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The Cauchy problem for the modified Novikov equation

Abstract

In this paper, we are concerned with the Cauchy problem for the modified Novikov equation. By using the transport equation theory and Littlewood-Paley decomposition as well as nonhomogeneous Besov spaces, we prove that the Cauchy problem for the modified Novikov equation is locally well posed in the Besov space B p , r s with 1p,r+ and s>max{1+ 1 p , 3 2 } and show that the Cauchy problem for the modified Novikov equation is locally well posed in the Besov space B 2 , 1 3 / 2 with the aid of Osgood lemma.

MSC: 35G25, 35L05, 35R25.

1 Introduction

Recently, Zhao and Zhou [1] considered the exact traveling wave solution to the following modified Novikov equation:

u t u t x x +4 u 4 u x =3u u x u x x + u 2 u x x x .
(1.1)

We recall that the Novikov equation

u t u t x x = u 2 u x x x +3u u x u x x 4 u 2 u x
(1.2)

was discovered by Vladimir Novikov [2] and it possesses the bi-Hamiltonian structure, infinite conservation laws. The well-posedness and blow-up of the Cauchy problem for the Novikov equation in Sobolev spaces and Besov spaces have been investigated by some authors [3]–[7]. The weak solution of the Cauchy problem for the Novikov equation has been investigated by some authors [4], [5], [8]. Recently, Li and Yan [9] considered the Cauchy problem for the KdV equation with higher dispersion.

We define P 1 (D)= x ( 1 x 2 ) 1 and P 2 (D)= ( 1 x 2 ) 1 . By using the fact that G(x)= 1 2 e | x | and G(x)f= ( 1 x 2 ) 1 f for all f L 2 (R) and Gy=u, we can rewrite (1.1) as follows:

u t + u 2 u x + P 1 (D) [ 4 5 u 5 1 3 u 3 + 3 2 u u x 2 ] + P 2 (D) [ 1 2 u x 3 ] =0,t>0.

Now we consider the following problem:

u t + u 2 u x + P 1 (D) [ 4 5 u 5 1 3 u 3 + 3 2 u u x 2 ] + P 2 (D) [ 1 2 u x 3 ] =0,t>0,
(1.3)
u(x,0)= u 0 (x),xR.
(1.4)

To the best of our knowledge, the well-posedness and blow-up of the Cauchy problem for (1.3) and (1.4) in Besov spaces are open up to now. More precisely, in this paper, motivated by [10], [11], using Littlewood-Paley decomposition and nonhomogeneous Besov spaces, we prove that the Cauchy problem for (1.4) is locally well posed in the Besov space B p , r s with s>max{1+ 1 p , 3 2 } and we give a blow-up criterion.

To introduce the main results, we define

E p , r s ( T ) = C ( [ 0 , T ] ; B p , r s ) C 1 ( [ 0 , T ] ; B p , r s 1 ) if  r < , E p , s ( T ) = L ( 0 , T ; B p , s ) Lip ( [ 0 , T ] ; B p , r s 1 ) .

The main results of this paper are as follows.

Theorem 1.1

Let1p,rands>max( 3 2 ,1+ 1 p )and u 0 B p , r s . Then there exists a timeT>0such that problem (1.3) and (1.4) has a unique solutionuin E p , r s (T). The map u 0 uis continuous from a neighborhood of u 0 in B p , r s intoC([0,T]; B p , r s ) C 1 ([0,T]; B p , r s 1 )for every s <s. Whenr<, the solution to problem (1.3) and (1.4) is continuous in E p , r s (T).

Theorem 1.2

When u 0 B 2 , 1 3 / 2 , (1.3) and (1.4) is locally well posed in the sense of Hadamard.

The remainder of this paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we establish local well-posedness of the Cauchy problem for the generalized Camassa-Holm equation in Besov spaces. In Section 4, we prove Theorem 1.2.

2 Preliminaries

In this section, the nonhomogeneous Besov spaces and the theory of transport equation which can be seen in [10]–[13] are presented.

Lemma 2.1

(Littlewood-Paley decomposition)

There exists a couple of smooth radial functions(χ,ϕ)valued in[0,1]such thatχis supported in the ballB={ξ R n ,|ξ| 4 3 }andϕis supported in the ringC={ξ R n , 3 4 |ξ| 8 3 }. Moreover,

and

Supp ϕ ( 2 q ) Supp ϕ ( 2 q ) = if  | q q | 2 , Supp χ ( ) Supp ϕ ( 2 q ) = if  | q | 1 .

Then, foru S (R), the nonhomogeneous dyadic blocks are defined as follows:

Δ q u = 0 if  q 2 , Δ 1 u = χ ( D ) u = F x 1 χ F x u , Δ q u = ϕ ( 2 q D ) = F x 1 ϕ ( 2 q ξ ) F x u if  q 0 .

Thusu= q Z Δ q uin S (R).

Remark

The low frequency cut-off S q is defined by

S q u= p = 1 q 1 Δu=χ ( 2 q D ) u= F x 1 χ ( 2 q ξ ) F x u,qN.

It is easily checked that

Δ p Δ q u 0 if  | p q | 2 , Δ q ( S p 1 u Δ p v ) 0 if  | p q | 5 , u , v S ( R )

as well as

Δ q u L p u L p , S q u L p C u L p ,1p+

with the aid of Young’s inequality, where C is a positive constant independent of q.

Definition

(Besov spaces)

Let sR, 1p+. The nonhomogeneous Besov space B p , r s ( R n ) is defined by

B p , r s ( R n ) = { f S ( R ) : f B p , r s = 2 q s Δ q f l r ( L p ) = ( 2 q s Δ q f L p ) q 1 l r < } .

In particular, B p , r = s R B p , r s . Let T>0, sR and 1p. Define E p , r s = T > 0 E p , r s (T).

Lemma 2.2

LetsR, 1p,r, p j , r j , j=1,2, then:

  1. (1)

    Topological properties: B p , r s is a Banach space which is continuously embedded in S (R).

  2. (2)

    Density: C c is dense in B p , r s 1p,r<.

  3. (3)

    Embedding: B p 1 , r 1 s B p 2 , r 2 s n ( 1 p 1 1 p 2 ) if p 1 p 2 and r 1 r 2 .

    B p , r 2 s 2 B p , r 1 s 1  locally compactif  s 1 < s 2 .
  1. (4)

    Algebraic properties: s>0, B p , r s L is a Banach algebra. B p , r s is a Banach algebra  B p , r s L s> 1 p or (s 1 p and r=1). In particular, B 2 , 1 1 / 2 is continuously embedded in B 2 , 1 / 2 L and B 2 , 1 / 2 L is a Banach algebra.

  2. (5)

    1-D Moser-type estimates:

  3. (i)

    For s>0,

    f g B p , r s C ( f B p , r s g L + f L g B p , r s ) .
  1. (ii)

    s 1 1 p < s 2 ( s 2 1 p if r=1) and s 1 + s 2 >0, we have

    f g B p , r s 1 C f B p , r s 1 g B p , r s 2 .
  1. (6)

    Complex interpolation:

    f B p , r θ s 1 + ( 1 θ ) s 2 f B p , r s 1 θ g B p , r s 2 1 θ ,f B p , r s 1 B p , r s 2 ,θ[0,1].
  1. (7)

    Real interpolation: θ(0,1), s 1 > s 2 , s=θ s 1 +(1θ) s 2 , there exists a constant C such that

    u B p , 1 s C ( θ ) s 1 s 2 u B p , s 1 θ u B p , s 2 1 θ ,u B p , s 1 .

In particular, for any0<θ<1, we have that

u B 2 , 1 1 / 2 u B 2 , 1 3 2 θ C(θ) u B 2 , 1 / 2 θ u B 2 , 3 / 2 1 θ .
(2.1)
  1. (8)

    Fatou lemma: if is bounded in B p , r s and u n u in S (R), then u B p , r s and

    u B p , r s lim inf n u n B p , r s .
  1. (9)

    Let mR and f be an S m -multiplier (i.e., f: R n R is smooth and satisfies that α N n , a constant C α , s.t. | α f(ξ)| C α ( 1 + | ξ | ) m | α | for all ξ R n ). Then the operator f(D) is continuous from B p , r s to B p , r s m . Notice that P 1 (D) is continuous from B p , r s to B p , r s 1 and P 2 (D) is continuous from B p , r s to B p , r s 2 .

  2. (10)

    The usual product is continuous from B 2 , 1 1 / 2 ×( B 2 , 1 / 2 L ) to B 2 , 1 / 2 .

  3. (11)

    There exists a constant C>0 such that the following interpolation inequality holds:

    f B 2 , 1 1 / 2 C f B 2 , 1 / 2 ln ( e + f B 2 , 3 / 2 f B 2 , 1 / 2 ) .

Lemma 2.3

(A priori estimates in Besov spaces)

Let1p,rands>min( 1 p ,1 1 p ). Assume that f 0 B p , r s , F L 1 (0,T; B p , r s )and x vbelongs to L 1 (0,T; B p , r s 1 )ifs>1+ 1 p or to L 1 (0,T; B p , r 1 / p L )otherwise. Iff L (0,T; B p , r s )C([0,T]; S (R))solves the following 1-D linear transport equation:

f t +v f x =F,
(2.2)
f(x,0)= f 0 ,
(2.3)

then there exists a constantCdepending only ons, p, rsuch that the following statements hold:

  1. (1)

    If r=1 or s1+ 1 p , then

    f B p , r s f 0 B p , r s + 0 t F ( τ ) B p , r s dτ+C 0 t V (τ) f ( τ ) B p , r s dτ

or hence,

f B p , r s e C V ( t ) ( f 0 B p , r s + 0 t e C V ( τ ) F ( τ ) B p , r s d τ )
(2.4)

withV(t)= 0 t v x ( τ ) B p , r 1 / p L dτifs<1+ 1 p andV(t)= 0 t v x ( τ ) B p , r s 1 dτelse.

  1. (2)

    If s1+ 1 p , f 0 L and f x L ((0,T)×R) and F x L 1 (0,T; L ), then

    f ( t ) B p , r s + f x ( t ) L e C V ( t ) ( f 0 B p , r s + f 0 x L + 0 t e C V ( τ ) [ F ( τ ) B p , r s + F x ( τ ) L ] d τ )

with

V(t)= 0 t x v ( τ ) B p , r 1 / p L .
  1. (3)

    If f=v, then for all s>0, (1) holds true when V(t)= 0 t v x ( τ ) L dτ.

  2. (4)

    If r<, then fC([0,T]; B p , r s ). If r=, then fC([0,T]; B p , 1 s ) for all s <s.

Lemma 2.4

(Existence and uniqueness)

Letp, r, s, f 0 andFbe as in the statement of Lemma  2.3. Assume thatv L ρ (0,T; B , M )for someρ>1andM>0and v x L 1 (0,T; B p , r s 1 )ifs>1+ 1 p ors=1+ 1 p andr=1and v x L 1 (0,T; B p , 1 / p L )ifs<1+ 1 p . Then problem (2.2) and (2.3) has a unique solutionf L (0,T; B p , r s )( s < s C([0,T]; B p , 1 s ))and the inequalities of Lemma  2.3can hold true. Moreover, ifr<, thenfC([0,T]; B p , r s ).

3 Proof of Theorem 1.1

By using the following six steps, we will complete the proof of Theorem 1.1.

First step: Approximate solution. We will construct a solution with the aid of a standard iterative process. Starting from u ( 0 ) :=0, by the inductive method and solving the following linear transport equation (3.1) and (3.2), we derive a sequence of smooth functions

[ t + ( u ( n ) ) 2 x ] u ( n + 1 ) = P 1 ( D ) [ 4 5 ( u ( n ) ) 5 1 3 ( u ( n ) ) 3 + 3 2 u ( n ) ( u x ( n ) ) 2 ] [ t + ( u ( n ) ) 2 x ] u ( n + 1 ) = P 2 ( D ) [ 1 2 ( u x ( n ) ) 3 ] ,
(3.1)
u ( n + 1 ) (x,0)= u 0 ( n + 1 ) = S n + 1 u 0 .
(3.2)

It is easily checked that S n + 1 u 0 B p , r , by using Lemma 2.4 and the inductive method, for all nN, we have that (3.1) and (3.2) has a global solution which belongs to C( R + , B p , r ).

Second step: Uniform bounds. We will prove

u ( n + 1 ) ( t ) B p , r s e C 0 t u ( n ) B p , r s 2 d τ u 0 B p , r s + C 2 0 t e C τ t u ( n ) B p , r s 2 d τ u ( n ) B p , r s 3 d τ + C 2 0 t e C τ t u ( n ) B p , r s 2 d τ u ( n ) B p , r s 5 d τ
(3.3)

for all nN.

Combining (2.4) of Lemma 2.3 with (3.1), we have

u ( n + 1 ) ( t ) B p , r s e C 0 t ( ( u ( n ) ) 2 ) ( t ) B p , r s d t u 0 B p , r s + 0 t e C τ t ( ( u ( n ) ) 2 ) ( t ) B p , r s d t F ( u ( n ) , u x ( n ) ) B p , r s d τ ,
(3.4)

where

F ( u n , u x n ) = P 1 (D) [ 4 5 ( u ( n ) ) 5 1 3 ( u ( n ) ) 3 + 3 2 u ( n ) ( u x ( n ) ) 2 ] + P 2 (D) [ 1 2 ( u x ( n ) ) 3 ] .
(3.5)

When s>max{1+ 1 p , 3 2 }, by using (4) in Lemma 2.2, we have

( ( u ( n ) ) 2 ) ( t ) B p , r s C ( u ( n ) ) ( t ) B p , r s 2 ,
(3.6)
F ( u ( n ) , u x ( n ) ) B p , r s C [ u ( n ) B p , r s 3 + u ( n ) B p , r s 5 ] .
(3.7)

Combining (3.6)-(3.7) with (3.4), we have (3.3).

Let T>0 satisfy

T<min { 1 16 C u 0 B p , r s 2 , 1 C } ,
(3.8)
u ( n ) ( t ) B p , r s 2 u 0 B p , r s ( 1 16 C u 0 B p , r s 2 t ) 1 / 2 .
(3.9)

By using (3.9), we have

e C τ t u ( n ) B p , r s 2 ( t ) d t e 1 4 τ t d ( 1 16 C u 0 B p , r s 2 t ) 1 16 C u 0 B p , r s 2 t = ( 1 16 C u 0 B p , r s 2 τ 1 16 C u 0 B p , r s 2 t ) 1 4 .
(3.10)

When τ=0 in (3.10), we have

e C U n ( t ) ( 1 1 16 C u 0 B p , r s 2 t ) 1 / 4 .
(3.11)

By using (3.10) and (3.9), we have

C 2 0 t e C τ t u ( n ) ( t ) B p , r s 2 d t u ( n ) ( τ ) B p , r s 5 d τ ( 1 16 C u 0 B p , r s 2 t ) 1 4 0 t 16 C u 0 B p , r s 5 ( 1 16 C u 0 B p , r s 2 τ ) 9 4 d τ = 4 5 u 0 B p , r s 3 ( 1 16 C u 0 B p , r s 2 t ) 1 4 [ ( 1 16 C u 0 B p , r s 2 t ) 5 4 1 ] .
(3.12)

With the aid of the mean value theorem, we have

[ ( 1 16 C u 0 B p , r s 2 t ) 5 4 1 ] =20C u 0 B p , r s 2 t ( 1 ξ ) 9 / 4 ,
(3.13)

where

16C u 0 B p , r s 2 t<ξ<1.

Combining (3.13) with (3.12), we have that

C 2 0 t e C τ t u ( n ) ( t ) B p , r s 2 d t u ( n ) ( τ ) B p , r s 5 dτ 16 u 0 B p , r s 5 C t ( 1 16 C u 0 B p , r s 2 t ) 5 / 2 .
(3.14)

Inserting (3.10)-(3.14) into (3.3) leads to

u ( n + 1 ) ( t ) B p , r s 2 u 0 B p , r s ( 1 16 C u 0 B p , r s 2 t ) 1 / 2 .
(3.15)

Consequently, ( u ( n ) ) n N is uniformly bounded in C([0,T]; B p , r s ). By using the fact that B p , r s 1 with s>max{1+ 1 p , 3 2 } is an algebra and B p , r s B p , r s 1 as well as the definition of the Besov spaces B p , r s , we derive that

( u ( n ) ) 2 u x ( n + 1 ) B p , r s 1 C ( u ( n ) ) 2 B p , r s 1 u x ( n + 1 ) B p , r s 1 C u ( n ) B p , r s 2 u ( n + 1 ) B p , r s 8 C u 0 B p , r s 3 ( 1 16 C u 0 B p , r s 2 t ) 3 2 .
(3.16)

Since s>max{1+ 1 p , 3 2 }, which leads to that B p , r s 1 is an algebra, by using the S 1 -multiplier property of P 1 (D) and the S 2 -multiplier property of P 2 (D) as well as (3.8), we have

F ( u ( n ) , u x ( n ) ) B p , r s 1 C 2 [ u ( n ) B p , r s 3 + u ( n ) B p , r s 5 ] 4 C [ u 0 B p , r s 3 ( 1 16 C u 0 B p , r s 2 t ) 3 2 + 4 u 0 B p , r s 5 ( 1 16 C u 0 B p , r s 2 t ) 5 2 ] .
(3.17)

Consequently, combining (3.1) with (3.16) and (3.17), we derive that

u t ( n + 1 ) B p , r s 1 4C [ 3 u 0 B p , r s 3 ( 1 16 C u 0 B p , r s 2 t ) 3 2 + 4 u 0 B p , r s 5 ( 1 16 C u 0 B p , r s 2 t ) 5 2 ] ,
(3.18)

which yields ( u ( n ) ) n C([0,T]; B p , r s ) C 1 ([0,T]; B p , r s 1 ).

Third step: Convergence. We will derive that ( u ( n ) ) n is a Cauchy sequence in C([0,T]; B p , r s 1 ).

For m,nN, from (3.1), we have

( u ( n + m + 1 ) u ( n + 1 ) ) t + ( u ( n + m ) ) 2 ( u ( n + m + 1 ) u ( n + 1 ) ) x = k = 1 5 T k ,
(3.19)

where

T 1 = 4 5 P 1 ( D ) [ ( u ( n + m ) ) 5 ( u ( n ) ) 5 ] , T 2 = 1 3 P 1 ( D ) [ ( u ( n + m ) ) 3 ( u ( n ) ) 3 ] , T 3 = 3 2 P 1 ( D ) [ ( u n + m u n ) ( u x ( n + m ) ) 2 ] , T 4 = 3 2 P 1 ( D ) [ ( u x ( n + m ) ) 2 ( u x ( n ) ) 2 ] , T 5 = 1 2 P 2 ( D ) [ ( u ( n + m ) ) x 3 ( u ( n ) ) x 3 ] , T 6 = [ ( u ( n + m ) ) 2 ( u ( n ) ) 2 ] u x ( n + 1 ) .

When s>max{1+ 1 p , 3 2 }, by using the S 1 multiplier property of P 1 (D), the S 2 multiplier property of P 2 (D) and B p , r s 1 B p , r s 2 , we have

T j B p , r s 1 C u ( n + m ) u ( n ) B p , r s 1 ,
(3.20)

where 1j6, jN. Since nN, we have

u ( n ) B p , r s 2 u 0 B p , r s ( 1 16 C u 0 B p , r s 2 t ) 1 2 .

By using (3.20), we have

j = 1 6 T j B p , r s 1 j = 1 6 T j B p , r s 1 C u ( n + m ) u ( n ) B p , r s 1 4 u 0 B p , r s 2 1 16 C u 0 B p , r s 2 t .
(3.21)

When s2+ 1 p , from Lemma 2.4 and (3.20), we have

( u ( n + m + 1 ) u ( n + 1 ) ( t ) B p , r s 1 e C W ( n + m ) ( t ) ( u ( n + m + 1 ) u ( n + 1 ) ) ( , 0 ) B p , r s 1 + C 0 t e C W ( n + m ) ( t ) C W ( n + m ) ( τ ) j = 1 6 T j B p , r s 1 d τ ,
(3.22)

where

W n + m (t)= 0 t x ( u ( n + m ) ) 2 ( τ ) B p , r 1 p L dτ
(3.23)

if s1<1+ 1 p and

W n + m (t)= 0 t x ( u ( n + m ) ) 2 ( τ ) B p , r s 2 dτ
(3.24)

if s1>1+ 1 p . From (3.23), if s1<1+ 1 p , by using B p , r s 1 L with s>1+ 1 p , we have

W n + m (t)C 0 t u ( n + m ) ( τ ) B p , r s 2 dτ.
(3.25)

From (3.25), if s1>1+ 1 p , we have

W n + m (t)C 0 t u ( n + m ) ( τ ) B p , r s 2 dτ.
(3.26)

It is easily showed that

q = n + 1 n + m Δ q u 0 B p , r s 1 C 2 n u 0 B p , r s 1 .
(3.27)

Inserting (3.25)-(3.27) into (3.22), we have

( u ( n + m + 1 ) u ( n + 1 ) ) ( t ) B p , r s 1 C T ( 2 n + 0 t ( u ( n + m ) u ( n ) ) ( τ ) B p , r s 1 d τ ) .
(3.28)

We define

W n , k (t)= ( u ( n + m ) u ( n ) ) ( t ) B p , r s 1 ,
(3.29)
W n (t)= sup m N W n , m (t),
(3.30)
W ˜ (t)= lim sup n W n (t).
(3.31)

Combining (3.28) with (3.29)-(3.30), we have that

W n + 1 (t)C 0 t W n (τ)dτ.
(3.32)

From (3.31) and (3.32), by using the Fatou lemma, we have that

W ˜ (t)C 0 t W ˜ (τ)dτ.
(3.33)

Applying the Gronwall inequality to (3.33), we have that

W ˜ (t) e C W ˜ (0).
(3.34)

From (3.31), we have that W ˜ (0)=0. Thus, W ˜ (t)=0. Consequently, ( u ( n ) ) n is a Cauchy sequence in C([0,T]; B p , r s 1 ); moreover, ( u ( n ) ) n is convergent to some limit function uC([0,T]; B p , r s 1 ).

When s=2+ 1 p , by using (6) of Lemma 2.2, we derive that

( u ( n + m + 1 ) u ( n + 1 ) ) ( t ) L T B p , r s 1 = ( u ( n + m + 1 ) u ( n + 1 ) ) ( t ) L T B p , r 1 + 1 p ( u ( n + m + 1 ) u ( n + 1 ) ) ( t ) L T B p , r s 1 θ ( u ( n + m + 1 ) u ( n + 1 ) ) ( t ) L T B p , r s 2 1 θ ( u ( n + m + 1 ) u ( n + 1 ) ) ( t ) B p , r 1 + 1 p θ [ u ( n + m + 1 ) B p , r 2 + 1 p + u ( n + 1 ) ( t ) B p , r 2 + 1 p ] 1 θ ( C T ) θ 2 θ n [ u ( n + m + 1 ) B p , r 2 + 1 p + u ( n + 1 ) ( t ) B p , r 2 + 1 p ] 1 θ ,
(3.35)

where

s 1 ( max ( 1 + 1 p , 3 2 ) 1 , 1 + 1 p ) , s 2 ( 1 + 1 p , 2 + 1 p ) .

Consequently, ( u ( n ) ) n is a Cauchy sequence in C([0,T]; B p , r s 1 ) and ( u ( n ) ) n converges to some limit function uC([0,T]; B p , r s 1 ).

Fourth step: Existence of solution in E p , r s (T). Existence of solution E p , r s (T) can be proved similarly to [11].

Fifth step: Uniqueness of solution. We consider case s2+ 1 p and case s=2+ 1 p , respectively. In fact, this can be proved similarly to [14].

Sixth step: Continuity with respect to the initial data. Continuity with respect to the initial data can be proved similarly to [6].

4 Proof of Theorem 1.2

Since B 2 , 1 3 / 2 and B 2 , 1 1 / 2 are Banach algebras, by using a proof similar to (3.3), we can prove that

u ( n + 1 ) ( t ) B 2 , 1 3 / 2 e C 0 t u ( n ) B 2 , 1 3 / 2 2 d τ u 0 B 2 , 1 3 / 2 + C 2 0 t e C τ t u ( n ) B 2 , 1 3 / 2 2 d τ u ( n ) B 2 , 1 3 / 2 3 d τ + C 2 0 t e C τ t u ( n ) B 2 , 1 3 / 2 2 d τ u ( n ) B 2 , 1 3 / 2 5 d τ
(4.1)

for all . We assume that

T<min { 1 16 C u 0 B 2 , 1 3 / 2 2 , 1 C } ,
(4.2)
u ( n ) ( t ) B 2 , 1 3 / 2 2 u 0 B 2 , 1 3 / 2 ( 1 16 C u 0 B 2 , 1 3 / 2 2 t ) 1 / 2 .
(4.3)

By a proof similar to (3.15), we can prove that

u ( n + 1 ) ( t ) B 2 , 1 3 / 2 2 u 0 B 2 , 1 3 / 2 ( 1 16 C u 0 B 2 , 1 3 / 2 2 t ) 1 / 2 .
(4.4)

Thus, ( u ( n ) ) n is uniformly bounded in B 2 , 1 3 / 2 . From (3.1), by using (4) in Lemma 2.2, we can prove that ( u t ( n ) ) n is uniformly bounded with respect to n in B 2 , 1 1 / 2 .

Consequently, ( u ( n ) ) n C([0,T]; B 2 , 1 3 / 2 ) C 1 ([0,T]; B 2 , 1 1 / 2 ).

We define

ρ n , k (t)= ( u ( n + m ) u ( n ) ) ( t ) B 2 , 1 / 2 ,
(4.5)
(4.7)
ρ ˜ (t)= lim sup n ρ n (t).
(4.7)

By a proof similar to [15], we derive that

ρ ˜ (t)=0,
(4.8)

and with the aid of (2.1), we can prove that ( u ( n ) ) n is a Cauchy sequence in C([0,T]; B 2 , 1 1 / 2 ).

The rest of Theorem 1.2 can be proved similarly to [13], [15].

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Acknowledgements

We would like to thank the reviewers for careful reading and valuable comments on the original draft.

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Correspondence to Xueping Hou.

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Hou, X., Zheng, Y. The Cauchy problem for the modified Novikov equation. Bound Value Probl 2014, 171 (2014). https://doi.org/10.1186/s13661-014-0171-x

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Keywords

  • Cauchy problem
  • modified Novikov equation