Estimées dispersives invariantes d’échelle pour le groupe de Korteweg-de Vries
Scaling-sharp dispersive estimates for the Korteweg-de Vries group
Raphaël Côte Luis Vega
Abstract
We prove weighted estimates on the linear KdV group, which are scaling sharp.
This kind of estimates are in the spirit of that used to prove small data scattering for the generalized KdV equations.
Nous prouvons des inégalités à poids pour le groupe linéaire de KdV, qui sont optimales vis-à-vis du changement d’échelle. Ce type d’inégalité suit l’esprit de celles utilisées pour montrer que les solutions des équations de KdV généralisées dont les données sont petites dispersent linéairement.
The purpose of this short note is to give a simple proof of two dispersive estimates which are heavily used in the proof of small data scattering for the generalized Korteweg- de Vries equations [2].
The proof of these estimates can be easily extended to other dispersive equations.
Denote U (t) the linear Korteweg-de Vries group, i.e v = U (t)φ is the solution to v
t+ v
xxx= 0,
v(t = 0) = φ, i.e. U \ (t)φ = e
itξ3/3φ ˆ or (U (t)φ)(x) = 1 t
1/3Z Ai
x − y t
1/3φ(y)dy,
where Ai is the Airy function
Ai(z) = 1 π
Z
∞0
cos ξ
33 + ξz
dξ.
Theorem 1. Let φ, ψ ∈ L
2, such that xφ, xψ ∈ L
2. Then
kU (t)φk
2L∞≤ 2k Ai k
2L∞t
−2/3kφk
L2kxφk
L2, (1) kU (t)φU (−t)ψ
xk
L∞≤ Ct
−1(kφk
L2kxψk
L2+ kψk
L2kxφk
L2). (2) Furthermore, the constant 2k Ai k
2L∞in the first estimate is optimal.
Remark 1. These estimates are often used with φ replaced by U (−t)φ : denoting J(t) = U (t)xU(−t), they take the form
kφk
2L∞≤ Ct
−2/3kφk
L2kJ (t)φk
L2, (3)
kφψ
xk
L∞≤ Ct
−1(kφk
L2kJ(t)ψk
L2+ kψk
L2kJ(t)φk
L2). (4)
Proof. Due to a scaling argument (and representation in term of the Airy function), we are reduced to show that
kU (1)φk
L∞≤ Ckφk
L2kxφk
L2, and similarly for the second inequality. Hence we consider
(U (1)φ)(x) = Z
Ai(x − y)φ(y)dy,
and we recall that the Airy function satisfies | Ai(x)| ≤ C(1 + |x|)
−1/4and | Ai
0(x)| ≤ C(1 + |x|)
1/4. Then
|U (1)φ|
2(x) = Z Z
Ai(x − y)φ(y) Ai(x − z) ¯ φ(z) y − z y − z dydz
= Z
Ai(x − y)yφ(y)
Z Ai(x − z) y − z φ(z)dz ¯
| {z }
Hz7→y(Ai(x−z)φ(z))(y)
dy
− Z
Ai(x − z)z φ(z) ¯
Z Ai(x − y)
y − z φ(y)dy
| {z }
−Hy7→z(Ai(x−y)φ(y))(z)
dz
= 2<
Z
Ai(x − y)yφ(y)H
z7→y(Ai(x − z)φ(z))(y)dy,
where H denotes the Hilbert transform (and with the slight abuse of notation
x1for vp
1x). As H : L
2→ L
2is isometric and hence continuous (with norm 1), and Ai ∈ L
∞, we get
|U (1)φ|
2(x) ≤ 2k Ai(x − y)yφ(y)k
L2(dy)kH(Ai(x − ·)φ)(y)k
L2(dy)(5)
≤ 2k Ai k
2L∞kyφk
L2kφk
L2. (6) This is the first inequality. Let us now prove that the constant is sharp.
First consider the minimizers in the following Cauchy-Schwarz inequality :
Z
yψ(y)H(ψ)(y)dy
≤ kyψ(y)k
L2(dy)kψk
L2. (7) There is equality if yψ(y) = λH(ψ)(y) for some λ ∈ C . Then a Fourier Transform shows that ∂
ξψ(ξ) = ˆ λ sgn ξ ψ(ξ), hence ˆ ψ(ξ) = ˆ C exp(−λ|x|), or equivalenty, one has equality in (7) as soon as
ψ(y) = C
1 + (y/λ)
2for some λ, C ∈ C . (8)
(Notice that all the functions involved lie in L
2)
We now go back to (6). Let x
0∈ R where | Ai | reaches its maximum. Now as
Ai(x
0) 6= 0, let ε > 0 such that for all y ∈ [−ε, ε], | Ai(x
0− y)| ≥ | Ai(x
0)|/2, and
consider the sequence of functions
Denote ψ
n(y) =
11+y|y|≤nε2. As Ai(x
0− y)φ
n(ny) = √
nψ
n(ny),
|U (1)φ
n|
2(x
0) = 2 Z
y √
nψ
n(ny)H
z→y( √
nψ
n(nz))(y)dy = 2 n
Z
yψ
n(ny)H(ψ
n)(ny)dy.
One easily sees that ψ
n(y) →
1+|y|1 2in L
2and yψ
n(y) →
1+|y|y 2in L
2, and hence, in view of (8), as H is homogeneous of degree 0 and L
2isometric, we have
|U (1)φ
n|
2(x
0) ∼ 2 n
Z y
1 + |y|
2H( 1
1 + | · |
2)(y)dy ∼ 2 n k y
1 + |y|
2k
L2k 1 1 + |y|
2k
L2∼ 2ky √
nψ
n(ny)k
L2k √
nψ
n(ny)k
L2∼ 2ky Ai(x
0− y)φ
n(y)k
L2k Ai(x
0− y)φ
n(y)k
L2. As φ
nconcentrates at point 0, we deduce
|U (1)φ
n|
2(x
0) ∼ 2| Ai(x
0)|
2kyφ
n(y)k
L2kφ
n(y)k
L2as n → ∞, (9) which proves that the sharp constant in the first inequality is 2k Ai k
2L∞.
For the second inequality (estimate of the derivative), we have as for the first in- equality :
(U (1)φU (1) ¯ ψ
x)(x)
= Z Z
Ai(x − y)φ(y)Ai
0(x − z) ¯ ψ(z) y − z y − z dydz
= Z
Ai
0(x − z) ¯ ψ(z)
Z Ai(x − y)yφ(y) y − z dy
dz −
Z
Ai
0(x − z)z ψ(z) ¯
Z Ai(x − y)φ(y) z − y dy
dz
= Z
Ai
0(x − z) ¯ ψ(z)H
y7→z(Ai(x − y)yφ(y))(z)dz − Z
Ai
0(x − z)z ψ(z)H ¯
y→z(Ai(x − y)φ(y))(z)dz.
Denote ω
x(z) = √
11+|x−z|
: ω
x−1∈ A
2(with the notation of [4]), so that there exists C not depending on x such that
∀v, Z
|Hv|
2ω
−1x≤ C Z
|v|
2ω
−1x.
Recall the well-known asymptotic |Ai
0(x)| ≤ C(1 + |x|
1/4). Then
Z
Ai
0(x − z) ¯ ψ(z)H
y→z(Ai(x − y)yφ(y))(z)dz
≤ Z
|Ai
0(x − z) ¯ ψ(z)|
2ω
x(z)dy
1/2Z
|H
y→z(Ai(x − y)yφ(y))(z)|
2ω
x−1(z)dz
1/2≤ Ckψk
L2Z
|Ai(x − y)yφ(y)|
2ω
x−1(y)dy
1/2≤ Ckψk
L2kyφ(y)k
L2.
In the same way,
Z
Ai
0(x − z)z ψ(z)H(Ai(x ¯ − ·)φ)(z)dz
≤ Z
|Ai
0(x − z)z ψ(z)| ¯
2ω
x(z)dz
1/2Z
|H(Ai(x − ·)φ)(z)|
2ω
x−1(z)dz
1/2≤ Ckzψ(z)k
L2Z
|Ai(x − y)φ(y)|
2ω
x−1(y)dy
1/2≤ Ckyψ(y)k
L2kφk
L2. Thus:
kU (1)φU(1) ¯ ψ
xk
L∞≤ C(kφk
L2kxψk
L2+ kψk
L2kxφk
L2).
Up to scaling and replacing ψ by ψ, this is the second inequality. ¯
Remark 2. This proof (especially (5)) is reminiscent of that in [3] (see also [1])
kφk
2L∞≤ kφk
L2kφ
0k
L2,
where the constant is sharp and minimizers are Ce
−λ|x|. This has application to the Schrödinger group U (t) (i.e U \ (t)φ = e
itξ2φ). We have the following Schrödinger version ˆ of estimate (3) (notice that U (t)xU (−t) = e
i|x|2 4t it
2
∂
xe
−i|x|2 4t
) :
kψk
2L∞= ke
i|x|2
4t
ψk
2L∞≤ ke
i|x|2
4t
ψk
L2ke
i|x|2 4t
∂
xe
i|x|2
4t
ψk
L2≤ 2
t kψk
L2kU (t)xU (−t)ψk
L2. From Theorem 1, we can easily obtain the optimal decay in a scaling sharp Besov like space. Let ϕ ∈ D( R ) be non-negative with support in ] −2, 2[ and such that ϕ equals 1 in a neighbourhood of [−1.5, 1.5]. Denote ψ(x) = ϕ(2x) − ϕ(x) and ψ
j(x) = ψ(x/2
j).
Finally introduce
kφk
Nt=
∞
X
j=−∞
2
j/2kψ
jU (−t)φk
L2.
Corollary 1. We have :
kφk
L∞≤ Ct
−1/3kφk
Nt. Proof. Notice that |xψ
j(x)| ≤ 2
j+1ψ
j(x). As φ = P
j
U (t)ψ
jU (−t)φ, we have : kφk
L∞≤ X
j
kU (t)ψ
jU (−t)φk
L∞≤ Ct
−1/3X
j
kU(t)ψ
jU (−t)φk
1/2L2kU (t)xU(−t)U (t)ψ
jU (−t)φk
1/2L2≤ Ct
−1/3X
j
kU(t)ψ
jU (−t)φk
1/2L2kxψ
jU (−t)φk
1/2L2≤ Ct
−1/3X
j
kU(t)ψ
jU (−t)φk
1/2L2
2
j/2kU (t)ψ
jU (−t)φk
1/2L2