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
1and Pavel I. Naumkin
21Department 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,x∈R, whereL∂tii∂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, x∈R, 1.1
whereλ∈C,L∂tii∂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
1.1is related to the Cauchy problem
vttv−vxxμv2, t, x∈R×R,
v0, x v0x, vt0, x v1x, x∈R, 1.2
wherev0andv1are the real-valued functions, andμ ∈R.Indeed we can putu 1/2v ii∂x−1vt,thenusatisfies
Lu i
2μi∂x−1uu2, t, x∈R×R, u0, x u0x, x∈R,
1.3
whereu0 1/2v0ii∂x−1v1.
There are a lot of works devoted to the study of the cubic nonlinear Klein-Gordon equation
vttv−vxxμv3, t, x∈R×R,
v0, x v0x, vt0, x v1x, x∈R 1.4
withμ ∈R.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
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 ∈ S;φLp < ∞}, with the normφLp
R|φx|pdx1/pif 1≤p <∞andφL∞ supx∈R|φx|ifp∞.The weighted Sobolev space is
Hm,sp
φ∈Lp;xsi∂xmφLp<∞
, 1.5
form, s ∈ R, 1 ≤ p ≤ ∞,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→ξφ ≡ φ 1/√
2π
Re−ixξφxdxthe Fourier transform of the function φ.Then the inverse Fourier transformation isF−1φF−1ξ→xφ 1/√
2π
Reixξφξdξ.
Our main result of this paper is the following.
Theorem 1.1. Letu0 ∈ H3,1and the norm u0H3,1 ε. Then there existsε0 > 0 such that for all 0< ε < ε0the Cauchy problem1.1has a unique global solution
ut∈C
0,∞;H3,1 1.6
satisfying the time decay estimate
utH1,0∞ ≤Cε1t−1/2. 1.7
Furthermore there exists a unique final stateW ∈H0,1∞ ∩H0,1such that ut−e−ii∂xtF−1We−2i|λ|2Ω|W|2logt
H1,0 ≤Cε3/2tγ−1/4, Feii∂xtut−We−2i|λ|2Ω|W|2logt
H0,1∞
≤Cε3/2tγ−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
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−2∂x; therefore the commutator L,J LJ − JL0, whereL∂tii∂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−1∂xL.
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, η−ξ
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
dτ
ReiτAv
τ, η−ξ v
τ,−η
dη. 1.12
Then we integrate by parts with respect toτ,taking into account1.11,
vt, ξ iλ
ReitAv t, η−ξ
v
t,−η dη
√2πξA ξ, η v0, ξ iλ
Rv
0, η−ξ v
0,−η dη
√2πξA ξ, η
−2i|λ|2 t
0
dτ
R
√ dη
2πξA
ξ, ηeiτA−ηv
τ, η−ξ Fx→η
i∂x−1u2 .
1.13
Returning to the functionut, x UtF−1ξ→xvF−1ξ→xe−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
ξ−η, η φ ξ−η
ψ
η
dη, 1.15
where
g
ζ, η
1
√2π
ζη ζη
ζ
η, 1.16
andζξ−η.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
iμ2G
u,i∂x−1uu2 iμ2G1
u,i∂x−1uu2 iμ2G2
uu,i∂x−1uu2 ,
1.18
where
Gj
φ, ψ
Fξ−1→x
Rgj
ξ−η, η φ ξ−η
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.
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
ξ−η, η φ ξ−η
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 x−y
ψx−zdydz, 2.4
where the kernelgy, zobeys the following estimate:
g
y, z≤C y−3
z−3ln2
1 1
z−y
2.5
for ally, z∈R, y /z.Moreover the following estimates are valid:
G φ, ψ
Lp ≤Cφ
Lqψ
Lr, PG
φ, ψ
Lp ≤CPφ
Lqψ
Lr CPψ
Lqφ
Lr
C∂tφ
Lqψ
Lr C∂tψ
Lqφ
Lr
2.6
for 1≤ p ≤ ∞,1 < q ≤ p/α,1 < r ≤ p/1−α, α ∈0,1, provided that the right-hand sides are bounded.
Proof. To prove representation2.4, we substitute the direct Fourier transforms
φζ 1
√2π
Re−ix−yζφ x−y
dy,
ψ
η 1
√2π
Re−ix−zηψx−zdz
2.7
into the definition of the operatorG. Then changingζξ−η, we find
G φ, ψ
2π−3/2
R2φ x−y
ψx−z
R2g ζ, η
eiyζizηdηdζ
dydz
R2g y, z
φ x−y
ψx−zdydz,
2.8
where the kernelgy, zis
g y, z
2π−3/2
R2g ζ, η
eiyζizηdηdζ
1 4π2
R2
eiyζizη ζη
ζ
η
ζηdηdζ.
2.9
Changing the variables of integrationζξ/2−ηandηξ/2ηthe prime we will omit, we get
g y, z
1 4π2
R2
eiyζizη ζη
ζ
η
ζηdηdζ 1
4π2
R
dξ
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
4π2
R
dξ ξeisξ
RB ξ, η
eiρηdη. 2.11
For the case of|ρ| ≤ 1,|s| ≤1 we integrate by parts using the identityeiρη Aη∂ηηeiρη, whereAη 1/1iρη.Then we get
g
s, ρ − 1
4π2
R
dξ ξeisξ
Reiρηη∂η
A η
B ξ, η
dη. 2.12
Note that
η∂ηB ξ, η
η ξ
ξ/2−η
ξ/2η2
η−ξ/2
η−ξ/2 ηξ/2 ξ/2η
. 2.13
Then
ξ−1η∂η
A η
B
ξ, η≤ C
ξ
1ρηξ
η. 2.14
Hence we can estimate the kernelgs, ρ as follows:
g
s, ρ≤C
R
dξ ξ
R
dη
1ρηξ η
≤C ∞
1
dη 1ρη
∞
1
dξ ξ
ηξ C ∞
1
dη 1ρη
1 η
ln ξ
ηξ ∞
1
dξ
≤C ∞
1
lnη 1ρη
ηdη≤C 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
Reiρη∂3ξ
ξ−1η∂η
A η
B
ξ, η dη. 2.16
Note that
∂3ξ
ξ−1η∂η A
η B
ξ, η ≤ C
ξ
1ρηξ
η. 2.17
Hence we can estimate the kernelgy, zas follows:
g
s, ρ≤C|s|−3
R
dξ ξ
R
dη
1ρηξ
η ≤C|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
Cρ−3
R
dξ ξeisξ
Reiρη∂3ηB ξ, η
dη. 2.19
Note that|∂3ηBξ, η| ≤Cξη−3.Hence
g
s, ρ≤Cρ−3
R
dξ ξ2
R
dη
η2 ≤Cρ−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
Reiρη∂3ξ∂3η ξ−1B
ξ, η dη. 2.21
Since
∂3ξ∂3η ξ−1B
ξ, η ≤ C
ξ2 ξ
η2, 2.22
then we can estimate the kernelgs, ρ as follows:
g
s, ρ≤Csρ−3
R
dξ ξ2
R
dη
η2 ≤Csρ−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
Lp ≤C
R
φ
x−y dy y3
Lpx1
Rln2
1 1
z−y
ψx−z dz z3
Lpx2L∞y
≤CφLqψLr,
2.24
where 1≤p≤ ∞,1< q≤p/α,1< r ≤p/1−α, α∈0,1.
We now estimate the operatorPt∂xx∂tas follows:
PG φ, ψ
G Pφ, ψ
R2yg y, z
ψx−z∂tφ x−y
dydz
G φ,Pψ
R2zg y, z
φ x−y
∂tψx−zdydz.
2.25
Then by virtue of estimate2.4applying the H ¨older and Young inequalities, we get PG
φ, ψ
Lp ≤G
Pφ, ψ
LpG
φ,Pψ
Lp
R2yg y, z
ψx−z∂tφ x−y
dydz Lp
R2zg y, z
φ x−y
∂tψx−zdydz Lp
≤CPφ
Lqψ
LrCPψ
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
Bφ 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 allξ∈R,sinceξx/ix ∈R andxξ/ξ ∈−1,1.We now introduce the operators Vt B−1MtD−1t F−1e−itξ,
Wt Mt1−θD−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
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≤Cφ
L2, ξρ∂ξVtφ
L2 ≤Cξα∂ξφ
L2Cξα−1φ
L2, 2.34
whereρ < α, α∈1,2,and
ξρ∂ξV−1tφ
L2≤Cξ1/4α∂ξφ
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 dξ
ξ3 1
−1
φx2dxφ2
L2−1,1. 2.36
HenceVtφL2 φL2.
Consider the estimate for the derivative∂ξVtφ. 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
dη−itξ−3
Re−itSφ η
ξ−η dη
ξ−3
Re−itS φ
η g
ξ, η φ
η gη
ξ, η dη,
2.37
wheregξ, η ξ−η/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 −ηξ ≤Cξ2 η2 η
ξ. 2.39
Since∂ξξaaξξa−2, we get
ξρ∂ξVtφξρ
it
2π∂ξξ−3/2
Re−itSφ η
dη
−3
2ξξρ−2Vtφξρ−9/2
it 2π
Re−itSφ η
g ξ, η
dη
ξρ−9/2
it 2π
Re−itSφ η
gη ξ, η
dη.
2.40
Consider theL2-estimate of the integral
√tξρ−9/2
Reitξη/ξψ η
g ξ, η
dη. 2.41
We have
√ t
Reitξη/ξψ η
g ξ, η
ξρ−9/2dη 2
L2
t
Rdηψ η
Rdζψζ
Reitξ/ξη−ζg ξ, η
gξ, ζξ2ρ−9dξ
Rdηψ η
RdζψζG η, ζ
,
2.42
where the kernel
G η, ζ
t
Reitξ/ξη−ζg ξ, η
gξ, ζξ2ρ−9dξ
t η
ζ
Reitξ/ξη−ζ
η ξ 1
η
ξ −ηξ ζξ
1ζξ −ζξξ2ρ−7dξ.
2.43
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
1tη−ζδ 2.46
forδ >1. Therefore we obtain the estimate
G
η, ζ≤ Ct ηα
ζα 1tη−ζδ
1
−1ix2α−ρ−δdx≤ Ct ηα
ζα
1tη−ζδ, 2.47
if we chooseρ < α, α∈1,2and 1< δ <1α−ρ.Thus we get √
t
Reitξη/ξψ η
g ξ, η
ξρ−9/2dη 2
L2
Rdηψ η
RdζψζG η, ζ
≤Cηα ψ
η
L2
Rdζψζζα t
1tη−ζδ
L2
≤Ct·αψ2
L2
R
dη
1tηδ ≤C·αψ2
L2
2.48
sinceδ >1.
In the same manner we consider the estimate of the integral √
t
Reitξη/ξψ η
gη
ξ, η
ξρ−9/2dη 2
L2
t
Rdηψ η
Rdζψζ
Reitξ/ξη−ζgη
ξ, η
gζξ, ζξ2ρ−9dξ
Rdηψ η
RdζψζG η, ζ
,
2.49
where the kernel G
η, ζ t
Reitξ/ξη−ζgη
ξ, η
gζξ, ζξ2ρ−9dξ
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, ζ≤ Ct ηα−1
ζα−1
1tη−ζδ 2.52
if we chooseρ < α, α∈1,2, and 1< δ <1α−ρ.Therefore we get √
t
Reitξη/ξψ η
gη ξ, η
ξρ−9/2dη 2
L2 ≤C ηα−1
ψ η2
L2
R
tdη
1tηδ ≤C·α−1ψ2
L2. 2.53
Henceξρ∂ξVtφL2≤Cξα∂ξφL2Cξα−1φL2.Thus we have the second estimate of the lemma.
Consider the estimate for the derivative∂ξV−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
ξρ∂ξV−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 g1η
ξ, η dη,
2.54
whereSη, ξ ξ −ξη1/ηandg1ξ, η 1ηξ −ηξ/ηξη1/2.The estimate is true
g1 ξ, η
1 η
ξ −ηξ η
ξ
η1/2≤C η3/2 ξ
η
ξ. 2.55
Consider theL2-estimate of the integral
ξρ−1√ t
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
dη 2
L2
t
Rdηψ η
Rdζψζ
Re−itξη/η−ζ/ζg1 ξ, η
g1ξ, ζξ2ρ−2dξ
t 1
−1dy iy3
ψ y
iy 1
−1dziz3ψ z
iz
×
Re−itξy−zg1
ξ, y iy
g1
ξ, z
iz
ξ2ρ−2dξ
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−2dξ
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
Since|ξ| x 2−y224x2y21/2∼ x and by2.55|1ξ/iy−yξ/iy/1/iy ξ| ≤Cξ1−αiy−αforα∈0,1, and also
e−itξy−z≤ C
1tx y−z2 ≤ C tx δy−zδ
y−z 2.60
forδ <1,we obtain the estimate G1
y, z≤ C
iy−1/2−α
iz−1/2−α y−zδ
y−z
Rx −2α2ρ−δdx
≤ C
iy−1/2−α
iz−1/2−α y−zδ
y−z ,
2.61
if we chooseρ < α,and 1> δ >1−2α2ρ.Therefore we get √
t
Reitξη/ξψ η
g ξ, η
ξρ−9/2dη 2
L2
1
−1dy iy5/2−α
ψ y
iy 1
−1dziz5/2−αψ z
iz
C y−zδ
y−z
≤C 1
−1dy
iy5/2−2α ψ
y iy
2
≤C 1
−1dξξ1/22αψξ2≤C·1/4αψ2
L2.
2.62
In the same manner we consider the estimate of the integral withφηg1ηξ, η.Hence ξρ∂ξV−1tφ
L2≤Cξ1/4α∂ξφ
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
L∞ξL2z 2.64
provided the right-hand side is finite, whereα∈R.
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 1/1−2itz2∂zze−itz2 ξαR
L∞ξ ≤C ξα
∞
0
1
1tz2Φξ, z−Φξ,0dz L∞ξ
C ξα
∞
0
z
1tz2Φzξ, zdz L∞ξ
≤CξαΦzξ, z
L∞ξL2z
√ z
1tz2 −1 L1z
CξαΦzξ, z
L∞ξL2z
z
1tz2 −1 L2z
≤Ct−3/4ξαΦzξ, zL∞
ξL2z,
2.67
since
|Φξ, z−Φξ,0| ≤ z
0
|Φzξ, 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
L∞≤Ct−1/4ξβ3/4∂ξφ
L2ξβ−1/4φ
L2 , ξβ−3/4
V−1t−1 φ
L∞ ≤Ct−1/4ξ3/2β∂ξφ
L2ξ1/2βφ
L2 ,
2.69
for allt≥1,where 0≤β≤3/2,provided the right-hand sides are finite.
Proof. We have the identities
S ξ, η
η
− ξ − ξ ξ
η−ξ
1 η
ξ −ηξ η
ξ2 ξ
η−ξ2
,
Sη
ξ, η η
η− ξ
ξ 1 η
ξ −ηξ ξ
η
η
ξ η−ξ
,
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 2π
Re−itSξ,ηφ η
dη
2ξ−3/2
it 2π
2 j1
∞
0
e−itz2φ ηjz
S
ξ, ηjz Sη
ξ, ηjzdz.
2.71
By a direct calculation we find
dz dη 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 2ξ−1
it 2π
2 j1
η φ
η
1ξ η
−ξη
ηηjz
. 2.74
Application ofLemma 2.3yields
Vtφφξ O
t−3/4Φzξ, zL∞
ξL2z . 2.75
By a direct calculation we have
Φzξ, z dz
dη
it 2π
2 j1
2 η3/2
φ η ξ3/4
1 η
ξ −ηξ3/4
it 2π
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
L∞ξL2z
ξβΦzξ, z dz
dη L∞ξL2η
≤C√ t
ξβ η9/4 η
ξ3/2φ η
L∞ξL2η
C√ t
ξβ η5/4 η
ξ3/2φ η
L∞ξL2η
≤C√ t
η3/4β φ
L2C√ t
ηβ−1/4 φ
L2
2.77
for 0 ≤ β≤ 3/2 sinceηξ ≥ Cξ2/3βη1−2/3β.This yields the first estimate of the lemma.
We now prove the second estimate. We have
V−1tφ
√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
η−ξ2
. 2.80
As above we changez
Sη, ξand represent
V−1tφx
√t
√2iπ
ReitSη,ξφ η
η−3/2 dη
√2t
√iπ 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
dη 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 iπ
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 dη 2√
√2t iπ
2 j1
1 η
ξ −ηξ3/4 η
ξ3/2 η9/4
φ η
2√
√ 2t iπ
2 j1
η5/4 1
η
ξ −ηξ1/4
η ξ
⎛
⎜⎝ η
1 η
ξ −ηξ η
ξ
⎞
⎟⎠
η
φ η
, 2.86