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Note that for the classical solution of (1.1) we have the conservation lawsR Ru(t, x)dx= R Ru0(x)dx andku(t)kL2 =ku0kL2

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Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 77, pp. 1–34.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

KDV TYPE ASYMPTOTICS FOR SOLUTIONS TO HIGHER-ORDER NONLINEAR SCHR ¨ODINGER EQUATIONS

PAVEL I. NAUMKIN, ISAHI S ´ANCHEZ-SU ´AREZ

Abstract. We consider the Cauchy problem for the higher-order nonlinear Schr¨odinger equation

i∂tua

3|∂x|3ub

4x4u=λi∂x(|u|2u), (t, x)R+×R, u(0, x) =u0(x), xR,

wherea, b >0,|∂x|α =F−1|ξ|αF andF is the Fourier transformation. Our purpose is to study the large time behavior of the solutions under the non-zero mass conditionR

u0(x)dx6= 0.

1. Introduction

We consider the Cauchy problem for the higher-order nonlinear Schr¨odinger equation

i∂tu−a

3|∂x|3u− b

4∂x4u=i∂x(|u|2u), (t, x)∈R+×R, u(0, x) =u0(x), x∈R,

(1.1) where a, b >0,|∂x|α=F−1|ξ|αF andF is the Fourier transformation defined by Fφ = 1

R

Re−ixξφdx, and its inverse by F−1φ = 1

R

Reixξφ(ξ)dξ. Note that we have the relationu(−t, x) =u(t,−x), so we only consider the caset >0. Note that for the classical solution of (1.1) we have the conservation lawsR

Ru(t, x)dx= R

Ru0(x)dx andku(t)kL2 =ku0kL2.

Equation (1.1) arises in the context of high-speed soliton transmission in long- haul optical communication system [13]. Also it can be considered as a particular form of the higher order nonlinear Schr¨odinger equation introduced by [42] to de- scribe the nonlinear propagation of pulses through optical fibers. This equation also represents the propagation of pulses by taking higher dispersion effects into account than those given by the Schr¨odinger equation (see [10, 11, 25, 28, 38, 32, 43, 49]).

Higher order nonlinear Schr¨odinger equations have been widely studied recently.

For the local and global well-posedness of the Cauchy problem we refer to [5, 6, 40]

and references cited therein. The dispersive blow-up was obtained in [2]. The existence and uniqueness of solutions to (1.1) were proved in [1, 19, 30, 33, 34, 39, 47, 50], and the smoothing properties of solutions were studied in [3, 8, 12, 30, 33,

2010Mathematics Subject Classification. 35B40, 35Q35.

Key words and phrases. Nonlinear Schr¨odinger equation; large time asymptotic behavior;

critical nonlinearity; self-similar solutions.

c

2020 Texas State University.

Submitted June 2, 2018. Published July 22, 2020.

1

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34, 36, 37, 48]. The blow-up effect for a special class of slowly decaying solutions of the Cauchy problem (1.1) was studied in [1].

In this article we are interested in the case of non zero total mass of the initial data

Z

R

u0(x)dx6= 0.

Then by (1.1) we obtain the non-zero total mass for the solution R

Ru(t, x)dx = R

Ru0(x)dx 6= 0 for all t > 0. We develop the factorization technique originated in our previous papers [22, 21, 26, 20, 44, 45, 46]. The case of zero total mass R

Ru(t, x)dx = R

Ru0(x)dx = 0 is easier and the corresponding results can be ob- tained following the approach in [24].

We denote the Lebesgue space byLp={φ∈S0:kφkLp<∞}, with norms kφkLp=Z

|φ(x)|pdx1/p

for 1≤p <∞andkφkL = supx∈R|φ(x)|. The weighted Sobolev space isHpm,s= {ϕ∈ S0 :kφkHm,sp =khxishi∂ximφkLp <∞}, where m, s∈R, 1 ≤p≤ ∞, hxi=

1 +x2, hi∂xi=p

1−∂x2. We also use the notations Hm,s=H2m,s, Hm=Hm,0 shortly, if it does not cause any confusion. LetC(I;B) be the space of continuous functions from an interval I to a Banach space B. Different positive constants might be denoted by the same letterC.

In [29], it was proved the existence of the self-similar solutions of the formu= t−1/3fm(xt−1/3) to the reduced equation

i∂tu−1

3|∂x|3u=i∂x(|u|2u), (1.2) which is defined by the mean valuem=R

Rfm(x)dx6= 0.

Now we state the main result of this article. We show that the asymptotic behavior of the solutions to (1.1) resembles that of the KdV equation, which was studied intensively (see [15, 18, 23]). We denote by µ(x) the root of the equation Λ0(ξ) =ξ|ξ|(a+b|ξ|) =xfor allx∈R.

Theorem 1.1. Suppose that R

Ru0(x)dx 6= 0, and that the initial data u0 ∈H1,1 have a sufficiently small norm ku0kH1,1 ≤ ε. Then there exists a unique global solutionu∈C([0,∞);H1,1)to the Cauchy problem (1.1). Furthermore there exists W+∈L such that the large time behavior satisfies

u(t, x) = M

pitΛ00(µ(xt))W+ µ(x t)

exp iµ(xt)

00(µ(xt))|

W+ µ(x t)

2logt +O(t−1/3ht1/3µ(x

t)i−3/4) +O(t13−δ)

on the domain |x| ≥ t13+2γ and u(t, x) =t−1/3fm(xt−1/3) +O(t13−δ) on the do- main|x| ≤t13+2γ, whereγ, δ >0 are small andt−1/3fm(xt−1/3)is the self-similar solution to equation (1.2)with m=R

Rfm(x)dx=R

Ru0(x)dx6= 0.

We organize the rest of the paper as follows. Section 2 is devoted to the fac- torization formulas and L2-boundedness of pseudodifferential operators. Then in Sections 3 and 4 we obtain estimates for the operator V and V, respectively, in

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the uniform norm and theL2-estimates of commutators. In Section 5, we prove a priori estimates of the solutionu(t) in the norm

kukXT = sup

t∈[1,T]

kϕkb L+t−γk∂xJu(t)kL2+t−γk∂−1x Ibu(t)kL2

+t−γk∂x−1Pbu(t)kL2+t−γku(t)kH1

,

where ϕb = F U(−t)u(t), U(t) = F−1e−itΛ(ξ)F, Λ(ξ) = a3|ξ|3+ b4ξ4, J = x− tΛ0(−i∂x),Ib=∂b+it∂bΛ(−i∂x),Pb = 3t∂t+∂xx+b∂b. Section 6 is devoted to the proof of Theorem 1.1.

2. Preliminaries

2.1. Factorization techniques. We define the free evolution group U(t) =F−1e−itΛ(ξ)F, Λ(ξ) =a3|ξ|3+b4ξ4. We write

U(t)F−1φ=Dt

r|t|

2π Z

R

eit(xξ−Λ(ξ))φ(ξ)dξ,

where Dtφ = |t|−1/2φ(xt) is the dilation operator. There is a unique station- ary point ξ = µ(x) in the integral R

Reit(xξ−Λ(ξ))φ(ξ)dξ, which is defined as the root of the equation Λ0(ξ) = ξ|ξ|(a+b|ξ|) = x for all x ∈ R. Thus we ob- tain Λ0(µ(x)) = x for all x ∈ R. Also Λ00(ξ) = 2|ξ|(a+ 32b|ξ|) > 0 for all ξ ∈ R. We have Λ0(ξ) = O(ξ2hξi), therefore µ(x) = O({x}1/2hxi1/3). Define the scaling operator (Bφ)(x) = φ(µ(x)). Hence U(t)F−1φ = DtBMVφ, where M =e−it(Λ(η)−ηΛ0(η)), the operatorV(t)φ=

q|t|

R

Re−itS(ξ,η)φ(ξ)dξ and the phase function S(ξ, η) = Λ(ξ)−Λ(η)−Λ0(η)(ξ−η). Denote Ak = Mk001(η)ηMk, k = 0,1. We have A1 =A0+iη, also A1V = Viξ, [iη,V] = −A0V, therefore we obtain the commutator∂ηV =−tΛ00(η)[iη,V]. Since∂ξS(ξ, η) = Λ0(ξ)−Λ0(η), we obtain the commutatorit[Λ0(η),V]φ=−V∂ξφ. Also we use the representation for the inverse evolution group F U(−t)φ=VMB−1Dt−1, where the inverse dilation operatorDt−1φ=|t|1/2φ(xt) and the inverse scaling operator (B−1φ)(η) =φ(Λ0(η)), and the conjugate operator

V(t)φ= r|t|

2π Z

R

eitS(ξ,η)φ(η)Λ00(η)dη.

We haveiξVφ=VA1φ. Hence [iξ,V] =VA0.

Define a new dependent variableϕb=F U(−t)u(t). SinceF U(−t)L=∂tF U(−t) withL=∂t+iΛ, whereΛ= Λ(−i∂x) =F−1Λ(ξ)F, applying the operatorF U(−t) to equation (1.1)Lu=∂x(|u|2u), we obtain

tϕb=F U(−t)Lu=F U(−t)∂x(|U(t)F−1ϕ|b2U(t)F−1ϕ)b

=iξt−1V(|Vϕ|b2Vϕ),b (2.1)

since the nonlinearity is gauge invariant. Also we mention that the operator J = U(t)xU(−t) =x−tΛ0, withΛ0= Λ0(−i∂x) =F−1Λ0(ξ)F, plays a crucial role in the large time asymptotic estimates. Note that J commutes with L, i.e. [J,L] = 0.

Also we see that the symbol Λ(ξ) = a3|ξ|3+ 4bξ4 satisfies the identities ξ∂ξΛ− b∂bΛ = 3Λ and ξ∂ξΛ +a∂aΛ = 4Λ. Hence we have the commutator relations [Pba, e−itΛ(ξ)] = [Pbb, e−itΛ(ξ)] = 0, withPba= 4t∂t−ξ∂ξ−a∂a,Pbb= 3t∂t−ξ∂ξ+b∂b.

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We also use the operators Pa = 4t∂t+∂xx−a∂a, Pb = 3t∂t+∂xx+b∂b and Ia = ∂a +it∂aΛ(−i∂x), Ib = ∂b +it∂bΛ(−i∂x), and the commutator relations [L,Ia] = [L,Ib] = 0 and [L,Pa] = 4L, [L,Pb] = 3L hold. Using the relation u(t) =U(t)F−1ϕb=F−1e−itΛ(ξ)ϕ, we obtainb

Pau=F−1Pbae−itΛ(ξ)ϕb=F−1[Pba, e−itΛ(ξ)]ϕb+F−1e−itΛ(ξ)Pcaϕb=U(t)F−1Pbaϕ,b andPbu=F−1Pbbe−itΛ(ξ)ϕb=F−1[Pbb, e−itΛ(ξ)]ϕb+F−1e−itΛ(ξ)Pcbϕb=U(t)F−1Pbbϕ.b Note thatIau=IaU(t)F−1ϕb=F−1(∂a+it∂aΛ(ξ))Eϕb=F−1E∂aϕb=U(t)F−1aϕ,b and Ibu = IbU(t)F−1ϕb = F−1(∂b +it∂bΛ(ξ))Eϕb = F−1E∂bϕb = U(t)F−1bϕ.b Hence kIaukL2 =k∂bϕkb L2 and kIbukL2 = k∂bϕkb L2. Also we have the identities Pa = 4tL+∂xJ −aIa andPb= 3tL+∂xJ +bIb.

2.2. Boundedness of pseudodifferential operators. Define the pseudodiffer- ential operator

a(x, D)φ≡ Z

R

eixξa(x, ξ)bφ(ξ)dξ.

There are many papers devoted to theL2-estimates of pseudodifferential operator a(x, D) (see [4, 7, 9, 27]). Below we will use the following results on the L2- boundedness of pseudodifferential operatora(x, D) (see [27]).

Lemma 2.1. Let the symbol a(x, ξ) be such that supx,ξ∈R|∂xkξla(x, ξ)| ≤ C for k, l= 0,1. Then ka(x, D)φkL2

x ≤CkφkL2.

Lemma 2.2. Let the symbol a(x, ξ) be such that supx∈Rk∂lξa(x, ξ)kL2

ξ ≤ C for l= 0,1. Then ka(x, D)φkL2x ≤CkφkL2.

Next we consider the time dependent pseudodifferential operator a(t, x, D)φ≡

Z

R

eixξa(t, x, ξ)bφ(ξ)dξ.

Lemma 2.3. Let the symbola(t, x, ξ) be such that sup

x,ξ∈R,t≥1

|{ξ}−νhξiν(ξ∂ξ)ka(t, x, ξ)| ≤C fork= 0,1,2, whereν∈(0,1). Thenka(t, x, D)φkL2

x≤CkφkL2 for allt≥1.

Proof. We define the kernelK(t, x, y) =R

Reiyξa(t, x, ξ)dξ, then we write a(t, x, D)φ=

Z

R

eixξa(t, x, ξ)bφ(ξ)dξ= Z

R

K(t, x, y)φ(x−y)dy.

Integrating two times by parts via the identityeiyξ =H∂ξ(ξeiyξ) withH = (1 + iξy)−1 we obtain K(t, x, y) = R

Reiyξξ∂ξ(Hξ∂ξ(Ha(x, ξ)))dξ. By the condition in this lemma, we find|ξ∂ξ(Hξ∂ξ(Ha(t, x, ξ)))| ≤ C{ξ}hξyiνhξi2−ν for allx, y, ξ∈R. Hence we obtain the estimate

|K(t, x, y)| ≤C Z 1

0

ξνdξ hξyi2 +C

Z

1

ξ−νdξ hξyi2

≤C|y|−1−ν Z |y|

0

ηνhηi−2dη+C|y|ν−1 Z

|y|

η−νhηi−2

≤C|y|ν−1hyi−2ν.

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Then by Young’s inequality we have ka(t, x, D)φkL2=k

Z

R

K(t, x, y)φ(x−y)dykL2

≤Ck Z

R

|y|ν−1hyi−2ν|φ(x−y)|dykL2≤C

|x|ν−1hxi−2ν

L1kφkL2 ≤CkφkL2.

The proof is complete.

Similarly, by considering the conjugate operator.

Lemma 2.4. Let the symbola(t, x, ξ) be such that sup

x,ξ∈R,t≥1

|{x}−νhxiν(x∂x)ka(t, x, ξ)| ≤C

fork= 0,1,2, whereν∈(0,1). Thenka(t, x, D)φkL2x≤CkφkL2 for allt≥1.

3. Estimates for the operator V

Let χ ∈ C4(R) be such that χ(x) = 1 for |x| ≤ 1 and χ(x) = 0 for |x| ≥ 2.

Define the cut off functionsχj(z)∈C4(R),j= 1,2,3, such thatχ123≡1, χ2(z) = 0 for z ≤ 13 or z ≥3 and χ2(z) = 1 for 23 ≤ z ≤ 32, χ1(z) = 1−χ2(z) for −32 < z < 23 and χ1(z) = 0 forz ≥ 23 or z ≤ −3 and χ3(z) = 1−χ2(z) for z > 32, χ3(z) = 1−χ1(z) for z < −32 and χ3(z) = 0 for −32 ≤ z ≤ 32. Denote Ψ1(t, ξ, η) = (1−χ(ηt1/3))χ1(ηξ), Ψ2(t, ξ, η) = (1−χ(ηt1/3))χ2(ξη), Ψ3(t, ξ, η) = (1−χ(ηt1/3))χ3(ηξ), Ψ4(t, ξ, η) =χ(ηt1/3)χ(13ξt1/3) and Ψ5(t, ξ, η) =χ(ηt1/3)(1− χ(13ξt1/3)) and consider the operators

Vj(t)φ= r|t|

2π Z

R

e−itS(ξ,η)Ψj(t, ξ, η)φ(ξ)dξ forj= 1,2,3,4,5.

3.1. Estimates for commutators. We first obtain theL2-estimate for the oper- atorV1.

Lemma 3.1. Let the weightP ∈C2(R\0) be such that ∂ηkP(η) =O(|η|α1−k)for k= 0,1,2, withα1≤2. Assume that the weight Q≡1 ifα2= 0, orQ∈C2(R\0) satisfies the estimate ∂ξkQ(ξ) = O(|ξ|α2−k), k = 0,1,2, if α2 ≥ 1. Suppose that 2≤α12<5/2. Then

00|1/2P tV1

L2 ≤Ck∂ξφkL2+C|t|13(52−α1−α2)|φ(0)|

for allt≥1.

Proof. Integrating by parts we obtain P tV1Qφ=C 1−χ(ηt1/3)

P(η)|t|1/2 Z

R

e−itS(ξ,η)ξ

Q(ξ)χ1(ξη)

ξS(ξ, η) φ(ξ) dξ

=C|t|1/2 Z

R

e−itS(ξ,η)q1(t, η, ξ)φξ(ξ)dξ +C|t|1/2

Z

R

e−itS(ξ,η)q2(t, η, ξ)φ(ξ)−φ(0)

ξ dξ

+C|t|1/2φ(0) Z

R

e−itS(ξ,η)q3(t, η, ξ)dξ,

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whereqk(t, η, ξ) = (1−χ(t1/3η))P(η)(ξ∂ξ)k−1(Q(ξ)χ1(ηξ) 1

ξS(ξ,η)),k= 1,2, q3(t, η, ξ) = (1−χ(t1/3η))P(η)∂ξ(Q(ξ)χ1

η) 1

ξS(ξ, η)).

Next we change η=µ(x) (recalling that the inverse scaling operator (B−1φ)(η) = φ(Λ0(η)) andµ(x) is the root of the equation Λ0(η) =x), we obtain

P tV1Qφ=MB−1|t|1/2 Z

R

eitxξq1(t, µ(x), ξ)e−itΛ(ξ)φξ(ξ)dξ +MB−1|t|1/2

Z

R

eitxξq2(t, µ(x), ξ)e−itΛ(ξ)φ(ξ)−φ(0)

ξ dξ

+φ(0)MB−1|t|1/2 Z

R

eitxξq3(t, µ(x), ξ)e−itΛ(ξ)dξ Also we change the variable of integrationξ=t−1/3ξ0; then

P tV1Qφ=MB−1D−1t2/3Z

R

eixξq1(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)φξ(ξ)dξ +

Z

R

eixξq2(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ dξ

+φ(0) Z

R

eixξq3(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)dξ ,

here D−1t2/3φ(x) = |t|1/3φ(xt2/3) and Dt1/3φ(ξ) = |t|−1/6φ(ξt−1/3). Define the pseudodifferential operators ak(t, x, D)φ ≡ R

Reixξak(t, x, ξ)φ(ξ)dξb with symbols ak(t, x, ξ) =qk(t, µ(xt−2/3), ξt−1/3). Then we obtain

P tV1Qφ=MB−1D−1t2/3

a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ) +a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ +φ(0)a3(t, x, D)F−1Dt1/3e−itΛ(ξ)

.

Let us prove theL2-boundedness of the operatorsak(t, x, D),k= 1,2. We obtain ak(t, x, ξ) = 1−χ t1/3µ(xt−2/3)

P(µ(xt−2/3))

×(ξ∂ξ)k−1Q(ξt−1/31(ξt−1/3/µ(xt−2/3)) Λ0(ξt13)−xt−2/3

.

Note that µ(x) =O(|x|1/2) for small |x|, therefore (1−χ(t1/3µ(xt−2/3)))6= 0 for

|x| ≥C >0,t≥1. Also we have χ1(z)6= 0 for−3< z <2/3, hence

0(ξt−1/3)−xt−2/3|=

Z ξt−1/3

µ(xt23)

Λ00(η)dη ≥

Z 23µ(xt−2/3)

µ(xt−2/3)

Λ00(η)dη

≥ 1

3|µ(xt−2/3)|Λ00(2

3µ(xt−2/3))

≥C|Λ0(µ(xt−2/3))|=C|x|t−2/3.

Note that |∂xkξlak(t, x, ξ)| ≤ C for all x, ξ ∈ R, t ≥ 1,k, l = 0,1, if 2 ≤ α1+ α2 ≤3. Therefore by Lemma 2.1 we havekak(t, x, D)φkL2x ≤CkφkL2. Thus the

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pseudodifferential operators ak(t, x, D) areL2-bounded fork= 1,2. Then we find that

k|Λ00|12MB−1D−1t2/3a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ)kL2 η

≤ ka1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ)kL2 x

≤CkF−1Dt1/3e−itΛ(ξ)φξ(ξ)kL2 =Ck∂ξφkL2

and by the Hardy inequality

00|12MB−1Dt−12/3a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0) ξ kL2η

≤ ka2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0) ξ

L2

x

≤CkF−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0) ξ kL2

ξ =Ckφ(ξ)−φ(0) ξ kL2

ξ ≤Ck∂ξφkL2. Now let us prove the L2 - L boundedness of the pseudodifferential operator a3(t, x, D). We have

a3(t, x, ξ) =q3 t, µ(xt−2/3), ξt−1/3

=

1−χ t1/3µ(xt−2/3)

P µ(xt−2/3)

×t1/3ξQ(ξt−1/31(ξt−1/3/µ(xt−2/3)) Λ0(ξt−1/3)−xt−2/3

.

We consider two symbolsa3,1(t, x, ξ) =a3(t, x, ξ)χ(ξ) anda3,2(t, x, ξ) =a3(t, x, ξ)(1−

χ(ξ)). Note that |hxiν(x∂x)jhξiδa3,1(t, x, ξ)| ≤C|t|13(3−α1−α2)for allx, ξ∈R,t≥ 1,j= 0,1,2 with someν ∈(0,1), ifα2= 0 orα2≥1,α1≤2, 2≤α12<5/2.

Therefore by Lemma 2.4 we find that ka3,1(t, x, D)φkL2

x≤C|t|13(3−α1−α2)khξi−δφkb L2 ≤C|t|13(3−α1−α2)kφkb L. Also we have |∂xkξlhξiδa3,2(t, x, ξ)| ≤C|t|13(3−α1−α2) for allx, ξ ∈R, t ≥1, k, l= 0,1, if 2≤α12<5/2,δ >1/2. Therefore by Lemma 2.1 we find that

ka3,2(t, x, D)φkL2

x≤C|t|13(3−α1−α2)khξi−δφkb L2 ≤C|t|13(3−α1−α2)kφkb L. Then

00|1/2φ(0)MB−1D−1

t23

a3(t, x, D)F−1D

t13e−itΛ(ξ) L2

η

≤ |φ(0)|ka3(t, x, D)F−1Dt1/3e−itΛ(ξ)kL2x

≤C|t|13(3−α1−α2)|φ(0)|kDt1/3e−itΛ(ξ)kLξ

≤C|t|13(52−α1−α2)|φ(0)|,

which yields the result of the lemma.

In the next lemma we estimate the commutator [h,V2].

Lemma 3.2. Let the weights P ∈ C1(R\0) and Q ∈ C2(R\0) be such that

ηkP(η) =O(|η|α1−k),k= 0,1, and∂ξkQ(ξ) =O(|ξ|α2−k),k= 0,1,2. Suppose that

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h(ξ)∈C4(R\0)is such that|∂ξkh(ξ)| ≤C|ξ|α3−k forξ∈R\0,0≤k≤4. Assume that α1, α2∈R,α3≥1,2≤α123<5/2. Then

k|Λ00|1/2P t[h,V2]QφkL2 ≤Ck∂ξφkL2+C|t|13(52−α1−α2−α3)|φ(0)|

for allt≥1.

Proof. Integrating by parts we obtain P t[h,V2]Qφ

=C

1−χ ηt1/3

P(η)|t|1/2 Z

R

e−itS(ξ,η)ξ

h(η)−h(ξ)

ξS(ξ, η)

Q(ξ)φ(ξ)χ2(ξ η)dξ

=C|t|1/2 Z

R

e−itS(ξ,η)q1(η, ξ)φξ(ξ)dξ+C|t|1/2 Z

R

e−itS(ξ,η)q2(η, ξ)φ(ξ)−φ(0)

ξ dξ

+Cφ(0)|t|1/2 Z

R

e−itS(ξ,η)q3(η, ξ)dξ,

where qk(η, ξ) = (1−χ(ηt1/3))P(η)(ξ∂ξ)k−1(Q(ξ)χ2(ηξ)h(η)−h(ξ)

ξS(ξ,η) ), q3(η, ξ) = (1− χ(ηt1/3))P(η)∂ξ(Q(ξ)χ2(ξη)h(η)−h(ξ)

ξS(ξ,η) ). Next we change η =µ(x) and the variable of integrationξ=t−1/3ξ0, then we obtain

P t[h,V2]Qφ=MB−1D−1t2/3Z

R

eixξq1(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)φξ(ξ)dξ +

Z

R

eixξq2 t, µ(xt−2/3), ξt−1/3

Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ dξ

+φ(0) Z

R

eixξq3 t, µ(xt−2/3), ξt−1/3

Dt1/3e−itΛ(ξ)dξ . We define the pseudodifferential operators ak(t, x, D)φ ≡ R

Reixξak(t, x, ξ)φ(ξ)dξb with symbolsak(t, x, ξ) =qk(t, µ(xt−2/3), ξt−1/3). Then we obtain

P t[h,V2]Qφ=MB−1D−1t2/3

a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ) +a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ +φ(0)a3(t, x, D)F−1Dt1/3e−itΛ(ξ)

.

We will prove theL2-boundedness of ak(t, x, D). Note thatχ2 ξt−1/3 µ(xt−2/3)

6= 0 for 1/3<t1/3µ(xtξ−2/3)<3. Using the identities Λ0(ξ)−x= (ξ−µ(x))R1

0 Λ00(ξ+(µ(x)−

ξ)τ)dτ, andh(µ(x))−h(ξ) = (µ(x)−ξ)R1

0 h0(ξ+ (µ(x)−ξ)τ)dτ, we obtain ak(t, x, ξ) =−

1−χ t1/3µ(xt−2/3)

P µ(xt−2/3)

(ξ∂ξ)k−1

χ2 ξt−1/3/µ(xt−2/3)

×Q(ξt−1/3) R1

0 h0(ξt−1/3+ (µ(xt−2/3)−ξt−1/3)τ)dτ R1

0 Λ00(ξt−1/3+ (µ(xt−2/3)−ξt−1/3)τ)dτ

for k = 1,2. We have |∂xkξlak(t, x, ξ)| ≤ C for allx, ξ ∈ R, t ≥ 1, k, l = 0,1, if α1, α2∈R,α3≥1, 2≤α123<5/2. Therefore by Lemma 2.1 we find that

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kak(t, x, D)φkL2x ≤CkφkL2. Thus the pseudodifferential operators ak(t, x, D) are L2-bounded fork= 1,2. Then as above we have

00|12MB−1Dt−12/3a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ)

L2η≤Ck∂ξφkL2

and

00|12MB−1D−1t2/3a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0) ξ

L2

η

≤Ck∂ξφkL2. Now let us prove the L2 - L boundedness of the pseudodifferential operator a3(t, x, D). We have

a3(t, x, ξ) =−

1−χ t1/3µ(xt−2/3)

P µ(xt−2/3) t1/3ξ

Q(ξt−1/3)

×χ2

ξt−1/3 µ(xt−2/3)

R1

0 h0(ξt−1/3+ (µ(xt−2/3)−ξt−1/3)τ)dτ R1

0 Λ00(ξt−1/3+ (µ(xt−2/3)−ξt−1/3)τ)dτ

. Note that|∂xkξlhξiδa3(t, x, ξ)| ≤C|t|13(3−α1−α2)for allx, ξ∈R,t≥1,k, l= 0,1, if α1, α2∈R,α3≥1, 2≤α123<5/2,δ >1/2. Therefore by Lemma 2.1, we have

ka3(t, x, D)φkL2x≤C|t|13(3−α1−α2−α3)khξi−δφkb L2 ≤C|t|13(3−α1−α2−α3)kφkb L. Then

00|1/2φ(0)MB−1D−1

t23

a3(t, x, D)F−1D

t13e−itΛ(ξ) L2

η

≤ |φ(0)|ka3(t, x, D)F−1Dt1/3e−itΛ(ξ)kL2 x

≤C|t|13(3−α1−α2−α3)|φ(0)|kDt1/3e−itΛ(ξ)kLξ

≤C|t|13(52−α1−α2−α3)|φ(0)|,

which yields the result of the lemma.

In the next lemma we estimate the operatorV3.

Lemma 3.3. Let the weights P ∈ C1(R\0) and Q ∈ C2(R\0) be such that

ηkP(η) =O(|η|α1−k),k= 0,1, and∂ξkQ(ξ) =O(|ξ|α2−k),k= 0,1,2. Assume that α1 ≥ 0, α2 ∈ R, 2 ≤α12 ≤5/2. Then

00|1/2P tV3

L2 ≤Ck∂ξφkL2+ C|t|13(52−α1−α2)|φ(0)|for all t≥1.

Proof. Integrating by parts we obtain P tV3Qφ=CP(η)|t|1/2

Z

R

e−itS(ξ,η)ξ

1

ξS(ξ, η)Q(ξ)φ(ξ)Ψ3(t, ξ, η)dξ

=C|t|1/2 Z

R

e−itS(ξ,η)q1(t, η, ξ)φξ(ξ)dξ +C|t|1/2

Z

R

e−itS(ξ,η)q2(t, η, ξ)φ(ξ)−φ(0)

ξ dξ

+Cφ(0)|t|1/2 Z

R

e−itS(ξ,η)q3(t, η, ξ)dξ,

(10)

whereqk(t, η, ξ) =P(η) ξ∂ξk−1

(Q(ξ)Ψ3(t, ξ, η) 1

ξS(ξ,η)), q3(t, η, ξ) = P(η)∂ξ(Q(ξ)Ψ3(t, ξ, η) 1

ξS(ξ,η)). Next we change η = µ(x), and the variable of integrationξ=t−1/3ξ0, then we obtain

P tV3Qφ=MB−1D−1t2/3Z

R

eixξq1(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)φξ(ξ)dξ +

Z

R

eixξq2(t, µ(xt23), ξt−1/3)Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ dξ

+φ(0) Z

R

eixξq3 t, µ(xt−2/3), ξt−1/3

Dt1/3e−itΛ(ξ)dξ . Define the pseudodifferential operatorsak(t, x, D)φ≡R

Reixξak(t, x, ξ)bφ(ξ)dξ with symbolsak(t, x, ξ) =qk t, µ(xt−2/3), ξt13

. Then we obtain P tV3Qφ=MB−1D−1t2/3

a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ) +a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ +φ(0)a3(t, x, D)F−1Dt1/3e−itΛ(ξ)

.

Let us prove theL2-boundedness of the operatorsak(t, x, D),k= 1,2. We have ak(t, x, ξ) =

1−χ t1/3µ(xt−2/3)

P(µ(xt−2/3))

×(ξ∂ξ)k−1Q(ξt−1/33(ξ/(t1/3µ(xt−2/3))) Λ0(ξt−1/3)−xt−2/3

.

Note thatµ(x) =O(|x|1/2) for small|x|, thereforeχ(t1/3µ(xt−2/3))6= 0 for|x| ≤C, t≥1. Alsoχ3(ξt−1/3/µ(xt−2/3))6= 0 for t1/3µ(xtξ−2/3)>3/2, hence

0(ξt−1/3)−xt−2/3|=|

Z ξt−1/3

µ(xt−2/3)

Λ00(η)dη| ≥ |

Z ξt−1/3

2 3ξt−1/3

Λ00(η)dη|

≥ 1

3|ξt−1/300 2 3ξt−1/3

≥C|Λ0(ξt−1/3)|.

Note that|∂xklξak(t, x, ξ)| ≤C for allx, ξ∈R,t≥1,k, l= 0,1, if 2≤α12≤3.

Therefore by Lemma 2.1 we find that kak(t, x, D)φkL2

x ≤ C|t|kφkL2. Thus the pseudodifferential operatorsak(t, x, D) areL2-bounded fork= 1,2. Then as above we have

00|12MB−1D−1t2/3a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ) L2

η

≤Ck∂ξφkL2, |Λ00|12MB−1D−1t2/3a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ L2

η

≤Ck∂ξφkL2. Now let us prove the L2 - L boundedness of the pseudodifferential operator a3(t, x, D). We have

a3(t, x, ξ) =

1−χ t1/3µ(xt−2/3)

P µ(xt−2/3) t1/3

×∂ξQ(ξt−1/33(ξ/(t1/3µ(xt−2/3))) Λ0(ξt−1/3)−xt−2/3

.

(11)

Note that|∂xkξlhξiδa3(t, x, ξ)| ≤C|t|13(3−α1−α2)for allx, ξ∈R,t≥1,k, l= 0,1, if 2≤α12+δ≤3,δ >1/2. Therefore by Lemma 2.1 we have

ka3(t, x, D)φkL2

x ≤C|t|13(3−α1−α2)khξi−δφkb L2 ≤C|t|13(3−α1−α2)kφkb L. Thus as above we obtain

k|Λ00|1/2φ(0)MB−1D−1

t23

a3(t, x, D)F−1D

t13e−itΛ(ξ)kL2η≤C|t|13(52−α1−α2)|φ(0)|.

The proof is complete.

Next we obtain theL2-estimate for the operatorV4.

Lemma 3.4. Let the weight P be such that P(η) = O(|η|α1), with α1 > −1.

Assume that the weightQsatisfies the estimateQ(ξ) =O(|ξ|α2), ifα2>−1. Then |Λ00|1/2PV4

Lp≤C|t|13(1212)−3p1(k∂ξφkL2+|t|16|φ(0)|) for allt≥1, where1≤p≤ ∞.

Proof. Using the estimate|φ(ξ)−φ(0)| ≤C|ξ|1/2k∂ξφkL2, we have k|Λ00|1/2PV4QφkLp

≤C|t|1/2k|Λ00|1/2P(η)χ(ηt1/3) Z

R

e−itS(ξ,η)χ(1

3ξt1/3)Q(ξ)φ(ξ)dξkLp

≤C|t|1/2k|Λ00|1/2P(η)χ(ηt1/3)kLpkχ(1

3ξt1/3)Q(ξ)φ(ξ)kL1

≤C|t|1/2k∂ξφkL2k|η|121kLp(|η|≤2t−1/3)k|ξ|122kL1(|ξ|≤6t−1/3)

+C|t|1/2|φ(0)|k|η|121kLp(|η|≤2t−1/3)k|ξ|α2kL1(|ξ|≤6t−1/3)

≤C|t|13(1212)−3p1 (k∂ξφkL2+|t|1/6|φ(0)|).

The proof is complete.

In the next lemma we estimate the operatorV5.

Lemma 3.5. Let the weights P andQ∈C3(R\0) be such that P(η) =O(|η|α1), and ∂kξQ(ξ) = O(|ξ|α2−k), k = 0,1,2,3. Assume that α1 ≥ 0, α2 ∈ R, −1 ≤ α12≤2. Then

00|1/2PV5

L2 ≤Ct1+α1 +3 α2(k∂ξφkL2+t16|φ(0)|) for allt≥1.

Proof. Integrating by parts we obtain PV5Qφ=Cχ(ηt1/3)P(η)|t|1/2

Z

R

e−itS(ξ,η)ξ

1

ξS(ξ, η)Q(ξ) 1−χ(1

3ξt1/3) φ(ξ)dξ

=C|t|1/2 Z

R

e−itS(ξ,η)q1(t, η, ξ)φξ(ξ)dξ +C|t|1/2

Z

R

e−itS(ξ,η)q2(t, η, ξ)φ(ξ)−φ(0)

ξ dξ

+Cφ(0)|t|1/2 Z

R

e−itS(ξ,η)q3(t, η, ξ)dξ,

(12)

whereqk(t, η, ξ) =t−1χ(ηt1/3)P(η)(ξ∂ξ)k−1(Q(ξ)(1−χ(13ξt1/3)) 1

ξS(ξ,η)),q3(t, η, ξ) = t−1χ(ηt1/3)P(η)∂ξ(Q(ξ)(1−χ(13ξt1/3)) 1

ξS(ξ,η)). Next we changeη =µ(x), and the variable of integrationξ=t−1/3ξ0, then we obtain

PV5Qφ=MB−1Dt−12/3Z

R

eixξq1(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)φξ(ξ)dξ +

Z

R

eixξq2(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ dξ

+φ(0) Z

R

eixξq3(t, µ(xt−2/3), ξt−1/3)Dt1/3e−itΛ(ξ)dξ . Define the pseudodifferential operatorsak(t, x, D)φ≡R

Reixξak(t, x, ξ)bφ(ξ)dξ with symbolsak(t, x, ξ) =qk(t, µ(xt−2/3), ξt13). Then we obtain

PV5Qφ=MB−1Dt−12/3

a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ) +a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0)

ξ +φ(0)a3(t, x, D)F−1Dt1/3e−itΛ(ξ)

.

Let us prove theL2-boundedness of the operatorsak(t, x, D),k= 1,2. We have ak(t, x, ξ) =t−1χ t1/3µ(xt−2/3)

P(µ(xt−2/3))(ξ∂ξ)k−1Q(ξt−1/3)(1−χ(13ξ)) Λ0(ξt−1/3)−xt−2/3

, Note thatµ(x) =O(|x|1/2) for small|x|, thereforeχ(t1/3µ(xt−2/3))6= 0 for|x| ≤C, t≥1. Alsoχ3(ξt−1/3/µ(xt−2/3))6= 0 for t1/3µ(xtξ−2/3)>3/2, hence

0(ξt−1/3)−xt−2/3|=|

Z ξt−1/3

µ(xt−2/3)

Λ00(η)dη| ≥ |

Z ξt−1/3

2 3ξt−1/3

Λ00(η)dη|

≥ 1

3|ξt−1/300(2

3ξt−1/3)≥C|Λ0(ξt−1/3)|.

Note that |hξiν(ξ∂ξ)jak(t, x, ξ)| ≤ Ct1+α1 +3 α2 for all x, ξ ∈ R, t ≥ 1, j = 0,1,2 with someν ∈(0,1), if α1≥0,α2 ∈R,−1≤α12 <2. Therefore by Lemma 2.3 we findkak(t, x, D)φkL2

x≤Ct1+α1 +3 α2kφkL2. Then as above we have |Λ00|1/2MB−1D−1t2/3a1(t, x, D)F−1Dt1/3e−itΛ(ξ)φξ(ξ)

L2η≤Ct1+α1 +3 α2k∂ξφkL2

and

00|1/2MB−1D−1t2/3a2(t, x, D)F−1Dt1/3e−itΛ(ξ)φ(ξ)−φ(0) ξ

L2

η

≤Ct1+α1 +3 α2k∂ξφkL2.

Now let us prove the L2 - L boundedness of the pseudodifferential operator a3(t, x, D). We have

a3(t, x, ξ) =t−1χ t1/3µ(xt−2/3)

P(µ(xt−2/3))t1/3ξ

Q(ξt−1/3)(1−χ(13ξ)) Λ0(ξt−1/3)−xt−2/3

, Note that |hξiν(ξ∂ξ)jhξiδa3(t, x, ξ)| ≤ C|t|1312) for all x, ξ ∈ R, t ≥ 1, j = 0,1,2 with some small ν ∈ (0,1), δ > 1/2. Therefore by Lemma 2.3 we find

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