Abstract

We consider the initial value problem for the reduced fifth-order KdV-type equation: 𝜕𝑡𝑢𝜕5𝑥𝑢10𝜕𝑥(𝑢3)+10𝜕𝑥(𝜕𝑥𝑢)2=0, 𝑡,𝑥, 𝑢(0,𝑥)=𝜙(𝑥), 𝑥. This equation is obtained by removing the nonlinear term 10𝑢𝜕3𝑥𝑢 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 𝜙𝐻𝑠()(𝑠>1/8) satisfies the condition 𝑘=0(𝐴𝑘0/𝑘!)(𝑥𝜕𝑥)𝑘𝜙𝐻𝑠<, 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).

1. Introduction

The KdV hierarchy is well known as the series of the Lax pair formulation [1, 2], which are presented as 1st-orderKdV𝜕𝑡𝑢𝜕𝑥𝑢=0,(1.1)03rd-orderKdV𝜕𝑡𝑢+𝜕3𝑥𝑢6𝑢𝜕𝑥𝑢=0,(1.1)15th-orderKdV𝜕𝑡𝑢𝜕5𝑥𝑢10𝜕𝑥𝑢3+10𝜕𝑥𝜕𝑥𝑢2+10𝑢𝜕3𝑥𝑢=0.(1.1)2 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: 𝜕𝑡𝑢+𝜕3𝑥𝑢+𝑎(𝑢)𝜕𝑥𝑢=0,𝑡,𝑥,(1.2) where 𝑎(𝜆) is the real analytic and the no growth rate function in 𝜆. 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 𝑗2, because the bilinear estimate of 𝑢𝜕𝑥2𝑗1𝑢 with 𝑗2 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𝜕𝑥2𝑗1𝑢 from the higher-order KdV equations (1.1)0 with 𝑗2. 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: 𝜕𝑡𝑢𝜕5𝑥𝑢=𝜕𝑥𝑢3+𝜕𝑥𝜕𝑥𝑢2,𝑡,𝑥,𝑢(0,𝑥)=𝜙(𝑥),𝑥,(1.3) 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𝜕3𝑥𝑢 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, define that 𝑋𝑠𝑏=𝑓𝒮2;𝑓𝑋𝑠𝑏<,(1.4) where 𝑓2𝑋𝑠𝑏=2||1+𝜏𝜉5||2𝑏||𝜉||1+2𝑠||𝑡,𝑥||𝑓(𝜏,𝜉)2𝑑𝜏𝑑𝜉,(1.5) and 𝑡,𝑥𝑓 is the Fourier transform of 𝑓 in both 𝑥 and 𝑡 variables; that is, 𝑡,𝑥𝑓(𝜏,𝜉)=(2𝜋)12𝑓(𝑡,𝑥)𝑒𝑖𝑡𝜏𝑖𝑥𝜉𝑑𝑡𝑑𝑥.(1.6)

Our main result is the following theorem.

Theorem 1.1. Let 𝑠>1/8 and let 𝑏(1/2,23/40). Then, for any 𝜙(𝑥)𝐻𝑠() such that 𝑥𝜕𝑥𝑘𝜙(𝑥)𝐻𝑠()(𝑘=0,1,2,),𝑘=0𝐴𝑘0𝑘!𝑥𝜕𝑥𝑘𝜙𝐻𝑠<,forsome0<𝐴0<1,(1.7) there exist a constant 𝑇=𝑇(𝜙)>0 and a unique solution 𝑢𝐶((𝑇,𝑇),𝐻𝑠)𝑋𝑠𝑏 of (1.3) satisfying 𝑃𝑘𝑢𝐶((𝑇,𝑇),𝐻𝑠)𝑋𝑠𝑏,𝑘=0𝐴𝑘0𝑃𝑘!𝑘𝑢𝑋𝑠𝑏<,(1.8) 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 constants 𝐶 and 𝐴1 such that ||𝜕𝑚𝑡𝜕𝑙𝑥𝑢||(𝑡,𝑥)𝐶𝐴1𝑚+𝑙(𝑚+𝑙)!(1.9) holds for all (𝑡,𝑥)(𝑇,0)(0,𝑇)× and 𝑙,𝑚=0,1,2,.

Remark 1.2. The initial data 𝜙(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 |𝑥|𝛾𝑒𝑥23with𝛾>8.(1.10)

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𝜕𝑡𝑢+𝜕𝑥2𝑗+1𝑢+𝑃𝑢,𝜕𝑥𝑢,,𝜕𝑥2𝑗1𝑢=0,(1.11) where 𝑗1 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 𝜕𝑡𝑢+𝜕5𝑥𝑢+𝜕𝑥𝑢𝜕2𝑥𝑢+𝑢𝜕3𝑥𝑢=0,(1.12) 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() 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 1𝑠>43,𝑠2𝑎2,where21<𝑎4.(1.13) Then, the local well-posedness for the IVP of (1.1)2 holds in 𝐻𝑠,𝑎(), where 𝐻𝑠,𝑎()𝑓𝒮();𝑓𝐻𝑠,𝑎𝜉𝑠𝑎||𝜉||𝑎𝑓𝐿2𝜉<.(1.14)(2) Let 1𝑠1,21𝑎4.(1.15) Then, the global well-posedness for the IVP of (1.1)2 holds in 𝐻𝑠,𝑎().

The plan of this paper is as follows. In Section 2, we give the existence and uniqueness of the local solution of (1.3), which is shown by the contraction argument consisting of Lemmas 2.32.5. 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. Let x be the Fourier transform in the x variable, and let 𝜉1 and 1𝜏,𝜉 be the Fourier inverse transform in the 𝜉 and (𝜏,𝜉) variables, respectively. The Riesz operator Dx and its fractional derivative Dxs are defined by 𝐷𝑥=𝜉1||𝜉||𝑥,𝐷𝑥𝑠=𝜉1𝜉𝑠𝑥,(1.16) respectively, where =(1+||). Similarly, 𝐷𝑡,𝑥𝑠 is defined by 𝐷𝑡,𝑥𝑠=1𝜏,𝜉||𝜉|||𝜏|+𝑠𝑡,𝑥.(1.17)[𝐴,𝐵] denotes the commutator relation of two operators given by 𝐴𝐵𝐵𝐴. 𝐿𝑝𝑡𝐿𝑞𝑥 denotes the space 𝐿𝑝(𝑡;𝐿𝑞(𝑥)) for 1𝑝,𝑞 with the norm 𝑓𝐿𝑝𝑡𝐿𝑞𝑥=||||𝑓(𝑡,𝑥)𝑝𝑑𝑡𝑞/𝑝𝑑𝑥1/𝑞.(1.18) We use Sobolev spaces with both time and space variables 𝐻𝑠𝑡,𝑥2=𝑢𝒮2𝐷𝑡,𝑥𝑠𝑢𝐿2𝑡𝐿2𝑥,(1.19) with the norm 𝐻𝑠𝑡,𝑥(2)=𝐷𝑡,𝑥𝑠𝐿2𝑡𝐿2𝑥. Moreover, 𝐿2𝑡(;𝐻𝑠𝑥) denotes the space 𝐿2(𝑡;𝐻𝑠(𝑥)) with the norm 𝐿2𝑡(;𝐻𝑠𝑥)=𝐷𝑥𝑠𝐿2𝑡𝐿2𝑥. The dual coupling is expressed as 𝑓,𝑔. The convolution of 𝑓 and 𝑔 with both space and time variables is denoted by 𝑓𝑔. For the constant 𝐴0 appearing in Theorem 1.1, we put 𝒜𝐴0𝑋𝑠𝑏=𝑓𝐟=0,𝑓1,;𝑓𝑘𝑋𝑠𝑏||𝐟||(𝑘=0,1,),𝒜𝐴0(𝑋𝑠𝑏),<(1.20) where ||||𝐟𝒜𝐴0(𝑋𝑠𝑏)𝑘=0𝐴𝑘0𝑓𝑘!𝑘𝑋𝑠𝑏.(1.21) For simplicity we make use of three notations: 𝐤=𝑘=𝑘1+𝑘2+𝑘3+𝑘4,𝐥=𝑙=𝑙1+𝑙2+𝑙3,𝐦=𝑚=𝑚1+𝑚2+𝑚3.(1.22)

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(𝑃+𝑙)𝑘𝜕𝑥=𝜕𝑥(𝑃+(𝑙1))𝑘,𝑘,𝑙=0,1,2,.(2.1) Using (2.1) and the following relations 𝜕𝑡𝜕5𝑥𝜕,𝑃=5𝑡𝜕5𝑥,𝜕𝑡𝜕5𝑥𝑃𝑘=(𝑃+5)𝑘𝜕𝑡𝜕5𝑥,(2.2) we have from (1.3)𝜕𝑡𝑢𝑘𝜕5𝑥𝑢𝑘=𝒩𝑘𝑢(𝑢),𝑡,𝑥,𝑘(0,𝑥)=𝜙𝑘(𝑥),𝑥,𝑘=0,1,2,,(2.3) where𝒩𝑘(𝑢)=𝜕𝑥(𝑃+4)𝑘𝑢3+𝜕𝑥(𝑃+4)𝑘𝜕𝑥𝑢2.(2.4) Using the Leibniz rule and (2.1), we can see that 𝒩𝑘(𝑢)=𝜕𝑥𝑘𝑙=0𝑘𝑙4𝑘𝑙𝑃𝑙𝑢3+𝜕𝑥𝑘𝑙=0𝑘𝑙3𝑘𝑙(𝑃+1)𝑙𝜕𝑥𝑢2=𝐤𝑘!𝑘1!𝑘2!𝑘3!𝑘4!4𝑘4𝜕𝑥𝑢𝑘1𝑢𝑘2𝑢𝑘3+𝐤𝑘!𝑘1!𝑘2!𝑘3!𝑘4!3𝑘4(1)𝑘3𝜕𝑥𝜕𝑥𝑢𝑘1𝜕𝑥𝑢𝑘2.(2.5) We will show the existence and uniqueness of the solution of (2.3).

Proposition 2.1. Let 1𝑠>41,𝑏2,12,+𝜎(2.6) where 𝜎=min{𝑠/5+1/20,3/16}. Then, for any 𝜙(𝜙0,𝜙1,) such that 𝜙𝑘𝐻𝑠()(𝑘=0,1,) and ||||𝜙𝒜𝐴0(𝐻𝑠)<,(2.7) there exist a constant 𝑇=𝑇(𝜙)>0 and a unique solution 𝑢𝑘𝐶((𝑇,𝑇),𝐻𝑠)𝑋𝑠𝑏 of (2.3) satisfying ||||𝐮𝒜𝐴0(𝑋s𝑏)𝑢<,𝐮0,𝑢1.,(2.8)

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 (Lemmas 2.3, 2.4 and 2.5), which play an important role in applying the contraction principle to the following system of the integral equations:𝜓(𝑡)𝑢𝑘=𝜓(𝑡)𝑒𝑡𝜕5𝑥𝜙𝑘+𝜓(𝑡)𝑡0𝑒(𝑡𝑡)𝜕5𝑥𝜓𝑇𝑡𝒩𝑘(𝑡𝑢)𝑑𝑡,(2.9) where 𝑒𝑡𝜕5𝑥𝑓𝜉1𝑒𝑖𝜉5𝑡𝑓(𝜉),(2.10)𝜓(𝑡) denotes a cut-off function in 𝐶0() satisfying 𝜓(𝑡)=1,if|𝑡|1,0,if|𝑡|>2,(2.11) and 𝜓𝑇(𝑡)=𝜓(𝑡/𝑇).

Lemma 2.3. Let 0<𝑇<1 and let 1𝑠,𝑏21,1,𝑎,𝑎0,2𝑎.<𝑎(2.12) Then 𝜓(𝑡)𝑒𝑡𝜕5𝑥𝜙(𝑥)𝑋𝑠𝑏𝐶0,𝑠,𝑏𝜙𝐻𝑠,(2.13)𝜓(𝑡)𝑡0𝑒(𝑡𝑡)𝜕5𝑥𝑡𝑑𝑡𝑋𝑠𝑏𝐶1,𝑠,𝑏𝑋𝑠𝑏1𝜓,(2.14)𝑇𝑋𝑠𝑎𝐶2,𝑠,𝑎,𝑎𝑇(𝑎a)/4(1𝑎)𝑋𝑠a,(2.15) 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 𝑒𝑡𝜕3𝑥 by 𝑒𝑡𝜕5𝑥 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 1𝑠>41,𝑏,𝑏2,12+𝜎𝑏𝑏,(2.16) where 𝜎=min{𝑠/5+1/20,3/16}. Then 𝜕𝑥𝜕𝑥𝑢𝜕𝑥𝑣𝑋𝑠𝑏1𝐶3,𝑠,𝑏,𝑏𝑢𝑋𝑠𝑏𝑣𝑋𝑠𝑏,(2.17) where 𝐶3,𝑠,𝑏,𝑏 is a constant depending on 𝑠, 𝑏, and 𝑏.

We give for the proof of this lemma, in Section 3.

Lemma 2.5. Let 1𝑠>4,𝑏,𝑏12,34(𝑏𝑏).(2.18) Then 𝜕𝑥(𝑢𝑣𝑤)𝑋𝑠𝑏1𝐶4,𝑠,𝑏,𝑏𝑢𝑋𝑠𝑏𝑣𝑋𝑠𝑏𝑤𝑋𝑠𝑏,(2.19) 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 where 𝑏=𝑏(1/2,3/4) [21].

Proof of Proposition 2.1. We define 𝑋𝑀0=𝐟𝒜𝐴0𝑋𝑠𝑏;||||𝐟𝒜𝐴0(𝑋𝑠𝑏)2𝐶0𝑀0,(2.20) where 𝐶0=𝐶0,𝑠,𝑏,𝑀0=||||𝜙𝒜𝐴0(𝐻𝑠).(2.21) We define a map Φ𝑋𝑀0𝑋𝑀0 by Φ(𝑢)=(Φ0(𝑢),Φ1(𝑢),) and Φ𝑘(𝑢)=𝜓(𝑡)𝑒𝑡𝜕5𝑥𝜙𝑘+𝜓(𝑡)𝑡0𝑒(𝑡𝑡)𝜕5𝑥𝜓𝑇𝑡𝒩𝑘(𝑡𝑢)𝑑𝑡.(2.22) Let 𝑏 and 𝑇 be positive constants satisfying 𝑏<𝑏<1/2+𝜎 and 𝑇<𝑚𝑖𝑛1,24𝐶20𝐶5𝑒4𝐴0𝑀20+8𝐶0𝐶6𝑒4𝐴0𝑀01/𝜇,(2.23) respectively, where 𝐶5=𝐶1,𝑠,𝑏𝐶2,𝑠,𝑏1,𝑏1𝐶4,𝑠,𝑏,𝑏𝐶6=𝐶1,𝑠,𝑏𝐶2,𝑠,𝑏1,𝑏1𝐶3,𝑠,𝑏,𝑏.(2.24) 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 𝑢𝒜𝐴0(𝑋𝑠𝑏)Φ𝑘(𝑢)𝑋𝑠𝑏𝐶0Φ𝑘𝐻𝑠+𝐶5𝑇𝜇𝐤𝑘!𝑘1!𝑘2!𝑘3!𝑘4!4𝑘4𝑢𝑘1𝑋𝑠𝑏𝑢𝑘2𝑋𝑠𝑏𝑢𝑘3𝑋𝑠𝑏+𝐶6𝑇𝜇𝐤𝑘!𝑘1!𝑘2!𝑘3!𝑘4!3𝑘4𝑢𝑘1𝑋𝑠𝑏𝑢𝑘2𝑋𝑠𝑏,(2.25) for any 𝑘0. Here 𝜇=(𝑏𝑏)/4𝑏>0. By taking a sum over 𝑘, we have ||(||Φ𝑢)𝒜𝐴0(𝑋𝑠𝑏)𝑘=0𝐴0𝑘Φ𝑘!𝑘(𝑢)𝑋𝑠𝑏𝐶0||||𝜙𝒜𝐴0(𝐻𝑠)+𝐶5𝑇𝜇𝑘4=04𝐴0𝑘4𝑘4!𝑘1=0𝐴𝑘10𝑘1!𝑢𝑘1𝑋𝑠𝑏𝑘2=0𝐴𝑘20𝑘2!𝑢𝑘2𝑋𝑠𝑏𝑘3=0𝐴𝑘30𝑘3!𝑢𝑘3𝑋𝑠𝑏+𝐶6𝑇𝜇𝑘4=03𝐴0𝑘4𝑘4!𝑘3=0𝐴𝑘30𝑘3!𝑘1=0𝐴𝑘10𝑘1!𝑢𝑘1𝑋𝑠𝑏𝑘2=0𝐴𝑘20𝑘2!𝑢𝑘2𝑋𝑠𝑏.(2.26) Since 𝐮𝑋𝑀0, we have from (2.23) ||||Φ(𝑢)𝒜𝐴0(𝑋𝑠𝑏)𝐶0||||𝜙𝒜𝐴0(𝐻𝑠)+𝐶5𝑒4𝐴0𝑇𝜇||||𝐮3𝒜𝐴0(𝑋𝑠𝑏)+𝐶6𝑒4𝐴0𝑇𝜇||||𝐮2𝒜𝐴0(𝑋𝑠𝑏)𝐶0𝑀0+8𝐶30𝐶5𝑒4𝐴0𝑇𝜇𝑀30+4𝐶20𝐶6𝑒4𝐴0𝑇𝜇𝑀2032𝐶0𝑀0,(2.27) which implies Φ(𝑢)𝑋𝑀0. Similarly, we have for 𝑢 and ̃𝑢𝒜𝐴0(𝑋𝑠𝑏)||||Φ(𝑢)Φ(̃𝑢)𝒜𝐴0(𝑋𝑠𝑏)𝐶5𝑒4𝐴0𝑇𝜇×||||𝐮2𝒜𝐴0𝑋𝑠𝑏+||||𝐮𝒜𝐴0(𝑋𝑠𝑏)||̃||𝐮𝒜𝐴0(𝑋𝑠𝑏)+||̃||𝐮2𝒜𝐴0𝑋𝑠𝑏||̃||𝐮𝐮𝒜𝐴0(𝑋𝑠𝑏)+𝐶6𝑒4𝐴0𝑇𝜇||||𝐮𝒜𝐴0(𝑋𝑠𝑏)+||̃||𝐮𝒜𝐴0(𝑋𝑠𝑏)||̃||𝐮𝐮𝒜𝐴0(𝑋𝑠𝑏)12𝐶20𝐶5𝑒4𝐴0𝑀20+4𝐶0𝐶6𝑒4𝐴0𝑀0𝑇𝜇||̃||𝐮𝐮𝒜𝐴0(𝑋𝑠𝑏)12||̃||𝐮𝐮𝒜𝐴0(𝑋𝑠𝑏).(2.28) Thus, the mapping Φ is contraction from 𝑋𝑀0 to itself. We obtain a unique fixed point 𝑢𝑘𝑋𝑠𝑏 satisfying 𝑢𝑘(𝑡)=𝜓(𝑡)𝑒𝑡𝜕5𝑥𝜙𝑘+𝜓(𝑡)𝑡0𝑒(𝑡𝑡)𝜕5𝑥𝜓𝑇𝑡𝒩𝑘(𝑡𝑢)𝑑𝑡(2.29) 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 Lemma 2.4

In this section we prove Lemma 2.4. To prove Lemma 2.4 we prepare the following lemma.

Lemma 3.1 (see [22]). (1) Let 𝛼,𝛽>0 and let 𝜅=min{𝛼,𝛽}. If 𝛼+𝛽>1+𝜅,(3.1) then 𝑑𝑥||||1+𝑥𝜁𝛼||||1+𝑥𝜂𝛽1/2𝐶7,𝛼,𝛽1||||1+𝜁𝜂𝜅1/2,forany𝜁,𝜂,(3.2) where 𝐶7,𝛼,𝛽 is a constant depending on 𝛼 and 𝛽.
(2) If 𝛾>1, then 𝑑𝑥||||1+𝑥+𝜂𝛾1/2𝐶8,𝛾,forany𝜂,(3.3) where 𝐶8,𝛾 is a constant depending on 𝛾.

Proof of Lemma 2.4. By duality, we have 𝜕𝑥𝜕𝑥𝑢𝜕𝑥𝑣𝑋𝑠𝑏1=𝜏𝜉5(𝑏1)𝜉𝑠𝜉𝑡,𝑥𝜕𝑥𝑢𝑡,𝑥𝜕𝑥𝑢𝐿2𝜏𝐿2𝜉=sup𝐿2𝜏𝐿2𝜉,𝐿2𝜏𝐿2𝜉1||||||𝜉𝑠𝜉𝜏𝜉51𝑏𝑡,𝑥𝜕𝑥𝑢𝑡,𝑥𝜕𝑥𝑣,𝐿2𝜏𝐿2𝜉||||||,(3.4) where (,)𝐿2𝜏𝐿2𝜉 is the inner product in 𝐿2(𝜏×𝜉). Setting 𝑓(𝜏,𝜉)=𝜏𝜉5𝑏𝜉𝑠𝑡,𝑥𝑢(𝜏,𝜉),𝑔(𝜏,𝜉)=𝜏𝜉5𝑏𝜉𝑠𝑡,𝑥𝑣(𝜏,𝜉),(3.5) we have 𝜉𝑠𝜉𝜏𝜉51𝑏𝑡,𝑥𝜕𝑥𝑢𝑡,𝑥𝜕𝑥𝑣,𝐿2𝜏𝐿2𝜉=2𝜉𝑠𝜉𝜏𝜉51𝑏×2𝜉1𝜉𝜉1𝑓𝜏1,𝜉1𝑔𝜏𝜏1,𝜉𝜉1𝜉1𝑠𝜉𝜉1𝑠𝜏1𝜉51𝑏𝜏𝜏1𝜉𝜉15𝑏𝑑𝜏1𝑑𝜉1(𝜏,𝜉)𝑑𝜏𝑑𝜉=𝐼Ω0.0.0+𝐼Ω𝑐0.0.0,(3.6) where 𝐼Ω0.0.0=Ω0.0.0𝜉𝑠𝜉𝜉1𝜉𝜉1𝜏(𝜏,𝜉)𝑓1,𝜉1𝑔𝜏𝜏1,𝜉𝜉1𝜏𝜉51𝑏𝜏1𝜉51𝑏𝜏𝜏1𝜉𝜉15𝑏𝜉1𝑠𝜉𝜉1𝑠𝑑𝜏1𝑑𝜉1𝐼𝑑𝜏𝑑𝜉,Ω𝑐0.0.0=Ω𝑐0.0.0𝜉𝑠𝜉𝜉1𝜉𝜉1𝜏(𝜏,𝜉)𝑓1,𝜉1𝑔𝜏𝜏1,𝜉𝜉1𝜏𝜉51𝑏𝜏1𝜉51𝑏𝜏𝜏1𝜉𝜉15𝑏𝜉1𝑠𝜉𝜉1𝑠𝑑𝜏1𝑑𝜉1Ω𝑑𝜏𝑑𝜉,(3.7)0.0.0=𝜏,𝜏1,𝜉,𝜉14||𝜉||,||𝜉1||,||𝜉𝜉1||5,Ω𝑐0.0.0=4Ω0.0.0.(3.8)
We split Ω𝑐0.0.0 into three regions, Ω1,Ω2,andΩ3: Ω1=𝜏,𝜏1,𝜉,𝜉1Ω𝑐0.0.0||𝜉||14||𝜉1||,Ω2=𝜏,𝜏1,𝜉,𝜉1Ω𝑐0.0.014||𝜉1||||𝜉||||𝜉41||,Ω3=𝜏,𝜏1,𝜉,𝜉1Ω𝑐0.0.0||𝜉41||||𝜉||,(3.9) and then, we split Ω𝑖(𝑖=1,2,3) into three regions: Ω𝑖.1=𝜏,𝜏1,𝜉,𝜉1Ω𝑖||𝜏1𝜉51||,|||𝜏𝜏1𝜉𝜉15|||||𝜏𝜉5||,Ω𝑖.2=𝜏,𝜏1,𝜉,𝜉1Ω𝑖||𝜏𝜉5||,|||𝜏𝜏1𝜉𝜉15|||||𝜏1𝜉51||,Ω𝑖.3=𝜏,𝜏1,𝜉,𝜉1Ω𝑖||𝜏𝜉5||,||𝜏1𝜉51|||||𝜏𝜏1𝜉𝜉15|||.(3.10) We further split Ω1.𝑗, Ω2.𝑗, and Ω3.𝑗(𝑗=1,2,3) into the following regions: Ω1.𝑗.1=𝜏,𝜏1,𝜉,𝜉1Ω1.𝑗||𝜉||,Ω11.𝑗.2=𝜏,𝜏1,𝜉,𝜉1Ω1.𝑗||𝜉||||𝜉||||𝜉1,1||4,Ω11.𝑗.3=𝜏,𝜏1,𝜉,𝜉1Ω1.𝑗||𝜉||||𝜉1||4,Ω12.1.1=𝜏,𝜏1,𝜉,𝜉1Ω2.1||𝜉𝜉1||,Ω12.1.2=𝜏,𝜏1,𝜉,𝜉1Ω2.1||𝜉𝜉1||||1,𝜉2𝜉1||||𝜉||3/2,Ω2.1.3=𝜏,𝜏1,𝜉,𝜉1Ω2.1||𝜉𝜉1||||1,𝜉2𝜉1||||𝜉||3/2,Ω2.2.1=𝜏,𝜏1,𝜉,𝜉1Ω2.2||𝜉𝜉1||,Ω12.2.2=𝜏,𝜏1,𝜉,𝜉1Ω2.2||𝜉𝜉1||||1,2𝜉𝜉1||||𝜉1||3/2,Ω2.2.3=𝜏,𝜏1,𝜉,𝜉1Ω2.2||𝜉𝜉1||||1,2𝜉𝜉1||||𝜉1||3/2,Ω2.3.1=𝜏,𝜏1,𝜉,𝜉1Ω2.3||𝜉𝜉1||||1,𝜉𝜉1||||𝜉1||4,Ω12.3.2=𝜏,𝜏1,𝜉,𝜉1Ω2.3||𝜉𝜉1||||𝜉1||4,Ω12.3.3=𝜏,𝜏1,𝜉,𝜉1Ω2.3||𝜉𝜉1||||1,𝜉+𝜉1||,Ω12.3.4=𝜏,𝜏1,𝜉,𝜉1Ω2.3||𝜉𝜉1||||1,𝜉𝜉1||3/2||𝜉+𝜉1||,Ω12.3.5=𝜏,𝜏1,𝜉,𝜉1Ω2.3||𝜉𝜉1||||1,𝜉+𝜉1||||𝜉𝜉1||3/2,Ω3.𝑗.1=𝜏,𝜏1,𝜉,𝜉1Ω3.𝑗||𝜉1||,Ω13.𝑗.2=𝜏,𝜏1,𝜉,𝜉1Ω3.𝑗||𝜉1||||𝜉1,1||||𝜉||4,Ω13.𝑗.3=𝜏,𝜏1,𝜉,𝜉1Ω3.𝑗||𝜉1||||𝜉||4,1(3.11) so that, we have ||𝐼Ω0.0.0+𝐼Ω𝑐0.0.0||||𝐼Ω0.0.0||+3𝑖,𝑗,𝑘=1|||𝐼Ω𝑖.𝑗.𝑘|||+||𝐼Ω2.3.4||+||𝐼Ω2.3.5||.(3.12) 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 34||𝜉1||||𝜉𝜉1||54||𝜉1||,inΩ1,34||𝜉||||𝜉𝜉1||54||𝜉||,inΩ3,||𝜉(3.13)1||4,inΩ1Ω2.2.3,||𝜉min1||,||𝜉||1,inΩ2Ω4Ω2.1.3Ω2.2.3,||𝜉min1||,||𝜉||3,inΩ4,||𝜉||4,inΩ3Ω2.1.3,(3.14) where Ω4=Ω2.1.1Ω2.2.1Ω2.3.1Ω2.3.2. Since (3.13) and (3.14) yield ||𝜉𝜉1||3,inΩ1,Ω3,(3.15) we have by (3.9) and (3.13)–(3.15) ||𝜉1||2||𝜉𝜉1||2𝜉12𝑠𝜉𝜉12𝑠𝐶29,𝑠||𝜉1||44𝑠,inΩ1,||𝜉||2||𝜉1||2𝜉2𝑠𝜉12𝑠𝐶29,𝑠||𝜉1||4,inΩ2,||𝜉||2||𝜉𝜉1||2𝜉2𝑠𝜉𝜉12𝑠𝐶29,𝑠||𝜉||4,inΩ3,(3.16) where 𝐶9,𝑠=4|𝑠|+1. Using (3.11), (3.14), and (3.15), we have 𝜉2𝑠22|𝑠|||𝜉||max1,2𝑠inΩ1,𝜉𝜉12𝑠22|𝑠|||max1,𝜉𝜉1||2𝑠inΩ2,𝜉12𝑠22|𝑠|||𝜉max1,1||2𝑠inΩ3.(3.17)
In Ω𝑖1.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 𝐿2𝜏𝐿2𝜉1 to have |||𝐼Ω𝑖111.𝑗.𝑘|||𝑓𝐿2𝜏1𝐿2𝜉1×||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2(𝜏,𝜉)𝑔𝜏𝜏1,𝜉𝜉1𝜉𝑠||𝜉||||𝜉𝜉1||𝜒Ω𝑖111.𝑗.𝑘𝜏,𝜏1,𝜉,𝜉1𝑑𝜉𝑑𝜏𝜏𝜉51𝑏𝜏𝜏1𝜉𝜉15𝑏𝜉𝜉1𝑠𝐿2𝜏1𝐿2𝜉1𝑓𝐿2𝜏𝐿2𝜉2||||(𝜏,𝜉)2||𝑔𝜏𝜏1,𝜉𝜉1||2𝑑𝜉𝑑𝜏1/2×||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2𝜉2𝑠||𝜉||2||𝜉𝜉1||2𝜒Ω𝑖111.𝑗.𝑘𝜏,𝜏1,𝜉,𝜉1𝑑𝜉𝑑𝜏𝜏𝜉52(1𝑏)𝜏𝜏1𝜉𝜉152𝑏𝜉𝜉12𝑠1/2𝐿2𝜏1𝐿2𝜉1𝑓𝐿2𝜏𝐿2𝜉𝑔𝐿2𝜏𝐿2𝜉×||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2𝜉2𝑠||𝜉||2||𝜉𝜉1||2𝜒Ω𝑖111.𝑗.𝑘𝜏,𝜏1,𝜉,𝜉1𝑑𝜉𝑑𝜏𝜏𝜉52(1𝑏)𝜏𝜏1𝜉𝜉152𝑏𝜉𝜉12𝑠1/2𝐿𝜏1𝐿𝜉1,(3.18) where 𝜒Ω𝑖111.𝑗.𝑘𝜏,𝜏1,𝜉,𝜉11,if𝜏,𝜏1,𝜉,𝜉1Ω𝑖1.𝑗1.𝑘1,0,if𝜏,𝜏1,𝜉,𝜉1Ω𝑖1.𝑗1.𝑘1.(3.19) 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 |||𝐼Ω𝑖222.𝑗.𝑘|||𝑓𝐿2𝜏𝐿2𝜉𝑔𝐿2𝜏𝐿2𝜉×𝜉𝑠||𝜉||𝜏𝜉5(1𝑏)2||𝜉1||2||𝜉𝜉1||2𝜒Ω𝑖222.𝑗.𝑘𝜏,𝜏1,𝜉,𝜉1𝑑𝜉1𝑑𝜏1𝜏1𝜉512𝑏𝜏𝜏1𝜉𝜉152𝑏𝜉12𝑠𝜉𝜉12𝑠1/2𝐿𝜏𝐿𝜉.(3.20)
In Ω2.3.2 we use the change of variables 𝜏2=𝜏1𝜏,𝜉2=𝜉1𝜉(3.21) to obtain Ω2.3.2𝜉𝑠𝜉𝜉1𝜉𝜉1𝜏(𝜏,𝜉)𝑓1,𝜉1𝑔𝜏𝜏1,𝜉𝜉1𝜏𝜉51𝑏𝜏1𝜉51𝑏𝜏𝜏1𝜉𝜉15𝑏𝜉1𝑠𝜉𝜉1𝑠𝑑𝜏1𝑑𝜉1=Ω𝑑𝜏𝑑𝜉2.3.2𝜉1𝜉2𝑠𝜉1𝜉1𝜉2𝜉2𝜏1𝜏2,𝜉1𝜉2𝑓𝜏1,𝜉1𝑔𝜏2,𝜉2𝜏1𝜏2𝜉𝜉151𝑏𝜏1𝜉51𝑏𝜏2𝜉52𝑏𝜉1𝑠𝜉1𝑠𝜉2𝑠𝑑𝜏1𝑑𝜉1𝑑𝜏2𝑑𝜉2Ω𝐽2.3.2,(3.22) where Ω2.3.2=𝜏1,𝜏2,𝜉1,𝜉2Ω𝑐0.0.014||𝜉1𝜉2||||𝜉1||||𝜉41𝜉2||,|||𝜏1𝜏2𝜉1𝜉25|||,||𝜏1𝜉51||||𝜏2𝜉52||,||𝜉2||||𝜉1||4.1(3.23) We integrate with respect to 𝜏2 and 𝜉2 first, then, we use the same way as in (3.18) to have |||𝐽Ω2.3.2|||𝑓𝐿2𝜏𝐿2𝜉𝑔𝐿2𝜏𝐿2𝜉×||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2𝜉1𝜉22𝑠||𝜉2||2||𝜉1𝜉2||2𝜒Ω2.3.2𝜏1,𝜏2,𝜉1,𝜉2𝑑𝜉2𝑑𝜏2𝜏1𝜏2𝜉1𝜉252(1𝑏)𝜏2𝜉522𝑏𝜉22𝑠1/2𝐿𝜏1𝐿𝜉1.(3.24) 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) 𝐼Ω𝑖33.3.𝑘Ω=𝐽𝑖33.3.𝑘,(3.25) where Ω𝑖3.3.𝑘3 is the region which is obtained from Ω𝑖3.3.𝑘3 by the change of variables 𝜏2=𝜏1𝜏 and 𝜉2=𝜉1𝜉. We integrate with respect to 𝜏1 and 𝜉1 first, then, we use the same way as in (3.18) to have ||||𝐽Ω𝑖33.3.𝑘||||𝑓𝐿2𝜏𝐿2𝜉𝑔𝐿2𝜏𝐿2𝜉×|𝜉2|𝜏2𝜉52𝑏𝜉2𝑠2𝜉1𝜉22𝑠||𝜉1||2||𝜉1𝜉2||2𝜒Ω𝑖33.3.𝑘𝜏1,𝜏2,𝜉1,𝜉2𝑑𝜉1𝑑𝜏1𝜏1𝜏2𝜉1𝜉252(1𝑏)𝜏1𝜉512𝑏𝜉12𝑠1/2𝐿𝜏2𝐿𝜉2.(3.26)
Now we will get bounds for ||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2𝜉2𝑠||𝜉||2||𝜉𝜉1||2𝜒Ω𝑖111.𝑗.𝑘𝜏,𝜏1,𝜉,𝜉1𝑑𝜉𝑑𝜏𝜏𝜉52(1𝑏)𝜏𝜏1𝜉𝜉152𝑏𝜉𝜉12𝑠1/2𝐿𝜏1𝐿𝜉1,𝜉𝑠||𝜉||𝜏𝜉5(1𝑏)2||𝜉1||2||𝜉𝜉1||2𝜒Ω𝑖222.𝑗.𝑘𝜏,𝜏1,𝜉,𝜉1𝑑𝜉1𝑑𝜏1𝜏1𝜉512𝑏𝜏𝜏1𝜉𝜉152𝑏𝜉12𝑠𝜉𝜉12𝑠1/2𝐿𝜏𝐿𝜉,||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2𝜉1𝜉22𝑠||𝜉2||2||𝜉1𝜉2||2𝜒Ω2.3.2𝑑𝜉2𝑑𝜏2𝜏1𝜏2(𝜉1𝜉2)52(1𝑏)𝜏2𝜉522𝑏𝜉22𝑠1/2𝐿𝜏1𝐿𝜉1,||𝜉2||𝜏2𝜉52𝑏𝜉2𝑠2𝜉1𝜉22𝑠||𝜉2||2||𝜉1𝜉2||2𝜒Ω𝑖33.3.𝑘𝑑𝜉1𝑑𝜏1𝜏1𝜏2(𝜉1𝜉2)52(1𝑏)𝜏1𝜉512𝑏𝜉12𝑠1/2𝐿𝜏2𝐿𝜉2.(3.27) By using the following methods, we estimate (3.27).
The Case of Ω0.0.0
Since 𝜏1+𝜉𝜉15𝜉5=𝜏1𝜉515𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉21,(3.28) it follows from (3.2) in Lemma 3.1 with 𝛼=2𝑏, 𝛽=𝜅=2(1𝑏) that ||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2𝜉2𝑠||𝜉||2||𝜉𝜉1||2𝜒Ω0.0.0𝜏,𝜏1,𝜉,𝜉1𝑑𝜉𝑑𝜏𝜏𝜉52(1𝑏)𝜏𝜏1𝜉𝜉152𝑏𝜉𝜉12𝑠1/2𝐿𝜏1𝐿𝜉1𝐶7,2(1𝑏),2𝑏×||𝜉1||𝜒Ω𝐵𝜉1𝜏1𝜉51𝑏𝜉1𝑠𝜉2𝑠||𝜉||2||𝜉𝜉1||2𝜒Ω1𝐴;𝜉(𝜉)𝜏1𝜉515𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212(1𝑏)𝜉𝜉12𝑠𝑑𝜉1/2𝐿𝜏1𝐿𝜉1,(3.29) where Ω𝐴;𝜉1={𝜉|𝜉|,|𝜉𝜉1|5} and Ω𝐵={𝜉1|𝜉1|5}. Since 𝜉𝑠max{1,6𝑠}, we have 𝐶7,2(1𝑏),2𝑏×||𝜉1||𝜒Ω𝐵𝜉1𝜏1𝜉51𝑏𝜉1𝑠𝜉2𝑠||𝜉||2||𝜉𝜉1||2𝜒Ω1𝐴;𝜉(𝜉)𝜏1𝜉515𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212(1𝑏)𝜉𝜉12𝑠𝑑𝜉1/2𝐿𝜏1𝐿𝜉1𝐶7,2(1𝑏),2𝑏max1,6𝑠52||𝜉||5||𝜉||2𝑑𝜉1/2𝑀1,𝑠,𝑏,𝑏,(3.30) where 𝑀1,𝑠,𝑏,𝑏 is some constant.
The Case of Ω1.𝑗.3,Ω3.𝑗.3Ω(𝑗=1,2),𝑖.3.3(𝑖=1,3) and Ω2.3.2
We consider Ω2.3.2. By (3.2), we have ||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠𝜉1𝜉22𝑠||𝜉2||2||𝜉𝜉1||2𝜒Ω2.3.2𝑑𝜉2𝑑𝜏2𝜏1𝜏2𝜉𝜉152(1𝑏)𝜏2𝜉522𝑏𝜉22𝑠1/2𝐿𝜏1𝐿𝜉1𝐶7,2(1𝑏),2𝑏𝐶9,𝑠2|𝑠|×||𝜉1||2𝜒Ω𝐶𝜉1𝜏1𝜉51𝑏||𝜉2||2𝜒Ω1𝐷;𝜉𝜉2𝜏1𝜉51+5𝜉1𝜉2𝜉1𝜉2𝜉21𝜉1𝜉2+𝜉222(1𝑏)𝑑𝜉21/2𝐿𝜏1𝐿𝜉1,(3.31) where Ω𝐶={𝜉1|𝜉1|4} and Ω𝐷;𝜉1={𝜉2|𝜉2||𝜉1|4}. Here we have used (3.16) and (3.17) with the change of variables 𝜉2=𝜉1𝜉. Noting 2𝑏>0,21𝑏>0,(3.32) we have ||𝜉1||2𝜒Ω𝐶𝜉1𝜏1𝜉51𝑏||𝜉2||2𝜒Ω1𝐷;𝜉𝜉2𝜏1𝜉51+5𝜉1𝜉2𝜉1𝜉2𝜉21𝜉1𝜉2+𝜉222(1𝑏)𝑑𝜉21/2𝐿𝜏1𝐿𝜉1||𝜉1||4𝜒Ω𝐶𝜉1|𝜉2||𝜉1|4||𝜉2||2𝑑𝜉21/2𝐿𝜉121/2||𝜉1||8𝜒Ω𝐶𝜉11/2𝐿𝜉121/2.(3.33) Thus, (3.27) is bounded by 𝑀2,𝑠,𝑏,𝑏=𝐶9,𝑠2|𝑠|+1/2𝐶max7,2(1𝑏),2𝑏,𝐶7,2𝑏,2𝑏(3.34) in Ω2.3.2. 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 Ω𝑖.3.3(𝑖=1,3).
The Case of Ω2.1.1, Ω2.2.1
We consider Ω2.2.1. Since ||𝜏𝜉5||+||𝜏1𝜉51||+|||𝜏𝜏1𝜉𝜉15||||||𝜏𝜉5𝜏1𝜉51𝜏𝜏1𝜉𝜉15|||||𝜉||||𝜉=51||||𝜉𝜉1||||𝜉2𝜉𝜉1+𝜉21||,(3.35) we obtain ||max𝜏𝜉5||,||𝜏1𝜉51||,|||𝜏𝜏1𝜉𝜉15|||53||𝜉||||𝜉1||||𝜉𝜉1||||𝜉2𝜉𝜉1+𝜉21||.(3.36) Noting that 2𝑏<0, we have 𝜏1𝜉512𝑏32𝑏5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212𝑏inΩ2.2.1.(3.37) Since (3.14) and |𝜉𝜉1|1 yield ||𝜉1||||3𝜉𝜉1||,(3.38) we have ||2𝜉𝜉1||||𝜉1||||2𝜉𝜉1||13||𝜉1||.(3.39) By (3.2), (3.16), and (3.37), we have |𝜉1|𝜏1𝜉51𝑏𝜉1𝑠2||𝜉|2||𝜉𝜉1|2𝜉2𝑠𝜒Ω2.2.1𝜏,𝜏1,𝜉,𝜉1𝑑𝜏𝑑𝜉𝜏𝜉52(1𝑏)𝜏𝜏1𝜉𝜉152𝑏𝜉𝜉12𝑠1/2𝐿𝜏1𝐿𝜉1𝐶7,2(1𝑏),2𝑏𝐶9,𝑠2|𝑠|3𝑏×5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212𝑏||𝜉1||4||𝜉𝜉1||2𝜒Ω𝐸;𝜉𝜉1𝜏1𝜉515𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212(1𝑏)𝑑𝜉1/2𝐿𝜏1𝐿𝜉1,(3.40) where Ω𝐸;𝜉={𝜉1|𝜉𝜉1|1,|𝜉1|4}. Using the change of variable 𝜇=5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉21(3.41) and (3.2), we have 5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212𝑏||𝜉1||4||𝜉𝜉1||2𝜒Ω𝐸;𝜉𝜉1𝜏1𝜉515𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212(1𝑏)𝑑𝜉1/2𝐿𝜏1𝐿𝜉1|𝜇|3|𝜏1𝜉51|𝜇2𝑏|𝜉1|4𝑑𝜇𝜏1𝜉51𝜇2(1𝑏)5||𝜉1||||2𝜉𝜉1||||𝜉212𝜉𝜉1+2𝜉2||1/2𝐿𝜏1𝐿𝜉1(5/6)1𝑑𝜇𝜇2𝑏𝜏𝜇1𝜉512(1𝑏)1/2𝐿𝜏1𝐿𝜉1𝐶7,2(1𝑏),2𝑏561/2,(3.42) where we have used 5||𝜉1||||2𝜉𝜉1||||𝜉212𝜉𝜉1+2𝜉2||56||𝜉1||4inΩ𝐸;𝜉,(3.43) which follows from (3.39) and ||2𝜉22𝜉𝜉1+𝜉21||||||21𝜉2𝜉12+12𝜉21||||12||𝜉1||2.(3.44) Hence (3.27) is bounded by 𝑀3,𝑠,𝑏,𝑏𝐶=max27,21𝑏,2𝑏,𝐶7,2(1𝑏),2𝑏𝐶7,2𝑏,2𝑏𝐶9,𝑠2|𝑠|3𝑏+31𝑏561/2(3.45) 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 of  Ω2.1.3, Ω2.2.3,  and  Ω2.3.5
We consider Ω2.2.3. Since (3.14) and |2𝜉𝜉1||𝜉1|3/2 yield ||2𝜉𝜉1||12||𝜉1||,(3.46) we have ||𝜉𝜉1||12||𝜉1||||2𝜉𝜉1||14||𝜉1||,||𝜉𝜉1||12||𝜉1||+||2𝜉𝜉1||34||𝜉1||,inΩ2.2.3.(3.47) By (3.9) and (3.47), we have 53||𝜉||||𝜉1||||𝜉𝜉1||||𝜉2𝜉𝜉1+𝜉21||5||𝜉481||5,(3.48) where we have used ||𝜉2𝜉𝜉1+𝜉21||=||||1𝜉2𝜉12+34𝜉21||||34||𝜉1||2.(3.49) Therefore using (3.36), we have 𝜏1𝜉512𝑏5482𝑏||𝜉1||5inΩ2.2.3.(3.50) By (3.2), (3.16), (3.17), (3.47), and (3.50), we have ||𝜉1||𝜏1𝜉51𝑏𝜉1𝑠2||𝜉||2||𝜉𝜉1||2𝜉2𝑠𝜒Ω2.2.1𝜏,𝜏1,𝜉,𝜉1𝑑𝜏𝑑𝜉𝜏𝜉52(1𝑏)𝜏𝜏1(𝜉𝜉1)52𝑏𝜉𝜉12𝑠1/2𝐿𝜏1𝐿𝜉1𝐶7,2(1𝑏),2𝑏𝐶9,𝑠2|𝑠|1max41𝑠,341𝑠548𝑏×||𝜉1||5||𝜉1||4||𝜉1||22𝑠𝜒Ω𝐺𝜉1𝜒Ω1𝐹;𝜉(𝜉)𝑑𝜉𝜏1𝜉515𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212(1𝑏)1/2𝐿𝜏1𝐿𝜉1,(3.51) where Ω𝐹;𝜉1={𝜉|2𝜉𝜉1||𝜉1|3/2} andΩ𝐺={𝜉1|𝜉1|4}. Since |2𝜉𝜉1||𝜉1|3/2, we have ||𝜉||12||𝜉1||||2𝜉𝜉1||12||𝜉1||||𝜉1||3/2,||𝜉||12||𝜉1||||2𝜉𝜉1||12||𝜉1||||𝜉1||3/2,inΩ𝐹;𝜉1.(3.52) Noting 1𝑠>4,21𝑏>0,(3.53) we have by (3.52) ||𝜉1||5||𝜉1||4||𝜉1||22𝑠𝜒Ω𝐺𝜉1𝜒Ω1𝐹;𝜉(𝜉)𝑑𝜉𝜏1𝜉515𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212(1𝑏)1/2𝐿𝜏1𝐿𝜉1||𝜉1||12𝑠𝜒Ω𝐺𝜉1Ω1𝐻;𝜉1𝑑𝜉1/2𝐿𝜏1𝐿𝜉11,(3.54) where Ω𝐻;𝜉1=1𝜉2||𝜉1||||𝜉1||3/2||𝜉||12||𝜉1||+||𝜉1||3/2.(3.55) Therefore, (3.27) is bounded by 𝑀4,𝑠,𝑏,𝑏=𝐶9,𝑠2|𝑠|𝐶max7,2(1𝑏),2𝑏,𝐶7,2𝑏,2𝑏1max41𝑠,341𝑠(3.56) 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 Ω2.3.5.
All the Other Cases
We consider Ω2.1.2. By (3.9), we have ||𝜉𝜉1||||𝜉||+||𝜉1||||𝜉||5.(3.57) Since 2𝑏<1+2𝜎 and 𝜎=min{𝑠/5+1/20,3/16} yield 32+42𝑏31<2+8𝜎<0,(3.58) we have by (3.16), (3.36), and (3.57) ||𝜉||2||𝜉1||2𝜉2𝑠𝜉12𝑠𝐶29,𝑠53/2||𝜉||11/28𝜎||𝜉𝜉1||3/2+8𝜎,𝜏𝜉52(𝑏1)5162(𝑏1)||𝜉𝜉1||||𝜉||41+2𝜎,inΩ2.1.2.(3.59) By (3.2), (3.17), and (3.59), we have 𝜉2𝑠||𝜉||2𝜏𝜉51𝑏2||𝜉1|2||𝜉𝜉1|2𝜒Ω2.1.2𝜏,𝜏1,𝜉,𝜉1𝑑𝜉1𝑑𝜏1𝜏1𝜉512𝑏𝜏𝜏1(𝜉𝜉1)52𝑏𝜉12𝑠𝜉𝜉12𝑠1/2𝐿𝜏𝐿𝜉𝐶7,2𝑏,2𝑏𝐶9,𝑠2|𝑠|53/4516(𝑏1)×||𝜉||(11/2)8𝜎||𝜉𝜉1||(1/2)2𝑠+8𝜎𝜒Ω𝐼;𝜉𝜉1𝑑𝜉1||𝜉𝜉1||||𝜉||412𝜎𝜏𝜉5+5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212𝑏1/2𝐿𝜏𝐿𝜉,(3.60) where Ω𝐼;𝜉={𝜉11|𝜉𝜉1|,|𝜉2𝜉1||𝜉|3/2}. Noting that 𝑠2𝑏>1,𝜎=min5+1,32016(3.61) and using (3.3) and the change of variables 𝜇=5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉21,(3.62) we have ||𝜉||11/28𝜎||𝜉𝜉1||1/22𝑠+8𝜎𝜒Ω𝐼;𝜉(𝜉1)||𝜉𝜉1||||𝜉||412𝜎𝜏𝜉5+5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212𝑏𝑑𝜉11/2𝐿𝜏𝐿𝜉=||𝜉||3/2||𝜉𝜉1||1/22𝑠+10𝜎𝜒Ω𝐼;𝜉𝜉1𝜏𝜉5+5𝜉𝜉1𝜉𝜉1𝜉2𝜉𝜉1+𝜉212𝑏𝑑𝜉11/2𝐿𝜏𝐿𝜉|𝜇|3|𝜏𝜉5|||𝜉||3/2𝜏𝜉5+𝜇2𝑏5||𝜉||||𝜉2𝜉1||||𝜉22𝜉𝜉1+2𝜉21||𝑑𝜇1/2𝐿𝜏𝐿𝜉521/2|𝜇|3|𝜏𝜉5|𝑑𝜇𝜏𝜉5+𝜇2𝑏1/2𝐿𝜏𝐿𝜉𝐶8,2𝑏521/2,(3.63) where we used the inequality 5||𝜉||||𝜉2𝜉1||||𝜉22𝜉𝜉1+2𝜉21||52||𝜉||3/4inΩ𝐼;𝜉,(3.64) which follows from ||𝜉22𝜉𝜉1+2𝜉21||12||𝜉||2,||𝜉2𝜉1||||𝜉||3/2.(3.65) Thus, (3.27) is bounded by 𝑀5,𝑠,𝑏,𝑏=53/2516(𝑏𝑏)1/22|𝑠|𝐶9,𝑠𝐶max7,2𝑏,2𝑏𝐶8,2𝑏,𝐶7,2(1𝑏),2𝑏3𝑏1/2(3.66) 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 𝐶3,𝑠,𝑏,𝑏=𝑀1,𝑠,𝑏,𝑏+7𝑀2,𝑠,𝑏,𝑏+2𝑀3,𝑠,𝑏,𝑏+3𝑀4,𝑠,𝑏,𝑏+17𝑀5,𝑠,𝑏,𝑏,(3.67) we have (2.17). This completes the proof of Lemma 2.4.

Remark 3.2. We briefly state the reason why the term 10𝑢𝜕3𝑥𝑢 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: 𝑢𝜕3𝑥𝑣𝑋𝑠𝑏1𝐶3,𝑠,𝑏,𝑏𝑢𝑋𝑠𝑏𝑣𝑋𝑠𝑏.(3.68) Unfortunately, we are not able to prove it, because our method can be used to estimate 𝜕𝑙𝑥(𝜕𝑚𝑥𝑢𝜕𝑛𝑥𝑣) in the case where 𝑙𝑚𝑛1, but not in the case where 𝑙𝑚𝑛=0. Therefore, it is necessarily for us to remove the term 10𝑢𝜕3𝑥𝑢 from (1.1)2 unavoidably.

4. Analyticity

In this section we prove the analyticity of the solution 𝑢𝑢0 given in Proposition 2.1. The proof is established by Propositions 4.54.9. To prove these propositions we prepare four lemmas (Lemmas 4.14.4).

Lemma 4.1 (see [21]). Let 7𝑠>41,𝑏2,12,+𝜎(4.1) where 𝜎=min{1/4,(4𝑠+11)/8,(𝑠+6)/5}. Then 𝜕𝑥(𝑢𝑣)𝑋𝑠𝑏1𝑀6,𝑠,𝑏𝑢𝑋𝑠𝑏𝑣𝑋𝑠𝑏,(4.2) where 𝑀6,𝑠,𝑏 is a constant depending on 𝑠 and 𝑏.

Lemma 4.2 (see [7]). Let 𝑓𝐻𝑠(𝑛),𝑔𝐻𝑟(𝑛).(4.3) Suppose that 𝑛0𝑠,𝑟2,𝑛2𝑠+𝑟.(4.4) Then, for any 𝜎1<𝑠+𝑟𝑛/2, 𝑓𝑔𝐻𝜎1(𝑛)𝑀7𝑓𝐻𝑠(𝑛)𝑔𝐻𝑟(𝑛)(4.5) 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)}×. (1)Suppose that 𝑏(0,1],𝑟(,0]. Then, for a sufficiently small 𝜀1>0 such that 𝜀1=𝜀49,if2𝜀<𝑟0,𝛼1with𝛼>29𝑟,if𝑟2,𝐷𝑡,𝑥5𝑏𝑔𝐿2𝑡(;𝐻𝑟𝑥())𝑀8,𝑟,𝑏,𝜀1𝑔𝑋𝑟𝑏1+𝑡𝜕5𝑥𝑔𝑋𝑟𝑏1+𝑃5𝑔𝑋𝑟𝑏1(4.6) holds for all 𝑔𝑋𝑟𝑏1 satisfying supp𝑔𝐵2𝜀1𝑡0,𝑥0,𝑡𝜕5𝑥𝑔,𝑃5𝑔𝑋𝑟𝑏1,(4.7) where 𝑀8,𝑟,𝑏,𝜀1=𝑀8,𝑟,𝑏,(𝑡0,𝑥0),𝜀1 depends on 𝑟, 𝑏, (𝑡0,𝑥0), and 𝜀.(2)Let 𝜇>0. Then, for a sufficiently small 𝜀2=𝜀4>0, 𝐷𝑡,𝑥𝜇𝑔𝐿2𝑡𝐿2𝑥𝑀9,𝜇,𝜀2𝑔𝐻𝜇5𝑡,𝑥(2)+𝑡𝜕5𝑥𝑔𝐻𝜇5𝑡,𝑥(2)+𝑃5𝑔𝐻𝜇5𝑡,𝑥()(4.8) holds for all 𝑔𝐻𝜇5𝑡,𝑥(2) satisfying supp𝑔𝐵2𝜀2𝑡0,𝑥0,𝑡𝜕5𝑥𝑔,𝑃5𝑔𝐻𝜇5𝑡,𝑥2,(4.9) where 𝑀9,𝜇,𝜀2=𝑀9,𝜇,(𝑡0,𝑥0),𝜀2 depends on 𝜇, (𝑡0,𝑥0), and 𝜀.

Let 𝜌(𝑡,𝑥) be a smooth cut-off function around the freezing point (𝑡0,𝑥0) such that 𝜌𝐶0(𝐵2𝜀4(𝑡0,𝑥0)).

Lemma 4.4. Let 𝑠,𝑏. Then 𝜌𝑓𝑋𝑠𝑏𝑀10,𝑠,𝑏,𝜌,𝜀3𝜀3|𝑠|9|𝑏|𝑓𝑋𝑏𝑠+4|𝑏|,(4.10) where 𝜀3=𝜀4 and 𝑀10,𝑠,𝑏,𝜌,𝜀3=10|𝑏|/2𝜀43𝜏𝜉5|𝑏|𝜉|𝑠|+4|𝑏|𝑡,𝑥𝜌(𝜏,𝜉)𝐿1𝜏𝐿1𝜉.

Proof. Lemmas 4.3 and 4.4 are proved by the same method as Lemmas  3.2 and 5.2 in [7].

Proposition 4.5. Let 𝑠>1/8, and let 𝑏(1/2,23/40). Then, for a sufficiently small 𝜀4>0, there exist positive constants 𝐾1,𝜌 and 𝐴1 such that 𝜌𝑃𝑘𝑢𝐻1/3𝑡,𝑥(2)+𝜌𝑃𝑘𝑢𝐿2𝑡(;𝐻1𝑥)𝐾1,𝜌𝐴𝑘1𝑘!(4.11) holds for all 𝑘=0,1,2,, where 𝐴1=2𝐴01 and 𝐾1,𝜌=𝐾1,𝑠,𝑏,(𝑡0,𝑥0),𝜀,𝜌 is a constant depending on 𝑠, 𝑏, (𝑡0,𝑥0), 𝜀 and 𝜌.

Proof. By Plancherel Theorem and Lemma 4.3 with 𝑔=𝜌𝑃𝑘𝑢, we have 𝜌𝑃𝑘𝑢𝐻1/3𝑡,𝑥(2)+𝜌𝑃𝑘𝑢𝐿2𝑡(;𝐻1𝑥)2(2𝜋)2𝐷𝑡,𝑥5𝑏𝜌𝑃𝑘𝑢𝐿2𝑡(;𝐻𝑟𝑥())2(2𝜋)2𝑀8,𝑟,𝑏,𝜀4𝜌𝑃𝑘𝑢𝑋𝑟𝑏1+𝑡𝜕5𝑥𝜌𝑃𝑘𝑢𝑋𝑟𝑏1+𝑃5𝜌𝑃𝑘𝑢𝑋𝑟𝑏1,(4.12) where 1𝑟=𝑠2,if8<𝑠2,158<𝑟0,if𝑠>2.(4.13) We note that 𝑟𝑠2 holds. Put 𝐾2,𝑠,𝑏=|𝐮|𝒜𝐴0(𝑋𝑠𝑏). Since (2.8) and Remark 2.2 yield 𝑃𝑘𝑢𝑋𝑠𝑏𝐾2,𝑠,𝑏𝐴01𝑘𝑘!,𝑘=0,1,2,,(4.14) it follows from Lemma 4.4 that 𝜌𝑃𝑘𝑢𝑋𝑟𝑏1𝑀10,𝑟,𝑏1,𝜌,𝜀4𝜀4|𝑟|36|𝑏1|𝑃𝑘𝑢𝑋𝑠𝑏1𝐾2,𝑠,𝑏1𝑀10,𝑟,𝑏1,𝜌,𝜀4𝜀4|𝑟|36|𝑏1|𝐴01𝑘𝑃𝑘!,(4.15)5𝜌𝑃𝑘𝑢𝑋𝑟𝑏15𝑙=05!𝑃(5𝑙)!𝑙!5𝑙𝜌𝑃𝑙+𝑘𝑢𝑋𝑟𝑏15𝑙=05!𝑀(5𝑙)!𝑙!10,𝑟,𝑏1,𝜌𝑙,𝜀4𝜀4|𝑟|36|𝑏1|𝑃𝑙+𝑘𝑢𝑋𝑠𝑏1max0𝑙5𝑀10,𝑟,𝑏1,𝜌𝑙,𝜀4𝜀4|𝑟|36|𝑏1|𝐾52,𝑠,𝑏1𝑙=05!(5𝑙)!𝑙!(𝑘+𝑙)!2𝑘𝐴𝑘!01𝑙2𝐴01𝑘𝑘!𝐾3𝐴𝑘1𝑘!,(4.16) where 𝜌𝑙=𝑃5𝑙𝜌, 𝐾3 is some constant and 𝐴1=(2𝐴01).
Now we estimate 𝑡𝜕5𝑥(𝜌𝑃𝑘𝑢)𝑋𝑟𝑏1. By using 𝑡𝜕5𝑥𝜌𝑃𝑘𝑢𝜕=𝑡𝜌5𝑥𝑃𝑘𝑢+5𝑡𝜕2𝑥𝜕2𝑥𝜌𝜕𝑥𝑃𝑘𝑢+5𝑡𝜕𝑥𝜕𝑥𝜌𝜕3𝑥𝑃𝑘𝑢𝜕+𝑡5𝑥𝜌𝑃𝑘𝜕𝑢,(4.17)5𝑥𝑃𝑘𝑢1=5𝑃𝑘+1𝑢𝑥𝜕𝑥𝑃𝑘𝑢+𝑡𝒩𝑘(𝑢),(4.18) we have 𝑡𝜕5𝑥𝜌𝑃𝑘𝑢𝑋𝑟𝑏115𝜌𝑃𝑘+1𝑢𝑋𝑟𝑏1+𝜌𝑥𝜕𝑥𝑃𝑘𝑢𝑋𝑟𝑏1+𝑡𝜌𝑁𝑘(𝑢)𝑋𝑟𝑏1𝜕+52𝑥𝑡𝜕2𝑥𝜌𝜕𝑥𝑃𝑘𝑢𝑋𝑟𝑏1𝜕+5𝑥𝑡𝜕𝑥𝜌𝜕3𝑥𝑃𝑘𝑢𝑋𝑟𝑏1+𝑡𝜕5𝑥𝜌𝑃𝑘𝑢𝑋𝑟𝑏1.(4.19) In the same manner as (4.15), we have 𝜌𝑃𝑘+1𝑢𝑋𝑟𝑏1𝑀10,𝑟,𝑏1,𝜌,𝜀4𝜀4|𝑟|36|𝑏1|𝑃𝑘+1𝑢𝑋𝑠𝑏1𝑀10,𝑟,𝑏1,𝜌,𝜀4𝜀4|𝑟|36|𝑏1|𝐾2,𝑠,𝑏1(𝑘+1)!2𝑘𝐴𝑘!012𝐴01𝑘𝑘!𝐾4𝐴𝑘1𝑘!,(4.20) where 𝐾4 is some constant. By Lemmas 4.1 and 4.4, we have 𝜌𝑥𝜕𝑥𝑃𝑘𝑢𝑋𝑟𝑏1𝜕𝑥𝜌𝑥𝑃𝑘𝑢𝑋𝑟𝑏1+𝜕𝑥𝑃(𝜌𝑥)𝑘𝑢𝑋𝑟𝑏1𝑀6,𝑟,𝑏𝜌𝑥𝑋𝑏𝑠2𝑃𝑘𝑢𝑋𝑏𝑠2+𝑀10,𝑟,𝑏1,𝜕𝑥(𝜌𝑥),𝜀4𝜀4|𝑟|36|𝑏1|𝑃𝑘𝑢𝑋𝑠𝑏1𝑀6,𝑟,𝑏𝜌𝑥𝑋𝑏𝑠2+𝑀10,𝑟,𝑏1,𝜕𝑥(𝜌𝑥),𝜀4𝜀4|𝑟|36|𝑏1|𝐾2,𝑠,𝑏𝐴01𝑘𝑘!.(4.21) By Lemmas 2.4, 2.5, and 4.4, we have 𝑡𝜌𝒩𝑘(𝑢)𝑋𝑟𝑏1𝑀10,𝑟,𝑏1,𝑡𝜌,𝜀4𝜀4|𝑟|36|𝑏1|×𝐶4,𝑠,𝑏𝐤𝑘!4𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑃𝑘1𝑢𝑋𝑠𝑏𝑃𝑘2𝑢𝑋𝑠𝑏𝑃𝑘3𝑢𝑋𝑠𝑏+𝐶3,𝑠,𝑏𝐤𝑘!3𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑃𝑘1𝑢𝑋𝑠𝑏𝑃𝑘2𝑢𝑋𝑠𝑏𝑀10,𝑟,𝑏1,𝑡𝜌,𝜀4𝜀4|𝑟|36|𝑏1|×𝐶4,𝑠,𝑏𝐾32,𝑠,𝑏𝐤4𝑘4𝑘4!𝐴01𝑘1+𝑘2+𝑘3+𝐶3,𝑠,𝑏𝐾22,𝑠,𝑏𝐤3𝑘4𝑘3!𝑘4!𝐴01𝑘1+𝑘2𝑘!𝑀10,𝑟,𝑏1,𝑡𝜌,𝜀4𝜀4|𝑟|36|𝑏1|𝑒4/𝐴01𝐶4,𝑠,𝑏𝐾32,𝑠,𝑏(𝑘+1)2𝑘+𝐶3,𝑠,𝑏𝐾22,𝑠,𝑏(𝑘+1)(𝑘+2)2𝑘2𝐴01𝑘𝑘!𝐾5𝐴𝑘1𝑘!,(4.22) where 𝐾5 is some constant. We also have 𝜕2𝑥𝑡𝜕2𝑥𝜌𝜕𝑥𝑃𝑘𝑢𝑋𝑟𝑏1𝜕𝑥𝑡𝜕2𝑥𝜌𝜕𝑥𝑃𝑘𝑢𝑋𝑠1𝑏1𝐶3,𝑠1,𝑏𝑡𝜕𝑥𝜌𝑋𝑏𝑠1𝑃𝑘𝑢𝑋𝑏𝑠1𝐶3,𝑠1,𝑏𝐾2,𝑠1,𝑏𝑡𝜕𝑥𝜌𝑋𝑏𝑠1𝐴01𝑘𝜎𝑘!,𝑥𝑡𝜕𝑥𝜌𝜕3𝑥𝑃𝑘𝑢𝑋𝑟𝑏1𝐶3,𝑠2,𝑏𝑡𝜌𝑋𝑏𝑠2𝜕2𝑥𝑃𝑘𝑢𝑋𝑏𝑠2𝐶3,𝑠2,𝑏𝑡𝜌𝑋𝑏𝑠2𝑃𝑘𝑢𝑋𝑠𝑏𝐶3,𝑠2,𝑏𝐾2,𝑠,𝑏𝑡𝜌𝑋𝑏𝑠2𝐴01𝑘𝑘!.(4.23) In the same manner as (4.15), we have 𝑡𝜕5𝑥𝜌𝑃𝑘𝑢𝑋𝑟𝑏1𝑀10,𝑟,𝑏1,𝑡𝜕5𝑥𝜌,𝜀4𝐾2,𝑠,𝑏1𝐴01𝑘𝑘!.(4.24) Hence 𝑡𝜕5𝑥𝜌𝑃𝑘𝑢𝑋𝑟𝑏1𝐾6𝐴𝑘1𝑘!,(4.25) where 𝐾6 is some constant. Putting 𝐾1,𝜌=32𝜋2𝑀8,𝑟,𝑏,𝜀4𝐾max2,𝑠,𝑏1𝑀10,𝑟,𝑏1,𝜌,𝜀4𝜀4|𝑟|36|𝑏1|,𝐾3,𝐾6,(4.26) we have (4.11).

Proposition 4.6. Under the same assumption as in Proposition 4.5, there exist 𝐾7 and 𝐴2 such that 𝜌4𝑃𝑘𝑢𝐻11/2𝑡,𝑥()𝐾7𝐴𝑘2𝑘!(4.27) holds for all 𝑘=0,1,2,, where 𝜌4 is a smooth cut-off function such that 𝜌4min𝜌,𝜌4,𝜌4𝑡1on0𝜀4,𝑡0+𝜀4×𝑥0𝜀4,𝑥0+𝜀4.(4.28)

Proof. At first, we show that there exists a constant 𝐾7,1/2 and 𝐴3 such that 𝜌4𝑃𝑘𝑢𝐻1/2𝑡,𝑥(2)𝐾7,1/2𝐴𝑘3𝑘!,𝑘=0,1,2,.(4.29) Applying Lemma 4.3 with 𝜇=1/2 and 𝑔=𝜌4𝑃𝑘𝑢, we have 𝐷𝑡,𝑥1/2𝜌4𝑃𝑘𝑢𝐿2𝑡𝐿2𝑥𝑀9,1/2,𝜀4𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)+𝑡𝜕5𝑥𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)+𝑃5𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2).(4.30) By Proposition 4.5, we have 𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)𝐾1,𝜌4𝐴𝑘1𝑃𝑘!,5𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)5𝑙=05!𝑃𝑙!(5𝑙)!5𝑙𝜌4𝐿𝑡,𝑥(2)𝜌𝑃𝑘+𝑙𝑢𝐿2𝑡,𝑥(2)𝐾1,𝜌max0𝑙5𝑃5𝑙𝜌4𝐿𝑡,𝑥(2)5𝑙=05!(5𝑙)!𝑙!(𝑘+𝑙)!2𝑘𝐴𝑘!11𝑙2𝐴11𝑘𝑘!𝐾1,𝜌𝐾82𝐴11𝑘𝑘!,(4.31) where 𝐾8 is some constant.
Now we estimate 𝑡𝜕5𝑥(𝜌4𝑃𝑘𝑢)𝐻9/2𝑡,𝑥(2). Using (4.18) and 𝜕5𝑥𝜌4𝑓=𝜌4𝜕5𝑥𝑓+5𝑙=1(1)𝑙15!𝜕𝑙!(5𝑙)!𝑥5𝑙𝜕𝑙𝑥𝜌4𝑃𝑘𝑢,(4.32) we obtain 𝑡𝜕5𝑥𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)15𝜌4𝑃𝑘+1𝑢𝐻9/2𝑡,𝑥(2)+𝜌4𝑥𝜕𝑥𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)+𝑡𝜌4𝒩𝑘(𝑢)𝐻9/2𝑡,𝑥(2)+5𝑙=15!𝑙!(5𝑙)!𝑡𝜕𝑥5𝑙𝜕𝑙𝑥𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2).(4.33) By Proposition 4.5, we have 15𝜌4𝑃𝑘+1𝑢𝐻9/2𝑡,𝑥(2)15𝜌4𝐿𝑡𝐿𝑥𝜌𝑃𝑘+1𝑢𝐿2𝑡𝐿2𝑥15𝐾1,𝜌𝜌4𝐿𝑡𝐿𝑥(𝑘+1)!2𝑘𝐴𝑘!112𝐴11𝑘𝑘!𝐾1,𝜌𝐾92𝐴11𝑘1𝑘!,5𝜌4𝑥𝜕𝑥𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)15𝑥𝜌4𝐿𝑡𝐿𝑥+𝜕𝑥𝑥𝜌4𝐿𝑡𝐿𝑥𝜌𝑃𝑘𝑢𝐿2𝑡𝐿2𝑥15𝑥𝜌4𝐿𝑡𝐿𝑥+𝜕𝑥𝑥𝜌4𝐿𝑡𝐿𝑥𝐾1,𝜌𝐴𝑘1𝑘!,(4.34) where 𝐾9 is some constant. By Sobolev embedding theorem and Proposition 4.5, we have 𝑡𝜌4𝜌4𝜕𝑥𝑃𝑘2𝑢𝑃𝑘3𝑢𝑃𝑘4𝑢𝐻9/2𝑡,𝑥(2)||𝜉|||𝜏|+7/2𝐿2𝜏𝐿2𝜉𝜕𝑥𝑡𝜌4𝐿𝑡𝐿𝑥+𝑡𝜌4𝐿𝑡𝐿𝑥𝜌𝐿𝑡𝐿𝑥×𝜌𝑃𝑘2𝑢𝐿3𝑡𝐿3𝑥𝜌𝑃𝑘3𝑢𝐿3𝑡𝐿3𝑥𝜌𝑃𝑘4𝑢𝐿3𝑡𝐿3𝑥𝐾10,7/2𝜌𝐿𝑡𝐿𝑥𝜌𝑃𝑘2𝑢𝐻1/3𝑡,𝑥(2)𝜌𝑃𝑘3𝑢𝐻1/3𝑡,𝑥(2)𝜌𝑃𝑘4𝑢𝐻1/3𝑡,𝑥(2)𝐾10,7/2𝜌𝐿𝑡𝐿𝑥𝐾31,𝜌𝐴𝑘2+𝑘3+𝑘41𝑘2!𝑘3!𝑘4!,𝑡𝜌4𝜌4𝜕𝑥((𝜕𝑥𝑃𝑘1𝑢)𝜕𝑥𝑃𝑘2𝑢))𝐻9/2𝑡,𝑥(2)𝐾10,7/2𝜌2𝜕𝑥𝑃𝑘1𝑢𝐿2𝑡𝐿2𝑥𝜌2𝜕𝑥𝑃𝑘2𝑢𝐿2𝑡𝐿2𝑥𝐾10,7/2𝜕𝑥𝜌𝐿𝑡𝐿𝑥+𝜌𝐿𝑡𝐿𝑥2𝜌𝑃𝑘1𝑢𝐿2𝑡(;𝐻1𝑥)𝜌𝑃𝑘2𝑢𝐿2𝑡(;𝐻1𝑥)𝐾11,7/2𝐾21,𝜌𝐴𝑘1+𝑘21𝑘1!𝑘2!,(4.35) where 𝐾10,7/2, 𝐾11,7/2 are some constants. Therefore, we have by (4.35) 𝑡𝜌4𝒩𝑘(𝑢)𝐻9/2𝑡,𝑥(2)𝑡𝜌4𝜌4𝒩𝑘(𝑢)𝐻9/2𝑡,𝑥(2)𝐤𝑘!4𝑘1𝑘1!𝑘2!𝑘3!𝑘4!𝑡𝜌4𝜌4𝜕𝑥(𝑃𝑘2𝑢𝑃𝑘3𝑢𝑃𝑘4𝑢))𝐻9/2𝑡,𝑥(2)+𝐤𝑘!3𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑡𝜌4𝜌4𝜕𝑥𝜕𝑥𝑃𝑘2𝑢𝜕𝑥𝑃𝑘1𝑢𝐻9/2𝑡,𝑥(2)𝐾10,7/2𝜌𝐿𝑡𝐿𝑥𝐾31,𝜌𝑘!𝐤4𝑘4𝑘4!𝐴𝑘1+𝑘2+𝑘31+𝐾11,7/2𝐾21,𝜌𝑘!𝐤3𝑘4𝑘3!𝐾4!𝐴𝑘1+𝑘21𝐾10,7/2𝜌𝐿𝑡𝐿𝑥𝐾31,𝜌+𝐾11,7/2𝐾21,𝜌𝑒4/𝐴1𝐴𝑘1𝑘!.(4.36) By Proposition 4.5, we have 5𝑙=15!𝜕𝑙!(5𝑙)!𝑥5𝑙𝑡𝜕𝑙𝑥𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)5𝑙=15!𝑙!(5𝑙)!𝑡𝜕𝑙𝑥𝜌4𝐿𝑡𝐿𝑥𝜌𝑃𝑘𝑢𝐿2𝑡𝐿2𝑥𝐾12𝐾1,𝜌𝐴𝑘1𝑘!,(4.37) where 𝐾12 is some constant. Hence 𝑡𝜕5𝑥𝜌4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)𝐾13,𝐾1,𝜌,𝐾10,7/2max2𝐴11,𝐴1𝑘𝑘!,(4.38) where 𝐾13,𝐾1,𝜌,𝐾10,7/2 is some constant. Putting 𝐴3=max{2𝐴11,𝐴1}, 𝐾7,1/2=𝐾7,1/2,𝐾1,𝜌,𝐾10,7/2=3𝑀9,1/2,𝜀4𝐾max1,𝜌4,𝐾1,𝜌𝐾8,𝐾14,𝐾1,𝜌,𝐾10,7/2,(4.39) we have (4.29). Similarly, we can prove by (4.8) with 𝜇=3/2 and (4.29) 𝜌4𝑃𝑘𝑢𝐻3/2𝑡,𝑥(2)𝐾7,3/2,𝐾7,1/2,𝐾10,5/2𝐴𝑘4𝑘!,(4.40) where 𝐴4=max{2𝐴31,𝐴3}. By (4.8) with 𝜇=5/2 and (4.40), we have 𝜌4𝑃𝑘𝑢𝐻5/2𝑡,𝑥(2)𝐾7,5/2,𝐾7,3/2,𝐾10,0𝐴𝑘5𝑘!,(4.41) where 𝐴5=max{2𝐴41,𝐴4} and 𝐾10,0=(𝜕𝑥𝑡𝜌4𝐿𝑡𝐿𝑥+𝑡𝜌4𝐿𝑡𝐿𝑥)𝜌𝐿𝑡𝐿𝑥. Repeating the same method as in above, we obtain 𝜌4𝑃𝑘𝑢𝐻7/2𝑡,𝑥(2)𝐾7,7/2,𝐾8,5/2,𝑀7𝐴𝑘6𝜌𝑘!,4𝑃𝑘𝑢𝐻9/2𝑡,𝑥(2)𝐾7,9/2,𝐾8,7/2,𝑀7𝐴𝑘7𝜌𝑘!,4𝑃𝑘𝑢𝐻11/2𝑡,𝑥(2)𝐾7,11/2,𝐾8,9/2,𝐾10,0𝐴𝑘2𝑘!,(4.42) where 𝐴6=max{2𝐴51,𝐴5}, 𝐴7=max{2𝐴61,𝐴6} and 𝐴2=max{2𝐴71,𝐴7}. Putting 𝐾7=𝐾7,11/2,𝐾8,9/2,𝐾10,0, we have (4.27).

Remark 4.7. When 𝜇=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 sup𝑡𝐼𝑡0𝑡1/5𝜕𝑥𝑙𝑃𝑘𝑢𝐻1(𝐼𝑥0)𝐾7𝐴8𝑘+𝑙(𝑘+𝑙)!(4.43) holds for all 𝑘,𝑙=0,1,2,, where 𝐼𝑡0=𝑡0𝜀4,𝑡0+𝜀4,𝐼𝑥0=𝑥0𝜀4,𝑥0+𝜀4,𝐴8||𝑡max0||+𝜀41/5,𝐴2,||𝑥0||+𝜀4||𝑡+10𝜀4||1/5||𝑡+1,0||+𝜀44/5𝐾7𝑒4/𝐴8||𝑡max0𝜀4||2/5,𝐾7.(4.44) 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 sup𝑡𝐼𝑡0(𝑡1/5𝜕𝑥)𝑙𝑃𝑘𝑢𝐻1(𝐼𝑥0)||𝑡0||+𝜀4𝑙/5𝜕𝑙𝑥𝑃𝑘𝑢𝐻3/2𝑡,𝑥(𝐼𝑡0×𝐼𝑥0)||𝑡0||+𝜀4𝑙/5𝜌4𝑃𝑘𝑢𝐻11/2𝑡,𝑥(2)𝐾7||𝑡0||+𝜀4𝑙/5𝐴𝑘2𝑘!𝐾7𝐴8𝑘+𝑙𝑘!.(4.45) We assume that (4.43) holds for any 𝑙5. Now we will prove sup𝑡𝐼𝑡0𝑡1/5𝜕𝑥𝑙+1𝑃𝑘𝑢𝐻1(𝐼𝑥0)𝐾7𝐴8𝑘+𝑙+1(𝑘+𝑙+1)!.(4.46) Since 𝑡1/5𝜕𝑥𝑙𝑃𝑘𝑢=𝑡(𝑙5)/5𝜕𝑥𝑙5𝑡𝜕5𝑥𝑃𝑘𝑢1=5𝑡(𝑙5)/5𝜕𝑥𝑙5𝑃𝑘+1𝑢𝑥𝜕𝑥𝑃𝑘𝑢+𝑡1/5𝑙𝜕𝑥𝑙5𝒩𝑘(𝑢),(4.47) we have sup𝑡𝐼𝑡0(𝑡1/5𝜕𝑥)𝑙+1𝑃𝑘𝑢𝐻1(𝐼𝑥0)15𝑡(𝑙4)/5𝜕𝑥𝑙4𝑃𝑘+1𝑢(𝑡)𝐻1(𝐼𝑥0)+15𝑡(𝑙4)/5𝜕𝑥𝑙4𝑥𝜕𝑥𝑃𝑘𝑢𝐻1(𝐼𝑥0)+𝑡1/5𝑙+1𝜕𝑥𝑙4𝒩𝑘(𝑢)𝐻1(𝐼𝑥0).(4.48) By (4.43), we have 15𝑡(𝑙4)/5𝜕𝑥𝑙4𝑃𝑘+1𝑢(𝑡)𝐻1(𝐼𝑥0)15𝐾7𝐴8𝑘+𝑙3(𝑘+𝑙3)!.(4.49) Since 𝑡(𝑙4)/5𝜕𝑥𝑙4𝑥𝜕𝑥=𝑥𝜕𝑥𝑡(𝑙4)/5𝜕𝑥𝑙4+(𝑙4)𝑡(𝑙4)/5𝜕𝑥𝑙4(𝑙=5,6,7,),(4.50) we have by (4.43) 15𝑡(𝑙4)/5𝜕𝑥𝑙4𝑥𝜕𝑥𝑃𝑘𝑢𝐻1(𝐼𝑥0)15𝑥𝜕𝑥𝑡(𝑙4)/5𝜕𝑥𝑙4𝑃𝑘𝑢𝐻1(𝐼𝑥0)𝑡+(𝑙4)(𝑙4)/5𝜕𝑥𝑙4𝑃𝑘𝑢𝐻1(𝐼𝑥0)15||𝑥0||+𝜀4||𝑡+10𝜀4||1/5𝑡(𝑙3)/5𝜕𝑥𝑙3𝑃𝑘𝑢𝐻1(𝐼𝑥0)+15(𝑙4)𝐾7𝐴8𝑘+𝑙41(𝑘+𝑙4)!5||𝑥0||+𝜀4||𝑡+10𝜀4||1/5𝐾7𝐴8𝑘+𝑙31(𝑘+𝑙3)!+5𝐾7𝐴8𝑘+𝑙31(𝑘+𝑙3)!5𝐾7𝐴8𝑘+𝑙2(𝑘+𝑙3)!.(4.51) Now we estimate (𝑡1/5)𝑙+1𝜕𝑥𝑙4𝒩𝑘(𝑢)𝐻1(𝐼𝑥0). We have 𝑡1/5𝑙+1𝜕𝑥𝑙4𝒩𝑘(𝑢)𝐻1(𝐼𝑥0)||𝑡0||+𝜀44/5×𝐤𝑘!4𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑡(𝑙3)/5𝜕𝑥𝑙3𝑃𝑘1𝑢𝑃𝑘2𝑢𝑃𝑘3𝑢𝐻1(𝐼𝑥0)+𝐤𝑘!3𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑡(𝑙3)/5𝜕𝑥𝑙3𝜕𝑥𝑃𝑘1𝑢𝜕𝑥𝑃𝑘2𝑢𝐻1(𝐼𝑥0).(4.52) Since 𝐥𝐤4𝐴81𝑘4𝑘4!𝑙1+𝑘1!𝑘1!𝑙1!𝑙2+𝑘2!𝑘2!𝑙2!𝑙3+𝑘3!𝑘3!𝑙3!(𝑙3)!𝑘!(𝑙+𝑘3)!𝑒4/𝐴8(𝑙+𝑘2),(4.53) we have ||𝑡0||+𝜀44/5𝐤𝑘!4𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑡(𝑙3)/5𝜕𝑥𝑙3𝑃𝑘1𝑢𝑃𝑘2𝑢𝑃𝑘3𝑢𝐻1(𝐼𝑥0)||𝑡0||+𝜀44/5𝐥𝐤(𝑙3)!𝑙1!𝑙2!𝑙3!𝑘!4𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑡1/5𝜕𝑥𝑙1𝑃𝑘1𝑢𝐻1(𝐼𝑥0)×𝑡1/5𝜕𝑥𝑙2𝑃𝑘2𝑢𝐻1(𝐼𝑥0)(𝑡1/5𝜕𝑥)𝑙3𝑃𝑘3𝑢𝐻1(𝐼𝑥0)||𝑡0||+𝜀44/5𝐾37(𝑙+𝑘3)!𝐴8𝑙+𝑘3×𝐥𝐤4𝐴81𝑘4𝑘4!𝑙1+𝑘1!𝑘1!𝑙1!𝑙2+𝑘2!𝑘2!𝑙2!𝑙3+𝑘3!𝑘3!𝑙3!(𝑙3)!𝑘!||𝑡(𝑙+𝑘3)!0||+𝜀44/5𝐾37𝐴8𝑙+𝑘3𝑒4/𝐴81(𝑙+𝑘2)!5𝐾7𝐴8𝑘+𝑙(𝑘+𝑙2)!.(4.54) Similarly, ||𝑡0||+𝜀44/5𝐤𝑘!3𝑘4𝑘1!𝑘2!𝑘3!𝑘4!𝑡𝑙3/5𝜕𝑥𝑙3𝜕𝑥𝑃𝑘1𝑢𝜕𝑥𝑃𝑘2𝑢𝐻1(𝐼𝑥0)||𝑡0||+𝜀44/5𝑙3=𝑙1+𝑙2𝐤(𝑙3)!𝑙1!𝑙2!3𝑘3𝑘!𝑘1!𝑘2!𝑘3!𝑘4!×(𝑡1/5𝜕𝑥)𝑙1𝜕𝑥𝑃𝑘1𝑢𝐻1(𝐼𝑥0)(𝑡1/5𝜕𝑥)𝑙2𝜕𝑥𝑃𝑘2𝑢𝐻1(𝐼𝑥0)||𝑡0||+𝜀44/5||𝑡0𝜀4||2/5𝑙1=𝑚1+𝑚2𝐤(𝑙3)!𝑚1𝑚2𝑚1!𝑚2!3𝑘3𝑘!𝑘1!𝑘2!𝑘3!𝑘4!×(𝑡1/5𝜕𝑥)𝑚1𝑃𝑘1𝑢𝐻1(𝐼𝑥0)(𝑡1/5𝜕𝑥)𝑚2𝑃𝑘2𝑢𝐻1(𝐼𝑥0),(4.55) where 𝑚1=𝑙1+1 and 𝑚2=𝑙2+1. By 𝑚21𝑚1 and 𝑚22𝑚2, we have 𝑚1𝑚2=12(𝑙1)2𝑚21+𝑚22(𝑙1)2𝑚1+𝑚2(𝑙2)(𝑙1).(4.56) Thus, we have by (4.56), (4.43) ||𝑡0||+𝜀44/5||𝑡0||+𝜀42/5𝑙1=𝑚1+𝑚2𝐤(𝑙3)!𝑚1!𝑚2!3𝑘3𝑘!𝑘1!𝑘2!𝑘3!𝑘4!×(𝑡1/5𝜕𝑥)𝑚1𝑃𝑘1𝑢𝐻1(𝐼𝑥0)(𝑡1/5𝜕𝑥)𝑚2𝑃𝑘2𝑢𝐻1(𝐼𝑥0)||𝑡0||+𝜀44/5||𝑡0𝜀4||2/5𝐾27(𝑙+𝑘1)!𝐴𝑔𝑙+𝑘1×𝑙1=𝑚1+𝑚2𝐤3𝐴81𝑘3!𝑘3!𝐴𝑘48!𝑘4!𝑚1+𝑘1!𝑘1𝑚1!𝑚2+𝑘2!𝑘2𝑚2!(𝑙1)!𝑘!||𝑡(𝑙+𝑘1)!0||+𝜀44/5||𝑡0𝜀4||2/5𝐾27𝑒4/𝐴𝑠(𝑙+𝑘)!𝐴8𝑙+𝑘115𝐾7𝐴8𝑘+𝑙+1(𝐾+𝑙)!.(4.57) 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 𝜀 such that sup𝑡𝐼𝑡0𝜕𝑚𝑡𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)𝐾7𝐴9𝑚+𝑙(𝑚+𝑙)!(4.58) holds for all 𝑚,𝑙=0,1,2,.

Proof. By induction on 𝑚, we prove (𝑥𝜕𝑥)𝑚𝜕𝑙𝑥𝑃𝑘𝑢𝐻1(𝐼𝑥0)𝐾7𝐴𝑘+𝑙+𝑚10𝐵𝑚1(𝑘+𝑙+𝑚)!𝑘,𝑙,𝑚=0,1,2,.(4.59) In the case 𝑚=0, we have by (4.43) sup𝑡𝐼𝑡0𝜕𝑙𝑥𝑃𝑘𝑢𝐻1(𝐼𝑥0)𝐾7𝐴𝑙10𝐴𝑘8(𝑘+𝑙)!𝐾7𝐴𝑘+𝑙10(𝑘+𝑙)!,𝑘,𝑙=0,1,2,,(4.60) where 𝐴10=max{𝐴8|𝑡0𝜀4|1/5,𝐴8}. We assume (4.59) is valid up to any 𝑚. Since 𝜕𝑥𝑥𝜕𝑥𝑚=𝑥𝜕𝑥+1𝑚𝜕𝑥(𝑚=0,1,2,),(4.61) we have 𝑥𝜕𝑥𝑚+1𝜕𝑙𝑥𝑃𝑘𝑢𝐻1(𝐼𝑥0)||𝑥0||+𝜀4+1𝑥𝜕𝑥+1𝑚𝜕𝑥𝑙+1𝑃𝑘𝑢𝐻1(𝐼𝑥0)||𝑥0||+𝜀4+1𝑚𝑚1=0𝑚!𝑚1!𝑚𝑚1!𝑥𝜕𝑥𝑚1𝜕𝑥𝑙+1𝑃𝑘𝑢𝐻1(𝐼𝑥0)||𝑥0||+𝜀4𝐾+17A𝑘+𝑚+𝑙+110𝐵𝑚1×(𝑘+𝑚+𝑙+1)!𝑚𝑚1=0𝐴10𝐵1(𝑚𝑚1)𝑚𝑚1!𝑚!𝑚1!𝑘+𝑚1!+𝑙+1||𝑥(𝑘+𝑚+𝑙+1)!0||+𝜀4𝑒+1𝐴10𝐵1𝐾7𝐴𝑘+𝑚+𝑙+110𝐵𝑚1(𝑘+𝑚+𝑙+1)!𝐾7𝐴𝑘+𝑚+𝑙+110𝐵1𝑚+1(𝑘+𝑚+𝑙+1)!,(4.62) where 𝐵1(|𝑥0|+𝜀4+1)𝑒𝐴10𝐵1. Since 𝑡𝜕𝑡=(𝑃𝑥𝜕𝑥)/5 and 𝑃𝑛1𝜕𝑛2𝑥=𝜕𝑛2𝑥𝑃𝑛2𝑛1𝑛1,𝑛2,=0,1,2,(4.63) it follows from (4.59) that 𝑡𝜕𝑡𝑚𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)5𝑚𝑚=𝑚1+𝑚2𝑚𝑚1!𝑚2!𝑥𝜕𝑥𝑚1𝑃𝑚2𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)5𝑚𝑚=𝑚1+𝑚2𝑚𝑚1!𝑚2!𝑥𝜕𝑥𝑚1𝜕𝑙𝑥(𝑃𝑙)𝑚2𝑢𝐻1(𝐼𝑥0)5𝑚𝐦𝑚!𝑚1!𝑚2!𝑚3!𝑙𝑚3𝑥𝜕𝑥𝑚1𝜕𝑙𝑥𝑃𝑚2𝑢𝐻1(𝐼𝑥0)5𝑚𝐾7(𝑙+𝑚)!𝐦𝑚!𝑚1!𝑚2!𝑚3!𝐴𝑙+𝑚𝑚310𝐵𝑚11𝑙𝑚3𝑚1+𝑚2!+𝑙(𝑙+𝑚)!5𝑚𝐾7𝐴𝑙+𝑚10𝐵𝑚1(𝑙+𝑚)!𝐦𝑚!𝑚1!𝑚2!𝑚3!𝐴11𝐵1𝑚3𝐵𝑚215𝑚𝐾7𝐴max10𝐵1,𝐵1𝑙+𝑚𝐴(𝑙+𝑚)!1+10𝐵11+𝐵11𝑚𝐾7𝐴𝑙+𝑚11(𝑙+𝑚)!,(4.64) where 𝐴11=max{1,51(1+(𝐴10𝐵1)1+𝐵11)}max{𝐴10𝐵1,𝐵1}. Thus, (𝑡𝜕𝑡)𝑚𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)𝐾7𝐴𝑙+𝑚11(𝑙+𝑚)!𝑙,𝑚=0,1,2,.(4.65) By induction on 𝑗 we prove that (4.65) implies 𝑡𝜕𝑡𝑚𝜕𝑗𝑡𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)𝐾7𝐴𝑗+𝑚+𝑙11𝐵𝑗2(𝑗+𝑚+𝑙)!𝑗,𝑙,𝑚=0,1,2,.(4.66) In the case 𝑗=0, we have by (4.65) sup𝑡𝐼𝑡0(𝑡𝜕𝑡)𝑚𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)𝐾7𝐴𝑙+𝑚11(𝑙+𝑚)!𝑘,𝑙=0,1,2,.(4.67) We assume that (4.66) is valid up to any 𝑗. Noting that 𝑡𝜕𝑡𝑚𝜕𝑡=𝜕𝑡𝑡𝜕𝑡1𝑚(𝑗=0,1,2,),(4.68) we have 𝑡𝜕𝑡𝑚𝜕𝑡𝑗+1𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)𝜕𝑡𝑡𝜕𝑡1𝑚𝜕𝑗𝑡𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)||𝑡0𝜀4||𝑚1𝑚1=0𝑚!𝑚1!𝑚𝑚1!𝑡𝜕𝑡𝑚1+1𝜕𝑗𝑡𝜕𝑙𝑥𝑢𝐻1(𝐼𝑥0)||𝑡0𝜀4||1𝐾7𝐴𝑗+𝑙+𝑚+111𝐵𝑗2×(𝑗+𝑙+𝑚+1)!𝑚𝑚1=0𝐴(𝑚𝑚1)11𝑚𝑚1!𝑚!𝑚1!𝑗+𝑙+𝑚1!+1||𝑡(𝑗+𝑙+𝑚+1)!0𝜀4||1𝑒𝐴11𝐾7𝐴𝑗+𝑙+𝑚+111𝐵𝑗2(𝑗+𝑙+𝑚+1)!𝐾7𝐴𝑗+𝑙+𝑚+111𝐵2𝑗+1(𝑗+𝑙+𝑚+1)!,(4.69) where 𝐵2|𝑡0𝜀4|1𝑒𝐴11. Thus (4.66) holds.
Choosing 𝑚=0 and 𝐴9=max{𝐴11𝐵2,𝐴11} in (4.66), we have (4.58). This completes the proof of Proposition 4.9.

Acknowledgment

The author is deeply grateful to the referees for having checked the manuscript carefully.