# Local Analyticity in the Time and Space Variables and the Smoothing Effect for the Fifth-Order KdV-Type Equation

Authors: Kyoko Tomoeda
Journal: Advances in Mathematical Physics
 (2011)

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## Abstract

We consider the initial value problem for the reduced fifth-order KdV-type equation:∂tu-∂x5u-10∂x(u3)+10∂x(∂xu)2=0, t,x∈ℝ, u(0,x)=ϕ(x), x∈ℝ. This equation is obtained by removing the nonlinear term 10u∂x3u from the fifth-order KdV equation. We show the existence of the local solution which is real analytic in both time and space variables if the initial data ϕ∈Hs(ℝ)(s>1/8) satisfies the condition ∑k=0∞(A0k/k!)∥(x∂x)kϕ∥Hs<∞, for some constant A0(0<A0<1). Moreover, the smoothing effect for this equation is obtained. The proof of our main result is based on the contraction principle and the bootstrap argument used in the third-order KdV equation (K. Kato and Ogawa 2000). The key of the proof is to obtain the estimate of ∂x(∂xu)2 on the Bourgain space, which is accomplished by improving Kenig et al.'s method used in (Kenig et al. 1996).


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## Body

## 1. Introduction

The KdV hierarchy is well known as the series of the Lax pair formulation [1, 2], which are presented as ((1.1)0)1st-orderKdV∂tu-∂xu=0,((1.1)1)3rd-orderKdV∂tu+∂x3u-6u∂xu=0,((1.1)2)5th-orderKdV∂tu-∂x5u-10∂x(u3)+10∂x(∂xu)2+10u∂x3u=0.⋮
These equations describe mathematical models of some water waves [3, 4]. We are interested in the existence theory of the analytic solution and the smoothing effect of the KdV hierarchy. There are some results concerning the analyticity for the third-order KdV equation (1.1)1. To. Kato and Masuda [5] considered the initial value problem of the following equation: (1.2)∂tu+∂x3u+a(u)∂xu=0,t,x∈R,
where a(λ) is the real analytic and the no growth rate function in λ∈ℝ. They showed that if the initial data is real analytic, then, the global solution of (1.2) is real analytic in the space variable. Hayashi [6] also considered (1.2) in which a(λ) is the polynomial. He showed that if the initial data is analytic and has an analytic continuation to a strip containing the real axis, then,  the local solution also has the same property. When a(u)=-6u, (1.2) becomes the third-order KdV equation (1.1)1. K. Kato and Ogawa [7] proved that (1.1)1 has not only the real analytic solution in both time and space variables but also the smoothing effect.Recently, it is shown that the nonlinear dispersive equations including the KdV hierarchy  have the local analytic solution in the space variable (see [8]). However, neither the existence of the real analytic solution in both time and space variables nor the smoothing effect is obtained for (1.1)j with j≥2, because the bilinear estimate of u∂x2j-1u with j≥2 cannot be obtained by their method used in [9].On the other hand, we may expect that the method used in [7] can work for the reduced equation given by removing u∂x2j-1u from the higher-order KdV equations (1.1)0 with j≥2. In this paper, as a starting point for this attempt, we consider the following initial value problem of the reduced fifth-order KdV-type equation: (1.3)∂tu-∂x5u=∂x(u3)+∂x(∂xu)2,t,x∈R,u(0,x)=ϕ(x),x∈R,
where we may take all coefficients of the nonlinear terms to be equal to 1 without loss of generality. This equation is obtained by removing the nonlinear term 10u∂x3u from the original fifth-order KdV equation (1.1)2. Our main purpose is to prove not only the existence of a local real analytic solution of (1.3) in both time and space variables but also the smoothing effect.Before stating the main result precisely, we introduce the function space introduced by Bourgain (see [10]): for s,b∈ℝ, define that (1.4)Xbs={f∈S′(R2);‖f‖Xbs<∞},
where  (1.5)‖f‖Xbs2=∬R2(1+|τ-ξ5|)2b(1+|ξ|)2s|Ft,xf(τ,ξ)|2dτdξ,
and ℱt,xf is the Fourier transform of f in both x and t variables; that is, (1.6)Ft,xf(τ,ξ)=(2π)-1∬R2f(t,x)e-itτ-ixξdtdx.Our main result is the following theorem.Theorem 1.1.
Lets>1/8 and let b∈(1/2,23/40). Then, for any ϕ(x)∈Hs(ℝ) such that
(1.7)(x∂x)kϕ(x)∈Hs(R)(k=0,1,2,…),∑k=0∞A0kk!‖(x∂x)kϕ‖Hs<∞,forsome0<A0<1,
there exist a constant T=T(ϕ)>0 and a unique solution u∈C((-T,T),Hs)∩Xbs of (1.3) satisfying
(1.8)Pku∈C((-T,T),Hs)∩Xbs,∑k=0∞A0kk!‖Pku‖Xbs<∞,
where P=5t∂t+x∂x is the generator of dilation for the linear part of the equation of (1.3).
Moreover this solution becomes real analytic in both time and space variables; that is, there exist the positive constantsC and A1 such that
(1.9)|∂tm∂xlu(t,x)|≤CA1m+l(m+l)!
holds for all (t,x)∈(-T,0)∪(0,T)×ℝ and l,m=0,1,2,….Remark 1.2.
The initial dataϕ(x) has to be analytic except for x=0 but is allowed to have Hs-singularity at x=0. It follows from (1.9) that the singularity of ϕ(x) disappears after time passes and the regularity of the local solution of (1.3) reaches real analyticity in both time and space variables; that is, the fifth-order KdV-type equation has the smoothing effect.Remark 1.3.
A typical example of the initial data satisfying the condition (1.7) is given by
(1.10)|x|γe-x2withγ>-38.The existence results of the higher-order KdV equation are studied by many authors. Saut [11] and Schwarz [12] proved that each equation of the KdV hierarchy has a unique global solution in the spatially periodic Sobolev space. Kenig, Ponce, and Vega studied the initial value problem of the higher-order dispersive equation(1.11)∂tu+∂x2j+1u+P(u,∂xu,…,∂x2j-1u)=0,
where j≥1 and P(·) is a polynomial having no constant or linear part. They proved the local well-posedness in the weighted Sobolev space [13, 14]. Recently, Kwon [15] studied the simplified fifth-order KdV-type equation (1.12)∂tu+∂x5u+∂xu∂x2u+u∂x3u=0,
which is obtained by removing the term 10∂x(u3) from (1.1)2. He showed the local well-posedness for the IVP of this equation in Hs(ℝ) with s>5/2. On the other hand, Ta. Kato [16] proved the following result for (1.1)2.Well-Posedness Theorem (Ta. Kato)
(1)
Let(1.13)s>-14,s≥-2a-2,where-32<a≤-14.
Then, the local well-posedness for the IVP of (1.1)2  holds in Hs,a(ℝ), where
(1.14)Hs,a(R)≡{f∈S′(R);‖f‖Hs,a≡‖〈ξ〉s-a|ξ|af̂‖Lξ2<∞}.(2)
Let(1.15)s≥1,-12≤a≤-14.
Then, the global well-posedness for the IVP of (1.1)2 holds in Hs,a(ℝ).The plan of this paper is as follows. In Section2, we give the existence and uniqueness of the local solution of (1.3), which is shown by the contraction argument consisting of Lemmas 2.3–2.5. In Section 3, we prove Lemma 2.4 which gives the bilinear estimate for ∂x(∂xu)2 in the Bourgain space. Kenig et al. [9] showed the bilinear estimate for u∂xu of (1.1)1 by estimating the potential which appears in an expression of the Bourgain norm of this term via duality. They divided the domain of integration of the potential into 17 subregions. However, their method of the domain decomposition is consistent with (1.1)2, but not with the fifth-order KdV-type equation. We divide this domain into 30 subregions to derive the bilinear estimate for ∂x(∂xu)2. In Section 4, we show the analyticity of the solution stated in Theorem 1.1 by the bootstrap argument. The result of this paper is announced in Proceedings of the Japan Academy [17].Notation
. Letℱx be the Fourier transform in the x variable, and let ℱξ-1 and ℱτ,ξ-1 be the Fourier inverse transform in the ξ and (τ,ξ) variables, respectively. The Riesz operator Dx and its fractional derivative 〈Dx〉s are defined by
(1.16)Dx=Fξ-1|ξ|Fx,〈Dx〉s=Fξ-1〈ξ〉sFx,
respectively, where 〈·〉=(1+|·|). Similarly, 〈Dt,x〉s is defined by
(1.17)〈Dt,x〉s=Fτ,ξ-1〈|τ|+|ξ|〉sFt,x.[A,B] denotes the commutator relation of two operators given by AB-BA. LtpLxq denotes the space Lp(ℝt;Lq(ℝx)) for 1≤p,q≤∞ with the norm
(1.18)‖f‖LtpLxq=(∫-∞∞(∫-∞∞|f(t,x)|pdt)q/pdx)1/q.
We use Sobolev spaces with both time and space variables
(1.19)Ht,xs(R2)={u∈S′(R2):〈Dt,x〉su∈Lt2Lx2},
with the norm ∥·∥Ht,xs(ℝ2)=∥〈Dt,x〉s·∥Lt2Lx2. Moreover, Lt2(ℝ;Hxs) denotes the space L2(ℝt;Hs(ℝx)) with the norm ∥·∥Lt2(ℝ;Hxs)=∥〈Dx〉s·∥Lt2Lx2. The dual coupling is expressed as 〈f,g〉. The convolution of f and g with both space and time variables is denoted by f*g. For the constant A0 appearing in Theorem 1.1, we put
(1.20)AA0(Xbs)={f=(f0,f1,…);fk∈Xbs(k=0,1,…),|f|AA0(Xbs)<∞},
where
(1.21)|‖f‖|AA0(Xbs)≡∑k=0∞A0kk!‖fk‖Xbs.
For simplicity we make use of three notations:
(1.22)∑k=∑k=k1+k2+k3+k4,∑l=∑l=l1+l2+l3,∑m=∑m=m1+m2+m3.


## 2. Existence and Uniqueness of the Solution

In this section, we give the proof of the existence and uniqueness of the solution of (1.3). Let uk=Pku and ϕk(x)=(x∂x)kϕ(x), and we derive the equation which uk and ϕk(x) satisfy. Since [x∂x,∂x]=-∂x, it follows that(2.1)(P+l)k∂x=∂x(P+(l-1))k,k,l=0,1,2,….
Using (2.1) and the following relations (2.2)[∂t-∂x5,P]=5(∂t-∂x5),(∂t-∂x5)Pk=(P+5)k(∂t-∂x5),
we have from (1.3)(2.3)∂tuk-∂x5uk=Nk(u),t,x∈R,uk(0,x)=ϕk(x),x∈R,k=0,1,2,…,
where(2.4)Nk(u)=∂x(P+4)k(u3)+∂x(P+4)k(∂xu)2.
Using the Leibniz rule and (2.1), we can see that (2.5)Nk(u)=∂x∑l=0k(kl)4k-lPl(u3)+∂x∑l=0k(kl)3k-l(P+1)l(∂xu)2=∑kk!k1!k2!k3!k4!4k4∂x(uk1uk2uk3)+∑kk!k1!k2!k3!k4!3k4(-1)k3∂x((∂xuk1)(∂xuk2)).
We will show the existence and uniqueness of the solution of (2.3).Proposition 2.1.
Let(2.6)s>-14,b∈(12,12+σ),
where σ=min{s/5+1/20,3/16}. Then, for any ϕ≡(ϕ0,ϕ1,…) such that ϕk∈Hs(ℝ)(k=0,1,…) and
(2.7)|‖ϕ‖|AA0(Hs)<∞,
there exist a constant T=T(ϕ)>0 and a unique solution uk∈C((-T,T),Hs)∩Xbs of (2.3) satisfying
(2.8)|‖u‖|AA0(Xbs)<∞,u≡(u0,u1,…).Remark 2.2.
The uniqueness of the solution of (2.3) yields uk=Pku for k=0,1,2,…. Moreover, u0 becomes a solution of (1.3), the uniqueness of which also follows.To prove this proposition we prepare three lemmas (Lemmas2.3, 2.4 and 2.5), which play an important role in applying the contraction principle to the following system of the integral equations:(2.9)ψ(t)uk=ψ(t)et∂x5ϕk+ψ(t)∫0te(t-t′)∂x5ψT(t′)Nk(u)(t′)dt′,
where (2.10)et∂x5f≡Fξ-1(eiξ5tf̂(ξ)),ψ(t) denotes a cut-off function in C0∞(ℝ) satisfying (2.11)ψ(t)={1,if|t|≤1,0,if|t|>2,
and ψT(t)=ψ(t/T).Lemma 2.3.
Let0<T<1 and let
(2.12)s∈R,b∈(12,1),a′,a∈(0,12)(a′<a).
Then
(2.13)‖ψ(t)et∂x5ϕ(x)‖Xbs≤C0,s,b‖ϕ‖Hs,(2.14)‖ψ(t)∫0te(t-t′)∂x5h(t′)dt′‖Xbs≤C1,s,b‖h‖Xb-1s,(2.15)‖ψTh‖X-as≤C2,s,-a,-a′T(a-a′)/4(1-a′)‖h‖X-a′s,
where C0,s,b, C1,s,b, and C2,s,-a,-a′ are constants depending on s, b, a, and a′.Proof.
Equations (2.13) and (2.14) are obtained by replacing the generator e-t∂x3 by et∂x5 in the argument given by Kenig et al. [18] (see [19]). For the proof of (2.15), we refer to Lemma  2.5 in the study by  Ginibre-Tsutsumi-Velo [20].Lemma 2.4.
Let(2.16)s>-14,b,b′∈(12,12+σ)(b≤b′),
where σ=min{s/5+1/20,3/16}. Then
(2.17)‖∂x((∂xu)(∂xv))‖Xb′-1s≤C3,s,b,b′‖u‖Xbs‖v‖Xbs,
where C3,s,b,b′ is a constant depending on s, b, and b′.We give for the proof of this lemma, in Section3.Lemma 2.5.
Let(2.18)s>-14,b,b′∈(12,34)(b≤b′).
Then
(2.19)‖∂x(uvw)‖Xb′-1s≤C4,s,b,b′‖u‖Xbs‖v‖Xbs‖w‖Xbs,
where C4,s,b,b′ is a constant depending on s, b, and b′.Proof.
This lemma is proved by improving Chen et al.’s argument used in the case whereb=b′∈(1/2,3/4) [21].Proof of Proposition2.1.
We define(2.20)XM0={f∈AA0(Xbs);|‖f‖|AA0(Xbs)≤2C0M0},
where
(2.21)C0=C0,s,b,M0=|‖ϕ‖|AA0(Hs).
We define a map Φ:XM0→XM0 by Φ(u)=(Φ0(u),Φ1(u),…) and
(2.22)Φk(u)=ψ(t)et∂x5ϕk+ψ(t)∫0te(t-t′)∂x5ψT(t′)Nk(u)(t′)dt′.
Let b′ and T be positive constants satisfying b<b′<1/2+σ and
(2.23)T<min{1,(24C02C5e4A0M02+8C0C6e4A0M0)-1/μ},
respectively, where
(2.24)C5=C1,s,bC2,s,b-1,b′-1C4,s,b,b′C6=C1,s,bC2,s,b-1,b′-1C3,s,b,b′.
We now show that Φ is a contraction mapping from XM0 to itself. According to Lemmas 2.3, 2.4, and 2.5, we have for u∈𝒜A0(Xbs)(2.25)‖Φk(u)‖Xbs≤C0‖Φk‖Hs+C5Tμ∑kk!k1!k2!k3!k4!4k4‖uk1‖Xbs‖uk2‖Xbs‖uk3‖Xbs+C6Tμ∑kk!k1!k2!k3!k4!3k4‖uk1‖Xbs‖uk2‖Xbs,
for any k≥0. Here μ=(b′-b)/4b′>0. By taking a sum over k, we have
(2.26)|‖Φ(u)‖|AA0(Xbs)≡∑k=0∞A0kk!‖Φk(u)‖Xbs≤C0|‖ϕ‖|AA0(Hs)+C5Tμ∑k4=0∞(4A0)k4k4!∑k1=0∞A0k1k1!‖uk1‖Xbs∑k2=0∞A0k2k2!‖uk2‖Xbs∑k3=0∞A0k3k3!‖uk3‖Xbs+C6Tμ∑k4=0∞(3A0)k4k4!∑k3=0∞A0k3k3!∑k1=0∞A0k1k1!‖uk1‖Xbs∑k2=0∞A0k2k2!‖uk2‖Xbs.
Since u∈XM0, we have from (2.23)
(2.27)|‖Φ(u)‖|AA0(Xbs)≤C0|‖ϕ‖|AA0(Hs)+C5e4A0Tμ|‖u‖|AA0(Xbs)3+C6e4A0Tμ|‖u‖|AA0(Xbs)2≤C0M0+8C03C5e4A0TμM03+4C02C6e4A0TμM02≤32C0M0,
which implies Φ(u)∈XM0. Similarly, we have for u and ũ∈𝒜A0(Xbs)(2.28)|‖Φ(u)-Φ(ũ)‖|AA0(Xbs)≤C5e4A0Tμ×(|‖u‖|AA0(Xbs)2+|‖u‖|AA0(Xbs)|‖ũ‖|AA0(Xbs)+|‖ũ‖|AA0(Xbs)2)|‖u-ũ‖|AA0(Xbs)+C6e4A0Tμ(|‖u‖|AA0(Xbs)+|‖ũ‖|AA0(Xbs))|‖u-ũ‖|AA0(Xbs)≤(12C02C5e4A0M02+4C0C6e4A0M0)Tμ|‖u-ũ‖|AA0(Xbs)≤12|‖u-ũ‖|AA0(Xbs).
Thus, the mapping Φ is contraction from XM0 to itself. We obtain a unique fixed point uk∈Xbs satisfying
(2.29)uk(t)=ψ(t)e-t∂x5ϕk+ψ(t)∫0te-(t-t′)∂x5ψT(t′)Nk(u)(t′)dt′
on the time interval [-T,T] and k=0,1,2,…. Uniqueness of the solution is also shown by using Bekiranov et al.’s argument in [22]. This completes the proof.


## 3. Proof of Lemma2.4

In this section we prove Lemma2.4. To prove Lemma 2.4 we prepare the following lemma.Lemma 3.1 (see [22]).
(1) Letα,β>0 and let κ=min{α,β}. If
(3.1)α+β>1+κ,
then
(3.2)(∫-∞∞dx(1+|x-ζ|)α(1+|x-η|)β)1/2≤C7,α,β(1(1+|ζ-η|)κ)1/2,foranyζ,η∈R,
where C7,α,β is a constant depending on α and β.
(2) Ifγ>1, then
(3.3)(∫-∞∞dx(1+|x+η|)γ)1/2≤C8,γ,foranyη∈R,
where C8,γ is a constant depending on γ.Proof of Lemma2.4.
By duality, we have(3.4)‖∂x((∂xu)(∂xv))‖Xb′-1s=‖〈τ-ξ5〉(b′-1)〈ξ〉sξ(Ft,x∂xu)*(Ft,x∂xu)‖Lτ2Lξ2=sup⁡h∈Lτ2Lξ2,‖h‖Lτ2Lξ2≤1|(〈ξ〉sξ〈τ-ξ5〉1-b′(Ft,x∂xu)*(Ft,x∂xv),h)Lτ2Lξ2|,
where (·,·)Lτ2Lξ2 is the inner product in L2(ℝτ×ℝξ). Setting
(3.5)f(τ,ξ)=〈τ-ξ5〉b〈ξ〉sFt,xu(τ,ξ),g(τ,ξ)=〈τ-ξ5〉b〈ξ〉sFt,xv(τ,ξ),
we have
(3.6)(〈ξ〉sξ〈τ-ξ5〉1-b′(Ft,x∂xu)*(Ft,x∂xv),h)Lτ2Lξ2=∬R2〈ξ〉sξ〈τ-ξ5〉1-b′×(∬R2ξ1(ξ-ξ1)f(τ1,ξ1)g(τ-τ1,ξ-ξ1)〈ξ1〉s〈ξ-ξ1〉s〈τ1-ξ15〉b〈τ-τ1-(ξ-ξ1)5〉bdτ1dξ1)h(τ,ξ)dτdξ=IΩ0.0.0+IΩ0.0.0c,
where
(3.7)IΩ0.0.0=∫∫∫∫Ω0.0.0〈ξ〉sξξ1(ξ-ξ1)h(τ,ξ)f(τ1,ξ1)g(τ-τ1,ξ-ξ1)〈τ-ξ5〉1-b′〈τ1-ξ15〉b〈τ-τ1-(ξ-ξ1)5〉b〈ξ1〉s〈ξ-ξ1〉sdτ1dξ1dτdξ,IΩ0.0.0c=∫∫∫∫Ω0.0.0c〈ξ〉sξξ1(ξ-ξ1)h(τ,ξ)f(τ1,ξ1)g(τ-τ1,ξ-ξ1)〈τ-ξ5〉1-b′〈τ1-ξ15〉b〈τ-τ1-(ξ-ξ1)5〉b〈ξ1〉s〈ξ-ξ1〉sdτ1dξ1dτdξ,(3.8)Ω0.0.0={(τ,τ1,ξ,ξ1)∈R4:|ξ|,|ξ1|,|ξ-ξ1|≤5},Ω0.0.0c=R4∖Ω0.0.0.
We splitΩ0.0.0c into three regions, Ω1,Ω2,andΩ3:
(3.9)Ω1={(τ,τ1,ξ,ξ1)∈Ω0.0.0c:|ξ|≤14|ξ1|},Ω2={(τ,τ1,ξ,ξ1)∈Ω0.0.0c:14|ξ1|≤|ξ|≤4|ξ1|},Ω3={(τ,τ1,ξ,ξ1)∈Ω0.0.0c:4|ξ1|≤|ξ|},
and then, we split Ωi(i=1,2,3) into three regions:
(3.10)Ωi.1={(τ,τ1,ξ,ξ1)∈Ωi:|τ1-ξ15|,|τ-τ1-(ξ-ξ1)5|≤|τ-ξ5|},Ωi.2={(τ,τ1,ξ,ξ1)∈Ωi:|τ-ξ5|,|τ-τ1-(ξ-ξ1)5|≤|τ1-ξ15|},Ωi.3={(τ,τ1,ξ,ξ1)∈Ωi:|τ-ξ5|,|τ1-ξ15|≤|τ-τ1-(ξ-ξ1)5|}.
We further split Ω1.j, Ω2.j, and Ω3.j(j=1,2,3) into the following regions:
(3.11)Ω1.j.1={(τ,τ1,ξ,ξ1)∈Ω1.j:|ξ|≥1},Ω1.j.2={(τ,τ1,ξ,ξ1)∈Ω1.j:|ξ|≤1,|ξ||ξ1|4≥1},Ω1.j.3={(τ,τ1,ξ,ξ1)∈Ω1.j:|ξ||ξ1|4≤1},Ω2.1.1={(τ,τ1,ξ,ξ1)∈Ω2.1:|ξ-ξ1|≤1},Ω2.1.2={(τ,τ1,ξ,ξ1)∈Ω2.1:|ξ-ξ1|≥1,|ξ-2ξ1|≥|ξ|-3/2},Ω2.1.3={(τ,τ1,ξ,ξ1)∈Ω2.1:|ξ-ξ1|≥1,|ξ-2ξ1|≤|ξ|-3/2},Ω2.2.1={(τ,τ1,ξ,ξ1)∈Ω2.2:|ξ-ξ1|≤1},Ω2.2.2={(τ,τ1,ξ,ξ1)∈Ω2.2:|ξ-ξ1|≥1,|2ξ-ξ1|≥|ξ1|-3/2},Ω2.2.3={(τ,τ1,ξ,ξ1)∈Ω2.2:|ξ-ξ1|≥1,|2ξ-ξ1|≤|ξ1|-3/2},Ω2.3.1={(τ,τ1,ξ,ξ1)∈Ω2.3:|ξ-ξ1|≤1,|ξ-ξ1||ξ1|4≥1},Ω2.3.2={(τ,τ1,ξ,ξ1)∈Ω2.3:|ξ-ξ1||ξ1|4≤1},Ω2.3.3={(τ,τ1,ξ,ξ1)∈Ω2.3:|ξ-ξ1|≥1,|ξ+ξ1|≥1},Ω2.3.4={(τ,τ1,ξ,ξ1)∈Ω2.3:|ξ-ξ1|≥1,|ξ-ξ1|-3/2≤|ξ+ξ1|≤1},Ω2.3.5={(τ,τ1,ξ,ξ1)∈Ω2.3:|ξ-ξ1|≥1,|ξ+ξ1|≤|ξ-ξ1|-3/2},Ω3.j.1={(τ,τ1,ξ,ξ1)∈Ω3.j:|ξ1|≥1},Ω3.j.2={(τ,τ1,ξ,ξ1)∈Ω3.j:|ξ1|≤1,|ξ1||ξ|4≥1},Ω3.j.3={(τ,τ1,ξ,ξ1)∈Ω3.j:|ξ1||ξ|4≤1},
so that, we have
(3.12)|IΩ0.0.0+IΩ0.0.0c|≤|IΩ0.0.0|+∑i,j,k=13|IΩi.j.k|+|IΩ2.3.4|+|IΩ2.3.5|.
Now we will estimate |IΩ0.0.0|, |IΩi.j.k|(i,j,k=1,2,3), and |IΩ2.3.l|(l=4,5). To estimate these terms, we prepare some estimates. By (3.8), (3.9) we obtain
(3.13)34|ξ1|≤|ξ-ξ1|≤54|ξ1|,inΩ1,34|ξ|≤|ξ-ξ1|≤54|ξ|,inΩ3,(3.14)|ξ1|≥4,inΩ1∪Ω2.2.3,min⁡{|ξ1|,|ξ|}≥1,inΩ2∖{Ω4∪Ω2.1.3∪Ω2.2.3},min⁡{|ξ1|,|ξ|}≥3,inΩ4,|ξ|≥4,inΩ3∪Ω2.1.3,
where Ω4=Ω2.1.1∪Ω2.2.1∪Ω2.3.1∪Ω2.3.2. Since (3.13) and (3.14) yield
(3.15)|ξ-ξ1|≥3,inΩ1,Ω3,
we have by (3.9) and (3.13)–(3.15)
(3.16)|ξ1|2|ξ-ξ1|2〈ξ1〉2s〈ξ-ξ1〉2s≤C9,s2|ξ1|4-4s,inΩ1,|ξ|2|ξ1|2〈ξ〉2s〈ξ1〉2s≤C9,s2|ξ1|4,inΩ2,|ξ|2|ξ-ξ1|2〈ξ〉2s〈ξ-ξ1〉2s≤C9,s2|ξ|4,inΩ3,
where C9,s=4|s|+1. Using (3.11), (3.14), and (3.15), we have
(3.17)〈ξ〉2s≤22|s|max⁡{1,|ξ|}2sinΩ1,〈ξ-ξ1〉-2s≤22|s|max⁡{1,|ξ-ξ1|}-2sinΩ2,〈ξ1〉-2s≤22|s|max⁡{1,|ξ1|}-2sinΩ3.
InΩi1.j1.k1 ((i1.j1.k1)=(0.0.0),(1.1.3),(3.2.1),(3.2.2),(i.2.k),(i=1,2,k=1,2,3)), we integrate with respect to τ and ξ first, then, we use Schwarz’s inequality, Fubini’s theorem, and note that ∥h∥Lτ2Lξ2≤1 to have
(3.18)|IΩi1.j1.k1|≤‖f‖Lτ12Lξ12×‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2h(τ,ξ)g(τ-τ1,ξ-ξ1)〈ξ〉s|ξ||ξ-ξ1|χΩi1.j1.k1(τ,τ1,ξ,ξ1)dξdτ〈τ-ξ5〉1-b′〈τ-τ1-(ξ-ξ1)5〉b〈ξ-ξ1〉s)‖Lτ12Lξ12≤‖f‖Lτ2Lξ2‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2〈ξ〉2s|ξ|2|ξ-ξ1|2χΩi1.j1.k1(τ,τ1,ξ,ξ1)dξdτ〈τ-ξ5〉2(1-b′)〈τ-τ1-(ξ-ξ1)5〉2b〈ξ-ξ1〉2s)2(∬R2|h(τ,ξ)|2|g(τ-τ1,ξ-ξ1)|2dξdτ)1/2×|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2〈ξ〉2s|ξ|2|ξ-ξ1|2χΩi1.j1.k1(τ,τ1,ξ,ξ1)dξdτ〈τ-ξ5〉2(1-b′)〈τ-τ1-(ξ-ξ1)5〉2b〈ξ-ξ1〉2s)1/2‖Lτ12Lξ12≤‖f‖Lτ2Lξ2‖g‖Lτ2Lξ2×‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2〈ξ〉2s|ξ|2|ξ-ξ1|2χΩi1.j1.k1(τ,τ1,ξ,ξ1)dξdτ〈τ-ξ5〉2(1-b′)〈τ-τ1-(ξ-ξ1)5〉2b〈ξ-ξ1〉2s)1/2‖Lτ1∞Lξ1∞,
where
(3.19)χΩi1.j1.k1(τ,τ1,ξ,ξ1)≡{1,if(τ,τ1,ξ,ξ1)∈Ωi1.j1.k1,0,if(τ,τ1,ξ,ξ1)∉Ωi1.j1.k1.
In Ωi2.j2.k2 ((i2.j2.k2)=(i.1.k),(i=1,2,3,k=1,2),(2.1.3),(3.1.3),(3.2.3)), we integrate with respect to τ and ξ first, then, we use the same way as in (3.18) to have
(3.20)|IΩi2.j2.k2|≤‖f‖Lτ2Lξ2‖g‖Lτ2Lξ2×‖〈ξ〉s|ξ|〈τ-ξ5〉(1-b′)(∬R2|ξ1|2|ξ-ξ1|2χΩi2.j2.k2(τ,τ1,ξ,ξ1)dξ1dτ1〈τ1-ξ15〉2b〈τ-τ1-(ξ-ξ1)5〉2b〈ξ1〉2s〈ξ-ξ1〉2s)1/2‖Lτ∞Lξ∞.
InΩ2.3.2 we use the change of variables
(3.21)τ2=τ1-τ,ξ2=ξ1-ξ
to obtain
(3.22)∫∫∫∫Ω2.3.2〈ξ〉sξξ1(ξ-ξ1)h(τ,ξ)f(τ1,ξ1)g(τ-τ1,ξ-ξ1)〈τ-ξ5〉1-b′〈τ1-ξ15〉b〈τ-τ1-(ξ-ξ1)5〉b〈ξ1〉s〈ξ-ξ1〉sdτ1dξ1dτdξ=∫∫∫∫Ω̃2.3.2〈ξ1-ξ2〉sξ1(ξ1-ξ2)(-ξ2)h(τ1-τ2,ξ1-ξ2)f(τ1,ξ1)g(-τ2,-ξ2)〈τ1-τ2-(ξ-ξ1)5〉1-b′〈τ1-ξ15〉b〈τ2-ξ25〉b〈ξ1〉s〈ξ1〉s〈ξ2〉sdτ1dξ1dτ2dξ2≡JΩ̃2.3.2,
where
(3.23)Ω̃2.3.2={(τ1,τ2,ξ1,ξ2)∈Ω0.0.0c:14|ξ1-ξ2|≤|ξ1|≤4|ξ1-ξ2|,|τ1-τ2-(ξ1-ξ2)5|,|τ1-ξ15|≤|τ2-ξ25|,|ξ2||ξ1|4≤114}.
We integrate with respect to τ2 and ξ2 first, then, we use the same way as in (3.18) to have
(3.24)|JΩ̃2.3.2|≤‖f‖Lτ2Lξ2‖g‖Lτ2Lξ2×‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2〈ξ1-ξ2〉2s|ξ2|2|ξ1-ξ2|2χΩ̃2.3.2(τ1,τ2,ξ1,ξ2)dξ2dτ2〈τ1-τ2-(ξ1-ξ2)5〉2(1-b′)〈τ2-ξ25〉2b〈ξ2〉2s)1/2‖Lτ1∞Lξ1∞.
In Ωi3.3.k3 ((i3.3.k3)=(i.3.k),(i=1,3,k=1,2,3),(2.3.1),(2.3.3),(2.3.4),(3.3.5)), we have by a similar argument to (3.22)
(3.25)IΩi3.3.k3=JΩ̃i3.3.k3,
where Ω̃i3.3.k3 is the region which is obtained from Ωi3.3.k3 by the change of variables  τ2=τ1-τ and ξ2=ξ1-ξ. We integrate with respect to τ1 and ξ1 first, then, we use the same way as in (3.18) to have
(3.26)|JΩ̃i3.3.k3|≤‖f‖Lτ2Lξ2‖g‖Lτ2Lξ2×‖|ξ2|〈τ2-ξ25〉b〈ξ2〉s(∬R2〈ξ1-ξ2〉2s|ξ1|2|ξ1-ξ2|2χΩ̃i3.3.k3(τ1,τ2,ξ1,ξ2)dξ1dτ1〈τ1-τ2-(ξ1-ξ2)5〉2(1-b′)〈τ1-ξ15〉2b〈ξ1〉2s)1/2‖Lτ2∞Lξ2∞.
Now we will get bounds for(3.27)‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2〈ξ〉2s|ξ|2|ξ-ξ1|2χΩi1.j1.k1(τ,τ1,ξ,ξ1)dξdτ〈τ-ξ5〉2(1-b′)〈τ-τ1-(ξ-ξ1)5〉2b〈ξ-ξ1〉2s)1/2‖Lτ1∞Lξ1∞,‖〈ξ〉s|ξ|〈τ-ξ5〉(1-b′)(∬R2|ξ1|2|ξ-ξ1|2χΩi2.j2.k2(τ,τ1,ξ,ξ1)dξ1dτ1〈τ1-ξ15〉2b〈τ-τ1-(ξ-ξ1)5〉2b〈ξ1〉2s〈ξ-ξ1〉2s)1/2‖Lτ∞Lξ∞,‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2〈ξ1-ξ2〉2s|ξ2|2|ξ1-ξ2|2χΩ̃2.3.2dξ2dτ2〈τ1-τ2-(ξ1-ξ2)5〉2(1-b′)〈τ2-ξ25〉2b〈ξ2〉2s)1/2‖Lτ1∞Lξ1∞,‖|ξ2|〈τ2-ξ25〉b〈ξ2〉s(∬R2〈ξ1-ξ2〉2s|ξ2|2|ξ1-ξ2|2χΩ̃i3.3.k3dξ1dτ1〈τ1-τ2-(ξ1-ξ2)5〉2(1-b′)〈τ1-ξ15〉2b〈ξ1〉2s)1/2‖Lτ2∞Lξ2∞.
By using the following methods, we estimate (3.27).
The Case ofΩ0.0.0
Since(3.28)(τ1+(ξ-ξ1)5)-ξ5=τ1-ξ15-5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12),
it follows from (3.2) in Lemma 3.1 with α=2b, β=κ=2(1-b′) that
(3.29)‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2〈ξ〉2s|ξ|2|ξ-ξ1|2χΩ0.0.0(τ,τ1,ξ,ξ1)dξdτ〈τ-ξ5〉2(1-b′)〈τ-τ1-(ξ-ξ1)5〉2b〈ξ-ξ1〉2s)1/2‖Lτ1∞Lξ1∞≤C7,2(1-b′),2b×‖|ξ1|χΩB(ξ1)〈τ1-ξ15〉b〈ξ1〉s(∫R〈ξ〉2s|ξ|2|ξ-ξ1|2χΩA;ξ1(ξ)〈τ1-ξ15-5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2(1-b′)〈ξ-ξ1〉2sdξ)1/2‖Lτ1∞Lξ1∞,
where ΩA;ξ1={ξ:|ξ|,|ξ-ξ1|≤5} and  ΩB={ξ1:|ξ1|≤5}. Since 〈ξ〉s≤max{1,6s}, we have
(3.30)C7,2(1-b′),2b×‖|ξ1|χΩB(ξ1)〈τ1-ξ15〉b〈ξ1〉s(∫R〈ξ〉2s|ξ|2|ξ-ξ1|2χΩA;ξ1(ξ)〈τ1-ξ15-5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2(1-b′)〈ξ-ξ1〉2sdξ)1/2‖Lτ1∞Lξ1∞≤C7,2(1-b′),2bmax⁡⁡{1,6s}52(∫|ξ|≤5|ξ|2dξ)1/2≤M1,s,b,b′,
where M1,s,b,b′ is some constant.The Case ofΩ1.j.3,Ω3.j.3(j=1,2),Ω̃i.3.3(i=1,3) and Ω̃2.3.2
We considerΩ̃2.3.2. By (3.2), we have
(3.31)‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R〈ξ1-ξ2〉2s|ξ2|2|ξ-ξ1|2χΩ̃2.3.2dξ2dτ2〈τ1-τ2-(ξ-ξ1)5〉2(1-b′)〈τ2-ξ25〉2b〈ξ2〉2s)1/2‖Lτ1∞Lξ1∞≤C7,2(1-b′),2bC9,s2|s|×‖|ξ1|2χΩ̃C(ξ1)〈τ1-ξ15〉b(∫R|ξ2|2χΩ̃D;ξ1(ξ2)〈τ1-ξ15+5ξ1ξ2(ξ1-ξ2)(ξ12-ξ1ξ2+ξ22)〉2(1-b′)dξ2)1/2‖Lτ1∞Lξ1∞,
where Ω̃C={ξ1:|ξ1|≥4} and  Ω̃D;ξ1={ξ2:|ξ2|≤|ξ1|-4}. Here we have used (3.16) and (3.17) with the change of variables  ξ2=ξ1-ξ. Noting
(3.32)2b>0,2(1-b′)>0,
we have
(3.33)‖|ξ1|2χΩ̃C(ξ1)〈τ1-ξ15〉b(∫R|ξ2|2χΩ̃D;ξ1(ξ2)〈τ1-ξ15+5ξ1ξ2(ξ1-ξ2)(ξ12-ξ1ξ2+ξ22)〉2(1-b′)dξ2)1/2‖Lτ1∞Lξ1∞≤‖{|ξ1|4χΩ̃C(ξ1)(∫|ξ2|≤|ξ1|-4|ξ2|2dξ2)}1/2‖Lξ1∞≤21/2‖(|ξ1|-8χΩ̃C(ξ1))1/2‖Lξ1∞≤21/2.
Thus, (3.27) is bounded by
(3.34)M2,s,b,b′=C9,s2|s|+1/2max⁡{C7,2(1-b′),2b,C7,2b,2b}
in Ω̃2.3.2. In the same manner as (3.31)-(3.33), (3.27) are bounded by M2,s,b,b′ in Ω1.j.3, Ω3.j.3(j=1,2), and Ω̃i.3.3(i=1,3).The Case ofΩ2.1.1, Ω2.2.1
We considerΩ2.2.1. Since
(3.35)|τ-ξ5|+|τ1-ξ15|+|τ-τ1-(ξ-ξ1)5|≥|τ-ξ5-(τ1-ξ15)-(τ-τ1-(ξ-ξ1)5)|=5|ξ||ξ1||ξ-ξ1||ξ2-ξξ1+ξ12|,
we obtain
(3.36)max⁡{|τ-ξ5|,|τ1-ξ15|,|τ-τ1-(ξ-ξ1)5|}≥53|ξ||ξ1||ξ-ξ1||ξ2-ξξ1+ξ12|.
Noting that -2b<0, we have
(3.37)〈τ1-ξ15〉-2b≤32b〈5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉-2binΩ2.2.1.
Since (3.14) and |ξ-ξ1|≤1 yield
(3.38)|ξ1|≥3|ξ-ξ1|,
we have
(3.39)|2ξ-ξ1|≥|ξ1|-2|ξ-ξ1|≥13|ξ1|.
By (3.2), (3.16), and (3.37), we have
(3.40)‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2|ξ|2|ξ-ξ1|2〈ξ〉2sχΩ2.2.1(τ,τ1,ξ,ξ1)dτdξ〈τ-ξ5〉2(1-b′)〈τ-τ1-(ξ-ξ1)5〉2b〈ξ-ξ1〉2s)1/2‖Lτ1∞Lξ1∞≤C7,2(1-b′),2bC9,s2|s|3b×‖(∫R〈5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉-2b|ξ1|4|ξ-ξ1|2χΩE;ξ(ξ1)〈τ1-ξ15-5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2(1-b′)dξ)1/2‖Lτ1∞Lξ1∞,
where ΩE;ξ={ξ1:|ξ-ξ1|≤1,|ξ1|≥4}. Using the change of variable
(3.41)μ=5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)
and (3.2), we have
(3.42)‖(∫R〈5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉-2b|ξ1|4|ξ-ξ1|2χΩE;ξ(ξ1)〈τ1-ξ15-5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2(1-b′)dξ)1/2‖Lτ1∞Lξ1∞≤‖(∫|μ|≤3|τ1-ξ15|〈μ〉-2b|ξ1|4dμ〈τ1-ξ15-μ〉2(1-b′)5|ξ1||2ξ-ξ1||ξ12-2ξξ1+2ξ2|)1/2‖Lτ1∞Lξ1∞≤‖(∫R(5/6)-1dμ〈μ〉2b〈μ-(τ1-ξ15)〉2(1-b′))1/2‖Lτ1∞Lξ1∞≤C7,2(1-b′),2b(56)-1/2,
where we have used
(3.43)5|ξ1||2ξ-ξ1||ξ12-2ξξ1+2ξ2|≥56|ξ1|4inΩE;ξ,
which follows from (3.39) and
(3.44)|2ξ2-2ξξ1+ξ12|≥|2(ξ-12ξ1)2+12ξ12|≥12|ξ1|2.
Hence (3.27) is bounded by
(3.45)M3,s,b,b′=max⁡{C7,2(1-b′),2b2,C7,2(1-b′),2bC7,2b,2b}C9,s2|s|(3b+31-b′)(56)-1/2
in Ω2.2.2. By a similar argument to (3.37)–(3.42), (3.27) is also bounded by M3,s,b,b′ in Ω2.2.1.The Case ofΩ2.1.3, Ω2.2.3,  and  Ω̃2.3.5
We considerΩ2.2.3. Since (3.14) and |2ξ-ξ1|≤|ξ1|-3/2 yield
(3.46)|2ξ-ξ1|≤12|ξ1|,
we have
(3.47)|ξ-ξ1|≥12(|ξ1|-|2ξ-ξ1|)≥14|ξ1|,|ξ-ξ1|≤12(|ξ1|+|2ξ-ξ1|)≤34|ξ1|,inΩ2.2.3.
By (3.9) and (3.47), we have
(3.48)53|ξ||ξ1||ξ-ξ1||ξ2-ξξ1+ξ12|≥548|ξ1|5,
where we have used
(3.49)|ξ2-ξξ1+ξ12|=|(ξ-12ξ1)2+34ξ12|≥34|ξ1|2.
Therefore using (3.36), we have
(3.50)〈τ1-ξ15〉-2b≤(548)-2b|ξ1|-5inΩ2.2.3.
By (3.2), (3.16), (3.17), (3.47), and (3.50), we have
(3.51)‖|ξ1|〈τ1-ξ15〉b〈ξ1〉s(∬R2|ξ|2|ξ-ξ1|2〈ξ〉2sχΩ2.2.1(τ,τ1,ξ,ξ1)dτdξ〈τ-ξ5〉2(1-b′)〈τ-τ1-(ξ-ξ1)5〉2b〈ξ-ξ1〉2s)1/2‖Lτ1∞Lξ1∞≤C7,2(1-b′),2bC9,s2|s|max⁡{(14)1-s,(34)1-s}(548)-b×‖(∫R|ξ1|-5|ξ1|4|ξ1|2-2sχΩG(ξ1)χΩF;ξ1(ξ)dξ〈τ1-ξ15-5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2(1-b′))1/2‖Lτ1∞Lξ1∞,
where ΩF;ξ1={ξ:|2ξ-ξ1|≤|ξ1|-3/2} andΩG={ξ1:|ξ1|≥4}. Since |2ξ-ξ1|≤|ξ1|-3/2, we have
(3.52)|ξ|≥12(|ξ1|-|2ξ-ξ1|)≥12(|ξ1|-|ξ1|-3/2),|ξ|≤12(|ξ1|-|2ξ-ξ1|)≤12(|ξ1|-|ξ1|-3/2),inΩF;ξ1.
Noting
(3.53)s>-14,2(1-b′)>0,
we have by (3.52)
(3.54)‖(∫R|ξ1|-5|ξ1|4|ξ1|2-2sχΩG(ξ1)χΩF;ξ1(ξ)dξ〈τ1-ξ15-5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2(1-b′))1/2‖Lτ1∞Lξ1∞≤‖{|ξ1|1-2sχΩG(ξ1)(∫ΩH;ξ11dξ)}1/2‖Lτ1∞Lξ1∞≤1,
where
(3.55)ΩH;ξ1={ξ:12(|ξ1|-|ξ1|-3/2)≤|ξ|≤12(|ξ1|+|ξ1|-3/2)}.
Therefore, (3.27) is bounded by
(3.56)M4,s,b,b′=C9,s2|s|max⁡{C7,2(1-b′),2b,C7,2b,2b}max⁡{(14)1-s,(34)1-s}
in Ω2.2.3. Using a similar argument to (3.47)–(3.54), we can get bounds of (3.27) in Ω2.1.3 and Ω̃2.3.5.All the Other Cases
We considerΩ2.1.2. By (3.9), we have
(3.57)|ξ-ξ1|≤|ξ|+|ξ1|≤5|ξ|.
Since 2b′<1+2σ   and   σ=min{s/5+1/20,3/16} yield
(3.58)-32+4(2b′-1)<-32+8σ<0,
we have by (3.16), (3.36), and (3.57)
(3.59)|ξ|2|ξ1|2〈ξ〉2s〈ξ1〉2s≤C9,s253/2|ξ|11/2-8σ|ξ-ξ1|-3/2+8σ,〈τ-ξ5〉2(b′-1)≤(516)2(b′-1)(|ξ-ξ1||ξ|4)-1+2σ,inΩ2.1.2.
By (3.2), (3.17), and (3.59), we have
(3.60)‖〈ξ〉2s|ξ|2〈τ-ξ5〉1-b′(∬R2|ξ1|2|ξ-ξ1|2χΩ2.1.2(τ,τ1,ξ,ξ1)dξ1dτ1〈τ1-ξ15〉2b〈τ-τ1-(ξ-ξ1)5〉2b〈ξ1〉2s〈ξ-ξ1〉2s)1/2‖Lτ∞Lξ∞≤C7,2b,2bC9,s2|s|53/4(516)(b′-1)×‖(∫R|ξ|(11/2)-8σ|ξ-ξ1|(1/2)-2s+8σχΩI;ξ(ξ1)dξ1(|ξ-ξ1||ξ|4)1-2σ〈τ-ξ5+5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2b)1/2‖Lτ∞Lξ∞,
where ΩI;ξ={ξ1:1≤|ξ-ξ1|,|ξ-2ξ1|≥|ξ|-3/2}. Noting that
(3.61)2b>1,σ=min⁡{s5+120,316}
and using (3.3) and the change of variables
(3.62)μ=5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12),
we have
(3.63)‖(∫R|ξ|11/2-8σ|ξ-ξ1|1/2-2s+8σχΩI;ξ(ξ1)(|ξ-ξ1||ξ|4)1-2σ〈τ-ξ5+5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2bdξ1)1/2‖Lτ∞Lξ∞=‖(∫R|ξ|3/2|ξ-ξ1|-1/2-2s+10σχΩI;ξ(ξ1)〈τ-ξ5+5ξξ1(ξ-ξ1)(ξ2-ξξ1+ξ12)〉2bdξ1)1/2‖Lτ∞Lξ∞≤‖(∫|μ|≤3|τ-ξ5||ξ|3/2〈τ-ξ5+μ〉2b5|ξ||ξ-2ξ1||ξ2-2ξξ1+2ξ12|dμ)1/2‖Lτ∞Lξ∞≤(52)-1/2‖(∫|μ|≤3|τ-ξ5|dμ〈τ-ξ5+μ〉2b)1/2‖Lτ∞Lξ∞≤C8,2b(52)-1/2,
where we used the inequality
(3.64)5|ξ||ξ-2ξ1||ξ2-2ξξ1+2ξ12|≥52|ξ|3/4inΩI;ξ,
which follows from
(3.65)|ξ2-2ξξ1+2ξ12|≥12|ξ|2,|ξ-2ξ1|≥|ξ|-3/2.
Thus, (3.27) is bounded by
(3.66)M5,s,b,b′=53/2(516)(b′-b)-1/22|s|C9,smax⁡{C7,2b,2bC8,2b,C7,2(1-b′),2b3b′-1/2}
in Ω2.1.2. In the other regions, we can get bounds of (3.27) by a similar argument to (3.57)–(3.63). Therefore we omit the proof.
Now (3.27) are shown to be bounded. Therefore, combining (3.4), (3.6), (3.12), and (3.18)–(3.26) and setting
(3.67)C3,s,b,b′=M1,s,b,b′+7M2,s,b,b′+2M3,s,b,b′+3M4,s,b,b′+17M5,s,b,b′,
we have (2.17). This completes the proof of Lemma 2.4.Remark 3.2.
We briefly state the reason why the term10u∂x3u is removed from (1.1)2. In order to show the existence of the solution of (1.1)2, we have to prove the following estimate:
(3.68)‖u∂x3v‖Xb′-1s≤C3,s,b,b′‖u‖Xbs‖v‖Xbs.
Unfortunately, we are not able to prove it, because our method can be used to estimate ∂xl(∂xmu∂xnv) in the case where lmn≥1, but not in the case where lmn=0. Therefore, it is necessarily for us to remove the term 10u∂x3u from (1.1)2 unavoidably.


## 4. Analyticity

In this section we prove the analyticity of the solutionu≡u0 given in Proposition 2.1. The proof is established by Propositions 4.5–4.9. To prove these propositions we prepare four lemmas (Lemmas 4.1–4.4).Lemma 4.1 (see [21]).
Let(4.1)s>-74,b∈(12,12+σ),
where σ=min{1/4,(4s+11)/8,(s+6)/5}. Then
(4.2)‖∂x(uv)‖Xb-1s≤M6,s,b‖u‖Xbs‖v‖Xbs,
where M6,s,b is a constant depending on s and b.Lemma 4.2 (see [7]).
Let(4.3)f∈Hs(Rn),g∈Hr(Rn).
Suppose that
(4.4)0≤s,r≤n2,n2≤s+r.
Then, for any σ1<s+r-n/2,
(4.5)‖fg‖Hσ1(Rn)≤M7‖f‖Hs(Rn)‖g‖Hr(Rn)
holds, where M7=M7,s,r,σ1,n is a constant depending on s, r, σ1, and n.Lemma 4.3.
Let(t0,x0) be an arbitrarily fixed point in {(-T,0)∪(0,T)}×ℝ.  (1)
Suppose thatb∈(0,1],r∈(-∞,0]. Then, for a sufficiently small ɛ1>0 such that
(4.6)ɛ1={ɛ4,if-92<r≤0,ɛαwithα>-12-r,ifr≤-92,‖〈Dt,x〉5bg‖Lt2(R;Hxr(R))≤M8,r,b,ɛ1{‖g‖Xb-1r+‖t∂x5g‖Xb-1r+‖P5g‖Xb-1r}
holds for all g∈Xb-1r satisfying
(4.7)supp⁡g⊂B2ɛ1(t0,x0),t∂x5g,P5g∈Xb-1r,
where M8,r,b,ɛ1=M8,r,b,(t0,x0),ɛ1 depends on r, b, (t0,x0), and ɛ.
(2)
Letμ>0. Then, for a sufficiently small ɛ2=ɛ4>0,
(4.8)‖〈Dt,x〉μg‖Lt2Lx2≤M9,μ,ɛ2{‖g‖Ht,xμ-5(R2)+‖t∂x5g‖Ht,xμ-5(R2)+‖P5g‖Ht,xμ-5(R)}
holds for all g∈Ht,xμ-5(ℝ2) satisfying
(4.9)supp⁡g⊂B2ɛ2(t0,x0),t∂x5g,P5g∈Ht,xμ-5(R2),
where M9,μ,ɛ2=M9,μ,(t0,x0),ɛ2 depends on μ, (t0,x0), and ɛ.Letρ(t,x) be a smooth cut-off function around the freezing point (t0,x0) such that ρ∈C0∞(B2ɛ4(t0,x0)).Lemma 4.4.
Lets,b∈ℝ. Then
(4.10)‖ρf‖Xbs≤M10,s,b,ρ,ɛ3ɛ3-|s|-9|b|‖f‖Xbs+4|b|,
where ɛ3=ɛ4 and M10,s,b,ρ,ɛ3=10|b|/2∥〈ɛ34τ-ξ5〉|b|〈ξ〉|s|+4|b|ℱt,xρ(τ,ξ)∥Lτ1Lξ1.Proof.
Lemmas4.3 and 4.4 are proved by the same method as Lemmas  3.2 and 5.2 in [7].Proposition 4.5.
Lets>1/8, and let b∈(1/2,23/40). Then, for a sufficiently small ɛ4>0, there exist positive constants K1,ρ and A1 such that
(4.11)‖ρPku‖Ht,x1/3(R2)+‖ρPku‖Lt2(R;Hx1)≤K1,ρA1kk!
holds for all k=0,1,2,…, where A1=2A0-1 and K1,ρ=K1,s,b,(t0,x0),ɛ,ρ is a constant depending on s, b, (t0,x0), ɛ and ρ.Proof.
By Plancherel Theorem and Lemma4.3 with g=ρPku, we have
(4.12)‖ρPku‖Ht,x1/3(R2)+‖ρPku‖Lt2(R;Hx1)≤2(2π)2‖〈Dt,x〉5bρPku‖Lt2(R;Hxr(R))≤2(2π)2M8,r,b,ɛ4{‖ρPku‖Xb-1r+‖t∂x5(ρPku)‖Xb-1r+‖P5(ρPku)‖Xb-1r},
where
(4.13)r=s-2,if18<s≤2,-158<r≤0,ifs>2.
We note that r≤s-2 holds. Put K2,s,b=|∥u∥|𝒜A0(Xbs). Since (2.8) and Remark 2.2 yield
(4.14)‖Pku‖Xbs≤K2,s,b(A0-1)kk!,k=0,1,2,…,
it follows from Lemma 4.4 that
(4.15)‖ρPku‖Xb-1r≤M10,r,b-1,ρ,ɛ4ɛ-4|r|-36|b-1|‖Pku‖Xb-1s≤K2,s,b-1M10,r,b-1,ρ,ɛ4ɛ-4|r|-36|b-1|(A0-1)kk!,(4.16)‖P5(ρPku)‖Xb-1r≤∑l=055!(5-l)!l!‖P5-lρPl+ku‖Xb-1r≤∑l=055!(5-l)!l!M10,r,b-1,ρl,ɛ4ɛ-4|r|-36|b-1|‖Pl+ku‖Xb-1s≤max⁡0≤l≤5M10,r,b-1,ρl,ɛ4ɛ-4|r|-36|b-1|K2,s,b-1∑l=055!(5-l)!l!(k+l)!2kk!(A0-1)l(2A0-1)kk!≤K3A1kk!,
where ρl=P5-lρ, K3 is some constant and A1=(2A0-1).
Now we estimate∥t∂x5(ρPku)∥Xb-1r. By using
(4.17)t∂x5(ρPku)=tρ(∂x5Pku)+5t∂x2((∂x2ρ)(∂xPku))+5t∂x((∂xρ)(∂x3Pku))+t(∂x5ρ)Pku,(4.18)(∂x5Pku)=-15{Pk+1u-x∂xPku}+tNk(u),
we have
(4.19)‖t∂x5(ρPku)‖Xb-1r≤15{‖ρPk+1u‖Xb-1r+‖ρx∂xPku‖Xb-1r}+‖tρNk(u)‖Xb-1r+5‖∂x2(t(∂x2ρ)(∂xPku))‖Xb-1r+5‖∂x(t(∂xρ)(∂x3Pku))‖Xb-1r+‖t(∂x5ρ)Pku‖Xb-1r.
In the same manner as (4.15), we have
(4.20)‖ρPk+1u‖Xb-1r≤M10,r,b-1,ρ,ɛ4ɛ-4|r|-36|b-1|‖Pk+1u‖Xb-1s≤M10,r,b-1,ρ,ɛ4ɛ-4|r|-36|b-1|K2,s,b-1(k+1)!2kk!(A0-1)(2A0-1)kk!≤K4A1kk!,
where K4 is some constant. By Lemmas 4.1 and 4.4, we have
(4.21)‖ρx∂xPku‖Xb-1r≤‖∂x(ρxPku)‖Xb-1r+‖(∂x(ρx))Pku‖Xb-1r≤M6,r,b‖ρx‖Xbs-2‖Pku‖Xbs-2+M10,r,b-1,∂x(ρx),ɛ4ɛ-4|r|-36|b-1|‖Pku‖Xb-1s≤{M6,r,b‖ρx‖Xbs-2+M10,r,b-1,∂x(ρx),ɛ4ɛ-4|r|-36|b-1|}K2,s,b(A0-1)kk!.
By Lemmas 2.4, 2.5, and 4.4, we have
(4.22)‖tρNk(u)‖Xb-1r≤M10,r,b-1,tρ,ɛ4ɛ-4|r|-36|b-1|×{C4,s,b∑kk!4k4k1!k2!k3!k4!‖Pk1u‖Xbs‖Pk2u‖Xbs‖Pk3u‖Xbs+C3,s,b∑kk!3k4k1!k2!k3!k4!‖Pk1u‖Xbs‖Pk2u‖Xbs}≤M10,r,b-1,tρ,ɛ4ɛ-4|r|-36|b-1|×{C4,s,bK2,s,b3∑k4k4k4!(A0-1)k1+k2+k3+C3,s,bK2,s,b2∑k3k4k3!k4!(A0-1)k1+k2}k!≤M10,r,b-1,tρ,ɛ4ɛ-4|r|-36|b-1|e4/A0-1{C4,s,bK2,s,b3(k+1)2k+C3,s,bK2,s,b2(k+1)(k+2)2k}(2A0-1)kk!≤K5A1kk!,
where K5 is some constant. We also have
(4.23)‖∂x2(t(∂x2ρ)(∂xPku))‖Xb-1r≤‖∂x(t(∂x2ρ)(∂xPku))‖Xb-1s-1≤C3,s-1,b‖t∂xρ‖Xbs-1‖Pku‖Xbs-1≤C3,s-1,bK2,s-1,b‖t∂xρ‖Xbs-1(A0-1)kk!,‖σx(t(∂xρ)(∂x3Pku))‖Xb-1r≤C3,s-2,b‖tρ‖Xbs-2‖∂x2Pku‖Xbs-2≤C3,s-2,b‖tρ‖Xbs-2‖Pku‖Xbs≤C3,s-2,bK2,s,b‖tρ‖Xbs-2(A0-1)kk!.
In the same manner as (4.15), we have
(4.24)‖t(∂x5ρ)Pku‖Xb-1r≤M10,r,b-1,t∂x5ρ,ɛ4K2,s,b-1(A0-1)kk!.
Hence
(4.25)‖t∂x5(ρPku)‖Xb-1r≤K6A1kk!,
where K6 is some constant. Putting
(4.26)K1,ρ=32π2M8,r,b,ɛ4max⁡{K2,s,b-1M10,r,b-1,ρ,ɛ4ɛ-4|r|-36|b-1|,K3,K6},
we have (4.11).Proposition 4.6.
Under the same assumption as in Proposition4.5, there exist K7 and A2 such that
(4.27)‖ρ4Pku‖Ht,x11/2(R)≤K7A2kk!
holds for all k=0,1,2,…, where ρ4 is a smooth cut-off function such that
(4.28)ρ4≤min⁡{ρ,ρ4},ρ4≡1on⁡(t0-ɛ4,t0+ɛ4)×(x0-ɛ4,x0+ɛ4).Proof.
At first, we show that there exists a constantK7,1/2 and A3 such that
(4.29)‖ρ4Pku‖Ht,x1/2(R2)≤K7,1/2A3kk!,k=0,1,2,….
Applying Lemma 4.3 with μ=1/2 and g=ρ4Pku, we have
(4.30)‖〈Dt,x〉1/2ρ4Pku‖Lt2Lx2≤M9,1/2,ɛ4{‖ρ4Pku‖Ht,x-9/2(R2)+‖t∂x5(ρ4Pku)‖Ht,x-9/2(R2)+‖P5(ρ4Pku)‖Ht,x-9/2(R2)}.
By Proposition 4.5, we have
(4.31)‖ρ4Pku‖Ht,x-9/2(R2)≤K1,ρ4A1kk!,‖P5(ρ4Pku)‖Ht,x-9/2(R2)≤∑l=055!l!(5-l)!‖P5-lρ4‖Lt,x∞(R2)‖ρPk+lu‖Lt,x2(R2)≤K1,ρmax⁡0≤l≤5‖P5-lρ4‖Lt,x∞(R2)∑l=055!(5-l)!l!(k+l)!2kk!(A1-1)l(2A1-1)kk!≤K1,ρK8(2A1-1)kk!,
where K8 is some constant.
Now we estimate∥t∂x5(ρ4Pku)∥Ht,x-9/2(ℝ2). Using (4.18) and
(4.32)∂x5(ρ4f)=ρ4∂x5f+∑l=15(-1)l-15!l!(5-l)!∂x5-l((∂xlρ4)Pku),
we obtain
(4.33)‖t∂x5(ρ4Pku)‖Ht,x-9/2(R2)≤15{‖ρ4Pk+1u‖Ht,x-9/2(R2)+‖ρ4x∂xPku‖Ht,x-9/2(R2)}+‖tρ4Nk(u)‖Ht,x-9/2(R2)+∑l=155!l!(5-l)!‖t∂x5-l((∂xlρ4)Pku)‖Ht,x-9/2(R2).
By Proposition 4.5, we have
(4.34)15‖ρ4Pk+1u‖Ht,x-9/2(R2)≤15‖ρ4‖Lt∞Lx∞‖ρPk+1u‖Lt2Lx2≤15K1,ρ‖ρ4‖Lt∞Lx∞(k+1)!2kk!(A1-1)(2A1-1)kk!≤K1,ρK9(2A1-1)kk!,15‖ρ4x∂xPku‖Ht,x-9/2(R2)≤15(‖xρ4‖Lt∞Lx∞+‖∂x(xρ4)‖Lt∞Lx∞)‖ρPku‖Lt2Lx2≤15(‖xρ4‖Lt∞Lx∞+‖∂x(xρ4)‖Lt∞Lx∞)K1,ρA1kk!,
where K9 is some constant. By Sobolev embedding theorem and Proposition 4.5, we have
(4.35)‖tρ4ρ4∂x(Pk2uPk3uPk4u)‖Ht,x-9/2(R2)≤‖〈|τ|+|ξ|〉-7/2‖Lτ2Lξ2(‖∂xtρ4‖Lt∞Lx∞+‖tρ4‖Lt∞Lx∞)‖ρ‖Lt∞Lx∞×‖ρPk2u‖Lt3Lx3‖ρPk3u‖Lt3Lx3‖ρPk4u‖Lt3Lx3≤K10,-7/2‖ρ‖Lt∞Lx∞‖ρPk2u‖Ht,x1/3(R2)‖ρPk3u‖Ht,x1/3(R2)‖ρPk4u‖Ht,x1/3(R2)≤K10,-7/2‖ρ‖Lt∞Lx∞K1,ρ3A1k2+k3+k4k2!k3!k4!,‖tρ4ρ4∂x((∂xPk1u)∂xPk2u))‖Ht,x-9/2(R2)≤K10,-7/2‖ρ2∂xPk1u‖Lt2Lx2‖ρ2∂xPk2u‖Lt2Lx2≤K10,-7/2(‖∂xρ‖Lt∞Lx∞+‖ρ‖Lt∞Lx∞)2‖ρPk1u‖Lt2(R;Hx1)‖ρPk2u‖Lt2(R;Hx1)≤K11,-7/2K1,ρ2A1k1+k2k1!k2!,
where K10,-7/2, K11,-7/2 are some constants. Therefore, we have by (4.35)
(4.36)‖tρ4Nk(u)‖Ht,x-9/2(R2)≤‖tρ4ρ4Nk(u)‖Ht,x-9/2(R2)≤∑kk!4k1k1!k2!k3!k4!‖tρ4ρ4∂x(Pk2uPk3uPk4u))‖Ht,x-9/2(R2)+∑kk!3k4k1!k2!k3!k4!‖tρ4ρ4∂x((∂xPk2u)(∂xPk1u))‖Ht,x-9/2(R2)≤K10,-7/2‖ρ‖Lt∞Lx∞K1,ρ3k!∑k4k4k4!A1k1+k2+k3+K11,-7/2K1,ρ2k!∑k3k4k3!K4!A1k1+k2≤(K10,-7/2‖ρ‖Lt∞Lx∞K1,ρ3+K11,-7/2K1,ρ2)e4/A1A1kk!.
By Proposition 4.5, we have
(4.37)∑l=155!l!(5-l)!‖∂x5-l((t∂xlρ4)Pku)‖Ht,x-9/2(R2)≤∑l=155!l!(5-l)!‖t∂xlρ4‖Lt∞Lx∞‖ρPku‖Lt2Lx2≤K12K1,ρA1kk!,
where K12 is some constant. Hence
(4.38)‖t∂x5(ρ4Pku)‖Ht,x-9/2(R2)≤K13,K1,ρ,K10,-7/2max⁡{2A1-1,A1}kk!,
where K13,K1,ρ,K10,-7/2 is some constant. Putting A3=max{2A1-1,A1},
(4.39)K7,1/2=K7,1/2,K1,ρ,K10,-7/2=3M9,1/2,ɛ4max⁡{K1,ρ4,K1,ρK8,K14,K1,ρ,K10,-7/2},
we have (4.29). Similarly, we can prove by (4.8) with μ=3/2 and (4.29)
(4.40)‖ρ4Pku‖Ht,x3/2(R2)≤K7,3/2,K7,1/2,K10,-5/2A4kk!,
where A4=max{2A3-1,A3}. By (4.8) with μ=5/2 and (4.40), we have
(4.41)‖ρ4Pku‖Ht,x5/2(R2)≤K7,5/2,K7,3/2,K10,0A5kk!,
where A5=max{2A4-1,A4} and K10,0=(∥∂xtρ4∥Lt∞Lx∞+∥tρ4∥Lt∞Lx∞)∥ρ∥Lt∞Lx∞. Repeating the same method as in above, we obtain
(4.42)‖ρ4Pku‖Ht,x7/2(R2)≤K7,7/2,K8,5/2,M7A6kk!,‖ρ4Pku‖Ht,x9/2(R2)≤K7,9/2,K8,7/2,M7A7kk!,‖ρ4Pku‖Ht,x11/2(R2)≤K7,11/2,K8,9/2,K10,0A2kk!,
where A6=max{2A5-1,A5}, A7=max{2A6-1,A6} and A2=max{2A7-1,A7}. Putting K7=K7,11/2,K8,9/2,K10,0, we have (4.27).Remark 4.7.
Whenμ=7/2, μ=9/2, we can obtain the similar estimates to (4.35) by using Lemma 4.2 and Sobolev embedding theorem.Proposition 4.8.
Suppose that (4.27) holds for all k=0,1,2,…. Then
(4.43)sup⁡t∈It0‖(t1/5∂x)lPku‖H1(Ix0)≤K7A8k+l(k+l)!
holds for all k,l=0,1,2,…, where
(4.44)It0=(t0-ɛ4,t0+ɛ4),Ix0=(x0-ɛ4,x0+ɛ4),A8≥max⁡{(|t0|+ɛ4)1/5,A2,(|x0|+ɛ4+1)(|t0-ɛ4|)-1/5+1,(|t0|+ɛ4)4/5K7e4/A8max⁡{|t0-ɛ4|-2/5,K7}}.Proof.
We prove (4.43) by induction on l. When l=0,1,2,3,4, we use the trace theorem and (4.27) to obtain
(4.45)sup⁡t∈It0‖(t1/5∂x)lPku‖H1(Ix0)≤(|t0|+ɛ4)l/5‖∂xlPku‖Ht,x3/2(It0×Ix0)≤(|t0|+ɛ4)l/5‖ρ4Pku‖Ht,x11/2(R2)≤K7(|t0|+ɛ4)l/5A2kk!≤K7A8k+lk!.
We assume that (4.43) holds for any l≥5. Now we will prove
(4.46)sup⁡t∈It0‖(t1/5∂x)l+1Pku‖H1(Ix0)≤K7A8k+l+1(k+l+1)!.
Since
(4.47)(t1/5∂x)lPku=t(l-5)/5∂xl-5(t∂x5Pku)=-15t(l-5)/5∂xl-5{Pk+1u-x∂xPku}+(t1/5)l∂xl-5Nk(u),
we have
(4.48)sup⁡t∈It0‖(t1/5∂x)l+1Pku‖H1(Ix0)≤15‖t(l-4)/5∂xl-4Pk+1u(t)‖H1(Ix0)+15‖t(l-4)/5∂xl-4x∂x(Pku)‖H1(Ix0)+‖(t1/5)l+1∂xl-4Nk(u)‖H1(Ix0).
By (4.43), we have
(4.49)15‖t(l-4)/5∂xl-4Pk+1u(t)‖H1(Ix0)≤15K7A8k+l-3(k+l-3)!.
Since
(4.50)t(l-4)/5∂xl-4(x∂x)=x∂xt(l-4)/5∂xl-4+(l-4)t(l-4)/5∂xl-4(l=5,6,7,…),
we have by (4.43)
(4.51)15‖t(l-4)/5∂xl-4x∂x(Pku)‖H1(Ix0)≤15{‖x∂xt(l-4)/5∂xl-4Pku‖H1(Ix0)+(l-4)‖t(l-4)/5∂xl-4Pku‖H1(Ix0)}≤15(|x0|+ɛ4+1)|t0-ɛ4|-1/5‖t(l-3)/5∂xl-3Pku‖H1(Ix0)+15(l-4)K7A8k+l-4(k+l-4)!≤15(|x0|+ɛ4+1)|t0-ɛ4|-1/5K7A8k+l-3(k+l-3)!+15K7A8k+l-3(k+l-3)!≤15K7A8k+l-2(k+l-3)!.
Now we estimate ∥(t1/5)l+1∂xl-4𝒩k(u)∥H1(Ix0). We have
(4.52)‖(t1/5)l+1∂xl-4Nk(u)‖H1(Ix0)≤(|t0|+ɛ4)4/5×{∑kk!4k4k1!k2!k3!k4!‖t(l-3)/5∂xl-3(Pk1uPk2uPk3u)‖H1(Ix0)+∑kk!3k4k1!k2!k3!k4!‖t(l-3)/5∂xl-3(∂xPk1u∂xPk2u)‖H1(Ix0)}.
Since
(4.53)∑l∑k(4A8-1)k4k4!(l1+k1)!k1!l1!(l2+k2)!k2!l2!(l3+k3)!k3!l3!(l-3)!k!(l+k-3)!≤e4/A8(l+k-2),
we have
(4.54)(|t0|+ɛ4)4/5∑kk!4k4k1!k2!k3!k4!‖t(l-3)/5∂xl-3(Pk1uPk2uPk3u)‖H1(Ix0)≤(|t0|+ɛ4)4/5∑l∑k(l-3)!l1!l2!l3!k!4k4k1!k2!k3!k4!‖(t1/5∂x)l1Pk1u‖H1(Ix0)×‖(t1/5∂x)l2Pk2u‖H1(Ix0)‖(t1/5∂x)l3Pk3u‖H1(Ix0)≤(|t0|+ɛ4)4/5K73(l+k-3)!A8l+k-3×∑l∑k(4A8-1)k4k4!(l1+k1)!k1!l1!(l2+k2)!k2!l2!(l3+k3)!k3!l3!(l-3)!k!(l+k-3)!≤(|t0|+ɛ4)4/5K73A8l+k-3e4/A8(l+k-2)!≤15K7A8k+l(k+l-2)!.
Similarly,
(4.55)(|t0|+ɛ4)4/5∑kk!3k4k1!k2!k3!k4!‖tl-3/5∂xl-3(∂xPk1u∂xPk2u)‖H1(Ix0)≤(|t0|+ɛ4)4/5∑l-3=l1+l2∑k(l-3)!l1!l2!3k3k!k1!k2!k3!k4!×‖(t1/5∂x)l1∂xPk1u‖H1(Ix0)‖(t1/5∂x)l2∂xPk2u‖H1(Ix0)≤(|t0|+ɛ4)4/5|t0-ɛ4|-2/5∑l-1=m1+m2∑k(l-3)!m1m2m1!m2!3k3k!k1!k2!k3!k4!×‖(t1/5∂x)m1Pk1u‖H1(Ix0)‖(t1/5∂x)m2Pk2u‖H1(Ix0),
where m1=l1+1 and m2=l2+1. By m12≥m1 and m22≥m2, we have
(4.56)m1m2=12{(l-1)2-(m12+m22)}≤{(l-1)2-(m1+m2)}≤(l-2)(l-1).
Thus, we have by (4.56), (4.43)
(4.57)(|t0|+ɛ4)4/5(|t0|+ɛ4)-2/5∑l-1=m1+m2∑k(l-3)!m1!m2!3k3k!k1!k2!k3!k4!×‖(t1/5∂x)m1Pk1u‖H1(Ix0)‖(t1/5∂x)m2Pk2u‖H1(Ix0)≤(|t0|+ɛ4)4/5|t0-ɛ4|-2/5K72(l+k-1)!Agl+k-1×∑l-1=m1+m2∑k(3A8-1)k3!k3!A8-k4!k4!(m1+k1)!k1m1!(m2+k2)!k2m2!(l-1)!k!(l+k-1)!≤(|t0|+ɛ4)4/5|t0-ɛ4|-2/5K72e4/As(l+k)!A8l+k-1≤15K7A8k+l+1(K+l)!.
Combining (4.48)–(4.57), we have (4.46). This completes the proof.Proposition 4.9.
Suppose that (4.43) holds for all k,l=0,1,2,…. Then, there exists A9>0 depending on A8, (t0,x0), and ɛ such that
(4.58)sup⁡t∈It0‖∂tm∂xlu‖H1(Ix0)≤K7A9m+l(m+l)!
holds for all m,l=0,1,2,….Proof.
By induction onm, we prove
(4.59)‖(x∂x)m∂xlPku‖H1(Ix0)≤K7A10k+l+mB1m(k+l+m)!k,l,m=0,1,2,….
In the case m=0, we have by (4.43)
(4.60)sup⁡t∈It0‖∂xlPku‖H1(Ix0)≤K7A10lA8k(k+l)!≤K7A10k+l(k+l)!,k,l=0,1,2,…,
where A10=max{A8|t0-ɛ4|-1/5,A8}. We assume (4.59) is valid up to any m. Since
(4.61)∂x(x∂x)m=(x∂x+1)m∂x(m=0,1,2,…),
we have
(4.62)‖(x∂x)m+1∂xlPku‖H1(Ix0)≤(|x0|+ɛ4+1)‖(x∂x+1)m∂xl+1Pku‖H1(Ix0)≤(|x0|+ɛ4+1)∑m1=0mm!m1!(m-m1)!‖(x∂x)m1∂xl+1Pku‖H1(Ix0)≤(|x0|+ɛ4+1)K7A10k+m+l+1B1m(k+m+l+1)!×∑m1=0m(A10B1)-(m-m1)(m-m1)!m!m1!(k+m1+l+1)!(k+m+l+1)!≤(|x0|+ɛ4+1)e-A10B1K7A10k+m+l+1B1m(k+m+l+1)!≤K7A10k+m+l+1B1m+1(k+m+l+1)!,
where B1≥(|x0|+ɛ4+1)e-A10B1. Since t∂t=(P-x∂x)/5 and
(4.63)Pn1∂xn2=∂xn2(P-n2)n1(n1,n2=0,1,2,…),
it follows from (4.59) that
(4.64)‖(t∂t)m∂xlu‖H1(Ix0)≤5-m∑m=m1+m2mm1!m2!‖(x∂x)m1Pm2∂xlu‖H1(Ix0)≤5-m∑m=m1+m2mm1!m2!‖(x∂x)m1∂xl(P-l)m2u‖H1(Ix0)≤5-m∑mm!m1!m2!m3!lm3‖(x∂x)m1∂xlPm2u‖H1(Ix0)≤5-mK7(l+m)!∑mm!m1!m2!m3!A10l+m-m3B1m1lm3(m1+m2+l)!(l+m)!≤5-mK7A10l+mB1m(l+m)!∑mm!m1!m2!m3!(A11B1)-m3B1-m2≤5-mK7max⁡{A10B1,B1}l+m(l+m)!(1+(A10B1)-1+B1-1)m≤K7A11l+m(l+m)!,
where A11=max{1,5-1(1+(A10B1)-1+B1-1)}·max{A10B1,B1}. Thus,
(4.65)‖(t∂t)m∂xlu‖H1(Ix0)≤K7A11l+m(l+m)!l,m=0,1,2,….
By induction on j we prove that (4.65) implies
(4.66)‖(t∂t)m∂tj∂xlu‖H1(Ix0)≤K7A11j+m+lB2j(j+m+l)!j,l,m=0,1,2,….
In the case j=0, we have by (4.65)
(4.67)sup⁡t∈It0‖(t∂t)m∂xlu‖H1(Ix0)≤K7A11l+m(l+m)!k,l=0,1,2,….
We assume that (4.66) is valid up to any j. Noting that
(4.68)(t∂t)m∂t=∂t(t∂t-1)m(j=0,1,2,…),
we have
(4.69)‖(t∂t)m∂tj+1∂xlu‖H1(Ix0)≤‖∂t(t∂t-1)m∂tj∂xlu‖H1(Ix0)≤|t0-ɛ4|-1∑m1=0mm!m1!(m-m1)!‖(t∂t)m1+1∂tj∂xlu‖H1(Ix0)≤|t0-ɛ4|-1K7A11j+l+m+1B2j(j+l+m+1)!×∑m1=0mA11-(m-m1)(m-m1)!m!m1!(j+l+m1+1)!(j+l+m+1)!≤|t0-ɛ4|-1e-A11K7A11j+l+m+1B2j(j+l+m+1)!≤K7A11j+l+m+1B2j+1(j+l+m+1)!,
where B2≥|t0-ɛ4|-1e-A11. Thus (4.66) holds.
Choosingm=0 and A9=max{A11B2,A11} in (4.66), we have (4.58). This completes the proof of Proposition 4.9.



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*Source: 238138-2011-04-17.xml*