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Advances in Mathematical Physics Volume 2010, Article ID 504324,35pages doi:10.1155/2010/504324

Research Article

The Initial Value Problem for the Quadratic Nonlinear Klein-Gordon Equation

Nakao Hayashi

1

and Pavel I. Naumkin

2

1Department of Mathematics, Graduate School of Science, Osaka University, Osaka, Toyonaka 560-0043, Japan

2Instituto de Matem´aticas, UNAM Campus Morelia, AP 61-3 (Xangari), Morelia CP 58089, Mexico

Correspondence should be addressed to Nakao Hayashi,nhayashi@math.sci.osaka-u.ac.jp Received 18 September 2009; Accepted 23 February 2010

Academic Editor: Dongho Chae

Copyrightq2010 N. Hayashi and P. I. Naumkin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We study the initial value problem for the quadratic nonlinear Klein-Gordon equationLu i∂x−1u2,t, x∈R×R,u0, x u0x,xR, whereLtii∂xandi∂x1−2x. Using the Shatah normal forms method, we obtain a sharp asymptotic behavior of small solutions without the condition of a compact support on the initial data which was assumed in the previous works.

1. Introduction

Let us consider the Cauchy problem for the nonlinear Klein-Gordon equation with a quadratic nonlinearity in one dimensional case

Luλi∂x−1u2, t, x∈R×R,

u0, x u0x, xR, 1.1

whereλC,Ltii∂x, andi∂x 1−2x.

Our purpose is to obtain the large time asymptotic profile of small solutions to the Cauchy problem1.1without the restriction of a compact support on the initial data which was assumed in the previous work1. One of the important tools of paper1was based on the transformation of the equation by virtue of the hyperbolic polar coordinates following to paper2. The application of the hyperbolic polar coordinates implies the restriction to the interior of the light cone, and therefore, requires the compactness of the initial data. Problem

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1.1is related to the Cauchy problem

vttvvxxμv2, t, x∈R×R,

v0, x v0x, vt0, x v1x, xR, 1.2

wherev0andv1are the real-valued functions, andμR.Indeed we can putu 1/2v ii∂x−1vt,thenusatisfies

Lu i

2μi∂x−1uu2, t, x∈R×R, u0, x u0x, xR,

1.3

whereu0 1/2v0ii∂x−1v1.

There are a lot of works devoted to the study of the cubic nonlinear Klein-Gordon equation

vttvvxxμv3, t, x∈R×R,

v0, x v0x, vt0, x v1x, xR 1.4

withμR.Whenμ < 0, the global existence of solutions to1.4can be easily obtained in the energy space, which is, however, insufficient for determining the large-time asymptotic behavior of solutions. The sharpL- time decay estimates of solutions and nonexistence of the usual scattering states for1.4were shown in3by using hyperbolic polar coordinates under the conditions that the initial data are sufficiently regular and have a compact support.

The initial value problem for the nonlinear Klein-Gordon equation with various cubic nonlinearities depending onv, vt, vx, vxx, vtx and having a suitable nonresonance structure was studied in 4–6, where small solutions were found in the neighborhood of the free solutions when the initial data are small and regular and decay rapidly at infinity. Hence the cubic nonlinearities are not necessarily critical; however the resonant nonlinear termv3 was excluded in these works. In paper4, the nonresonant nonlinearities were classified into two types, one of them can be treated by the nonlinear transformation which is different from the method of normal forms7and the other reveals an additional time decay rate via the operatorx∂tt∂xwhich was used in2. This nonlinear transformation was refined in8and applied to a system of nonlinear Klein-Gordon equations in one or two space dimensions with nonresonant nonlinearities. It seems that the method of normal forms is very useful in the case of a single equation; however it does not work well in the case of a system of nonlinear Klein-Gordon equations. Some sufficient conditions on quadratic or cubic nonlinearities were given in1, which allow us to prove global existence and to find sharp asymptotics of small solutions to the Cauchy problem including1.2with small and regular initial data having a compact support. Moreover it was proved that the asymptotic profile differs from that of the linear Klein-Gordon equation. See also9,10in which asymptotic behavior of solutions to1.4was studied as in1by using hyperbolic polar coordinates. Compactness condition on the data was removed in11in the case of the cubic nonlinearityv3 and a real-valued solution. Final value problem with the cubic nonlinearity was studied in12for a real-valued solution. As far as we know the problem of finding the large-time asymptotics is still open

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for the case of the cubic nonlinearityv3and the complex valued initial data. When the initial data are complex-valued, global existence andL-time decay estimates of small solutions to the Klein-Gordon equation with cubic nonlinearity|v|2vwere obtained in paper13under the conditions that the initial data are smooth and have a compact support.

The scattering problem and the time decay rates of small solutions to1.4with super- critical nonlinearities|v|p−1vand|v|pwithp > 3 were studied in papers of14,15. Finally, we note that the Klein-Gordon equation 1.4 with quadratic nonlinearities in two space dimensions was studied in 16, where combining the method of the normal forms of7 and the time decay estimate through the operatorx∂tt∂xof17, it was shown that every quadratic nonlinearity is nonresonant.

We denote the Lebesgue space byLp {φ ∈ SLp < ∞}, with the normφLp

R|φx|pdx1/pif 1≤p <∞andφL supx∈R|φx|ifp∞.The weighted Sobolev space is

Hm,sp

φLp;xsi∂xmφLp<

, 1.5

form, sR, 1p ≤ ∞,wherex √

1x2.For simplicity we writeHm,s Hm,s2 . The index 0 we usually omit if it does not cause a confusion. We denote by Fφ Fx→ξφφ 1/√

Re−ixξφxdxthe Fourier transform of the function φ.Then the inverse Fourier transformation isF−1φF−1ξxφ 1/√

Reixξφξdξ.

Our main result of this paper is the following.

Theorem 1.1. Letu0H3,1and the norm u0H3,1 ε. Then there existsε0 > 0 such that for all 0< ε < ε0the Cauchy problem1.1has a unique global solution

utC

0,∞;H3,1 1.6

satisfying the time decay estimate

utH1,0Cε1t−1/2. 1.7

Furthermore there exists a unique final stateWH0,1H0,1such that ute−ii∂xtF−1We−2i|λ|2Ω|W|2logt

H1,03/2tγ−1/4, Feii∂xtutWe−2i|λ|2Ω|W|2logt

H0,1

3/2tγ−1/4,

1.8

whereγ∈0,1/4,Ωξ≡ ξ2/2ξ2ξ2ξ.

An important tool for obtaining the time decay estimates of solutions to the nonlinear Klein-Gordon equation is implementation of the operator

Ji∂xe−ii∂xtxeii∂xtF−1ξe−iξti∂ξeiξtFi∂xxit∂x, 1.9

(4)

which is analogous to the operatorxit∂x e−it/2∂2xxeit/2∂2x in the case of the nonlinear Schr ¨odinger equations used in 18. The operator J was used previously in paper of 15 for constructing the scattering operator for nonlinear Klein-Gordon equations with a supercritical nonlinearity. We have x,i∂xα αi∂xα−2x; therefore the commutator L,J LJ − JL0, whereLtii∂xis a linear part of1.1. SinceJis not a purely differential operator, it is apparently difficult to calculate the action ofJon the nonlinearity in1.1. So, instead we use the first-order differential operator

Pt∂xx∂t 1.10

which is closely related toJby the identityPLx−iJand acts easily on the nonlinearity.

Moreover, it almost commutes withL, sinceL,P −ii∂x−1xL.

Also we use the method of normal forms of7by which we transform the quadratic nonlinearity into a cubic one with a nonlocal operator. We multiply both sides of equation 1.1by the free Klein-Gordon evolution groupFU−t Feiti∂x eitξFand putvt, ξ eitξuto get

vtt, ξ λeitξF

i∂x−1u2 λ

√2πξ

ReitAv t, ηξ

v t,−η

dη, 1.11

whereAξ, η ξξ−ηη.Integrating1.11with respect to time, we find

vt, ξ v0, ξ λ

√2πξ t

0

ReiτAv

τ, ηξ v

τ,−η

dη. 1.12

Then we integrate by parts with respect toτ,taking into account1.11,

vt, ξ iλ

ReitAv t, ηξ

v

t,−η

√2πξA ξ, η v0, ξ iλ

Rv

0, η−ξ v

0,−η

√2πξA ξ, η

−2i|λ|2 t

0

R

2πξA

ξ, ηeiτA−ηv

τ, ηξ Fx→η

i∂x−1u2 .

1.13

Returning to the functionut, x UtF−1ξ→xvF−1ξxe−itξvt, ξ, we obtain the following equation:

LuiλGu, u −2i|λ|2G

u,i∂x−1u2 , 1.14

with the symmetric bilinear operator

G φ, ψ

F−1ξ→x

Rg

ξη, η φ ξη

ψ

η

dη, 1.15

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where

g

ζ, η

1

√2π

ζη ζη

ζ

η, 1.16

andζξη.Our main point in this paper is to show that the right-hand side of1.14can be decomposed into two terms; one of them is a cubic nonlinearity

−2i|λ|21

tUtF−1Ωξ|FU−tut, ξ|2FU−tut, ξ, 1.17 and the other one is a remainder term with an estimate likeOt−5/4FU−tut3H1,3. Remark 1.2. We believe that all quadratic nonlinear terms u2, u2,|u|2 of problem 1.3also could be considered by this approach. In the same way as in the derivation of1.14we get from1.3

L

u i

2μGu, u G1u, u 2G2u, u

2G

u,i∂x−1uu2 2G1

u,i∂x−1uu2 2G2

uu,i∂x−1uu2 ,

1.18

where

Gj

φ, ψ

Fξ−1→x

Rgj

ξη, η φ ξη

ψ

η dη,

gj ζ, η

1

√2πAj

ζη, η ζη

1.19

withA1ξ, η ξ−ξ−η−η, A2ξ, η ξ−ξ−ηη.Some more regularity conditions are necessary to treat the bilinear operatorsGj. Also we have to show that

G

u,i∂x−1

u22|u|2

, G1

u,i∂x−1

u2u2 , G2

u,i∂x−1

u2u2

, G2

u,i∂x−1

u22|u|2 1.20

are the nonresonant termsi.e., remaindersand to remove the resonant terms G

u,i∂x−1u2 , G1

u,i∂x−1|u|2 , G2

u,i∂x−1|u|2 , G2

u,i∂x−1u2 1.21

by an appropriate phase function. We will dedicate a separate paper to this problem.

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We prove our main result inSection 3. In the next section we prove several lemmas used in the proof of the main result.

2. Preliminaries

First we give some estimates for the symmetric bilinear operator

G φ, ψ

F−1ξ→x

Rg

ξη, η φ ξη

ψ

η

dη, 2.1

where

g

ζ, η

1

√2π

ζη ζη

ζ

η. 2.2

Denote the kernel as follows:

g y, z

1 4π2

R2

eiyζizη ζη

ζ

η

ζηdηdζ. 2.3

Lemma 2.1. The representation is true G

φ, ψ

R2g y, z

φ xy

ψxzdydz, 2.4

where the kernelgy, zobeys the following estimate:

g

y, zC y−3

z−3ln2

1 1

zy

2.5

for ally, zR, y /z.Moreover the following estimates are valid:

G φ, ψ

Lp

Lqψ

Lr, PG

φ, ψ

LpC

Lqψ

Lr C

Lqφ

Lr

C∂tφ

Lqψ

Lr C∂tψ

Lqφ

Lr

2.6

for 1p ≤ ∞,1 < qp/α,1 < rp/1α, α ∈0,1, provided that the right-hand sides are bounded.

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Proof. To prove representation2.4, we substitute the direct Fourier transforms

φζ 1

√2π

Re−ix−yζφ xy

dy,

ψ

η 1

√2π

Re−ix−zηψxzdz

2.7

into the definition of the operatorG. Then changingζξη, we find

G φ, ψ

−3/2

R2φ xy

ψxz

R2g ζ, η

eiyζizηdηdζ

dydz

R2g y, z

φ xy

ψxzdydz,

2.8

where the kernelgy, zis

g y, z

−3/2

R2g ζ, η

eiyζizηdηdζ

1 4π2

R2

eiyζizη ζη

ζ

η

ζηdηdζ.

2.9

Changing the variables of integrationζξ/2ηandηξ/2ηthe prime we will omit, we get

g y, z

1 4π2

R2

eiyζizη ζη

ζ

η

ζηdηdζ 1

2

R

ξeiyz/2ξ

RB ξ, η

eiz−yηdη,

2.10

whereBξ, η 1/ξξ/2−ηξ/2η.We changesyz/2 andρ z−yand denote

g

s, ρ 1

2

R

ξeisξ

RB ξ, η

eiρηdη. 2.11

For the case of|ρ| ≤ 1,|s| ≤1 we integrate by parts using the identityeiρη Aη∂ηηeiρη, where1/1iρη.Then we get

g

s, ρ − 1

2

R

ξeisξ

Reiρηη∂η

A η

B ξ, η

dη. 2.12

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Note that

η∂ηB ξ, η

η ξ

ξ/2η

ξ/2η2

ηξ/2

ηξ/2 ηξ/2 ξ/2η

. 2.13

Then

ξ−1η∂η

A η

B

ξ, ηC

ξ

1ρηξ

η. 2.14

Hence we can estimate the kernelgs, ρ as follows:

g

s, ρC

R

ξ

R

1ρηξ η

C

1

1ρη

1

ξ

ηξ C

1

1ρη

1 η

ln ξ

ηξ

1

C

1

lnη 1ρη

ηdηC 1/|ρ|

1

lnηdη

η C

ρ

1/|ρ| lnηdη η2

Cln2

1 1 ρ

2.15

for the case of|ρ| ≤1,|s| ≤1.For the case of|ρ| ≤1,|s| ≥1 we integrate three times by parts with respect toξ

g

s, ρ

C|s|−3

Rdξeisξ

Reiρη3ξ

ξ−1η∂η

A η

B

ξ, η dη. 2.16

Note that

3ξ

ξ−1η∂η A

η B

ξ, ηC

ξ

1ρηξ

η. 2.17

Hence we can estimate the kernelgy, zas follows:

g

s, ρC|s|−3

R

ξ

R

1ρηξ

ηC|s|−3ln2

1 1 ρ

2.18

for all|ρ| ≤1,|s| ≥1.For the case|ρ| ≥1,|s| ≤1 we integrate by parts three times with respect toη

g

s, ρ

−3

R

ξeisξ

Reiρη3ηB ξ, η

dη. 2.19

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Note that|∂3ηBξ, η| ≤Cξη−3.Hence

g

s, ρ−3

R

ξ2

R

η2−3. 2.20

Finally for the case of|ρ| ≥1,|s| ≥1 we integrate by parts three times with respect toξandη

g

s, ρ

Csρ−3

Rdξeisξ

Reiρη3ξ3η ξ−1B

ξ, η dη. 2.21

Since

3ξ3η ξ−1B

ξ, ηC

ξ2 ξ

η2, 2.22

then we can estimate the kernelgs, ρ as follows:

g

s, ρCsρ−3

R

ξ2

R

η2Csρ−3 2.23

for all|ρ| ≥1,|s| ≥1.Hence estimate2.5is true.

By virtue of estimate2.5applying the H ¨older inequality with 1/p1/p11/p2and the Young inequality with 1/p111/q1/q1and 1/p211/r1/r1, we find

G φ, ψ

LpC

R

φ

xy dy y3

Lpx1

Rln2

1 1

zy

ψx−z dz z3

Lpx2Ly

LqψLr,

2.24

where 1≤p≤ ∞,1< qp/α,1< rp/1α, α∈0,1.

We now estimate the operatorPt∂xx∂tas follows:

PG φ, ψ

G Pφ, ψ

R2yg y, z

ψxz∂tφ xy

dydz

G φ,

R2zg y, z

φ xy

tψx−zdydz.

2.25

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Then by virtue of estimate2.4applying the H ¨older and Young inequalities, we get PG

φ, ψ

Lp ≤G

Pφ, ψ

LpG

φ,

Lp

R2yg y, z

ψxz∂tφ xy

dydz Lp

R2zg y, z

φ xy

tψxzdydz Lp

C

Lqψ

LrC

Lqφ

LrC∂tφ

Lqψ

Lr

C∂tψ

Lqφ

Lr.

2.26

Lemma 2.1is proved.

We now decompose the free Klein-Gordon evolution groupUt e−ii∂xtF−1EtF, whereEt e−itξsimilarly to the factorization of the free Schr ¨odinger evolution group. We denote the dilation operator by

Dωφ 1

iωφx

ω , Dω−1iD1/ω. 2.27

Define the multiplication factorMt e−itixθx,whereθx 1 for|x|<1 andθx 0 for

|x| ≥1.We introduce the operator

θx ix3/2φ

x ix

. 2.28

The inverse operatorB−1acts on the functionsφxdefined on−1,1as follows:

B−1φ 1 ξ3/2φ

ξ ξ

2.29

for allξR,sinceξx/ix ∈R andxξ/ξ ∈−1,1.We now introduce the operators Vt B−1MtD−1t F−1e−itξ,

Wt Mt1θD−1t F−1e−itξ,

2.30

so that we have the representation for the free Klein-Gordon evolution group UtF−1e−iti∂xF−1F−1e−itξDtMtBVt Wt

DtMtBDtMtBVt−1 DtMtWt. 2.31

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The first term DtMtBφ of the right-hand side of 2.31 describes inside the light cone the well-known leading term of the large-time asymptotics of solutions of the linear Klein- Gordon equation Lu 0 with initial data φ. The second term of the right-hand side of 2.31is a remainder inside of the light cone, whereas the last term represents the large time asymptotics outside of the light cone which decays more rapidly in time. We also have

FU−t Feiti∂xeitξFV−1tB−1MtD−1t W−1tDt−1 B−1MtDt−1

V−1t−1 B−1MtD−1t W−1tDt−1,

2.32

where the right-inverse operators are

V−1t EtFDtMtB,

W−1t EtFDt1−θ, 2.33

whereEt e−itξ.

In the next lemma we state the estimates of the operatorsVt B−1MtD−1t F−1e−itξ. Lemma 2.2. The estimates hold as follows:

Vtφ

L2

L2, ξρξVtφ

L2Cξαξφ

L2Cξα−1φ

L2, 2.34

whereρ < α, α∈1,2,and

ξρξV−1

L2Cξ1/4αξφ

L2Cξα−3/4φ

L2, 2.35

whereρ < α, α∈0,1,provided the right-hand sides are finite.

Proof. Changing the variable of integrationxξξ−1, we see that B−1φ2

L2

R

φ ξ

ξ 2

ξ3 1

−1

φx2dxφ2

L2−1,1. 2.36

HenceVtφL2 φL2.

Consider the estimate for the derivativeξVtφ. DefineSξ, η η −ξη1/ξ.

Note that∂S/∂ηη/η −ξ/ξ 1ηξ −ηξ/ξηηξη−ξand∂S/∂ξ ξ−η/ξ3.Hence integrating by parts one time yields

ξ

Re−itSφ η

−itξ−3

Re−itSφ η

ξη

ξ−3

Re−itS φ

η g

ξ, η φ

η gη

ξ, η dη,

2.37

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wheregξ, η ξη/ξ/ξ −η/η ξηηξ/1ηξ −ηξ. Also we have 1

η

ξ −ηξ−1

C ξ η η

ξ2. 2.38

Hence the estimate is true

g ξ, η

ξ

η

η ξ 1

η

ξ −ηξCξ2 η2 η

ξ. 2.39

Sinceξξaaξξa−2, we get

ξρξVtφξρ

it

ξξ−3/2

Re−itSφ η

−3

2ξξρ−2Vtφξρ−9/2

it

Re−itSφ η

g ξ, η

ξρ−9/2

it

Re−itSφ η

gη ξ, η

dη.

2.40

Consider theL2-estimate of the integral

ρ−9/2

Reitξη/ξψ η

g ξ, η

dη. 2.41

We have

t

Reitξη/ξψ η

g ξ, η

ξρ−9/2 2

L2

t

Rdηψ η

Rdζψζ

Reitξ/ξη−ζg ξ, η

gξ, ζξ2ρ−9

Rdηψ η

RdζψζG η, ζ

,

2.42

where the kernel

G η, ζ

t

Reitξ/ξη−ζg ξ, η

gξ, ζξ2ρ−9

t η

ζ

Reitξ/ξη−ζ

η ξ 1

η

ξ −ηξ ζξ

1ζξ −ζξξ2ρ−7dξ.

2.43

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After a changeξx/ix, we find

G η, ζ

t η

ζ 1

−1eitxη−ζ

η

1/ix 1

η

/ix −ηx/ix ζ1/ix

1ζ/ix −ζx/ixix4−2ρdx.

2.44

We now change the contourzxiy∈Γof integration inGasysignη−ζix2; then

G η, ζ

t η

ζ

Γeitzη−ζ

η

1/iz 1

η

/iz −ηz/iz ζ1/iz

1ζ/iz −ζz/iziz4−2ρdz.

2.45

Since|iz| ix2y224x2y21/4 ∼ ix, using2.38and the inequalityηξ ≥ ηα−1ξ2−αforα∈1,2,we get|η1/iz/1η/iz −ηz/iz| ≤Cηα−1ixα−2; also we have

eitzη−ζe−ix2t|η−ζ|Cix−2δ

1ζδ 2.46

forδ >1. Therefore we obtain the estimate

G

η, ζCt ηα

ζα 1ζδ

1

−1ix2α−ρ−δdxCt ηα

ζα

1ζδ, 2.47

if we chooseρ < α, α∈1,2and 1< δ <1αρ.Thus we get √

t

Reitξη/ξψ η

g ξ, η

ξρ−9/2 2

L2

Rdηψ η

RdζψζG η, ζ

α ψ

η

L2

Rdζψζζα t

1ζδ

L2

Ct·αψ2

L2

R

1δC·αψ2

L2

2.48

sinceδ >1.

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In the same manner we consider the estimate of the integral √

t

Reitξη/ξψ η

gη

ξ, η

ξρ−9/2 2

L2

t

Rdηψ η

Rdζψζ

Reitξ/ξη−ζgη

ξ, η

gζξ, ζξ2ρ−9

Rdηψ η

RdζψζG η, ζ

,

2.49

where the kernel G

η, ζ t

Reitξ/ξη−ζgη

ξ, η

gζξ, ζξ2ρ−9

t

Reitξ/ξη−ζ

ηξ η/

η ξ 1

η

ξ −ηξ ζξ ζ/ζξ

1ζξ −ζξ ξ2ρ−7dξ,

2.50

since by a direct calculation

η η

η ξ 1

η

ξ −ηξ ηξ η/

η ξ

1 η

ξ −ηξ . 2.51

In the same way as in2.47we have G

η, ζCt ηα−1

ζα−1

1ζδ 2.52

if we chooseρ < α, α∈1,2, and 1< δ <1αρ.Therefore we get √

t

Reitξη/ξψ η

gη ξ, η

ξρ−9/2 2

L2C ηα−1

ψ η2

L2

R

tdη

1δC·α−1ψ2

L2. 2.53

HenceξρξVtφL2αξφL2α−1φL2.Thus we have the second estimate of the lemma.

Consider the estimate for the derivativeξV−1tφ. Note that∂Sη, ξ/∂ξ ξ/ξ − η/η 1ηξ −ηξ/ξηηξξ−ηand∂Sη, ξ/∂η η−ξ/η3.Hence integrating by parts one time yields

ξρξV−1tφ−

t

√2iπξρ−1

R

1 η

ξ −ηξ η

ξ η1/2

φ η

deitSη,ξ

t

√2iπξρ−1

ReitSη,ξ φ

η g1

ξ, η φ

η g

ξ, η dη,

2.54

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whereSη, ξ ξ −ξη1/ηandg1ξ, η 1ηξ −ηξ/ηξη1/2.The estimate is true

g1 ξ, η

1 η

ξ −ηξ η

ξ

η1/2C η3/2 ξ

η

ξ. 2.55

Consider theL2-estimate of the integral

ξρ−1t

Re−itξη/ηψ η

g1 ξ, η

dη. 2.56

We have, changingη/ηyandζ/ζz, ξρ−1

t

Re−itξη/ηψ η

g1 ξ, η

2

L2

t

Rdηψ η

Rdζψζ

Re−itξη/η−ζ/ζg1 ξ, η

g1ξ, ζξ2ρ−2

t 1

−1dy iy3

ψ y

iy 1

−1dziz3ψ z

iz

×

Re−itξy−zg1

ξ, y iy

g1

ξ, z

iz

ξ2ρ−2

1

−1dy iy3

ψ y

iy 1

−1dziz3ψ z

iz

G1

y, z ,

2.57

where the kernel G1

y, z t

Re−itξy−zg1

ξ, y

iy

g1

ξ, z

iz

ξ2ρ−2

t iy−1/2

iz−1/2

Re−itξy−z1ξ/

iy

yξ/

iy 1/

iy ξ

1ξ/iz −zξ/iz

1/izξ ξ2ρ−2dξ.

2.58

We now change the contourξxiy∈Γof integration inGasysigny−zx 2; then G1

y, z t

iy−1/2

iz−1/2

Γe−itξy−z1ξ/

iy

yξ/

iy 1/

iy ξ

1ξ/iz −zξ/iz

1/izξ ξ2ρ−2dξ.

2.59

(16)

Since|ξ| x 2y224x2y21/2x and by2.55|1ξ/iy−yξ/iy/1/iy ξ| ≤1−αiy−αforα∈0,1, and also

e−itξy−zC

1tx yz2C tx δyzδ

yz 2.60

forδ <1,we obtain the estimate G1

y, zC

iy−1/2−α

iz−1/2−α yzδ

yz

Rx −2α2ρ−δdx

C

iy−1/2−α

iz−1/2−α yzδ

yz ,

2.61

if we chooseρ < α,and 1> δ >1−2α2ρ.Therefore we get √

t

Reitξη/ξψ η

g ξ, η

ξρ−9/2 2

L2

1

−1dy iy5/2−α

ψ y

iy 1

−1dziz5/2−αψ z

iz

C yzδ

yz

C 1

−1dy

iy5/2−2α ψ

y iy

2

C 1

−1dξξ1/22αψξ2C·1/4αψ2

L2.

2.62

In the same manner we consider the estimate of the integral withφηgξ, η.Hence ξρξV−1

L2Cξ1/4αξφ

L2Cξα−3/4φ

L2, 2.63

whereα > ρ.Thus we have the second estimate of the lemma.Lemma 2.2is proved.

In the next lemma we prove an auxiliary asymptotics for the integral

0e−itz2Φξ, zdz.

Lemma 2.3. The estimate holds as follows:

ξα

0

e−itz2Φξ, zdz−Φξ,0 π

4it

Ct−3/4ξαΦzξ, z

LξL2z 2.64

provided the right-hand side is finite, whereαR.

(17)

Proof. We represent the integral

0

e−itz2Φξ, zdz Φξ,0

0

e−itz2dzR Φξ,0 π

4itR, 2.65

where the remainder term is

R

0

e−itz2Φξ, z−Φξ,0dz. 2.66

In the remainder term we integrate by parts via identitye−itz2 1/1−2itz2zze−itz2 ξαR

LξC ξα

0

1

1tz2Φξ, z−Φξ,0dz Lξ

C ξα

0

z

1tz2Φzξ, zdz Lξ

CξαΦzξ, z

LξL2z

z

1tz2 −1 L1z

CξαΦzξ, z

LξL2z

z

1tz2 −1 L2z

Ct−3/4ξαΦzξ, zL

ξL2z,

2.67

since

|Φξ, z−Φξ,0| ≤ z

0

zξ, z|dz≤C

|z|Φzξ, zL

ξL2z. 2.68

Thus we have the estimate for the remainder in the asymptotic formula.Lemma 2.3is proved.

In the next lemma we obtain the asymptotics for the operator Vt B−1MtD−1t F−1e−itξand the right-inverse operatorV−1t EtFDtMtB.

Lemma 2.4. The estimates hold as follows:

ξβVt−1φ

LCt−1/4ξβ3/4ξφ

L2ξβ−1/4φ

L2 , ξβ−3/4

V−1t−1 φ

LCt−1/4ξ3/2βξφ

L2ξ1/2βφ

L2 ,

2.69

for allt≥1,where 0β≤3/2,provided the right-hand sides are finite.

(18)

Proof. We have the identities

S ξ, η

η

ξ − ξ ξ

ηξ

1 η

ξ −ηξ η

ξ2 ξ

ηξ2

,

Sη

ξ, η η

ηξ

ξ 1 η

ξ −ηξ ξ

η

η

ξ ηξ

,

2.70

and Sηηξ, η η−3.We now change z

Sξ, η.We denote The inverse functions by ηjz,so thatη1z:0,∞ → ξ,∞andη2z:0,∞ → −∞, ξ.

Thus the stationary point z 0 transforms into η ξ. Hence we can write the representation

Vtφξ−3/2

it

Re−itSξ,ηφ η

−3/2

it

2 j1

0

e−itz2φ ηjz

S

ξ, ηjz Sη

ξ, ηjzdz.

2.71

By a direct calculation we find

dz Sη

ξ, η

S ξ, η

1

η

ξ −ηξ η

ξ1/2 . 2.72

Therefore we obtain

Vtφ

0

e−itz2Φξ, zdz, 2.73

where

Φξ, z −1

it

2 j1

η φ

η

η

ξη

ηηjz

. 2.74

Application ofLemma 2.3yields

Vtφφξ O

t−3/4Φzξ, zL

ξL2z . 2.75

(19)

By a direct calculation we have

Φzξ, z dz

it

2 j1

2 η3/2

φ η ξ3/4

1 η

ξ −ηξ3/4

it

2 j1

η

ηξξ η

/ η

ξ ξ3/4

η1/2 1

η

ξ −ηξ3/4φ η

.

2.76

By2.38we get the estimate

ξβΦzξ, z

LξL2z

ξβΦzξ, z dz

LξL2η

Ct

ξβ η9/4 η

ξ3/2φ η

LξL2η

Ct

ξβ η5/4 η

ξ3/2φ η

LξL2η

Ct

η3/4β φ

L2Ct

ηβ−1/4 φ

L2

2.77

for 0 ≤ β≤ 3/2 sinceηξ ≥ 2/3βη1−2/3β.This yields the first estimate of the lemma.

We now prove the second estimate. We have

V−1

t

√2iπ 1

−1eitξ−ξy−iy iy−3/2

φ y

iy

dy, 2.78

then changingyηη−1, we find

V−1tφx

t

√2iπ

R

eitSη,ξφ η

η−3/2

dη, 2.79

where

S η, ξ

ξ −ξη1

η 1 η

ξ −ηξ η

ξ2 η

ηξ2

. 2.80

(20)

As above we changez

Sη, ξand represent

V−1tφx

t

√2iπ

ReitSη,ξφ η

η−3/2

√2t

2 j1

0

eitz2φ ηjz

ηjz−3/2

S

ηjz, ξ Sη

ηjz, ξdz.

2.81

By a direct calculation we findSηη, ξ η−ξ/η3 and

dz

Sη η, ξ 2

S

η, ξ

η ξ 2

η5/2 1

η

ξ −ηξ

. 2.82

Therefore we obtain

V−1tφx

0

eitz2Φξ, zdz, 2.83

where

Φξ, z 2√

√ 2t

2 j1

η 1

η

ξ −ηξ η

ξ φ η

ηηjz

. 2.84

Application ofLemma 2.3yields

V−1tφx φξ O

t−3/4Φzξ, zL

ξL2z . 2.85

By a direct calculation we have

Φzξ, z

dz 2√

√2t

2 j1

1 η

ξ −ηξ3/4 η

ξ3/2 η9/4

φ η

2√

√ 2t

2 j1

η5/4 1

η

ξ −ηξ1/4

η ξ

⎜⎝ η

1 η

ξ −ηξ η

ξ

⎟⎠

η

φ η

, 2.86

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