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

Boundary Value Problems20142014:123

https://doi.org/10.1186/1687-2770-2014-123

©  Ma et al.; licensee Springer. 2014

Received: 19 February 2014

Accepted: 6 May 2014

Published: 20 May 2014

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.

Keywords

weakly dissipative generalized periodic Degasperis-Procesi equation blow-up global existence blow-up rate

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 + 3 μ ( 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 ) d x 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 y d x , H 1 = 1 2 ∫ S u 2 d x , H 2 = ∫ S ( 3 2 μ ( u ) ( A − 1 ∂ x u ) 2 + 1 6 u 3 ) d x ,

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 ) d x 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 + 4 u 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 λ y = 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 ) d x − ∫ 0 x f ( y ) d y + ∫ 0 1 ∫ 0 x f ( y ) d y d x
(1.5)
and
( A − 1 ∂ x 2 f ) ( x ) = − f ( x ) + ∫ 0 1 f ( x ) d x .
(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 ) d x = 0 , then we have
max x ∈ S f 2 ( x ) ≤ 1 12 ∫ S f x 2 ( x ) d x .

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 , ⋅ ) 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 ) = − λ μ ( 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 ) d x .
(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 + 3 μ ( u ) A − 1 ∂ x u + λ u = 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 d x = ∫ S u 0 2 d x ⋅ 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 + λ u ( 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 , ⋅ ) , 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 , ⋅ ) 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 − 2 λ ( 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 d x ≤ ( 5 N − 2 λ ) ∫ S y 2 d x .
Applying Gronwall’s inequality then yields for t ∈ [ 0 , T )
∫ S y 2 d x ≤ e ( 5 N − 2 λ ) t ∫ S y 2 ( 0 , x ) d x .
Note that
∫ S y 2 d x = μ 2 ( u ) + ∫ S u x x 2 d x ≥ ∥ 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 + 3 μ ( 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 + 3 μ ( 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 ) + λ m ( t ) ≤ K a.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 d x = 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.

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics, Tianshui Normal University
(2)
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Faculty of Science, King Abdulaziz University
(3)
Department of Mathematics, Quaid-I-Azam University
(4)
Department of Mathematics, Zhejiang Normal University

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© Ma et al.; licensee Springer. 2014

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