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Large time asymptotics of solutions for the modified

KdV equation with a fifth order dispersive term

Nakao Hayashi and Pavel I. Naumkin

(Received Received November 7, 2017; Revised May 8, 2018)

Abstract. We study the large time asymptotics of small amplitude solutions to

the Cauchy problem for the modified KdV equation with a fifth order dispersive term.

AMS 2010 Mathematics Subject Classification. 35Q35, 81Q05.

Key words and phrases. Generalized KdV equation, modified KdV equation,

large time of solutions, dispersive equation

§1. Introduction

We study the large time asymptotics of small amplitude solutions to the Cauchy problem for the modified Korteweg-de Vries equation with a fifth order dispersive term

(1.1) { ∂tv−a3∂x3v +b5∂ 5 xv = ∂xv3, t > 0, x∈ R, v (0, x) = v0(x) , x∈ R,

where a, b > 0 or a, b < 0. The case of a, b > 0 is called the positive dispersion, whereas the case of a, b < 0 is called the negative dispersion in the study of the solitary waves (see [14]). In this paper we assume that a, b > 0 since the case of a, b < 0 is considered in the same way. In the case of a, b > 0 it is expected that the main term of the large time asymptotics of solutions to (1.1) is located in the positive half-line. We also assume that the initial data are real-valued functions. The problem of large time asymptotics of solutions to (1.1) is open for the complex-valued initial data.

Korteweg-de Vries (KdV) equation

(1.2) ∂tv− a 3∂ 3 xv = ∂xv2 7

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was introduced in [17] to describe a model for unidirectional propagation of nonlinear dispersive long waves. In [13], it was shown that the system of equations for magneto-acoustic waves of small finite amplitude propagating in a cold collision-free plasma can be reduced to KdV equation. It was also found that the coefficient a depends on masses of ion and electron and has a possibility to be zero. Therefore the higher order dispersive terms should be taken into account. Indeed in [12] it was shown that, when a = 0, then the system of equations can be reduced to a simple non linear dispersive equation of the form ∂tv + b 5∂ 5 xv = ∂xv2.

Thus, when a is close to 0, we obtain a generalized KdV equation

(1.3) ∂tv− a 3∂ 3 xv + b 5∂ 5 xv = ∂xv2.

In [14], the steady travelling wave solutions of (1.3) were investigated numer-ically to reveal how the solitary wave solutions of (1.2) are modified by the fifth order dispersive term. In [12], it was also pointed out that the Alfv´en waves are described by the modified KdV equation

(1.4) ∂tv−

a 3∂

3

xv = ∂xv3.

Thus we arrive to equation (1.1). Concerning the solitary wave solutions of higher order dispersive equations, there are a lot of works (see, e.g., [11], [15]). However the large time asymptotic behavior of small solutions was not studied well. On the other hand, the Cauchy problem (1.3) was considered in papers [2], [3], [20] and problem (1.1) was studied in [19]. In [2] and [19], the local existence in Hr with r > 1/4 and global existence in H2 were obtained for (1.3) and (1.1), respectively by using the Strichartz estimates. In [3], it was shown via the bilinear estimates that problem (1.3) has local solutions for Hr data with r >−1. This result combined with the L2- conservation law yields

global existence of solutions to (1.3) in L2. In [20] the order of Sobolev space r was reduced to r ≥ −7/5 by improving the bilinear estimates. And also the global existence in Hr with r >−1/2 was obtained by using the high-low frequency method from paper [1] developed for (1.2). As far as we know the large time asymptotics of solutions to (1.3) is still an open problem.

For the modified KdV equation (1.4) we studied the large time asymptotics of small solutions and showed that the small amplitude solutions are stable in the neighborhood of the self-similar solutions (see [9]).

In the case of the fourth order nonlinear Schr¨odinger equation

(1.5)

{

i∂tu +12∂x2u−14∂x4u = λ|u|

2u, t > 0, x∈ R,

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we developed the factorization technique used in [7] for the nonlinear Klein-Gordon equation to obtain the large time asymptotics of small solutions. Since the symbol of equation (1.5) K (ξ) =−2iξ2− 4iξ4 is inhomogeneous, it seems difficult to apply the operator

J1 = etK(−i∂x)xe−tK(−i∂x)= x + it

(

∂x− ∂x3 ) to (1.5) directly, so we defined the new operator

A1= M1 √ K′′(ξ)∂ξ M1 √ K′′(ξ), M1= e −t(ξK′(ξ)−K(ξ)) .

In the case of the nonlinear Schr¨odinger equation i∂tu +12∂x2u = λ|u|2u, the symbol is−2iξ2, and we have

A2 = e− it 2ξ 2 ∂ξe it 2ξ 2 = ∂ξ+ itξ

= −iF (x + it∂x)F−1=−iFJ2F−1.

Hence the operator A1 is a generalization of A2, and J2 = e

it 2∂ 2 xxe−it2∂ 2 x =

x + it∂x was widely used in the study of the large time behavior of solutions to the nonlinear Schr¨odinger equations (see [10]). Here we denote by Fϕ or ˆϕ the Fourier transform of the function ϕ

ˆ ϕ(ξ) = √1 2π ∫ R e−ixξϕ(x)dx,

then the inverse Fourier transformation F−1 is given by F−1ϕ = √1

2π ∫

R

eixξϕ(ξ)dξ.

Multiplying both sides of (1.5) by e−itK(ξ) and using the factorization tech-nique in [8], we obtained for the gauge invariant nonlinearity

√ K′′(ξ) √bu (ξ)K′′(ξ) 2 bu (ξ) √ K′′(ξ).

Then we can see that the operatorA1 works well for the gauge invariant

non-linearity in (1.5). In the case of equations (1.1) or (1.4) we encounter another difficulty. By the factorization technique the nonlinearity can be decomposed as a combination of the following four terms bu3, |bu|2bu, bu3, |bu|2bu with some additional oscillating factors, except the term |bu|2bu. (Note that for the case of the complex-valued solution, we do not have the same representation for the nonlinearity). Due to these oscillation factors the operator J does not

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work on the terms bu3, bu3 and |bu|2bu. To avoid this difficulty in [9] we applied for (1.4) the dilation operator x∂x+ 3t∂t. However the dilation operators only work well for dispersive equations with homogeneous symbols such as (1.4). So we can not overcome this difficulty by introducing the modified dilation operator due to the inhomogeneous symbol in equation (1.3).

In our previous paper [9] we also used the property that the difference between the total mass of the solution and the self-similar solution is equal to zero. However we do not know the existence of the self-similar solution for (1.1). So we assume in this paper that the total mass of the initial data is zero, i.e. ∫v0(x) dx = 0. Then by the equation we get

∫

v (t, x) dx = 0 for all t > 0. Under this condition, taking u (t, x) =∫−∞x v (t, x) dx we can rewrite (1.1) in the potential form

(1.6)

{

∂tu−a3∂x3u + b5∂x5u = (ux)3, t > 0, x∈ R, u (0, x) = u0(x) , x∈ R.

This is our equation which we study in this paper.

To state our results precisely we introduce Notation and Function Spaces. We denote the Lebesgue space by Lp ={ϕ ∈ S′;∥ϕ∥Lp<∞}, where the norm

∥ϕ∥Lp = (∫ |ϕ (x)|p dx) 1 p for 1 ≤ p < ∞ and ∥ϕ∥ L∞ = ess.supx∈R|ϕ (x)| for p =∞. The weighted Sobolev space is

Hk,sp = { φ∈ S′;∥ϕ∥Hk,s p = ⟨x⟩s⟨i∂ x⟩kϕ Lp <∞ } , k, s ∈ R, 1 ≤ p ≤ ∞, ⟨x⟩ = √1 + x2, ⟨i∂ x⟩ = √ 1− ∂2

x. We also use the notations Hk,s= Hk,s2 , Hk = Hk,0 shortly, if it does not cause any confusion. Let C(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 letter C. We define the free evolution group U (t) = e−itΛ(−i∂x) = F−1EF, where the

multiplication factor E (t, ξ) = e−itΛ(ξ), and Λ (ξ) = a3ξ3+5bξ5 is the dispersion relation for equation (1.6).

We are now in a position to state our main result. Define the Heaviside function θ (x) = 1 for x≥ 0 and θ (x) = 0 for x < 0.

Theorem 1.1. Assume that the initial data u0 ∈ H3∩ H1,1 are real-valued

with a sufficiently small norm ∥u0∥H3∩H1,1 ≤ ε. Then there exists a unique global solution eitΛ(−i∂x)u∈ C([0,∞) ; H3∩ H1,1)of the Cauchy problem (1.6)

satisfying the time decay estimates

∥∂xu (t)∥L∞+ ∂x2u (t) L∞ ≤ Cεt−

1 2.

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asymptotics ∂xu (t) = 2Ret− 1 2 Λ′′ (x t ) −1 2 eit(xtΛ′( x t)−Λ( x t)) ×θ(x t ) ix t W+ (x t ) exp ( 3i Λ′′ (x t ) −1(x t )3 W+ (x t ) 2 log t−π 4i ) +O ( εt−12−δ ) (1.7)

is valid for t→ ∞ uniformly with respect to x ∈ R, where δ ∈(0,16) is a small constant.

We introduce the factorization formula for equation (1.6). We have

U (t) F−1ϕ = F−1Eϕ = √1 2π ∫ R eit(xtξ−Λ(ξ))ϕ (ξ) dξ = Dt √ it 2π ∫ R eit(xξ−Λ(ξ))ϕ (ξ) dξ, where Dtϕ = (it)− 1

2 ϕ(xt−1). Consider the stationary point defined by the equation Λ′(ξ) = x. Since Λ′(ξ) = aξ2 + bξ4 is monotonous in the do-main ξ ∈ R+, there exists a stationary point ξ = η (x) , where η (x) =

1

√

2b

√√

4bx + a2− a, such that Λ′(η (x)) = x for x > 0. This fact means

that the main term of the large time asymptotics of the solutions lies in the positive half-line. We extend η (x) for all x∈ R by

η (x) = √ x 2b|x|

√√

4b|x| + a2− a.

Define the operator (Bϕ) (x) = |Λ′′(η (x))|−12 ϕ (η (x)) . Since x = η

|η|Λ′(η) , we get U (t) F−1ϕ =Dt √ it 2π ∫ R e−it(Λ(ξ)−xξ)ϕ (ξ) dξ = Dt √ it 2π ∫ R e−it ( Λ(ξ)−|η|ηΛ′(η)ξ ) ϕ (ξ) dξ = DtB √ it|Λ′′(η)| 2π ∫ R e−it ( Λ(ξ)−|η|η Λ′(η)ξ) ϕ (ξ) dξ,

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b φ (−ξ) = bφ (ξ). Hence U (t) F−1φ =b DtB √ it|Λ′′(η)| 2π ∫ ∞ 0 e−it ( Λ(ξ)−|η|ηΛ′(η)ξ) b φ (ξ) dξ +DtB √ it|Λ′′(η)| 2π ∫ ∞ 0 e−it ( Λ(ξ)−|η|ηΛ′(η)ξ) b φ (ξ) dξ = 2ReDtBM √ it|Λ′′(η)| 2π ∫ ∞ 0 e−itS(ξ,η)φ (ξ) dξ = 2Reb DtBMV bφ (1.8)

for η ∈ R, where the multiplication factor M (t, η) = eit(ηΛ′(η)−Λ(η))θ(η) = eitH(η), H (η) = (ηΛ′(η)− Λ (η)) θ (η) , the phase function

S (ξ, η) = Λ (ξ)− Λ (η) θ (η) − η

|η|Λ′(η) (ξ− ηθ (η)) and the operator

Vϕ = √ it|Λ′′(η)| 2π ∫ ∞ 0 e−itS(ξ,η)ϕ (ξ) dξ.

Also we need the representation for the inverse evolution group FU (−t) for all ξ≥ 0 FU (−t) ϕ = EFϕ = √ 1 2π ∫ R eitΛ(ξ)−ixξϕ (x) dx = √ t 2πi ∫ R eit(Λ(ξ)−xξ)Dt−1ϕ (x) dx = √ t 2πi ∫ R eit ( Λ(ξ)−|η|ηΛ′(η)ξ)( B−1D−1 t ϕ ) (η) Λ′′(η) 1 2 dη = √ t 2πi ∫ R eitS(ξ,η)M(B−1Dt−1ϕ)(η) Λ′′(η) 1 2 dη =QMB−1D−1 t ϕ, (1.9) where D−1t ϕ = (it) 1 2 ϕ (xt) , (B−1ϕ)(η) = Λ′′(η) 1 2 ϕ ( η |η|Λ′(η) ) , and the operator

Qϕ = √ t 2πi ∫ R eitS(ξ,η)ϕ (η) Λ′′(η) 1 2 dη

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Since FU (−t) L = ∂tFU (−t) with L = ∂t − a3∂x3 + b5∂x5, applying the operator FU (−t) to equation (1.6) we get for the new dependent variable

b

φ =FU (−t) u (t)

∂tφ = ∂b tFU (−t) u = FU (−t) Lu = FU (−t) u3x. Then by (1.8) and (1.9) we find the following factorization property

FU (−t) u3 x =QMB−1D−1t u3x = QMB−1D−1t ( DtBMV (iξ) bφ +DtBMV (iξ) bφ )3 = QMB−1 1 (it) ( BMV (iξ) bφ + (it) 1 2 (−it)12 BMV (iξ) bφ )3 = (it)−1QMB−1 (

BMV (iξ) bφ + iBMV (iξ) bφ )3 = (it)−1Q Λ′′ −1M ( MV (iξ) bφ + iMV (iξ) bφ )3 . Hence FU (−t) u3 x

= (it)−1Q Λ′′ −1M2(Viξ bφ)3+ 3t−1Q Λ′′ −1|Viξ bφ|2Viξ bφ −3 (it)−1Q Λ′′ −1M2|Viξ bφ|2Viξ bφ− t−1Q Λ′′ −1M4

( Viξ bφ

)3

. We need to calculate the commutator ofQ and M since ∂ξQM yields undesir-able time growth, when we wish to estimate the derivative ∂ξφ. Note that inb the case of the usual nonlinear Schr¨odinger equation with cubic nonlinearity u3, we have FU (−t) u3 = (it)−1QM2(V bφ)3 with Qϕ = √ t 2πi ∫ R eit2(ξ−η) 2 ϕ (η) dη =F−1e−2ti|x| 2 Fϕ, Vϕ = √ it 2π ∫ R e−it2(ξ−η) 2 ϕ (ξ) dξ =Fe2ti|ξ| 2 F−1ϕ. We can see that M = eit2η

2

yields an additional time growth since ∂ξQM2 has a time growing summand like −itξQM2.. We encounter a similar difficulty in our case.

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We have for α̸= −1 it (S (ξ, η) + αH (η)) = it ( Λ(ξ)− Λ (η) θ (η) − η |η|Λ′(η) (ξ− ηθ (η)) + α(ηΛ′(η) θ (η)− Λ (η) θ(η)) ) = it ( Λ(ξ)− (1 + α) Λ ( ξ 1 + α )) +i (1 + α) t ( Λ ( ξ 1 + α ) − Λ (η) θ (η) −|η|η Λ′(η) ( ξ 1 + α− ηθ (η) )) = it ( Λ (ξ)− (1 + α) Λ ( ξ 1 + α )) + i (1 + α) tS ( ξ 1 + α, η ) . Hence Q (t) Mαϕ = eit(Λ(ξ)−(1+α)Λ( ξ 1+α))i 1 2D1+αQ (t (1 + α)) ϕ.

This fact implies that our method does not work for the nonlinearity |u|2, namely for the case of equations (1.2) or (1.3) with a quadratic nonlinear-ity. Thus we obtain from (1.6) the following equation for the new dependent variableφ =b FU (−t) u (t) ∂tφ =b −i 3 2t−1eitΩD3Q (3t) 1 |Λ′′|(Viξ bφ)3 +3t−1Q (t) 1 |Λ′′||Viξ bφ|2Viξ bφ +3i32t−1D−1Q (−t) 1 |Λ′′||Viξ bφ|2Viξ bφ −i1 2t−1eitΩD−3Q (−3t) 1 |Λ′′| ( Viξ bφ )3 , (1.10) where Ω = Λ (ξ)− 3Λ ( ξ 3 ) .

It is well known that the operator J = U (t) xU (−t) is a useful tool for obtaining the L∞- time decay estimates of solutions and has been used widely for the studying the asymptotic behavior of solutions to various nonlinear dispersive equations. We have

J = U (t) xU (−t) = F−1e−itΛ(ξ)i∂ ξeitΛ(ξ)F = F−1(i∂ξ− tΛ′(ξ) ) F = x − tΛ′(−i∂ x) ,

where Λ′(−i∂x) =−a∂x2+b∂x4. Note that the commutators are true [J , L] = 0, [J , ∂x] =−1, where L = ∂t+iΛ (−i∂x) = ∂t−a3∂x3+5b∂

5

x. However, it seems that J does not work well on the nonlinear terms. Then, instead of the operators

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J and the dilation operator used in [5] we apply the modified dilation operator defined by P = t∂t+ 1 5x∂x− 2 5a∂a.

This is our idea in this paper. Note thatP acts well on the nonlinear terms as the first order differential operator. AlsoJ and P are related via the identity P = tL +1

5J ∂x+ 2a

5I, where I = −∂a+ 1

3t∂x3. In order to get the estimate of J ∂xu, we will show the a-priori estimates of Pu, tLu and Iu. Different point compared to the previous works is to consider the estimate of Iu since Iu contains the third order derivatives t∂x3u with the additional time growth (if a̸= 0). Note that the commutators [P, L] = −L and [P, ∂x] =−15∂x are true. Also we have [I, L] = 0.

Our main task is to estimate each term of the right-hand side of equation (1.10) in L∞ and H1 norms. We organize the rest of our paper as follows. In Section 2, we state the uniform estimates for the decomposition operator V (Lemma 2.2). The uniform estimates of Q (t) and Q (−t) are also obtained in Lemma 2.3. We prove L2 - estimates of the derivatives of Q (Lemma 3.1) and V (Lemma 3.5) in Section 3. Lemma 3.2 and Lemma 3.3 are prepared for proving Lemma 3.5. In Section 4 we divide the nonlinear term into the main term and the remainder terms (Lemma 4.1). In Section 5 we obtain the estimates of ∥FU (−t) u∥L∞ and ∥J u (t)∥H1. Section 6 is devoted to the proof of Theorem 1.1.

§2. Estimates in the uniform norm

2.1. Estimates for two kernels

Define two kernels, which describe the main term of the large time asymptotics of the operatorsV and Q, respectively,

A (t, η) = √ it 2π Λ ′′(η) 1 2 ∫ ∞ 0 e−itS(ξ,η)χ(ξη−1)dξ and A∗(t, ξ) = √ t 2πi ∫ ∞ 0 eitS(ξ,η) Λ′′(η) 1 2 χ(ξη−1)dη, for ξ, η > 0, where S (ξ, η) = Λ (ξ)− Λ (η) − Λ′(η) (ξ− η) , Λ (ξ) = a3ξ3+b5ξ5, and the cut off function χ (z) ∈ C2(R) is such that χ (z) = 0 for z ≤ 13, χ (z) = 1 for 23 ≤ z ≤ 32, and χ (z) = 0 for z≥ 3.

Lemma 2.1. The estimate

∥A (t)∥L∞(R+)+∥A

∗(t)∥

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is true for all t > 0. Moreover the asymptotics are valid A (t, η) = 1 + O((tη3)−1 ) for t13η→ ∞ and A∗(t, ξ) = 1 + O((tξ3)−1 ) for t13ξ→ ∞.

Proof. We use the identity

(2.1) e−itS(ξ,η) = H1∂ξ (

(ξ− η) e−itS(ξ,η) )

with H1= (1− it (ξ − η) Sξ(ξ, η))−1 in the domain ξ, η > 0, and integrate by parts to get A (t, η) =− √ it 2π Λ ′′(η) 1 2 ∫ ∞ 0 e−itS(ξ,η)(ξ− η) ∂ξ ( H1χ ( ξη−1))dξ. Hence in view of the estimate

(2.2) (ξ− η) ∂ξ ( H1χ ( ξη−1)) ≤C ( 1 + tη⟨η⟩2(ξ− η)2 )−1 in the domain η3 ≤ ξ ≤ 3η, we obtain

|A (t, η)| ≤ Ct12 Λ′′(η) 1 2 ∫ 3η η 3 dξ 1 + tη⟨η⟩2(ξ− η)2 ≤ Ct1 2η 3 2 ⟨η⟩ ∫ 3 1 3 dy 1 + tΛ (η) (y− 1)2 ≤ C |tΛ (η)| 1 2 ⟨tΛ (η)⟩− 1 2 ≤ C. In the same manner we use the identity

(2.3) eitS(ξ,η)= H2∂η (

(η− ξ) eitS(ξ,η) )

with H2= (1 + it (η− ξ) Sη(ξ, η))−1 in the domain ξ, η > 0, and integrate by parts A∗(t, ξ) = √ t 2πi ∫ ∞ 0 eitS(ξ,η) Λ′′(η) 1 2 χ(ξη−1)dη = − √ t 2πi ∫ ∞ 0 eitS(ξ,η)(ξ− η) ∂η ( H2 Λ′′(η) 1 2 χ(ξη−1))dη.

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Then using the estimate (2.4) (ξ − η) ∂η ( H2 Λ′′(η) 1 2χ(ξη−1)) ≤ Cξ12⟨ξ⟩ ( 1 + tξ⟨ξ⟩2(ξ− η)2 )−1 in the domain ξ3 ≤ η ≤ 3ξ, we get as above

|A∗(t, ξ)| ≤ Ct1 2ξ 1 2 ⟨ξ⟩ ∫ 3ξ ξ 3 dη 1 + tξ⟨ξ⟩2(ξ− η)2 ≤ Ct1 2ξ 3 2 ⟨ξ⟩ ∫ 3 1 3 dy 1 + tΛ (ξ) (y− 1)2 ≤ C |tΛ (ξ)| 1 2 ⟨tΛ (ξ)⟩− 1 2 ≤ C. To compute the asymptotics of the functions A (t, η) and A∗(t, ξ) for large ηt13 and ξt13 we apply the stationary phase method (see [4], p. 163). We have the asymptotics (2.5) ∫ R eirG(x)f (x) dx = eirG(x0)f (x 0) √ 2π r|G′′(x0)| eiπ4sgnG′′(x0)+ O ( r−32 ) for r → +∞, where the stationary point x0 is defined by G′(x0) = 0. We

change ξ = xη, then we get A (t, η) = √ it 2π|Λ′′(η)| 1 2 ∫∞ 0 e−itS(ξ,η)χ ( ξη−1)dξ = η √ it 2π 2η ( a + 2bη2) 1 2 ∫∞ 0 e−itη 3(a 3x 3+b 5x 5η2−a 3− b 5η 2−(a+bη2)(x−1) )χ (x) dx = √ ir 2π 2 ( a + 2bη2) 1 2 ∫∞ 0 e irG(x,η)χ (x) dx,

where r = tη3. By virtue of formula (2.5) with r = tη3, f (x) = χ (x) , x0= 1 and G (x, η) =− ( a 3x 3+ b 5x 5η2−a 3 − b 5η 2−(a + bη2)(x− 1) ) , we get A (t, η) = 1 + O((tη3)−1 )

for t13η → +∞. Also changing η = xξ we obtain A∗(t, ξ) = √ t 2πi ∫ ∞ 0 eitS(ξ,η) Λ′′(η) 1 2 χ(x−1)dη = ξ32 √ t 2πi ∫ ∞ 0 eitξ3(a3+ b 5ξ 2−a 3x 3−b 5x 5ξ2−(ax2+bx4ξ2)(1−x) )χ(x−1) × 2ax + 4bx3ξ2 1 2 dx = √ r 2πi ∫ ∞ 0 eirG(x,η)χ(x−1) (2x(a + 2bx2ξ2)) 1 2 dx,

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where r = tξ3. Then by virtue of formula (2.5) with r = tξ3, f (x) = χ(x−1) (2x(a + 2bx2ξ2)) 1 2 , x 0= 1 and G (x, ξ) = a 3+ b 5ξ 2−a 3x 3− b 5x 5ξ2−(ax2+ bx4ξ2)(1− x) , we get A∗(t, ξ) = 1 + O((tξ3)−1 ) as t13ξ → +∞. Lemma 2.1 is proved.

2.2. Estimates for the operator V

We estimate the operator

Vξjϕ = √ it|Λ′′(η)| 2π ∫ ∞ 0 e−itS(ξ,η)ϕ (ξ) ξjdξ

in the uniform norm. Denote the norm

∥ϕ∥Z1 =∥ϕ∥L∞(R+)+∥⟨ξ⟩ ϕξ∥L2(R +). Note that |Vϕ| ≤ Ct1 2 Λ′′ 1 2 ⟨ξ⟩−2 L1(R +) ⟨ξ⟩2ϕ L∞(R+) ≤ Ct1 2 Λ′′(η) 1 2 ⟨ξ⟩2ϕ L∞(R+) .

Next lemma says that the main term of (Vϕ) (η) is A (t, η) ϕ (η) .

Lemma 2.2. Let j = 0, 1, 2, 3. Then the estimates are valid for all t≥ 1

|η|α⟨η⟩β(Vξjϕ− A (t, η) ηjϕ (η)) L∞(R+) ≤ C max(t−14, t− j+α 3 ) ∥ϕ∥Z1 with α≥ −12, β ≤ 74 − j − α, and |η|α⟨η⟩βVξjϕ L∞(R−)≤ C max ( t−12 log⟨t⟩ , t− j+α 3 ) ∥ϕ∥Z1 with α≥ −12, β ≤ 52 − j − α. When j = 2, 3, the norm ∥⟨ξ⟩ ϕξ∥L2(R

+) can be replaced by∥ξϕξ∥L2(R

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Proof. By the definition ofV and A (t, η), we can write Vξjϕ− A (t, η) ηjϕ (η) = √ it|Λ′′(η)| 2π ∫ 3η η 3 e−itS(ξ,η)(ϕ (ξ) ξj− ϕ (η) ηj)χ(ξη−1)dξ + √ it|Λ′′(η)| 2π ∫ 2η 3 0 e−itS(ξ,η)(1− χ(ξη−1))ϕ (ξ) ξjdξ + √ it|Λ′′(η)| 2π ∫ ∞ 3η 2 e−itS(ξ,η)(1− χ(ξη−1))ϕ (ξ) ξjdξ = √ it|Λ′′(η)| 2π (I1+ I2+ I3)

for η > 0. We integrate by parts via the identity (2.1)

I1 = − ∫ 3η η 3 e−itS(ξ,η)(ξ− η) χ(ξη−1)H1ϕ (ξ) jξj−1dξ − ∫ 3η η 3 e−itS(ξ,η)(ϕ (ξ) ξj− ϕ (η) ηj)(ξ− η) ∂ξ ( χ(ξη−1)H1 ) dξ − ∫ 3η η 3 e−itS(ξ,η)(ξ− η) χ(ξη−1)H1ξjϕξ(ξ) dξ. Using the estimates

(2.6) |H1| = |1 − it (ξ − η) Sξ(ξ, η)|−1≤ C (

1 + tη⟨η⟩2(ξ− η)2 )−1 and (2.2) in the domain η3 < ξ < 3η, we obtain for j = 0, 1, 2, 3

|I1| ≤ C ∥ϕ∥L∞(R+)jη j−1 ∫ 3η η 3 |ξ − η| dξ 1 + tη⟨η⟩2(ξ− η)2 +Cηj ∫ 3η η 3 |ϕ (ξ) − ϕ (η)| dξ 1 + tη⟨η⟩2(ξ− η)2 +Cηj ∫ 3η η 3 |ξ − η| |ϕξ(ξ)| dξ 1 + tη⟨η⟩2(ξ− η)2. Since |ϕ (ξ) − ϕ (η)| ≤ C ∫ η ξ ⟨z⟩−1⟨z⟩ |∂zϕ (z)| dz ≤ C (∫ η ξ ⟨z⟩−2dz )1 2 ∥⟨ξ⟩ ϕξ∥L2(R +)≤ C ⟨η⟩ −1|ξ − η|1 2 ∥ϕ∥ Z1

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in the domain η3 < ξ < 3η, we get |I1| ≤ C ∥ϕ∥L∞(R+)jη j−1 ∫ 3η η 3 |ξ − η| dξ 1 + tη⟨η⟩2(ξ− η)2 +C∥ϕ∥Z 1η j⟨η⟩−1 ∫ 3η η 3 |ξ − η|12 dξ 1 + tη⟨η⟩2(ξ− η)2 +C∥ϕ∥Z 1η j⟨η⟩−1    ∫ 3η η 3 (ξ− η)2dξ ( 1 + tη⟨η⟩2(ξ− η)2 )2    1 2

for j = 0, 1, 2, 3. Note that ∫ 3η η 3 |ξ − η| dξ 1 + tη⟨η⟩2(ξ− η)2 ≤ Cη2 ∫ 3 1 3 |y − 1| dy 1 + tΛ (η) (y− 1)2 ≤ C ⟨tΛ (η)⟩−1η2 ∫ 2t√Λ(η) −2 3t √ Λ(η) |y| dy 1 + y2 ≤ Cη 2⟨tΛ (η)⟩−1log⟨tΛ (η)⟩ , ∫ 3η η 3 |ξ − η|12dξ 1 + tη⟨η⟩2(ξ− η)2 ≤ Cη 3 2 ∫ 3 1 3 |y − 1|12 dy 1 + tΛ (η) (y− 1)2 ≤ Cη 3 2⟨tΛ (η)⟩− 3 4 and    ∫ 3η η 3 (ξ− η)2dξ ( 1 + tη⟨η⟩2(ξ− η)2 )2    1 2 ≤ Cη3 2    ∫ 3 1 3 (y− 1)2dy ( 1 + tΛ (η) (y− 1)2 )2    1 2 ≤ Cη3 2 ⟨tΛ (η)⟩− 3 4.

Hence we get for j = 0, 1, 2, 3

t12 Λ′′(η) 1 2 |η|α⟨η⟩β|I 1| ≤ Ct1 2 ∥ϕ∥ Z1 ( jηj+32+α⟨η⟩β+1⟨tΛ (η)⟩−1log⟨tΛ (η)⟩ +ηj+2+α⟨η⟩β⟨tΛ (η)⟩−34 ) ≤ C(t−14 + t− j+α 3 ) ∥ϕ∥Z1 (2.7)

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with j + α + β≤ 74, j + 2 + α≥ 0, since ηj+32+α⟨η⟩β+1⟨tΛ (η)⟩−1log⟨tΛ (η)⟩ ≤ Cηj+32+α⟨t13η⟩−3log⟨t13η⟩ ≤ C { t−j+α3 − 1 2, 3 > j +3 2 + α≥ 0 t−1log⟨t⟩ , j + 32 + α≥ 3 and ηj+2+α⟨η⟩β⟨tΛ (η)⟩−34 ≤ Cηj+2+α ⟨ t13η ⟩−9 4 ≤ C { t−j+α+23 , 9 4 ≥ j + 2 + α ≥ 0 t−34, j + 2 + α≥ 9 4 for η < 1, also ηj+32+α⟨η⟩β+1⟨tΛ (η)⟩−1log⟨tΛ (η)⟩ ≤ Cηj+52+α+β⟨t15η⟩−5log⟨t15η⟩≤ Ct−1log⟨t⟩ and ηj+2+α⟨η⟩β⟨tΛ (η)⟩−34 ≤ Cηj+2+α+β ⟨ t15η ⟩−15 4 ≤ Ct−3 4 for η > 1 and j + α + β ≤ 74. Next we consider I2 = ∫ 2η 3 0 e−itS(ξ,η)(1− χ(ξη−1))ϕ (ξ) ξjdξ for η > 0. We integrate by parts via the identity

(2.8) e−itS(ξ,η) = H3∂ξ

(

ξe−itS(ξ,η) ) with H3= (1− itξ (Λ′(ξ)− Λ′(η)))−1 to get

I2 = − ∫ 2η 3 0 e−itS(ξ,η)ϕ (ξ) ξ∂ξ ( H3 ( 1− χ(ξη−1))ξj)dξ − ∫ 2η 3 0 e−itS(ξ,η)H3 ( 1− χ(ξη−1))ξj+1ϕξ(ξ) dξ. Using the estimates|H3| ≤ C (1 + tξΛ′(η))−1 and

ξ∂ξ( H3

(

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in the domain 0 < ξ < 2η3 , we obtain |I2| ≤ C ∥ϕ∥L∞(R+) ∫ 2η 3 0 ξjdξ 1 + tξΛ′(η) + C ∫ 2η 3 0 ξj+12 ξ 1 2ϕξ dξ 1 + tξΛ′(η) . We have ∫ 2η 3 0 ξjdξ 1 + tξΛ′(η) ≤ Cη 1+j ∫ 2 3 0 yjdy 1 + tΛ (η) y ≤ Cη 1+j⟨tΛ (η)⟩−1 and ∫ 2η 3 0 ξ2j+1dξ (1 + tξΛ′(η))2 ≤ Cη 2j+2 ∫ 2 3 0 y2j+1dy (1 + tΛ (η) y)2 ≤ Cη 2j+2⟨tΛ (η)⟩−2. Hence we get t12 Λ′′(η) 1 2 |η|α⟨η⟩β|I 2| ≤ Ct1 2 ∥ϕ∥ Z1η α+j+32 ⟨η⟩β+1⟨tΛ (η)⟩−1 ≤ C(t−12 + t− α+j 3 ) ∥ϕ∥Z1 (2.9) with α + j +32 ≥ 0, j + α + β ≤ 52, since ηα+j+32 ⟨η⟩β+1⟨tΛ (η)⟩−1 ≤ Cηα+j+ 3 2 ⟨ t13η ⟩−3 ≤ C { t−α+j3 − 1 2, 3≥ α + j +3 2 ≥ 0 t−1, α + j +32 ≥ 3 for η < 1, and ηα+j+32 ⟨η⟩β+1⟨tΛ (η)⟩−1 ≤ Cηj+52+α+β⟨t15η⟩−5≤ Ct−1, for η > 1 and j + α + β ≤ 52. Finally we estimate I3= ∫ ∞ 3η 2 e−itS(ξ,η)(1− χ(ξη−1))ϕ (ξ) ξjdξ for η > 0. We integrate by parts via the identity (2.8)

I3 = − ∫ ∞ 3η 2 e−itS(ξ,η)ϕ (ξ) ξ∂ξ ( H3 ( 1− χ(ξη−1))ξj)dξ − ∫ ∞ 3η 2 e−itS(ξ,η)H3 ( 1− χ(ξη−1))ξj+1ϕξ(ξ) dξ.

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Using the estimates|H3| ≤ C (1 + tΛ (ξ))−1 and ξ∂ξ ( H3 ( 1− χ(ξη−1))ξj) ≤Cξj(1 + tΛ (ξ))−1 in the domain ξ > 3η2 , we obtain

|I3| ≤ C ∥ϕ∥L∞(R+) ∫ ∞ 3η 2 ξjdξ 1 + tΛ (ξ)+ C ∫ ∞ 3η 2 ξj+12 ξ 1 2ϕξ dξ 1 + tΛ (ξ) . Since ∫ ∞ 3η 2 ξjdξ 1 + tΛ (ξ) ≤ Ct −1∫ ∞ 3η 2 ξj−5dξ≤ Ct−1ηj−4 for η > 1, t≥ 1, and for 0 < η ≤ 1, we find

∫ ∞ 3η 2 ξjdξ 1 + tΛ (ξ) ≤ ∫ 3 2 3η 2 ξjdξ 1 + tξ3 + Ct −1∫ ∞ 3 2 ξj−4dξ ≤ Ct−j+13 ∫ 3 2t 1 3 3η 2 t 1 3 ξjdξ 1 + ξ3 + Ct −1 ≤ Ct−1log⟨t⟩ + Ct−j+1 3 ⟨ t13η ⟩j−2 for j = 0, 1, 2 and for j = 3

∫ ∞ 3η 2 ξjdξ 1 + tΛ (ξ) ≤ ∫ 3 2 3η 2 ξjdξ 1 + tξ3 + Ct −1∫ ∞ 3 2 ξj−5dξ ≤ Ct−j+13 ∫ 3 2t 1 3 3η 2t 1 3 ξjdξ 1 + ξ3 + Ct−1≤ Ct−1. Similarly ∫ ∞ 3η 2 ξ2j+1dξ (1 + tΛ (ξ))2 ≤ Ct −2∫ ∞ 3η 2 ξ2j−9dξ≤ Ct−2η2j−8 for η > 1. For 0 < η≤ 1 ∫ ∞ 3η 2 ξ2j+1dξ (1 + tΛ (ξ))2 ≤ ∫ 3 2 3η 2 ξ2j+1dξ (1 + tξ3)2 + Ct −2∫ ∞ 3 2 ξ2j−9dξ ≤ Ct−2j+2 3 ∫ 3 2t 1 3 3η 2 t 1 3 ξ2j+1dξ (1 + ξ3)2 + Ct −2 ≤ Ct−2log⟨t⟩ + Ct−2j+2 3 ⟨ t13η ⟩2j−4

for j = 0, 1, 2, and for j = 3 ∫ ∞ 3η 2 ξ2j+1dξ (1 + tΛ (ξ))2 ≤ Ct −2∫ 3 2 3η 2 dξ + Ct−2 ∫ ∞ 3 2 ξ2j−9dξ≤ Ct−2.

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Hence we get t12 Λ′′(η) 1 2 |η|α⟨η⟩β|I 3| ≤ C ∥ϕ∥Z1⟨η⟩ j−3+β|η|12+α ( t−12 log⟨t⟩ + t 1 2− j+1 3 ⟨ t13η ⟩j−2) ≤ C(t−12 log⟨t⟩ + t− j+α 3 ) ∥ϕ∥Z1 for j = 0, 1, 2, and for j = 3

t12 Λ′′(η) 1 2|η|α⟨η⟩β|I 3| ≤ Ct− 1 2 ∥ϕ∥ Z1⟨η⟩ β|η|12+α≤ Ct−1 2 ∥ϕ∥ Z1 with j + β + α≤ 52, α≥ −12. Finally we estimate Vξjϕ = √ it|Λ′′(η)| 2π ∫ ∞ 0 e−itS(ξ,η)ϕ (ξ) ξjdξ

for η < 0. We integrate by parts via the identity

(2.10) e−itS(ξ,η) = H4∂ξ (

ξe−itS(ξ,η) )

with H4= (1− itξ (Λ′(ξ) + Λ′(η)))−1 to get

Vξjϕ = − √ it|Λ′′(η)| 2π ∫ ∞ 0 e−itS(ξ,η)ϕ (ξ) ξ∂ξ ( H4ξj ) dξ − √ it|Λ′′(η)| 2π ∫ ∞ 0 e−itS(ξ,η)ξj+1H4ϕξ(ξ) dξ.

Using the estimates|H4| ≤ C (1 + tξ (Λ′(ξ) + Λ′(η)))−1 and

ξ∂ξ ( H4ξj) ≤Cξj ( 1 + tξ(Λ′(ξ) + Λ′(η)))−1 we obtain Vξjϕ ≤ C∥ϕ∥L∞(R+)t 1 2 Λ′′(η) 1 2 ∫ ∞ 0 ξjdξ 1 + tξ (Λ′(ξ) + Λ′(η)) +Ct12 Λ′′(η) 1 2 ∫ ∞ 0 ξj+12 ξ 1 2ϕξ dξ 1 + tξ (Λ′(ξ) + Λ′(η)).

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As above we get ∫ ∞ 0 ξjdξ 1 + tξ (Λ′(ξ) + Λ′(η))+ (∫ ∞ 0 ξ2j+1dξ (1 + tξ (Λ′(ξ) + Λ′(η)))2 )1 2 ≤ ∫ |η| 0 ξjdξ 1 + tξΛ′(η) + ∫ ∞ |η| ξjdξ 1 + tΛ (ξ) + (∫ |η| 0 ξ2j+1dξ (1 + tξΛ′(η))2 )1 2 + (∫ ∞ |η| ξ2j+1dξ (1 + tΛ (ξ))2 )1 2 ≤ Cη1+j⟨tΛ (η)⟩−1+ C⟨η⟩j−4 { t−1log⟨t⟩ + t−j+13 ⟨ t13η ⟩j−2 , j = 0, 1, 2, t−1, j = 3. Then we find |η|α⟨η⟩βVξjϕ L∞(R−)≤ C ( t−12log⟨t⟩ + t− α+1 3 ) ∥ϕ∥Z1 if α≥ −12, α + β + j ≤ 52. Lemma 2.2 is proved.

2.3. Estimates for the operator Q

In this subsection we consider the estimate of the operator Qϕ = Q (t) ϕ = √ t 2πi ∫ R eitS(ξ,η)ϕ (η) Λ′′(η) 1 2 dη

in the uniform norm. Note that ∥Qϕ∥L∞(R+)≤ Ct 1 2 ∫ R |ϕ (η)| Λ′′(η) 1 2 dη≤ Ct12 Λ′′(η) 1 2 ϕ L1(R). In particular we find ∥Q (1 − θ) ϕ∥L∞(R+)+∥D−1Q (−t) (1 − θ) ϕ∥L∞(R+) ≤ Ct1 2 Λ′′(η) 1 2 ϕ L1(R−). (2.11)

The next lemma says that the main term of the asymptotics of Qϕ in the positive half-line is A∗(t) ϕ.

Lemma 2.3. Let α > −2, α1 > −32, β ≤ 34 − α, β1 < 52 − α1. Then the

estimate ∥Qθϕ − θA∗(t) ϕ∥ L∞(R+)+∥Q (−t) θϕ∥L∞(R−) ≤ C max(t−α13 , t− 1 2 log⟨t⟩) η−α1⟨η⟩−β1ϕ L∞(R+) +C max ( t−1+2α6 , t− 1 4) η−α⟨η⟩−β∂ηϕ L2(R +)

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is valid for all t≥ 1.

Proof. For the case of ξ≥ 0, we write Qθϕ − θA∗(t) ϕ = √ t 2πi ∫ ∞ 0 eitS(ξ,η)(ϕ (η)− ϕ (ξ)) Λ′′(η) 1 2 χ(ξη−1)dη + √ t 2πi ∫ 2 3ξ 0 eitS(ξ,η)ϕ (η) Λ′′(η) 1 2 (1− χ(ξη−1))dη + √ t 2πi ∫ ∞ 3 2ξ eitS(ξ,η)ϕ (η) Λ′′(η) 1 2 (1− χ(ξη−1))dη = I1+ I2+ I3.

In the first integral I1 by using the identity (2.3) we get

I1 =− √ t 2πi ∫ 3ξ ξ 3 eitS(ξ,η)(ϕ (η)− ϕ (ξ)) (η − ξ) × ∂η ( H2 Λ′′(η) 1 2 χ(ξη−1))dη − √ t 2πi ∫ 3ξ ξ 3 eitS(ξ,η)(η− ξ) Λ′′(η) 1 2 χ(ξη−1)H 2ϕη(η) dη.

Then using (2.4) and the estimates

|ϕ (η) − ϕ (ξ)| ≤ Cξα⟨ξ⟩β|η − ξ|12 η−α⟨η⟩−β ∂ηϕ L2(R +) and |H2| ≤ C ( 1 + tξ⟨ξ⟩2(ξ− η)2 )−1 in the domain ξ3 ≤ η ≤ 3ξ, we find

|I1| ≤ Ct 1 2 η−α⟨η⟩−β∂ηϕ L2(R +) ξ12+α⟨ξ⟩1+β ×     ∫ 3ξ ξ 3 |η − ξ|12 dη 1 + tξ⟨ξ⟩2(ξ− η)2 +    ∫ 3ξ ξ 3 (η− ξ)2dη ( 1 + tξ⟨ξ⟩2(ξ− η)2 )2    1 2     . Changing η = ξy we get

∫ 3ξ ξ 3 |η − ξ|12 dη 1 + tξ⟨ξ⟩2(ξ− η)2 ≤ Cξ 3 2 ∫ 3 1 3 |y − 1|12 dy 1 + tΛ (ξ) (1− y)2 ≤ Cξ3 2 ⟨tΛ (ξ)⟩− 3 4

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and ∫ 3ξ ξ 3 (η− ξ)2dη ( 1 + tξ⟨ξ⟩2(ξ− η)2 )2 ≤ Cξ 3 ∫ 3 1 3 (y− 1)2dy ( 1 + tΛ (ξ) (1− y)2 )2 ≤ Cξ3⟨tΛ (ξ)⟩−32. Then using t12ξ2+α⟨ξ⟩1+β⟨tΛ (ξ)⟩− 3 4 ≤ Ct12ξ2+α ⟨ t13ξ ⟩−9 4 ≤ C(t−2α+16 + t− 1 4 ) for 0 < ξ < 1, if α≥ −2 and t12ξ2+α⟨ξ⟩1+β⟨tΛ (ξ)⟩− 3 4 ≤ Ct12ξ3+α+β ⟨ t15ξ ⟩−15 4 ≤ Ct−1 4 for ξ≥ 1, if β ≤ 34 − α, we obtain |I1| ≤ C η−α⟨η⟩−β∂ηϕ L2(R +) t12ξ2+α⟨ξ⟩1+β⟨tΛ (ξ)⟩− 3 4 ≤ C max(t−2α+16 , t− 1 4) η−α⟨η⟩−β∂ηϕ L2(R +) . In the second integral I2 using the identity eitS(ξ,η)= H5∂η

(

ηeitS(ξ,η))with H5 = (1 + itηSη(ξ, η))−1 we integrate by parts

I2 =− √ t 2πi ∫ 2 3ξ 0 eitS(ξ,η)ϕ (η) η∂η ( H5 Λ′′(η) 1 2 (1− χ(ξη−1)))dη − √ t 2πi ∫ 2 3ξ 0 eitS(ξ,η)ηH5 Λ′′(η) 1 2 (1− χ(ξη−1))ϕ η(η) dη. Since Sη(ξ, η) =−Λ′′(η) (ξ− η) for η > 0, we have the estimates (2.12) ηH5 Λ′′(η) 1 2 (1− χ(ξη−1)) ≤ Cη32⟨η⟩ ( 1 + t (ξ− η) η2⟨η⟩2 )−1 , and (2.13) η∂η ( H5 Λ′′(η) 1 2 (1− χ(ξη−1))) ≤ Cη12 ⟨η⟩ ( 1 + t (ξ− η) η2⟨η⟩2 )−1 . Hence in the domain 0 < η < 23ξ we get

|I2| ≤ Ct 1 2 η−α1⟨η⟩−β1ϕ L∞(R+) ∫ 2 3ξ 0 η12+α1⟨η⟩1+β1dη 1 + tξη2⟨η⟩2 +Ct12 η−α⟨η⟩−β∂ηϕ L2(R +)    ∫ 2 3ξ 0 η3+2α⟨η⟩2+2βdη ( 1 + tξη2⟨η⟩2)2    1 2 .

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We get ∫ 2 3ξ 0 η12+α1⟨η⟩1+β1dη 1 + tξη2⟨η⟩2 ≤ C ∫ ∞ 0 η12+α1⟨η⟩1+β1dη 1 + tΛ (η) ≤ C ∫ 1 0 η12+α1dη 1 + tη3 + Ct −1∫ ∞ 1 η−72+α+βdη ≤ Ct−1 2−α13 ∫ t13 0 η12+α1dη 1 + η3 + Ct−1≤ C ( t−12−α13 + t−1log⟨t⟩ ) , if α1 >−32, α1+ β1< 52. Similarly ∫ 2 3ξ 0 η3+2α⟨η⟩2+2βdη ( 1 + tξη2⟨η⟩2)2 ≤ C ∫ ∞ 0 η3+2α⟨η⟩2+2βdη (1 + tΛ (η))2 ≤ C ∫ 1 0 η3+2αdη 1 + t2η6 + Ct−2 ∫ ∞ 1 η−5+2α+2βdη ≤ Ct−4 3− 2α 3 ∫ t13 0 η3+2αdη 1 + η6 + Ct −2≤ C(t−4 3− 2α 3 + t−2log⟨t⟩ ) , where α >−2, α + β < 2. Therefore |I2| ≤ C max ( t−α13 , t− 1 2 log⟨t⟩) η−α1⟨η⟩−β1ϕ L∞(R+) +C max ( t−1+2α6 , t− 1 2log⟨t⟩) η−α⟨η⟩−β∂ηϕ L2(R +) . In the third integral I3 using the identity eitS(ξ,η) = H5∂η

(

ηeitS(ξ,η)) with H5 = (1 + itηSη(ξ, η))−1 we integrate by parts

I3 =− √ t 2πi ∫ ∞ 3 2ξ eitS(ξ,η)ϕ (η) η∂η ( H5 Λ′′(η) 1 2 (1− χ(ξη−1)))dη − √ t 2πi ∫ ∞ 3 2ξ eitS(ξ,η)ηH5 Λ′′(η) 1 2(1− χ(ξη−1))ϕ η(η) dη.

Since 32ξ < η implies 13η < (η− ξ) , then by (2.12) and (2.13) we get |I3| ≤ Ct 1 2 η−α1⟨η⟩−β1ϕ L∞(R+) ∫ ∞ 3 2ξ η12+α1⟨η⟩1+β1dη 1 + tΛ (η) +Ct12 η−α⟨η⟩−β∂ηϕ L2(R +) (∫ ∞ 3 2ξ η3+2α⟨η⟩2+2βdη (1 + tΛ (η))2 )1 2 .

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In the same way as in the proof of the estimate for I2 we obtain |I3| ≤ C max ( t−α13 , t− 1 2 log⟨t⟩) η−α1⟨η⟩−β1ϕ L∞(R+) +C max ( t−1+2α6 , t− 1 2log⟨t⟩) η−α⟨η⟩−β∂ηϕ L2(R +) .

Next consider Q (−t) θϕ for ξ < 0. Using the identity e−itS(ξ,η) = H6∂η

(

ηe−itS(ξ,η) )

with H6= (1− itηSη(ξ, η))−1, we integrate by parts Q (−t) θϕ = − √ it 2π ∫ ∞ 0 e−itS(ξ,η)ϕ (η) η∂η ( H6 Λ′′(η) 1 2 ) dη − √ it 2π ∫ ∞ 0 e−itS(ξ,η)ηH6 Λ′′(η) 1 2ϕ η(η) dη.

Since Sη(ξ, η) =−Λ′′(η) (ξ− η) , ξ < 0, η > 0, then we have the estimates ηH6 Λ′′(η) 1 2 ≤ Cη32 ⟨η⟩ (1 + tΛ (η))−1, and η∂η ( H6 Λ′′(η) 1 2) ≤ Cη12⟨η⟩ (1 + tΛ (η))−1. Hence we get |Q (−t) θϕ| ≤ Ct1 2 η−α1⟨η⟩−β1ϕ L∞(R) ∫ ∞ 0 |η|12+α1⟨η⟩1+β1dη 1 + tΛ (η) +Ct12 η−α⟨η⟩−β∂ηϕ L2(R) (∫ ∞ 0 |η|3+2α⟨η⟩2+2β dη (1 + tΛ (η))2 )1 2 .

Then in the same way as in the proof of the estimate for I2 we obtain

|Q (−t) θϕ| ≤ C max(t−α13 , t− 1 2 log⟨t⟩) η−α1⟨η⟩−β1ϕ L∞(R+) +C max ( t−1+2α6 , t− 1 2 log⟨t⟩) η−α⟨η⟩−β∂ηϕ L2(R +)

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§3. Estimates in the L2 - norm

3.1. Estimate for derivatives of Q

First we note that ∥Qϕ∥L2(R +) = Ct 1 2 ∫ R e−it η |η|Λ′(η)ξM ϕ (η) Λ′′(η) 12 dη L2(R +) = Ct12 ∫ R e−itxξM ϕ (η (x)) Λ′′(η (x)) − 1 2 dx L2(R +) ≤ C ∫ R |ϕ (η (x))|2 Λ′′(η (x)) −1dx = C∥ϕ∥L2(R). (3.1)

Also we have the relation

iξQϕ = QAϕ, where A = M t|Λ′′(η)|12 ∂η M |Λ′′(η)|1 2 =A0+ iηθ (η) with A0 = 1 t|Λ′′(η)|12∂η 1

|Λ′′(η)|12. In the next lemma we obtain the estimates through the operatorA0.

Lemma 3.1. The estimates

⟨ξ⟩−4t∂tQ (t) ϕ L2(R +) ≤ C 4 ∑ l=0 tA0ηlϕ L2(R +) + Ct Λ′ϕ L2(R −). and ⟨ξ⟩−4t∂tQ (t) ϕ L2(R +) ≤ tA2 0ϕ L2(R +)+∥ηtA0ϕ∥L2(R+) + 4 ∑ l=1 tA0ηlϕ L2(R +) + Ct Λ′ϕ L2(R −) are true.

Proof. By a direct calculation we find

t∂tQϕ = (itΛ (ξ) Qϕ − itQΛθϕ)

+(itQηΛ′θϕ− itξQΛ′θϕ)+ itξQ (1 − θ) Λ′ϕ. Applying the commutator

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we obtain

itQηΛ′θϕ− itξQΛ′θϕ =−QtA

0Λ′θϕ.

By the relation A = A0+ iηθ (η) we get

(iξ)nQϕ − Q (iη)nθϕ = n−1 ∑ l=0 (iξ)lQA0(iη)n−1−lθϕ. Therefore itΛ (ξ)Qϕ − itQΛθϕ = ita 3 2 ∑ l=0 (iξ)lQA0(iη)2−lθϕ +itb 5 4 ∑ l=0 (iξ)lQA0(iη)4−lθϕ.

Then by inequality (3.1) we obtain the first estimate of the lemma. The second estimate is proved in the same way. Lemma 3.1 is proved.

3.2. Auxiliary estimates in L2

In the next two lemmas we estimate some integrals to prove the estimate of the derivative of Vϕ. First, we consider the integral

I1ϕ = Λ′′(η) 1 2 t12 ∫ ∞ 0 e−itS(ξ,η)ϕ (ξ) ψ (ξ, η) dξ.

Lemma 3.2. Let β > α > 0, δ = 0 or β > α = 0, δ > 0. Suppose that

|η|−α⟨η⟩β⟨ξ⟩δ

(η∂η)kψ (ξ, η) ≤ C for all k = 0, 1, 2, η∈ R, ξ > 0. Then the estimate

∥I1ϕ∥L2(R)≤ C ∥ϕ∥L2(R +)

{

1 for α > 0 log⟨t⟩ for α = 0 is true for all t≥ 1.

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Proof. We have ∥I1ϕ∥2L2(R)= Ct ∫ R dη Λ′′(η) ∫ ∞ 0 eitS(ξ,η)ϕ (ξ)ψ (ξ, η)dξ × ∫ ∞ 0 e−itS(ζ,η)ϕ (ζ) ψ (ζ, η) dζ = Ct ∫ ∞ 0 eitΛ(ξ)ϕ (ξ)dξ ∫ ∞ 0 e−itΛ(ζ)ϕ (ζ) dζ × ∫ R dη Λ′′(η) ψ (ξ, η)ψ (ζ, η) e−it η |η|Λ′(η)(ξ−ζ) = C ∫ ∞ 0 dξeitΛ(ξ)ϕ (ξ) ∫ ∞ 0 dζe−itΛ(ζ)ϕ (ζ) K (t, ξ, ζ) , where K (t, ξ, ζ) = t ∫ R e−it η |η|Λ′(η)(ξ−ζ) Λ′′(η) ψ (ξ, η)ψ (ζ, η) dη.

We integrate two times by parts via the identity e−it

η

|η|Λ′(η)(ξ−ζ) = H6∂η(ηe−it|η|ηΛ′(η)(ξ−ζ))

with H6= (1− itη |Λ′′(η)| (ξ − ζ))−1 to get

K (t, ξ, ζ) = t ∫ R e−it η |η|Λ′(η)(ξ−ζ)η∂η(H6η∂η(H6 Λ′′(η) ψ (ξ, η)ψ (ζ, η)))dη.

Then using the estimate η∂η ( H6η∂η ( H6 Λ′′(η) ψ (ξ, η)ψ (ζ, η))) ≤ C|η|1+2α⟨η⟩2−2β⟨ξ⟩−δ⟨ζ⟩−δ (1 + t|Λ′(η)| |ξ − ζ|)2 , we get |K (t, ξ, ζ)| ≤ Ct ∫ R η∂η ( H6η∂η ( H6 Λ′′(η) ψ (ξ, η)ψ (ζ, η))) dη ≤ Ct ⟨ξ − ζ⟩−δ (∫ 1 0 η1+2αdη (1 + tη2|ξ − ζ|)2 + ∫ ∞ 1 η3+2α−2βdη (1 + tη4|ξ − ζ|)2 ) ≤ Ct ⟨ξ − ζ⟩−δ|(ξ − ζ) t|−1−α ∫ (t|ξ−ζ|)12 0 y1+2αdy (1 + y2)2 + Ct|(ξ − ζ) t|β−α2 −1 ∫ ∞ (t|ξ−ζ|)14 y3+2α−2βdy (1 + y4)2 ≤ Ct ⟨ξ − ζ⟩−δ⟨(ξ − ζ) t⟩−1−α + Ct ( |(ξ − ζ) t|β−α2 −1+ 1 ) ⟨(ξ − ζ) t⟩−β−α2 −1

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if β > α≥ 0. Then by the Young inequality for convolutions we obtain ∥I1ϕ∥2L2(R)≤ C ∥ϕ∥L2(R +) ∫ ∞ 0 |K (t, ξ, ζ)| |ϕ (ζ)| dζ L2 ≤ C ∥ϕ∥L2(R +) ∫ ∞ 0 t⟨ξ − ζ⟩−δ⟨(ξ − ζ) t⟩−1−α|ϕ (ζ)| dζ L2 + C∥ϕ∥L2(R +) ∫ ∞ 0 tβ−α2 |ξ − ζ| β−α 2 −1 ( |(ξ − ζ) t|β−α2 −1+ 1 ) |ϕ (ζ)| dζ L2 ≤ C ∥ϕ∥2 L2(R +) ( t ⟨ξt⟩−1−α⟨ξ⟩−δ L1+ t(|ξt|β−α2 −1+ 1 ) ⟨ξt⟩−β−α2 −1 L1 ) ≤ C ∥ϕ∥2 L2(R +) { 1 for α > 0 log⟨t⟩ for α = 0 , since in the case of α > 0, δ = 0 we have

t ⟨ξt⟩−1−α⟨ξ⟩−δ L1 ≤ C ∫ t 0 dξ (1 + ξ)1+α + C ∫ ∞ 1 ξ−1−αdξ≤ C, and in the case of α = 0, δ > 0 we find

t ⟨ξt⟩−1−α⟨ξ⟩−δ L1 ≤ C ∫ t 0 dξ 1 + ξ + C ∫ ∞ 1 ξ−1−δdξ≤ C log ⟨t⟩ . Lemma 3.2 is proved.

In the next lemma we prove another estimate of the integral I2ϕ = t 1 2 ∫ ∞ 0 e−itS(ξ,η)ϕ (ξ) ψ (ξ, η) dξ in L2.

Lemma 3.3. Let k = 0, 1, 2, k− 1 ≤ α < k + 1 − β. Suppose that

|ψ (ξ, η)| + |(ξ − ηθ (η)) ∂ξψ (ξ, η)| ≤ |ξ + |η||−k|η|α⟨η⟩β for ξ > 0, η∈ R. Suppose that ϕ (0) = 0. Then the estimate

∥I2ϕ∥L2(R)≤ C max ( 1, t−16− α−k 3 ) ∥⟨ξ⟩ ϕξ∥L2(R +) is true for all t≥ 1.

Proof. Consider η > 0. Using identity (2.1) we integrate by parts I2ϕ = −t 1 2 ∫ ∞ 0 e−itS(ξ,η)ϕ (ξ) (ξ− η) ∂ξ(H1ψ (ξ, η)) dξ −t12 ∫ ∞ 0 e−itS(ξ,η)(ξ− η) H1ψ (ξ, η) ϕξ(ξ) dξ.

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By the estimates |H1| ≤ C ( 1 + t (ξ− η)2(η + ξ)(1 + η2+ ξ2))−1, and |(ξ − η) ∂ξ(H1ψ (ξ, η))| ≤ C|ξ + |η||−k|η|α⟨η⟩β 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) we obtain |I2ϕ| ≤ Ct 1 2 ∫ ∞ 0 |ϕ (ξ) − ϕ (0)| |ξ + |η||−k|η|α⟨η⟩β dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) +Ct12 ∫ ∞ 0 |ξ − η| |ξ + |η||−k|η|α⟨η⟩β|∂ ξϕ| dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2). Since |ϕ (ξ) − ϕ (0)| ≤ Cξ12 ⟨ξ⟩− 1 2∥⟨ξ⟩ ϕ ξ∥L2(R +) we get |I2ϕ| ≤ Ct 1 2 ∥⟨ξ⟩ ϕξ∥ L2(R +) ∫ ∞ 0 |ξ + |η||−kξ12 ⟨ξ⟩− 1 2 |η|α⟨η⟩βdξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) +Ct12 ∫ ∞ 0 |ξ − η| |ξ + |η||−k|η|α⟨η⟩β|∂ ξϕ| dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2). (3.2)

We estimate the first summand in (3.2) as ∫ ∞ 0 |ξ + |η||−kξ12⟨ξ⟩− 1 2|η|α⟨η⟩βdξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) ≤ C ∫ 2η 0 |η|α−k+12⟨η⟩βdξ 1 + tη⟨η⟩2(ξ− η)2 + C|η| α⟨η⟩β ∫ ∞ η ξ12−k⟨ξ⟩− 1 2 dξ 1 + tΛ (ξ) . (3.3) Then we find ∫ 2η 0 |η|α−k+1 2 ⟨η⟩βdξ 1 + tη⟨η⟩2(ξ− η)2 ≤ C ∫ 2 0 |η|α−k+3 2 ⟨η⟩βdy 1 + tΛ (η) (y− 1)2 ≤ C |η|α−k+32⟨η⟩β⟨tΛ (η)⟩−12 and t12 |η|α−k+ 3 2⟨η⟩β⟨tΛ (η)⟩− 1 2 L2(R +) ≤ Ct1 2 (∫ 1 0 ⟨ t13η ⟩−3 η2α−2k+3dη )1 2 +C (∫ ∞ 1 η2α+2β−2k−2dη )1 2 ≤ C max(1, t−16− α−k 3 ) (3.4)

if α > k− 2, α + β − k < 1. Also the second term of the right-hand side of (3.3) is estimated as |η|α⟨η⟩β ∫ ∞ η ξ12−k⟨ξ⟩− 1 2 dξ 1 + tΛ (ξ) ≤ Ct −1ηα+β ∫ ∞ η ξ−k−5dξ≤ Ct−1⟨η⟩α+β−k−4

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for η > 1, t≥ 1, and |η|α⟨η⟩β ∫ ∞ η ξ12−k⟨ξ⟩− 1 2 dξ 1 + tΛ (ξ) ≤ C ∫ 1 η ξα−k+12dξ 1 + tξ3 + Ct−1 ∫ ∞ 1 ξ−k−5dξ ≤ Ct−12− α−k 3 ∫ t13 ηt13 ξα−k+12dξ 1 + ξ3 + Ct−1 ≤ Ct−12− α−k 3 max ( 1, ⟨ t13η ⟩α−k−32) + Ct−1log t for 0 < η≤ 1. Hence |η|α⟨η⟩β ∫ ∞ η ξ12−k⟨ξ⟩− 1 2 dξ 1 + tΛ (ξ) L2(R +) ≤ Ct−1 2− α−k 3 (∫ 1 0 max ( 1, ⟨ t13η ⟩2α−2k−3) dη )1 2 + Ct−1log t +Ct−1 (∫ ∞ 1 ⟨η⟩2α+2β−2k−8dη )1 2 ≤ C max(t−1log t, t−23−α−k3 ) (3.5) for α + β− k < 72. Thus by (3.3)-(3.5) (3.6) t12 ∫ ∞ 0 |ξ + |η||−kξ12⟨ξ⟩− 1 2 |η|α⟨η⟩βdξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) L2(R +) ≤ C max(1, t−16− α−k 3 ) if α > k− 2, α + β − k < 1.

Consider the second summand in (3.2) for 0 < η < 1 ∫ ∞ 0 |ξ − η| |ξ + |η||−k|η|α⟨η⟩β|∂ ξϕ| dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) ≤ C ∫ 2η 0 |ξ − η|α+1−k|∂ ξϕ| dξ 1 + t|η − ξ|3 + C∥⟨ξ⟩ ϕξ∥L2(R +) (∫ ∞ 2η ξ2α+2−2kdξ (1 + tΛ (ξ))2 )1 2 . (3.7)

if α + 1− k ≥ 0. By the Young inequality ∫ 2η 0 |ξ − η|α+1−k|∂ ξϕ| dξ 1 + t|η − ξ|3 L2 ≤ C ∥ϕξ∥L2(R +) |η|α+1−k⟨t1 3η ⟩−3 L1(R) ≤ C max(t−1log t, t−α−k+23 ) ∥ϕξ∥L2(R +). (3.8)

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Let us consider the second term of the right hand side of (3.7). We have ∫ ∞ 2η ξ2α+2−2kdξ (1 + tΛ (ξ))2 ≤ ∫ 2 2η ξ2α+2−2kdξ (1 + tξ3)2 + Ct −2∫ ∞ 2 ξ2α−2k−8dξ ≤ Ct−2α−2k+33 ∫ 2t13 2ηt13 ξ2α+2−2kdξ (1 + ξ3)2 + Ct −2 ≤ Ct−1−2α−2k3 max ( 1, ⟨ t13η ⟩2α−2k−3) + Ct−2 if α− k < 72 and then (∫ ∞ 2η ξ2α+2−2kdξ (1 + tΛ (ξ))2 )1 2 L2(0,1) ≤ Ct−1 2− α−k 3 (∫ 1 0 max ( 1, ⟨ t13η ⟩2α−2k−3) dη )1 2 + Ct−1 ≤ C max(t−12, t− 2 3− α−k 3 ) . (3.9) By (3.7)-(3.9) t12 ∫ ∞ 0 |ξ − η| |ξ + |η||−k|η|α⟨η⟩β|∂ ξϕ| dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) L2(0,1) ≤ C ∥ϕξ∥L2(R +)max ( 1, t−16− α−k 3 ) . (3.10) For η > 1 we have ∫ ∞ 0 |ξ − η| |ξ + |η||−k|η|α⟨η⟩β|∂ ξϕ| dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) ≤ Cηα+β−k ∫ 2η 0 |ξ − η| |∂ξϕ| dξ 1 + tη3(ξ− η)2 + Ct −1ηα+β ∫ ∞ η ξ−k−4|∂ξϕ| dξ ≤ C ∥ϕξ∥L2(R +)η α+β−k    ∫ 2η 0 (ξ− η)2dξ ( 1 + tη3(ξ− η)2)2    1 2 +Ct−1∥ϕξ∥L2(R +)η α+β (∫ ∞ η ξ−2k−8dξ )1 2 ≤ C ∥ϕξ∥L2(R +)t −3 4ηα+β−k− 9 4,

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and then ∫ ∞ 0 |ξ − η| |ξ + |η||−k|η|α⟨η⟩β|∂ ξϕ| dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) L2(1,∞) ≤ Ct−34 ∥ϕξ∥ L2(R +) ηα+β−k−94 L2(1,∞)≤ Ct −3 4 ∥ϕξ∥ L2(R +) (3.11) if α + β− k < 74. Therefore by (3.10) and (3.11) t12 ∫ ∞ 0 |ξ − η| |ξ + |η||−k|η|α⟨η⟩β|∂ ξϕ| dξ 1 + t (ξ− η)2(η + ξ) (1 + η2+ ξ2) L2(R +) ≤ C max(1, t−16− α−k 3 ) ∥⟨ξ⟩ ϕξ∥L2(R +). (3.12) By (3.6) and (3.12) ∥I2ϕ∥L2(R +)≤ C max ( 1, t−16− α−k 3 ) ∥⟨ξ⟩ ϕξ∥L2(R +).

To estimate I2ϕ for η < 0, we integrate by parts via the identity (2.10)

I2ϕ =−t 1 2 ∫ ∞ 0 e−itS(ξ,η)(ϕ (ξ)− ϕ (0)) ξ∂ξ(H4ψ (ξ, η)) dξ −t1 2 ∫ ∞ 0 e−itS(ξ,η)ξH4ψ (ξ, η) ϕξ(ξ) dξ.

Using the estimates

|H4| ≤ C ( 1 + tξ ( ξ2⟨ξ⟩2+ η2⟨η⟩2 ))−1 ,|ψ (ξ, η)| ≤ C |ξ + |η||−k|η|α⟨η⟩β and |ξ∂ξ(H4ψ (ξ, η))| ≤ C |ξ + |η||−k|η|α⟨η⟩β ( 1 + tξ ( ξ2⟨ξ⟩2+ η2⟨η⟩2 ))−1 ,

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