Open Access

Global attractor for the generalized hyperelastic-rod equation

Boundary Value Problems20142014:209

https://doi.org/10.1186/s13661-014-0209-0

Received: 24 June 2014

Accepted: 26 August 2014

Published: 25 September 2014

Abstract

In this paper, we investigate the dynamical behavior of the initial boundary value problem for a class of generalized hyperelastic-rod equations. Under certain conditions, the existence of a global solution in H 3 is proved by using some prior estimates and the Galerkin method. Moreover, the existence of an absorbing set and a global attractor in H 2 is obtained.

Keywords

generalized hyperelastic-rod equationglobal solutionglobal attractor

1 Introduction

Camassa and Holm [1] first proposed a completely integrable dispersive shallow water equation as follows:
u t u x x t + 3 u u x + k u x = 2 u x u x x + u u x x x .
(1.1)

The C-H equation (1.1) was obtained by using an asymptotic expansion directly in the Hamiltonian for the Euler equations in the shallow water regime and possessed a bi-Hamiltonian structure and an infinite number of conservation laws in involution. Research on the C-H equation becomes a hot field due to its good properties [2]–[4] since it was proposed in 1993. Some equations also have similar characters to the C-H equation, which are called C-H family equations. Because of the wide applications in applied sciences such as physics, the C-H family equations have attracted much attention in recent years.

In 1998, Dai [5] derived the following hyperelastic-rod wave equation for finite-length and finite-amplitude waves in 1998 when doing research on hyperelastic compressible material:
v τ + σ 1 v v ξ + σ 2 v ξ ξ τ + σ 3 ( 2 v ξ v ξ ξ + v v ξ ξ ξ ) = 0 ,
(1.2)

where v ( ξ , τ ) represents the radial stretch relative to a pre-stressed state. The three coefficients σ 1 , σ 2 , and σ 3 are constants determined by the pre-stress and the material parameters, σ 1 0 , σ 2 < 0 , σ 3 0 .

If τ = 3 σ 2 σ 1 t and ξ = σ 2 x , then the following equation can be obtained by (1.2):
u t u x x t + 3 u u x = γ ( 2 u x u x x + u u x x x ) , γ = 3 σ 3 σ 1 σ 2 .
(1.3)

The constant γ is called the pre-stressed coefficient of the material rod.

There have been many research results as regards the hyperelastic-rod equation (1.3) [6]–[12], such as traveling-wave solutions, blow-up of solutions, well-posedness of solutions, the existence of weak solutions, the global solutions of Cauchy problem, the periodic boundary value problem, etc.

In 2005, Coclite et al.[13], [14] studied the following extension of (1.3):
u t u x x t + g ( u ) u x = γ ( 2 u x u x x + u u x x x ) , g ( 0 ) = 0 .
(1.4)

The existence of a global weak solution to (1.4) for any initial function u 0 belonging to H 1 ( R ) was obtained. They showed stability of the solution when a regularizing term vanishes based on a vanishing viscosity argument and presented a ‘weak equals strong’ uniqueness result.

It is easy to observe that if γ = 0 and g ( u ) = 2 k u + a , (1.4) becomes the BBM equation (1.5) [15], [16],
u t u x x t + a u x + k ( u 2 ) x = 0 .
(1.5)

Here γ = 1 and g ( u ) = 3 u + k , (1.4) is transformed into the C-H equation (1.1).

If γ = 1 and g ( u ) = ( b + 1 ) u , (1.4) can be changed to the D-P equation (1.6) [17]–[20],
u t u x x t + ( b + 1 ) u u x = 2 u x u x x + u u x x x .
(1.6)
Actually, the KdV equation [21], the C-H equation, the hyperelastic-rod wave equation etc. are all considered as special cases of the generalized hyperelastic-rod equation. So many researchers focused on this class of equations [22]–[24]. Among them, Holden and Raynaud [22] studied the following generalized hyperelastic-rod equation:
u t u x x t + f ( u ) x f ( u ) x x x + ( g ( u ) + 1 2 f ( u ) ( u x ) 2 ) x = 0 .
(1.7)

They considered the Cauchy problem of (1.7) and proved the existence of global and conservative solutions. It was shown that the equation was well-posed for initial data in H 1 ( R ) if one included a Radon measure corresponding to the energy of the system with the initial data.

However, there are few works with respect to the global asymptotical behaviors of solutions and the existence of global attractors, which are important for the study of the dynamical properties of general nonlinear dissipative dynamical systems [25]–[27]. Motivated by the references cited above, the goal of the present paper is to investigate the initial boundary problem of the following equation:
{ u t u x x t + [ G ( u ) ] x ( 4 ) = 2 u x u x x + u u x x x , t > 0 , x Ω , u ( 0 , x ) = u 0 ( x ) , x Ω ,
(1.8)

where Ω = [ 0 , L ] . We will study the dynamics behavior of (1.8) and discuss the existence of the global solution and the global attractor under the periodic boundary condition when G ( u ) satisfies the particular conditions.

The rest of this paper is organized as follows: Section 2 describes the main definitions used in this paper. The existence of the global solution is discussed in Section 3. The existence of the absorbing set is detailed in Section 4. Section 5 shows the existence of the global attractor.

2 Preliminaries

In this work, ( , ) stands for the inner product in the usual sense and represents the norm determined by the inner product, u H m ( Ω ) = D m u L 2 ( Ω ) . Apparently, this norm is equal to the natural norm in H m ( Ω ) . The following signs are adopted in this paper to express the norms of different spaces: u L 2 ( Ω ) = Δ | u | , D u L 2 ( Ω ) = Δ u , D m u L 2 ( Ω ) = Δ | D m u | .

The notion of bilinear operator is introduced, B ( u , v ) = u v , where is called a first order differential operator. Then we can get b ( u , v , ω ) = ( B ( u , v ) , ω ) = Ω ( u v ) ω d x .

The generalized hyperelastic-rod equation we studied is one-dimensional, and the operator acting on u ( x , t ) is not identically vanishing, so b ( u , v , ω ) = 0 cannot be found. However, the following formulas can be derived by the periodic boundary condition and formula of integration by parts:
( B ( u , v ) , ω ) = ( B ( u , ω ) , v ) ( B ( ω , u ) , v ) , ( B ( v , u ) , ω ) = ( B ( ω , v ) , u ) ( B ( v , ω ) , u ) ,

furthermore, ( B ( u , v ) , u ) = 2 ( B ( u , u ) , v ) , ( B ( u , v ) , u ) = 2 ( B ( v , u ) , u ) , so we get ( B ( u , u ) , v ) = ( B ( v , u ) , u ) and ( B ( u , u ) , u ) = 0 .

Suppose A = Δ is a second order differential operator, v = u + A u , then A is a self-adjoint operator, which possesses the eigenvalues like ( k 1 2 + k 2 2 ) ( 2 π L ) 2 , where k 1 , k 2 N 0 and k 1 2 + k 2 2 0 . λ 1 represents the smallest eigenvalue of A.

Based on the above statements, the initial boundary value problem of (1.8) under the periodic boundary condition can be rewritten as follows:
d v d t + [ G ( u ) ] x ( 4 ) + B ( u , v ) + 2 B ( v , u ) 3 B ( u , u ) = 0 ,
(2.1)
u ( x , 0 ) = u 0 ,
(2.2)
u ( 0 , t ) = u ( L , t ) .
(2.3)

In this work, we assume that H = { u u L 2 ( Ω )  and  u ( 0 , t ) = u ( L , t ) } , V = { u u L 2 ( Ω )  and  u ( 0 , t ) = u ( L , t ) } , G u ( u ) g 0 > 0 and | G u ( k ) ( u ) | C | u | 5 k , k = 1 , 2 , 3 , 4 , C is a constant.

3 The existence of global solution

Theorem 1

If u 0 V , G u ( u ) g 0 > 0 , and | G u ( k ) ( u ) | C | u | 5 k , k = 1 , 2 , 3 , 4 , then (2.1)-(2.3) possess the global solution u = u ( , u 0 ) C ( [ 0 , ) ; H 3 ( R ) ) C 1 ( [ 0 , ) ; H 2 ( R ) ) .

Proof

The Galerkin method is adopted to prove this theorem. Assume that { ϕ i } j = 1 is an orthogonal basis of H constituted by the eigenvectors of the operator A, H m = span { ϕ 1 , ϕ 2 , , ϕ m } , P m is the orthogonal projection from H to H m . Through the Galerkin method, we can obtain the following ordinary differential equations by (2.1), (2.2):
d v m d t + [ G ( u m ) ] x ( 4 ) + P m B ( u m , v m ) + 2 P m B ( v m , u m ) 3 P m B ( u m , u m ) = 0 ,
(3.1)
u m ( 0 ) = P m u ( 0 ) ,
(3.2)

where v m = u m + A u m . Considering the expressions of B ( u m , v m ) , B ( v m , u m ) , B ( u m , u m ) , according to the qualitative theories of ordinary differential equations, (3.1)-(3.2) have a unique solution u m in ( 0 , T m ) . In order to prove the existence of a global solution, we need to do some prior estimates as regards u m .

Taking the inner product of (3.1) with u m in Ω, we have
( d v m d t , u m ) + ( [ G ( u m ) ] x ( 4 ) , u m ) + P m ( B ( u m , v m ) , u m ) + 2 P m ( B ( v m , u m ) , u m ) 3 P m ( B ( u m , u m ) , u m ) = 0 .
By using integration by parts and the periodic boundary conditions, we get
( d v m d t , u m ) = 1 2 d d t ( Ω ( u m 2 + u m x 2 ) d x ) = 1 2 d d t ( | u m | 2 + u m 2 ) , P m ( B ( u m , v m ) , u m ) + 2 P m ( B ( v m , u m ) , u m ) 3 P m ( B ( u m , u m ) , u m ) = 0 , ( [ G ( u m ) ] x ( 4 ) , u m ) = Ω [ G ( u m ) ] x ( 4 ) u m d x = Ω G u m ( u m ) u m x u m x x x d x .
Moreover, in terms of Ω u m x u m x x x d x = Ω u m x x 2 d x 0 and G u ( u ) g 0 > 0 , we have
( [ G ( u m ) ] x ( 4 ) , u m ) g 0 Ω u m x u m x x x d x , 1 2 d d t ( | u m | 2 + u m 2 ) g 0 R u m x u m x x x d x 0 .
Employing Ω u m x u m x x x d x = Ω u m x x 2 d x again, the following formula can be obtained:
d d t ( | u m | 2 + u m 2 ) + 2 g 0 | A u m | 2 0 .
By the Poincaré inequality, | A u m | 2 > λ 1 u m 2 , we have
d d t ( | u m | 2 + u m 2 ) + g 0 λ 1 u m 2 + g 0 | A u m | 2 0 .
Let g 1 = min { g 0 λ 1 , g 0 } , then
d d t ( | u m | 2 + u m 2 ) + g 1 ( u m 2 + | A u m | 2 ) 0 .
(3.3)
Using the Poincaré inequality again, u m 2 > λ 1 | u m | 2 and | A u m | 2 > λ 1 u m 2 , (3.3) can be changed to
d d t ( | u m | 2 + u m 2 ) + g 1 λ 1 ( | u m | 2 + u m 2 ) 0 .
So we can obtain
| u m | 2 + u m 2 ( | u m ( 0 ) | 2 + u m ( 0 ) 2 ) exp { g 1 λ 1 t } | u m ( 0 ) | 2 + u m ( 0 ) 2 = Δ r 1 .
Integrating (3.3) over the interval [ t , t + r ] ,
g 1 t t + r ( u m ( s ) 2 + | A u m ( s ) | 2 ) d s r 1 .
(3.4)
Taking the inner product of (3.1) with A u m in Ω, we have
( d v m d t , A u m ) + ( [ G ( u m ) ] x ( 4 ) , A u m ) + P m ( B ( u m , v m ) , A u m ) + 2 P m ( B ( v m , u m ) , A u m ) 3 P m ( B ( u m , u m ) , A u m ) = 0 .
By using integration by parts and the periodic boundary conditions, we get
( d v m d t , A u m ) = 1 2 d d t ( u m 2 + | A u m | 2 ) , ( [ G ( u m ) ] x ( 4 ) , A u m ) = Ω [ G ( u m ) ] x ( 4 ) A u m d x = Ω G u m ( u m ) u m x u m x x x x x d x .
Moreover, Ω u m x u m x x x x x d x = Ω u m x x x 2 d x 0 and G u ( u ) g 0 > 0 . So
1 2 d d t ( u m 2 + | A u m | 2 ) + g 0 R u m x u m x x x x x d x + P m ( B ( u m , v m ) , A u m ) + 2 P m ( B ( v m , u m ) , A u m ) 3 P m ( B ( u m , u m ) , A u m ) 0 .
Employing Ω u m x u m x x x x x d x = Ω u m x x x 2 d x again, we obtain
1 2 d d t ( u m 2 + | A u m | 2 ) + g 0 | A u m | 2 + P m ( B ( u m , v m ) , A u m ) + 2 P m ( B ( v m , u m ) , A u m ) 3 P m ( B ( u m , u m ) , A u m ) 0 .
(3.5)
By computing, we have
P m ( B ( u m , v m ) , A u m ) + 2 P m ( B ( v m , u m ) , A u m ) 3 P m ( B ( u m , u m ) , A u m ) = 2 P m ( B ( A u m , u m ) , A u m ) + P m ( B ( u m , A u m ) , A u m ) .
According to the Agmon inequality when n = 1 , φ L c φ L 2 1 2 φ H 1 1 2 , where c is a constant which only depends on Ω. Furthermore, we can get
P m ( B ( A u m , u m ) , A u m ) u m L | A u m | 2 c 1 u m 1 2 | A u m | 5 2 , P m ( B ( u m , A u m ) , A u m ) 1 2 u m L | A u m | 2 c 2 2 u m 1 2 | A u m | 5 2 .
So the following inequality can be gotten by (3.5):
1 2 d d t ( u m 2 + | A u m | 2 ) + g 0 | A u m | 2 c 1 u m 1 2 | A u m | 5 2 + c 2 2 u m 1 2 | A u m | 5 2 .
By the Poincaré inequality, | A u m | 2 > λ 1 | A u m | 2 , together with g 1 = min { g 0 λ 1 , g 0 } , we have
1 2 d d t ( u m 2 + | A u m | 2 ) + g 1 2 ( | A u m | 2 + | A u m | 2 ) c 1 u m 1 2 | A u m | 5 2 + c 2 2 u m 1 2 | A u m | 5 2 .
By the Young inequality, the following inequality can be obtained:
1 2 d d t ( u m 2 + | A u m | 2 ) + g 1 2 ( | A u m | 2 + | A u m | 2 ) 1 2 g 1 λ 1 ( u m 2 + | A u m | 2 ) + c 3 u m | A u m | ( u m 2 + | A u m | 2 ) ,
(3.6)
where
c 3 = [ min { c 1 , c 2 2 } ] 2 2 g 1 λ 1 ,
and, by using the Poincaré inequality, we have
d d t ( u m 2 + | A u m | 2 ) 2 c 3 u m | A u m | ( u m 2 + | A u m | 2 ) .
Using the Young inequality again, we can further get
d d t ( u m 2 + | A u m | 2 ) c 3 ( u m 2 + | A u m | 2 ) 2 .
(3.7)
Denoting y = u m ( s ) 2 + | A u m ( s ) | 2 , g = c 3 ( u m ( s ) 2 + | A u m ( s ) | 2 ) . According to (3.4),
t t + r y d s r 1 g 1 , t t + r g d s c 3 r 1 g 1 .
Based on the uniform Grownwall inequality, we have
u m 2 + | A u m | 2 r 1 r g 1 exp { c 3 r 1 g 1 } = Δ r 2 , t > t 0 + r ,
(3.8)

where r, r 1 , and c 3 are nonnegative constants.

Integrating (3.6) over the interval [ t , t + r ] to obtain
1 2 g 1 t t + r ( | A u m | 2 + | A u m | 2 ) d s t t + r ( 1 2 g 1 λ 1 ( u m 2 + | A u m | 2 ) + c 3 2 ( u m 2 + | A u m | 2 ) 2 ) d s + ( u m 2 + | A u m | 2 ) 1 2 ( g 1 λ 1 r 2 + c 3 r 2 2 ) r + r 2 = Δ r 3 .
(3.9)
Taking the inner product of (3.1) with A 2 u m in Ω, together with integration by parts, the periodic boundary conditions, and G u ( u ) g 0 > 0 , we have
1 2 d d t ( | A u m | 2 + | A u m | 2 ) + g 0 | A 2 u m | 2 + P m ( B ( u m , v m ) , A 2 u m ) + 2 P m ( B ( v m , u m ) , A 2 u m ) 3 P m ( B ( u m , u m ) , A 2 u m ) 0 .
Through the Young inequality, the Hölder inequality and the Poincaré inequality, we deduce that
1 2 d d t ( | A u m | 2 + | A u m | 2 ) + g 1 2 ( | A u m | 2 + | A 2 u m | 2 ) 1 2 g 1 λ 1 ( | A u m | 2 + | A u m | 2 ) + c 4 u m | A u m | ( | A u m | 2 + | A u m | 2 ) .
According to the Poincaré inequality and the Young inequality again, we have
d d t ( | A u m | 2 + | A u m | 2 ) 2 c 4 u m | A u m | ( | A u m | 2 + | A u m | 2 ) c 4 ( u m 2 + | A u m | 2 ) ( | A u m | 2 + | A u m | 2 ) .
From (3.4) and (3.9), we get
c 4 t t + r ( u m ( s ) 2 + | A u m ( s ) | 2 ) d s r 1 c 4 g 1 , t t + r ( | A u m | 2 + | A u m | 2 ) d s 2 r 3 g 1 .
Based on the uniform Grownwall inequality, we have
| A u m | 2 + | A u m | 2 2 r 3 r g 1 exp { r 1 c 4 g 1 } = Δ r 4 , t > t 0 .
(3.10)

Overall, | u m | 2 r 1 , u m 2 r 2 , | A u m | 2 r 3 , | A u m | 2 r 4 , that is, | v m | 2 r 1 + r 3 , v m 2 r 2 + r 4 .

According to the qualitative theories of ordinary differential equations, (3.1)-(3.2) have a global solution u m .

From the above discussion, we have
| P m B ( u m , v m ) | | u m | v m ( r 1 ( r 2 + r 4 ) ) 1 2 = Δ r 5 , | P m B ( v m , u m ) | | v m | u m ( r 2 ( r 1 + r 3 ) ) 1 2 = Δ r 6 , | P m B ( u m , u m ) | | u m | u m ( r 1 r 2 ) 1 2 = Δ r 7 .
Then (3.1) can be rewritten as
d v m d t = 3 P m B ( u m , u m ) P m B ( u m , v m ) 2 P m ( v m , u m ) [ G ( u m ) ] x ( 4 ) .
Because of | G u ( k ) ( u ) | C | u | 5 k , k = 1 , 2 , 3 , 4 ,
| d v m d t | 3 | P m B ( u m , u m ) | + | P m B ( u m , v m ) | + 2 | P m B ( v m , u m ) | + | P m B ( [ G ( u m ) x ] u , u m ) | 3 r 7 + r 5 + 2 r 6 + h ( C , r 1 , r 2 , r 3 , r 4 ) u m 3 r 7 + r 5 + 2 r 6 + h r 2 1 2 = Δ k ,
(3.11)

where h is a constant which depends on C, r 1 , r 2 , r 3 , r 4 .

According to the Aubin compactness theorem, we conclude that there is a convergent subsequence u m , so that u m u , or equivalently v m v . Suppose that u m and v m are replaced by u m and v m , then we need to prove that u, v satisfy (2.1).

Selecting ω D ( A ) randomly, | ω | is bounded as we see from the above discussion. By the ordinary differential equation (3.1), we have
( v m ( t ) , ω ) + t 0 t ( G ( u m ( s ) ) , A 2 ω ) d s + t 0 t ( B ( u m ( s ) , v m ( s ) ) , P m ω ) d s + 2 t 0 t ( B ( v m ( s ) , u m ( s ) ) , P m ω ) d s 3 t 0 t ( B ( u m ( s ) , u m ( s ) ) , P m ω ) d s = ( v m ( t 0 ) , ω ) .
Obviously, lim m + | P m ω ω | = 0 , lim m + | P m A 2 ω A 2 ω | = 0 , according to the convergence,
lim m + t 0 t ( G ( u m ( s ) ) , A 2 ω ) d s = t 0 t ( G ( u ( s ) ) , A 2 ω ) d s , | t 0 t ( B ( u m ( s ) , v m ( s ) ) , P m ω ) d s t 0 t ( B ( u ( s ) , v ( s ) ) , ω ) d s | | t 0 t ( B ( u m ( s ) , v m ( s ) ) , P m ω ω ) d s | + | t 0 t ( B ( u m ( s ) u ( s ) , v m ( s ) ) , ω ) d s | + | t 0 t ( B ( u ( s ) , v m ( s ) v ( s ) ) , ω ) d s | ,
where
I m ( 1 ) = | t 0 t ( B ( u m ( s ) , v m ( s ) ) , P m ω ω ) d s | t 0 t | B ( u m ( s ) , v m ( s ) ) | | P m ω ω | d s .
Considering the boundness of | B ( u m ( s ) , v m ( s ) ) | , so I m ( 1 ) 0 ,
I m ( 2 ) = | t 0 t ( B ( u m ( s ) u ( s ) , v m ( s ) ) , ω ) d s | t 0 t | B ( u m ( s ) u ( s ) , v m ( s ) ) | | ω | d s t 0 t | u m ( s ) u ( s ) | v m ( s ) | ω | d s 0 , I m ( 3 ) = | t 0 t ( B ( u ( s ) , v m ( s ) v ( s ) ) , ω ) d s | t 0 t | B ( u ( s ) , v m ( s ) v ( s ) ) | | ω | d s t 0 t | u ( s ) | v m ( s ) v ( s ) | ω | d s 0 , | t 0 t ( B ( v m ( s ) , u m ( s ) ) , P m ω ) d s t 0 t ( B ( v ( s ) , u ( s ) ) , ω ) d s | | t 0 t ( B ( v m ( s ) , u m ( s ) ) , P m ω ω ) d s | + | t 0 t ( B ( v m ( s ) v ( s ) , u m ( s ) ) , ω ) d s | + | t 0 t ( B ( v ( s ) , u m ( s ) u ( s ) ) , ω ) d s | = I m ( 4 ) + I m ( 5 ) + I m ( 6 ) ,
where
I m ( 4 ) = | t 0 t ( B ( v m ( s ) , u m ( s ) ) , P m ω ω ) d s | t 0 t | B ( v m ( s ) , u m ( s ) ) | | P m ω ω | d s 0 , I m ( 5 ) = | t 0 t ( B ( v m ( s ) v ( s ) , u m ( s ) ) , ω ) d s | t 0 t | B ( v m ( s ) v ( s ) , u m ( s ) ) | | ω | d s t 0 t | v m ( s ) v ( s ) | u m ( s ) | ω | d s 0 , I m ( 6 ) = | t 0 t ( B ( v ( s ) , u m ( s ) u ( s ) ) , ω ) d s | t 0 t | B ( v ( s ) , u m ( s ) u ( s ) ) | | ω | d s t 0 t | v ( s ) | u m ( s ) u ( s ) | ω | d s 0 , | t 0 t ( B ( u m ( s ) , u m ( s ) ) , P m ω ) d s t 0 t ( B ( u ( s ) , u ( s ) ) , ω ) d s | | t 0 t ( B ( u m ( s ) , u m ( s ) ) , P m ω ω ) d s | + | t 0 t ( B ( u m ( s ) u ( s ) , u m ( s ) ) , ω ) d s | + | t 0 t ( B ( u ( s ) , u m ( s ) u ( s ) ) , ω ) d s | = I m ( 7 ) + I m ( 8 ) + I m ( 9 ) 0 .
From the above discussion, we can deduce that u, v satisfy the following equation:
( v ( t ) , ω ) + t 0 t ( G ( u ( s ) ) , A 2 ω ) d s + t 0 t ( B ( u ( s ) , v ( s ) ) , ω ) d s + 2 t 0 t ( B ( v ( s ) , u ( s ) ) , ω ) d s 3 t 0 t ( B ( u ( s ) , u ( s ) ) , ω ) d s = ( v ( t 0 ) , ω ) .

Above all, u is the solution of (2.1)-(2.3), that is, their global solution exists. □

4 The existence of the absorbing set

Theorem 2

If u 0 V , the semi-group of the solution to (2.1)-(2.3), i.e. S ( t ) : H 2 ( Ω ) H 2 ( Ω ) , u ( t ) = S ( t ) u 0 , has an absorbing set.

Proof

Taking the inner product of (2.1) with u in Ω we obtain
( d v d t , u ) + ( A 2 G ( u ) , u ) + 2 ( B ( v , u ) , u ) + ( B ( u , v ) , u ) 3 ( B ( u , u ) , u ) = 0 .
Because of G u ( u ) g 0 , g 0 > 0 , we have
1 2 d d t ( | u | 2 + u 2 ) + g 0 | A u | 2 0 .
By the Poincaré inequality, | A u | 2 > λ 1 u 2 , we get
d d t ( | u | 2 + u 2 ) + g 0 λ 1 u 2 + g 0 | A u | 2 0 .
Let g 1 = min { g 0 λ 1 , g 0 } , then
d d t ( | u | 2 + u 2 ) + g 1 ( u 2 + | A u | 2 ) 0 .
(4.1)
Using the Poincaré inequality, u 2 > λ 1 | u | 2 and | A u | 2 > λ 1 u 2 , (4.1) is changed to
d d t ( | u | 2 + u 2 ) + g 1 λ 1 ( | u | 2 + u 2 ) 0 .
By the Grownwall inequality, we obtain
| u | 2 + u 2 ( | u ( 0 ) | 2 + u ( 0 ) 2 ) exp { g 1 λ 1 t } .
(4.2)

It is easy to see that | u ( x , t ) | and u ( x , t ) are uniformly bounded from (4.2). In other words, the semi-group S ( t ) is uniformly bounded in L 2 ( Ω ) and H 1 ( Ω ) .

Integrating (4.1) over the interval [ t , t + r ] , we have
lim s + t t + r ( u ( x , s ) 2 + | A u ( x , s ) | 2 ) d s 1 g 1 ( | u 0 | 2 + u 0 2 ) , t t + s ( u ( x , s ) 2 + | A u ( x , s ) | 2 ) d s 1 g 1 ( | u 0 | 2 + u 0 2 ) ρ 0 g 1 .

If B ( 0 , ρ ) is an open ball in L 2 ( Ω ) and H 1 ( Ω ) whose radius is ρ, it is easy to calculate that S ( t ) u 0 B ( 0 , ρ ) when t t 0 , t 0 = max ( 1 g 1 λ 1 ln ρ ρ 0 , 0 ) .

We will make a uniform estimate of (2.1)-(2.3) in H 2 ( Ω ) .

Taking the inner product of (2.1) with Au in Ω, and denoting F ( u , A u ) = ( B ( u , v ) , A u ) + 2 ( B ( v , u ) , A u ) 3 ( B ( u , u ) , A u ) , we have
1 2 d d t ( u 2 + | A u | 2 ) + g 0 | A u | 2 + F ( u , A u ) 0 .
(4.3)
By computing, F ( u , A u ) = 2 ( B ( A u , u ) , A u ) + ( B ( u , A u ) , A u ) , through the Agmon inequality, we get
| ( B ( A u , u ) , A u ) | u L ( Ω ) A u L 2 ( Ω ) 2 c 5 u 1 2 | A u | 5 2 , | ( B ( u , A u ) , A u ) | 1 2 u L ( Ω ) A u L 2 ( Ω ) 2 c 6 2 u 1 2 | A u | 5 2 .
So we have
| F ( u , A u ) | 2 | ( B ( A u , u ) , A u ) | + | ( B ( u , A u ) , A u ) | c 7 u 1 2 | A u | 1 2 ( u 2 + | A u | 2 ) 1 2 g 1 λ 1 ( u 2 + | A u | 2 ) + c 8 u | A u | ( u 2 + | A u | 2 ) ,
where
c 7 = max { 2 c 5 , c 6 2 } , c 8 = c 7 2 2 g 1 λ 1 .
By the Poincaré inequality, | A u | 2 > λ 1 | A u | 2 , g 1 = min { g 0 λ 1 , g 0 } , and (4.3) it can be deduced that
1 2 d d t ( u 2 + | A u | 2 ) + g 1 2 ( | A u | 2 + | A u | 2 ) 1 2 g 1 λ 1 ( u 2 + | A u | 2 ) + c 8 u | A u | ( u 2 + | A u | 2 ) .
(4.4)
Employing the Poincaré inequality again, we obtain
d d t ( u 2 + | A u | 2 ) 2 c 8 u | A u | ( u 2 + | A u | 2 ) .
Using the Young inequality, the following inequality can be gotten:
d d t ( u 2 + | A u | 2 ) c 8 ( u 2 + | A u | 2 ) 2 .
By denoting y = u 2 + | A u | 2 , g = c 8 ( u 2 + | A u | 2 ) ,
t t + r y ( s ) d s 1 g 1 ( | u 0 | 2 + u 0 2 ) = α 1 , t t + r g ( s ) d s c 8 g 1 ( | u 0 | 2 + u 0 2 ) = c 8 α 1 .
According to the uniform Grownwall inequality, we get
u 2 + | A u | 2 α 1 r exp { c 8 α 1 } , t > t 0 + r ,

where r, α 1 , c 8 are nonnegative constants. Let ρ 1 = α 1 r exp { c 8 α 1 } , and then | A u | 2 ρ 1 . In other words, B ( 0 , ρ 1 ) is the attracting set of S ( t ) in H 2 ( Ω ) . This completes the proof of Theorem 2. □

5 The existence of global attractor

Theorem 3

If u 0 V , the semi-group of the solution S ( t ) to (2.1)-(2.3) has a global attractor in H 2 ( Ω ) .

Proof

Based on the proof of Theorem 2, we only need to prove that S ( t ) is a completely continuous operator, thus the existence of global attractor can be proved.

Taking the inner product of (2.1) with t 2 Δ A u in Ω, furthermore, according to integration by parts and the Green formula, we have
( d v d t , t 2 Δ A u ) + ( A 2 G ( u ) , t 2 Δ A u ) + ( B ( u , v ) , t 2 Δ A u ) + 2 ( B ( v , u ) , t 2 Δ A u ) 3 ( B ( u , u ) , t 2 Δ A u ) = 0 , ( d v d t , t 2 Δ A u ) = 1 2 d d t ( | t A u | 2 + | t A u | 2 ) + ( | t 1 2 A u | 2 + | t 1 2 A u | 2 ) .
(5.1)
By the assumption of G u ( u ) g 0 , g 0 > 0 , we can get
( A 2 G ( u ) , t 2 Δ A u ) g 0 | t Δ A u | 2 , ( B ( u , v ) , t 2 Δ A u ) + 2 ( B ( v , u ) , t 2 Δ A u ) 3 ( B ( u , u ) , t 2 Δ A u ) = ( B ( u , A u ) , t 2 Δ A u ) + 2 ( B ( A u , u ) , t 2 Δ A u ) = 5 2 Ω t 2 u x u x x x 2 d x
and through the Agmon inequality and the Poincaré inequality, we obtain
| ( B ( u , v ) , t 2 Δ A u ) + 2 ( B ( v , u ) , t 2 Δ A u ) 3 ( B ( u , u ) , t 2 Δ A u ) | = | 5 2 Ω t 2 u x u x x x 2 d x | 5 2 | Ω t 2 u x u x x x 2 d x + Ω t 2 u x u x x 2 d x | 5 2 c 9 u 1 2 | A u | 1 2 ( | t A u | 2 + | t A u | 2 ) .
By (5.1), we can get the following inequality:
d d t ( | t A u | 2 + | t A u | 2 ) + 2 g 0 | t Δ A u | 2 5 c 9 u 1 2 | A u | 1 2 ( | t A u | 2 + | t A u | 2 ) + 2 ( | t 1 2 A u | 2 + | t 1 2 A u | 2 ) .
Based on the Poincaré inequality: | t Δ A u | 2 > λ 1 | t A u | 2 , | t A u | 2 > λ 1 | t A u | 2 , g 1 = min { g 0 λ 1 , g 0 } , and the Young inequality, we have
d d t ( | t A u | 2 + | t A u | 2 ) + g 1 λ 1 ( | t A u | 2 + | t A u | 2 ) 5 c 9 u 1 2 | A u | 1 2 ( | t A u | 2 + | t A u | 2 ) + 2 ( | t 1 2 A u | 2 + | t 1 2 A u | 2 ) g 1 λ 1 ( | t A u | 2 + | t A u | 2 ) + c 10 u | A u | ( | t A u | 2 + | t A u | 2 ) + c 11 ( | A u | 2 + | A u | 2 ) ,
(5.2)
where
c 10 = 25 c 9 2 4 g 1 λ 1 , c 11 = 8 g 1 λ 1 .
By (4.4), the following inequality can be obtained:
d d t ( u 2 + | A u | 2 ) + g 1 ( | A u | 2 + | A u | 2 ) g 1 λ 1 ( u 2 + | A u | 2 ) + c 8 ( u 2 + | A u | 2 ) 2 , t t 0 .
Integrating the above inequality over the interval [ t , t + r ] , we get
t t + r ( | A u ( x , s ) | 2 + | A u ( x , s ) | 2 ) d s ( λ 1 ρ 1 + c 8 ρ 1 2 g 1 ) r + ρ 1 g 1 .
Equation (5.2) can be rewritten as follows:
d d t ( | t A u | 2 + | t A u | 2 ) c 10 u | A u | ( | t A u | 2 + | t A u | 2 ) + c 11 ( | A u | 2 + | A u | 2 ) .
By denoting ( λ 1 ρ 1 + c 8 ρ 1 2 g 1 ) r + ρ 1 g 1 = α 2 ( λ 1 , ρ 1 , g 1 ) , we have
t t + r c 10 u ( x , s ) | A u ( x , s ) | d s c 10 2 t t + r ( u ( x , s ) 2 + | A u ( x , s ) | 2 ) d s c 10 ρ 0 2 g 1 = Δ α 3 ( ρ 0 , g 1 ) , t t + r ( | s A u ( x , s ) | 2 + | s A u ( x , s ) | 2 ) ( t + r ) 2 α 2 = Δ α 4 ( λ 1 , ρ 1 , g 1 ) .
By the uniform Gronwall inequality, we have
| t A u | 2 + | t A u | 2 ( α 4 r + c 11 α 2 ) exp ( α 3 ) .

Let ( α 4 r + c 11 α 2 ) exp ( α 3 ) = E ( λ 1 , ρ 1 , g 1 , t ) , then we can obtain | A u | < E ( λ 1 , ρ 1 , g 1 , t ) t .

Therefore, we can conclude that S ( t ) is equicontinuous. From the Ascoli-Arzela theorem, S ( t ) is a completely continuous operator. Thus, we have proved that S ( t ) has a global attractor in H 2 ( Ω ) . □

Declarations

Acknowledgements

This work was supported by the National Natural Science Foundation of China (61374194), the National Natural Science Foundation of China (61403081), the Natural Science Foundation of Jiangsu Province (BK20140638), the China Postdoctoral Science Foundation (2013M540405) and the Special Program of China Postdoctoral Science Foundation (2014T70454).

Authors’ Affiliations

(1)
School of Automation, Southeast University, Nanjing, China
(2)
Key Laboratory of Measurement and Control of CSE, Ministry of Education, Southeast University, Nanjing, China
(3)
Faculty of Science, Jiangsu University, Zhenjiang, China
(4)
National Laboratory of Industrial Control Technology, Institute of Cyber-Systems and Control, Zhejiang University, Hangzhou, China

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