Analyticity and Smoothing Effect for the Coupled System
of Equations of Korteweg-de Vries Type with a Single
Point Singularity
Margareth S. Alves· Bianca M.R. Calsavara · Jaime E. Muñoz Rivera· Mauricio Sepúlveda · Octavio Vera Villagrán
Received: 3 December 2008 / Accepted: 6 September 2010 / Published online: 18 September 2010 © Springer Science+Business Media B.V. 2010
Abstract Using Bourgain spaces and the generator of dilation P= 3t∂t+x∂x, which almost commutes with the linear Korteweg-de Vries operator, we show that a solution of the initial value problem associated for the coupled system of equations of Korteweg-de Vries type which appears as a model to describe the strong interaction of weakly nonlinear long waves, has an analyticity in time and a smoothing effect up to real analyticity if the initial data only have a single point singularity at x= 0.
Keywords Evolution equations· Bourgain space · Smoothing effect
Mathematics Subject Classification (2000) 35Q53
M.S. Alves
Departamento de Matemática, Universidade Federal de Viçosa-UFV, 36570-000 Viçosa, MG, Brazil e-mail:malves@ufv.br
B.M.R. Calsavara
Universidade Estadual de Campinas-Unicamp, Limeira, SP, Brazil e-mail:biancamrc@yahoo.com
J.E. Muñoz Rivera
National Laboratory for Scientific Computation, Rua Getulio Vargas 333, Quitandinha-Petrópolis, 25651-070 Rio de Janeiro, RJ, Brazil
e-mail:rivera@lncc.br M. Sepúlveda (
)CI²MA and Departamento de Ingeniería Matemática, Universidad de Concepción, Casilla 160-C, Concepción, Chile
e-mail:mauricio@ing-mat.udec.cl O.V. Villagrán
Departamento de Matemática, Universidad del Bío-Bío, Collao 1202, Casilla 5-C, Concepción, Chile e-mail:overa@ubiobio.cl
1 Introduction
In this work, we consider the following nonlinear coupled system of equations of Korteweg-de Vries type
ut+ uxxx+ a3vxxx+ uux+ a1vvx+ a2(uv)x= 0, x, t ∈ R, (1.1) b1vt+ vxxx+ b2a3uxxx+ vvx+ b2a2uux+ b2a1(uv)x= 0, (1.2)
u(x,0)= u0(x), v(x,0)= v0(x) (1.3)
where u= u(x, t), v = v(x, t) are real-valued functions of the variables x and t and a1, a2, a3, b1, b2are real constants with b1>0 and b2>0. This system has the structure of a pair of Korteweg-de Vries equations coupled through both dispersive and nonlinear effects. It was derived by Gear and Grimshaw [9] as a model to describe the strong interaction of weakly nonlinear, long waves. Mathematical results on the system (1.1)–(1.3) were given by J. Bona et al. [5]. They proved that the system is globally well posed in Hs(R) × Hs(R) for any s≥ 1 provided that |a3|
√
b2<1. This system has been intensively studied by several authors (see [1,3–5,8] and the references therein). It have the following conservation laws
E1(u)= Rudx, E2(v)= Rvdx, E3(u, v)= R(b2u 2+ b 1v2)dx.
The time-invariance of the functionalsE1 and E2 expresses the property that the mass of each mode separately is conserved during interaction, while that ofE3is an expression of the conservation of energy for the system of two models taken as a whole. Solutions of (1.1)–(1.3) satisfy an additional conservation law which is revealed by the time-invariance of the functional E4= R b2u2x+ v 2 x+ 2b2a3uxvx− b2 u3 3 − b2a2u 2v− b 2a2u2v− b2a1uv2− v3 3 dx. The functionalE4is a Hamiltonian for the system (1.1)–(1.3) and if b2a32<1,E4will be seen to provide an a priori estimate for the solutions (u, v) of (1.1)–(1.3) in the space H1(R) × H1(R). Furthermore, the linearization of (1.1)–(1.3) about the rest state can be reduced to two, linear Korteweg-de Vries equations by a process of diagonalization (see Sect.2). Using this remark and the smoothing properties (in both the temporal and spatial variables) for the linear Korteweg-de Vries derived by Kato [11–13], Kenig, Ponce and Vega [16,17] it will be shown that the system is locally well-posed in Hs(R) × Hs(R) for any s ≥ 1 whenever
√
b2a3= 1. Later on, this result was improved by J.M. Ash et al. [2], showing that the system (1.1)–(1.3) is globally well-posed in L2(R) × L2(R) with√b
2a3= 1. In 2004, F. Linares and M. Panthee [18] improved this result showing that the system (1.1)–(1.3) is locally well-posed in Hs(R) × Hs(R) for s > −3/4 and globally well-posed in Hs(R) × Hs(R) for s >−3/10 under some conditions on the coefficients, indeed for a3= 0 and b1= b2. In 2001, O. Vera [19], following the idea of W. Craig et al. [7] shown that C∞ solutions (u(·, t), v(·, t)) to (1.1)–(1.3) are obtained for t > 0 if the initial data (u0(x), v0(x))belong to a suitable Sobolev space satisfying reasonable conditions as|x| → ∞. In 2000, K. Kato and T. Ogawa [10] shown that a solution of the Korteweg-de Vries equation has a smoothing effect up to real analyticity if the initial data only has a single point singularity at x= 0. It is shown that for Hs(R) (s > −3/4) data satisfying the condition
∞ k=0 Ak 0 k! (x∂x) ku(x,0) Hs(R)<+∞,
the solution is analytic in both space and time variables. This result generalized the result given by W. Craig et al. [7]. On the other hand, because (1.1)–(1.3) is a coupled system of Korteweg-de Vries equations, it is natural to ask whether it has a smoothing effect up to real analyticity if the initial data only has a single point singularity at x= 0 similar to those results obtained by K. Kato and T. Ogawa [10] for the Korteweg-de Vries equation. In this paper, we give an answer to this question which will improve those results obtained in O. Vera [19] (see also [1]). For this reason, first we will decouple the dispersive term in the system (1.1)–(1.3) and we will consider the associated linear system. Our main tool to obtain the main result is to use the generator of dilation P= 3t∂t+ x∂x, which almost commutes with the linear Korteweg-de Vries operator L= ∂t+ ∂x3. A typical example of initial data satisfying the assumption of Theorem1.1is the Dirac delta measure, since (xk∂x)kδ(x)= (−1)kk!δ(x). The other example of the data is the Cauchy principal value p.v. (x1). Linear combination of those distributions with analytic Hs(R) data satisfying the assumption is also possible. In this sense, the Dirac delta measure adding to the soliton initial data can be taken as an initial datum.
This paper is organized as follow: In Sect.2we have the reduction of the problem, no-tation, terminology to be used subsequently and results that will be used several times. In Sect.3we prove a theorem of existence and well-posedness of the solutions. In Sect.4we prove the following theorem:
Theorem 1.1 Suppose that the initial data (u0, v0)∈ Hs(R) × Hs(R), s > −3/4 and A0, A1>0 such that ∞ k=0 Ak 0 k! (x∂x) ku 0Hs(R)<+∞ and ∞ k=0 Ak 1 k! (x∂x) kv 0Hs(R)<+∞. (1.4) Then for some b∈ (1/2, 7/12), there exist T = T (u0Hs(R),v0Hs(R))and a unique solu-tion (u, v)∈ C((−T , T ) : Hs(R)) ∩ Xs
b× C((−T , T ) : Hs(R)) ∩ X s
bto the coupled system (1.1)–(1.3) in a certain time (−T , T ) and the solution (u, v) is time locally well-posed,
i.e., the solution continuously depends on the initial data. Moreover the solution (u, v) is analytic at each point (x, t)∈ R × (−T , T ), t = 0.
As a consequence of this theorem is the following Gevrey 3 regularity result
Corollary 1.1 Let s >−3/4, b ∈ (1/2, 7/12). Suppose that the initial data (u0, v0)∈ Hs(R) × Hs(R), and A
0, A1>0 such that (1.4) is verified. Then there exists a unique
so-lution (u, v)∈ C((−T , T ), Hs(R)) ∩ Xs
b× C((−T , T ), Hs(R)) ∩ Xsbto the coupled system
of Korteweg-de Vries equations (1.1)–(1.3) for a certain (−T , T ) and for any t ∈ (−T , T ), t= 0, the pair (u, v) are analytic functions in the variable space and for x ∈ R, u(x, ·) and v(x,·) is Gevrey 3 as function of the time variable.
Remark 1.1 In Theorem1.1and Corollary1.1, the assumption on the initial data implies analyticity and Gevrey 3 regularity except at the origin respectively. In this sense, those results are stating that the singularity at the origin immediately disappears after t > 0 or t <0, up to analyticity.
Remark 1.2 The crucial part for obtaining a full regularity is to gain the L2(R2)regularity of the solutions (uk, vk)from the negative order Sobolev space. This part is considered in
Proposition 4.1 in Sect.4. We utilize a three steps recurrence argument for treating the nonlinearity appearing in the right hand side of
t ∂x3uk= −1 3P uk+ 1 3x∂xuk+ tB 1 k(u, u)+ tB 2 k(v, v)+ tB 3 k(u, v), (1.5) t ∂x3vk= −1 3P vk+ 1 3x∂xvk+ tC 1 k(u, u)+ tC 2 k(v, v)+ tC 3 k(u, v). (1.6) Then step by step, we obtain the pointwise analytic estimates
sup t∈[t0−,t0+] ∂m t ∂xluH1(x 0−,x0+)≤ cA m+l 1 (m+ l)!, l, m = 0, 1, 2, . . . , sup t∈[t0−,t0+] ∂tm∂xlvH1(x 0−,x0+)≤ cA m+l 2 (m+ l)!, l, m = 0, 1, 2, . . . .
Since initially we do not know whether the solution belongs to even L2(R2) we should mention that the local well-posedness is essentially important for our argument and therefore it merely satisfies the coupled system equations in the sense of distribution.
2 Reduction of the Problem and Preliminary Results
If a3= 0 in the system (1.1)–(1.3), there is no coupling in the dispersive terms. Let us assume that a3= 0. We are interested into decouple the dispersive terms in the system (1.1)–(1.3). For this, let a2
3b2= 1. We consider the associated linear system Wt+ AWxxx= 0, W(x, 0) = W0(x) where W= u v , A= 1 a3 a3b2/b1 1/b1 . The eigenvalues α+and α−of A are given by
α±=1 2 ⎛ ⎝1 + 1 b1 ± 1− 1 b1 2 +4b2a32 b1 ⎞ ⎠
which are distinct since b1>0, b2>0 and a3= 0. Our assumption a32b2= 1 guarantees that α±= 0. Thus we can write the system (1.1)–(1.3) in a matrix form as in [18]. After we make
the change of scale
u(x, t )= u(α+−1/3x, t ) and v(x, t )= v(α−−1/3x, t ), (2.1) obtaining the coupled system of equations
ut+ α+uxxx+ auux+ bvvx+ c(uv)x= 0, x, t ∈ R, (2.2)
vt+ α−vxxx+a uux+ bvvx+c(uv)x= 0, (2.3)
where
u(x, t )= u(α1/3− x, t ) and v(x, t )= v(α+1/3x, t ), (2.5) and a, b, c anda, b,care constant.
Remark 2.1 Notice that the nonlinear terms involving the functionsuandvare not evaluated at the same point. Therefore, those terms are not local anymore. Hence we should be careful in making the estimates (see for instance [18]). We remark additionally, that thank to the property of homogeneity of the generator of dilation P , we have uk= Pku= Pku= uk. Thus, for the sake of simplicity, while there is no ambiguity we will drop ‘’ and ‘’ using the notation u, v, u0, v0in the system (2.2)–(2.4).
For s, b∈ R define the following Bourgain spaces Xs
band Xsb−1to be the completion of
the Schwartz spaceS(R2)with respect to the norms
uXs b:= R R(1+ |τ − ξ 3|)2b(1+ |ξ|)2s|u(ξ, τ )|2dξ dτ 1/2
and uXsb−1, where X
s
b= {u ∈S (R2): uXbs <∞} (see [6]). Let Fx and Fx,t be the Fourier transforms in the x and (x, t) variables respectively. The Riesz operator Dx is de-fined by Dx=Fξ−1|ξ|Fx. The fractional derivative is defined by
Dx s:=Fξ−1 ξ sF
x=Fξ−1(1+ |ξ| 2)s/2Fx,
Dx,t s:=Fξ,τ−1 |ξ| + |τ| sFx,t. For ·, · = (1 + | · |2)1/2, we have the following notation
(i) · Hb(R:Hr(R))= Dt b Dx rL2x,t(R2) (ii) Hs(R) = {u ∈S (R) : D
x su∈ L2(R)} (iii) · Hs(R)= Dx sL2(R).
Remark 2.2 With the above notation we have
(a) uHxs(R)= ξ su L2(R) (b) uL2 t(R:Hxr(R))= ξ ru L2(R2) (c) Dx suL2(R)= uHs(R) (d) Dt b Dx ruL2 x,t(R2)= uHtb(R:Hxr(R)) (e) Dx,t suL2t(R:Hxr(R))= ξ r |ξ| + |τ| su(ξ, τ ) L2(R2).
Let L= ∂t+ ∂x3; we introduce the generator of dilation P = 3t∂t+ x∂x for the linear part of the coupled system (2.2)–(2.4) and the localized dilation operator P0= 3t0∂t+ x0∂x. By employing a localization argument, we look the operator P as a vector field P0= 3t0∂t+ x0∂x near a fixed point (x0, t0)∈ R × (−T , T ), t0= 0. Since P0 is a directional derivative towards (x0, t ), we introduce another operatorL30= t0∂x3which plays the role of a non-tangential vector field to P0. Since P0andL0are linearly independent, the space and time derivative can be covered by those operators. The main reason why we chooseL0is because the corresponding variable coefficients operatorL3= t∂3
x can be treated via (1.5)– (1.6) and a cut-off procedure enables us to handle the right hand side of those equations.
Remark 2.3 For L and P we have the following properties: (a) [L, P ] ≡ LP − P L = 3L; (b) LPk= (P + 3)kL; (c) (P+ 3)k∂x= ∂ x(P+ 2)k; (d) (P+ 3)k∂3 x= ∂x3Pk; (e) P0P= P P0+ 3P0− 2x0∂x.
Notation 2.1 The summationk=k1+k2+k3
0≤k1,k2,k3≤k is simply abbreviated byk=k 1+k2+k3. Let Pku= u k, then ∂t(Pku)+ ∂x3(Pku)= LPku= (P + 3)kLu= (P + 3)k(∂tu+ ∂x3u) = −(P + 3)k a 2∂x(u 2)+b 2∂x(v 2)+ c∂x(uv) = −a 2(P+ 3) k∂x(u2)−b 2(P+ 3) k∂x(v2)− c(P + 3)k∂x(uv) = −a 2∂x(P+ 2) k(u2)−b 2∂x(P+ 2) k(v2)− c∂x(P+ 2)k(uv). Noting that (P+ 2)ku=k j=0 k j 2k−jPju.Hence Bk1(u, u)= − a 2∂x(P+ 2) k(u2)= −a 2 k=k1+k2+k3 k! k1!k2!k3! 2k1∂x(uk 2uk3), Bk2(v, v)= −b 2∂x(P+ 2) k(v2)= −b 2 k=k 1+k 2+k 3 k! k 1!k2 !k3 !2 k 1∂ x(vk 2vk 3), Bk3(u, v)= c∂x(P+ 2)k(uv)= −c k=k 1+k2 +k 3 k! k 1!k 2!k3 !2 k 1∂ x(uk2 vk3 ). Therefore ∂t(Pku)+ ∂x3(Pku) = −a 2 k=k1+k2+k3 k! k1!k2!k3! 2k1∂x(uk 2uk3)− b 2 k=k 1+k 2+k 3 k! k1 !k2 !k3 !2 k1 ∂x(v k2 vk3 ) − c k=k 1+k 2+k 3 k! k 1!k 2!k 3!2 k 1∂x(u k2 vk3 ) = B1 k(u, u)+ B 2 k(v, v)+ B 3 k(u, v). (2.6)
Performing similar calculations as above we obtain ∂t(Pkv)+ ∂x3(Pkv) = −a 2 k=k1+k2+k3 k! k1!k2!k3! 2k1∂x(uk 2uk3)− b 2 k=k 1+k 2+k 3 k! k1 !k2 !k3 !2 k1 ∂x(v k2 vk3 )
−c k=k1 +k 2+k 3 k! k 1!k 2!k 3! 2k 1∂x(uk 2vk3 ) = C1 k(u, u)+ C 2 k(v, v)+ C 3 k(u, v).
The above nonlinear terms maintain the bilinear structure like that of the original coupled system of equations of KdV type, since Leibniz’s rule can be applicable to operations of P . Now each ukand vksatisfies the following system of equations
∂tuk+ ∂x3uk= Bk1(u, u)+ Bk2(v, v)+ Bk3(u, v)≡ Bk, (2.7) ∂tvk+ ∂x3vk= Ck1(u, u)+ Ck2(v, v)+ Ck3(u, v)≡ Ck, (2.8) uk(x,0)= (x∂x)ku0(x)≡ uk0(x), vk(x,0)= (x∂x)kv0(x)≡ vk0(x). (2.9) In order to obtain a well-posedness result for the system (2.7)–(2.9) we use Duhamel’s principle and we study the following system of integral equations equivalent to the system (2.7)–(2.9) ψ (t )uk= ψ(t)V (t)uk0− ψ(t) t 0 V (t− t )ψT(t )Bk(t )dt , ψ (t )vk= ψ(t)V (t)vk0− ψ(t) t 0 V (t− t )ψT(t )Ck(t )dt
where V (t)= e−t∂x3 is the unitary groups associated with the linear problems and ψ(t)∈ C0∞(R), 0 ≤ ψ ≤ 1 is a cut-off function such that
ψ (t )=
1, if|t| < 1
0, if|t| > 2 and ψT(t )= ψ(t/T ).
The following results are going to be used several times in the rest of this paper.
Lemma 2.1 ([14,15]) Let s∈ R, a, a ∈ (0, 1/2), b ∈ (1/2, 1) and δ < 1. Then for any k= 0, 1, 2, . . . we have ψδφkXs−a≤ cδ(a−a )/4(1−a ) φkXs−a , ψδV (t )φkXsb≤ cδ 1/2−bφ kHs(R), ψδ t 0 V (t− t )Fk(t )dt Xsb ≤ cδ1/2−bF kXsb−1.
Lemma 2.2 ([10]) Let s >−3/4, b, b ∈ (1/2, 7/12) with b < b . Then for any k, l= 0, 1, 2, . . . we have
∂x(ukvl)Xs
b −1≤ cvkXsbvlXsb.
Lemma 2.3 ([10]) Let s < 0, b∈ (1/2, 7/12) and ψ = ψ(x, t) be a smooth cut-off function
t0)/). Then for f∈ Xbs, we have
ψfXsb≤ c
−|s|−5|b|ψ
X|s|+2|b|
|b| f Xbs+2|b|,
where the constant c is independent of and f .
Lemma 2.4 ([10]) Let P be the generator of the dilation and Dx,t be an operator defined
byFξ,τ−1 |τ| + |ξ| Fx,t. We fix an arbitrary point (x0, t0)∈ R × (−T , T ), t = 0. Then 1. Suppose that b∈ (0, 1], r ∈ (−∞, 0] and g ∈ Xr
b−1 with supp g⊂ B2(x0, t0)and t∂x3g, P3g∈ Xr
b−1. If > 0 is sufficiently small, then we have
Dx,t 3bgL2(R:Hr(R))≤ c
gXbr−1+ t∂x3gXrb−1+ P3gXrb−1
(2.10)
where the constant c= c(x0, t0; ). 2. If g∈ Hμ−3(R2)with suppg⊂ B
2(x0, t0)and t∂x3g, P3g∈ Hμ−3(R2). Then for small ,
we have Dx,t μgL2(R2)≤ c gHμ−3(R2)+ t∂x3gHμ−3(R2)+ P3gHμ−3(R2) (2.11)
where the constant c= c(x0, t0; ).
Lemma 2.5 ([10]) Let 0≤ s, r ≤ n/2 with n/2 ≤ s + r and suppose f ∈ Hs(Rn)and g∈ Hr(Rn). Then for any σ < s+ r − n/2, we have fg ∈ Hσ(Rn)and
fgHσ(Rn)≤ c()f Hs(Rn)gHr(Rn), where = s + r − n/2 − σ .
Corollary 2.1 ([10]) For 1/2 < b < 1 and−3/4 < s < 0, we have ψf Xs−1
b−1≤ cf Xsb−1, where ψ∈ C0∞(R2)and c is independent of f .
Lemma 2.6 ([10]) Let ψ(x) be a smooth cut-off function in C0∞((−2, 2)) with ψ(x) = 1 on (−1, 1). We set ψ= ψ(x/) for 0 < < 1. Then for r ≤ 0, and f ∈ Hr, we have
ψfHr(R)≤
c−δf Hr(R) if − 1/2 ≤ r ≤ 0, c1/2+rf Hr(R) if r <−1/2 where δ > 0 is an arbitrary small constant and c is independent of .
Proposition 2.1 Let s >−3/4, b, b ∈ (1/2, 7/12) with b < b . Then for any k, l = 0, 1, 2, . . . we have
∂x(ukvl)Xsb −1≤ cukXsbvlXsb, (2.12)
∂x(ukvl)Xsb −1≤ cukXsbvlXsb, (2.13) whereu,v, uand vare defined by the change of scales (2.1) and (2.5), from u and v.
Proof The proof of (2.12)–(2.13) follows the same argument used in [14,15] to prove
∂x(ukvl)Xsb −1≤ cukXbsvlXbs.
It is not difficult to prove the same inequality since the only changes coming out are from the contributions given by the constants α+and α−(see for example [18, Remark 2.2]). Throughout this paper c is a generic constant, not necessarily the same at each occasion (it will change from line to line), which depends in an increasing way on the indicated quantities.
3 Existence of the Solutions
In this section, we first solve the following (slightly general) system of equations ∂tuk+ ∂x3uk= Bk1(u, u)+ B 2 k(v, v)+ B 3 k(u, v)≡ Bk, (3.1) ∂tvk+ ∂x3vk= C1k(u, u)+ C 2 k(v, v)+ C 3 k(u, v)≡ Ck, (3.2) uk(x,0)= (x∂x)ku0(x)≡ uk0(x), vk(x,0)= (x∂x)kv0(x)≡ v0k(x) (3.3) where Bkand Ckare as above.
Definition 3.1 Let f = (f0, f1, . . . , fk)denotes the infinity series of distributions and define
AA0(X s b)≡ f= (f0, f1, . . . , fk), fi∈ Xbs (i= 0, 1, 2 . . .) such that f AA0(X s b)<+∞ where f AA0(Xs b)≡ ∞ k=0 Ak 0 k!fkXbs.
Similarly, for u0= {u00, u10, . . . , uk0, . . .} and v0= {v00, v01, . . . , v0k, . . .} we set
u0AA0(Hs(R))≡ ∞ k=0 Ak 0 k!u k 0Hs(R) and v0AA 0(Hs(R))≡ ∞ k=0 Ak 0 k!v k 0Hs(R) respectively.
Remark 3.1 The corresponding of the solution to the coupled system of Korteweg-de Vries
equations is accompanied by the following estimate (for k= 0, 1, 2, . . .)
Pku
Xbs≤ cAk0k!, and PkvXsb≤ cAk1k!.
Theorem 3.1 Let s >−3/4, b ∈ (1/2, 7/12). Suppose that uk 0, v k 0∈ Hs(R) (k = 0, 1, 2, . . .) and satisfy u0AA0(Xs b)= ∞ k=0 Ak 0 k!u k 0Hs(R)<+∞ and v0AA 0(X s b)= ∞ k=0 Ak 0 k!v k 0Hs(R)<+∞.
Then there exist T = T (uk
0Hs(R),vk
0Hs(R))and a unique solution u= (u0, u1, . . .)and v= (v0, v1, . . .)of the system (3.1)–(3.3) with uk, vk∈ C((−T , T ) : Hs(R)) ∩ Xsband
∞ k=0 Ak 0 k!ukX s b(R)<+∞ and ∞ k=0 Ak 0 k!vkX s b(R)<+∞. Moreover, the map (uk
0, v0k)→ (u(t), v(t)) is Lipschitz continuous, i.e.,
u(t) − u(t)AA0(Xsb)+ u(t) − u(t)C((−T ,T ):Hs(R))≤ c(T )u0− u
0AA0(Hs(R)),
v(t) − v(t)AA0(Xs
b)+ v(t) − v(t)C((−T ,T ):Hs(R))≤ c(T )v0− v0AA0(H s(R)),
for two initial conditions (u0, v0)and (u0, v0)in (3.3), and its respective solutions (u, v)
and (u, v) of (3.1)–(3.3).
Proof For given (u0, v0)∈AA0(H
s(R)) ×A A0(H
s(R)) and b > 1/2, let us define,
HR1,R2= (u, v)∈AA0(X s b)×AA0(X s b): uAA0(Xs b)≤ R1,vAA0(X s b)≤ R2 where R1= 2c0u0AA0(Hs(R))and R2= 2c0v0AA
0(Hs(R)). ThenHR1,R2is a complete met-ric space with norm
(u, v)HR1,R2= uAA0(Xsb)+ vAA0(X s b).
Without loss of generality, we may assume R1>1 and R2>1. For (u, v)∈HR1,R2, let us define the maps,
ku 0(u, v)= ψ(t)V (t)u k 0− ψ(t) t 0 V (t− t )ψT(t )Bk(t )dt , vk0(u, v)= ψ(t)V (t)v k 0− ψ(t) t 0 V (t− t )ψT(t )Ck(t )dt .
We prove that × maps HR1,R2 into HR1,R2 and it is a contraction. In fact, using Lemma2.1, Lemma2.2, Proposition2.1and the fact that u and v are not evaluated at the same point (see Remark2.1and proof of Proposition2.1) we have
k u0(u, v)Xbs = ψ(t)V (t)u k 0Xbs+ ψ(t)0tV (t− t )ψT(t )Bk(t )dt Xs b ≤ c0uk0Hs(R)+ c1TμBkXs b −1 ≤ c0uk0Hs(R)+ c1Tμa 2 k=k1+k2+k3 k! k1!k2!k3! 2k1u k2Xsbuk3Xbs + c1Tμ b 2 k=k 1+k2 +k3 k! k 1!k 2!k 3!2 k 1v k 2Xsbvk3 Xbs + c1Tμc k=k1 +k 2+k3 k! k1 !k2 !k 3!2 k1 u k2 Xsbvk3 Xsb.
Applying a sum over k we have ∞ k=0 Ak0 k! k u0(u, v)Xsb ≤ c0 ∞ k=0 Ak0 k!u k 0Hs(R)+ c1Tμa 2 ∞ k=0 Ak0 k! k=k1+k2+k3 k! k1!k2!k3! 2k1u k2Xsbuk3Xbs + c1Tμ b 2 ∞ k=0 Ak0 k! k=k1 +k2 +k3 k! k1 !k 2!k 3!2 k 1v k 2Xsbvk 3Xsb + c1Tμc ∞ k=0 Ak 0 k! k=k 1+k 2+k3 k! k1 !k2 !k3 !2 k1 u k 2Xs bvk 3Xsb ≤ c0u0AA0(Hs(R))+ c1Tμ a 2 ∞ k1=0 2k1A k1 0 k1! ∞ k2=0 Ak2 0 k2! uk2Xbs ∞ k3=0 Ak3 0 k3! uk3Xbs + c1Tμ b 2 ∞ k1 =0 2k1 A k1 0 k1 ! ∞ k2 =0 Ak2 0 k2 !vk2 X s b ∞ k3 =0 Ak3 0 k3 !vk3 X s b + c1Tμc ∞ k 1=0 2k 1A k1 0 k1 ! ∞ k 2=0 Ak 2 0 k2 ! uk2 Xsb ∞ k3 =0 Ak 3 0 k 3! vk 3Xbs = c0u0AA0(Hs(R))+ c1Tμ a 2e 2A0u2 AA0(Xs b) + c1Tμ b 2e 2A0v2 AA0(Xs b)+ c1T μce2A0u AA0(Xsb)vAA0(X s b).
Hence, choosing d= max{a/2, b/2, c} we have
u0(u, v)AA0(Xbs)≤ c0u0AA0(H s(R)) + c1Tμde2A0 u2 AA0(Xsb)+ v 2 AA0(Xsb)+ uAA0(X s b)vAA0(X s b) ≤ c0u0AA0(Hs(R))+3 2c1dT μe2A0 u2 AA0(Xsb)+ v 2 AA0(Xsb) . (3.4)
In a similar way, choosing d= max{a/2,b/2,c} we have
v0(u, v)AA0(Xbs)≤ c0v0AA0(Hs(R)) +3 2c2dT μe2A0 u2 AA0(Xsb)+ v 2 AA0(Xsb) . (3.5) Choosing Tμ≤ 1
3 max{c1,c2}(R1+R2)2,we obtain from (3.4) and (3.5), the following estimates
u0(u, v)AA0(X s
b)≤ R1 and v0(u, v)AA0(X s b)≤ R2, that is, ( u0, v0)∈HR1,R2.
Now, we will show that u0× v0: (u, v) → ( u0(u, v), v0(u, v))is a contraction. Let (u, v), (u, v)∈HR1,R2, then as above we get for d= max{a/2, b/2, c,a/2,b/2,c}
u0(u, v)− u0(u, v)AA0(Xsb) ≤3 2c1dT μe2A0(R 1+ R2) u − uAA0(Xs b)+ v − vAA0(X s b) . Choosing Tμ≤ 1 6 max{c1,c2}(R1+R2)2, we obtain u0(u, v)− u0(u, v)AA0(Xsb)≤ 1 4 u − uAA0(Xsb)+ v − vAA0(X s b) . In a similar way v0(u, v)− v0(u, v)AA0(Xsb)≤ 1 4 u − uAA0(Xsb)+ v − vAA0(X s b) .
Therefore, the map u0× v0 is a contraction and we obtain a unique fixed point (u, v) which solves the initial value problem (3.1)–(3.3) for T < Tμ. The remainder of the proof
follows a standard argument.
Corollary 3.1 Let s >−3/4, b ∈ (1/2, 7/12). Suppose that (x∂x)ku
0, (x∂x)kv0∈ Hs(R),
with k= 0, 1, 2, . . . and satisfy
∞ k=0 Ak 0 k!u k 0Hs(R)<+∞ and ∞ k=0 Ak 0 k!v k 0Hs(R)<+∞. Then there exist T = T (uk
0Hs(R),vk
0Hs(R))and a unique solution (u, v) of the coupled system equations KdV type (3.1)–(3.3) with u, v∈ C((−T , T ) : Hs(R)) ∩ Xs
band ∞ k=0 Ak 1 k!P ku Xsb(R)<+∞, ∞ k=0 Ak 1 k!P kv Xsb(R)<+∞.
Moreover, the map (u0, v0)→ (u(t), v(t)) is Lipschitz continuous in the following sense:
u(t) − u(t)Xsb+ u(t) − u(t)C((−T ,T ):Hs(R))≤ c(T )
∞ k=0 Ak 0 k!(x∂x) k(u 0− u0)Hs(R), v(t) − v(t)Xbs+ v(t) − v(t)C((−T ,T ):Hs(R))≤ c(T ) ∞ k=0 Ak0 k!(x∂x) k(v 0− v0)Hs(R).
4 The Main Result
In this section we prove analyticity of the solution obtained in the previous section. We treat the solution uk≡ Pkuand vk≡ Pkvas if it satisfies the coupled system of (3.1)–(3.3) in the classical sense. This can be justified by a proper approximation procedure. The following results are going to be used in this section. Let (x0, t0)be arbitrarily taken inR × (−T , T ),
t= 0. By ψ(x, t) we denote a smooth cut-off function in C0∞(B1(0)) and ψ= ψ((x − x0)/, (t− t0)/).
Let ψ be a smooth cut-off function around the freezing point (x0, t0) with supp ψ⊂ C0∞(B(x0, t0)).
Proposition 4.1 For the cut-off function ψ defined above, there exist positive constant c and Asuch that for k= 0, 1, 2, . . .
ψPku L2 x,t(R2)≤ cA k(k!)2, ψPkv L2 x,t(R2)≤ cA k(k!)2.
Proof Using (2.10) with r= s − 1, we obtain
Dx,t 3bψ PkuL2 t(R:Hxs−1(R)) ≤ cψukXs−1 b−1+ t∂ 3 x(ψ uk)Xbs−1−1+ P 3(ψ uk) Xsb−1−1 . (4.1)
Each term in (4.1) is estimated separately. For first term in the right hand side we use Lemma2.3. Indeed,
ψukXsb−1−1≤ ψukXbs−1≤ cψX|s|+2|b−1||b−1| ukXbs≤ c(ψ)A k
1k!, k = 0, 1, 2, . . . . (4.2) The third term is estimated again using Corollary2.1
P3(ψ uk) Xsb−1−1≤ 3 l=0 3! l(l− 3)!P 3−lψ Pluk Xbs−1 ≤ c(ψ) 3 l=0 3! l(l− 3)!P luk Xsb ≤ c 3 l=0 3! l(l− 3)!P k+lu Xsb = c 3 l=0 Ak1+l(k+ l)! ≤ cAk2k!, (4.3)
for k= 0, 1, 2, . . . . For the second term, we use (3.1) and reduce the third derivative in space to the dilation operator P . On the other hand, from the generator of dilation P uk= 3t∂tuk+ x∂xukwe remark that
t ∂tuk=1 3P uk− 1 3x∂xuk. (4.4) Multiplying (3.1) by ψt , we have ψ t ∂tuk+ ψt∂x3uk= ψtBk. (4.5)
Replacing (4.4) in (4.5) we obtain ψt∂3 xuk= − 1 3ψ P uk+ 1 3ψ x∂xuk+ ψtBk.Hence ψt∂3 xukXsb−1−1= 1 3ψP ukXsb−1−1+ 1 3ψx∂xukXbs−1−1+ ψtBkXbs−1−1 = F1+ F2+ F3. (4.6)
Using the assumption in theorem, we have F1= 1 3ψP ukXbs−1−1≤ cψX−s1−bPk+1uXsb−1≤ cPk+1uXsb ≤ cAk+1 3 (k+ 1)! ≤ cA k 4k!. (4.7) Similarly, we obtain F2= 1 3ψx∂xukXsb−1−1≤ 1 3∂x(ψ xuk)Xbs−1−1+ 1 3∂x(ψ x)ukXbs−1−1 ≤1 3∂x(ψ xvk)Xbs−1+ c∂x(ψ x)X−s1−bukXs b−1 ≤1 3ψxXbsukXbs+ c∂x(ψ x)X−s1−bukXbs−1 ≤ cψxXsb+ ∂x(ψ x)X1−s−b Ak5k! ≤ cAk6k!. (4.8) Using Lemmas2.2,2.3, and Proposition2.1, we have
F3= ψtBkXs−1 b−1≤ cψX−sb−1B 1 k+ B 2 k+ B 3 kXs b−1 ≤ cB1 kXs b−1+ B 2 kXs b−1+ B 3 kXs b−1 ≤ c k=k1+k2+k3 k! k1!k2!k3! 2k1u k2Xsbuk3Xbs+ c k=k1 +k2 +k3 k! k1 !k 2!k 3! 2k 1v k 2Xsbvk 3Xsb + c k=k1 +k 2+k 3 k! k 1!k 2!k 3! 2k 1u k2 Xsbvk3 Xbs ≤ c k=k1+k2+k3 k! k1!k2!k3! 2k1Ak2 7k2!A k3 7k3! + c k=k 1+k 2+k 3 k! k1 !k2 !k3 ! 2k1 Ak 2 8k2 !A k3 8k 3! + c k=k1 +k 2+k 3 k! k 1!k 2!k 3! 2k 1Ak 2 9 k 2!A k3 10k3 ! ≤ ck! ⎛ ⎝Ak 7 k k1=0 k−k1 k2=0 1 k1! 2k1A−k1 7 + Ak8 k k 1=0 k−k 1 k 2=0 1 k1 !2 k1 A−k1 8 + k=k1 +k2 +k 3 1 k 1!2 k 1Ak 2 9 A k 3 10 ⎞ ⎠ ≤ ck! ⎛ ⎝Ak 7 k k1=0 k−k1 k2=0 (A2 7) k1 k1! + Ak 8 k k 1=0 k−k 1 k2 =0 (A2 8) k1 k1 ! + k=k1 +k 2+k3 1 k1 !2 k1 Ak2 9 A k 3 10 ⎞ ⎠
≤ ce2/A7Ak 7k! + ce 2/A8Ak 8k! + ck! k=k 1+k2 +k3 1 k1 !2 k 1Ak2 9A k3 10 ≤ ce2/A7+ e2/A8A 11k! + ck! k=k1 +k 2+k 3 1 k 1!2 k 1Ak 2 9 A k 3 10, k= 0, 1, 2, . . . . (4.9)
Hence, replacing (4.7), (4.8) and (4.9) in (4.6) we obtain that there exist positive constants c, A9, A10and A11, such that
ψt∂3 xukXsb−1−1≤ cA11k! + ck! k=k 1+k2 +k 3 1 k1 !2 k 1Ak 2 9A k3 10, k= 0, 1, 2, . . . . (4.10)
On the other hand, using ∂3
x(ψf )= ψ∂x3f+ 3∂x2(∂xψf )− 3∂x(∂x2ψf )+ ∂x3ψf we have t∂3 x(ψ uk)Xsb−1−1≤ tψ∂ 3 xukXsb−1−1+ 3∂ 2 x(t ∂xψ uk)Xsb−1−1 +3∂x(t ∂x2ψ uk)Xs−1 b−1+ t∂ 3 xψ ukXsb−1−1. (4.11) Using Lemma2.3and Proposition2.2, we obtain
∂2 x(t ∂xψ uk)Xsb−1−1≤ ∂x(t ∂xψ uk)Xbs−1≤ ct∂xψXsbukXsb ≤ cAk 10k!, (4.12) ∂x(t ∂x2ψ uk)Xsb−1−1≤ ∂x(t ∂ 2 xψ uk)Xs b−1≤ ct∂ 2 xψXs bukXsb ≤ cAk 11k!, (4.13) t∂3 xψ ukXsb−1−1≤ c Dx,t 3/2t ∂x3ψX|s|+2|b−1|1−b ukXsb−1≤ cukXbs ≤ cAk 12k!. (4.14)
Hence, replacing (4.10), (4.12), (4.13) and (4.14) in (4.11) we obtain that there exist con-stants c and A14such that
t∂3 x(ψ uk)Xbs−1−1≤ cA k 14k! + ck! k=k 1+k 2+k3 1 k1 ! 2k1 Ak2 9 A k 3 10, k= 0, 1, 2, . . . . (4.15)
Therefore, replacing (4.2), (4.3) and (4.15) in (4.1) we obtain that there exist constants c and A15such that Dx,t 3bψ ukL2 t(R:Hxs−1(R)) ≤ cAk 15k! + ck! k=k 1+k 2+k3 1 k1 !2 k1 Ak2 9 A k 3 10, k= 0, 1, 2, . . . . (4.16)
Dx,t 3bψ vkL2 t(R:Hxs−1(R)) ≤ cAk 16k! + ck! k=k 1+k 2+k3 1 k1 !2 k1 Ak2 9 A k 3 10, k= 0, 1, 2, . . . . (4.17)
Adding (4.16) and (4.17) we have
Dx,t 3bψ ukL2 t(R:Hxs−1(R))+ Dx,t 3bψ vk L2t(R:Hxs−1(R)) ≤ cAk 15k! + cAk16k! + ck! k=k1 +k 2+k 3 1 k 1! 2k1 2Ak 2 9 A k 3 10 ≤ c(Ak 15+ Ak16)k! + ck! k=k1 +k 2+k 3 1 k 1! 2k1 2Ak 2 9 A k 3 10 ≤ cAk 17k! + ck! k=k1 +k 2+k 3 1 k 1!2 k1 2Ak 2 9 A k 3 10. (4.18)
We estimate now the last term in the right hand side of (4.18)
k=k 1+k 2+k3 1 k1 !2 k 12Ak2 9 A k 3 10= k m=0 m j=0 1 (m− j)!2 (m−j)2Aj 9A k−m 10 ≤ Ak 10 k m=0 m j=0 1 (m− j)!2 A9 2 j 2 A10 m ≤ Ak 10 k m=0 m j=0 A29 4 j + 4 A2 10 m ≤ Ak 10 k m=0 m j=0 j!( A29 4) j j! + A k 10 k m=0 m j=0 m! ( 4 A210) m m! ≤ Ak 10k! k m=0 m j=0 (A29 4) j j! + A k 10k! k m=0 m j=0 (A42 10 )m m! ≤ eA29/4Ak 10k! + e 4/A210Ak 10k! ≤ cA k 10k!. (4.19) Replacing (4.19) in (4.18) we obtain Dx,t 3bψ ukL2 t(R:Hxs−1(R))+ Dx,t 3bψ vk L2 t(R:Hxs−1(R)) ≤ cAk 17k! + cA k 19(k!) 2≤ cAk 17(k!) 2+ cAk 19(k!) 2≤ cAk 20(k!) 2 (4.20)
Remark 4.1
1. For simplicity, we only illustrate the conclusion for the case s≥ −1/2−δ with b = 1/2+ δ/3 (for small δ > 0) and the case s= −3/4 + δ and b = 7/12 − δ/3. If s = −1/2 − δ with b= 1/2 + δ/3, the initial data can involve Dirac’s delta measure δ0and the latter is the critical case of the local well-posedness.
2. The following inequality is simple to verify in both cases,
ψukL2 x,t(R2)≤ Dx 3b(ψ uk) L2t(R:Hxs−1(R))≤ c Dx,t 3b(ψ uk) L2t(R:Hxs−1(R)).
Proposition 4.2 Under the same assumptions as in Proposition 4.1, there exist positive
constants c and A such that for k= 0, 1, 2, . . . ψPku
H7/2(R2)≤ cAk(k!)2 and ψPkvH7/2(R2)≤ cAk(k!)2. Proof We apply Lemma2.4to ψuk≡ ψPkuwith b= 1 and r = 0.
Dx,t 3ψ PkuL2(R:L2 x(R))
≤ cψukL2(R:Lx2(R))+ t∂x3(ψ uk)L2(R:L2x(R))+ P3(ψ uk)L2(R:L2x(R))
.(4.21) Therefore, if we wish to estimate the second term in the right hand side of (4.21) with the aid of (2.7) ψ t ∂x3uk= − 1 3ψ P uk+ 1 3ψ x∂xuk+ tψBk
it is necessary to estimateψukL2t(R:Hx1(R))which is not yet obtained. Hence, we start from the lower regularity setting, i.e., apply (2.11) in Lemma2.4to ψukwith μ= 1/2. Let ψ1be a smaller size of smooth cut-off function with ψ1≤ ψ and ψ1= 1 around (x0, t0). Applying (2.11) a ψuk= ψPkuwith μ= 1/2 we have
Dx,t 3ψ1PkuH−5/2(R2)≤ c Dx,t 3ψ1PkuL2(R2)
≤ cψ1ukH−5/2(R2)+ t∂x3(ψ1uk)H−5/2(R2)+ P3(ψ1uk)H−5/2(R2)
. (4.22) The first term on the right hand side of (4.22) has already been estimated. For the third term we have P3(ψ 1uk)H−5/2(R2)≤ P3(ψ1uk)L2x,t(R2) = 3 l=0 3! l!(3 − l)!P 3−lψ 1PlukL2x,t(R2) ≤ 3 l=0 3! l!(3 − l)!P 3−lψ 1L∞x,t(R2)P luk L2x,t(R2) ≤ c 3 l=0 3! l!(3 − l)!P k+lu L2x,t(R2)
≤ c
3
l=1
Ak1+lk! ≤ cAk2k! ≤ cAk2(k!)2. (4.23) For the second term on the right side hand we use the same idea of the above remark, using the dilation operator P . Indeed,
t∂3
x(ψ1uk)H−5/2(R2)≤ ψ1t ∂x3ukH−5/2(R2)+ 3∂x2(t ∂xψ1uk)H−5/2(R2)
+ 3∂x(t ∂x2ψ1uk)H−5/2(R2)+ t∂x3ψ1ukH−5/2(R2). (4.24) The last three term are bounded by the following:
c 3 l=1 ∂l xψ1L∞x,t(R2)ψukL2x,t(R2)≤ cA k 3k! ≤ cA k 3(k!) 2. (4.25)
On the other hand, using
ψ1t ∂x3ukH−5/2(R2)≤ 1 3ψ1P ukL2(R:L2x(R))+ 1 3xψ1∂xukH−5/2(R2)+ tψ1BkH−5/2(R2) = F1+ F2+ F3. (4.26) Thus F1≤ cψ1L∞x,t(R2)ψP k+1u L2x,t(R2)≤ cψP k+1v L2x,t(R2) ≤ cAk+1 4 (k+ 1)! ≤ cA k 5k! ≤ cA k 5(k!) 2, F2≤ xψ1∂xvkL2(R:Hx−1(R)) ≤ ∂x(xψ1vk)L2(R:Hx−1(R))+ ∂x(xψ1)ψ vkL2(R:Hx−1(R)) ≤ xψ1vkL2 x,t(R2)+ ∂x(xψ1)L∞x,t(R2)ψvkL2x,t(R2) ≤xψ1L∞x,t(R2)+ ∂x(xψ1)L∞x,t(R2) ψvkL2x,t(R2) ≤ cAk 6k! ≤ cAk6(k!) 2.
Using Lemma2.5(case σ= −5/2, s = 5, r = −5/2) and replacing Bkby (2.6), we have F3= tψ1BkH−5/2(R2)≤ c1ψ1H5(R2)ψ2BxH−5/2(R2) ≤ c1|a| 2 k=k1+k2+k3 2k1k! k1!k2!k3! ψuk2ψ uk3H−3/2(R2) + c1 |b| 2 k=k 1+k 2+k3 2k 1k! k 1!k 2!k3 !ψvk 2ψ vk3 H−3/2(R2) + c1|c| k=k1 +k 2+k 3 2k 1k! k1 !k 2!k 3!ψuk 2ψ vk 3H−3/2(R2) ≤ c1 |a| 2 k=k1+k2+k3 2k1k! k1!k2!k3! ψuk2L2(R2)ψuk3L2(R2)
+ c1 |b| 2 k=k 1+k 2+k3 2k 1k! k 1!k 2!k3 !ψvk 2L2(R2)ψvk 3L2(R2) + c1|c| k=k1 +k 2+k 3 2k 1k! k1 !k 2!k 3!ψuk 2L2(R2)ψvk3 L2(R2) ≤ c1k! |a| 2 k=k1+k2+k3 2k1 k1! Ak2+k3 7 + |b| 2 k=k 1+k 2+k 3 2k 1 k1 !A k2 +k3 8 + |c| k=k1 +k 2+k 3 2k 1 k1 !A k 2 9 A k 3 10 = c1k! |a| 2 A k 7 k k1=0 k−k1 k2=0 2k1 k1! A−k1 7 + |b| 2 A k 8 k k1=0 k−k1 k 2=0 2k1 k 1!A −k 1 8 + |c| k=k1 +k 2+k 3 2k 1 k1 !A k 2 9 A k 3 10 ≤ c1 |a| 2 e 2/A7Ak 7(k+ 1)! + c1 |b| 2 e 3/A8Ak 8(k+ 1)! + c1|c|k! k=k1 +k 2+k 3 2k 1 k1 ! Ak 2 9 A k 3 10. (4.27)
Replacing (4.25), (4.27) and (4.24) in (4.26) we obtain
ψ1t ∂x3ukH−5/2(R2) ≤ c2Ak11k! + c1|c|k! k=k1 +k 2+k 3 2k 1 k1 ! Ak 2 9 A k 3 10, k= 0, 1, 2, . . . . (4.28) Replacing (4.25) and (4.28) in (4.24) t∂3 x(ψ1uk)H−5/2(R2) ≤ c3Ak12k! + c1|c|k! k=k1 +k 2+k 3 2k 1 k1 !A k 2 9 A k 3 10, k= 0, 1, 2, . . . . (4.29)
Now replacing (4.23) and (4.29) in (4.22) we obtain
Dx,t 3ψ ukH−5/2(R2)≤ c4Ak13k! + c1|c|k! k=k 1+k2 +k 3 2k 1 k1 !A k 2 9 A k 3 10, for k= 0, 1, 2, . . . . In particular ψukH1/2(R2)≤ c5Ak14k! + c1|c|k! k=k1 +k 2+k 3 2k 1 k 1!A k2 9 A k 3 10,
for k= 0, 1, 2, . . . . Using similar argument as above for Dx,t 3ψ PkuH−3/2(R2)with μ= 3/2 in (2.11) and replacing the support of the cut-off function ψwe obtain
ψukH3/2(R2)≤ c5Ak14k! + c1|c|k! k=k1 +k 2+k 3 2k 1 k 1!A k2 9 A k 3 10, (4.30)
for k= 0, 1, 2, . . . . In a similar way we have
ψvkH3/2(R2)≤ c5Ak15k! + c1|c|k! k=k 1+k2 +k 3 2k 1 k1 ! Ak 2 9A k3 10, (4.31)
for k= 0, 1, 2, . . . . Adding (4.30) with (4.31) and performing straightforward calculations as (4.19) we obtain
ψukH3/2(R2)+ ψvkH3/2(R2)≤ CAk(k!)2, k= 0, 1, 2, . . . . Remark 4.2 To obtain the estimate forψPku
H7/2(R2)andψPkvH7/2(R2)we repeat the above method with μ= 7/2.
Proposition 4.3 Suppose that
ψukH7/2(R2)≤ cAk1(k!)2 and ψvkH7/2(R2)≤ cAk2(k!)2 for k= 0, 1, 2, . . . . Then we have
sup t∈[t0−,t0+] (t1/3∂x)Pku H1(x 0−,x0+)≤ c1A k+l 3 [(k + l)!] 2, sup t∈[t0−,t0+] (t1/3∂x)Pkv H1(x 0−,x0+)≤ c1A k+l 4 [(k + l)!] 2
for k, l= 0, 1, 2, . . . where > 0 is so small that ψ ≡ 1 near I = (x0− , x0+ ) × (t0− , t0+ ).
Proof Let It0= (t0− , t0+ ) and Ix0= (x0− , x0+ ), then we have I = Ix0× It0. For any fixed t∈ Ix0, letL= t
1/3∂
x. We show that for some positive constants c and A0the following inequality holds
LlPku
Hx1(Ix0)≤ cA k+l
0 [(k + l)!]
2, ∀k, ∀l = 0, 1, 2, . . . . (4.32)
Induction over l. By the trace theorem, we have
LlPku Hx1(Ix0)≤ t l/3∂l xP ku(t ) Hx1(Ix0) ≤ (t0+ )l/3∂xlP ku H3/2(I x0×It0) ≤ (t0+ )l/3PkuH7/2(I x0×It0) ≤ (t0+ )l/3ψPkuH7/2(R2) ≤ (t0+ )l/3c1Ak1k!
≤ (t0+ )l/3c1Ak0+l(k+ l)
≤ (t0+ )l/3c1Ak0+l[(k + l)!] 2,
where we take c= (t0+ )l/3c1and A0= max{1, A1}. Hence, in the case l = 0, 1, 2, it is easy to show that (4.32) follows directly from the assumption.
Now, we assume that (4.32) is true for l≥ 2. Applying Pkto the (2.2) equation we have ∂t(Pku)+ ∂x3(Pku)= LPku = (P + 3)kLu = (P + 3)k(∂ tu+ ∂x3u) = −(P + 3)k a 2∂x(u 2)+b 2∂x(v 2)+ c∂x(uv) = −a 2(P+ 3) k∂x(u2)−b 2(P+ 3) k∂x(v2)− c(P + 3)k∂x(uv) = −a 2∂x(P+ 2) k(u2)−b 2∂x(P+ 2) k(v2)− c∂x(P+ 2)k(uv) such that t ∂t(Pku)+ t∂x3(Pku)= −a 2t ∂x(P+ 2) k(u2)−b 2t ∂x(P + 2) k(v2) − ct∂x(P+ 2)k(uv). (4.33) Moreover, P = 3t∂t+ x∂x. Then t ∂t(Pku)=1 3P k+1u−1 3x∂x(P ku). (4.34) Replacing (4.34) in (4.33) we obtain L3Pku= t∂3 x(P ku) = −1 3P k+1u+1 3x∂x(P ku) −a 2t ∂x(P+ 2) k(u2 )−b 2t ∂x(P + 2) k(v2 )− ct∂x(P+ 2)k(uv). Hence, applyingLl−2we have
Ll+1Pku Hx1(Ix0)= L l−2L3Pku Hx1(Ix0) ≤ 1 3L l−2Pk+1u Hx1(I x0)+ 1 3L l−2x∂x(Pku) Hx1(I x0) +|a| 2 tL l−2∂x(P+ 2)k(u2) Hx1(Ix0) +|b| 2 tL l−2∂x(P + 2)k(v2) Hx1(Ix0) + |c|tLl−2∂x(P+ 2)k(uv) Hx1(Ix0) = F1+ F2+ F3+ F4+ F5.
Using the induction assumption, we obtain F1≤ 1 3c1A k+l+1 14 (k+ l + 1)!. (4.35)
We estimate the termLl−2(x∂x)for l≥ 3. Let r = l − 2, then we estimateLr(x∂x)for r≥ 1. ∂xr(x∂x)= r k=0 r k ∂xr−k(x)∂xk(∂x). (4.36) Replacing ∂xr−k(x)= 1 if k= r − 1, 0 if k≤ r − 2 in (4.36), we obtain ∂xr(x∂x)= r∂xr−1(∂x)+ x∂xr(∂x)= r∂xr+ x∂x(∂xr) = (l − 2)∂(l−2) x + x∂x(∂x(l−2)), thusLl−2(x∂x)= x∂
xLl−2+ (l − 2)Ll−2for l≥ 3. For F2we have F2≤ x∂xLl−2PkuHx1(Ix0)+ (l − 2)L l−2Pku Hx1(Ix0) ≤ xt−1/3Ll−1Pku Hx1(Ix0)+ (l − 2)L l−2Pku Hx1(Ix0) ≤ c(t0− )(|x0| + + 1)Ll−1PkuHx1(Ix0)+ (l − 2)L l−2Pku Hx1(Ix0) ≤ (t0− )−1/3(|x0+ + 1)c1A14k+l−1(k+ l − 1)! + c1Ak14+l−1(l− 2)(k + l − 1)! ≤ 1 3c1A k+l+1 14 (k+ l + 1)! (4.37)
where we take A14larger than (t0− )−1/3(|x0| + + 1) and 3. Using thatL= t1/3∂x3, and then tLl−2∂x= tt(l−2)/3∂(l−2) x ∂x= tt−1/3t(l−1)/3∂x(l−1)= t2/3Ll−1,we have F3= |a| 2 t 2/3Ll−1(P+ 2)k(u2) Hx1(I x0) ≤|a| 2 (t0+ ) 2/3 l−1=l1+l2 k=k1+k2+k3 (l− 1)! l1!l2! k! k1!k2!k3! 2k3 × c2Ll1Pk1uHx1(Ix0)L l2Pk2u Hx1(Ix0). Using the induction assumption
F3≤ |a| 2 (t0+ ) 2/3 l−1=l1+l2 k=k1+k2+k3 c2c13k!(l − 1)! 2k3 k3! ×(l1+ k1)! l1!k1! (l2+ k2)! l2!k2! Ak14+l−1 ≤|a| 2 (t0+ ) 2/3c 2c31(l+ k − 1)!A k+l−1 14 l−1=l1+l2 k=k1+k2+k3 2k3 k3!
