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Blow-up phenomena and global existence for the weakly dissipative generalized periodic Degasperis-Procesi equation

Abstract

In this paper, we investigate the Cauchy problem of a weakly dissipative generalized periodic Degasperis-Procesi equation. The precise blow-up scenarios of strong solutions to the equation are derived by a direct method. Several new criteria guaranteeing the blow-up of strong solutions are presented. The exact blow-up rates of strong solutions are also determined. Finally, we give a new global existence results to the equation.

MSC:35G25, 35Q35, 58D05.

1 Introduction

Recently, the following generalized periodic Degasperis-Procesi equation (μ DP) was introduced and studied in [1–3]

μ ( u ) t − u t x x +3μ(u) u x =3 u x u x x +u u x x x ,

where u(t,x) is a time-dependent function on the unite circle S=R/Z and μ(u)= ∫ S u(t,x)dx denotes its mean. The μ DP equation can be formally described as an evolution equation on the space of tensor densities over the Lie algebra of smooth vector fields on the circle S. In [2], the authors verified that the periodic μ DP equation describes the geodesic flows of a right-invariant affine connection on the Fréchet Lie group Diff ∞ (S) of all smooth and orientation-preserving diffeomorphisms of the circle S.

Analogous to the generalized periodic Camassa-Holm (μ CH) equation [4–6], μ DP equation possesses bi-Hamiltonian form and infinitely many conservation laws. Here we list some of the simplest conserved quantities:

H 0 =− 9 2 ∫ S ydx, H 1 = 1 2 ∫ S u 2 dx, H 2 = ∫ S ( 3 2 μ ( u ) ( A − 1 ∂ x u ) 2 + 1 6 u 3 ) dx,

where y=μ(u)− u x x , A=μ− ∂ x 2 is an isomorphism between H S and H s − 1 . Moreover, it is easy to see that ∫ S u(t,x)dx is also a conserved quantity for the μ DP equation.

Obviously, under the constraint of μ≡0, the μ DP equation is reduced to the μ Burgers equation [7].

It is clear that the closest relatives of the μ DP equation are the DP equation [8–11]

u t − u t x x +4u u x =3 u x u x x +u u x x x ,

which was derived by Degasperis and Procesi in [8] as a model for the motion of shallow water waves, and its asymptotic accuracy is the same as for the Camassa-Holm equation.

Generally speaking, energy dissipation is a very common phenomenon in the real world. It is interesting for us to study this kind of equation. Recently, Wu and Yin [12] considered the weakly dissipative Degasperis-Procesi equation. For related studies, we refer to [13] and [14]. Liu and Yin [15] discussed the blow-up, global existence for the weakly dissipative μ-Hunter-Saxton equation.

In this paper, we investigate the Cauchy problem of the following weakly dissipative periodic Degasperis-Procesi equation [16]:

{ μ ( u ) t − u t x x + 3 μ ( u ) u x = 3 u x u x x + u u x x x − λ ( μ ( u ) − u x x ) , t > 0 , x ∈ R , u ( 0 , x ) = u 0 ( x ) , x ∈ R , u ( t , x + 1 ) = u ( t , x ) , t ≥ 0 , x ∈ R ,
(1.1)

the constant λ is a nonnegative dissipative parameter and the term λy=λ(μ(u)− u x x ) models energy dissipation. Obviously, if λ=0 then the equation reduces to the μ DP equation. we can rewrite the system (1.1) as follows:

{ y t + u y x + 3 u x y + λ y = 0 , t > 0 , x ∈ R , y = μ ( u ) − u x x , t > 0 , x ∈ R , u ( 0 , x ) = u 0 ( x ) , x ∈ R , u ( t , x + 1 ) = u ( t , x ) , t ≥ 0 , x ∈ R .
(1.2)

Let G(x):= 1 2 x 2 − 1 2 |x|+ 13 12 , x∈R be the associated Green’s function of the operator A − 1 , then the operator can be expressed by its associated Green’s function,

A − 1 f(x)=(G∗f)(x),f∈ L 2 ,

where ∗ denotes the spatial convolution. Then equation (1.1) takes the equivalent form of a quasi-linear evolution equation of hyperbolic type:

{ u t + u u x + 3 μ ( u ) A − 1 ∂ x u + λ u = 0 , t > 0 , x ∈ R , u ( 0 , x ) = u 0 ( x ) , x ∈ R , u ( t , x + 1 ) = u ( t , x ) , t ≥ 0 , x ∈ R .
(1.3)

It is easy to check that the operator A=μ− ∂ x 2 has the inverse

( A − 1 f ) ( x ) = ( 1 2 x 2 − 1 2 x + 13 12 ) μ ( f ) + ( x − 1 2 ) ∫ 0 1 ∫ 0 y f ( s ) d s d y − ∫ 0 x ∫ 0 y f ( s ) d s d y + ∫ 0 1 ∫ 0 y ∫ 0 s f ( r ) d r d s d y .
(1.4)

Since A − 1 and ∂ x commute, the following identities hold:

( A − 1 ∂ x f ) (x)= ( x − 1 2 ) ∫ 0 1 f(x)dx− ∫ 0 x f(y)dy+ ∫ 0 1 ∫ 0 x f(y)dydx
(1.5)

and

( A − 1 ∂ x 2 f ) (x)=−f(x)+ ∫ 0 1 f(x)dx.
(1.6)

The paper is organized as follows. In Section 2, we briefly give some needed results, including the local well-posedness of equation (1.1), and some useful lemmas and results which will be used in subsequent sections. In Section 3, we establish the precise blow-up scenarios and blow-up criteria of strong solutions. In Section 4, we give the blow-up rate of strong solutions. In Section 5, we give two global existence results of strong solutions.

Remark 1.1 Although blow-up criteria and global existence results of strong solutions to equation (1.1) are presented in [16], our blow-up results improve considerably earlier results.

2 Preliminaries

In this section we recall some elementary results which we want to use in this paper. We list them and skip their proofs for conciseness. Local well-posedness for equation (1.1) can be obtained by Kato’s theory [17], in [16] the authors gave a detailed description on well-posedness theorem.

Theorem 2.1 [16]

Let s>3/2 and u 0 ∈ H s (S); then there is a maximal time T and a unique solution

u∈C ( [ 0 , T ) ; H s ( S ) ) ∩ C 1 ( [ 0 , T ) ; H s − 1 ( S ) )

of the Cauchy problems (1.1) which depends continuously on the initial data, i.e. the mapping

H s (S)→C ( [ 0 , T ) ; H s ( S ) ) ∩ C 1 ( [ 0 , T ) ; H s − 1 ( S ) ) , u 0 ↦u(⋅, u 0 ),

is continuous.

Remark 2.1 The maximal time of existence T>0 in Theorem 2.1 is independent of the Sobolev index s>3/2.

Next we present the Sobolev-type inequalities, which play a key role to obtain blow-up results for the Cauchy problem (1.1) in the sequel.

Lemma 2.2 [18]

If f∈ H 1 (S) is such that ∫ S f(x)dx=0, then we have

max x ∈ S f 2 (x)≤ 1 12 ∫ S f x 2 (x)dx.

Lemma 2.3 [19]

If r>0, let Λ= ( 1 − ∂ x 2 ) 1 / 2 , then

∥ [ Λ r , f ] g ∥ L 2 ≤c ( ∥ ∂ x f ∥ L ∞ ∥ Λ r − 1 g ∥ L 2 + ∥ Λ r f ∥ L 2 ∥ g ∥ L ∞ ) ,

where c is a constant depending only on r.

Lemma 2.4 [20]

Let t 0 >0 and v∈ C 1 ([0, t 0 ); H 2 (R)), then for every t∈[0, t 0 ) there exists at least one point ξ(t)∈R with

m(t):= inf x ∈ R v x (t,x)= v x ( t , ξ ( t ) ) ,

and the function m is almost everywhere differentiable on (0, t 0 ) with

d d t m(t)= v t x ( t , ξ ( t ) ) a.e. on (0, t 0 ).

We also need to introduce the classical particle trajectory method which is motivated by McKean’s deep observation for the Camassa-Holm equation in [21]. Suppose u(x,t) is the solution of the Camassa-Holm equation and q(x,t) satisfies the following equation:

{ q t = u ( q , t ) , 0 < t < T , x ∈ R , q ( x , 0 ) = x , x ∈ R , q ( x + 1 , t ) = x , 0 < t < T , x ∈ R ,
(2.1)

where T is the maximal existence time of solution, then q(t,â‹…) is a diffeomorphism of the line. Taking the derivative with respect to x, we have

d q x d t = q t x = u x (q,t) q x ,t∈(0,T).

Hence

q x (x,t)=exp ( ∫ 0 t u x ( q , s ) d s ) >0, q x (x,0)=1,
(2.2)

which is always positive before the blow-up time.

In addition, integrating both sides of the first equation in equation (1.1) with respect to x on S, we obtain

d d t μ(u)=−λμ(u),

it follows that

μ(u)=μ( u 0 ) e − λ t := μ 0 e − λ t ,
(2.3)

where

μ 0 :=μ( u 0 )= ∫ S u 0 (x)dx.
(2.4)

3 Blow-up solutions

In this section, we are able to derive an import estimate for the L ∞ -norm of strong solutions. This enables us to establish precise blow-up scenario and several blow-up results for equation (1.1).

Lemma 3.1 Let u 0 ∈ H s , s>3/2 be given and assume the T is the maximal existence time of the corresponding solution u to equation (1.1) with the initial data u 0 . Then we have

∥ u ( t , x ) ∥ L ∞ ≤ e − λ t ( 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ + ∥ u 0 ∥ L ∞ ) ,∀t∈[0,T).
(3.1)

Proof The first equation of the Cauchy problem (1.1) is

u t +u u x +3μ(u) A − 1 ∂ x u+λu=0.

In view of equation (1.5), we have

| A − 1 ∂ x u|≤ 1 2 | μ 0 | e − λ t +2 ( ∫ S u 2 d x ) 1 2 .

A direct computation implies that

d d t ∫ S u 2 d x = 2 ∫ S 2 u u t d x = − 2 ∫ S 2 u ( u u x + 3 μ ( u ) A − 1 ∂ x u + λ u ) d x = − 2 λ ∫ S u 2 d x .

It follows that

∫ S u 2 dx= ∫ S u 0 2 dx⋅ e − 2 λ t := μ 2 2 e − 2 λ t .
(3.2)

So we have

| A − 1 ∂ x (u)|≤ ( 1 2 | μ 0 | + 2 μ 2 ) e − λ t .

In view of equation (2.1) we have

d u ( t , q ( t , x ) ) d t = u t ( t , q ( t , x ) ) + u x ( t , q ( t , x ) ) d q ( t , x ) d t =( u t +u u x ) ( t , q ( t , x ) ) .

Combing the above relations, we arrive at

| d u ( t , q ( t , x ) ) d t +λu ( t , q ( t , x ) ) |≤3| μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) e − 2 λ t .

Integrating the above inequality with respect to t<T on [0,t] yields

| e λ t u ( t , q ( t , x ) ) − u 0 (x)|≤ 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ .

Thus

|u ( t , q ( t , x ) ) |≤ ∥ u ( t , q ( t , x ) ) ∥ L ∞ ≤ e − λ t ( 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ + ∥ u 0 ∥ L ∞ ) .

In view of the diffeomorphism property of q(t,â‹…), we can obtain

|u(t,x)|≤ ∥ u ( t , x ) ∥ L ∞ ≤ e − λ t ( 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ + ∥ u 0 ∥ L ∞ ) .

This completes the proof of Lemma 3.1. □

Theorem 3.2 Let u 0 ∈ H s , s>3/2 be given and assume that T is the maximal existence time of the corresponding solution u(t,x) to the Cauchy problem (1.1) with the initial data u 0 . If there exists M>0 such that

∥ u x ( t , ⋅ ) ∥ L ∞ ≤M,t∈[0,T),

then the H s -norm of u(t,â‹…) does not blow up on [0,T).

Proof We assume that c is a generic positive constant depending only on s. Let Λ= ( 1 − ∂ x 2 ) 1 / 2 . Applying the operator Λ s to the first one in equation (1.3), multiplying by Λ s u, and integrating over S, we obtain

d d t ∥ u ∥ H s 2 =−2 ( u u x , u ) H s −6 ( u , A − 1 ∂ x ( μ ( u ) u ) ) H s −2λ ( u , u ) H s .
(3.3)

Let us estimate the first term of the above equation,

| ( u u x , u ) H s | = | ( Λ s ( u u x ) , Λ s u ) L 2 | = | ( [ Λ s , u ] u x , Λ s u ) L 2 + ( u Λ s u x , Λ s u ) L 2 | ≤ ∥ [ Λ s , u ] u x ∥ L 2 ∥ Λ s u ∥ L 2 + 1 2 | ( u x Λ s u , Λ s u ) L 2 | ≤ 2 ∥ ( u , v ) ∥ H 1 × H 1 2 ( 2 ∥ ( u , v ) ∥ H 1 × H 1 2 ) ≤ c ∥ u x ∥ L ∞ ∥ u ∥ H s 2 ,
(3.4)

where we used Lemma 2.3 with r=s. Furthermore, we estimate the second term of the right hand side of equation (3.3) in the following way:

| ( u , A − 1 ∂ x ( μ ( u ) u ) ) H s | = | ( u , A − 1 ∂ x ( e − λ t μ 0 u ) ) H s | ≤ e − λ t | μ 0 | ∥ u ∥ H s ∥ A − 1 ∂ x u ∥ H s ≤ c | μ 0 | ∥ u ∥ H s 2 .
(3.5)

Combing equations (3.4) and (3.5) with equation (3.3) we arrive at

d d t ∥ u ∥ H s 2 ≤c ( | μ 0 | + ∥ u x ∥ L ∞ + 2 λ ) | ∥ u ∥ H s 2 .

An application of Gronwall’s inequality and the assumption of the theorem yield

∥ u ∥ H s 2 ≤ e c ( | μ 0 | + M + 2 λ ) t ∥ u 0 ∥ H s 2 .

This completes the proof of the theorem. □

The following result describes the precise blow-up scenario. Although the result which is proved in [16], our method is new, concise, and direct.

Theorem 3.3 Let u 0 ∈ H s , s>3/2 be given and assume that T is the maximal existence time of the corresponding solution u(t,x) to the Cauchy problem (1.1) with the initial data u 0 . Then the corresponding solution blows up in finite time if and only if

lim inf t → T { inf x ∈ S u x ( t , x ) } =−∞.

Proof Since the maximal existence time T is independent of the choice of s by Theorem 2.1, applying a simple density argument, we only need to consider the case s=3. Multiplying the first one in equation (1.2) by y and integrating over S with respect to x yield

d d t ∫ S y 2 d x = 2 ∫ S y y t d x = − 2 ∫ S y ( u y x + 3 u x y + λ y ) d x = − 2 ∫ S u y y x d x − 6 ∫ S u x y 2 d x − 2 λ ∫ S y 2 d x = − 5 ∫ S u x y 2 d x − 2 λ ∫ S y 2 d x .

If u x is bounded from below on [0,T)×S, then there exists N>λ>0 such that

u x (t,x)≥−N,∀(t,x)∈[0,T)×S,

then

d d t ∫ S y 2 dx≤(5N−2λ) ∫ S y 2 dx.

Applying Gronwall’s inequality then yields for t∈[0,T)

∫ S y 2 dx≤ e ( 5 N − 2 λ ) t ∫ S y 2 (0,x)dx.

Note that

∫ S y 2 dx= μ 2 (u)+ ∫ S u x x 2 dx≥ ∥ u x x ∥ L 2 2 .

Since u x ∈ H 2 ⊂ H 1 and ∫ S u x =0, Lemma 2.2 implies that

∥ u x ∥ L ∞ ≤ 1 2 3 ∥ u x x ∥ L 2 ≤ e ( 5 N − 2 λ ) t 2 ∥ y ( 0 , x ) ∥ L 2 .

Theorem 3.1 ensures that the solution u does not blow up in finite time. On the other hand, by the Sobolev embedding theorem it is clear that if

lim inf t → T { inf x ∈ S u x ( t , x ) } =−∞,

then T<∞. This completes the proof of the theorem. □

We now give first sufficient conditions to guarantee wave breaking.

Theorem 3.4 Let u 0 ∈ H s , s>3/2 and T be the maximal time of the solution u(t,x) to equation (1.1) with the initial data u 0 . If

inf x ∈ S u 0 ′ (x)<− 1 2 λ− 1 2 λ 2 + 4 α ,

then the corresponding solution to equation (1.1) blow up in finite time in the following sense: there exists T 0 satisfying

0< T 0 ≤ 1 λ 2 + 4 α ln ( 2 inf x ∈ S u 0 ′ ( x ) + λ − λ 2 + 4 α 2 inf x ∈ S u 0 ′ ( x ) + λ + λ 2 + 4 α ) ,

where α=3| μ 0 |( 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ + ∥ u 0 ∥ L ∞ ), such that

lim inf t → T 0 { inf x ∈ S u x ( t , x ) } =−∞.

Proof As mentioned early, we only need to consider the case s=3. Let

m(t):= inf x ∈ S [ u x ( t , x ) ] ,t∈[0,T)

and let ξ(t)∈S be a point where this minimum is attained by using Lemma 2.4. It follows that

m(t)= u x ( t , ξ ( t ) ) .

Differentiating the first one in equation (1.3) with respect to x, we have

u t x + u x 2 +u u x x +3μ(u) A − 1 ∂ x 2 u+λ u x =0.

From equation (1.6) we deduce that

u t x =− u x 2 −u u x x +3μ(u)(u− μ 0 )−λ u x .
(3.6)

Obviously u x x (t,ξ(t))=0 and u(t,⋅)∈ H 3 (S)⊂ C 2 (S). Substituting (t,ξ(t)) into equation (3.6), we get

d m ( t ) d t = − m 2 ( t ) − λ m ( t ) + 3 μ ( u ) u ( t , ξ ( t ) ) − 3 μ 2 ( u ) = − m 2 ( t ) − λ m ( t ) + 3 μ 0 e − λ t u ( t , ξ ( t ) ) − 3 μ 0 2 e − 2 λ t ≤ − m 2 ( t ) − λ m ( t ) + 3 | μ 0 | ( 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ + ∥ u 0 ∥ L ∞ ) .

Set

α=3| μ 0 | ( 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ + ∥ u 0 ∥ L ∞ ) .

Then we obtain

d m ( t ) d t ≤ − m 2 ( t ) − λ m ( t ) + α ≤ − 1 4 ( 2 m ( t ) + λ + λ 2 + 4 α ) ( 2 m ( t ) + λ − λ 2 + 4 α ) .

Note that if m(0)<− 1 2 λ− 1 2 λ 2 + 4 α , then m(t)<− 1 2 λ− 1 2 λ 2 + 4 α for all t∈[0,T). From the above inequality we obtain

2 m ( 0 ) + λ + λ 2 + 4 α 2 m ( 0 ) + λ − λ 2 + 4 α e λ 2 + 4 α t −1≤ 2 λ 2 + 4 α 2 m ( t ) + λ − λ 2 + 4 α ≤0.

Since

0< 2 m ( 0 ) + λ + λ 2 + 4 α 2 m ( 0 ) + λ − λ 2 + 4 α <1,

then there exists T 0 ,

0< T 0 ≤ 1 λ 2 + 4 α ln ( 2 m ( 0 ) + λ − λ 2 + 4 α 2 m ( 0 ) + λ + λ 2 + 4 α )

such that lim t → T 0 m(t)=−∞. Theorem 3.3 implies that the solution u blows up in finite time. □

We give another blow-up result for the solutions of equation (1.1).

Theorem 3.5 Let u 0 ∈ H s , s>3/2 and T be the maximal time of the solution u(t,x) to equation (1.1) with the initial data u 0 . If u 0 is odd satisfies u 0 ′ <−λ, then the corresponding solution to equation (1.1) blows up in finite time.

Proof By μ(u(t,−x))= μ 0 (t,−x) e − λ t =− μ 0 (t,x) e − λ t =−μ(u(t,x)), we can check the function

v(t,x):=−u(t,−x),t∈[0,T),x∈R,

is also a solution of equation (1.1), therefore u(x,t) is odd for any t∈[0,T). By continuity with respect to x of u and u x x , we get

u(t,0)= u x x (t,0)=0,∀t∈[0,T).

Define h(t):= u x (t,0) for t∈[0,T). From equation (3.6), we obtain

d h ( t ) d t = − h 2 ( t ) − λ h ( t ) − 3 μ 2 ( u ) ≤ − h 2 ( t ) − λ h ( t ) = − h ( t ) ( h ( t ) + λ ) .

Note that if h(0)<−λ, then h(t)<−λ for all t∈[0,T). From the above inequality we obtain

( 1 + λ h ( 0 ) ) e λ t −1≤ λ h ( t ) ≤0.

Since

0< h ( 0 ) + λ h ( 0 ) <1,

there exists T 0 ,

0< T 0 ≤ 1 λ ln h ( 0 ) h ( 0 ) + λ

such that lim t → T 0 m(t)=−∞. Theorem 3.3 implies that the solution u blows up in finite time. □

4 Blow-up rate

In this section, we consider the blow-up profile; the blow-up rate of equation (1.1) with respect to time can be shown as follows.

Theorem 4.1 Let u 0 ∈ H s , s>3/2 and T be the maximal time of the solution u(t,x) to equation (1.1) with the initial data u 0 . If T is finite, then

lim t → T { ( T − t ) min x ∈ S u x ( x , t ) } =−1.

Proof It is inferred from Lemma 2.4 that the function

m(t):= min x ∈ S u x (x,t)= u x ( t , ξ ( t ) )

is locally Lipschitz with m(t)<0, t∈[0,T). Note that u x x =0, a.e. t∈[0,T). Then we deduce that

| m ′ ( t ) + m 2 ( t ) + λ m ( t ) | = | 3 μ ( u ) u ( t , ξ ( t ) ) − 3 μ 2 ( u ) | = | 3 μ 0 e − λ t u ( t , ξ ( t ) ) − 3 μ 0 2 e − 2 λ t | ≤ 3 | μ 0 | ( 3 | μ 0 | ( 1 2 | μ 0 | + 2 μ 2 ) λ + ∥ u 0 ∥ L ∞ + | μ 0 | ) : = K .

It follows that

−K≤ m ′ (t)+ m 2 (t)+λm(t)≤Ka.e. on (0,T).
(4.1)

Thus,

−K− 1 4 λ 2 ≤ m ′ (t)+ ( m ( t ) + 1 2 λ ) 2 ≤K+ 1 4 λ 2 a.e. on (0,T).

Now fix any ε∈(0,1). In view of Theorem 3.1, there exists t 0 ∈(0,T) such that m( t 0 )<− ( K + 1 4 λ 2 ) ( 1 + 1 ε ) − 1 2 λ. Being locally Lipschitz, the function m(t) is absolutely continuous on [0,T). It then follows from the above inequality that m(t) is decreasing on [ t 0 ,T) and satisfies

m(t)<− ( K + 1 4 λ 2 ) ( 1 + 1 ε ) − 1 2 λ,t∈[ t 0 ,T).

Since m(t) is decreasing on [ t 0 ,T), it follows that

lim t → T m(t)=−∞.

It is found from equation (4.1) that

1−ε≤ d d t ( m ( t ) + 1 2 λ ) − 1 =− m ′ ( t ) ( m ( t ) + 1 2 λ ) 2 ≤1+ε.
(4.2)

Integrating both sides of equation (4.2) on (t,T), we obtain

(1−ε)(T−t)≤− 1 ( m ( t ) + 1 2 λ ) ≤(1+ε)(T−t),t∈[ t 0 ,T),
(4.3)

that is,

1 ( 1 + ε ) −≤ ( m ( t ) + 1 2 λ ) (T−t)≤ 1 ( 1 − ε ) ,t∈[ t 0 ,T).
(4.4)

By the arbitrariness of ε∈(0, 1 2 ), we have

lim t → T (T−t) ( m ( t ) + λ ) =−1.
(4.5)

This completes the proof of the theorem. □

5 Global existence

In this section, we will present some global existence results. Let us now prove the following lemma.

Lemma 5.1 Let u 0 ∈ H s , s>3/2 be given and assume that T>0 is the maximal existence time of the corresponding solution u(t,x) to the Cauchy problem (1.1). Let q∈ C 1 ([0,T)×R;R) be the unique solution of equation (2.1). Then we have

y ( t , q ( t , x ) ) q x 3 = y 0 (x) e − λ t ,

where y=μ(u)− u x x .

Proof By the first one in equation (1.2) and equation (2.1) we have

d d t y ( t , q ( t , x ) ) q x 3 = ( y t + y x q t ) q x 3 + 3 y q x q x t = ( y t + y x u ) q x 3 + 3 y q x q x t = ( y t + u y x + 3 y u x y x u ) q x 3 = − λ y q x 3 .

Therefore

y ( t , q ( t , x ) ) q x 3 = y 0 (x) e − λ t .

 □

Lemma 5.1 and equation (2.2) imply that y and y 0 have the same sign.

Theorem 5.2 Let u 0 ∈ H s , s>3/2. If y 0 = μ 0 − u 0 , x x ∈ H 1 does not change sign, then the corresponding solution u(t,x) to equation (1.1) with the initial data u 0 exists globally in time.

Proof By equation (2.1), we know that q(t,⋅) is diffeomorphism of the line and the periodicity of u with respect to spatial variable x, given t∈[0,T), there exists a ξ(t)∈S such that u x (t,ξ(t))=0.

We first consider the case that y 0 ≥0 on S, in which case Lemma 5.1 ensures that y≥0. For x∈[ξ(t),ξ(t)+1], we have

− u x ( t , x ) = − ∫ ξ ( t ) x u x x ( t , x ) d x = ∫ ξ ( t ) x ( y − μ ( u ) ) d x = ∫ ξ ( t ) x y d x − μ ( u ) ( x − ξ ( t ) ) ≤ ∫ S y d x − μ ( u ) ( x − ξ ( t ) ) = μ ( u ) ( 1 − x + ξ ( t ) ) ≤ | μ 0 | .

It follows that u x (t,x)≥−| μ 0 |.

On the other hand, if y 0 ≤0 on S, then Lemma 5.1 ensures that y≤0. Therefore, for x∈[ξ(t),ξ(t)+1], we have

− u x ( t , x ) = − ∫ ξ ( t ) x u x x ( t , x ) d x = ∫ ξ ( t ) x ( y − μ ( u ) ) d x = ∫ ξ ( t ) x y d x − μ ( u ) ( x − ξ ( t ) ) ≤ − μ ( u ) ( x − ξ ( t ) ) ≤ | μ 0 | .

It follows that u x (t,x)≥−| μ 0 |. By using Theorem 3.2, we immediately conclude that the solution is global. This completes the proof of the theorem. □

Corollary 5.3 If the initial value u 0 ∈ H 3 such that

∥ ∂ x 3 u 0 ∥ L 2 ≤2 3 | μ 0 |,

then the corresponding solution u of the initial value u 0 exists globally in time.

Proof Since ∫ S ∂ x 2 u 0 dx=0, by Lemma 2.2, we obtain

∥ ∂ x 2 u 0 ∥ L ∞ ≤ 1 2 3 ∥ ∂ x 3 u 0 ∥ L 2 .

If μ 0 ≥0, we have

y 0 = μ 0 − ∂ x 2 u 0 ≥ μ 0 − 1 2 3 ∥ ∂ x 3 u 0 ∥ L 2 ≥ μ 0 −| μ 0 |=0.

If μ 0 <0, we have

y 0 = μ 0 − ∂ x 2 u 0 ≤ μ 0 + ∥ ∂ x 2 u 0 ∥ L ∞ ≤ μ 0 + 1 2 3 ∥ ∂ x 3 u 0 ∥ L 2 ≤ μ 0 +| μ 0 |=0.

 □

Thus the theorem is proved by using Theorem 5.2.

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Acknowledgements

This work is partially supported by the NSFC (Grant No. 11101376) the HiCi Project (Grant No. 27-130-35-HiCi).

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Ma, C., Alsaedi, A., Hayat, T. et al. Blow-up phenomena and global existence for the weakly dissipative generalized periodic Degasperis-Procesi equation. Bound Value Probl 2014, 123 (2014). https://doi.org/10.1186/1687-2770-2014-123

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