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 JAMP  Vol.8 No.10 , October 2020
Blow-Up for a Periodic Two-Component Camassa-Holm Equation with Generalized Weakly Dissipation
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.

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.

Cite this paper: 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.
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