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The global solution and blow-up phenomena to a modified Novikov equation

Abstract

A modified Novikov equation with symmetric coefficients is investigated. Provided that the initial value u 0 ∈ H s (R) (s> 3 2 ), (1− ∂ x 2 ) u 0 does not change sign and the solution u itself belongs to L 1 (R), the existence and uniqueness of the global strong solutions to the equation are established in the space C([0,∞); H s (R))∩ C 1 ([0,∞); H s − 1 (R)). A blow-up result to the development of singularities in finite time for the equation is acquired.

MSC:35G25, 35L05.

1 Introduction

Many scholars have paid attention to the integrable equation

u t − u t x x +4 u 2 u x =3u u x u x x + u 2 u x x x ,
(1)

which was derived by Novikov [1]. Well-posedness of the Novikov equation in the Sobolev spaces on the torus was first done by Tiglay in [2], and was completed on both the line and the circle by Himonas and Holliman in [3]. Its Hölder continuity properties were studied in Himonas and Holmes [4]. The periodic and the non-periodic Cauchy problem for Eq. (1) and continuity results for the data-to-solution map in the Sobolev spaces are discussed in Grayshan [5]. A matrix Lax pair for Eq. (1) is acquired in [6] and is shown to be related to a negative flow in the Sawada-Kotera hierarchy. The scattering theory is applied to find non-smooth explicit soliton solutions with multiple peaks for Eq. (1) in [7]. Sufficient conditions on the initial data to guarantee the formation of singularities in finite time for Eq. (1) are given in Jiang and Ni [8]. This multiple peak property is common with the Camassa-Holm and Degasperis-Procesi equations [9–11]. Mi and Mu [12] established many dynamic results for a modified Novikov equation with peak solution. It is shown in Ni and Zhou [13] that the Novikov equation associated with initial value has locally well-posedness in a Sobolev space H s with s> 3 2 by using the abstract Kato theorem. Two results about the persistence properties of the strong solution for Eq. (1) are established in [13]. Using the Littlewood-Paley decomposition and nonhomogeneous Besov spaces, Yan et al. [14] proved the global existence and blow-up phenomena for the weakly dissipative Novikov equation. For other methods to handle the Novikov equation and the related partial differential equations, the reader is referred to [15–22] and the references therein.

Observing the coefficients of the Novikov equation (1), we see that the coefficient of u 2 u x is equal to the coefficient of u u x u x x plus the coefficient of u 2 u x x x . That is,

4=3+1.

Indeed, this relationship among the coefficients plays important roles in the study of the essential dynamical properties of the Novikov model [1, 2, 11–13]. This motivates us to study the following equation:

u t − u t x x +(a+b) u 2 u x =au u x u x x +b u 2 u x x x ,
(2)

where a>0 and b>0 are arbitrary constants. Clearly, letting a=3 and b=1, Eq. (2) becomes the Novikov equation (1). The essential difference between Eq. (2) and the Novikov equation (1) is that Eq. (2) does not conform with the following conservation law:

∫ R ( u 2 + u x 2 ) dx= ∫ R ( u 0 2 + u 0 x 2 ) dx,

which results in the bounds of ∥ u ( t , ⋅ ) ∥ L ∞ ( R ) for Eq. (1).

Making use of u 0 ∈ H s (R), s> 3 2 , the assumption that (1− ∂ x 2 ) u 0 does not change sign, and the assumption that the solution of Eq. (2) satisfies u∈ L 1 (R), we prove the global existence theorem of Eq. (2) in the Sobolev space,

u(t,x)∈C ( [ 0 , ∞ ) ; H s ( R ) ) ∩ C 1 ( [ 0 , ∞ ) ; H s − 1 ( R ) ) .

The objective of this work is to investigate Eq. (2). Since a>0 and b>0 are arbitrary constants, we cannot obtain the boundedness of the solution u for Eq. (2) although the initial data satisfy the sign condition. To overcome this, assuming that the solution itself satisfies u∈ L 1 (R) and the initial data satisfy the sign condition, we adopt the methods used in Rodriguez-Blanco [16] to derive that ∥ ∂ u ( t , x ) ∂ x ∥ L ∞ ( R ) possesses bounds for any time t>0. This leads us to establish the well-posedness of the global strong solutions to Eq. (2). Parts of the main results in [17, 18] are extended. In addition, we acquire a blow-up result to the development of singularities in finite time, which includes the blow-up result in [12].

The rest of this paper is organized as follows. Section 2 states the main results of this work. Section 3 proves the global existence result. The proof of a blow-up result is given in Section 4.

2 Main results

We let L p = L p (R) (1≤p<+∞) be the space of all measurable functions h such that ∥ h ∥ L p p = ∫ R | h ( t , x ) | p dx<∞. We define L ∞ = L ∞ (R) with the standard norm ∥ h ∥ L ∞ = inf m ( e ) = 0 sup x ∈ R ∖ e |h(t,x)|. For any real number s, we let H s = H s (R) denote the Sobolev space with the norm defined by

∥ h ∥ H s = ( ∫ R ( 1 + | ξ | 2 ) s | h ˆ ( t , ξ ) | 2 d ξ ) 1 2 <∞,

where h ˆ (t,ξ)= ∫ R e − i x ξ h(t,x)dx. Here we note that the norms ∥ ⋅ ∥ L p p , ∥ ⋅ ∥ L ∞ and ∥ ⋅ ∥ H s depend on variable t.

For T>0 and nonnegative number s, C([0,T); H s (R)) denotes the Frechet space of all continuous H s -valued functions on [0,T). We set Λ= ( 1 − ∂ x 2 ) 1 2 .

In order to study the existence of solutions for Eq. (2), we consider its Cauchy problem in the form

{ u t − u t x x + ( a + b ) u 2 u x = a u u x u x x + b u 2 u x x x , u ( 0 , x ) = u 0 ( x ) ,
(3)

which is equivalent to

{ u t + b u 2 u x = Λ − 2 [ ( − a u 2 u x + a − 6 b 2 ( u u x 2 ) x + 2 b − a 2 u x 3 ] , u ( 0 , x ) = u 0 ( x ) ,
(4)

where a>0 and b>0 are arbitrary constants. Now we give the main results for problem (3).

Theorem 1 Assume that the solution of problem (3) satisfies u(t,x)∈ L 1 (R) and let u 0 (x)∈ H s , s> 3 2 and (1− ∂ x 2 ) u 0 ≥0 for all x∈R (or equivalently (1− ∂ x 2 ) u 0 ≤0 for all x∈R). Then problem (3) has a unique solution satisfying

u(t,x)∈C ( [ 0 , ∞ ) ; H s ( R ) ) ∩ C 1 ( [ 0 , ∞ ) ; H s − 1 ( R ) ) .

Theorem 2 Assume that u 0 (x)∈ H s (R) with s> 3 2 . If a=b, then every solution of problem (3) exists globally in time. If a>b, then the solution blows up in finite time if and only if u u x becomes unbounded from below in finite time. If a<b, then the solution blows up in finite time if and only if u u x becomes unbounded from above in finite time.

3 Global strong solutions

For proving the global existence for problem (3), we cite the local well-posedness result presented in [18].

Lemma 3.1 ([18])

Let u 0 (x)∈ H s (R) with s> 3 2 . Then the Cauchy problem (3) has a unique solution u(t,x)∈C([0,T); H s (R))∩ C 1 ([0,T); H s − 1 (R)) where T>0 depends on ∥ u 0 ∥ H s ( R ) .

Assume u 0 ∈ H s (R) with s> 3 2 . Then there exists a unique solution u(t,x) to problem (3) and

u(t,x)∈C ( [ 0 , T ) ; H s ( R ) ) ∩ C 1 ( [ 0 , T ) ; H s − 1 ( R ) )

with the maximal existence time T>0. First, we study the differential equation

{ p t = b u 2 ( t , p ) , t ∈ [ 0 , T ) , p ( 0 , x ) = x .
(5)

Lemma 3.2 Let u 0 ∈ H s , s>3 and let T>0 be the maximal existence time of the solution to problem (3). Then problem (5) has a unique solution p∈ C 1 ([0,T)×R,R). Moreover, the map p(t,⋅) is an increasing diffeomorphism of R with p x (t,x)>0 for (t,x)∈[0,T)×R.

Proof From Lemma 3.1, we have u∈ C 1 ([0,T); H s − 1 (R)) and H s − 1 ∈ C 1 (R). Thus we conclude that both functions u(t,x) and u x (t,x) are bounded, Lipschitz in space and C 1 in time. Using the existence and uniqueness theorem of ordinary differential equations derives that problem (5) has a unique solution p∈ C 1 ([0,T)×R,R).

Differentiating Eq. (5) with respect to x yields

{ d d t p x = 2 b u u x ( t , p ) p x , t ∈ [ 0 , T ) , b ≠ 0 , p x ( 0 , x ) = 1 ,
(6)

which leads to

p x (t,x)=exp ( ∫ 0 t 2 b u u x ( τ , p ( τ , x ) ) d τ ) .
(7)

For every T ′ <T, using the Sobolev imbedding theorem yields

sup ( τ , x ) ∈ [ 0 , T ′ ) × R | u u x ( τ , x ) | <∞.

It is inferred that there exists a constant K 0 >0 such that p x (t,x)≥ e − K 0 t for (t,x)∈[0,T)×R. It completes the proof. □

Lemma 3.3 Let u 0 ∈ H s with s>3, and let T>0 be the maximal existence time of the problem (3). We have

y ( t , p ( t , x ) ) p x 2 (t,x)= y 0 (x) e − ( a − 4 b ) ∫ 0 t u u x d τ ,
(8)

where (t,x)∈[0,T)×R and y:=u− u x x .

Proof Using Eqs. (2) and (6)-(8), we have

d d t [ y ( t , p ( t , x ) ) p x 2 ( t , x ) ] = y t p x 2 + 2 y p x p x t + y x p t p x 2 = y t p x 2 + 4 b y u u x p x 2 + b u 2 y x p x 2 = ( u t − u t x x + a u u x ( u − u x x ) + b u 2 ( u x − u x x x ) ) p x 2 − a u u x y p x 2 + 4 b u u x y p x 2 = ( u t − u t x x + ( a + b ) u 2 u x − a u u x u x x − b u 2 u x x x ) p x 2 − ( a − 4 b ) u u x y p x 2 = − ( a − 4 b ) u u x y p x 2 .
(9)

Using p x (0,x)=1 and solving the above equation, we complete the proof of the lemma. □

Remark 1 From Lemma 3.3, we conclude that, if u 0 − u 0 x x =(1− ∂ x 2 ) u 0 ≥0, then (1− ∂ x 2 )u(t,x)≥0. Since the operator ( 1 − ∂ x 2 ) − 1 preserves positivity, we get u≥0. Similarly, if (1− ∂ x 2 ) u 0 ≤0, we have (1− ∂ x 2 )u≤0 and u≤0.

Lemma 3.4 If u 0 ∈ H s , s> 3 2 , such that (1− ∂ x 2 ) u 0 ≥0 (or (1− ∂ x 2 ) u 0 ≤0) and ∫ R |u|dx<∞, then there exists a constant K>0 such that the solution of problem (3) satisfies ∥ u x ∥ L ∞ ≤K.

Proof We will prove this lemma to assume u 0 ∈ H ∞ which results in u∈ H ∞ from Lemma 3.1. For (1− ∂ x 2 ) u 0 ≥0, from Lemma 3.3, we have (1− ∂ x 2 )u≥0. Then u≥0 does not change sign. From the assumption ∫ R |u|dx<∞ one derives

− u x + ∫ − ∞ x udx= ∫ − ∞ x (u− u x x )dx≤ ∫ − ∞ ∞ (u− u x x )dx=c,
(10)

where c is a positive constant. Then

− u x ≤c− ∫ − ∞ x udx≤c+ ∫ − ∞ x udx≤2c.
(11)

On the other hand, we have

u x + ∫ x ∞ udx= ∫ x ∞ (u− u x x )dx≤ ∫ − ∞ ∞ (u− u x x )dx=c,
(12)

which results in

u x ≤c− ∫ x ∞ udx≤c+ ∫ x ∞ udx≤2c.
(13)

We conclude from Eqs. (11) and (13) that ∥ u x ∥ L ∞ ≤K. To complete the proof, we use a simple density argument [16]. Setting u 0 ε = e ε ∂ x 2 u 0 , we have u 0 ε ∈ H ∞ and ∥ u x ε ∥ L ∞ ≤2 ∫ R |u|dx<K. Applying ∥ u x ε − u x ∥ L ∞ ≤ sup [ 0 , T ] ∥ u x ε − u x ∥ H s →0 when ε→0, we have ∥ u x ∥ L ∞ ≤K. □

Using the first equation of system (3) one derives

d d t ∫ R ( u 2 + u x 2 ) dx+2(a−3b) ∫ R u u x 3 dx=0,

from which we have the conservation law

∫ R ( u 2 + u x 2 ) dx+2(a−3b) ∫ 0 t ∫ R u u x 3 dx= ∫ R ( u 0 2 + u 0 x 2 ) dx.
(14)

Lemma 3.5 (Kato and Ponce [23])

If r≥0, then H r ∩ L ∞ is an algebra. Moreover

∥ u v ∥ r ≤c ( ∥ u ∥ L ∞ ∥ v ∥ r + ∥ u ∥ r ∥ v ∥ L ∞ ) ,

where c is a constant depending only on r.

Lemma 3.6 (Kato and Ponce [23])

Let r>0. If u∈ H r ∩ W 1 , ∞ and v∈ H r − 1 ∩ L ∞ , then

∥ [ Λ r , u ] v ∥ L 2 ≤c ( ∥ ∂ x u ∥ L ∞ ∥ Λ r − 1 v ∥ L 2 + ∥ Λ r u ∥ L 2 ∥ v ∥ L ∞ ) .

Lemma 3.7 Let s> 3 2 and the function u(t,x) is a solution of problem (3) and the initial data u 0 (x)∈ H s (R). Then the following results hold:

∥ u ∥ L ∞ ≤ ∥ u ∥ H 1 ≤ ∥ u 0 ∥ H 1 ( R ) e | a − 3 b | 2 ∫ 0 t ∥ u x ∥ L ∞ ( R ) 2 d τ .
(15)

For q∈(0,s−1], there is a constant c only depending on a and b such that

∫ R ( Λ q + 1 u ) 2 d x ≤ ∫ R [ ( Λ q + 1 u 0 ) 2 ] d x + c ∫ 0 t ∥ u ∥ H q + 1 2 ( ∥ u x ∥ L ∞ ∥ u ∥ L ∞ + ∥ u x ∥ L ∞ 2 ) d τ .
(16)

Proof Using |2u u x |≤( u 2 + u x 2 ), the Gronwall inequality and Eq. (14), one derives Eq. (15).

Using ∂ x 2 =− Λ 2 +1 and the Parseval equality gives rise to

∫ R Λ q u Λ q ∂ x 3 ( u 3 ) dx=−3 ∫ R ( Λ q + 1 u ) Λ q + 1 ( u 2 u x ) dx+3 ∫ R ( Λ q u ) Λ q ( u 2 u x ) dx.

For q∈(0,s−1], applying ( Λ q u) Λ q to both sides of the first equation of system (3) and integrating with respect to x by parts, we have the identity

1 2 d d t ∫ R [ ( Λ q u ) 2 + ( Λ q u x ) 2 ] d x = − a ∫ R ( Λ q u ) Λ q ( u 2 u x ) d x − b ∫ R ( Λ q + 1 u ) Λ q + 1 ( u 2 u x ) d x − 2 b ∫ R Λ q u Λ q u x 3 d x + ( a − 6 b ) ∫ R Λ q u Λ q ( u u x u x x ) d x .
(17)

We will estimate the terms on the right-hand side of Eq. (17) separately. For the first term, by using the Cauchy-Schwartz inequality and Lemmas 3.5 and 3.6, we have

| ∫ R ( Λ q u ) Λ q ( u 2 u x ) d x | = | ∫ R ( Λ q u ) [ Λ q ( u 2 u x ) − u 2 Λ q u x ] d x + ∫ R ( Λ q u ) u 2 Λ q u x d x | ≤ c ∥ u ∥ H q ( 2 ∥ u ∥ L ∞ ∥ u x ∥ L ∞ ∥ u ∥ H q + ∥ u x ∥ L ∞ ∥ u ∥ L ∞ ∥ u ∥ H q ) + ∥ u ∥ L ∞ ∥ u x ∥ L ∞ ∥ Λ q u ∥ L 2 2 ≤ c ∥ u ∥ H q 2 ∥ u ∥ L ∞ ∥ u x ∥ L ∞ .
(18)

Using the above estimate to the second term yields

| ∫ R ( Λ q + 1 u ) Λ q + 1 ( u 2 u x ) d x | ≤c ∥ u ∥ H q + 1 2 ∥ u ∥ L ∞ ∥ u x ∥ L ∞ .
(19)

Using the Cauchy-Schwartz inequality and Lemma 3.5, we obtain

| ∫ R ( Λ q u x ) Λ q ( u u x 2 ) d x | ≤ ∥ Λ q u x ∥ L 2 ∥ Λ q ( u u x 2 ) ∥ L 2 ≤ c ∥ u ∥ H q + 1 ( ∥ u ∥ L ∞ ∥ u x 2 ∥ H q + ∥ u ∥ H q ∥ u x 2 ∥ L ∞ ) ≤ c ∥ u ∥ H q + 1 2 ( ∥ u ∥ L ∞ ∥ u x ∥ L ∞ + ∥ u x ∥ L ∞ 2 ) .
(20)

For the last term in Eq. (17), using u ( u x 2 ) x = ( u u x 2 ) x − u x u x 2 results in

| ∫ R ( Λ q u ) Λ q ( u u x u x x ) d x | ≤ 1 2 | ∫ R Λ q u x Λ q ( u u x 2 ) d x | + 1 2 | ∫ R Λ q u Λ q [ u x u x 2 ] d x | = K 1 + K 2 .
(21)

For K 1 , it follows from Eq. (20) that

K 1 ≤c ∥ u ∥ H q + 1 2 ( ∥ u ∥ L ∞ ∥ u x ∥ L ∞ + ∥ u x ∥ L ∞ 2 ) .
(22)

For K 2 , applying Lemma 3.5 derives

K 2 ≤ c ∥ u ∥ H q ∥ u x u x 2 ∥ H q ≤ c ∥ u ∥ H q ( ∥ u x ∥ L ∞ ∥ u x 2 ∥ H q + ∥ u x ∥ H q ∥ u x 2 ∥ L ∞ ) ≤ c ∥ u ∥ H q + 1 2 ∥ u x ∥ L ∞ 2 .
(23)

It follows from Eqs. (18)-(23) that there exists a constant c such that

1 2 d d t ∫ R [ ( Λ q u ) 2 + ( Λ q u x ) 2 ] dx≤c ∥ u ∥ H q + 1 2 ( ∥ u x ∥ L ∞ ∥ u ∥ L ∞ + ∥ u x ∥ L ∞ 2 ) .
(24)

Integrating both sides of the above inequality with respect to t results in inequality (16). □

Proof of Theorem 1 Using Eq. (16) with q=s−1, we obtain

∥ u ∥ H s 2 ≤ ∥ u 0 ∥ H s 2 +c ∫ 0 t ∥ u ∥ H s 2 ( ∥ u ∥ L ∞ ∥ u x ∥ L ∞ + ∥ u x ∥ L ∞ 2 ) dτ.
(25)

Applying the Gronwall inequality, we get

∥ u ∥ H s 2 ≤ ∥ u 0 ∥ H s 2 e 2 c ∫ 0 t ( ∥ u ∥ L ∞ ∥ u x ∥ L ∞ + ∥ u x ∥ L ∞ 2 ) d τ .
(26)

Using Eq. (15) and Lemma 3.4, we complete the proof of Theorem 1. □

Remark 2 In fact, using ∥ u x ∥ L ∞ ≤ ∥ u ∥ H s with s> 3 2 , Eqs. (15) and (26), we derive that the solution of Eq. (2) in space H s (R) blows up in finite time if and only if ∥ u x ∥ L ∞ =+∞.

4 Proof of Theorem 2

Multiplying Eq. (2) by y=u− u x x and integrating by parts, we get

1 2 d d t ∫ R y 2 d x = ∫ R y y t d x = ∫ R y ( a u u x u x x + b u 2 u x x x − ( a + b ) u 2 u x ) d x = ∫ R y ( a u u x ( u − y ) + b u 2 ( u x − y x ) − ( a + b ) u 2 u x ) d x = ∫ R y ( − a u u x y − b u 2 y x ) d x = ( b − a ) ∫ R u u x y 2 d x .
(27)

When a=b, from Eq. (27), we derive ∥ u x ∥ L ∞ is bounded. From Lemma 3.7 and Remark 2, we see that problem (3) has a global solution in the space

C ( [ 0 , ∞ ) ; H s ( R ) ) ∩ C 1 ( [ 0 , ∞ ) ; H s − 1 ( R ) ) .

Assume that the solution u=u(⋅, u 0 ) of problem (3) blows up in finite time in the space H s (R) with s> 3 2 . If b−a<0, we assume that u u x is bounded from below on [0,T)×R, i.e., there exists a constant M>1 such that

(b−a)u u x (t,x)≤Mon [0,T)×R.

From Eq. (27), we get

∥ u ∥ H 2 ≤c ∥ u 0 ∥ H 2 e M t ,
(28)

from which we derive that the H 2 norm of the solution to problem (3) does not blow up in finite time. From Remark 2, we know that this is impossible. Therefore, we have lim t → T inf{ inf x ∈ R u u x (t,x)}=−∞.

Similar to the above, we know that if b−a>0, the solution of problem (3) blows up if and only if lim t → T inf{ inf x ∈ R u u x (t,x)}=∞.

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Acknowledgements

This work is supported by both the Fundamental Research Funds for the Central Universities (JBK130401, JBK120504) and the Applied and Basic Project of Sichuan Province (2012JY0020).

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Lai, S., Yan, H. & Li, N. The global solution and blow-up phenomena to a modified Novikov equation. Bound Value Probl 2014, 16 (2014). https://doi.org/10.1186/1687-2770-2014-16

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