Electronic Journal of Differential Equations, Vol. 2004(2004), No. 13, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
LOCAL WELL-POSEDNESS FOR A HIGHER ORDER
NONLINEAR SCHRÖDINGER EQUATION IN SOBOLEV
SPACES OF NEGATIVE INDICES
XAVIER CARVAJAL
Abstract. We prove that the initial value problem associated with
∂t u + iα∂x2 u + β∂x3 u + iγ|u|2 u = 0,
x, t ∈ R,
is locally well-posed in H s for s > −1/4.
1. Introduction
In this work, we study a particular case of the initial value problem (IVP)
∂t u + iα∂x2 u + β∂x3 u + F (u) = 0,
x, t ∈ R,
u(x, 0) = u0 (x) .
(1.1)
Here u is a complex valued function, F (u) = iγ|u|2 u + δ|u|2 ∂x u + u2 ∂x u, γ, δ, ∈ C
and α, β ∈ R are constants.
Hasegawa and Kodama [10, 14] proposed (1.1) as a model for propagation of
pulse in optical fiber. We will study the IVP (1.1) in Sobolev space H s (R) under
the condition δ = = 0, β 6= 0 (see case (iv) in Theorem 1.1 below). When
γ, δ, ∈ R, it was shown in [16] that the flow associated to the IVP (1.1) leaves the
following quantity
Z
I1 (v) =
|v|2 (x, t) dx,
(1.2)
R
conserved in time. Also, when δ − γ = 6= 0 we have the following quantity
conserved:
Z
Z
Z
I2 (v) = c1 |∂x v|2 (x, t) dx + c2 |v|4 (x, t)dx + c3 v(x, t)∂x v(x, t)dx, (1.3)
R
R
R
where c1 = 3β , c2 = −( + δ)/2 and c3 = i(α( + δ) − 3βγ). These quantities
were used in [16] toRestablish global well-posedness for (1.1) in H s (R), s ≥ 1. Note
that the quantity i R v(x, t)∂x v(x, t)dx in (1.3) is real since
Z
Z
∂t ( i v(x, t)∂x v(x, t)dx) = 2 Im( [v(x, t)∂x v(x, t)]2 dx).
R
R
2000 Mathematics Subject Classification. 35Q58, 35Q60.
Key words and phrases. Schrödinger equation, Korteweg-de Vries equation,
trilinear estimate, Bourgain spaces.
c
2004
Texas State University - San Marcos.
Submitted July 30, 2003. Published January 23, 2004.
Partially supported by FAPESP, Brazil.
1
2
XAVIER CARVAJAL
EJDE-2004/13
We say that the IVP (1.1) is locally well-posed in X (Banach space) if the solution
uniquely exists in certain time interval [−T, T ] (unique existence), the solution
describes a continuous curve in X in the interval [−T, T ] whenever initial data
belongs to X (persistence), and the solution varies continuously depending upon
the initial data (continuous dependence) i.e. continuity of application u0 7→ u(t)
from X to C([−T, T ]; X). We say that the IVP (1.1) is globally well-posed in X if
the same properties hold for all time T > 0. If some hypothesis in the definition of
local well-posed fails, we say that the IVP is ill-posed.
Particular cases of (1.1) are the following:
• Cubic nonlinear Schrödinger equation (NLS), (α = ∓1, β = 0, γ = −1, δ = = 0).
iut ± uxx + |u|2 u = 0,
x, t ∈ R.
(1.4)
s
The best known local result for the IVP associated to (1.4) is in H (R), s ≥ 0,
obtained by Tsutsumi [26].
• Nonlinear Schrödinger equation with derivative (α = −1, β = 0, γ = 0, δ = 2).
iut + uxx + iλ(|u|2 u)x = 0,
x, t ∈ R.
(1.5)
The best known local result for the IVP associated to (1.5) is in H s (R), s ≥ 1/2,
obtained by Takaoka [24].
• Complex modified Korteweg-de Vries (mKdV) equation (α = 0, β = 1, γ = 0,
δ = 1, = 0).
ut + uxxx + |u|2 ux = 0, x, t ∈ R.
(1.6)
If u is real, (1.6) is the usual mKdV equation and Kenig et al. [11] proved the IVP
associated to it is locally well-posed in H s (R), s ≥ 1/4.
• When α 6= 0 is real and β = 0, we obtain a particular case of the well-known
mixed nonlinear Schrödinger equation
ut = iαuxx + λ(|u|2 )x u + g(u),
x, t ∈ R,
(1.7)
where g satisfies some appropriate conditions. Ozawa and Tsutsumi in [19] proved
that for any ρ > 0, there is a positive constant T (ρ) depending only on ρ and g,
such that the IVP (1.7) is locally well-posed in H 1/2 (R), whenever the initial data
satisfies
ku0 kH1/2 ≤ ρ.
There are other dispersive models similar to (1.1). The interested readers can
see the following works and the references therein [1, 7, 20, 21, 23].
Laurey [17, 16] proved that the IVP associated to (1.1) is locally well-posed in
H s (R), s > 3/4. Staffilani [22] improved this result by proving the IVP associated
to (1.1) is locally well-posed in H s (R), s ≥ 1/4.
When α, β are functions of t, we proved in [2, 3] local well-posedness in H s (R),
s ≥ 1/4. Also we studied in [2, 5] the unique continuation property for the solution
of (1.1). Regarding the ill-posedness of the IVP (1.1), we proved in [4] the following
theorem.
Theorem 1.1. The mapping data-solution u0 7→ u(t) for the IVP (1.1) is not C 3
at origin in the following cases:
(i) β = 0, α 6= 0, δ = = 0, γ 6= 0 for s < 0.
(ii) β = 0, α 6= 0, δ 6= 0 or 6= 0 for s < 1/2.
(iii) β 6= 0, δ 6= 0 or 6= 0 for s < 1/4.
(iv) β 6= 0, δ = = 0, γ 6= 0 for s < −1/4.
EJDE-2004/13
LOCAL WELL-POSEDNESS
3
In this work, we consider the case (iv) and prove the following result.
Theorem 1.2. Let β 6= 0 real and γ 6= 0 complex, then the following IVP
∂t u + iα∂x2 u + β∂x3 u + iγ|u|2 u = 0,
x, t ∈ R,
u(x, 0) = u0 ,
(1.8)
is locally well-posed in H s (R), s > −1/4.
The following trilinear estimate will be fundamental in the proof of Theorem 1.2.
Theorem 1.3. Let −1/4 < s ≤ 0, b > 7/12, b0 < s/3, then we have
kuvwkXs,b0 ≤ CkukXs,b kvkXs,b kwkXs,b ,
(1.9)
where
kukXs,b = khξis hτ − φ(ξ)ib ûkL2ξ L2τ ,
hξi = 1 + |ξ|,
φ(ξ) = αξ 2 + βξ 3 .
Theorem 1.4. The trilinear estimate (1.9) fails if s < −1/4 and b ∈ R.
Remarks. • When γ ∈ R, as (1.1) preserves L2 norm, Theorem 1.2 permits to
obtain global existence in L2 .
• From Lemma 2.3 we note that b = 7/12+ is the best possible for s = −1/4+, in
the trilinear estimate (1.9).
• The trilinear estimate is valid for all s > 0, as it can be seen by combining
hξis ≤ hξ − (ξ2 − ξ1 )is hξ2 is hξ1 is and the estimate (1.9) for s = 0.
• We will use the notation kuk{s,b} := kukX s,b .
• When α = 0, β = 1, we have the usual bilinear estimate due to Kenig et al. [12],
k(uv)x k{−3/4+,−1/2+} ≤ Ckuk{−3/4+,1/2+} kvk{−3/4+,1/2+}.
Also we have the 1/4 trilinear estimate due to Tao [25],
k(uvw)x k{1/4,−1/2+} ≤ Ckuk{1/4,1/2+} kvk{1/4,1/2+} kwk{1/4,1/2+} .
2. Proofs of Main Result
Proof of Theorem 1.4. As in [12] consider the set
B := {(ξ, τ ); N ≤ ξ ≤ N + N −1/2 , |τ − φ(ξ)| ≤ 1},
where φ(ξ) = αξ 2 + βξ 3 . We have |B| ∼ N −1/2 . Let us consider v̂ = χB , it is not
difficult to see that kvk{s,b} ≤ N s |B|1/2 . Moreover
F(|v|2 v) := χB ∗ χB ∗ χ−B &
1
χA ,
N
where A is a rectangle contained in B such that |A| ∼ N −1/2 .
Therefore
0
1
k |v|2 vk{s,b0 } = khξis hτ − φ(ξ)ib F(|v|2 v)kL2ξ L2τ & N s N −1/4 = N s−5/4 .
N
As a consequence, for large N the trilinear estimate fails if 3(s − 1/4) < s − 5/4,
i.e. if s < −1/4.
4
XAVIER CARVAJAL
EJDE-2004/13
Proof of Theorem 1.3. In Lemma 2.1 below, we gather some elementary estimates needed in the proof of Theorem 1.3, we need the following results from
elementary calculus.
Lemma 2.1.
(1) If b > 1/2, a1 , a2 ∈ R then
Z
dx
1
∼
.
2b hx − a i2b
hx
−
a
i
ha
−
a2 i2b
1
2
1
R
(2) If 0 < c1 , c2 < 1, c1 + c2 > 1, a1 6= a2 , then
Z
dx
1
.
.
c1 |x − a |c2
(c1 +c2 −1)
|x
−
a
|
|a
−
a
|
1
2
1
2
R
(2.1)
(2.2)
(3) Let a ∈ R, c1 ≤ c2 , then
|x|c1
1
≤ c1 .
haxic2
|a|
(4) Let a, η ∈ R, b > 1/2, then
Z
R
ha(x2
dx
1
.
.
2
2b
− η )i
|aη|
(2.3)
(2.4)
Now, let f (ξ, τ ) = hξis hτ − ξ 3 ib û, g(ξ, τ ) = hξis hτ − ξ 3 ib v̂, h(ξ, τ ) = hξis ,
hτ − ξ 3 ib ŵ, η = (ξ, τ ), x = (ξ1 , τ1 ), y = (ξ2 , τ2 ). We have
Z
kuvwk{s,b0 } =k
f (η + x − y)g(y)h(x)K(η, x, y)dxdykL2η
R4
2 kf kL2 kgkL2 khkL2 ,
≤kK(η, x, y)kL∞
η Lx,y
where
K(η, x, y) =
r(ξ, τ )hτ1 − φ(ξ1
hξ + ξ1 − ξ2 iρ hξ2 iρ hξ1 iρ
b
b
2 − φ(ξ2 )i hτ + τ1 − τ2 − φ(ξ + ξ1 − ξ2 )i
)ib hτ
0
and r(ξ, τ ) = hξiρ hτ − φ(ξ)i−b , ρ = −s. Using (2.1) we obtain
Z
Gρ (ξ, ξ1 , ξ2 ) dξ1 dξ2
1
I(ξ, τ ) := kKk2L2x,y ∼
2
r(ξ, τ ) R2 hτ − φ(ξ + ξ1 − ξ2 ) − φ(ξ2 ) + φ(ξ1 )i2b
Z
1
Gρ (ξ, ξ1 , ξ2 ) dξ1 dξ2
=
r(ξ, τ )2 R2 hτ − φ(ξ) + g(ξ, ξ1 , ξ2 )i2b
where
Gρ (ξ, ξ1 , ξ2 ) := hξ + ξ1 − ξ2 i2ρ hξ1 i2ρ hξ2 i2ρ ,
g(ξ, ξ1 , ξ2 ) = (ξ1 − ξ2 )(ξ + ξ2 )(2α + 3β(ξ − ξ1 )).
(2.5)
Assuming y = τ − φ(ξ), to get Theorem 1.3 it is sufficient to prove the following
lemma.
Lemma 2.2. Let 0 ≤ ρ < 1/4, b > 7/12, b0 < −ρ/3. Then
Z
1
Gρ (ξ, −ξ1 , −ξ2 )dξ1 dξ2
I(ξ, y) :=
≤ C(ρ, b, b0 ) < ∞,
hξi2ρ hyi−2b0 R2 hy + g(ξ, ξ1 , ξ2 )i2b
where C(ρ, b, b0 ) is a constant independent of ξ and y.
To prove Lemma 2.2 we need to prove the following lemmas.
EJDE-2004/13
LOCAL WELL-POSEDNESS
Lemma 2.3. Let ρ < 1/4. Then
(
Z
C(ρ, b) < ∞
Gρ (0, −ξ1 , −ξ2 )dξ1 dξ2
I(0, 0) =
=
2b
hg(0,
ξ
,
ξ
)i
∞
1 2
R2
5
if ρ + 1/3 < b
if ρ + 1/3 ≥ b,
where C(ρ, b) is a constant.
We have that if b = 7/12 and ρ = −s ≥ 1/4 then I(0, 0) = ∞, therefore Lemma
2.3 shows that b = 7/12+ is the best possible when s = −1/4+.
Lemma 2.4. Let 0 ≤ ρ < 1/4, b > 7/12. Then
Z
1
Gρ (ξ, −ξ1 , −ξ2 )dξ1 dξ2
I(ξ, 0) =
≤ C(ρ, b),
2ρ
hξi
hg(ξ, ξ1 , ξ2 )i2b
R2
where C(ρ, b) is a constant independent of ξ.
For clarity in exposition, we consider the case α = 0, β = 1, i.e. φ(ξ) = ξ 3 (see
the observation at the end of the proof of Lemma 2.2).
In the definition of I(ξ, y) if we make the change of variables ξ − ξ1 := ξξ1 ,
ξ + ξ2 := ξξ2 and y = ξ 3 z, then I(ξ, y) becomes
Z
Hρ (ξ, ξ1 , ξ2 )dξ1 dξ2
I(ξ, z) = p(ξ, z)
,
(2.6)
3 (z + F (ξ , ξ ))i2b
hξ
2
1 2
R
0
where p(ξ, z) = ξ 2 hξ 3 zi2b hξi−2ρ , F (ξ1 , ξ2 ) = (2 − (ξ1 + ξ2 ))ξ1 ξ2 and
Hρ (ξ, ξ1 , ξ2 ) = hξ(1 − (ξ1 + ξ2 ))i2ρ hξ(1 − ξ1 )i2ρ hξ(1 − ξ2 )i2ρ .
From here onwards we will suppose z > 0, because if z < 0 we can obtain the
same result by symmetry (see Remark after Proposition 1).
Proof of Lemma 2.3. By symmetry it is sufficient to prove that the integrals
Z ∞Z ∞
Gρ (0, −ξ1 , −ξ2 )dξ1 dξ2
I1 (0, 0) :=
,
hg(0, ξ1 , ξ2 )i2b
0
0
Z
Z
∞
∞
Gρ (0, −ξ1 , ξ2 )dξ1 dξ2
I2 (0, 0) :=
hg(0, ξ1 , −ξ2 )i2b
0
0
are finite. We will prove that I1 (0, 0) is finite only; the same proof works for I2 (0, 0).
Also, by symmetry we can suppose that 0 ≤ ξ2 ≤ ξ1 . We have
Z ∞
Z ξ1
Z ∞
Z ξ1 /2
Z ∞
Z ξ1
Gρ (0, −ξ1 , −ξ2 )
dξ1
dξ2
=
dξ
dξ
+
dξ
dξ2
1
2
1
hg(0, ξ1 , ξ2 )i2b
(2.7)
1
0
1
0
1
ξ1 /2
=I1,1 + I1,2 .
Since 0 ≤ ξ2 ≤ ξ1 , we have Gρ (0, −ξ1 , −ξ2 ) ≤ hξ1 i4ρ hξ2 i2ρ . In I1,1 we have ξ1 /2 <
ξ1 − ξ2 < ξ1 , therefore if b > ρ + 1/3,
Z ∞
Z ξ1 /2
hξ2 i2ρ dξ2
4ρ
I1,1 .
hξ1 i dξ1
h3ξ12 ξ2 i2b
1
0
Z 3ξ13 /2
Z ∞
1
1
x2ρ dx 1
.
hξ1 i4ρ 2 + 2+4ρ + 2+4ρ
dξ1
ξ1
(1 + x)2b
ξ1
ξ1
1
1
=C(ρ, b) < ∞.
6
XAVIER CARVAJAL
EJDE-2004/13
Analogously we can prove that I1,1 = ∞ if b ≤ ρ + 1/3. In I1,2 we have ξ1 /2 ≤
ξ2 ≤ ξ1 , so
Z ∞
Z ξ1
hξ1 − ξ2 i2ρ dξ2
4ρ
I1,2 .
hξ1 i dξ1
2 2b
1
ξ1 /2 h(ξ1 − ξ2 )ξ1 i
Z ∞
Z ξ1 /2
hxi2ρ dx
=
hξ1 i4ρ dξ1
hξ12 xi2b
1
0
. C(ρ, b), b > ρ + 1/3.
The propositions will be useful for proving Lemmas 2.2 and 2.4.
Proposition 1. Let 0 ≤ ρ < 1/4, b > 1/3 + 2ρ/3, then we have
Z
dξ1 dξ2
J1 = ξ 2+4ρ
≤ C,
3
2b
R2 hξ (z + F )i
where C is a constant independent of ξ and z.
Proof. If ξ1 ≤ 0, ξ2 ≤ 0, then |z + F | ≥ |ξ1 + ξ2 ||ξ1 ξ2 |. Therefore by Lemma 2.3 and
by symmetry, it is enough to consider ξ1 ≥ 0. We have |z + F | = |ξ1 ||(ξ2 + (ξ1 −
2)/2)2 − (ξ1 − 2)2 /4 − z/ξ1 |. Let l2 = (ξ1 − 2)2 /4 + z/ξ1 , c(ρ) = (2 + 4ρ)/3, then
making change of variable η = ξ2 + (ξ1 − 2)/2 and using (2.2) and (2.3) we have
Z ∞
Z
dη
J1 =ξ 2+4ρ
dξ1
3 ξ (η 2 − l2 )i2b
hξ
1
R
Z ∞ 0Z
l dx
.
dξ1
2 2
c(ρ)
0
R [|ξ1 |l |x − 1|]
Z
Z ∞
dx
dξ1
.
|ξ1 |c(ρ) |ξ1 − 2|(1+8ρ)/3 R |x2 − 1|c(ρ)
0
. 2−(20ρ+1)/3 ,
0 < ρ < 1/4.
The case ρ = 0 follows from the case 0 < ρ < 1/4, taking the limit.
Remark. When z < 0, we make ξ1 := −ξ1 , ξ2 := −ξ2 then |z + F | = |ξ1 ||(ξ2 +
(ξ1 + 2)/2)2 − (ξ1 + 2)2 /4 + z/ξ1 | and the proof is similar.
Proposition 2. Let |ξ| > 1, b > 1/2, 0 ≤ ρ < 1/4. Then
Z ∞
Z
dξ2
4ρ
2+4ρ
J2 = ξ
ξ1 dξ1
≤ C,
3 (z + F )i2b
hξ
0
R
where C is a constant independent of ξ and z.
Proof. By Proposition 1 we can suppose ξ1 > 4, so (ξ1 − 2) > ξ1 /2. Using (2.4)
and making change of variables as above, we have
Z
ξ 2+4ρ ∞ ξ14ρ
J2 .
dξ1 ≤ C.
|ξ|3 4 ξ1 l
Proof of Lemma 2.4. Case |ξ| ≤ 1. Let
A1 = {(ξ1 , ξ2 )/|ξ1 | > 2, |ξ2 | > 2},
A2 = {(ξ1 , ξ2 )/|ξ1 | ≤ 2, |ξ2 | ≤ 2},
A3 = {(ξ1 , ξ2 )/|ξ1 | ≤ 2, |ξ2 | > 2},
A4 = {(ξ1 , ξ2 )/|ξ1 | > 2, |ξ2 | ≤ 2} .
EJDE-2004/13
LOCAL WELL-POSEDNESS
7
P4
Consider I(ξ, 0) = j=1 Ij (ξ, 0), where Ij (ξ, 0) is defined in the region Aj . Obviously I2 (ξ, 0) ≤ C. In A1 we have |ξ − ξ1 | > |ξ1 |/2 and |ξ + ξ2 | > |ξ2 |/2, therefore
Lemma 2.3 gives I1 (ξ, 0) ≤ C. In A3 we have |ξ + ξ2 | > |ξ2 |/2, and consequently
Z
hξ2 i4ρ dξ1 dξ2
1
I3 (ξ, 0) . 2ρ
hξi
h(ξ1 − ξ2 )ξ2 (ξ − ξ1 )i2b
ZA3
Z
1
1
= 2ρ
+ 2ρ
hξi
hξi
A3 ∩{|ξ1 −ξ2 |>|ξ2 |}
A3 ∩{|ξ1 −ξ2 |≤|ξ2 |}
=I3,1 (ξ, 0) + I3,2 (ξ, 0).
In the first integral, for ρ < 1/4, b > 1/2 we have
Z
Z
dξ1
1
4ρ
hξ
i
dξ
I3,1 (ξ, 0) .
2
2
2 (ξ − ξ )i2b
hξi2ρ |ξ2 |>2
hξ
1
|ξ1 |≤2
2
Z
4ρ
1
hξ2 i dξ2
.
≤ C.
hξi2ρ |ξ2 |>2
ξ22
To estimate I3,2 (ξ, 0) we make the change of variables η2 = ξ1 − ξ2 , η1 = ξ1 and as
|ξ1 | ≤ 2 we obtain the same estimate as that for I3,1 (ξ, 0).
By symmetry we can estimate I4 (ξ, 0) in the same manner as I3 (ξ, 0).
Case |ξ| > 1. Let us consider I(ξ, 0) in the form (2.6) and let B1 = {|ξ1 + ξ2 | > 4}
and B2 = {|ξ1 + ξ2 | ≤ 4}, then I(ξ, 0) = I1 (ξ) + I2 (ξ), where Ij (ξ) is defined in Bj .
In B1 we have
|2 − (ξ1 + ξ2 )| > |ξ1 + ξ2 |/2,
|1 − (ξ1 + ξ2 )| ≤ 5|ξ1 + ξ2 |/4,
(2.8)
moreover B1 ⊂ {|ξ1 | ≥ 2} ∪ {|ξ2 | ≥ 2} =: B1,1 ∪ B1,2 and therefore I1 (ξ) ≤
I1,1 (ξ) + I1,2 (ξ), where I1,j (ξ) is defined in B1,j ∩ B1 . In B1,1 we have |ξ1 |/2 ≤
|1 − ξ1 | ≤ 3|ξ1 |/2, therefore using (2.8), we obtain that I1,1 (ξ) . I(0, 0) ≤ C if
ρ < 1/4, ρ + 1/3 < b. In similar manner we have I1,2 (ξ) . I(0, 0) ≤ C.
From definition of B2 we have Hρ . hξi2ρ hξ + ξ|ξ1 |i4ρ , so using symmetry and
Propositions 1 and 2, we have I2 (ξ) ≤ C < ∞ if 0 ≤ ρ < 1/4, b > ρ + 1/3.
Proof of Lemma 2.2. Let 0 ≤ ρ < 1/4, b > 7/12, b0 < −ρ/3. Using symmetry and
Lemma 2.4 it is sufficient to prove
Z ∞Z
Hρ (ξ, ξ1 , ξ2 )dξ1 dξ2
J = p(ξ, z)
≤ C < ∞.
3
2b
0
R hξ (z + F (ξ1 , ξ2 )i
By Lemma 2.4 we can suppose |ξ|3 z ≥ 1; since if |ξ|3 z < 1,
hξ 3 (z + F )i−2b ≤ 22b hξ 3 F i−2b .
Also by symmetry we can suppose |ξ2 | ≤ |ξ1 |. Therefore
Hρ (ξ, ξ1 , ξ2 ) . 1 + |ξ|6ρ + |ξ|6ρ |ξ1 |6ρ .
Using Proposition 1 we can suppose |ξ1 | > 4 (l−1 ≤ |ξ1 |−1 ).
Case |ξ||ξ1 | ≤ 1. We have Hρ . hξi6ρ and therefore J ≤ C < ∞, by Proposition
1.
Case |ξ||ξ1 | > 1.
i) If |ξ1 |3 ≤ z, |ξ1 | ≤ z 1/3 , we have Hρ (ξ, ξ1 , ξ2 ) . 1 + |ξ|6ρ + |z|2ρ/3 |ξ|6ρ |ξ1 |4ρ .
8
XAVIER CARVAJAL
EJDE-2004/13
Therefore using (2.4), in this region we have
ξ 2+6ρ |z|2ρ/3
hξ 3 zi−2b0
Z
|z|1/3
4ρ
Z
|ξ1 | dξ1
1/|ξ|
R
ξ 2+6ρ |z|2ρ/3
dη
. 3 −2b0 3
3
2
2
2b
hξ ξ1 (η − l )i
hξ zi
|ξ|
Z
∞
1/|ξ|
|ξ1 |4ρ dξ1
|ξ1 |2
(|ξ|3 z)2ρ/3
. 3 −2b0
hξ zi
≤C.
ii) If |ξ1 |3 ≥ z, |ξ1 | ≥ z 1/3 , we can proceed as follows. By Lemma 2.4 we can
suppose |z + F | ≤ |F |/2, so |F | ≤ 2z, |(2 − (ξ1 + ξ2 ))ξ1 ξ2 | ≤ 2z. This implies that
|1 − ξ2 ||1 − (ξ1 + ξ2 )| . 1 + |ξ1 | + z 2/3 . Therefore
Hρ .(hξi4ρ + |ξ|6ρ ) + |ξ|4ρ |ξ1 |4ρ + |ξ|6ρ |ξ1 |2ρ + |ξ|4ρ |ξ1 |2ρ + |ξ|6ρ |ξ1 |4ρ
4ρ 4ρ/3
+ |ξ| z
6ρ 4ρ/3
+ |ξ| z
6ρ 4ρ/3
+ |ξ| z
2ρ
|ξ1 |
=:
8
X
lj .
j=1
We have,
|ξ|6ρ
≤ |ξ|4ρ .
hξi2ρ
(2.9)
To estimate the term that contains l1 = hξi4ρ + |ξ|6ρ , we use (2.9) and Proposition
1.
For terms lj , j = 2, . . . , 5, we use (2.9) and Propositions 1 and 2 if |ξ| > 1. If
|ξ| < 1, we integrate in the region ξ1 > 1/|ξ| as above.
In l6 = |ξ|4ρ z 4ρ/3 , we have
Z ∞
dξ1
1
|ξ|2 |ξ|4ρ z 4ρ/3
.
≤ C.
0
2
3
−2b
3
2ρ
3
(1−4ρ)/3
hξ zi
|ξ| hξi
(|ξ| z)
z 1/3 ξ1
We estimate l7 = |ξ|6ρ z 4ρ/3 , as in l6 using (2.9). Finally in l8 = |ξ|6ρ z 4ρ/3 |ξ1 |2ρ , we
have
Z ∞
|ξ|2+6ρ z 4ρ/3
|ξ1 |2ρ dξ1
(|ξ|3 z)(6ρ−1)/3
.
≤ C.
0
2
3
−2b
2ρ
3
hξ zi
hξi |ξ| z1/3
ξ1
hξ 3 zi−2b0
Remark. In the case α 6= 0 under little modifications, the proofs of Propositions
1 and 2 and the proofs of Lemmas 2.2, 2.3 and 2.4 are similar to the case α = 0.
For example in order to prove Lemma 2.3 with α 6= 0 we proceed as follows
In (2.5) we have g(ξ, ξ1 , ξ2 ) = (ξ1 − ξ2 )ξ2 (2α − 3βξ1 )). In order to obtain symmetry in ξ1 and ξ2 , we consider the change of variable 2α − 3βξ1 := 3βξ1 . In this
way we have
Z
α
hξ1 + ξ2 i2ρ hξ1 i2ρ hξ2 i2ρ dξ1 dξ2
I(0, 0) . C
(2.10)
D E2ρ .
β
2α − (ξ + ξ ) ξ ξ
β
2
1
2
1
2
R
β
Now using symmetry, the rest of the proof is the same as that of Lemma 2.3, if we
replace the lower limit 1 in the integrals in (2.7) by 4α/3β.
EJDE-2004/13
LOCAL WELL-POSEDNESS
9
Proof of Theorem 1.2
Consider a cut-off function ψ ∈ C ∞ , such that 0 ≤ ψ ≤ 1,
(
1 if |t| ≤ 1
ψ(t) =
0 if |t| ≥ 2,
and let ψT (t) := ψ(t/T ). To prove Theorem 1.2 we need the following result.
Proposition 3. Let −1/2 < b0 ≤ 0 ≤ b ≤ b0 + 1, T ∈ [0, 1]. Then
kψ1 (t)U (t)u0 k{s,b} = Cku0 kHs ,
Z
kψT (t)
t
0
U (t − t0 )F (t0 , ·))dt0 k{s,b} ≤ CT 1−b+b kF (u)k{s,b0 } ,
(2.11)
(2.12)
0
where F (u) := iγ|u|2 u.
The proof of (2.11) is obvious, and the proof of (2.12) is practically done in [8].
Let us consider (1.8) in its equivalent integral form:
Z t
u(t) = U (t)u0 −
U (t − t0 )F (u)(t0 , ·)dt0 .
(2.13)
0
Note that, if for all t ∈ R, u(t) satisfies:
t
Z
U (t − t0 )F (u)(t0 , ·)dt0 ,
u(t) = ψ1 (t)U (t)u0 − ψT (t)
0
then u(t) satisfies (2.13) in [−T, T ]. Let a > 0 and
Xa = {v ∈ X s,b ; kvk{s,b} ≤ a}.
For v ∈ Xa fixed, let us define
Z
Φ(v) = ψ1 (t)U (t)u0 − ψT (t)
t
U (t − t0 )F (v)(t0 , ·)dt0 .
0
Let = 1 − b + b0 > 0, b − 1 < b0 < s/3 (this implies 7/12 < b < 11/12) using
Proposition 3 and Theorem 1.3 we obtain
kΦ(v)ks,b ≤ Cku0 kHs + CT kF (v)ks,b0 ≤ Cku0 kHs + CT a3 ≤ a,
where a = 2Cku0 kHs and T ≤ 1/(2Ca2 ).
We can prove that Φ is a contraction in an analogous manner. The proof of
Theorem 1.2 follows by using a standard argument, see for example [11, 12].
Acknowledgment. The author wants to thank the anonymous referee for his/her
valuable suggestions, also to M. Panthee and J. Salazar for their help in improving
the presentation of this article.
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Xavier Carvajal
IMECC-UNICAMP, Caixa Postal:6065, 13083-859, Barão Geraldo, Campinas, SP, Brazil
E-mail address: carvajal@ime.unicamp.br