Blow-Up for a Periodic Two-Component Camassa-Holm Equation with Generalized Weakly Dissipation

DOI: 10.4236/jamp.2020.810167   PDF   HTML   XML   46 Downloads   136 Views  

Abstract

In this paper we study a periodic two-component Camassa-Holm equation with generalized weakly dissipation. The local well-posedness of Cauchy problem is investigated by utilizing Kato’s theorem. The blow-up criteria and the blow-up rate are established by applying monotonicity. Finally, the global existence results for solutions to the Cauchy problem of equation are proved by structuring functions.

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Li, Y. , Liu, J. and Zhu, X. (2020) Blow-Up for a Periodic Two-Component Camassa-Holm Equation with Generalized Weakly Dissipation. Journal of Applied Mathematics and Physics, 8, 2223-2240. doi: 10.4236/jamp.2020.810167.

1. Introduction

In this paper, we consider the Cauchy problem of periodic two-component Camassa-Holm equation with a generalized weakly dissipation:

{ u t u x x t + k u x + 3 u u x 2 u x u x x u u x x x + λ ( u u x x ) + σ ρ ρ x = 0 , t > 0 , x R , ρ t + ( ρ u ) x = 0 , t > 0 , x R , u ( 0 , x ) = u 0 ( x ) ; ρ ( 0 , x ) = ρ 0 ( x ) , x R , u ( t , x ) = u ( t , x + 1 ) ; ρ ( t , x ) = ρ ( t , x + 1 ) , t 0 , x R , (1.1)

where λ 0 and k is a fixed constant; σ is a free parameter.

It is well known that the two-component integrable Camassa-Holm equation is

{ u t u x x t + k u x + 3 u u x 2 u x u x x u u x x x ρ ρ x = 0 , t > 0 , x R ρ t + ( ρ u ) x = 0 , t > 0 , x R (1.2)

which is a model for wave motion on shallow water, where u ( t , x ) standing for the fluid velocity at time t 0 in the spatial x direction [1], ρ ( t , x ) is in connection with the horizontal deviation of the surface from equilibrium (i.e. amplitude). Equation (1.2) possesses a bi-Hamiltonian structure [2] and the solution interaction of peaked travelling waves and wave breaking [1] [2] [3]. It is completely integrable [3] and becomes the Camassa-Holm equation when ρ = 0 .

Equation (1.2) was derived physically by Constantin and Ivanov [4] in the context of shallow water theory. As soon as this equation was put forward, it attracted attention of a large number of researchers. Escher et al. [5] established the local well-posedness and present the blow-up scenarios and several blow-up results of strong solutions to Equation (1.2). Constantin and Ivanov [6] investigated the global existence and blow-up phenomena of strong solutions of Equation (1.2). Guan and Yin [7] obtained a new global existence result for strong solutions to Equation (1.2) and several blow-up results, which improved the results in [6]. Gui and Liu [8] established the local well-posedness for Equation (1.2) in a range of the Besov spaces, they also characterized a wave breaking mechanism for strong solutions. Hu and Yin [9] [10] studied the blow-up phenomena and the global existence of Equation (1.2).

Dissipation is an inevitable phenomenon in real physical word. It is necessary to study periodic two-Camassa-Holm equation with a generalized weakly dissipation. Hu and Yin [11] study the blow-up of solutions to a weakly dissipative periodic rod equation. Hu considered global existence and blow-up phenomena for a weakly dissipative two-component Camassa-Holm system [12] [13]. The purpose of this paper is to study the blow-up phenomenon of the solutions of Equation (1.1). The results show that the behavior of solutions to the periodic two-component Camassa-Holm equation with a generalized weakly dissipation is similar to Equation (1.2) and the blow-up rate of Equation (1.1) is not affected by the dissipative term when σ > 0 .

The paper is organized as follows. Section 2 gives the local well-posedness of the Cauchy problem associated with Equation (1.1). The blow-up criteria for solutions and two conditions for wave breaking in finite time are given in Section 3. Furthermore, we also learn the blow-up rate of solutions. In Section 4, we address the global existence of Equation (1.1).

2. Local Well-Posedness

Let us introduce some notations, the S = R / Z is the circle of unit length, the [ x ] stands for the integer part of x R , the stands for the convolution, the X is used to represent the norm of Banach space X.

In this section, we investigate the local well-posedness for the Cauchy problem of Equation (1.1) by applying Kato’s theory [14] in H s ( S ) × H s 1 ( S ) , s 2 .

For convenience we recall the Kato’s theorem in the suitable form for our purpose. Consider the following abstract quasilinear evolution equation:

{ d z d t + A ( z ) z = f ( z ) , t 0 z ( 0 ) = z 0 (2.1)

There are two Hilbert’s spaces X and Y, Y is continuously and densely embedded in X and Q : Y X is a topological isomorphism, the L ( Y , X ) stands for the space of all bounded linear operator from Y to X.

Theorem 2.1 [14] 1) A ( y ) L ( Y , X ) , for y X with

( A ( y ) A ( z ) ) w X μ 1 y z X w Y (2.2)

where z , y , w Y , A ( y ) G ( X , 1 , β ) , i.e. A ( y ) is quasi-m-accretive, uniformly on bounded sets in Y.

2) Q A ( y ) Q 1 = A ( y ) + B ( y ) , where B ( y ) L ( X ) is uniformly bounded on a bounded sets in Y

( B ( y ) B ( z ) ) w X μ 2 y z Y w X (2.3)

where z , y Y , w X .

3) f : Y Y is a bounded map on bounded sets in Y

f ( y ) f ( z ) Y μ 3 y z Y (2.4)

f ( y ) f ( z ) X μ 4 y z X (2.5)

where z , y Y , μ 1 , μ 2 , μ 3 , μ 4 are constants which only depending { y Y , z Y } .

If the 1), 2), 3) hold, given u 0 Y , there is a maximal T > 0 depending only on u 0 Y and a unique solution u of Equation (2.1) such that

u = u ( , u 0 ) C ( [ 0 , T ) ; Y ) C 1 ( [ 0 , T ) ; X ) (2.6)

Moreover, the map u u ( , u 0 ) is continuous from Y to C ( [ 0 , T ) ; Y ) C 1 ( [ 0 , T ) ; X ) .

Note that g ( x ) : = cosh ( x [ x ] 1 2 ) 2 sinh 1 2 , x R , ( 1 x 2 ) 1 f = g f for all f L 2 ( S ) and g ( u u x x ) = u . Then Equation (1.1) can be rewritten as

{ u t + u u x = x g ( u 2 + 1 2 u x 2 + k u + σ 2 ρ 2 ) λ u ρ t + ( ρ u ) x = 0 u ( 0 , x ) = u 0 ( x ) ; ρ ( 0 , x ) = ρ 0 ( x ) u ( t , x ) = u ( t , x + 1 ) ; ρ ( t , x ) = ρ ( t , x + 1 ) (2.7)

Theorem 2.2 Let z 0 = ( u 0 , ρ 0 1 ) H s × H s 1 with s 2 , there exists a maximal time T > 0 which is independent on s and exists a unique solution ( u , ρ ) of Equation (1.1) in the interval [ 0 , T ) with initial data z 0 , such that the solution depends continuously on the initial data.

The remainder of this section is devoted the proof of Theorem 2.2. Let z = ( u ρ ) , T = H s × H s , X = H s 1 × H s 1 , = ( 1 x 2 ) 1 2 , Q = ( 0 0 ) , and

A ( z ) = ( u x 0 0 u x ) (2.8)

The [15] shows that Q is an isomorphism from H s × H s onto H s 1 × H s 1 . It is sufficiently to verify A ( z ) , B ( z ) , f ( z ) satisfy 1), 2), 3) to prove the theorem 2.2. For this purpose, the following lemmas are necessary.

Lemma 2.1 [15] The operator A ( z ) is defined in (2.8) with z H s × H s , s > 3 2 belongs to G ( L 2 × L 2 , 1 , β ) .

Lemma 2.2 [15] The operator A ( z ) is defined in (2.8) with z H s × H s , s > 3 2 belongs to G ( H s 1 × H s 1 , 1 , β ) .

Lemma 2.3 [15] The operator A ( z ) is defined in (2.8) with z H s × H s , s > 3 2 belongs to L ( H s × H s , H s 1 × H s 1 ) , moreover,

( A ( y ) A ( z ) ) w H s 1 × H s 1 μ 1 y z H s × H s w H s × H s (2.9)

where y , z , w H s × H s .

Lemma 2.4 [15] Let B ( z ) = Q A ( z ) Q 1 A ( z ) with z H s × H s , s > 3 2 , then the operator B ( z ) L ( H s 1 × H s 1 ) and

( B ( y ) B ( z ) ) w H s 1 × H s 1 μ 2 y z H s × H s w H s 1 × H s 1 (2.10)

for y , z H s × H s , and w H s 1 × H s 1 .

Lemma 2.5 Let z H s × H s , s > 3 2 , and

f ( z ) = ( x ( 1 x 2 ) 1 ( u 2 + 1 2 u x 2 + k u + σ 2 ρ 2 ) + λ u ρ u x )

Then f is bounded on bounded sets in H s × H s and satisfies

1) f ( y ) f ( z ) H s × H s μ 3 y z H s × H s , y , z H s × H s (2.11)

2) f ( y ) f ( z ) H s 1 × H s 1 μ 4 y z H s 1 × H s 1 , y , z H s × H s (2.12)

Proof: For any z , y H s × H s , s > 3 2 ,

f ( y ) f ( z ) H s × H s x ( 1 x 2 ) 1 [ ( y 1 2 u 2 ) + 1 2 ( y 1 x 2 u x 2 ) + k ( y 1 u ) + σ 2 ( y 2 2 ρ 2 ) ] H s + λ ( y 1 u ) H s + u x ρ y 1 x y 2 H s

( y 1 2 u 2 ) + 1 2 ( y 1 x 2 u x 2 ) + k ( y 1 u ) H s 1 + | σ | 2 y 2 2 ρ 2 H s 1 + | λ | y 1 u H s + ( u x y 1 x ) ρ H s + y 1 x ( ρ y 2 ) H s

( y 1 u ) ( y 1 + u ) H s 1 + 1 2 ( y 1 x u x ) ( y 1 x + u x ) H s 1 + | k | y 1 u H s 1 + | λ | y 1 u H s + | σ | 2 y 2 ρ H s 1 y 2 + ρ H s 1 + y 1 u H s ρ H s + y 1 H s ρ y 2 H s

( y 1 u ) ( y 1 + u ) H s 1 + 1 2 ( y 1 x u x ) ( y 1 x + u x ) H s 1 + | k | y 1 u H s 1 + | λ | y 1 u H s + | σ | 2 y 2 ρ H s 1 y 2 + ρ H s 1 + y 1 u H s ρ H s + y 1 H s ρ y 2 H s

Let μ 3 = 5 + | σ | 2 y H s × H s + 3 + | σ | 2 z H s × H s + | k | + | λ | , then

f ( y ) f ( z ) H s × H s μ 3 y z H s × H s , y , z H s × H s

Making y = 0 in the above inequality, it shows that f is bounded on bounded sets in H s × H s , the proof of 1) is complete.

Similarly, the inequality (2.12) also can be proved.

Proof of Theorem 2.2: The 1) is true for A ( z ) from the inequality (2.9), the 2) is true for B ( z ) from the inequality (2.10), the 3) is true for f ( z ) from the inequalities (2.11) (2.12). According to the Theorem 2.1, the proof of the Theorem 2.2 is complete.

3. Blow-Up

This section will establish a blow-up criterion for solution of Equation (1.1) when σ > 0 .

Theorem 3.1 [8] [16] Let σ 0 and ( u , ρ ) be the solution of (1.1) with initial data ( u 0 , ρ 0 1 ) H s × H s 1 , s > 3 2 , T is the maximal time of existence of the solution, then

T < 0 T u x ( τ ) L d τ = (3.1)

Consider the following equation of trajectory:

{ d q ( t , x ) d t = u ( t , q ( t , x ) ) , t [ 0 , T ) q ( 0 , x ) = x , x S (3.2)

The (3.2) shows q ( t , ) : S S is the differential homeomorphism for every t [ 0 , T )

q x ( t , x ) = e 0 t u x ( τ , q ( τ , x ) ) d τ > 0 , ( t , x ) [ 0 , T ) × S (3.3)

Hence

v ( t , ) L = v ( t , q ( t , ) ) L (3.4)

Lemma 3.1 [17] Let T > 0 and v C 1 ( [ 0 , T ) ; H 1 ( R ) ) , then for every t [ 0 , T ) , there exists at least one point ξ ( t ) R with

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

The function m ( t ) is absolutely continuous in ( 0 , T ) with

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

Lemma 3.2 Let z 0 = ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) with s 2 , there exist a maximal time T > 0 and a unique solution ( u , ρ ) of Equation (1.1) with initial data z 0 , then we have

u H 1 2 + σ ρ 1 L 2 2 u 0 H 1 2 + σ ρ 0 1 L 2 2 (3.5)

Proof: Multiply the first equation of Equation (1.1) by u and integrate

d d t S ( u 2 + u x 2 ) d x + 2 λ S ( u 2 + u x 2 ) d x + 2 σ S ρ ρ x u d x = 0 (3.6)

The second equation of Equation (1.1) can be rewritten as

( ρ 1 ) t + ρ x u + ρ u x = 0

Multiply the above equation by ( ρ 1 ) and integrate

d d t S ( ρ 1 ) 2 d x + 2 S u ρ ρ x d x 2 S u ρ x d x + 2 S u x ρ 2 d x 2 S u x ρ d x = 0 (3.7)

According to (3.6) and (3.7)

d d t S ( u 2 + u x 2 + σ ( ρ 1 ) 2 + 2 λ 0 t ( u 2 + u x 2 ) d τ ) d x = 0

Then

S ( u 2 + u x 2 + σ ( ρ 1 ) 2 + 2 λ 0 t ( u 2 + u x 2 ) d τ ) d x = S ( u 0 2 + u 0 x 2 + σ ( ρ 0 1 ) 2 ) d x = u 0 H 1 2 + σ ρ 0 1 L 2 2

Notice that 2 λ 0 t ( u 2 + u x 2 ) d x 0 , then

u H 1 2 + σ ρ 1 L 2 2 = S ( u 2 + u x 2 + σ ( ρ 1 ) 2 ) d x u 0 H 1 2 + σ ρ 0 1 L 2 2

Lemma 3.3 [18] [19] 1) For every f H 1 ( S ) , we have

max x [ 0 , 1 ] f 2 ( x ) e + 1 2 ( e 1 ) f H 1 2 (3.8)

where the constant e + 1 2 ( e 1 ) is the best constant.

2) For every f H 3 ( S ) , we have

max x [ 0 , 1 ] f 2 ( x ) c f H 1 2 (3.9)

where the best constant c is e + 1 2 ( e 1 ) .

3) For every f H 3 ( S ) , we have

max x [ 0 , 1 ] f x 2 ( x ) 1 12 f H 2 2 (3.10)

Lemma 3.4 Suppose σ > 0 , and ( u , ρ ) be the solution of Equation (1.1) with initial data ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) , s 2 , and T be the maximal time of existence, then

sup x S u x ( t , x ) u 0 x L + λ 2 + σ ρ 0 L 2 + C 1 2

where C 1 = ( 3 σ + 2 ) ( e + 1 ) 2 ( e 1 ) + ( e + 1 e 1 + k 2 + 1 2 ) ( u 0 H 1 2 + σ ρ 0 1 L 2 2 ) .

Proof: The theorem 2.2 and a density argument imply that it is sufficient to prove the desired estimates for s = 3 .

Differentiate the first equation of Equation (2.7) with respect to x

u t x = u 2 1 2 u x 2 λ u x + σ 2 ρ 2 k x 2 g u g ( u 2 + 1 2 u x 2 + σ 2 ρ 2 ) u u x x (3.11)

Define

m ¯ ( t ) = u x ( t , η ( t ) ) = sup x S ( u x ( t , x ) ) , m ( t ) = inf x S ( u x ( t , x ) ) (3.12)

From the Fermat’s lemma, we know

u x x ( t , η ( t ) ) ) = 0 , a . e . t [ 0 , T )

there exists x 1 ( t ) S such that

q ( t , x 1 ( t ) ) = η ( t ) , t [ 0 , T ) (3.13)

Set

ζ ¯ ( t ) = ρ ( t , q ( t , x 1 ) ) , t [ 0 , T ) (3.14)

From (3.11) and the second equation of Equation (1.1), we obtain

{ m ¯ ( t ) = 1 2 m ¯ 2 ( t ) λ m ¯ ( t ) + σ 2 ζ ¯ 2 ( t ) + f ( t , q ( t , x 1 ) ) ζ ¯ ( t ) = ζ ¯ ( t ) m ¯ ( t ) (3.15)

where f = u 2 k x 2 g u g ( u 2 + 1 2 u x 2 + σ 2 ρ 2 ) .

Notice that x 2 g u = x g x u , then

f = u 2 k x 2 g u g ( u 2 + 1 2 u x 2 ) σ 2 g ( ρ 2 ) = u 2 k x 2 g u g ( u 2 + 1 2 u x 2 ) σ 2 g 1 σ g ( ρ 1 ) σ 2 g ( ρ 1 ) 2 u 2 + k | x g x u | + σ 2 | g 1 | + σ | g ( ρ 1 ) |

From (3.8) (3.9) and (3.10), we have

u 2 e + 1 2 ( e 1 ) u H 1 2

k | x g x u | k g x L 2 u x L 2 e + 1 2 ( e 1 ) + 1 4 k 2 u x L 2 2

| g ( u 2 + 1 2 u x 2 ) | e + 1 2 ( e 1 ) u L 2 2 + e + 1 4 ( e 1 ) u x L 2 2

σ 2 | g 1 | σ 2 g L σ ( e + 1 ) 4 ( e 1 )

σ | g ( ρ 1 ) | σ g L 2 ρ 1 L 1 σ ( e + 1 ) 2 ( e 1 ) + σ 4 ρ 1 L 2 2

σ 2 | g ( ρ 1 ) 2 | σ 2 g L ( ρ 1 ) L 1 σ ( e + 1 ) 4 ( e 1 ) ρ 1 L 2 2

Therefore we get the upper bound of f

f ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( e + 1 2 ( e 1 ) + k 2 4 ) u H 1 2 + 1 4 σ ρ 1 L 2 2 ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( e + 1 2 ( e 1 ) + k 2 + 1 4 ) ( u H 1 2 + σ ρ 1 L 2 2 ) ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( e + 1 2 ( e 1 ) + k 2 + 1 4 ) ( u 0 H 1 2 + σ ρ 0 1 L 2 2 ) = 1 2 C 1 2 (3.16)

Similarly, we turn to the lower bound of f

f u 2 + k | x g x u | + | g ( u 2 + 1 2 u x 2 ) | + σ 2 | g 1 | + σ | g ( ρ 1 ) | + σ 2 | g ( ρ 1 ) 2 | ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + e + 1 e 1 u L 2 2 + ( 3 ( e + 1 ) 4 ( e 1 ) + k 2 4 ) u x L 2 2 + ( e + 1 4 ( e 1 ) + 1 4 ) σ ρ 1 L 2 2 ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( 7 ( e + 1 ) 4 ( e 1 ) + k 2 + 1 4 ) ( u H 1 2 + σ ρ 1 L 2 2 ) ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( 7 ( e + 1 ) 4 ( e 1 ) + k 2 + 1 4 ) ( u 0 H 1 2 + σ ρ 0 1 L 2 2 ) (3.17)

According to (3.16) and (3.17)

| f | ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( 7 ( e + 1 ) 4 ( e 1 ) + k 2 + 1 4 ) ( u 0 H 1 2 + σ ρ 0 1 L 2 2 ) (3.18)

From Sobolev’s embedding theorem, we have u C 0 1 ( S ) , due to the periodic of Equation (1.1), then

inf x S u x ( t , x ) 0 , sup x S u x ( t , x ) 0 , t [ 0 , T ) (3.19)

hence

m ¯ ( t ) 0 , t [ 0 , T ) (3.20)

From the second Equation of (3.15), we have

ζ ¯ ( t ) = ζ ¯ ( 0 ) e 0 t m ¯ ( τ ) d τ

then

| ρ ( t , q ( t , x 1 ) ) | = | ζ ¯ ( t ) | | ζ ¯ ( 0 ) | ρ 0 L

For any given x S , define

P 1 ( t ) = m ¯ ( t ) u 0 x L λ 2 + σ ρ 0 L 2 + C 1 2 (3.21)

then P 1 ( t ) is C 1 -function in [ 0 , T ) and satisfies

P 1 ( 0 ) = m ¯ ( 0 ) u 0 x L λ 2 + σ ρ 0 L 2 + C 1 2 m ¯ ( 0 ) u 0 x L 0

Next, we will show P 1 ( t ) 0 , t [ 0 , T ) .

By contradictory arguement, there exists t 0 [ 0 , T ) such that P 1 ( t 0 ) > 0 . Making t 1 = max { t < t 0 : P 1 ( t ) = 0 } , we have

P 1 ( t 1 ) = 0 , P 1 ( t 1 ) 0

then

m ¯ ( t 1 ) = u 0 x L + λ 2 + σ ρ 0 L 2 + C 1 2 .

From (3.21), we know

m ¯ ( t 1 ) = P 1 ( t 1 ) 0 (3.22)

On the other hand, from the first Equation of (3.15), we have

m ¯ ( t 1 ) = 1 2 m ¯ 2 ( t 1 ) λ m ¯ ( t 1 ) + σ 2 ζ ¯ 2 ( t 1 ) + f ( t 1 , q ( t 1 , x 1 ) ) 1 2 ( m ( t 1 ) + λ ) 2 + 1 2 λ 2 + σ 2 ρ 0 L 2 + 1 2 C 1 2 1 2 ( u 0 x L + λ 2 + σ ρ 0 L 2 + C 1 2 + λ ) 2 + σ 2 ρ 0 L 2 + 1 2 C 1 2 < 0

It yields a contradiction, then the proof of the Lemma 3.4 is complete.

Lemma 3.5 Suppose σ > 0 , and ( u , ρ ) be the solution of Equation (1.1) with initial data ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) , s 2 , and T is the maximal time of the solution. If there exists M 0 such that

inf ( t , x ) [ 0 , T ) × S u x M (3.23)

then

ρ ( t , ) L ( S ) ρ 0 L ( S ) e M t (3.24)

Proof: For any given x S , define

U ( t ) = u x ( t , q ( t , x 1 ) ) , γ ( t ) = ρ ( t , q ( t , x ) )

the second equation of Equation (1.1) becomes

γ ( t ) = γ U

then

γ ( t ) = γ ( 0 ) e 0 t U ( τ ) d τ .

From (3.23), we know U ( t ) M , t [ 0 , T ) . Hence

| ρ ( t , q ( t , x ) ) | = | γ ( t ) | | γ ( 0 ) | e 0 t U ( τ ) d τ | γ ( 0 ) | e M t ρ 0 L e M t

which together with (3.4), then the proof of lemma 3.5 is complete.

Theorem 3.2 Suppose σ > 0 , and ( u , ρ ) be the solution of Equation (1.1) with initial data ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) , s 2 , and T is the maximal time of existence of the solution, then the solution of Equation (1.1) blows up in finite time if and only if

lim t T inf x S u x ( t , x ) = (3.25)

Proof: Suppose that T < and (3.25) is invalid, then there exists M > 0 satisfies

u x ( t , x ) M , ( t , x ) [ 0 , T ) × S

The Lemma 3.4 shows that u x ( t , x ) is bounded on [ 0 , T ) , i.e. | u x ( t , x ) | C , where C = C ( k , M , σ , λ , ( u 0 , ρ 0 1 ) H s × H s 1 ) . Then from the Theorem 3.1, we have T = , which contradicts the assumption T < .

On the other hand, Sobolev embedding theorem H s L with s > 1 2 implies that if (3.25) holds, then the corresponding solution blows up in finite time, the proof of Theorem 3.2 is complete.

Next we give two blow-up conditions in finite time.

Theorem 3.3 Suppose σ > 0 , and ( u , ρ ) be the solution of Equation (1.1) with initial data ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) , s 2 , and T is the maximal time of existence of the solution. If there exists x 0 S satisfies

ρ 0 ( x 0 ) = 0 , u 0 x ( x 0 ) = inf x R u 0 x ( x ) (3.26)

and

u 0 H 1 2 + σ ρ 0 1 L 2 2 ( ( 8 σ 1 ) ( e + 1 ) 36 ( e 1 ) 1 2 λ 2 ) 4 ( e + 1 ) 3 ( e + 1 ) + 18 ( k 2 + 1 ) ( e 1 ) (3.27)

then the corresponding solution u of Equation (1.1) blows up in finite time when 0 < T < T , where

T = 2 ( 1 + | u 0 x ( x 0 ) | ) ( 8 σ 1 ) ( e + 1 ) 18 ( e 1 ) + ( 3 ( e + 1 ) + 18 ( k 2 + 1 ) ( e 1 ) 2 ( e 1 ) ) ( u 0 H 1 2 + σ ρ 0 1 L 2 2 ) λ 2 ) + 2 1 λ

Proof: Without loss of generality, assume s = 3 , and choose x 2 ( t ) such that q ( t , x 2 ( t ) ) = ξ ( t ) , t [ 0 , t ) , along the trajectory q ( t , x 2 ) , we rewrite the transport Equation of ρ in (2.7) as

d ρ ( t , ξ ( t ) ) d t = ρ ( t , ξ ( t ) ) u x ( t , ξ ( t ) ) (3.28)

From (3.26), we have

m ( 0 ) = u x ( 0 , ξ ( 0 ) ) = inf x S u 0 x ( x ) = u 0 x ( x 0 )

Let x 0 = ξ ( 0 ) , then ρ 0 ( ξ ( 0 ) ) = ρ 0 ( x 0 ) , from (3.28)

ρ ( t , ξ ( t ) ) = 0 , t [ 0 , T ) (3.29)

From (3.11), (3.29) and u x x ( t , ξ ( t ) ) = 0 , we obtain

m ¯ ( t ) = 1 2 m 2 ( t ) λ m ( t ) + u 2 ( t , ξ ( t ) ) k x g x u g ( u 2 + 1 2 u x 2 + σ 2 ρ 2 ) ( t , ξ ( t ) ) = 1 2 m 2 ( t ) λ m ( t ) + f ( t , q ( t , x 2 ) ) = 1 2 ( m ( t ) + λ ) 2 + 1 2 λ 2 + f ( t , q ( t , x 2 ) ) (3.30)

where

f = u 2 ( t , ξ ( t ) ) k x g x u g ( u 2 + 1 2 u x 2 + σ 2 ρ 2 ) ( t , ξ ( t ) ) (3.31)

Modify the estimates:

k | x g x u | k g x L 2 u x L 2 e + 1 36 ( e 1 ) + 9 2 k 2 u x L 2 2

σ | g ( ρ 1 ) | σ g L 2 ρ 1 L 2 ( e + 1 36 ( e 1 ) + 9 2 ) σ ρ 1 L 2 2

The similar process to (3.16) leads to

f ( 1 8 σ ) ( e + 1 ) 36 ( e 1 ) + 3 ( e + 1 ) + 18 ( 1 + k 2 ) ( e 1 ) 4 ( e 1 ) ( u 0 H 1 2 + σ ρ 1 L 2 2 ) = C 2

From the above inequality and (3.27), we have 1 2 λ 2 C 2 < 0 , then

m ( t ) 1 2 ( m ( t ) + λ ) 2 + 1 2 λ 2 C 2 1 2 λ 2 C 2 < 0 , t [ 0 , T ) (3.32)

So m ( t ) is strictly decreasing in [ 0 , T ) .

If there exist global solutions, we will show that this leads to a contradiction. Let

t 1 = 2 ( 1 + | u 0 x ( x 0 ) | ) 2 C 2 λ 2

integrating (3.32) over [ 0 , t 1 ] yields

m ( t 1 ) = m ( 0 ) + 0 t 1 m ( t ) d t | u 0 x ( x 0 ) | + ( 1 2 λ 2 C 2 ) t 1 = 1 (3.33)

Hence we know m ( t ) m ( t 1 ) 1 , t [ t 1 , T ) .

From (3.32), we have

m ( t ) 1 2 ( m ( t ) + λ ) 2 (3.34)

Integrating (3.34) over [ t 1 , T ) and knowing m ( t 1 ) 1 , we get

1 m ( t ) + λ + 1 λ 1 1 m ( t ) + λ + 1 λ + m ( t 1 ) 1 2 ( t t 1 ) , t [ t 1 , T )

then

m ( t ) 1 1 2 ( t t 1 ) + 1 λ 1 λ , as t t 1 + 2 1 λ

Thus T t 1 + 2 1 λ is a contradiction with T = .

The proof of the Theorem 3.3 is complete.

Theorem 3.4 Let σ > 0 , and ( u , ρ ) be the solution of Equation (1.1) with initial data ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) , s 2 , and T is the maximal time of existence of the solution. If there exists x 0 S satisfies

ρ 0 ( x 0 ) = 0 , u 0 x ( x 0 ) = inf x R u 0 x ( x ) (3.35)

and

u 0 x ( x 0 ) < λ 2 + C 1 2 λ (3.36)

then the corresponding solution u of Equation (1.1) blows up in finite time when 0 < T < T ,where

T = 2 ( λ + u 0 x ( x 0 ) ) ( λ + u 0 x ( x 0 ) ) 2 ( λ 2 + C 1 2 )

Proof: From (3.16), we have

m ( t ) 1 2 ( m ( t ) + λ ) 2 + 1 2 ( λ 2 + C 1 2 ) , t [ 0 , T )

From (3.36), we have m ( 0 ) < 0 , m ( t ) is strictly decreasing on [ 0 , T ) and set

δ = 1 2 1 ( λ + u 0 x ( x 0 ) ) 2 ( 1 2 λ 2 + 1 2 C 1 2 ) ( 0 , 1 2 ) .

Because m ( t ) < m ( 0 ) = u 0 x ( x 0 ) < λ , then

m ( t ) 1 2 ( m ( t ) + λ ) 2 + 1 2 ( λ 2 + C 1 2 ) δ ( m ( t ) + λ ) 2

Similar discussion of the Theorem 3.3

m ( t ) λ + u 0 x ( x 0 ) 1 + δ t ( λ + u 0 x ( x 0 ) ) λ , t 1 λ δ + δ u 0 x ( x 0 )

Hence

0 < T < 2 ( λ + u 0 x ( x 0 ) ) ( λ + u 0 x ( x 0 ) ) 2 ( λ 2 + C 1 2 ) .

The proof of the theorem 3.4 is complete.

Next we will show the blow-up rate of solutions and the result shows: the blow-up rate is not affected by the weakly dissipation.

Theorem 3.5 (blow-up rate) Let σ > 0 , and ( u , ρ ) be the solution of Equation (1.1) with initial data ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) , s 2 , and T is the maximal time of existence of the solution. If T < , then

lim t T ( inf x S u x ( t , x ) ( T t ) ) = 2

Proof: Without loss of generality, assume s = 3 .

Set

M = ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( 7 ( e + 1 ) 4 ( e 1 ) + 1 + k 2 4 ) ( u 0 H 1 2 + σ ρ 1 L 2 2 ) (3.37)

From (3.30), we have

1 2 ( m ( t ) + λ ) 2 1 2 λ 2 M m ( t ) 1 2 ( m ( t ) + λ ) 2 + 1 2 λ 2 + M (3.38)

Because of lim t T ( m ( t ) + λ ) = , there exists t 0 ( 0 , T ) satisfies m ( t 0 ) + λ < 0 and ( m ( t 0 ) + λ ) 2 > 1 ε ( λ 2 + M ) , ε ( 0 , 1 2 ) . Since m is locally Lipschitz, m is absolutely continuous. We deduce that m is decreasing in [ t 0 , T ) and

( m ( t ) + λ ) 2 > 1 ε ( λ 2 + M ) (3.39)

According to (3.38) and (3.39)

1 2 ε d d t ( 1 m ( t ) + λ ) 1 2 + ε , t ( t 0 , T )

Integrating (3.39) over ( t , T ) with respect to t [ t 0 , T ) , notice that lim t T ( m ( t ) + λ ) = , then

( 1 2 ε ) ( T t ) ( 1 m ( t ) + λ ) ( 1 2 + ε ) ( T t ) , t ( t 0 , T )

Since ε is arbitrary, so

lim t T { m ( t ) ( T t ) + λ ( T t ) } = 2

That is lim t T m ( t ) ( T t ) = 2 , the blow-up rate of solutions of Equation (1.1) is not effected by the weakly dissipation.

4. Global Existence

In this section, we provide a sufficient condition for the global solution of Equation (1.1) in the case 0 < σ < 2 .

Theorem 4.1 Let 0 < σ < 2 , ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) with s > 3 2 , there exist a maximal time T > 0 and a unique solution ( u , ρ ) of Equation (1.1) with initial data. Assume that inf x S ρ 0 ( x ) > 0 , then

1) when 0 < σ 1 ,

| inf x S u x ( t , x ) | 1 inf x S ρ 0 ( x ) C 4 e C 3 t

| inf x S u x ( t , x ) | 1 inf x S ρ 0 σ 2 σ ( x ) C 4 1 2 σ e C 3 t 2 σ

2) when 1 < σ < 2 ,

| inf x S u x ( t , x ) | 1 inf x S ρ 0 σ 2 σ ( x ) C 4 1 2 σ e C 3 t 2 σ

| inf x S u x ( t , x ) | 1 inf x S ρ 0 ( x ) C 4 e C 3 t

where

C 3 = 1 + ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( 7 ( e + 1 ) 4 ( e 1 ) + 1 + k 2 4 ) ( u 0 H 1 2 + σ ρ 1 L 2 2 )

C 4 = 1 + u 0 x L 2 + ρ 0 L 2

Proof: It is sufficient to prove the desired results for s = 3 .

1) We will estimate the | inf x S u x ( t , x ) | .

From (3.22), we have

m ( t ) 0 , t [ 0 , T ) (4.1)

Let ζ ( t ) = ρ ( t , ξ ( t ) ) , thus we have

{ m ( t ) = 1 2 m 2 ( t ) λ m ( t ) + σ 2 ζ 2 ( t ) + f ( t , q ( t , x 2 ) ) ζ ( t ) = ζ ( t ) m ¯ ( t ) (4.2)

where f is defined as (3.15). The second Equation of (3.15) shows that ζ ( t ) and ζ ( 0 ) have the same sign. Hence ζ ( 0 ) = ρ ( 0 , ξ ) ( 0 ) > 0 .

Suppose 0 < σ 1 , define the function

w 1 ( t ) = ζ ( 0 ) ζ ( t ) + ζ ( 0 ) ζ ( t ) ( 1 + m 2 ( t ) ) (4.3)

which is positive on t [ 0 , T ) .

Differentiate w 1 ( t )

w 1 ( t ) = ζ ( 0 ) ζ ( t ) + ζ ( 0 ) ζ 2 ( t ) ( 1 + m 2 ( t ) ) ζ ( t ) + 2 ζ ( 0 ) ζ ( t ) m ( t ) m ( t ) = ζ ( 0 ) ζ ( t ) m ( t ) + ζ ( 0 ) ζ 2 ( t ) ( 1 + m 2 ( t ) ) ζ ( t ) m ( t ) + 2 ζ ( 0 ) ζ ( t ) m ( t ) ( 1 2 m 2 ( t ) λ m ( t ) + σ 2 ζ 2 ( t ) + f ) = ( σ 1 ) ζ ( 0 ) ζ ( t ) m ( t ) + ζ ( 0 ) ζ ( t ) m ( t ) 2 λ ζ ( 0 ) ζ ( t ) m 2 ( t ) + 2 ζ ( 0 ) ζ ( t ) m ( t ) f

( σ 1 ) ζ ( 0 ) ζ ( t ) m ( t ) + ζ ( 0 ) ζ ( t ) m ( t ) + 2 ζ ( 0 ) ζ ( t ) m ( t ) f = 2 ζ ( 0 ) ζ ( t ) m ( t ) ( 1 2 + f + σ 1 2 ζ 2 ( t ) ) ζ ( 0 ) ζ ( t ) ( 1 + m 2 ( t ) ) ( 1 + | f | ) C 3 w 1 ( t ) (4.4)

where C 3 = 1 + ( 3 σ + 2 ) ( e + 1 ) 4 ( e 1 ) + ( 7 ( e + 1 ) 4 ( e 1 ) + 1 + k 2 4 ) ( u 0 H 1 2 + σ ρ 1 L 2 2 ) .

Then

w 1 ( t ) w 1 ( 0 ) e C 3 t = ( ζ 2 ( 0 ) + 1 + m 2 ( 0 ) ) e C 3 t ( 1 + u 0 x L 2 + ρ 0 L 2 ) e C 3 t = C 4 e C 3 t (4.5)

where C 4 = 1 + u 0 x L 2 + ρ 0 L 2 .

From (4.3), we have

ζ ( 0 ) ζ ( t ) w 1 ( t ) , | ζ ( 0 ) | m ( t ) w 1 ( t ) (4.6)

then

| inf x S u x ( t , x ) | = | m ( t ) | w 1 ( t ) | ζ ( 0 ) | 1 inf x S ρ 0 ( x ) C 4 e C 3 t , t [ 0 , T )

Suppose 1 < σ < 2 , define the function

w 2 ( t ) = ζ σ ( 0 ) ζ 2 ( t ) + 1 + m 2 ( t ) ζ σ ( t ) (4.7)

Differentiate w 2 ( t )

w 2 ( t ) = 2 ζ σ ( 0 ) ζ σ ( t ) m ( t ) ( ( σ 1 ) ζ 2 ( t ) + σ 1 2 m 2 ( t ) λ m ( t ) + f + σ 2 ) ζ σ ( 0 ) ζ σ ( t ) ( 1 + m 2 ( t ) ) ( | f | + σ 2 ) ζ σ ( 0 ) ζ σ ( t ) ( 1 + m 2 ( t ) ) ( | f | + 1 ) C 3 w 2 ( t ) (4.8)

then

w 2 ( t ) w 2 ( 0 ) e C 3 t = ( ζ 2 ( 0 ) + 1 + m 2 ( 0 ) ) e C 3 t ( 1 + u 0 x L 2 + ρ 0 L 2 ) e C 3 t = C 4 e C 3 t (4.9)

Here we apply Young’s inequality a b a p p + b q q , for p = 2 σ , q = 2 2 σ .

w 2 ( t ) ζ σ ( 0 ) = ( ζ σ ( 2 σ ) 2 ) 2 σ + ( ( 1 + m 2 ) 2 σ 2 ζ σ ( 2 σ ) 2 ) 2 2 σ σ 2 ( ζ σ ( 2 σ ) 2 ) 2 σ + 2 σ 2 ( ( 1 + m 2 ) 2 σ 2 ζ σ ( 2 σ ) 2 ) 2 2 σ ( 1 + m 2 ) 2 σ 2 | m ( t ) | 2 σ

Hence

| inf x S u x ( t , x ) | ( w 2 ( t ) | ζ σ ( 0 ) | ) 1 2 σ 1 inf x S ρ 0 σ 2 σ ( x ) C 4 1 2 σ e C 3 t 2 σ , t [ 0 , T )

2) Next we control | sup x S u x ( t , x ) | .

Similarly,

{ m ¯ ( t ) = 1 2 m ¯ 2 ( t ) λ m ¯ ( t ) + σ 2 ζ ¯ 2 ( t ) + f ( t , q ( t , x 1 ) ) ζ ¯ ( t ) = ζ ¯ ( t ) m ¯ ( t )

Suppose 0 < σ 1 , define the function

w ¯ 1 ( t ) = ζ ¯ σ ( 0 ) ζ ¯ 2 ( t ) + 1 + m ¯ 2 ( t ) ζ ¯ σ ( t ) (4.10)

From (3.20) and (4.8), we obtain w ¯ 1 ( t ) C 3 w ¯ 1 ( t ) , then w ¯ 1 ( t ) C 4 e C 3 t .

Similarly, we get

w ¯ 1 ( t ) ζ ¯ σ ( 0 ) | m ¯ ( t ) | 2 σ

then

| sup x S u x ( t , x ) | ( w ¯ 1 ( t ) | ζ ¯ σ ( 0 ) | ) 1 2 σ 1 inf x S ρ 0 σ 2 σ ( x ) C 4 1 2 σ e C 3 t 2 σ , t [ 0 , T )

Suppose 1 < σ < 2 , define the function

w ¯ 2 ( t ) = ζ ¯ ( 0 ) ζ ¯ ( t ) + ζ ¯ ( 0 ) ζ ¯ ( t ) ( 1 + m ¯ 2 ( t ) ) (4.11)

From (3.20) and (4.4), we have w ¯ 2 ( t ) C 3 w ¯ 2 ( t ) , then w ¯ 2 ( t ) C 4 e C 3 t .

Hence

| sup x S u x ( t , x ) | = | m ¯ ( t ) | w ¯ 2 ( t ) | ζ ¯ ( 0 ) | 1 inf x S ρ 0 ( x ) C 4 e C 3 t , t [ 0 , T )

Theorem 4.2 Let 0 < σ < 2 , ( u 0 , ρ 0 1 ) H s ( S ) × H s 1 ( S ) with s 2 , there exist a maximal time T > 0 and a unique solution ( u , ρ ) of Equation (1.1) with initial data. If inf x S ρ 0 ( x ) > 0 , then T = and the the solution ( u , ρ ) is global.

Proof: By contradictory arguement, assume T < and the solution blows up. The Theorem 3.1 shows

0 T | u x ( t , x ) | L d t = (4.12)

The assumptions and the Theorem 4.1 show

| u x ( t , x ) | <

For all ( t , x ) [ 0 , T ) × S , that is a contradiction to (4.12).

The proof of Theorem 4.2 is complete.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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