Electronic Journal of Qualitative Theory of Differential Equations 2005, No. 20, 1-21;http://www.math.u-szeged.hu/ejqtde/

SPATIAL ANALYTICITY OF SOLUTIONS OF A NONLOCAL PERTURBATION OF THE KDV EQUATION

BORYS ALVAREZ SAMANIEGO

IMECC-UNICAMP, C.P. 6065, 13083-970, Campinas, SP, Brazil.

Abstract. LetHdenote the Hilbert transform andη≥0. We show that if
the initial data of the following problemsut+uux+uxxx+η(Hux+Huxxx) =
0, u(·,0) =φ(·) andvt+^{1}_{2}(vx)^{2}+vxxx+η(Hvx+Hvxxx) = 0, v(·,0) =ψ(·) has
an analytic continuation to a strip containing the real axis, then the solution
has the same property, although the width of the strip might diminish with
time. Whenη >0 and the initial data is complex-valued we prove local well-
posedness of the two problems above in spaces of analytic functions, which
implies the constancy over time of the radius of the strip of analyticity in the
complex plane around the real axis.

1. Introduction

We are interested in studying spatial analyticity of solutions of the following problems:

ut+uux+uxxx+η(Hux+Huxxx) = 0, u(·,0) =φ(·), (1) vt+1

2(vx)^{2}+vxxx+η(Hvx+Hvxxx) = 0, v(·,0) =ψ(·), (2)
where H denotes the Hilbert transform given byHf(x) = _{π}^{1}PR∞

−∞

f(y)

y−xdy forf ∈
S(R) the Schwartz space of rapidly decreasingC^{∞}(R) functions, P represents the
principal value of the integral and the parameter η is an arbitrary nonnegative
number. It is known that([Hf)(ξ) =isgn(ξ) ˆf(ξ), for allf ∈H^{s}(R), where

sgn (ξ) =

−1, ξ <0, 1, ξ >0.

Equation (1) was derived by Ostrovskyet.al. (see [9] for more details) to describe the radiational instability of long non-linear waves in a stratified fluid caused by internal wave radiation from a shear layer; the fourth term corresponds to thewave amplification and the fifth term represents damping. It models the motion of a homogeneous finite-thickness fluid layer with densityδ1, which moves at a constant speed U, slipping over an immobile infinitely deep stratified fluid with a density δ2 > δ1. The upper boundary of the layer is supposed to be rigid and the lower one is contiguous to the infinitely deep fluid. Here u(x, t) is the deviation of the

2000Mathematics Subject Classification. 35Q53, 35B30, 35A07.

Key words and phrases. Spatial analyticity, Hilbert transform, KdV equation.

Research supported by FAPESP/Brazil under grant No. 2002/02522-0.

EJQTDE, 2005 No. 20, p. 1

interface from its equilibrium position. Let us remark that some numerical results for periodic and solitary-wave solutions of equation (1) were obtained by Bao-Feng Feng and T. Kawahara [4].

The Cauchy problems associated to (1) and (2) were studied in [1], where it
was proved that problems (1) and (2) are globally well-posed in H^{s}(R) for s≥1,
considering real-valued solutions.

In this paper we are interested in proving that if the initial condition of the problem (1) (resp. (2)) is analytic and has an analytic continuation to a strip containing the real axis, then the solution of (1) (resp. (2)) has the same property.

Section 2 is devoted to studying the case when the solutions are real-valued on the real axis at any time, and η ≥ 0. Hence, the results obtained in [1] about the initial value problems associated to (1) and (2) will be helpful. We use the method developed by Kato&Masuda [8] which estimates certain families of Liapunov functions for the solutions, to prove global spatial analyticity of the solutions, but the width of the strip might decrease with time.

Section 3 shows that problems (1) and (2) admit a Gevrey-class analysis. For
η >0, we prove local well-posedness of problem (1) (resp. (2)) in X^{σ,s} forσ >0
ands >1/2 (resp. s >3/2); here, the initial data can be complex-valued. So, if the
initial data of problem (1) (resp. (2)) is analytic and has an analytic continuation
to a strip containing the real axis, then the solution of (1) (resp. (2)) has the same
property, maintaining the width of the strip in time. It should be mentioned that
it was recently proved by Gruji´c&Kalisch [5] a result on local well-posedness of the
generalized KdV equation (KdV is an abbreviation for Korteweg-de Vries) in spaces
of analytic functions on a strip containing the real axis without shrinking the width
of the strip in time; their proof uses space-time estimates and Bourgain-type spaces.

Here we do not make use of Bourgain spaces, we mainly use some properties of the
Semigroup associated to the linear part of problem (1), namely Lemmas 3.1 and 3.2,
to prove local well-posedness of problem (1) in X^{σ,s}. Moreover, proceeding as in
[2], where Bona&Gruji´c studied some KdV-type equations, we prove for real-valued
solutions and η ≥ 0 that if the initial state belongs to a Gevrey class, then the
solution of (1) (resp. (2)) remains in this class for all time but the width of the
strip of analyticity may diminish as a function of time.

Finally, in Section 4 we consider η ≥ 0 in (1) and complex-valued initial data in Xr-spaces forr >0. Similar as in [6], where analyticity of solutions of the KdV equation was studied, we use Banach’s fixed point theorem in a suitable function space in order to find a local solution of problem (1) that is analytic and has an EJQTDE, 2005 No. 20, p. 2

analytic extension to a strip around the real axis although the radius of the strip of analyticity in the complex plane around the real axis may decrease with time.

Notation:

• fˆ=Ff : the Fourier transform off (F^{−}^{1}: the inverse of the Fourier transform),
where fˆ(ξ) = ^{√}^{1}_{2π}R

e^{−}^{iξx}f(x)dx forf ∈L^{1}(R).

• k · ks, (·,·)s: the norm and the inner product respectively in H^{s}(R) (Sobolev
space of ordersofL^{2} type), s∈R. kfk^{2}s≡R

(1 +|ξ|^{2})^{s}|fˆ(ξ)|^{2}dξ.

k · k=k · k^{0}: the L^{2}(R) norm. (·,·) denotes the inner product onL^{2}(R).

H^{∞}(R)≡ ∩H^{s}(R).

• H: the Hilbert transform.

• B(X, Y): set of bounded linear operators onX toY. IfX =Y we writeB(X).

k · kB(X,Y): the operator norm inB(X, Y).

• S(r) ={x+iy∈C; x∈R, |y|< r}, forr >0.

A(r): the set of all analytic functionsf onS(r) such thatf ∈L^{2}(S(r^{0})) for each
0< r^{0}< r and thatf(x)∈Rforx∈R.

• A= (I−∂_{x}^{2})^{1/2},X^{σ,s}=D(A^{s}e^{σA}) the domain of the operatorA^{s}e^{σA}.

• L^{p}={f;f is measurable onR,kfk^{L}^{p}<∞}, where kfk^{L}^{p}= R

|f(x)|^{p}dx1/p

if
1≤p <+∞, and kfk^{L}^{∞} = ess sup_{x}_{∈R}|f(x)|,f is an equivalence class.

• Lr={f ∈L^{2};kfk^{2}Lr = (f, f)Lr = (cosh(2rξ) ˆf ,fˆ)<+∞}.

• Xr=

f∈L^{2};kfk^{2}Xr = (f, f)Xr =P1

j=0(ξ^{2j}(cosh(2rξ)+ξsinh(2rξ)) ˆf ,f)ˆ <∞ .

• Yr=

f ∈L^{2};∂xf ∈L^{2},kfk^{2}Yr = (f, f)Yr =P1

j=0(ξ^{2j+2}cosh(2rξ) ˆf ,fˆ)<+∞ .

• kfk^{p}m,p=Pm

j=0k∂_{x}^{j}fk^{p}L^{p}, 1≤p <+∞. kfk^{m,}∞=Pm

j=0k∂_{x}^{j}fk∞.

• H^{p}(r): the analytic Hardy space on the strip S(r).

H^{p}(r) ={F;F is analytic onS(r),kFkH^{p}(r)= sup_{|}_{y}_{|}_{<r}kF(·+iy)k^{L}^{p}<∞}.

• H^{m,p}(r) ={F ∈H^{p}(r);kFk^{p}H^{m,p}(r)=Pm

j=0k∂_{z}^{j}Fk^{p}H^{p}(r)<∞}.

• C(I;X) : set of continuous functions on the intervalI into the Banach spaceX.

• C^{ω}(I;X) : the set of weakly continuous functions from I to X.

• <(z): the real part of the complex number z.

2. Real-valued initial data.

We deduce in this Section global analyticity (in space variables) of solutions of
problems (1) and (2) when the initial data and the corresponding solution take real
values on the real axis and supposing moreover that the initial data has an analytic
continuation that is analytic in a strip containing the real axis. We will use the fact
that problems (1) and (2) are globally well possed in H^{s}(R) fors ≥ 1, when the
solution of the two previously mentioned problems take real values on the real axis
EJQTDE, 2005 No. 20, p. 3

at any time. More precisely we have the following two theorems (for real-valued solutions) which can be found in [1].

Theorem 2.1. Let s ≥ 1. If φ ∈ H^{s}(R), then for each η > 0 there exists a
unique u = uη ∈ C([0,∞);H^{s}(R)) solution to the problem (1) such that ∂tu ∈
C([0,∞);H^{s}^{−}^{3}(R)).

Proof. See Theorem 4.2 in [1].

Theorem 2.2. Let s ≥ 1. If ψ ∈ H^{s}(R), then for each η > 0 there exists a
unique v = vη ∈ C([0,∞);H^{s}(R)) solution to the problem (2) such that ∂tv ∈
C([0,∞);H^{s}^{−}^{3}(R)).

Proof. See Theorem 4.1 in [1].

Theorem 2.3 (resp. 2.4) states that if the initial state has an analytic continu-
ation that belongs to A(r0) for somer0 >0 then the solutionu(t) (resp. v(t)) of
problem (1) (resp. (2)), with η≥0, also has an analytic continuation belonging to
A(r1) for allt ∈ [0, T], wherer1 might decrease with time. Theorem 2.3, below,
is an application of the method developed by Kato&Masuda in [8] to study global
analyticity (in space variables) of some partial differential equations. Similar as in
the proof of Theorem 2 in [8] we consider H^{m+5}(R) ≡ Z ⊂X ≡ H^{m+2}(R) and
Φσ;m(v)≡ ^{1}2kvk^{2}σ,2;mdefined on an appropriate open setO⊂Z, where

kfk^{2}σ,s;m≡
Xm
j=0

e^{2jσ}

(j!)^{2}k∂_{x}^{j}fk^{2}s and kfk^{2}σ,s≡
X∞
j=0

e^{2jσ}

(j!)^{2}k∂_{x}^{j}fk^{2}s, s∈R.
Lemma 2.1. LetF(v)be defined by F(v)≡ −vvx−vxxx−η(Hvx+Hvxxx). Then
there exist constants c, γ >0such that for every v∈H^{m+5}(R),

hF(v), DΦσ;m(v)i ≤c(η+kvk2)Φσ;m(v) +γq

Φσ;m(v)∂σΦσ;m(v). (3) Proof. It is not difficult to see that DΦσ;m(v) =Pm

j=0
e^{2jσ}

(j!)^{2}(−∂x)^{j}A^{4}∂_{x}^{j}v, where
A= (1−∂_{x}^{2})^{1/2}. Then

hF(v), DΦσ;m(v)i= Xm j=0

e^{2jσ}
(j!)^{2}

− ∂_{x}^{j}(v∂xv), ∂_{x}^{j}v

2−η ∂_{x}^{j}(H∂xv+H∂_{x}^{3}v), ∂^{j}_{x}v

2

= Xm j=0

e^{2jσ}
(j!)^{2}

− v∂_{x}^{j+1}v, ∂_{x}^{j}v

2−Qj(v)−η ∂_{x}^{j}(H∂xv+H∂_{x}^{3}v), ∂_{x}^{j}v

2

, where Qj(v) = Pj

k=1 j k

∂_{x}^{k}v∂_{x}^{j}^{−}^{k+1}v, ∂_{x}^{j}v

2, and Q0(v) ≡ 0. By using Kato’s inequality (K) in the Appendix we have that

|(v∂_{x}^{j+1}v, ∂_{x}^{j}v)2| ≤ck∂xvk1k∂_{x}^{j}vk^{2}2≤ckvk2k∂_{x}^{j}vk^{2}2.

EJQTDE, 2005 No. 20, p. 4

Moreover

− ∂_{x}^{j}(H∂xv+H∂_{x}^{3}v), ∂_{x}^{j}v

2 =

Z

(1 +ξ^{2})^{2}ξ^{2j}(|ξ| − |ξ|^{3})|v(ξ)ˆ |^{2}dξ

≤ Z

(1 +ξ^{2})^{2}ξ^{2j}|ˆv(ξ)|^{2}dξ=k∂^{j}_{x}vk^{2}2. (4)
Then

hF(v), DΦσ;m(v)i ≤c(kvk^{2}+η)Φσ;m(v)−
Xm
j=0

e^{2jσ}

(j!)^{2}Qj(v). (5)
Now, using the Schwarz inequality and the formulakf gk2≤γ kfk2kgk1+kfk1kgk2
we get

|Qj(v)| ≤γ Xj k=1

j k

k∂_{x}^{j}vk^{2} k∂_{x}^{k}vk^{1}k∂_{x}^{j}^{−}^{k+1}vk^{2}+k∂^{k}_{x}vk^{2}k∂_{x}^{j}^{−}^{k+1}vk^{1}
. (6)
We denote as in [8], bj ≡ ^{e}j!^{jσ}k∂^{j}_{x}vk^{2}, B^{2} ≡ Pm

j=0b^{2}_{j} = 2Φσ;m(v) and ˜B^{2} ≡
Pm

j=1jb^{2}_{j} = ∂σΦσ;m(v). By using (6) it follows, as a particular case of Lemma
3.1 in [8], that

Xm j=0

e^{2jσ}

(j!)^{2}|Qj(v)| ≤ γ
Xm
j=1

Xj k=1

bj

bk−1

k (j−k+ 1)bj−k+1+bjbkbj−k

= γ

Xm k=1

bk−1

k Xm j=k

bj(j−k+ 1)bj−k+1 +γ

Xm k=1

bk

Xm j=k

bjbj−k

≤ γB˜^{2}
Xm
k=1

1
k^{2}

1/2 Xm k=1

b^{2}_{k}_{−}_{1}1/2

+γBB˜ Xm k=1

bk

√k

≤ 2γBB˜^{2}+γBB˜
Xm
k=1

1
k^{2}

1/2 Xm k=1

kb^{2}_{k}1/2

≤4γBB˜^{2}. (7)
By replacing the inequality (7) into (5), the Lemma follows.

Next, we enunciate Lemma 2.4 in [8] and we give, for expository completeness, a proof of this trivial result.

Lemma 2.2. Letφn ∈A(r), n= 1,2, ...be a sequence withkφnkσ,2bounded, where
e^{σ}< r. Ifφn→0in H^{∞} asn→ ∞, thenkφnk^{σ}^{0}^{,2}→0for each σ^{0}< σ.

Proof. Letσ^{0}< σ. ChooseMsuch thatkφnk^{σ,2}≤Mfor alln. Since^{e}^{2jσ}

0

(j!)^{2}k∂_{x}^{j}φnk^{2}2≤

M^{2}

e^{2j(σ−σ}^{0}^{)}, it follows, by using the dominated convergence theorem, that

nlim→∞kφnk^{2}σ^{0},2=
X∞
j=0

e^{2jσ}^{0}
(j!)^{2}( lim

n→∞k∂_{x}^{j}φnk^{2}2) = 0.

EJQTDE, 2005 No. 20, p. 5

Theorem 2.3. Let η ≥ 0 and T > 0. Let u ∈ C([0, T];H^{∞}(R)) be a solution
of (1). If u(0) = φ ∈ A(r0) for some r0 > 0, there exists r1 > 0 such that
u∈C([0, T];A(r1)).

Proof. Let us remark that φ ∈ A(r0) implies, by Lemma 2.2 in [8], that φ ∈
H^{∞}(R). So, it follows from Theorem 2.1 and from Corollary 4.7 in [7] that u ∈
C([0, T];H^{∞}(R)). We proceed as in the proof of Theorem 2 in [8], to prove that
Φσ;mis a Liapunov family for (1) onO=Z =H^{m+5}(R), considering the functions:

α(r) = γ√

r, β(r) =c(η+M)r ≡Kr, for r ≥ 0, where M = maxt∈[0,T]ku(t)k2,
and γ, c >0 are constants given by Lemma 2.1,ρ(t) = ^{1}_{2}kφk^{2}b,2e^{Kt} where b < σ0,
e^{σ}^{0} < r0, and

σ(t) =b−

√2γ

K kφk^{b,2}(e^{K}^{2}^{t}−1), t∈[0, T]. (8)
We have thatu(t)∈A(e^{σ(t)}), for allt∈[0, T]. Thenu(t)∈A(r1) for allt∈[0, T],
where r1 = e^{σ(T)}. The continuity of u, as in the proof of Theorem 2 in [8], is a

consequence of Lemma 2.2 above.

Theorem 2.4. Let η ≥ 0 and T > 0. Let v ∈ C([0, T];H^{∞}(R)) be a solution
of (2). If v(0) = ψ ∈ A(r0) for some r0 > 0, there exists r1 > 0 such that
v∈C([0, T];A(r1)).

Proof. We remark thatu≡vx∈C([0, T];H^{∞}(R)) is a solution of (1) withu(0) =
ψ^{0}. Since ψ^{0} ∈ H^{∞}(R) and kψ^{0}kσ,0 ≤ kψkσ,1 for each σ such that e^{σ} < r0 it
follows, as a consequence of Lemma 2.2 in [8], that ψ^{0} ∈ A(r0). So, by Theorem
2.3, there existsr1 >0 such that vx∈C([0, T];A(r1)). Now, since v(t)∈H^{∞}(R)
andkv(t)k^{2}σ,0≤sup_{t}_{∈}_{[0,T]}kv(t)k^{2}+e^{2σ}sup_{t}_{∈}_{[0,T]}ku(t)k^{2}σ,0<∞for eachσsuch that
e^{σ} < r1 and for all t ∈ [0, T], it follows that v(t)∈ A(r1) for all t ∈ [0, T]. The

continuity ofv follows as in the previous Theorem.

3. Gevrey Class Regularity.

Now, we make a Gevrey-class analysis of problems (1) and (2). Let us remark
that, since each function f ∈ X^{σ,s} = D(A^{s}e^{σA}) with σ > 0 and s ≥ 0, satisfies
kfk^{2}Lσ ≤ R

e^{2σ}^{|}^{ξ}^{|}|fˆ(ξ)|^{2}dξ ≤ kfk^{2}X^{σ,s} < ∞, it follows from Theorem 1 in [6] that
f has an analytic extensionF ∈H^{2}(σ), where σ > 0 is the radius of the strip of
analyticity in the complex plane around the real axis. Theorem 3.1 (resp. Theorem
3.2) states that problem (1) (resp. (2)), forη >0 and complex-valued initial data,
is locally well-posed in X^{σ,s} where σ > 0 is a fixed number and s is a suitable
nonnegative number. Theorem 3.1 (resp. 3.2)) implies that if the initial condition
of problem (1) (resp. (2)) is analytic and has an analytic continuation to a strip
EJQTDE, 2005 No. 20, p. 6

containing the real axis, then the solution of (1) (resp. (2)) has the same property, without reducing the width of the strip in time.

Fort≥0 andξ∈R, let

Fη(t, ξ) = e^{(iξ}^{3}^{+η(}^{|}^{ξ}^{|−|}^{ξ}^{|}^{3}^{))t}

Eη(t)f = F^{−}^{1}(Fη(t,·) ˆf), f∈L^{2}(R). (9)
It is not difficult to see that|Fη(t, ξ)| ≤e^{ηt}, for allt≥0 andξ∈R.

Lemma 3.1. Letη > 0. Then (Eη(t))t≥0 is a C^{0}-semigroup on X^{σ,s} for σ >0
ands∈R. Moreover,

kEη(t)kB(X^{σ,s})≤e^{ηt}. (10)
Whenη= 0, kEη(t)kB(X^{σ,s})= 1 for allt≥0.

Proof. Similar to the proof of Lemma 2.1 in [1].

Lemma 3.2. Let t >0, λ≥0, η >0, σ > 0and s∈R be given. Then Eη(t)∈
B(X^{σ,s}, X^{σ,s+λ}). Moreover,

kEη(t)φkX^{σ,s+λ} ≤cλ

e^{ηt}+ 1
(ηt)^{λ/3}

kφk^{X}^{σ,s}, (11)
whereφ∈X^{σ,s} andcλ is a constant depending only onλ.

Proof.

kEη(t)φk^{2}X^{σ,s+λ} ≤ cλ Z

(1 +ξ^{2})^{s}e^{2σ(1+ξ}^{2}^{)}^{1/2}|Fη(t, ξ)|^{2}|φ(ξ)ˆ |^{2}dξ
+

Z

ξ^{2λ}(1 +ξ^{2})^{s}e^{2σ(1+ξ}^{2}^{)}^{1/2}e^{2ηt(}^{|}^{ξ}^{|−|}^{ξ}^{|}^{3}^{)}|φ(ξ)ˆ |^{2}dξ

≤ cλ

e^{2ηt}+ sup

ξ∈R

ξ^{2λ}e^{−}^{2ηt(}^{|}^{ξ}^{|}^{3}^{−|}^{ξ}^{|}^{)}

kφk^{2}X^{σ,s}. (12)
On the other hand,

sup

ξ∈R|ξ|^{λ}e^{−}^{ηt(}^{|}^{ξ}^{|}^{3}^{−|}^{ξ}^{|}^{)}= sup

ξ≥0

ξ^{λ}e^{−}^{ηt(ξ}^{3}^{−}^{ξ)}≤√

2^{λ}e^{ηt}+cλ 1

(ηt)^{λ/3}. (13)

The lemma follows immediately from (12) and (13).

Next theorem proves, without using Bourgain-type spaces, local well-posedness to
problem (1) withη >0 inX^{σ,s}forσ >0,s >1/2, and complex-valued initial data.

Theorem 3.1. Letη > 0, σ > 0and s > 1/2 be given. If φ ∈ X^{σ,s}, then there
exist T =T(s,σ,η,kφkXσ,s) >0 and a unique function u∈C([0, T];X^{σ,s}) satisfying
the integral equation

u(t) =Eη(t)φ−1 2

Z t 0

Eη(t−t^{0})∂x(u^{2}(t^{0}))dt^{0}. (14)
EJQTDE, 2005 No. 20, p. 7

Proof. LetM,T >0 be fixed but arbitrary. Let us consider the map Af(t) =Eη(t)φ−1

2 Z t

0

Eη(t−t^{0})∂x(f^{2}(t^{0}))dt^{0},
defined on the complete metric space

Θs,σ,η(T) ={f ∈C([0, T];X^{σ,s}); sup

t∈[0,T]kf(t)−Eη(t)φkX^{σ,s} ≤M},

whereT >0 will be suitably chosen later. First, we prove that iff ∈Θs,σ,η(T) then
Af ∈C([0, T];X^{σ,s}). Without loss of generality, we may assume thatτ > t > 0.

Then

kAf(t)−Af(τ)kX^{σ,s} ≤ k(Eη(t)−Eη(τ))φkX^{σ,s} +F(t, τ) +G(t, τ),
whereF(t, τ) andG(t, τ) will be estimated below.

G(t, τ) ≡ Z τ

t

kEη(τ−t^{0})∂x(f^{2}(t^{0}))kX^{σ,s}dt^{0}

≤ cs

η(M+e^{ηT}kφk^{X}^{σ,s})^{2}(e^{η(τ}^{−}^{t)}−1 + 3

2(η(τ −t))^{2}^{3})→0 as τ ↓t,
where in the last inequality we have used Lemmas 3.1 and 3.2 (withλ= 1) and the
fact thatX^{σ,s} is a Banach algebra fors >1/2 andσ ≥0 (Lemma 6 in [2]). Since
τ −t^{0} ≥t−t^{0}, for all t^{0} ∈[0, t], it follows from Lemma 3.2 and from the triangle
inequality that

k(Eη(t−t^{0})−Eη(τ−t^{0}))∂x(f^{2}(t^{0}))k^{X}^{σ,s}≤cs(M+e^{ηT}kφk^{X}^{σ,s})^{2} e^{η(T}^{−}^{t}^{0}^{)}+(η(t−t^{0}))^{−}^{1}^{3}
,

and the expression on the right hand side of the last inequality belongs toL^{1}([0, t], dt^{0}).

The fact thatk(Eη(t−t^{0})−Eη(τ−t^{0}))∂x(f^{2}(t^{0}))k^{X}^{σ,s} →0 asτ ↓tis a consequence of
the dominated convergence theorem. So, by using again the dominated convergence
theorem, we have that

F(t, τ)≡ Z t

0 k(Eη(t−t^{0})−Eη(τ−t^{0}))∂x(f^{2}(t^{0}))kX^{σ,s}dt^{0} →0 as τ ↓t.

Now, we prove thatA(Θs,σ,η(T))⊂Θs,σ,η(T), forT = ˜T >0 sufficiently small. Let u∈Θs,σ,η(T). Then

kAu(t)−Eη(t)φkX^{σ,s} ≤
Z t

0 kEη(t−t^{0})∂x(u^{2}(t^{0}))kX^{σ,s}dt^{0}

≤ cs

η (M+e^{ηT}kφkX^{σ,s})^{2}h(T), (15)
where h(T)≡e^{ηT} −1 +^{3}_{2}(ηT)^{2/3}. By choosingT = ˜T >0 sufficiently small, the
right hand side of (15) is less thanM. Finally , we claim that there exists ˆT ∈(0,T˜]
EJQTDE, 2005 No. 20, p. 8

such thatAis a contraction on Θs,σ,η( ˆT). Lett∈[0,T˜],u, v∈Θs,σ,η( ˜T). Then
kAu(t)−Av(t)k^{X}^{σ,s} ≤ c1

Z t 0

e^{η(t}^{−}^{t}^{0}^{)}+ 1
(η(t−t^{0}))^{1/3}

k∂x(u^{2}(t^{0})−v^{2}(t^{0}))kX^{σ,s−1}dt^{0}

≤ cs(M+e^{ηt}kφk^{X}^{σ,s}) sup

t^{0}∈[0,T]˜

ku(t^{0})−v(t^{0})k^{X}^{σ,s}

· Z t

0

e^{η(t}^{−}^{t}^{0}^{)}+ 1
(η(t−t^{0}))^{1/3}

dt^{0}

≤ cs

η(M+e^{η}^{T}^{˜}kφk^{X}^{σ,s})h( ˜T) sup

t^{0}∈[0,T]˜

ku(t^{0})−v(t^{0})k^{X}^{σ,s}.
So, by choosing ˆT ∈(0,T] such that˜ ^{c}_{η}^{s}(M+e^{η}^{T}^{ˆ}kφkX^{σ,s})h( ˆT)<1, the claim follows.

Hence A has a unique fixed point u∈Θs,σ,η( ˆT), which satisfies (14). Uniqueness
of the solutionu∈C([0,Tˆ];X^{σ,s}) follows from Proposition 3.2 in [1], which implies
uniqueness of the solution in the classC([0,T];ˆ H^{s}(R)) fors >1/2.

Proposition 3.1. Problem (1) is equivalent to the integral equation (14). More
precisely, if u ∈ C([0, T];X^{σ,s}), s > 1/2 is a solution of (1) then u satisfies
(14). Reciprocally, if u ∈ C([0, T];X^{σ,s}), s > 1/2 is a solution of (14) then
u∈C^{1}([0, T];X^{σ,s}^{−}^{3})and satisfies (1).

Proof. Similar to the proof of Proposition 3.1 in [1]. However, here we use Lemmas

3.1 and 3.2.

Theorem 3.2. Let η >0, σ >0and s >3/2be given. Letψ∈X^{σ,s}. Then there
exist T =T(s,σ,η,kψkXσ,s)>0and a uniquev∈C([0, T];X^{σ,s})solution of (2).

Proof. Sinceψ∈X^{σ,s}, it follows thatφ≡ψ^{0}∈X^{σ,s}^{−}^{1}. By Theorem 3.1 and Propo-
sition 3.1, there exist T = T(s,σ,η,kψkXσ,s) >0 and a unique u∈C([0, T];X^{σ,s}^{−}^{1})
satisfying (14) and (1). Let us define

v(t)≡Eη(t)ψ−1 2

Z t 0

Eη(t−t^{0})u^{2}(t^{0})dt^{0}, t∈[0, T]. (16)
It follows easily from (16) and from the uniqueness of the solution of (14) that

∂xv(t) =u(t). Sinceu∈C([0, T];X^{σ,s}^{−}^{1}), it follows from Lemmas 3.1 and 3.2 that
v∈C([0, T];X^{σ,s}). Now, by similar calculations as in the proof of Proposition 3.1
in [1], we have thatv∈C^{1}([0, T];X^{σ,s}^{−}^{3}) and satisfies (2).

Theorems 3.3 and 3.4, below, consider the case of real-valued solutions (on the real axis) to problems (1) and (2) respectively, forη≥0. So, Theorems 2.1 and 2.2 will be required again. The following theorem is proved similarly to Theorem 11 in [2], where the rate of decrease of the uniform radius of analyticity for KdV-type equations of the formut+G(u)ux−Lux = 0 was also studied, whereu=u(x, t), EJQTDE, 2005 No. 20, p. 9

forx, t∈R,Gis a function that is analytic at least in a neighborhood of zero inC,
but real-valued on the real axis, andLis a homogeneus Fourier multiplier operator
defined by Lu(ξ) =c |ξ|^{µ}u(ξˆ ), for someµ > 0. In the case of problem (1) we have
thatLu(ξ) = [ξc ^{2}−ηi(sgn(ξ)−ξ|ξ|)]ˆu(ξ).

Theorem 3.3. Letη ≥0 andT > 0. Suppose thatφ ∈X^{σ}^{0}^{,s}, for some σ0 > 0
and s > 5/2. Suppose moreover that φ(x) ∈ R for x ∈ R. Then the solution u
of problem (1) satisfies u ∈ C([0, T];X^{σ(T),s}), where σ(t) is a positive monotone
decreasing function given by (24).

Proof. Letη, T, σ0, andsbe as in the hypothesis of the theorem. Letr≡s−1>3/2.

By the Remark at the end of this Section we have that φ ∈A(σ0). So, by using
Lemma 2.2 in [8], we have that φ ∈ H^{∞}(R). Then u ∈ C([0, T];H^{s}(R)), which
follows from Theorem 2.1 and from Corollary 4.7 in [7]. Letv≡ux, then

vt+v^{2}+vxxx+η(Hvx+Hvxxx) +uvx= 0, v(0) =φ^{0}. (17)
Letσ∈C^{1}([0, T];R) be a positive function such thatσ^{0}<0 andσ(0) =σ0. Then

1 2

d

dtkv(t)k^{2}_{X}^{σ(t),r}−σ^{0}(t)kv(t)k^{2}_{X}^{σ(t),r+1/2} =<

Z

(1+ξ^{2})^{r}e^{2σ(t)(1+ξ}^{2}^{)}^{1/2}∂tˆv(t, ξ)ˆv(t, ξ)dξ.

It follows from the last expression and from (17) that 1

2 d

dtkv(t)k^{2}X^{σ(t),r} −σ^{0}(t)kv(t)k^{2}X^{σ(t),r+1/2} ≤I1+I2+ηkv(t)k^{2}X^{σ(t),r}, (18)

where

I1≡ |(A^{r}e^{σ(t)A}v^{2}(t), A^{r}e^{σ(t)A}v(t))|,
I2≡ |(A^{r}e^{σ(t)A}(uvx)(t), A^{r}e^{σ(t)A}v(t))|.

I1andI2are particular cases of the corresponding ones in [2], takingG(u) =u. For the sake of completeness we estimate them here. Sincer >1/2, by using Lemmas 6 and 9 in [2], we have that

I1 ≤ crkA^{r}e^{σ(t)A}v(t)k^{3}≤crkA^{r}v(t)k^{3}+ ˜crσ(t)kA^{r+1/3}e^{σ(t)A}v(t)k^{3}

≤ crkA^{r}v(t)k^{3}+ ˜crσ(t)kA^{r}e^{σ(t)A}v(t)kkA^{r+1/2}e^{σ(t)A}v(t)k^{2}. (19)
Sincer >3/2, by using Lemma 10 in [2], we obtain

I2≤crkA^{r+1}u(t)kkA^{r}v(t)k^{2}+ ˜crσ(t)kA^{r+1}e^{σ(t)A}u(t)kkA^{r+1/2}e^{σ(t)A}v(t)k^{2}. (20)
Sinceu∈C([0, T];H^{s}(R)), it follows thatkA^{r+1}u(t)k ≤c(r,T). Moreover

kA^{r+1}e^{σ(t)A}u(t)k^{2}=
Z

(1 +ξ^{2})^{r}e^{2σ(t)(1+ξ}^{2}^{)}^{1/2}|u(ξ, t)ˆ |^{2}dξ+kA^{r}e^{σ(t)A}v(t)k^{2}

≤e^{2}^{√}^{2σ}^{0}ku(t)k^{2}r+ 2kA^{r}e^{σ(t)A}v(t)k^{2}≤c(r,T,σ0)+ 2kA^{r}e^{σ(t)A}v(t)k^{2}, (21)
EJQTDE, 2005 No. 20, p. 10

wherec(r,T,σ0) is a positive constant depending only onr, T andσ0. Replacing the
last inequalities into (20) and sincev∈C([0, T];H^{r}(R)), we get

I2≤c(r,T)+ ˜crσ(t) c(r,T,σ0)+√

2kA^{r}e^{σ(t)A}v(t)k

kA^{r+1/2}e^{σ(t)A}v(t)k^{2}. (22)
Let us remark thatc(r,T)andc(r,T,σ0)are positive, continuous, non-decreasing func-
tions of the variableT∈[0,+∞). Now, by using (19) and (22) into (18), we obtain

1 2

d

dtkv(t)k^{2}X^{σ(t),r} −σ^{0}(t)kv(t)k^{2}X^{σ(t),r+1/2}

≤c(r,T)+c(r,T,σ0)σ(t) 1 +kv(t)kX^{σ(t),r}

kv(t)k^{2}X^{σ(t),r+1/2}+ηkv(t)k^{2}X^{σ(t),r}.
Then

d

dtkv(t)k^{2}X^{σ(t),r} + 2 −σ^{0}(t)−c(r,T,σ0)σ(t)(1 +kv(t)kX^{σ(t),r})

kv(t)k^{2}X^{σ(t),r+1/2}

≤c(r,T)+ 2ηkv(t)k^{2}X^{σ(t),r}. (23)

Inequality (23) implies thatv(t)∈X^{σ(t),r} for allt∈[0, T], where

σ(t) =σ0e^{−}^{Kt}, (24)

K=c(r,T,σ0)(1 +c(r,T,σ0,φ)e^{ηT}), andc(r,T,σ0,φ)= kA^{r}e^{σ}^{0}^{A}φ^{0}k^{2}+c(r,T)T1/2

. More precisely we have that

kv(t)kX^{σ(t),r} ≤c(r,T,σ0,φ)e^{ηT}, t∈[0, T]. (25)
Now, we will prove assertion (25). In fact let

T^{∗}≡sup{T >0;∃!u∈C([0, T];X^{σ(T}^{),s}) solution of (1), sup

t∈[0,T]ku(t)kX^{σ(t),s} <∞}.
We claim that T^{∗}= +∞. Suppose by contradiction thatT^{∗}<∞. It follows from
Theorem 3.1, Proposition 3.1 and Theorem 1 in [5] (for η = 0) that T^{∗} >0. Let
T < T˜ ^{∗}. We have that

sup

t∈[0,T]˜

kv(t)kX^{σ(t),r} ≤ sup

t∈[0,T˜]

ku(t)kX^{σ(t),s} ≡M_{( ˜}_{T}_{)}≡M.

By choosingσ(t) =σ0e^{−}^{Kt}^{˜} fort∈[0,T˜], where ˜K≡c_{(r,}T ,σ˜ 0)(1 +M), we have that

−σ^{0}(t)−c_{(r,}T ,σ˜ 0)σ(t)(1 +kv(t)kX^{σ(t),r}) =σ(t)c_{(r,}T ,σ˜ 0)(M− kv(t)kX^{σ(t),r})≥0,
for allt∈[0,T]. So, it follows from (23) that˜

d

dtkv(t)k^{2}X^{σ(t),r} ≤c_{(r,}T˜)+ 2ηkv(t)k^{2}X^{σ(t),r},

for allt∈[0,T˜]. Now, Gronwall´s inequality implies thatkv(t)kX^{σ(t),r} ≤c_{(r,}T ,σ˜ 0,φ)e^{ηt},
for allt∈[0,T˜], wherec_{(r,}T ,σ˜ 0,φ)≡(kφ^{0}k^{2}X^{σ}^{0}^{,r}+c_{(r,}T˜)T˜)^{1/2}is a continuous, positive,
non-decreasing function of ˜T ∈[0,∞). So, we can replace above the upper bound
M byc_{(r,}T ,σ˜ 0,φ)e^{η}^{T}^{˜} and take σ(t) = σ0e^{−}^{Kt}, K ≡c_{(r,}T ,σ˜ 0)(1 +c_{(r,}T ,σ˜ 0,φ)e^{η}^{T}^{˜}), for
t∈[0,T˜] . Sincec_{(r,}T ,σ˜ 0)andc_{(r,}T ,σ˜ 0,φ)are continuous, non-decreasing functions of
EJQTDE, 2005 No. 20, p. 11

T˜∈[0,∞), we can chooseσ(t) =σ0e^{−}^{Kt}^{ˆ} , where ˆK≡c(r,T^{∗},σ0)(1 +c(r,T^{∗},σ0,φ)e^{ηT}^{∗})
for allt∈[0, T^{∗}], and applying again the local theory we obtain a contradiction. Fi-
nally, it follows from (21) and (25) thatu(t)∈X^{σ(t),s}⊂X^{σ(T}^{),s}for allt∈[0, T].

Theorem 3.4. Letη ≥ 0and T >0. Suppose that ψ ∈ X^{σ}^{0}^{,s}, for some σ0 > 0
and s > 7/2. Suppose moreover that ψ(x) ∈ R for x ∈ R. Then the solution v
of problem (2) satisfies v ∈ C([0, T];X^{σ(T}^{),s}), where σ(t) is a positive monotone
decreasing function given by (24).

Proof. Since ψ^{0} ∈ X^{σ}^{0}^{,s}^{−}^{1}, and s−1 > 5/2, it follows from Theorem 3.3 that
u≡vx∈C([0, T];X^{σ(T),s}^{−}^{1}), whereσis given by (24). Making similar calculations
to (21), we see thatv(t)∈X^{σ(t),s} for allt∈[0, T]. Finally, since

kv(t)−v(τ)k^{2}X^{σ(T),s} ≤e^{2}^{√}^{2σ(T}^{)}kv(t)−v(τ)k^{2}s−1+ku(t)−u(τ)k^{2}X^{σ(T),s}^{−}^{1},

it follows thatv∈C([0, T];X^{σ(T),s}).

Remark: Forr >0, we have

X^{r,s}⊂Lr, if s≥0.

X^{r,s}⊂Xr, if s≥3/2.

Xr⊂Lr.

If f ∈Lr, and f(x)∈R for x∈R, then f ∈A(r).

(26)

The first statement above was already proved at the beginning of this Section. The
second part in (26) follows from the inequalitykfk^{2}Xr ≤4kfk^{2}X^{r,s} fors≥3/2. The
fact thatXr⊂Lris obvious. Finally, suppose thatf ∈Lrandf(x)∈Rforx∈R;
then we already know (see the first paragraph of this Section) thatfhas an analytic
extensionF ∈H^{2}(r); moreover, for every 0< r^{0}< r we have

Z r^{0}

−r^{0}

Z +∞

−∞ |F(x+iy)|^{2}dxdy ≤
Z r^{0}

−r^{0}

sup

|y|<r^{0}

Z +∞

−∞ |F(x+iy)|^{2}dx

dy≤2rkFk^{2}H^{2}(r).
4. Analyticity of Local Solutions of (1) in Xr-Spaces.

In this section we show that if the initial dataφof (1), with η ≥0, is analytic and has an analytic continuation to a strip containing the real axis, then there exists aT >0 such that the solutionu(t) of (1) has the same property for all t∈[0, T], but the width of the strip may decrease as a function of time. Similar to the proof of Theorem 2 in [6], we establish in Theorem 4.1 existence and analyticity of the solution of problem (1) simultaneously including the case when the initial condition is complex-valued on the real axis, more precisely whenφ∈Xσ0 for someσ0 >0.

The following lemma, which states a close relation between spacesXr andYr, will
be mainly used to prove thatu∈C^{ω}([0, T];Xσ(T)) in Theorem 4.1.

EJQTDE, 2005 No. 20, p. 12

Lemma 4.1. Yr is a dense subset of Xr, for r >0.

Proof. Letf ∈Yr. We have that Z

cosh(2rξ)|fˆ(ξ)|^{2}dξ ≤ cosh(2r)kfk^{2}+
Z

|ξ|>1

cosh(2rξ)|fˆ(ξ)|^{2}dξ

≤ cosh(2r)(kfk^{2}+kfk^{2}Yr),
Z

ξsinh(2rξ)|fˆ(ξ)|^{2}dξ ≤ sinh(2r)kfk^{2}+
Z

|ξ|>1

ξ^{2}cosh(2rξ)|fˆ(ξ)|^{2}dξ

≤ cosh(2r)(kfk^{2}+kfk^{2}Yr),
and similarlyR

ξ^{3}sinh(2rξ)|fˆ(ξ)|^{2}dξ ≤cosh(2r)(kfk^{2}+kfk^{2}Yr). So, using the last
three inequalities, it is not difficult to prove that

kfk^{2}Xr ≤4 cosh(2r)(kfk^{2}+kfk^{2}Yr). (27)
It follows from the last inequality thatYr⊂Xr. Now, let us consider the set

L2,0(R)=

g∈L^{2}(R); ∃K >0, |ˆg(ξ)| ≤Ka.e. inR,g(ξ) = 0 a.e. inˆ {ξ;|ξ|> K} .
Letg∈L2,0(R), thenP1

j=0

R ξ^{2j+2}cosh(2rξ)|ˆg(ξ)|^{2}dξ≤(K^{2}+K^{4}) cosh(2rK)kgk^{2},
sog∈Yr. Now we will prove thatL2,0(R) is a dense set inXr. Letf be an element
ofXr. For eachn∈N, take fn∈L2,0(R) given by

fcn(ξ) =

fˆ(ξ), if|ξ| ≤nand|fˆ(ξ)| ≤n 0, otherwise.

It follows easily from the definition of fn that |cfn(ξ)| ≤ |f(ξ)ˆ |, for all n ∈N and
ξ ∈ R. Moreover, cfn(ξ) → fˆ(ξ) as n→ ∞, for all ξ ∈R. So, by the dominated
convergence theorem, we have thatkfn−fk^{X}r →0 asn→ ∞. This concludes the

proof.

Lemma 4.2. (i.) Suppose thatF ∈H^{2,2}(r). Letf be the trace of F on the real
line. Thenf ∈Yr and

kfkYr≤√

2kFkH^{2,2}(r). (28)

(ii.) Conversely, suppose that f ∈Yr. Then f has an analytic extension F ∈
H^{2,2}(r)and

kFkH^{2,2}(r)≤p

10 cosh(2r)(kfk+kfkYr). (29)
Proof. (i.) Since F ∈ H^{2,2}(r), we see that ∂zF, ∂_{z}^{2}F ∈ H^{2}(r). Then, by using
Theorem 1 in [6], we have that f^{0}, f^{00}∈Lrand moreover

kfk^{2}Yr =kf^{0}k^{2}Lr+kf^{00}k^{2}Lr ≤2(k∂zFk^{2}H^{2}(r)+k∂_{z}^{2}Fk^{2}H^{2}(r))≤2kFk^{2}H^{2,2}(r).
EJQTDE, 2005 No. 20, p. 13

(ii.) By Lemma 4.1, Yr ⊂ Xr. So, it follows from Lemma 2.1 in [6] that
f has an analytic extension F on S(r) such that k∂zFkH^{1,2}(r) ≤ √

2kfkYr and
kFkH^{1,2}(r)≤√

2kfkXr. Now, using the last inequalities, we have that
kFk^{2}H^{2,2}(r) = kFk^{2}H^{1,2}(r)+k∂_{z}^{2}Fk^{2}H^{2}(r)≤ kFk^{2}H^{1,2}(r)+k∂zFk^{2}H^{1,2}(r)

≤ 2(kfk^{2}Xr+kfk^{2}Yr)≤10 cosh(2r)(kfk^{2}+kfk^{2}Yr),

where in the last inequality we have used (27).

Lemma 4.3. Letr >0. Suppose that F, G∈H^{2,2}(r). Then F G∈H^{2,2}(r) and
kF GkH^{2,2}(r)≤ckFkH^{2,2}(r)kGkH^{2,2}(r). (30)
Proof. Using Leibniz’s rule and Sobolev’s inequality (kGkH^{∞}(r)≤ckGkH^{1,2}(r)) we
have that

k∂_{z}^{2}(F G)kH^{2}(r) ≤ k∂_{z}^{2}FkH^{2}(r)kGkH^{∞}(r)+ 2k∂zFkH^{∞}(r)k∂zGkH^{2}(r)

+kFkH^{∞}(r)k∂_{z}^{2}GkH^{2}(r).

≤ ckFkH^{2,2}(r)kGkH^{2,2}(r).

Moreover, by Lemma 2.3 in [6], we have thatkF GkH^{1,2}(r)≤ckFkH^{1,2}(r)kGkH^{1,2}(r).
SincekF Gk^{2}H^{2,2}(r)=kF Gk^{2}H^{1,2}(r)+k∂_{z}^{2}(F G)k^{2}H^{2}(r), the proof of the Lemma follows

from the last two inequalities.

Lemma 4.4 and Corollary 4.1, below, are particular cases of Lemma 2.4 and its Corollary in Hayashi [6] respectively.

Lemma 4.4. There exists a polynomial ˜aof which coefficients are all nonnegative with the following property: If r >0and f, g, v∈Xr∩Yr, then

|(f ∂xf−g∂xg, v)Xr| ≤˜a(kfk^{X}r,kgk^{X}r) kfk^{Y}rkf−gk^{X}r+kf−gk^{Y}r

kvk^{X}r+kvk^{Y}r

. Corollary 4.1. There exists a polynomiala˜1with nonnegative coefficients such that iff, v∈Xr∩Yr, then

|(f ∂xf, v)Xr| ≤a˜1(kfk^{X}r)kfk^{Y}r kvk^{X}r+kvk^{Y}r

. (31)

Now, we state the main theorem of this Section.

Theorem 4.1. Let η ≥0. If φ ∈ Xσ0 for some σ0 > 0, then there exist aT =
T(kφk^{X}σ0, η, σ0) > 0 and a positive monotone decreasing function σ(t) satisfying
σ(0) = σ0 and such that (1) has a unique solution u ∈ C^{ω}([0, T];Xσ(T)). When
η= 0, u∈C([0, T];Xσ(T)).

EJQTDE, 2005 No. 20, p. 14

Proof. Letη ≥0, σ0 >0, and φ∈Xσ0. Letσ(t) =σ0e^{−}^{At/σ}^{0}, where the positive
constantAwill be conveniently chosen later. ForT >0, consider the space
B(T) =

f : [0, T]×R7→C;kfk^{2}B(T)= sup

0≤t≤Tkf(t)k^{2}Xσ(t)+A
Z T

0 kf(t)k^{2}Yσ(t)dt <+∞
and

Bρ(T) ={f ∈B(T);kfkB(T)≤ρ}.

Letρ= 4kφk^{X}σ0. Forv ∈Bρ(T), we define the mapping M byu=M v, whereu
is the solution of the linearized problem

∂tu+∂_{x}^{3}u+η(H∂xu+H∂_{x}^{3}u) =−v∂xv

u(0) =φ. (32)

More precisely ucan be obtained as follows. Take the Fourier transform to (32), then

∂tu(t, ξ)ˆ −iξ^{3}u(t, ξ) +ˆ η(−|ξ|+|ξ|^{3})ˆu(t, ξ) =−dvvx(t, ξ).

Now, integrating the last expression between 0 andt, it follows that ˆ

u(t, ξ) =e^{(iξ}^{3}^{+η(}^{|}^{ξ}^{|−|}^{ξ}^{|}^{3}^{))t}φ(ξ)ˆ −
Z t

0

e^{(iξ}^{3}^{+η(}^{|}^{ξ}^{|−|}^{ξ}^{|}^{3}^{))(t}^{−}^{τ)}vvdx(τ, ξ)dτ, (33)
where the last integral is well defined, since

Z t

0 |dvvx(τ, ξ)|dτ = Z t

0

dv(τ)∗v[x(τ)

(ξ)dτ = Z t

0

Z ˆ

v(τ, ξ−ω)ωˆv(τ, ω)dωdτ

≤ Z t

0 kv(τ)kkv(τ)k^{1}dτ ≤ρ^{2}t,
and in the last inequality we have used the fact that

kv(τ)k1≤ kv(τ)kXσ(τ) ≤ kvkB(T)≤ρ.

Let us chooseA, T >0 such that

e^{−}^{AT /σ}^{0} > 1

2. (34)

Let v ∈ Bρ(T). It will be proved that u = M v ∈ Bρ(T), for suitably chosen T, A >0. Now, it is not difficult to see that

d

dtku(t)k^{2}Xσ(t) = 2σ^{0}(t)
X1
j=0

Z

ξ^{2j} ξsinh(2σ(t)ξ) +ξ^{2}cosh(2σ(t)ξ)

|u(t, ξ)ˆ |^{2}dξ

+2<

X1 j=0

Z

ξ^{2j} cosh(2σ(t)ξ) +ξsinh(2σ(t)ξ)

∂tu(t, ξ)ˆˆ u(t, ξ)dξ.

EJQTDE, 2005 No. 20, p. 15

Taking the Fourier transform to (32), multiplying both sides of the obtained equa-
tion byξ^{2j} cosh(2σ(t)ξ) +ξsinh(2σ(t)ξ)

ˆ

u(t, ξ), integrating with respect toξ, sum- ming the terms for j= 0,1 and finally taking the real part we get

1 2

d

dtku(t)k^{2}X_{σ(t)}−σ^{0}(t)
X1
j=0

Z

ξ^{2j} ξsinh(2σ(t)ξ) +ξ^{2}cosh(2σ(t)ξ)

|u(t, ξ)ˆ |^{2}dξ

+η X1 j=0

Z

ξ^{2j} cosh(2σ(t)ξ) +ξsinh(2σ(t)ξ)

(−|ξ|+|ξ|^{3})|u(t, ξ)ˆ |^{2}dξ

=−<(vvx, u)Xσ(t).

Sinceσ^{0}(t) =−Ae^{−}^{At}^{σ}^{0} and|ξ| − |ξ|^{3}≤1, for allξ∈R, it follows that
1

2 d

dtku(t)k^{2}Xσ(t)+Ae^{−}^{At}^{σ}^{0}ku(t)k^{2}Yσ(t) ≤ −<(vvx, u)Xσ(t)+ηku(t)k^{2}Xσ(t).
Using (34) and Corollary 4.1, it follows from the last inequality that

1 2

d

dtku(t)k^{2}Xσ(t)+A

2ku(t)k^{2}Yσ(t) ≤a(ρ)kv(t)k^{Y}σ(t)(ku(t)k^{X}σ(t)+ku(t)k^{Y}σ(t))+ηku(t)k^{2}Xσ(t),
where a(·) is a polynomial with nonnegative coefficients. Now integrating the last
inequality from 0 tot we get

sup

0≤t≤Tku(t)k^{2}Xσ(t)+A
Z T

0 ku(t)k^{2}Yσ(t)dt≤2

kφk^{2}Xσ0+ 2a(ρ)
Z T

0 kv(t)k^{2}Yσ(t)dt^{1/2}

· Z ^{T}

0 ku(t)k^{2}Xσ(t)dt1/2

+ Z T

0 ku(t)k^{2}Yσ(t)dt1/2
+ 2η

Z T

0 ku(t)k^{2}Xσ(t)dt
.

Then,

kuk^{2}B(T) ≤ ρ^{2}

8 +4ρa(ρ)

√A

√T+ 1

√A

kukB(T)+ 4ηTkuk^{2}B(T)

≤ ρ^{2}
8 +1

2

4ρa(ρ)

√A

√T+ 1

√A
^{2}

+ (1

2+ 4ηT)kuk^{2}B(T).
By choosingT >0 small enough such that

ηT < 1

12, (35)

we have that

kuk^{2}B(T)≤ 3

4+48a^{2}(ρ)
A

√T+ 1

√A 2

ρ^{2}.

Now we takeA, T >0 such that a(ρ)√

A

√T+ 1

√A < 1

8√

3. (36)

So, choosingA, T >0 such that (34)-(36) are satisfied, it follows that u=M v ∈ Bρ(T).

Now let us prove that the mapping M is a contraction, defined from Bρ(T) into EJQTDE, 2005 No. 20, p. 16