WELL-POSEDNESS FOR THE ZK EQUATION IN A CYLINDER
AND ON THE BACKGROUND OF A KDV SOLITON
FELIPE LINARES1 , ADEMIR PASTOR1∗ AND JEAN-CLAUDE SAUT2
1
IMPA - Instituto Nacional de Matemática Pura e Aplicada,
Estrada Dona Castorina 110, CEP 22460-320, Rio de Janeiro, RJ, Brazil.
2
Université de Paris-Sud,
UMR de Mathématiques, Bât. 425, 91405 Orsay Cedex, France.
Abstract. We establish the well-posedness of two Cauchy problems for the
two-dimensional Zakharov-Kuznetsov equation motivated by the study of transverse stability properties of the N -soliton φN of the Korteweg-de Vries equation. They differ by the functional setting: Sobolev spaces H s (R × T), s > 23
for the first one, φN + H 1 (R2 ) for the second one. In the latter case, the
solution is shown to be global.
1. Introduction
The Zakharov-Kuznetsov (ZK) equation
ut + u ∂x u + ∂x ∆u = 0,
u = u(x, y, z, t),
(x, y, z) ∈ Rd , t ∈ R, d = 2, 3
(1.1)
was introduced as an asymptotic model in [18] (see also [9]) to describe the propagation of nonlinear ionic-sonic waves in a magnetized plasma.
The ZK equation is a natural multi-dimensional extension of the Korteweg-de
Vries (KdV) equation, quite different from the well-known Kadomtsev-Petviasvili
(KP) equation though. In particular, it admits as a solution the well-known KdV
solitary wave φω (x, t) = φω (x − ωt), where
φω (ξ) = 3ω sech2
√
ω
ξ ,
2
ω > 0.
(1.2)
More generally, the N -soliton φN of the KdV equation (see [2], [12]) is also
a particular solution of the ZK equation which is smooth and bounded together
with its time and space derivatives and behaves as a sum of solitons of velocities
4n2 , 1 ≤ n ≤ N when t → ∞.
For instance (see [2]), the 2-soliton φ2 is given by
φ2 (x, t) = 72
3 + 4 cosh(2x − 8t) + cosh(4x − 64t)
.
{3 cosh(x − 28t) + cosh(3x − 36t)}2
(1.3)
∗ Current address:IMECC-UNICAMP, Rua Sérgio Buarque de Holanda 651, 13083-859, Campinas, SP, Brazil.
1
2
F. LINARES, A. PASTOR AND J.-C. SAUT
A fundamental issue is that of the transverse stability/instability of those onedimensional “localized” solutions of the KdV equation (such as the solitary wave)
with respect to transverse perturbations governed by the ZK equation. This question was rigorously addressed recently by Rousset and Tzvetkov ([14], [15]) who
developed a general theory which applies in particular to one-dimensional transverse perturbations of the KdV solitary wave.
The aim of the present paper is to present a functional framework for the Cauchy
problem which should be suitable to describe the aforementioned transverse perturbations. This framework cannot be (as for instance in [1], [5], [10], [11]) the
classical Sobolev spaces H s (Rd ) since the KdV soliton or multi-soliton do not belong to this class of spaces. A natural space to study the transverse stability of
localized one-dimensional solutions should contain those solutions.
A first possibility consists in functions which are “localized” in x and periodic in
y. This leads to our study of the Cauchy problem for the ZK equation in H s (R×T).
A second possibility (which has been carried out in [13] for similar problems for
the KP-I equation) is to consider two-dimensional “localized” perturbations of the
one-dimensional solution φ. This motivates the study of the Cauchy problem (1.5)
below.
Let T = R/2πZ be the one-dimensional torus. We will thus consider the initial
value problem (IVP) associated to the ZK equation in a cylinder
(
∂t u + ∂x ∆u + u∂x u = 0, (x, y) ∈ R × T, t > 0,
(1.4)
u(x, y, 0) = u0 (x, y)
and the IVP,
ut + ∂x ∆u + u∂x u + ∂x (φu) = 0,
u(x, y, 0) = u0 (x, y).
(x, y) ∈ R2 , t > 0
(1.5)
where ∆ = ∂x2 + ∂y2 is the Laplace operator and φ is the KdV solitary wave or
more generally any N -soliton of the KdV equation. Actually, we will only use
that φ = φ(x, t) is a solution of the KdV equation which is smooth and bounded
together with its time and space derivatives, and furthermore belongs to the space
L2x L∞
T (see the Notations below). Those assumptions are obviously satisfied by the
N -soliton solution of the KdV equation.
The paper is organized as follows. We will first consider (1.4) and prove that
this Cauchy problem is locally well-posed for data in H s (R × T) where s > 32 . We
adapt the method used by Ionescu and Kenig [3] to the KP-I equation in the same
situation.
Then we turn to the perturbed problem (1.5). Using suitable dispersive estimates on the linear group we prove the local well-posedness for data in H 1 (R2 ).
The solution is shown to be global from energy identities inherited from the two
conservation laws of the unperturbed ZK equation.
Before leaving this section we would like to point out that recently Ionescu,
Kenig and Taturu [4] put forward a method to establish global well-posedness in
the energy space for the Cauchy problem associated to the KP-I equation. We do
not know whether or not this method can be extended to show a similar result for
the IVP (1.4).
ZK EQUATION
3
Notation. Let Ω = R × R or Ω = R × T. Given f ∈ S 0 (Ω), fb(ξ, η) denotes its
Fourier transform; here and throughout it is understood that (ξ, η) ∈ Λ, where Λ
denotes the “dual” set of Ω, that is, Λ = R × R if Ω = R × R and Λ = R × Z if
Ω = R × T. As usual in periodic settings, we replace the variable η by n, in the
case η ∈ Z.
Let s ∈ R. The
space H s (Ω) denotes the usual Sobolev space of order s
T linear
∞
s
and H (Ω) = s H (Ω). On S 0 (R × T), define the operators Jxs , Jys and J s , via
Fourier transform, by
s g(ξ, n) = (1 + ξ 2 )s/2 g
Jd
b(ξ, n);
x
2 s/2
s
Jd
gb(ξ, n);
y g(ξ, n) = (1 + n )
s g(ξ, n) = (1 + ξ 2 + n2 )s/2 g
Jd
b(ξ, n).
For any set A let χA be its characteristic function. Given a Banach space Y, a
measurable function u : R → Y and p ∈ [1, ∞], we define
kukLp Y = kku(t)kY kLp .
If I ⊂ R is a measurable set and u : I → Y is a measurable function, we set
kukLpI Y = kχI (t)ukLp Y .
For short, if T > 0, we denote kukLp[−T ,T ] Y simply by kukLpT Y .
For integers k = 0, 1, 2, . . . we define the operators Qkx and Qky on H ∞ (R × T) by
d
0 g(ξ, n) = χ
Q
g (ξ, n),
[0,1) (|ξ|)b
x
d
k g(ξ, n) = χ k−1 k (|ξ|)b
Q
g (ξ, n) if k ≥ 1,
[2
,2 )
x
d
0 g(ξ, n) = χ
Q
g (ξ, n),
[0,1) (|n|)b
y
d
k g(ξ, n) = χ k−1 k (|n|)b
Q
g (ξ, n) if k ≥ 1.
[2
,2 )
y
and
2. ZK on a cylinder
We will follow the arguments of Ionescu and Kenig [3] to establish local wellposedness for the IVP (1.4). The first ingredient in their analysis is a localized
version of the Strichartz estimates.
2.1. Localized Strichartz Estimates. Consider the solution of the linear IVP
(
∂t u + ∂x ∆u = 0,
(x, y) ∈ R × T, t ∈ R,
(2.6)
u(x, y, 0) = u0 (x, y),
that is, u(x, y, t) = W (t)u0 (x, y), where
XZ
3
2
W (t)u0 (x, y) =
ei(xξ+yn+t(ξ +ξn )) u
b0 (ξ, n) dξ.
n∈Z
(2.7)
R
Our main theorem in this subsection reads as follows.
Theorem 2.1. Assume φ ∈ S(R × T). Then, for any > 0,
kW (t)Qky Qjx φkL2−k−j L∞
≤ C 2(−1/2+)j kQky Qjx φkL2xy .
xy
2
(2.8)
4
F. LINARES, A. PASTOR AND J.-C. SAUT
Proof. The proof is similar to the proofs of Theorems 9.3.1 and 9.3.2 in [3]. So, here
we give only the main steps. Define g(ξ, n) = (Qxy Qjx φ)∧ (ξ, n). Let ψ0 , ψ1 : R →
[0, 1] denote smooth even functions such that supp ψ0 ⊂ [−2, 2], supp ψ1 ⊂ {r; |r| ∈
[1/4, 4]}; ψ0 ≡ 1 in the interval [−1, 1] and ψ1 ≡ 1 in the set {r; |r| ∈ [1/2, 2]}.
Then we can write
Z X
k j
W (t)Qy Qx φ(x, y) =
g(ξ, n)ψ1 (ξ/2j )ψ0 (n/2k )ei(ξx+ny+F (ξ,n)t) dξ,
R n∈Z
where
F (ξ, n) = ξ 3 + ξn2 .
(2.9)
Using the same argument as in the proof of Theorem 9.3.2 of [3], we see that (2.8)
reduces to proving that
Z
X
(2.10)
ψ12 (ξ/2j )ψ02 (n/2k )ei(ξx+ny+F (ξ,n)t) dξ ≤ C2l−j
R
n∈Z
for any x ∈ R, y ∈ [0, 2π), |t| ∈ [2−l , 2−l+1 ] and l ≥ k + j. The case j = 0 is
immediate, because we have a size estimate and l ≥ k + j. If j > 0, we use the
Poisson summation formula (see e.g., [17, Chapter VII]):
X
X
b
G(n) =
G(2πν),
G ∈ S(R),
n∈Z
ν∈Z
to estimate the summation in n first. Since n ∼ 2k , ξ ∼ 2j and |t| ∼ 2−l ≤ 2−k−j ,
we use (2.9) and replace the sum in n with a sum of C integrals, modulo acceptable
errors. So that it suffices to prove that
Z
Z
ψ12 (ξ/2j )ei(ξx+ξ3 t) ψ02 (η/2k )ei(ηy0 +ξη2 t) dηdξ ≤ C2l−j
(2.11)
R
0
−l
R
−l+1
for any x, y ∈ R and t ∈ [2 , 2
]. But since
Z
Z
0
2
C
c2 (η /2k )e−i(y0 +η1 /2k )2 /(4ξt) dη ,
ψ
ψ02 (η/2k )ei(ηy +ξη t) dη =
1
0 1
1/2
|ξt|
R
R
we see that it suffices to show that
Z
1
2
j i(ξx+ξ 3 t)
(2.12)
dξ ≤ C2l−j .
|tξ|1/2 ψ1 (ξ/2 )e
R
Since ξ ∼ 2j and |t| ∼ 2−l , for (2.12), we only have to prove that
Z
ψ12 (ξ/2j )ei(ξx+ξ3 t) dξ ≤ C2(l−j)/2 .
(2.13)
R
But (2.13) follows immediately from Van-der Corput’s lemma (see e.g. [16, Chapter
8]). This proves the theorem.
2.2. Key estimates. We will show that the initial value problem (1.4) is locally
well-posed in the functional space X s defined by the norm
|||f |||X s := kJxs f kL2 (R×T) + kJys f kL2 (R×T)
(2.14)
for s > 3/2.
Remark 2.2. Using Kato’s theory we can prove that the IVP is locally well posed
for data in H s (R × T), s > 2.
Our main key estimate is given next.
ZK EQUATION
5
Lemma 2.3. Let u ∈ C([0, T ] : H ∞ (R × T)) ∩ C 1 ([0, T ] : H ∞ (R × T)), F ∈
C([0, T ] : H ∞ (R × T)), T ∈ [0, 1], and
∂t u + ∂x ∆u = ∂x F
R × T × [0, T ].
on
(2.15)
Then, for any s1 > 0 and s2 > 1/2,
1/2
s1 s2
s1
∞ L2
1
kukL1T L∞
≤
c
T
kJ
J
uk
+
kJ
F
k
2
s
,
s
L
LT Lxy .
x y
x
1
2
xy
T
xy
(2.16)
Proof. It will be sufficient to show that for j ≥ 0, k ≥ 1,
s
s1 s2
−(j+k)/2 1/2
∞ L2 +kJ 1 F k 1
kJ
J
uk
kQky Qjx ukL1T L∞
≤
c
2
T
2
s
,
s
L
LT Lxy . (2.17)
x
x y
1
2
xy
T
xy
To prove (2.17), divide the interval [0, T ] into 2j+k equal intervals of length T 2−j−k ,
denoted by [aj,m , aj,m+1 ), m = 1, . . . , 2j+k .
The Cauchy-Schwarz inequality yields
kQky Qjx ukL1T L∞
xy
≤
j+k
2X
kχ[aj,m ,aj,m+1 ) (t)Qky Qjx ukL1T L∞
xy
m=1
≤ c 2−(j+k)/2 T 1/2
j+k
2X
(2.18)
kχ[aj,m ,aj,m+1 ) (t)Qky Qjx ukL2T L∞
.
xy
m=1
Using the Duhamel formula, for t ∈ [aj,m , aj,m+1 ], we obtain
Zt
u(t) = W (t − aj,m )[u(aj,m )] +
W (t − s)[∂x F (s)] ds,
(2.19)
aj,m
Thus, from (2.8) it follows that
kχ[aj,m ,aj,m+1 ) (t)Qky Qjx ukL2T L∞
≤ c()2(−1/2+/2)j kQky Qjx u(aj,m )kL2xy
xy
+c()2(−1/2+/2)j 2j kχ[aj,m ,aj,m+1 ) (t)Qky Qjx F kL1T L2xy .
(2.20)
Now, from (2.18) and (2.20), we deduce
kQky Qjx ukL1T L∞
xy
≤ c()2
−(j+k)/2
T
1/2
j+k
2X
−j−k
2
kJxs1 Jys2 u(aj,m )kL2xy + kJxs1 F kL1T L2xy .
m=1
(2.21)
6
F. LINARES, A. PASTOR AND J.-C. SAUT
Indeed, the left hand side of (2.17) is bounded by
−(j+k)/2
c2
T
1/2
j+k
2X
h
2(−1/2+/2)j kQky Qjx u(aj,m )kL2xy
i
m=1
h
i
+ 2(−1/2+/2)j 2j kχ[aj,m ,aj,m+1 ) (t)Qky Qjx F kL1T L2xy
≤ c2
−(j+k)/2
+ c2
T
−(j+k)/2
1/2 (−1/2+/2)j
2
T
j+k
2X
kQky Qjx u(aj,m )kL2xy
m=1
1/2 (−1/2+/2)j j
2
2
(2.22)
kQky Qjx F kL1T L2xy
j+k
2X
0
k2(1/2+ )k Qky 2j Qjx u(aj,m )kL2xy
≤ c2−(j+k)/2 T 1/2 2−j−k
m=1
+
ckQky
2j Qjx F kL1T L2xy ,
where in the last term we used 2(−1)k/2 ≤ 1 for k ≥ 1. The inequality (2.21) then
follows.
Before going on, let us recall the periodic version of the Kato–Ponce commutator.
Lemma 2.4. Let s ≥ 1 and f, g ∈ H ∞ (R × T). Then
h
i
kJys (f g) − f Jys gkL2y ≤ Cs kJys f kL2y kgkL∞
+ (kf kL∞
+ k∂y f kL∞
)kJys−1 gkL2y .
y
y
y
Proof. See Lemma 9.A.1 in [3].
Lemma 2.5. Let u be a solution of the IVP (1.4) with u0 ∈ H ∞ (R × T). Then for
any s ≥ 1, we have, for any T ∈ [0, 1], that
sup |||u(t)|||X s ≤ cs exp cs (kukL1T L∞
+k∂x ukL1T L∞
+k∂y ukL1T L∞
) ku0 kX s . (2.23)
xy
xy
xy
0<t<T
Proof. We apply Jxs to the equation in (1.4), multiply by Jxs u, integrate by parts
and use the Cauchy-Schwarz inequality to obtain
Z
Z
1
1 d s 2
kJx ukL2xy = −
∂x u (Jxs u)2 − [Jxs , u]∂x uJxs u
2 dt
2
(2.24)
s
2
s
s
2
2
≤ k∂x ukL∞
kJ
uk
+
k[J
,
u]∂
uk
kJ
uk
.
2
x
Lxy
Lxy
x
x
x
Lxy
xy
Using the Kato-Ponce commutator estimate [6], we obtain
k[Jxs , u]∂x ukL2xy ≤ ck∂x ukL∞
kJxs ukL2xy .
xy
Combining (2.24) and (2.25), we have
1 d s 2
kJ uk 2 ≤ c k∂x ukL∞
kJxs uk2L2xy .
xy
2 dt x Lxy
A similar argument shows us that
1 d s 2
s
2
s
s
2 k[J , u]∂x ukL2
kJy ukL2xy ≤ c k∂x ukL∞
kJ
uk
+
kJ
uk
.
2
L
y
y
y
Lxy
xy
xy
xy
2 dt
Now, an application of Lemma 2.4 yields
k[Jys , u]∂x ukL2xy
≤ Cs k∂x ukL∞
kJys ukL2xy+(kukL∞
+ k∂y ukL∞
)kJys−1 ∂x ukL2xy .
xy
xy
xy
(2.25)
(2.26)
(2.27)
(2.28)
ZK EQUATION
7
Moreover, by using the Young inequality, we see that
kJys−1 ∂x ukL2xy ≤ c kJxs ukL2xy + kJys ukL2xy .
(2.29)
Thus, from (2.27), (2.28) and (2.29), we deduce
1 d s 2
∞ + k∂y ukL∞
kJy ukL2xy ≤ c kukL∞
+
k∂
uk
|||u|||2X s .
(2.30)
x
L
xy
xy
xy
2 dt
Gathering together (2.26) and (2.30), and using Gronwall’s inequality we obtain
(2.23).
2.3. A priori estimate. First we recall the continuous Leibniz rule.
Lemma 2.6. Let 1 < p < ∞ and 0 < s < 1. Then,
kJxs (f g)kLpx ≤ c kJxs f kL∞
kJxs gkLpx .
kgkLpx + kf kL∞
x
x
Proof. See Kenig, Ponce and Vega [8, Theorem A.12].
We also need the following lemma.
Lemma 2.7. For every 0 < s < 1 there exists a constant C such that for every
∞
u ∈ L∞
xy satisfying ∂x u ∈ Lxy , one has
∞ + k∂x ukL∞
kJxs ukL∞
≤
C
kuk
.
L
xy
xy
xy
Proof. See Molinet, Saut and Tzvetkov [13, Lemma 7].
∞
Proposition 2.8. Let u be a solution of (1.4) with u0 ∈ H (R × T) defined in
[0, T ]. Let
|||u0 |||X s = kJxs u0 kL2 (R×T) + kJys u0 kL2 (R×T) .
Then for any s > 3/2, there exists T = T (ku0 kX s , s) and a constant cT (ku0 kX s , s)
such that
ZT f (T ) :=
ku(t)kL∞
+ k∂x u(t)kL∞
+ k∂y u(t)kL∞
dt ≤ cT .
(2.31)
xy
xy
xy
0
Proof. We first note that u, ∂x u and ∂y u satisfy Lemma 2.3 with F given, respectively, by u2 /2, ∂x (u2 )/2 and ∂y (u2 )/2. Hence, for any s1 > 0 and s2 > 1/2, we
obtain
s1 s2
s1 s2
2
2 +kJ
2 +kJ
f (T ) ≤ c T 1/2 kJxs1 Jys2 ukL∞
x Jy ∂x ukL∞
x Jy ∂y ukL∞
T Lxy
T Lxy
T Lxy
ZT
+
kJxs1 (u2 )kL2xy
ZT
dt +
0
kJxs1 ∂x (u2 )kL2xy dt+
0
ZT
(2.32)
kJxs1 ∂y (u2 )kL2xy dt .
0
Obviously, we have
kJxs1 Jys2 ukL2xy ≤ c kJxs ukL2xy + kJys ukL2xy .
(2.33)
Moreover, since s > 3/2, by choosing s1 > 0 small enough and s2 > 1/2 sufficiently
close to 1/2, using the Young inequality (with p = 3/2 and p0 = 3), we can estimate
i2
X Z h
kJxs1 Jys2 ∂x uk2L2xy ≤ c
(1 + |ξ|)(1+s1 )3/2 + (1 + |n|)3s2 |b
u(ξ, n)| dξ
n
≤c
R
kJxs uk2L2xy
+ kJys uk2L2xy .
(2.34)
8
F. LINARES, A. PASTOR AND J.-C. SAUT
Also, using the Young inequality (with p = (1 + s1 )/s1 and p0 = 1 + s1 ), we obtain
i2
X Z h
u(ξ, n)| dξ
kJxs1 Jys2 ∂y ukL2xy ≤ c
(1 + |ξ|)(1+s1 ) + (1 + |n|)(s2 +1)(s1 +1) |b
R
n
≤c
kJxs ukL2xy
+ kJys ukL2xy .
(2.35)
Now, applying Lemma 2.6, we get
Z T
Z T
2
kJxs1 u2 kL2xy ≤ c
kukL∞
kJxs1 ukL2xy ≤ ckukL1T L∞
kJxs1 ukL∞
xy
T Lxy
xy
0
0
(2.36)
s,
≤ ckukL1T L∞
kukL∞
T X
xy
T
Z
kJxs1 (u∂x u)kL2xy ≤ c
0
T
Z
0
h
kJxs1 ukL∞
k∂x ukL2xy + kukL∞
kJxs1 ∂x ukL2xy
xy
xy
i
2
2
k∂x ukL∞
+ ckukL1T L∞
kJxs1 ∂x ukL∞
≤ c kJxs1 ukL1T L∞
x Lxy
T Lxy
xy
xy
(2.37)
s + ckuk 1
s
kukL∞
kukL∞
≤ ckJxs1 ukL1T L∞
LT L∞
T X
T X
xy
xy
and
Z
0
T
kJxs1 (u∂y u)kL2xy
Z
≤ c
0
T
i
h
s1
∞ kJ
2
2
∂
uk
k∂
uk
+
kuk
kJxs1 ukL∞
y
y
L
L
L
x
xy
xy
xy
xy
2
2
k∂y ukL∞
+ ckukL1T L∞
kJxs1 ∂y ukL∞
≤ ckJxs1 ukL1T L∞
x Lxy
T Lxy
xy
xy
(2.38)
s + ckuk 1
s.
kukL∞
kukL∞
≤ ckJxs1 ukL1T L∞
LT L∞
T X
T X
xy
xy
Gathering together (2.33)–(2.38), we obtain from (2.32),
o
n
s1
s
1
1
uk
f (T ) ≤ cT 1/2 kukL∞
1
+
kuk
+
kJ
∞
∞
X
LT Lxy .
LT Lxy
x
T
So, from Lemma 2.7,
n
o
s
1 L∞ + k∂x ukL1 L∞
f (T ) ≤ cT 1/2 kukL∞
1
+
kuk
.
X
L
T
T xy
T xy
(2.39)
Hence, an application of Lemma 2.5 gives us
f (T ) ≤ cs T 1/2 |||u0 |||X s exp cs (kukL1T L∞
+ k∂x ukL1T L∞
+ k∂y ukL1T L∞
)
xy
xy
xy
n
o
× 1 + kukL1T L∞
+ k∂x ukL1T L∞
.
xy
xy
Therefore, if T ≤ T0 (s, |||u0 |||X s ) is small enough,
f (T ) ≤ cT (s, |||u0 |||X s ).
This completes the proof of the proposition.
(2.40)
¿From this point on, one can use similar arguments as in Kenig [7] or Linares
and Saut [11] for proving the following result.
ZK EQUATION
9
Theorem 2.9. Given u0 ∈ X s , s > 3/2, there exist T = T (|||u0 |||X s ) and a unique
solution u of the IVP (1.4), such that u ∈ C([0, T ] : X s ) and u, ∂x u, ∂y u ∈ L1T L∞
xy .
Moreover, the map data-solution u0 ∈ X s 7→ u ∈ C([0, T ] : X s ) is continuous.
Remark 2.10. We can prove a similar result as in Proposition 2.8 (and consequently as in Theorem 2.9) replacing space X s with the space X r1 ,r2 defined by the
norm
ku0 kX r1 ,r2 := kJxr1 + u0 kL2xy + kJyr2 + u0 kL2xy ,
where r1 > 1 and r2 ≥ 3/2 are such that
1
1
+
= 1.
r1
2r2
Indeed, except by estimate (2.34) the proof is the same. An estimate similar to
(2.34) in space X r1 ,r2 can be proved using the Young inequality with p = r1 and
p0 = 2r2 so that
(1 + |ξ|)1+s1 (1 + |n|)s2 ≤ (1 + |ξ|)(1+s1 )r1 + (1 + |n|)2s2 r2
≤ (1 + |ξ|)r1 + + (1 + |n|)r2 + .
Observe that in case r1 = r2 = 3/2, we obtain the estimate in Proposition 2.8.
Remark 2.11. We also note that in spaces X r1 ,r2 we cannot reach indices r2 + ≤
3/2 in the y-direction, just by using the Young inequality. This can be seen just
taking a look at estimate (2.35). In fact, from Young’s inequality it follows that
(1 + |ξ|)s1 (1 + |n|)1+s2 ≤ (1 + |ξ|)s1 p + (1 + |n|)(1+s2 )p
0
with 1 < p, p0 < ∞ and 1/p + 1/p0 = 1. Hence, since s2 > 1/2 we always have
(1 + s2 )p0 > 3/2.
3. The perturbed IVP
Now we consider the IVP (1.5). We recall that φ = φ(x, t) is a solution of
the KdV equation which is smooth and bounded together with its space and time
derivatives. Typical examples are the N -soliton solutions to the KdV equation.
3.1. Preliminary Results. Consider the linear initial value problem
ut + ∂x ∆u = 0,
(x, t) ∈ R2 , t ∈ R,
u(x, y, 0) = u0 (x, y).
(3.41)
The solution of (3.41) is given by the unitary group {U (t)}∞
t=−∞ such that
Z
3
2
u(t) = U (t)u0 (x, y) =
ei(t(ξ +ξη )+xξ+yη) u
b0 (ξ, η)dξdη.
(3.42)
R2
Here, for α ∈ C, the operators Dxα and Dyα are defined via Fourier transform
αb
αb
α
α
[
[
by D
x f (ξ, η) = |ξ| f (ξ, η) and Dy f (ξ, η) = |η| f (ξ, η), respectively. The mixed
space-time norm is defined by (for 1 ≤ p, q, r < ∞)
1/p


!q/r p/q
Z +∞ Z +∞ Z T



|f (x, y, t)|r dt
dy  dx .
kf kLpx Lqy LrT = 
−∞
−∞
0
with usual modifications if either p = ∞, q = ∞ or r = ∞.
Next we give some suitable properties of solution (3.42).
10
F. LINARES, A. PASTOR AND J.-C. SAUT
Proposition 3.1. Let 0 ≤ ε < 1/2 and 0 ≤ θ ≤ 1. Then the group {U (t)}∞
t=−∞
satisfies
(3.43)
kDxθε/2 U (t)f kLqt Lpxy ≤ ckf kL2xy ,
Z ∞
(3.44)
U (t − t0 )g(·, t0 )dt0 kLqt Lpxy ≤ ckgkLq0 Lp0
kDxθε
t
−∞
and
kDxθε
Z
∞
−∞
where
1
p
+
1
p0
=
1
q
+
1
q0
xy
U (t)g(·, t)dtkL2xy ≤ ckgkLq0 Lp0 ,
t
= 1, p =
2
1−θ
and
2
q
=
xy
(3.45)
θ(2+ε)
.
3
Proof. See Linares and Pastor [10].
Lemma 3.2. Let 0 ≤ ε < 1/2. Then the group {U (t)}∞
t=−∞ satisfies
kU (t)f kL2T L∞
≤ cT γ kDx−ε/2 f kL2xy .
xy
(3.46)
where γ = (1 − ε)/6. In particular, if 0 < T ≤ 1,
kU (t)f kL2T L∞
≤ ckf kL2xy .
xy
Proof. The proof is a consequence of Hölder’s inequality (in t) and Proposition
3.1.
Lemma 3.3. The following statements hold:
(i) Let u0 ∈ L2 (R2 ). Then
2
k∂x U (t)u0 kL∞
≤ cku0 kL2xy .
x LyT
(ii) Let u0 ∈ H s (R2 ), s > 3/4. Then,
s ,
kU (t)u0 kL2x L∞
≤ c(s, T )ku0 kHxy
yT
where c(s, T ) is a constant depending on s and T .
Proof. See Faminskii [5].
3.2. Main Theorem. Our main theorem concerning IVP (1.5) reads as follows.
Theorem 3.4. For any u0 ∈ H 1 (R2 ), there exist a unique solution of the IVP
(1.5), defined in the interval R+ , such that for any T > 0
u ∈ C(R+ ; H 1 (R2 )),
2
k∂x2 ukL∞
x LyT
(3.47)
2
+ k∂y ∂x ukL∞
< ∞,
x LyT
k∂x ukL2T L∞
< ∞,
xy
(3.48)
(3.49)
and
kukL2x L∞
< ∞.
yT
(3.50)
1
2
Moreover, there exists a neighborhood W of u0 in H (R ) such that the map u
e0 7→
u
e(t) from W into the class defined by (3.47)–(3.50) is smooth. One has the two
energy identities
Z
Z
1
1 d
2
u (x, y, t)dxdy +
u2 (x, y, t)φ(x, t)dxdy = 0
(3.51)
2 dt R2
2 R2
ZK EQUATION
Z
i
1 dh
1
|∇u|2 (x, y, t) − u3 (x, y, t) − u2 (x, y, t)φ(x, t)dxdy
2 dt R2
3
Z
1
u2 (x, y, t)φt (x, t)dxdy = 0.
+
2 R2
11
(3.52)
Proof. We will prove first the local well-posedness of the IVP (1.5) and consider
the (equivalent) associated integral equation, that is,
Z t
u(t) = U (t)u0 +
U (t − t0 )(uux + ∂x (φu))(t0 )dt0 .
0
As usual, the idea is to apply the Contraction Principle for the integral operator
Z t
Φ(u)(t) = Φu0 (u)(t) = U (t)u0 +
U (t − t0 )(uux + ∂x (φu))(t0 )dt0
0
in the metric space defined by (3.47)–(3.50). We let
XT = {u ∈ C([0, T ]; H 1 (R2 )); |||u||| < ∞}
and
XTa = {u ∈ XT ; |||u||| ≤ a}
with
2
1 + kux kL2 L∞ + k∂ ukL∞ L2
2
|||u||| := kukL∞
+ k∂y ∂x ukL∞
+ kukL2x L∞
,
x
T Hxy
yT
x
x LyT
T xy
yT
where a, T > 0 will be chosen later.
¿From now on, we assume T ≤ 1. To begin with, we estimate the H 1 -norm of
Φ(u). Let u ∈ XT . Then, by using Minkowski’s inequality and group properties,
we get
Z T
Z T
Z T
kΦ(u)(t)kL2xy ≤ ku0 kL2xy+
kuux kL2xy dt0+
kφ0 ukL2xy dt0+
kφux kL2xy dt0
(3.53)
0
0
0
1 + I1 + I2 + I3 .
≤ ku0 kHxy
Now, from Hölder’s inequality, we have
2
I1 ≤ cT 1/2 kuux kL2xyT ≤ cT 1/2 kux kL2T L∞
kukL∞
≤ cT 1/2 |||u|||2 ,
T Lxy
xy
T
Z
I2
kkφ0 ukL2x kL2y dt0 ≤
=
0
Z
0
(3.54)
T
kkφ0 kL∞
kukL2x kL2y dt0
x
(3.55)
T
Z
≤
0
1
kφ0 kL∞
kukL2xy dt0 ≤ T kφ0 kL∞
kukL∞
≤ cT 1/2 |||u|||
x
xT
T Hxy
and, similarly,
1
I3 ≤ T kφkL∞
kukL∞
≤ cT 1/2 |||u|||
xT
T Hxy
(3.56)
where c > 0 is a constant independent of T (because φ(x, t) and φ0 (x, t) are uniformly bounded in x, t). Thus, from (3.53)–(3.56),
1/2
1 + cT
kΦ(u)(t)kL2xy ≤ ku0 kHxy
|||u|||2 + cT 1/2 |||u|||.
(3.57)
12
F. LINARES, A. PASTOR AND J.-C. SAUT
Also,
Z
k∂x Φ(u)(t)kL2xy ≤ k∂x u0 kL2xy +
0
T
k∂x (uux )kL2xy dt0 +
Z
T
0
k∂x (φ0 u + φux )kL2xy dt0
≤ ku0 kH 1 + J1 + J2 .
But, from Hölder’s inequality,
Z T
Z T
J1 ≤
kux ux kL2xy +
kuuxx kL2xy
0
0
≤ T 1/2 kux ux kL2xyT + T 1/2 kuuxx kL2xyT
≤T
1/2
≤ cT
2
kux kL2T L∞
kux kL∞
+T
T Lxy
xy
1/2
1/2
(3.58)
2
kuxx kL∞
kukL2x L∞
yT
x LyT
2
|||u|||
and
T
Z
J2 ≤
0
kφ00 u + 2φ0 ukL2xy +
Z
T
0
kφuxx kL2xy .
The first integral can be estimated as in (3.55). Thus,
J2
≤ cT 1/2 |||u||| + cT 1/2 kφuxx kL2xyT
≤ cT 1/2 |||u||| + cT 1/2 kkkφkL∞
kuxx kL2T kL2y kL2x
T
(3.59)
= cT 1/2 |||u||| + cT 1/2 kkφkL∞
kuxx kL2yT kL2x
T
2
≤ cT 1/2 |||u||| + cT 1/2 kφkL2x L∞
kuxx kL∞
≤ cT 1/2 |||u|||.
T
x LyT
(Here we have used that φ(x, t) ∈ L2x L∞
T ).
Hence, from (3.58) and (3.59), we obtain
1/2
1 + cT
k∂x Φ(u)(t)kL2xy ≤ ku0 kHxy
|||u|||2 + cT 1/2 |||u|||.
(3.60)
Using similar estimates we can show that
1/2
1 + cT
k∂y Φ(u)(t)kL2xy ≤ ku0 kHxy
|||u|||2 + cT 1/2 |||u|||.
(3.61)
Therefore, from (3.57), (3.60) and (3.61), we get
1/2
1
1 + cT
kΦ(u)kL∞
≤ ku0 kHxy
(|||u|||2 + |||u|||).
T Hxy
(3.62)
Now, from Lemma 3.2 and group properties, using similar arguments as above,
we obtain
k∂x Φ(u)kL2T L∞
xy
Z t
0
0 ≤ kU (t)∂x u0 kL2T L∞
+ U (t)
U (−t )∂x (uux + ∂x (φu))dt xy
0
≤ ck∂x u0 kL2xy + c
Z
1 + c
≤ c ku0 kHxy
1 + cT
≤ cku0 kHxy
0
T
0
1/2
L2T L∞
xy
T
Z
k∂x (uux + ∂x (φu))kL2xy dt0
k∂x (uux + φ0 u + φux )kL2xy dt0
(|||u|||2 + |||u|||).
(3.63)
ZK EQUATION
13
Applying Lemma 3.3 (i), group properties, Minkowski and Hölder’s inequalities,
we obtain
2
2
k∂x ∂x Φ(u)kL∞
≤ ck∂x U (t)∂x u0 kL∞
x LyT
x LyT
Z t
0
0 U (−t )∂x (uux + ∂x (φu))dt + ∂x U (t)
2
L∞
x LyT
0
T
Z
0
≤ ck∂x u0 kL2xy + c
1 + cT
≤ cku0 kHxy
(3.64)
0
k∂x (uux + φ u + φux )kL2xy dt
0
1/2
(|||u|||2 + |||u|||)
and
2
2
≤ ck∂x U (t)∂y u0 kL∞
k∂y ∂x Φ(u)kL∞
x LyT
x LyT
Z t
0
0 U (−t )∂y (uux + ∂x (φu))dt + ∂x U (t)
2
L∞
x LyT
0
T
Z
0
≤ ck∂y u0 kL2xy + c
1 + cT
≤ cku0 kHxy
0
1/2
(3.65)
0
k∂y (uux + φ u + φux )kL2xy dt
(|||u|||2 + |||u|||).
Finally, an application of Lemma 3.3 (ii), Minkowski’s inequality and group
properties yield
Z t
0
0
kΦ(u)kL2x L∞
≤ ckU (t)u0 kL2x L∞
U (−t )(uux + ∂x (φu))dt + U (t)
yT
yT
0
Z
≤ c(T )ku0 kH 1 + c(T )
0
1 dt
kuux + φ0 u + φukHxy
0
1/2
1 + c(T )T
≤ c(T )ku0 kHxy
L2x L∞
yT
T
(|||u|||2 + |||u|||).
(3.66)
Therefore, from (3.62)–(3.66) it follows that
|||Φ(u)||| ≤ c(T )ku0 kH 1 + c(T )T 1/2 (|||u||| + 1)|||u|||.
By choosing a = 2c(T )ku0 kH 1 and T such that
1
,
2
we see that Φ : XTa 7→ XTa is well defined. Similar arguments show that Φ is also
a contraction. From the contraction mapping principle we deduce the existence of
a unique fixed point for Φ, which solves our problem. To finish the proof we use
standard arguments thus we omit the details here.
The identities (3.51), (3.52) are obtained formally by multiplying the first equa2
tion in (1.5) by u (resp. u2 + ∆u + uφ). The formal computations are as usual
justifed by a smoothing procedure.
Recalling that φ and its space and time derivatives are bounded, those identities
imply an a priori H 1 bound which proves that the solution is global. This in turn
implies that the estimates (3.48), (3.49), (3.50) are valid for any T > 0.
c(T )T 1/2 (a + 1) ≤
14
F. LINARES, A. PASTOR AND J.-C. SAUT
Acknowledgements. The authors acknowledge support from the Brazilian-French
Program for Mathematics. F. L. was partially supported by CNPq-Brazil, A. P.
was supported by CNPq-Brazil under grant no. 152234/2007-1 and J.-C. S. was
partially supported by the project ANR-07-BLAN-0250 of the Agence Nationale de
la Recherche and by the Wolfgang Pauli Institute in Vienna.
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E-mail address:
FL: [email protected]
AP: [email protected]
JCS: [email protected]