
1Q L
1uk L
2. (4.5) On the other hand, we clearly have
(L ∧ L 1 ) 1/2 (L ∨ L 1 ) 1/4 N 1/4 + 1
. (L ∨ L 1 ) −ε/2 N −
3 4
+ε
1 (L + N 2 ) 1/2 (L 1 + N 1 2 ) 1/2 Inserting this into (4.5) we deduce
I 3 . X
1≪N
1.N
X
L∨L
1.N
2N
1(LL 1 ) −ε/4 N −
1 4
+2ε
1 kP N Q N
2N
1vk
X
−1,12,1× kP N Q L wk
X
1,12,∞kP N
1Q L
1uk
X
−1−ε,12,1. (4.6) Now either N 1 ≤ N or N 1 ∼ N . In the first case we have kP N Q N
2N
1vk
X
−1,12,1= kP N Q N
2N
1vk
X
−1,1 2,1 ε
which shows that (4.6) is acceptable for 0 < ε < 1/8.
In the second case, we have N 1 −ε kP N Q N
2N
1vk
X
−1,12,1≤ kP N Q N
2N
1vk
X
−1,1 2,1 ε
which shows that (4.6) is acceptable for 0 < ε < 1/12.
4.2 High-High interactions We perform the decomposition
P ≪N
1∂ x (P N
1uP N
1v) = X
N≪N
1X
L,L
1,L
2P N Q L ∂ x (P N
1Q L
1uP N
1Q L
2v).
By symmetry we can assume that L 1 ≥ L 2 . Then (4.3) ensures that max(L, L 1 ) & N N 1 2 .
4.2.1 N 1 2 N . L 1 ≤ N 1 3 .
We can set L 1 ∼ 2 l N 1 2 N with l ≥ 0. Using the Y −1,−
12part of N ε −1 , we want to estimate
I 4 := X
N≪N1 l≥0
P N ∂ x (P N
1Q 2
lN
12N uP N
1v)
Y
−1,−12. X
N≪N1 l≥0
kP N (P N
1Q 2
lN
12N uP N
1v)k 2 L
1t
L
2x1/2
. X
N≪N1 l≥0
[N 1/2 kP N
1Q 2
lN
12N uk L
2kP N
1vk L
2] 2 1/2
.
According to (2.7) and (2.8), this leads for ε ≤ 1/2 to I 4 . X
N≪N1 l≥0
2 −l X
L
1∼2
lN
12N≤N
13[N 1 −1 L 1/2 1 kP N
1Q L
1uk L
2] 2 1/2
kP N
1vk S
−1ε
. X
L1≤N3 1 l≥0
2 −l X
N∼2
−lL
1N
1−2[N 1 −1 L 1/2 1 kP N
1Q L
1uk L
2] 2 1/2
kP N
1vk S
ε−1. X
L
1≤N
13[N 1 −1 L 1/2 1 kP N
1Q L
1uk L
2] 2 1/2
kP N
1vk S
ε−1. kP N
1uk S
ε−1kP N
1vk S
−1ε.
4.2.2 L 1 ≥ N 1 3
We can set L 1 = 2 l N 1 3 with l ≥ 0. We proceed by duality as in Subsection 4.1.2 to get
I 5 . X
1.N ≪N
1X
l≥0 L
2,L
2 −l/2 N −
1 2
+ε
1 (N 1 −1−ε L 1/2 1 kP N
1Q 2
lN
13uk L
2) ∂ x P ^ N Q L w⋆ P N ^
1Q ˇ L
2v
L
2.
By virtue of Lemma 4.1, we have the bound
∂ x P ^ N Q L w ⋆ P N ^
1Q ˇ L
2v
L
2. (L ∧ L 2 ) 1/2 (L ∨ L 2 ) 1/4 N 1 1/4 + 1
k∂ x P N Q L wk L
2kP N
1Q L
2vk L
2. (L ∨ L 2 ) −ε/2 N
1 4
+2ε
1 kP N Q L wk
X
1,12,∞kP N
1Q L
2vk
X
−1−ε,12,1. Thus it is enough to check that
X
1.N≪N
1X
l≥0 L
2,L
2 −l/2 (LL 2 ) −ε/4 N −
1 4
+3ε
1 . 1,
but this is easily verified for ε < 1/12.
4.2.3 L & N 1 2 N and N 1 2 N 1−ε ≤ L 2 ≤ L 1 ≪ N 1 2 N
Then, by the resonance relation (4.3) we must have L ∼ N 1 2 N . We set L 2 ∼ 2 q N 1 2 N 1−ε and L 1 = 2 p L 2 with q ≥ 0 and p ≥ 0. Since N ≪ N 1 we are in the region L & N 3 . However since X −1,−
12,1 ֒ → X −1−ε,−
12,1 ∩ Z −1,−
12, it suffices to show that
I 6 . kP N
1uk S
−1ε
kP N
1vk S
−1ε
X
N
kP N Q N
21
N wk 2
X
1,12,∞1/2
where
I 6 := X
N ≪N
1X
p≥0,q≥0
P N Q N
21
N w, ∂ x (P N
1Q L
1uP N
1Q L
2v)
L
2.
Using Lemma 4.1 we get I 6 . X
N ≪N
1X
p≥0,q≥0
kP N
1Q L
1uk L
2N 1 −1 kP N Q N
21
N wk
X
1,12,∞kP N
1Q L
2vk
X
0,12,1. X
N ≪N
1X
p≥0,q≥0
2 −q/2 2 −p/2 N −
12+
2εkP N
1Q 2
p2
qN
12N
1−εQ ≤N
31
uk
X
−1,12,1× kP N Q N
21
N wk
X
1,12,∞kP N
1Q 2
qN
12N
1−εQ ≤N
31
vk
X
−1,12,1which is acceptable as soon as ε < 1.
4.2.4 L & N 1 2 N , L 1 ≪ N 1 2 N and L 2 ≤ N 1 2 N 1−ε
Since N ≪ N 1 we are in the region L > N 3 . It thus suffices to estimate both the X −1−ε,−
12,1 and the Z −1,−
12norms. Let us start by estimating the first one. Note that in this region we can replace P N
1u and P N
1v by P N
1Q ≤N
31
u and P N
1Q ≤N
31
v. Taking into account the gain of ε in the definition of the space, we get
k∂ x (uv)k
X
−1−ε,−12,1. X
1.N≪N
1N −ε N 1 −1 N −1/2 kP N (P N
1Q ≤N
31
uP N
1Q ≤N
31
v)k L
2. X
1.N≪N
1N −ε N 1 −1 kP N
1Q ≤N
31
u P N
1Q ≤N
31
vk L
2t
L
1x. X
1.N≪N
1N −ε (N 1 −1 kP N
1Q ≤N
31
uk L
∞tL
2x)kP N
1Q ≤N
31
vk L
2, which is acceptable as soon as ε > 0. It remains to estimate the Z −1,−
12- norm. By duality we have to estimate
I 7 := X
N≪N
1N 1 −2 N −1
P [ N wχ hσi∼N
21
N , P N ^
1Q ≤N
31
u ⋆ P N
1Q ^ ≤N
21
N
1−εv
L
2where w only depends on k and with σ = τ − k 3 (recall that we can assume that the space-time Fourier transforms of u and w are non negative real- valued functions). We follow an idea that can be found in [4]. First we notice that for any fixed k,
χ hσi∼L . χ hσi∼L ⋆ τ ( 1
L χ hσi≤L ) and thus the above scalar product can be rewritten as
P [ N wχ hσi∼N
21
N , P N ^
1Q ≤N
31
u ⋆ P N
1Q ^ ≤N
21
N
1−εv ⋆ τ ( 1
N 1 2 N χ hσi≤N
21
N )
L
2where F −1 h P N
1Q ^ ≤N
21
N
1−εv ⋆ τ ( N 1
21
N χ hσi≤N
21
N ) i
is of the form P N
1Q .N
21
N v ′ with
kP N
1Q .N
21
N v ′ k L
2. N −ε/2 kP N
1Q ≤N
21
N
1−εvk L
2. (4.7) Indeed the linear operator T K,K
2: v 7→ K 1 v(·)χ {h·i≤K
2} ⋆ χ {h·i≤K} is a con- tinuous endomorphism of L 1 ( R ) and L ∞ ( R ) with
kT K
,K
2vk L
∞( R ) ≤ sup
x∈ R
1 K
Z
R v(y)χ {hyi≤K
2} χ {hx−yi≤K} dy . min(K, K 2 )
K kvk L
∞( R )
and
kT K
,K
2vk L
1( R ) ≤ 1
K kvk L
1( R ) kχ {h·i≤K} k L
1( R ) . kvk L
1( R ) .
Therefore, by Riesz interpolation theorem T K,K
2is a continuous endomor- phism of L 2 ( R ) with
kT K
,K
2vk L
2( R ) . min 1, K 2
K 1/2
kvk L
2( R ) .
Hence, by Sobolev in k and (4.7), I 7 . X
N≪N
1N 1 −2 N −1 N 1/2 k P [ N wχ hσi∼N
21
N k L
2kP N
1Q ≤N
31
uk L
∞t
L
2xkP N
1Q .N
21
N v ′ k L
2. X
N≪N
1N −ε/2 kP N wk L
2( T ) (N 1 −1 kP N
1Q ≤N
31
uk L
∞tL
2x)kP N
1vk L
2,
which is acceptable as soon as ε > 0.
5 Well-posedness
In this section, we prove the well-posedness result. The proof follows exactly the same lines as in [14]. Using a standard fixed point procedure, it is clear that the bilinear estimate (4.1) allows us to show local well-posedness but for small initial data only. This is because H −1 appears as a critical space for KdV-Burgers. Indeed, on one hand, we cannot get any contraction factor by restricting time. On the other hand, a dilation argument does not work here since the reduction of the H −1 -norm of the dilated initial data would be exactly compensated by the diminution of the dissipative coefficient in front of u xx (that we take equal to 1 in (1.1)) in the equation satisfied by the dilated solution. In order to remove the size restriction on the data, we change the metric on our resolution space.
For 0 < ε < 1/12 and β ≥ 1, let us define the following norm on S g ε −1 , kuk Z
β= inf
u=u
1+u
2u
1∈ S g
ε−1,u
2∈ S f
ε0ku 1 k g
S
ε−1+ 1 β ku 2 k S f
0ε