### Existence and Polynomial Growth of

### periodic solutions to KdV-type Equations

Yung-fu Fang

Abstract. We establish local and global existence of periodic solutions for KdV type equations, employing Fourier series and a fixed point argument. We also investigate the polynomial growth of the solutions.

1. Introduction

In this paper, we study the existence and the polynomial bound of periodic solutions for the nonlinear dispersive equation of the Korteweg-de Vries type:

(1)

½

*ut* *+ ∂xαu + uk∂xu = 0,* *(t, x) ∈ R*+*× T;*
*u(0, x) = φ(x),*

*where φ is a real function, α a real number, k a positive integer and ∂α*
*x* the

fractional derivative defined by, via Fourier transform,

(2) *∂*c*α*

*x* *= i|ξ|αsgnξ.*

*The function u considered here is a real-valued and space-periodic function.*
The method used here is the fixed point argument applied to the
correspond-ing integral equation

(3) *u(t) = W (t)φ −*

Z _{t}

0

*W (t − τ )w(τ )dτ,*
*1991 Mathematics Subject Classification. 35D05.*

*Key words and phrases. global existence, fixed point argument, periodic solution, KdV*

type equation, Polynomial growth..

*where W (t) = e−t∂xα* *and w = uk∂ _{x}u, see [B2] and [FG].*

The original KdV equation,

(4a) *∂tu + ∂x*3*u + u∂xu = 0,*

was derived in 1895 by Korteweg and de Vries as an approximate model of
shallow water waves, see [KdV]. It also has been derived in plasma physics
and in the studies of anharmonic lattices, see [MGKr]. Some generalizations
of KdV equation has been used to describe certain physical problems, e.g.
KdV-type equations in certain crystalline lattices, see [ABFS]. In 1975, P.
Lax [L] constructed a large class of special solutions of the KdV equation
which are periodic in space and almost periodic in time. In 1993, Bourgain
[B2] proved existence of periodic solutions for generalized KdV equations,
(4b) *∂tu + ∂x*3*u + uk∂xu = 0.*

In 1995, Bourgain [B3] extended the result of local solutions to more general KdV equation,

(4c) *∂tu + ∂x2j+1u + F (u, lower order terms) = 0.*

On the other hand, some fifth order (even 7th order) KdV-type equations,
(4d) *∂tu + up∂xu + ∂x*3*u + ∂x*5*u = 0,*

also has been considered, see [K]. In 1996, Bourgain [B4] obtained a
polyno-mial bound of higher Sobolev norm of solutions for generalized KdV
equa-tions. In 1997, Staffilani [S] improved the existence result and the polynomial
*bound of solution for equation (4b). In 2004, Colliander eta [CKSTT] gave*
multilinear estimates for for periodic case and their applications.

It is well known that the KdV equation and some KdV-type equations
possess solitary waves and infinitely many conservation laws, see [L] and
[MGKr]. For the equation (1), there are three quantities are conserved,
namely,
(5)
R
T*u(t)dx,*
R
T*u*2*(t)dx,* and
R
T
1
2*(∂*
*α−1*
2
*x* *u)*2*(t)dx −*
R
T
1
*(k+1)(k+2)uk+2(t)dx.*

In the nonperiodic case there have been some good results on questions of existence and regularity, see [KPV] and [BKPSV].

The outline of this paper is that we first show the local existence result
*for the initial value problem (1) with k = 1. The essence of the proof is an a*
priori estimate inspired by work of J. Bourgain, see [B2] and [B3]. It can be
*understood as a multiplier estimate on the set of R×Z. However the proof of*
the estimate presented here is different from those of [B2]. It essentially relies
on an idea of Zygmund [Zy]. Once the local existence is proved, we invoke a
conservation law to get global existence. Next we discuss the existence results
for the initial value problem (1), (hereafter we write IVP), with higher order
*nonlinearity k ≥ 2. In section 4, we will give a straightforward proof of the*
a priori estimate. Finally we will discuss the polynomial bound for solution
of IVP (1). The main results of this paper are the following theorems.
*Theorem A. Let α ≥ 3. If the initial data of (1) is in L*2 _{for k = 1 and in}*Hα−1*2 *(and small) for k ≥ 2, then the initial value problem of (1) is globally*

*well-posed.*

*Theorem B. Let α ≥ 3 and* 3
2 *≤*

*α − 1*

2 *< s. If the initial data is in H*

*s*
*and small, then the global solution u satisfies*

(6) *ku(t)kHs* *≤ C|t|2s.*

2. Existence Results

Throughout this paper we call

(7) *A(ξ) = |ξ|αsgnξ* and *S = |τ − A(ξ)| + 1;*

denote by e*g(t, ξ) =* 1
*2π*
Z * _{2π}*
0

*e−ixξg(t, x) dx and by bg(τ, ξ) =*Z

*R*

*e−itτ*e

*g(t, ξ) dt*

the Fourier coefficients and the Fourier transforms with respect to the space
variable and to both the space-time variables, respectively. First we state
*the local existence result for IVP (1) with k = 1:*

*Theorem 1. If α ≥ 3 and the initial data φ ∈ L*2 _{(H}s_{, s ≥ 0), then the}*IVP of (1) is locally well-posed.*

To prove the above theorem, we use a fixed point argument and the fol-lowing a priori estimate whose proof will be given later.

*Theorem 2. If α ≥ 2, then we have the following estimates*

(8a) *kf kL*4_{(R×T)}*≤ C*

°
*°S1+α*

*4α* *f k*b _{L}_{2}_{(R×Z)}

*and its dual*

(8b) *k bf /S1+α4α* *k _{L}*

_{2}

_{(R×Z)}*≤ Ckf k*

*L*43*(R×T).*

Before proving Theorem 1, we consider the corresponding linear problem: (9)

½

*ut* *+ ∂xαu + w = 0,* *(t, x) ∈ R*+*× S*1;
*u(0, x) = φ(x).*

The periodic solution of (9) can be expressed in integral form as follows.
(10) *u(t, x) =*X
*ξ*
b
*φ(ξ)ei(xξ+tA) _{+ 2i}*X

*ξ*

*eixξ*Z

*eitτ*

_{− e}itA*τ − A*

*w(τ, ξ)dτ.*b

*Call U (t, x) and V (t, x) the linear and nonlinear parts of u respectively,*

(11)

(

*U (t, x) =*P* _{ξ}φ(ξ)e*b

*i(xξ+tA)*

_{;}

*V (t, x) = 2i*P

*R*

_{ξ}eixξ*eitτ−eitA*

*τ −A* *w(τ, ξ)dτ.*b

We want to study the nonlinear part first. Choose cut-off functions b*a and bb*

such that b*a + bb = 1, supp ba ⊂ [−2R, 2R] and suppbb ⊂ {x : |x| ≥ R}. Make*

*a decomposition of V (t, x) in the following way.*

(12) *V (t, x) = H(t, x) + Ψ*1*(t, x) + Ψ*2*(t, x),*
where
(13)
b
*H(τ, ξ) =* b*b(τ −A) _{τ −A}*

*w(τ, ξ),*b b Ψ1

*(τ, ξ) = δ(τ − A)*R b

_{b(λ−A)}*λ−A*

*w(λ, ξ)dλ,*b b Ψ2

*(τ, ξ) =*P

*kδ(k)(τ − A) bGk(ξ),*b

*Gk(ξ) =*

*i*

*k*

_{(2R)}k−1*k!*R ¡

_{λ−A}*2R*¢

*k−1*b

*a(λ − A) bw(λ, ξ)dλ,*

*where δ(τ ) is the delta function and δ(k)* _{is its k-th derivative.}

Since the solution does not decay in time, it is necessary to localize it in
*time. We assume that ψ is a cutoff function supported in a neighborhood of*
*0 and denote ψδ(t) = ψ(t/δ), where δ is a small number to be determined*

later. Let

(15) *uδ(t, x) = ψδ(t)(Ψ*1+ Ψ2*)(t, x) + F (t, x).*

The norm used here is defined by

(16) *N (u) = kS*12*uk*b _{L}_{2}_{(R×Z)}.

We want to prove the following result first.

*Theorem 3. Let uδ* *be defined as in (15), we have the estimate*

(17) *N (uδ) ≤ C*
°
° b*w*
*S*12
°
°
*L*2* _{(R×Z)}+ C*
X

*ξ*¯ ¯ Z

*| bw|*

*S*

*dλ*¯ ¯2 1 2

*.*

*Proof. For the term H, since* *S*

2¯*¯bb(τ − A)*¯¯2
*(τ − A)*2 *≤ 1, we get*
*kS*12*Hk*b 2
*L*2 *≤ C*
°
° b*w*
*S*12
°
°
*L*2*.*

For the term Ψ1, since

Z
*S| bψδ(τ − A)|*2*dτ ≤ C(ψ), we have*
*kS*12( b*ψ _{δ}∗ b*Ψ

_{1}

*)k*2

_{L}*≤ C* X

*ξ*¯ ¯Z

*| bw|*

*S*

*dλ*¯ ¯2 1 2

*.*

For the last term, using the facts that

*kS*12*t*d*kψ _{δ}k_{L}*

_{2}

*≤ C(ψ)(2δ)k*

*and k bG*

_{k}k*L*(

_{Z)}*≤ C*

*(2R)k*

*k!*° ° b

*w*

*S*12 ° °

*L*2

*,*we obtain

*kS*12( b

*ψ*Ψ

_{δ}∗ b_{2}

*)k*

_{L}_{2}

*≤ C(ψ)e4Rδ*° ° b

*w*

*S*12 ° °

*L*2

*. ¤*

We divide the proof for Theorem 1 into several steps. First we state and
*prove two lemmas. Notice that now w = u∂xu.*

Lemma 4.
(18) °° b*w*
*S*12
°
°
*L*2_{(R×Z)}*≤ Cδ*
*α−1*
*8α* *N (u)*2*.*
Lemma 5.
(19)
X
*ξ*
¯
¯
Z
*| bw|*
*S* *dλ*
¯
¯2
1
2
*≤ Cδα−18α* *N (u)*2*.*

*Proof of Lemma 4. Observe that | bw(τ, ξ)| is bounded by*

(20) *|ξ|*X

*η*

Z

*|bu(λ, η)||bu(τ − λ, ξ − η)|.*

*To cancel out the factor |ξ|, notice that*
¯

*¯(τ − A(ξ)) − [(λ − A(η)) + (τ − λ − A(ξ − η))]*¯¯
=¯*¯ − A(ξ) + A(η) + A(ξ − η)*¯*¯ ≥ C|ξ|α−1,*

(21)

*provided ξ 6= 0, η 6= 0 and ξ 6= η. Also observe that bw(τ, 0) = 0. Assume the*

*average of u is zero, i.e. bu(τ, 0) = 0, temporarily so that we have (21).*

( This assumption will be removed later.) For the sake of convenience, we denote

(22) (

*C(λ, η) = (|λ − A(η)| + 1)*12*|bu(λ, η)| = S*12*|bu(λ, η)|,*

b

*F (λ, η) = |bu(λ, η)|* and *G(λ, η) = C(λ, η) = S*b 12*|bu(λ, η)|*

Thus we can bound *w*b

*S*12

by

(24) Z X

*η*

*|ξ|C(λ, η)C(τ − λ, ξ − η)*

*(|τ − A(ξ)| + 1)*12*(|λ − A(η)| + 1)*12*(|τ − λ − A(ξ − η)| + 1)*12

*dλ*

From (21) one of the following cases happens.

(25)
*|τ − A(ξ)| ≥* *C*_{3}*|ξ|α−1 _{,}*

*|λ − A(η)| ≥*

*C*

_{3}

*|ξ|α−1*

_{,}_{and}

*|τ − λ − A(ξ − η)| ≥*

*C*

_{3}

*|ξ|α−1*

_{.}For the first case of (25), we have
Z X
*η*
*C(λ, η)C(τ − λ, ξ − η)*
*|ξ|α−3*2 *(|λ − A(η)| + 1)*12*(|τ − λ − A(ξ − η)| + 1|*12
*dλ ≤ cF*2_{(τ, ξ).}

*Taking L*2 _{norm on c}* _{F}*2

_{and applying Theorem 2, we get}

(26) *k cF*2_{k}*L*2 *≤ N (u)*
*α+1*
*α* *kuk*
*α−1*
*α*
*L*2 *.*

*Assume that u is supported by [−δ, δ] × T , since* *α + 1*
*4α* *<*
1
2, we have
(27) *kukL*2 *≤ δ*
1
4*kuk _{L}*4
which implies
(28)

*kF kL*4

*≤ Cδ*

*α−1*

*8α*

*N (u).*

For the second case of (25), we have
Z X
*η*
*C(λ, η)C(τ − λ, ξ − η)*
*(|τ − A(ξ)| + 1)*12*|ξ|*
*α−3*
2 *(|τ − λ − A(ξ − η)| + 1|*12
*dλ ≤* *F G(τ, ξ)*d
*S*12
*.*

*Taking L*2 _{norm on d}* _{F G/S}*1

_{2}

_{, we have}

°
°*F G(τ, ξ)*d
*S*12
*kL*2 *≤ Cδ*
*α−1*
*8α* *N (u)*2*.*

The proof of the last case of (25) is similar to the second one. ¤ Remark. To remove the condition that the solution is of zero average, b

*u(τ, 0) = 0, we may modify the problem (1) by replacing φ by φ*1 *+ φ*0

*and u by u*1 *+ φ*0*, where φ*0 =

Z

*φ(x)dx =*

Z

*u(t, x)dx. All arguments go*

*through if A(ξ) is replaced by A(ξ) − φ*0*ξ.*

Proof of Lemma 5. Observe that *| bw|*

*S* is bounded by

(29) Z X

*η*

*|ξ|C(λ, η)C(τ − λ, ξ − η)*

*(|τ − A(ξ)| + 1)(|λ − A(η)| + 1)*12*(|τ − λ − A(ξ − η)| + 1|*12

We use the notations denoted in the previous Lemma and distinguish again the cases in (25).

For the first case of (25), we have
X
*ξ*
¡Z *| bw(τ, ξ)|*
*|τ − A| + 1dτ*
¢2
1
2
*∼*
X
*ξ*
¡Z *|ξ| cF*2_{(τ, ξ)}*|τ − A| + |ξ|α−1dλ*
¢2
1
2
*.*

*Let a(ξ) be a nonnegative sequence with unit l*2_{-norm, i.e.} X
*ξ*

*a*2*(ξ) = 1.*
Using the first one in (25) and

(30)
Z
*ξ*2
*(|τ − A| + |ξ|α−1*_{)}2*dτ ≤ C,*
we can estimate
(31) X
*ξ*
Z
*a(ξ)|ξ| cF*2_{(τ, ξ)}*|τ − A| + |ξ|α−1dτ ≤ Cδ*
*α−1*
*4α* *N (u)*2*.*

Use a duality argument, we get (18). For the second case of (25), we have

Z X
*η*
*C(λ, η)C(τ − λ, ξ − η)*
*(|τ − A(ξ)| + 1)|ξ|α−3*2 *(|τ − λ − A(ξ − η)| + 1|*
1
2
*dλ ≤* *F G(τ, ξ)*d
*S* *.*

*Taking l*2 _{norm on the integral}

Z
d
*F G/Sdτ, we get*
(32)
X
*ξ*
¡Z d*F G(τ, ξ)*
*S* *dτ*
¢2
1
2
*≤ Cδα−18α* *N (u)*2*.*

The proof of the last case of (25) is again similar to the second one. ¤ Here we come to the stage that we can prove Theorem 1.

Proof of Theorem 1. First we combine the results of Theorem 3 and
*Lemmas 1 and 2 to get , for the nonlinear part V (t, x) of the solution,*

(33) *kS*12*u*b_{δ}k_{L}_{2} *≤ Cδ*

*α−1*

*8α* *N (u)*2*.*

Define the map by

(34) *T u(t, x) = ψδ(t)U (t, x) + ψδ(t)V (t, x).*

*Thus the N norm of T u is bounded by*

(35) *C*
³
*kφkL*2 *+ δ*
*α−1*
*8α* *N (u)*2
´
*.*

*By choosing sufficiently large M , we have, for suitable δ and R,*

(36) *N (u) ≤ M* *=⇒* *N (T u) ≤ M,*

*provided that C(φ) + δα−18α* *M*2 *≤ M.*

*Next we estimate the difference of T u and T v and get*

(37) *N (T u − T v) ≤ Cδα−18α* ¡*N (u) + N (v)*¢*N (u − v).*

*Therefore, again for suitable δ and R, we obtain*

(38) *N (T u − T v) ≤* 1

2*N (u − v),*

*provided that Cδα−18α*

¡

*N (u) + N (v)*¢*≤* 1

2 which can be satisfied by choosing

*δ small for given M . By Picard’s theorem, the map T is a contraction with*

*respect to the norm N (u), hence it has a unique fixed point.* ¤
*Remarks. The nonlinear term can be replaced by ∂ _{x}γu*2

*, but 1 ≤ γ ≤*

*α − 1*

2 .
*To get global existence we need a conservation law, i.e.ku(t)kL*2 is constant

*Theorem 6. Let α ≥ 3. If the initial data of (1) is in L*2 _{(H}s_{, s ≥ 0),}*then there is a unique periodic solution for the IVP of (1) which exists for*
*all time.*

Remarks. The method used here can be applied to the following extension of equation (1)

(40) *ut+ ∂xαu + ∂xβu + u∂xu = 0,*

*where 1 < β < α and 3 ≤ α. (See [K] for a particular case called the fifth*
order KdV-type equation.)

3. Further Results

*In this section, we want to discuss the IVP of (1) for k ≥ 2. First we*
*consider the case k = 2, then k ≥ 3.*

(41)

½

*ut+ ∂xαu + u*2*∂xu = 0,* *(t, x) ∈ R*+*× T,*
*u(0, x) = φ(x),*

*where α ≥ 3.*

*Theorem 7. For k = 2, the IVP of (1) is locally well-posed for data in H*1

*(Hs, s ≥ 1), and for specified*

Z

*T*
*φ*2*dx.*

To prove the theorem 7, we need the followings.
Lemma 8. (Bourgain [B2])
³
*f*2*−*
Z
T
*f*2*dx*
´
*∂xf =*
1
3
X
*η+ζ6=0*
*ξ−η6=0*
*ξ−ζ6=0*
*ξ bf (η) bf (ζ) bf (ξ − η − ζ)eiξx−*X
*ξ*
b
*f (ξ)*2*f (−ξ)e*b *iξx.*

*To estimate w we introduce the following norm and notation.*
*|||u|||*2 =X*(1 + |ξ|*2)
Z
*S|bu(τ, ξ)|*2*dτ +*X*(1 + |ξ|*2)
³ Z
*|bu(τ, ξ)|dτ*
´_{2}
;
*S = 1 + |τ − B(ξ)| = 1 + |τ − A(ξ) + cξ|.*

*Proposition 9. For uδ, we can estimate it as follows.*

*|||uδ|||*2 *≤*
X
*(1 + |ξ|*2)
Z
*| bw(τ, ξ)|*2
*S* *dτ +*
X
*(1 + |ξ|*2)*³ Z | bw(τ, ξ)|*
*S* *dτ*
´2
*.*

This proposition can be proved in a similar manner as that in [B2]. Proof. Due to the conservation law,

Z
T
*u*2*(t, x)dx =*
Z
T
*φ*2*(x)dx, we denote*
(42) *c =*
Z
T
*φ*2*(x)dx*

and consider the IVP

(43)

½

*ut* *+ ∂xαu + c∂xu = 0,*
*u(0, x) = ψ(x)*

for which the solution can be written as

(44) *u(t, x) = Stψ(x) =*

X

*ξ*

b

*ψ(ξ)ei(ξx+(A−cξ)t).*

Consider the integral equation

(45) *u(t) = Stφ +*
Z * _{t}*
0

*S(t − τ )w(τ )dτ,*

*where w = [*Z T

*u*2*dx − u*2*]∂xu, which is equivalent to the IVP*

(46)
(
*ut+ ∂xα* *+ c∂xu =*
³ R
T*u*2*dx − u*2
´
*∂xu,*
*u(0, x) = S*0*φ = φ.*

*We construct a sequence of functions {uk} by*
(47) *uk+1* =
X
*ξ*
b
*φ(ξ)ei(ξx+Bt)* _{+}X
*ξ*
*eiξx*
Z
b
*wk(ξ, τ )*
*eitτ* _{− e}iBt*τ − B* *dτ*
*where wk* = [
Z
T

*u*2* _{k}dx − u*2

_{k}]∂xuk, and B = A − c. Observe that¯
¯
*¯(τ − B(ξ)) − [(λ − B(η)) + (θ − B(ζ)) + (τ − λ − θ − B(ξ − η − ζ))]*
¯
¯
¯
*∼*
¯
¯
*¯|ξ|α− |η|α− |ζ|α* *− |ξ − η − ζ|α*
¯
¯
¯
(48)

*To find a lower bound of (48), assume that η +ζ 6= 0, ξ −η 6= 0, and ξ −ζ 6= 0.*
*Case I, if one or two of |ξ|, |η|, |ζ| are larger than the others, then*

*(48) ≥*
³
*|η| + |ζ| + |ξ − η − ζ|*
´*α−1*
*.*
*Case II, if |η| ∼ |ζ| ∼ |ξ − η − ζ|,*
*(48) ≥*
³
*|η| + |ζ| + |ξ − η − ζ|*
´*α−2*
*.*

*Apply Bourgain’s lemma and use the notation Ω(ξ) = {(η, ζ) ∈ Z*2 :

*η + ζ 6= 0, ξ − η 6= 0, ξ − ζ 6= 0}, we can rewrite*
(51)
b
*w(τ, ξ) =*1
3
X
*η,ζ∈Ω(ξ)*
*ξ*
Z
b

*u(η, λ)bu(ζ, θ)bu(ξ − η − ζ, τ − λ − θ)dλdθ*

*− ξ*

Z b

*u(ξ, λ)bu(ξ, θ)bu(−ξ, τ − λ − θ)dλdθ.*

Call
b
*w*1*(τ, ξ) =*
*|ξ|*
3
X
*η,ζ∈Ω(ξ)*
Z

*|bu|(η, λ)|bu|(ζ, θ)|bu|(ξ − η − ζ, τ − λ − θ)dλdθ*

b

*w*2*(τ, ξ) = |ξ|*

Z

*So it is sufficient to estimate, for j = 1, 2,*
X
*(1 + |ξ|*2)
Z
*| bwj(τ, ξ)|*2
*S* *dτ* and
X
*(1 + |ξ|*2)*³ Z | bwj(τ, ξ)|*
*S* *dτ*
´2
*,*

For the case I, we distinguish four cases,

(53)
*|τ − B(ξ)| ≥ |ξ|α−1 _{,}*

*|λ − B(η)| ≥ |ξ|α−1*

_{,}*|θ − B(ζ)| ≥ |ξ|α−1*

_{,}*|τ − λ − θ − B(ξ − η − ζ)| ≥ |ξ|α−1*

_{.}*For the case II, we employ the inequality (1 + |ξ|)|ξ| < C|η||ζ|.*
*We can control the solution u by the norm ||| · ||| and get*
(54) *|||T u||| ≤ Cδα−14α* *|||u|||*3*.*

Fixed point argument ensures the existence and uniqueness of the

solu-tion. ¤

To get global existence we need conservation laws. We first discuss briefly
how to derive those conserved quantities given in (5). For the first one, it
is straightforward to get that R_{T}*u(t)dx =* R_{T}*φdx. The second one can be*

*proved as follows. Multiplying the equation (1) by u and integrating by*
parts, we get
Z
1
2*∂t(u*
2* _{)dx +}*
Z

*u∂*

_{x}αudx = 0.*Using the identity |eu(t, −ξ)| = |eu(t, ξ)|, we can prove that the second integral*

*above is 0 which implies that ku(t)kL*2 is conserved. For the last one, we first

*take the integral operator ∂ _{x}−1*

*on the equation, multiply by ut*and then

integrate by parts.

*Next we apply those conservation laws to obtain the boundedness of Hα−1*2

*norm of solution. We first use interpolation inequalities to bound H*1_{-norm}

*of u, cf [L]. Let us assume that u is a smooth periodic function temporarily*
and denote by
(55)
R
T*u*2*dx = F*0*,* *max |u(x)| = M,*
R
T
³
*∂α−1*2
*x* *u*
´2
*−* *2uk+2*
*(k+1)(k+2)dx = F*1*,* and
R
T*u*2*xdx = S.*

*Since u is smooth, there exists a point x*0 such that

(56) *u*2*(x*0) =

Z

*T*

*u*2*(x)dx = F*0*,*

*we apply the identity u(x) = u(x*0) +

Z _{x}*x*0
*uxdx to get*
*u*2*(x) ≤ 2u*2*(x*0) + 2
Z
*u*2* _{x}dx ≤ 2F*0

*+ 2S.*

*This implies that M*2 *≤ 2F*0 *+ 2S. On the other hand, we can bound S as*

follows.
*S =*
Z
T
*u*2* _{x}dx ≤ C*
µZ
T
³

*∂α−1*2

*x*

*u*´

_{2}

*dx +*Z

*T*

*u*2

*dx*¶

*≤ C*³

*F*1

*+ F*0+

*2M*

*k*

*(k + 1)(k + 2)F*0

*dx*´

*.*Hence we have (57)

*M*2

*≤ 2F*0

*+ 2C(F*1

*+ F*0) +

*2CF*0

*(k + 1)(k + 2)M*

*k*

_{.}*Thus we can deduce that M is bounded by some constant C = C(F*0*, F*1),

*provided that F*0 *and F*1 are small. Also we have

(58)
Z
T
³
*∂α−1*2
*x* *u*
´_{2}
*dx ≤ F*1+
*2C(F*0*, F*1)*k*
*(k + 1)(k + 2)F*0*.*

*Another approach to bound the Hα−1*2 *-norm of solution u is that we *

*inter-polate between the L*2 _{and H}α−1

2 -norms, see [B1]. Using H¨older and Sobolev

inequalities, we have
Z
T
³
*∂α−1*2
*x* *u*
´2
*dx = F*1+
Z
T
*2uk+2*
*(k + 1)(k + 2)dx*
*≤ F*1*+ CkukL*2*kukk+1*
*L2(k+1)* *≤ F*1*+ CkφkL*2*kuk*
*k+1*
*Hα−1*2
*.*

*This implies that if kφk*

*Hα−1*2 is small, then we have

(60) *ku(t)k*

*Hα−1*2 *≤ C* *for all t .*

*Theorem 10. For k = 2, the IVP of (1) is globally well-posed for small*

*data in Hα−1*2 *(HS, s ≥* *α − 1*

2 *), and for specified*
Z

*T*
*φ*2*dx.*

*For the case k ≥ 3, besides ideas in [B2], we use those in [S] as well.*
*Definition. i) The space Ys,b _{, s, b ≥ 0, is the closure of the Schwartz }*

*func-tions S(T × R), with respect to the norm*

(61) *kf k _{Y}s,b* = max

*i=1,2,3ν*

*(s,b)*

*i*

*(f ),*where (62)

*ν*

_{1}

*(s,b)(f )*2

_{=}P

*ξ(1 + |ξ|)2s*³ R R

*| bf |(τ, ξ)dτ*´2

*ν*

_{2}

*(s,b)(f )*2

_{=}P

*ξ(1 + |ξ|)2s*R R

*| bf |*2

*(τ, ξ)(1 + |τ − A(ξ)|)2bdτ*

*ν*

_{3}

*(s,b)(f )*2

_{=}P

*ξ(1 + |ξ|)2s−2*R R

*| bf |*2

*(τ, ξ)(1 + |τ − A(ξ)|)2b+1dτ.*

*Denote the space Ys,b _{[−δ, δ] of functions defined on T × [−δ, δ] with the}*

restriction norm

(63) *kf kYs,b _{[−δ,δ]}*

*= inf kF k*

_{Y}s,b,*where the infimum is taken over all the extensions F of f on T × R.*

*ii) The space Ys,b, s, b ≥ 0, is the closure of the Schwartz functions S(T ×*
R), with respect to the norm

(64) *kf k _{Y}s,b* = max

*i=1,2,3,4µ*

*(s,b)*

*i*

*(f ),*where (65)

*µ(s,b)*

_{1}

*(f )*2

_{=}P

*ξ(1 + |ξ|)2s*³ R

*|τ −A(ξ)>|ξ|| bf |(τ, ξ)dτ*´

_{2}

*µ(s,b)*

_{2}

*(f )*2

_{=}P

*ξ(1 + |ξ|)2s*R

*|τ −A(ξ)>|ξ|| bf |*2

*(1 + |τ − A(ξ)|)2bdτ*

*µ(s,b)*

_{3}

*(f )*2

_{=}P

*ξ(1 + |ξ|)2s−2*R R

*| bf |*2

*(τ, ξ)(1 + |τ − A(ξ)|)4bdτ*

*µ(s,b)*

_{4}

*(f )*

*= k∂s*

*xf kL∞*

*t*

*L*2

*x.*

*As in i), we have the space Ys,b[−δ, δ].*

*iii) Let f and g be functions on T × [−δ, δ] and F and G be the extensions*
*on T × R. Denote*
(66)
(
*βF(t)* =
R
T*Fk(t, x)dx*
*F(F )(τ, ξ) =*R_{R}*e−itτ _{e}iξ*R

_{0}

*tβF(σ)dσF (t, ξ)dt*e (67)

*ds*1

*(F, G)*2 = P

*ξ(1 + |ξ|)2s*³ R R

*|F(F ) − F(G)|(τ, ξ)dτ*´

_{2}

*ds*2

*(F, G)*2 = P

*ξ(1 + |ξ|)2s*R R

*|F(F ) − F(G)|*2

*(1 + |τ − A|)2bdτ*

*ds*3

*(F, G)*2 = P

*ξ(1 + |ξ|)2s*R R

*|F(F ) − F(G)|*2

*(1 + |τ − A|)4bdτ.*

*Then denote the metric space by X _{k}s,b[−δ, δ] with respect to the metric*

(68) *ds _{∗}(F, G) = inf*

*F G*n X

*i*

*ds*o

_{i}(F, G)*.*

*The space X _{k}s,b[−δ, δ] is a complete metric space, s ≥* 1

_{2}

*, b ≥ 0, see [S].*

*Theorem 11. Consider the IVP (1) for k ≥ 3. If φ ∈ Hs*

_{, s ≥}*α−1*

2 *, then*
*there exists δ = δ(kφk*

*Hα−1*2 *) and a unique solution u in the space X*

*s,b*
*k* *[−δ, δ]*
*such that*

(70) *ds*

*∗(u, 0) ≤ CkφkHs.*

To prove Theorem 11, we consider the associated problem of IVP (1),

(71)
½
*vt+ ∂xαv + (vk−*
R
T*vkdx)∂xv = 0,* *(t, x) ∈ R*+*× T;*
*v(0, x) = φ(x),*
Consider
e
*u(t, ξ) = eiξ*R_{0}*tβv(σ)dσ*_{e}_{v(t, ξ).}

*The importance of the IVP (71) is that if v is a solution for the problem,*
*then u given by above is a solution for IVP (1).*

*Proposition 12. Let φ ∈ Hs* *and s ≥* *α − 1*

2 *. Then there exists δ =*

*δ(kφk*

*Hα−1*2 *) such that the problem (71) is well posed in Y*

*s,*1

2*[−δ, δ] and the*

*solution satisfies the bound*

(72) *kvk*

*Ys, 1*2*[−δ,δ]* *≤ CkφkHs.*

The proof relies on Bourgain’s ideas and following lemma.
*Lemma 13. (Bourgain, [B2]) If w ∈ Ys,*1
2*, s ≥ 1 and denote*
(73) *P (τ, ξ) = [ψδ(wk− βw)∂xw]∼(τ, ξ),*
(74)
³ P
*ξ*
R
R*(1 + |ξ|)2s−2|P (τ, ξ)|*2*dτ*
´1
2
*≤ Cδ² _{kwk}*

*Y1, 1*2

*kwk*

*k*

*Ys, 1*2 ³ P

*ξ*R

*τ −A(ξ)≤|ξ|2*

_{200}

*(1 + |ξ|)*

*2s*2

_{|P |}*´ 1 2*

_{dτ}*≤ Cδ²*

_{kwk}*Y1, 1*2

*kwk*

*k*

*Ys, 1*2

*,*

*for some ² > 0.*

*Proof of Proposition 12. Define the map T on Ys,*1

2*[−δ, δ] such that*
]
*T (v)(t, ξ) = ψ(t)e−itA(ξ)φ(ξ)+*b
*ψδ(t)*
Z * _{t}*
0

*e−i(t−s)A(ξ)F*¡

*(vk− βv)∂xv*¢

*(s, ξ)ds.*(75)

*We want to show that the map T is a contraction.*

*As in Theorem 1, we first split T (v) into linear and nonlinear parts and*
*denoted by U and V respectively.*

(76)
(
e
*U (t, ξ) = ψ(t)e−itA(ξ) _{φ(ξ)}*b
e

*V (t, ξ) = ψδ(t)*R

*0*

_{t}*e−i(t−s)A(ξ)F*¡

*(vk*

_{− β}*v)∂xv*¢

*(s, ξ)ds.*

*To estimate U , we follow arguments in [KPV1] and [S] obtain, for j =*
*1, 2, 3,*

*To estimate V , we follow Bourgain’s argument, and use Lemma (13) and*
*k∂ _{X}s*

*wk*2

_{L}∞*t*

*L*2

*x*

*≤*X

*ξ*³ Z R

*(1 + |ξ|)s| ew(τ, ξ)dτ*´2

*.*

*We have, for j = 1, 2, 3,*(78)

*ν*2

_{j}s(V )*≤ Cδ2γkvk*2

*Y1, 1*2

*kvk*

*2k*

*Ys, 1*2

*.*Hence we obtain (80)

*kT (v)k*

_{Y}s, 1_{2}

_{[−δ,δ]}*≤ CkφkHs*

*+ δγkvk*

_{Y}1, 1_{2}

_{[−δ,δ]}kvkk*Ys, 1*2

*[−δ,δ].*

*Thus if δ = δ(kφkH*1*) is small, then, for R = C(kφk _{H}s), T is a contraction*

*from a ball BR* into itself.

Next we observe that

*(81) F(T u − T v)(t, ξ) = ψδ(t)*

Z _{t}

0

*e−i(t−s)A(ξ)·*

*F([(u − v)Pk−1(u, v) − (βu* *− βv)]∂xv)F((uk− βu)∂x(u − v)) ds*

which suggest that we can consider the integral equation

(82) *w(t, ξ) = ψ*e *δ(t)*

Z _{t}

0

*e−i(t−s)A(ξ)·*

*F([wPk−1(u, v) − θ(s)]∂xv)F(uk− βu)∂xw) ds,*

*where Pk−1(u, v) is a polynomial of degree k−1 and θ(t) =*

R

T*wPk−1(u, v)dx.*

*Let Φ be the operator defined on Ys,*1

2*[−δ*_{1}*, δ*_{1}*], δ*_{1} *< δ, by the above integral*

*equation. We can show that there exists δ*1 *= δ*1*(kuk _{Y}s, 1*

_{2}

*, kvk*

_{Y}s, 1_{2}) such that

*Φ is a contraction from a ball Bρ* *into itself, for arbitrary ρ. By uniqueness,*

*we have u = v almost everywhere on [−δ*1*, δ*1]. Repeating the argument finite

times, we conclude the proof. ¤

*Theorem 14. For k > 2, the IVP of (1) is globally well-posed for data in*

*Hα−1*2 *(Hs, s ≥* *α − 1*

2 *), with sufficiently small H*

*α−1*

2 *-norm.*

4. Proof of A priori Estimate

In this part, we want to prove Theorem 2.

*Theorem 2. If α ≥ 2, then we have the following estimates*

(83)
*kf kL*4* _{(R×T)}≤ C*
°

*°S1+α*

*4α*

*f*b ° °

*L*2

*° °*

_{(R×Z)},*b*

_{f}*S1+α4α*° °

*L*2

*4*

_{(R×Z)}≤ Ckf k_{L}_{3}

_{(R×T)}.*Proof. First we split the function f into positive and negative parts with*
respect to the dual of space variable and denote

(84) *f = f*+*+ f−* =X

*ξ≥0*

*eixξf (t, ξ) +*e X
*ξ<0*

*eixξf (t, ξ).*e

*It suffices to prove that f*+ _{and f}−_{both satisfy the estimate. We will only}

*prove the case of f*+ _{since the proof for f}−* _{is similar. Hence we replace f}*+

*by f and decompose the function in the following way. Choose a smooth*
function b*a with support in [2−1 _{, 2]. Let b}_{a}*

*j(τ ) = ba(2−jτ ) and ba*0 *= 1 −*
X
b
*aj.*
Consider
(86) *f (t, x) =* X
*j*
*fj(t, x),* where *f*b*j(τ, ξ) = baj(τ − |ξ|α*) b*f (τ, ξ).*
Thus we have
(88) *kf k*2* _{L}*4

*≤*X

*j,k*

*kfjfkkL*2

*.*

*Observe that (fjfk)(t, x) can be written as*

(90) Z Z X

*ξ*1*ξ*2

*ei(t(τ*1*+τ*2*)+x(ξ*1*+ξ*2))* _{f}*b

We choose a change of variables
(91)
½
*τ = τ*1*+ τ*2*,* *ξ = ξ*1*+ ξ*2*,*
*p = p*1*+ p*2*,* *q = p*2*,*
where
(92)
½
*p*1 *= τ*1*− |ξ*1*|α* *∈ ∆j* = [2*j−1, 2j+1],*
*p*2 *= τ*2*− |ξ*2*|α* *∈ ∆k* = [2*k−1, 2k+1].*

*(Without loss of generality, we assume that p*1 *and p*2 are both positive. The

*case of negative p*1 *and p*2 *can be treated in the same manner.) Thus, fjfk*

can be rewritten as follows.
(93) *(fjfk)(t, x) =*
Z X
*ξ*
*ei(tτ +xξ)G*b*jk(τ, ξ)dτ,*
where
(94)
(
b
*Gjk(τ, ξ) =*
R
∆*k*
P
*p∈Λj*( b*fj*
b
*fk)(τ, ξ, q, p)dq and*
Λ*j(τ, ξ, q) =*
©
*p ∈ ∆j+ q :* *ξ*1*, ξ*2 *∈ Z*+
ª
*.*

Applying Plancherel’s Theorem, we have
(95) *kfjfkkL*2 *= k bG _{jk}k_{L}*2

*.*

*Without loss of generality we may assume that j > k. Observe that*

(96) *k bGjkk*2*L*2 =
Z X
*ξ*
¯
¯
¯
¯
¯
¯
Z
∆*k*
X
*p∈Λj*
( b*fjf*b*k)(τ, ξ, q, p)dq*
¯
¯
¯
¯
¯
¯
2
*dτ.*
Claim:
(97) sup
*τ,ξ,q*
¯
¯Λ*j*
¯
*¯ ≤ C2j*
*α.*

Assuming the claim, we get

(98) *k bGjkk*2* _{L}*2

*≤*1 2

*α−12α*

*(j−k)*°

*°S1+α*

*4α*

*f*b

*° °2*

_{j}*L*2 °

*°S1+α*

*4α*

*f*b

*° °2*

_{k}*L*2

*.*

*The case of j < k can be treated in a similar fashion. Thus we have*
(100) *kf*+*k*2* _{L}*4

*≤*X

*jk*1 2

*α−12α*

*|j−k|*°

*°S1+α*

*4α*

*f*b ° °2

*L*2

*.*Therefore we obtain (101)

*kf k*2

*4*

_{L}*≤*°

*°S1+α*

*4α*

*f*b ° °2

*L*2 X

*jk*1 2

*α−12α*

*|j−k|*

*which implies that f satisfies the estimate.* ¤

Proof of the Claim:. Since
(102) Λ*j(τ, ξ, q) =*

©

*p ∈ ∆j* *+ q :* *ξ*1*, ξ*2 *∈ Z*+

ª

*,*

we can deduce that

(103) *τ − q − 2j+1* *≤ ξ*_{1}*α+ ξ*_{2}*α* *≤ τ − q − 2j−1.*

Denote

(104) ½

*A = τ − q − 2j+1 _{,}*

_{M = 3 · 2}j−1_{,}*d(a, b) = |a − b| : the distance between point a and point b.*

Thus we can rewrite the above inequality as
(105) *A ≤ ξ*_{1}*α+ ξ*_{2}*α* *≤ A + M,*

*and distinguish the cases, A À M , A ∼ M and A ¿ M .*

*Let C*1 *and C*2 *be the graphs of level curves of |ξ*1*|α+|ξ*2*|α* *at A and A+M*

respectively.

(106)

½

*C*1 *= {(ξ*1*, ξ*2*) : |ξ*1*|α+ |ξ*2*|α* *= A};*
*C*2 *= {(ξ*1*, ξ*2*) : |ξ*1*|α+ |ξ*2*|α* *= A + M }.*

Notice that we can only consider the first quadrant. It can be shown easily that, along each level curve, the farthest point to the origin is on the line

**a****b*** l*
1
2
3
2
1

**C**

**C**

**C****FIGURE 1.**1

**l***2*

**C**

**b**

**a***2 1*

**C****FIGURE 2.**

### ξ

### ξ

### ξ

### ξ

can be interpreted as the number of lattice points which lie on the straight
*line ξ*1*+ ξ*2 *= ξ and fall into the region between curves C*1 *and C*2.

*For the case A À M , let C*3 *be a circumscribed circle to the curve C*2,

(107) *C*3 *= {(ξ*1*, ξ*2*) : ξ*12*+ ξ*22 = 2

*α*

r

*(A + M )*2

4 *},*

*then the largest possible line segment in the region is on the line l,*

(108) *l = {(ξ*1*, ξ*2*) : ξ*1*+ ξ*2 = 2*α*

r

*A*

2*},*

*which is tangent to the curve C*1*, see Fig. 1. Let a and b be the intersections*

*of the line l and the circle C*3, then we get

(110) *d*2 *∼* *α*
r
*A*2
4 *−*
Ã
2*α*
r
*A*2
4 *−*
*α*
r
*(A + M )*2
4
!
*≤ C√αM*2_{.}

*For the case A ∼ M , the previous argument goes through.*

*For the case A ¿ M , since C*1 *is small, we can take the line segment l*

*between two intercepts of C*2,

(111) *l = {(ξ*1*, ξ*2*) : ξ*1*+ ξ*2 = *α*
*√*

*A + M },*

see Fig. 2, and estimate

(112) *d ∼* *√α*

¤
*Remark. It is known that the L*6_{-norm estimate is not true. In fact, }

Bour-gain proved the estimate
(113)
°
°
° X
*|n|<N*
*anei(nx+n*
3* _{t)}*°°
°
6

*¿ N*

*²*¡ X

_{|a}*n|*2 ¢1 2

*in his paper [B2]. The optimal estimate should be a Lp* _{estimate for 4 ≤ p <}

6, see [B] and [FG].

5. The Polynomial Bound

*In the final part, we discuss a polynomial bound for Hs*_{-norm of the global}

solution. First we recall two technical lemmas.

*Lemma 15. (Kenig-Ponce-Vega, [KPV3]) Assume that 0 ≤ ρ ≤* 1_{2} *and*
*² > 0 is small. Assume also that ν(−ρ+²,*12)

2 *(u) and ν*

*(−ρ+²,*1
2)

2 *(v) are bounded*
*and* R_{T}*u(t, x)dx =* R_{T}*v(t, x)dx = 0. Then*

(114) *ν(−ρ−²,−1*2 *+²)*
2 *(∂x(uv)) ≤ Cν(−ρ+²,*
1
2)
2 *(u)ν*
*(−ρ+²,*1
2)
2 *(v).*

*Lemma 16. (Staffilani, [S]) Assume that ρ ≥ 0, ² > 0 is small and k ≥ 3.*

*Then*

(115) *ν(ρ+²,*12*−²)*

2 (X*[0,1]uk) ≤ Ckuk _{Y}1+ρ, 1*

_{2}

*kuk*

*k−1*
*Y1, 1*2*.*

Instead of proving Theorem B, we state and prove a more general result.
*Theorem 17. Consider IVP (1) and assume that there exists an a-priori*

*bound for the Hα−1*2 *-norm of u. Then if φ ∈ Hs* *and s ≥* *α−1*

2 *, the global*
*solution satisfies the bounds*

(116)
(
*ku(t)kHs* *≤ C|t|*
*s*
*ρ* *provided ρ + 1 ≤* *α−1*
2 *< S;*
*ku(t)kHs* *≤ C|t|*
*4s*
*2ρ+(α−3)* *provided* *α−1*
2 *< ρ + 1 < S.*

*Proof. It is sufficient to show that, for all t ∈ [0,δ*_{2}],
(117) *k∂ _{x}su(t)kL*2

*x*

*≤ k∂*

*s*

*xφkL*2

*x*

*+ Ck∂*

*s*

*xφk1−δL*2

*x*

*,*

*where δ−1* _{is the exponents in Theorem 9. Since}

*k∂s*
*xu(t)k*2*L*2
*x* *= k∂*
*s*
*xφk*2*L*2
*x* +
Z * _{t}*
0

*d*

*dσk∂*

*s*

*xu(σ)k*2

*L*2

*xdσ*

*= k∂*2

_{x}sφk*2*

_{L}*x*

*−*Z R Z T

X*[0,t]uk∂x(∂xsu)*2*dxdσ + lower order terms.*

Call
(118) *J =*
Z
R
Z
T
X*[0,t]uk∂x(∂xsu)*2*dxdσ*
and set
(120) *w(t, ξ) = e*e *iξ*R0*tβu(σ)dσu(t, ξ).*e

Taking Fourier transform with respect to space variable, multiplying by

*eiξ*R0*tβudσ*, then taking Fourier transform with respect to time variable, we

have
*J =*X
*ξ*
Z
R
\
X*[0,t]wk(τ, ξ) \∂x(∂xsw)*2*(τ, ξ)dτ*
*≤*X
*ξ*
Z
R
*| \*X*[0,t]wk|(τ, ξ)(1 + |ξ|)ρ+²(1 + |τ − A(ξ))*
1
2*−²·*
*| \∂x(∂xsw)*2*|(τ, ξ)(1 + |ξ|)−ρ−²(1 + |τ − A(ξ))−*
1
2*+²dτ*
*≤ν(ρ+²,*12*−²)*
2 (X*[0,1]uk)ν*
*(−ρ−²,−1*_{2} *+²)*
2 *(∂x(∂xsu)*2*).*

*Employing Lemmas, Theorem 7, and interpolation between the Hα−1*2 and

*the Hs* _{norms, we obtain}
*J ≤ Ckwk*
*Y1+ρ, 1*2*kwk*
*k−1*
*Y1, 1*2*[ν*
*(−ρ+²,*1
2)
2 *(∂xsw)]*2
*≤ Cd1+ρ _{∗}*

*(u, 0)d*1

_{∗}(u, 0)k−1ds−ρ+²_{∗}*(u, 0)*2

*For ρ + 1 ≤* *α−1*_{2} *< S, interpolation gives*
(121) *J ≤ C(kφk*
*Hα−1*2 *)kφk*
*2(1− _{2s−(α−1)}2(ρ−²)* )

*Hs*

*.*

*If we choose ² = ρα−1*

*2s*, we have (122)

*ku(t)kHs*

*≤ C|t|*

*s*

*ρ.*

For *α−1*_{2} *< ρ + 1 < S, interpolation gives*

(123) *J ≤ C(kφk*
*Hα−1*2 *)kφk*
*2(1−ρ+ α−3*2 *−2²*
*2s−(α−1)* )
*Hs* *.*
*Choose ² =* *(α−1)(ρ+α−3*2 )
*4s* , we have
(124) *ku(t)kHs* *≤ C|t|*
*4s*
*2ρ+(α−3).*
¤

Acknowledgment: I want to express my gratitude toward Professor M. Grillakis for his inspiring and helpful conversations.

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Department of Mathematics, National Cheng Kung Univ, Tainan, Taiwan

*E-mail address: fang@math.ncku.edu.tw*

. *U RL : http://math.ncku.edu.tw/∼fang*

. *P hone : 886-06-275-7575 ext 65131*