This paper investigates the following primal value problems of a system of generalized Kirchhoff-type coupled equations:

{ u t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m u + β ( − Δ ) 2 m u t + g 1 ( u , v ) = f 1 ( x ) , (1) v t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m v + β ( − Δ ) 2 m v t + g 2 ( u , v ) = f 2 ( x ) ,     (2) u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x ∈ Ω ,   (3) v ( x , 0 ) = v 0 ( x ) , v t ( x , 0 ) = v 1 ( x ) , x ∈ Ω ,   (4) ∂ i u ∂ n i = 0 , ∂ i v ∂ n i = 0   ( i = 0 , 1 , 2 , ⋯ , 2 m ) .     (5)

where Ω is a bounded region with a smooth boundary in R n , ∂ Ω represents the boundary of Ω , u 0 ( x ) , u 1 ( x ) and v 0 ( x ) , v 1 ( x ) are known functions, where g j ( u , v ) , f j ( u , v )   ( j = 1 , 2 ) are nonlinear terms and external interference terms, respectively, and are known functions on Ω × ( 0 , T ) , β is the normal number, M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) is a non-negative first-order continuous derivative function, and m > 1 is the normal number, ‖ D m u ‖ p p = ∫ Ω | D m u | p d x .The innovation of this article is that the rigid term is changed from M ( ‖ D m u ‖ 2 + ‖ D m v ‖ 2 ) to M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) , and we mainly make appropriate assumptions about this, and then use the Holder’s inequality, the Young’s inequality, the Poincare’s inequality, interpolation inequality, and the Gronwall’s inequality to obtain the required a priori estimate.

Recently, Yang Zhijian [1] studied the long-term behavior of kirchhoff-type equations with strong damping on R n , demonstrating that the related continuous semigroup has a connected, fractal dimension and Hausdorff dimension of the global attractor in the equation

u t t − M ( ‖ ∇ u ‖ 2 ) Δ u − Δ u t + u + u t + g ( u , v ) = f ( x ) ,   in   R N × R + .

At the same time, Yang Zhijian [2] processed a class of Kirchhoff-type global attractors and Hausdorff dimensions, and obtained the global attractors, regularities and Hausdorff-dimensional equations of the Kirchhoff type produced in a class of elastoplastic flows

u t t − div { σ ( | ∇ u | 2 ) ∇ u } − Δ u t + Δ 2 u + h ( u t ) + g ( u ) = f ( x ) ,   in   Ω × R + .

In addition, Xiaoming Fan, and Shengfan Zhou [3] also demonstrated the presence of a tight-core section during the nonlinear vibration of the nonlinear elastic string in which the non-degraded Kirchhoff type strong damping wave equation simulates, and obtained an accurate estimate of the upper boundary of the Kirchhoff type of the kernel section in the equation

u t t − α Δ u t − ( β + γ ( ∫ Ω | ∇ u | 2 d x ) ρ ) Δ u + h ( u t ) + f ( u , t ) = g ( x , t ) , x ∈ Ω , t > τ .

In addition, Lin Guoguang and Gao Yunlong [4] studied the long-term behavior of a class of strongly damped high-order Kirchhoff-type equations for solving the initial edge value problem

u t t + ( − Δ ) m u t + ( α + β ‖ ∇ m u ‖ 2 ) q ( − Δ ) m u + g ( u ) = f ( x ) , ( x , t ) ∈ Ω × [ 0 , + ∞ ) .

They used the Galerkin method to obtain the understanding of the uniqueness of existence, and based on the attractor theorem to obtain the existence of the global attractor at H 0 m ( Ω ) × L 2 ( Ω ) , and established an estimate of the Hausdorff dimension of the attractor.

In paper [5], Lin Guoguang and Zhou Chunmeng studied a class of high-order strong-damped Kirchhoff equations on the initial edge value problems

{ u t t + M ( ‖ ∇ m u ‖ p p ) ( − Δ ) 2 m u + β ( − Δ ) 2 m u t + | u | ρ u = f ( x ) , α > 0 , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x ∈ Ω , u ( x , t ) = ∂ j u ∂ v j , j = 1 , 2 , ⋯ , 2 m − 1 , x ∈ ∂ Ω .

Wherein m > 0 p ≥ 2 , Ω ⊂ R n is a bounded region with a smooth boundary ∂ Ω , β > 0 is a dissipation coefficient, β ( − Δ ) 2 m u t is a strong dissipation term, | u | ρ u is a nonlinear term, and ρ ≥ − 1 , f ( x ) is an external force interference term, when studying rigid term M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) , the method in the literature is referred to.

Guoguang Lin and Lingjuan Hu [6] studied a class of nonlinearly coupled Kirchhoff equations with strong damping

{ u t t + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m u + β ( − Δ ) m u t + g 1 ( u , v ) = f 1 ( x ) , v t t + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m v + β ( − Δ ) m v t + g 2 ( u , v ) = f 2 ( x ) , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x ∈ Ω , v ( x , 0 ) = v 0 ( x ) , v t ( x , 0 ) = v 1 ( x ) , x ∈ Ω , ∂ i u ∂ n i = 0 , ∂ i v ∂ n i = 0 ( i = 0 , 1 , 2 , ⋯ , 2 m − 1 ) x ∈ ∂ Ω .

where Ω is a bounded region with a smooth boundary in R n , ∂ Ω represents the boundary of Ω , g j ( u , v )   ( j = 1 , 2 ) is a nonlinear source term, f 1 ( x ) , f 2 ( x ) is an external force interference term, and β ( − Δ ) m u , β ( − Δ ) m v ( β ≥ 0 ) is a strong dissipation terms.

More research on the Kirchhoff equations see [7] - [13].

Using the Rellich-Kondrachov compact embedding theorem, it is obtained that the solution semigroup S ( t ) generated by the Kirchhoff equation has a family of global attractors in space E k = V 2 m + k × V k × V 2 m + k × V k ( k = 0 , 1 , ⋯ , 2 m ) ; then proves that the solution semigroup S ( t ) has Fréchet differentiability on space E k ; Dimensional estimation of the family of global attractors yields that both the Hausdorff and Fractal dimensions are finite, and that the Fractal dimension does not exceed twice the Hausdorff dimension.

For narrative convenience, we introduce the following symbols and assumptions:

Set ∇ = D . Consider the Hilbert space V α = D ( ( − Δ ) α / 2 ) , α ∈ R , whose inner product and norm are ( • , • ) V α = ( ( − Δ ) α / 2 , ( − Δ ) α / 2 ) and ‖ • ‖ V α = ‖ ( − Δ ) α / 2 ‖ , respectively. Apparently

V 0 = L 2 ( Ω ) , V 2 m = H 2 m ( Ω ) ∩ H 0 1 ( Ω ) , V 2 m + k = H 2 m + k ( Ω ) ∩ H 0 1 ( Ω ) , V k = H k ( Ω ) ∩ H 0 1 ( Ω ) , E 0 = V 2 m × V 0 × V 2 m × V 0 , E k = V 2 m + k × V k × V 2 m + k × V k , ( k = 1 , 2 , ⋯ , 2 m ) .

The assumption is as follows:

(H1) Let M ( s ) be a continuous function on interval D 1 ( D 1 ∈ Ω ) , and M ( s ) ∈ C 1 ( R + ) :

1) 1 ≤ μ 0 ≤ M ( s ) ≤ μ 1 , set M ( s ) = M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) .

2) { When   d d t ‖ D 2 m u ‖ 2 ≥ 0 ,   1 2 M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) d d t ‖ D 2 m u ‖ 2 ≥ 1 2 μ 0 d d t ‖ D 2 m u ‖ 2 , When   d d t ‖ D 2 m u ‖ 2 < 0 ,   1 2 M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) d d t ‖ D 2 m u ‖ 2 ≥ 1 2 μ 1 d d t ‖ D 2 m u ‖ 2 .

3) { When   d d t ‖ Δ 2 m u ‖ 2 ≥ 0 ,   1 2 M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) d d t ‖ Δ 2 m u ‖ 2 ≥ 1 2 μ 0 d d t ‖ Δ 2 m u ‖ 2 , When   d d t ‖ Δ 2 m u ‖ 2 < 0 ,   1 2 M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) d d t ‖ Δ 2 m u ‖ 2 ≥ 1 2 μ 1 d d t ‖ Δ 2 m u ‖ 2 .

(H2) For any u , v ∈ V , J ( u , v ) = ∫ Ω [ G 1 ( u , v ) + G 2 ( u , v ) ] d x , G 1 ( u , v ) = ∫ 0 u g 1 ( ς , v ) d ς , G 2 ( u , v ) = ∫ 0 v g 2 ( u , η ) d η .

Then for any μ ≥ 0 , the existence of c > 0 , c μ ≥ 0 , c ′ μ ≥ 0 , makes

( g 1 ( u , v ) , u ) + ( g 2 ( u , v ) , v ) − c J ( u , v ) + μ ( ‖ D m u ‖ 2 + ‖ D m v ‖ 2 ) ≥ − c μ ;

2 J ( u , v ) + ε 2 ( ‖ u ‖ 2 + ‖ v ‖ 2 ) ≥ − 2 c ′ μ .

(H3) g j ( u , v ) ( j = 1 , 2 ) ∈ C 1 ( R ) is a differentiable non-subtractive function that makes

| g j ( u , v ) | ≤ c ( 1 + | u | r j + | v | r j ) , | g j u ( u , v ) | ≤ c ( 1 + | u | r j − 1 + | v | r j ) , | g j v ( u , v ) | ≤ c ( 1 + | u | r j + | v | r j − 1 ) .

(H4) Let 0 < κ 1 , κ 2 < 1 , for E k , there are constants l k = l k ( E k ) , l ′ k = l ′ k ( E k ) , such that

‖ g 1 i ( u ˜ , v ˜ ) − g 1 i ( u , v ) ‖ ≤ l k ‖ ( u ˜ , v ˜ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 1 , ‖ g 2 i ( u ˜ , v ˜ ) − g 2 i ( u , v ) ‖ ≤ l ′ k ‖ ( u ˜ , v ˜ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 2 , ( i = u , v ) , ∀ ( u ˜ , v ˜ ) , ( u , v ) ∈ V 2 m + k × V 2 m + k , ‖ ( u ˜ , v ˜ ) ‖ V 2 m + k × V 2 m + k ≤ c ( E k ) , ‖ ( u , v ) ‖ V 2 m + k × V 2 m + k ≤ c ( E k ) .

where ‖ ( u ˜ , v ˜ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k l = ‖ D 2 m + k ( u ˜ − u ) ‖ l + ‖ D 2 m + k ( v ˜ − v ) ‖ l .

Make a priori estimates as following:

Lemma 1 Assumes that (H1) - (H2) holds, and ( u 0 , z 0 , v 0 , q 0 ) ∈ E 0 , f 1 ( x ) , f 2 ( x ) ∈ L 2 ( Ω ) , then Equations (1)-(5) have solutions ( u , z , v , q ) and have the following properties

1) ( u , z , v , q ) ∈ L ∞ ( ( 0 , + ∞ ) ; E 0 ) ;

2) E 0 ( t ) ≤ 1 k 1 ( 2 C μ C + C 2 ε 0 2 ( ‖ f 1 ‖ 2 + ‖ f 2 ‖ 2 ) ) ,

where E 0 ( t ) = ‖ D 2 m u ‖ 2 + ‖ z ‖ 2 + ‖ D 2 m v ‖ 2 + ‖ q ‖ 2 ≤ c ( R 0 ) .

3) There are positive constants c ( R 0 ) and t 0 = t 0 ( Ω ) > 0 , such that

‖ ( u , z , v , q ) ‖ E 0 2 = ‖ D 2 m u ‖ 2 + ‖ z ‖ 2 + ‖ D 2 m v ‖ 2 + ‖ q ‖ 2 ≤ c ( R 0 ) .

Proof: z = u t + ε u and the Equation (1) as the inner product, that is

( u t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m u + β ( − Δ ) 2 m u t + g 1 ( u , v ) , z ) = ( f 1 ( x ) , z ) .

where ( u t t , z ) = 1 2 d d t ‖ z ‖ 2 − ε ‖ z ‖ 2 + ε 3 ‖ u ‖ 2 + ε 2 2 d d t ‖ u ‖ 2 ;

( M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m u , z ) = M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( 1 2 d d t ‖ D 2 m u ‖ 2 + ε ‖ D 2 m u ‖ 2 ) ≥ μ 2 d d t ‖ D 2 m u ‖ 2 + ε μ 0 ‖ D 2 m u ‖ 2 ;

( β ( − Δ ) 2 m u t , z ) = β ( D 2 m u t , D 2 m z ) = β ‖ D 2 m z ‖ 2 − ε β ( D 2 m u , D 2 m z ) ≥ β 2 ‖ D 2 m z ‖ 2 − β ε 2 2 ‖ D 2 m u ‖ 2 ;

( g 1 ( u , v ) , z ) = d d t ∫ Ω ∫ 0 u g 1 ( ς , v ) d ς d x + ε ( g 1 ( u , v ) , u ) .

Similarly, q = v t + ε v and the Equation (2) as the inner product, that is

( v t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m v + β ( − Δ ) 2 m v t + g 2 ( u , v ) , q ) = ( f 2 ( x ) , q ) .

where,

( v t t , q ) = 1 2 d d t ‖ q ‖ 2 − ε ‖ q ‖ 2 + ε 3 ‖ v ‖ 2 + ε 2 2 d d t ‖ v ‖ 2 ;

( M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m v , q ) = M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( 1 2 d d t ‖ D 2 m v ‖ 2 + ε ‖ D 2 m v ‖ 2 ) ≥ μ 2 d d t ‖ D 2 m v ‖ 2 + ε μ 0 ‖ D 2 m v ‖ 2 ;

( β ( − Δ ) 2 m v t , q ) = β ( D 2 m v t , D 2 m q ) = β ‖ D 2 m q ‖ 2 − ε β ( D 2 m v , D 2 m q ) ≥ β 2 ‖ D 2 m q ‖ 2 − β ε 2 2 ‖ D 2 m v ‖ 2 ;

( g 2 ( u , v ) , q ) = d d t ∫ Ω ∫ 0 v g 2 ( u , η ) d η d x + ε ( g 2 ( u , v ) , v ) .

The above result and other product terms are collated, using the Holder’s inequality, the Young’s inequality

1 2 d d t ‖ z ‖ 2 − ε ‖ z ‖ 2 + ε 3 ‖ u ‖ 2 + ε 2 2 d d t ‖ u ‖ 2 + 1 2 d d t ‖ q ‖ 2 − ε ‖ q ‖ 2 + ε 3 ‖ v ‖ 2 + ε 2 2 d d t ‖ v ‖ 2 + μ 2 d d t ‖ D 2 m u ‖ 2 + ε μ 0 ‖ D 2 m u ‖ 2 + μ 2 d d t ‖ D 2 m v ‖ 2 + ε μ 0 ‖ D 2 m v ‖ 2 + β 2 ‖ D 2 m z ‖ 2 − β ε 2 2 ‖ D 2 m u ‖ 2 + β 2 ‖ D 2 m q ‖ 2 − β ε 2 2 ‖ D 2 m v ‖ 2 + d d t ∫ Ω ∫ 0 u g 1 ( ς , v ) d ς d x + ε ( g 1 ( u , v ) , u ) + d d t ∫ Ω ∫ 0 v g 2 ( u , η ) d η d x + ε ( g 2 ( u , v ) , v ) ≤ ( f 1 , z ) + ( f 2 , q ) ≤ 1 4 ε ( ‖ f 1 ‖ 2 + ‖ f 2 ‖ 2 ) + ε ( ‖ z ‖ 2 + ‖ q ‖ 2 ) .

Again by hypothesis (H2), obtained

− ε ( g 1 ( u , v ) , u ) − ε ( g 2 ( u , v ) , v ) ≤ − ε c J ( u , v ) + ε μ ( ‖ D m u ‖ 2 + ‖ D m v ‖ 2 ) + ε c μ ≤ − ε c J ( u , v ) + ε μ c * 2 ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) + ε c μ .

Reuse the Poincare’s inequality to obtain

d d t ( μ ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) + ‖ z ‖ 2 + ‖ q ‖ 2 + 2 J ( u , v ) + ε 2 ( ‖ u ‖ 2 + ‖ v ‖ 2 ) )   + 2 ε ( μ 0 − β ε 2 − μ c ∗ 2 ) ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) + ( β c ∗ 2 − 4 ε ) ( ‖ z ‖ 2 + ‖ q ‖ 2 )   + 2 ε c J ( u , v ) + 2 ε 3 ( ‖ u ‖ 2 + ‖ v ‖ 2 ) ≤ 2 ε c μ + 1 2 ε ( ‖ f 1 ‖ 2 + ‖ f 2 ‖ 2 ) .

Again by hypothesis (H2), obtained 2 J ( u , v ) + ε 2 ( ‖ u ‖ 2 + ‖ v ‖ 2 ) + 2 c ′ μ ≥ 0 ,

At this point, Order y ¯ ( t ) = μ ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) + ‖ z ‖ 2 + ‖ q ‖ 2 + 2 J ( u , v ) + ε 2 ( ‖ u ‖ 2 + ‖ v ‖ 2 ) .

Then there is a constant of k 1 = min { 1 , μ } , such that y ¯ ( t ) + 2 c ′ μ ≥ k 1 E 0 ( t ) ≥ 0 , of which, E 0 ( t ) = ‖ D 2 m u ‖ 2 + ‖ z ‖ 2 + ‖ D 2 m v ‖ 2 + ‖ q ‖ 2 . Since c > 0 ,

Then Order ε 0 = min { 2 ε μ ( μ 0 − β ε 2 − μ c ∗ 2 ) , β c ∗ 2 − 4 ε , c ε , 2 ε } .

Here to denote y ( t ) = y ¯ ( t ) + 2 c ′ μ , using the Gronwall’s inequality, that is

d d t y ( t ) + ε 0 y ( t ) ≤ 2 ε c μ + 1 2 ε ( ‖ f 1 ‖ 2 + ‖ f 2 ‖ 2 ) ,

y ( t ) ≤ y ( 0 ) e − ε 0 t + 2 c μ c + c 2 ε 0 2 ( ‖ f 1 ‖ 2 + ‖ f 2 ‖ 2 ) .

Therefore, E 0 ( t ) ≤ 1 k 1 ( y ¯ ( t ) + 2 c ′ μ ) = y ( t ) k 1 = 1 k 1 y ( 0 ) e − ε 0 t + 1 k 1 ( 2 c μ c + c 2 ε 0 2 ( ‖ f 1 ‖ 2 + ‖ f 2 ‖ 2 ) ) .

Well, E 0 ( t ) ≤ 1 k 1 ( 2 c μ c + c 2 ε 0 2 ( ‖ f 1 ‖ 2 + ‖ f 2 ‖ 2 ) ) , that is, there are positive constants c ( R 0 ) and t 0 = t 0 ( Ω ) > 0 , such that for t > t 0 , ‖ ( u , z , v , q ) ‖ E 0 2 = ‖ D 2 m u ‖ 2 + ‖ z ‖ 2 + ‖ D 2 m v ‖ 2 + ‖ q ‖ 2 ≤ c ( R 0 ) .

Lemma 1 is proved.

Lemma 2 Assumes that (H1) - (H3) holds, and f 1 ( x ) , f 2 ( x ) ∈ H k ( Ω ) , | ( f 1 , u ) + ( f 2 , v ) | ≤ 2 c , then Equations (1)-(5) have solutions ( u , z , v , q ) and have the following properties

1) ( u , z , v , q ) ∈ L ∞ ( ( 0 , + ∞ ) ; E k ) ;

2) E k ( t ) ≤ y ( 0 ) e − k 1 t + c 3 k 1 ;

3) There are positive constants c ( E k ) and t k = t k ( Ω ) > 0 , such that

‖ ( u , z , v , q ) ‖ E k 2 = ‖ D 2 m + k u ‖ 2 + ‖ D k z ‖ 2 + ‖ D 2 m + k v ‖ 2 + ‖ D k q ‖ 2 ≤ c ( E k ) .

Prove: ( − Δ ) k z and the Equation (1) as the inner product, that is

( u t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m u + β ( − Δ ) 2 m u t + g 1 ( u , v ) , ( − Δ ) k z ) = ( f 1 ( x ) , ( − Δ ) k z ) .

where, ( u t t , ( − Δ ) k z ) = 1 2 d d t ‖ D k z ‖ 2 − ε ‖ D k z ‖ 2 + ε 3 ‖ D k u ‖ 2 + ε 2 2 d d t ‖ D k u ‖ 2 ;

( M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m u , ( − Δ ) k z ) = M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( 1 2 d d t ‖ D 2 m + k u ‖ 2 + ε ‖ D 2 m + k u ‖ 2 ) ≥ μ 2 d d t ‖ D 2 m + k u ‖ 2 + ε μ 0 ‖ D 2 m + k u ‖ 2 ;

( β ( − Δ ) 2 m u t , ( − Δ ) k z ) = β ( Δ 2 m u t , Δ k z ) = β ‖ D 2 m + k z ‖ 2 − ε β ( ( − Δ ) 2 m u , ( − Δ ) k z ) ≥ β 2 ‖ D 2 m + k z ‖ 2 − β ε 2 2 ‖ D 2 m + k u ‖ 2 .

Similarly, ( − Δ ) k q and the Equation (2) as the inner product, that is

( v t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m v + β ( − Δ ) 2 m v t + g 2 ( u , v ) , ( − Δ ) k q ) = ( f 2 ( x ) , ( − Δ ) k q ) .

where,

( v t t , ( − Δ ) k q ) = 1 2 d d t ‖ D k q ‖ 2 − ε ‖ D k q ‖ 2 + ε 3 ‖ D k v ‖ 2 + ε 2 2 d d t ‖ D k v ‖ 2 ;

( M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m v , ( − Δ ) k q ) = M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( 1 2 d d t ‖ D 2 m + k v ‖ 2 + ε ‖ D 2 m + k v ‖ 2 ) ≥ μ 2 d d t ‖ D 2 m + k v ‖ 2 + ε μ 0 ‖ D 2 m + k v ‖ 2 ;

( β ( − Δ ) 2 m v t , ( − Δ ) k q ) = β ( Δ 2 m v t , Δ k q ) = β ‖ D 2 m + k q ‖ 2 − ε β ( ( − Δ ) 2 m v , ( − Δ ) k q ) ≥ β 2 ‖ D 2 m + k q ‖ 2 − β ε 2 2 ‖ D 2 m + k v ‖ 2 .

The above result and other internal product terms are sorted out

1 2 d d t ‖ D k z ‖ 2 − ε ‖ D k z ‖ 2 + ε 3 ‖ D k u ‖ 2 + ε 2 2 d d t ‖ D k u ‖ 2 + 1 2 d d t ‖ D k q ‖ 2 − ε ‖ D k q ‖ 2   + ε 3 ‖ D k v ‖ 2 + ε 2 2 d d t ‖ D k v ‖ 2 + μ 2 d d t ‖ D 2 m + k u ‖ 2 + ε μ 0 ‖ D 2 m + k u ‖ 2 + μ 2 d d t ‖ D 2 m + k v ‖ 2   + ε μ 0 ‖ D 2 m + k v ‖ 2 + β 2 ‖ D 2 m + k z ‖ 2 − β ε 2 2 ‖ D 2 m + k u ‖ 2 + β 2 ‖ D 2 m + k q ‖ 2   − β ε 2 2 ‖ D 2 m + k v ‖ 2 + ( g 1 ( u , v ) , ( − Δ ) k z ) + ( g 2 ( u , v ) , ( − Δ ) k q ) ≤ ( f 1 ( x ) , ( − Δ ) k z ) + ( f 2 ( x ) , ( − Δ ) k q ) = ( D k f 1 , D k z ) + ( D k f 2 , D k q ) .

Again by hypothesis (H3), obtained

‖ g 1 ( u , v ) ‖ 2 = ∫ Ω | g 1 ( u , v ) | 2 d x ≤ ∫ Ω | c ( 1 + | u | r 1 + | v | r 1 ) | 2 d x = c ∫ Ω | ( 1 + 2 | u | r 1 + 2 | v | r 1 + | u | 2 r 1 + | v | 2 r 1 + 2 | u | r 1 | v | r 1 ) | d x ≤ c ∫ Ω ( 1 + 1 + | u | 2 r 1 + 1 + | v | 2 r 1 + 2 | u | 2 r 1 + 2 | v | 2 r 1 ) d x = 3 c ∫ Ω ( 1 + | u | 2 r 1 + | v | 2 r 1 ) d x ≤ 3 c ( 1 + ‖ u ‖ 2 r 1 2 r 1 + ‖ v ‖ 2 r 1 2 r 1 ) .

Similarly, ‖ g 2 ( u , v ) ‖ 2 ≤ 3 c ( 1 + ‖ u ‖ 2 r 2 2 r 2 + ‖ v ‖ 2 r 2 2 r 2 ) .

By interpolation inequality, there is ‖ u ‖ 2 r j ≤ c ‖ D 2 m u ‖ n ( r j − 1 ) 4 m r j ‖ u ‖ 4 m r j − n ( r j − 1 ) 4 m r j ,

This can be concluded { ‖ u ‖ 2 r j 2 r j ≤ c ‖ D 2 m u ‖ n ( r j − 1 ) 2 m ‖ u ‖ 4 m r j − n ( r j − 1 ) 2 m ‖ v ‖ 2 r j 2 r j ≤ c ‖ D 2 m v ‖ n ( r j − 1 ) 2 m ‖ v ‖ 4 m r j − n ( r j − 1 ) 2 m

where, 0 < r j ≤ 2 n / [ n − 4 m ] + ( n = 4 m , 0 < r j < + ∞ ) ,

Therefore, ‖ g 1 ( u , v ) ‖ 2 + ‖ g 2 ( u , v ) ‖ 2 ≤ c 1 .

Reuse the Young’s inequality and the Sobolev-Poincare’s inequality, thus

( g 1 ( u , v ) , ( − Δ ) k z ) + ( g 2 ( u , v ) , ( − Δ ) k q ) ≥ − c 1 2 ε − ε 2 ( ‖ ( − Δ ) k z ‖ 2 + ‖ ( − Δ ) k q ‖ 2 ) ,

β − ε 2 ( ‖ ( − Δ ) k z ‖ 2 + ‖ ( − Δ ) k q ‖ 2 ) ≥ β − ε 2 c ∗ 2 ( ‖ D k z ‖ 2 + ‖ D k q ‖ 2 ) .

In summary, according to the Holder’s inequality, the Young’s inequality is obtained

d d t ( μ ( ‖ D 2 m + k u ‖ 2 + ‖ D 2 m + k v ‖ 2 ) + ‖ D k z ‖ 2 + ‖ D k q ‖ 2 + ε 2 ( ‖ D k u ‖ 2 + ‖ D k v ‖ 2 ) )   + 2 ε ( μ 0 − β ε 2 ) ( ‖ D 2 m + k u ‖ 2 + ‖ D 2 m + k v ‖ 2 ) + ( β − ε c ∗ 2 ) ( ‖ D k z ‖ 2 + ‖ D k q ‖ 2 )   + 2 ε 3 ( ‖ D k u ‖ 2 + ‖ D k v ‖ 2 ) ≤ c 1 ε + 2 [ ( D k f 1 , D k z ) + ( D k f 2 , D k q ) ] ,

that is

d d t ( μ ( ‖ D 2 m + k u ‖ 2 + ‖ D 2 m + k v ‖ 2 ) + ‖ D k z ‖ 2 + ‖ D k q ‖ 2 + ε 2 ( ‖ D k u ‖ 2 + ‖ D k v ‖ 2 ) )   + 2 ε ( μ 0 − β ε 2 ) ( ‖ D 2 m + k u ‖ 2 + ‖ D 2 m + k v ‖ 2 ) + ( β − ε c ∗ 2 − 4 ε ) ( ‖ D k z ‖ 2 + ‖ D k q ‖ 2 )   + 2 ε 3 ( ‖ D k u ‖ 2 + ‖ D k v ‖ 2 ) ≤ c 1 ε + 1 2 ε ( ‖ D k f 1 ‖ 2 + ‖ D k f 2 ‖ 2 )

At this point, Order y ′ ( t ) = μ ( ‖ D 2 m + k u ‖ 2 + ‖ D 2 m + k v ‖ 2 ) + ‖ D k z ‖ 2 + ‖ D k q ‖ 2 + ε 2 ( ‖ D k u ‖ 2 + ‖ D k v ‖ 2 ) ,

So there are d d t y ′ ( t ) + k 1 y ′ ( t ) ≤ c 3 , k 1 = min { 2 ε μ ( μ 0 − β ε 2 ) , β − ε c ∗ 2 − 4 ε , 2 ε } ,

Reuse the Gronwall’s inequality,

y ′ ( t ) ≤ y ′ ( 0 ) e − k 1 t + c 3 k 1 , t ≥ 0.

E k ( t ) = ‖ D 2 m + k u ‖ 2 + ‖ D k z ‖ 2 + ‖ D 2 m + k v ‖ 2 + ‖ D k q ‖ 2 ≤ μ ( ‖ D 2 m + k u ‖ 2 + ‖ D 2 m + k v ‖ 2 ) + ‖ D k z ‖ 2 + ‖ D k q ‖ 2 ≤ μ ( ‖ D 2 m + k u ‖ 2 + ‖ D 2 m + k v ‖ 2 ) + ‖ D k z ‖ 2 + ‖ D k q ‖ 2 + ε 2 ( ‖ D k u ‖ 2 + ‖ D k v ‖ 2 ) = y ′ ( t ) ≤ y ′ ( 0 ) e − k 1 t + c 3 k 1 .

that is E k ( t ) ≤ c 3 k 1 .

Thus, there is a positive constant, there is a positive constant c ( E k ) and t k = t k ( Ω ) > 0 , so that for t > t k , there are ‖ ( u , z , v , q ) ‖ E k 2 = ‖ D 2 m + k u ‖ 2 + ‖ D k z ‖ 2 + ‖ D 2 m + k v ‖ 2 + ‖ D k q ‖ 2 ≤ c ( E k ) .

Lemma 2 is proved.

Theorem 1 (The existence and uniqueness of solution) Suppose (H1) - (H3) holds, and ( u 0 , z 0 , v 0 , q 0 ) ∈ E k , f 1 ( x ) , f 2 ( x ) ∈ H k ( Ω ) , then Equations (1) - (5) have an unique solution

( u ( x , t ) , z ( x , t ) , v ( x , t ) , q ( x , t ) ) ∈ L ∞ ( ( 0 , + ∞ ) ; E k ) .

Proof: Using the Galerkin method, combining lemmas 1 and 2, where the first a priori estimation has been proved; the second step: approximate solution.

We can take sequences w 1 , w 2 , ⋯ , w r , ⋯ , w i ∈ H 0 2 m + k ( Ω ) ∩ L 2 ( Ω ) ( ∀ i ) , ∀ r , w 1 , w 2 , ⋯ , w r , w i = ( u i , v i ) , where the linear combination of w i is dense in H 0 2 m + k ( Ω ) ∩ L 2 ( Ω ) , so w j represents the eigenvalue function corresponding to the eigenvalue, and w 1 , w 2 , ⋯ , w r is the standard orthogonal basis that constitutes H 2 m + k ; and λ j is the eigenvalue of ( − Δ ) with a homogeneous Dirichlet boundary condition on Ω , then there is ( − Δ ) 2 m + k w j = λ j 2 m + k w j ( k = 0 , 1 , 2 , ⋯ , 2 m ) .

Set the approximate solution of the initial edge value problem (1)-(5), u r = u r ( t ) = ∑ j = 1 r g j r ( t ) w j , v r = v r ( t ) = ∑ j = 1 r f j r ( t ) w j .

Easy ( u r , v r ) is dense in H 2 m + k × H 2 m + k and satisfies the following conditions

{ ( u ″ r ( t ) , u j ) + ( M ( ‖ D m u r ( t ) ‖ p p + ‖ D m v r ( t ) ‖ p p ) ( − Δ ) 2 m u r ( t ) , u j ) + ( β ( − Δ ) 2 m u ′ r ( t ) , u j ) + ( g ( u ′ r ( t ) , v ′ r ( t ) ) , u j ) = ( f 1 ( x ) , u j ) , 1 ≤ j ≤ r , ( v ″ r ( t ) , v j ) + ( M ( ‖ D m u r ( t ) ‖ p p + ‖ D m v r ( t ) ‖ p p ) ( − Δ ) ) 2 m v r ( t ) , v j ) + ( β ( − Δ ) 2 m v ′ r ( t ) , v j ) + ( g ( u ′ r ( t ) , v ′ r ( t ) ) , v j ) = ( f 2 ( x ) , v j ) , 1 ≤ j ≤ r . (6)

And the above nonlinear system of ordinary differential Equations (6) satisfies the initial conditions:

u r ( 0 ) = u 0 r = ∑ j = 1 r ( u 0 , w j ) w j → u 0 , ( r → ∞ ) ,     in   H 0 2 m + k ( Ω ) ∩ L 2 ( Ω ) (7)

u ′ r ( 0 ) = u 1 r = ∑ j = 1 r ( u 1 , w j ) w j → u 1 , ( r → ∞ ) ,     in   H 0 k ( Ω ) ∩ L 2 ( Ω ) (8)

v r ( 0 ) = v 0 r = ∑ j = 1 r ( v 0 , w j ) w j → v 0 , ( r → ∞ ) ,     in   H 0 2 m + k ( Ω ) ∩ L 2 ( Ω ) (9)

v ′ r ( 0 ) = v 1 r = ∑ j = 1 r ( v 1 , w j ) w j → v 1 , ( r → ∞ )     in   H 0 k ( Ω ) ∩ L 2 ( Ω ) (10)

The general conclusion of the system of nonlinear ordinary differential equations is easy to know, which ensures that the approximate solution of the problem (6)-(10) exists on the interval [ 0 , t r ] .

Known z = u t + ε u , q = v t + ε v , binding lemma 1, lemma 2, in space E k , and we can pick subsequence { u s } from sequence { u h } and subsequence { v s } from sequence { v h } , such that ( u s , z s , v s , q s ) → ( u , z , v , q ) is weak * convergence in

L ∞ ( [ 0 , + ∞ ) ; E k ) . (11)

and z r , q r is bounded on L 2 ( ( 0 , T ) ; E 0 2 m + k ) .

By the Rellich-Kondrachov compact embedding theorem, E k is compactly embedding in E 0 , ( u s , z s , v s , q s ) → ( u , z , v , q ) is strong convergence almost everywhere.

This can be obtained from the above assumptions and lemmas

M ( ‖ D m u m ( t ) ‖ p p + ‖ D m v m ( t ) ‖ p p ) ( − Δ ) 2 m u r ( t ) → M ( ‖ D m u m ( t ) ‖ p p + ‖ D m v m ( t ) ‖ p p ) ( − Δ ) 2 m u ( t ) weak converges in

L ∞ ( 0 , T ; H 0 2 m + k ( Ω ) ) , and β ( − Δ ) 2 m u ′ r ( t ) → β ( − Δ ) 2 m u ′ ( t ) weak converges in L ∞ ( 0 , T ; H 0 2 m + k ( Ω ) ) , g ( u r ( t ) , v r ( t ) ) → g ( u ( t ) , v ( t ) ) weak converges in L ∞ ( 0 , T ; H 0 2 m + k ( Ω ) ) .

Thus it is possible to take r = μ in (1), (2), and take the limit. To the fixed j and μ ≥ j , get

( u ″ μ , u j ) + ( M ( ‖ D m u r ( t ) ‖ p p + ‖ D m v r ( t ) ‖ p p ) ( − Δ ) 2 m u μ , u j ) + ( β ( − Δ ) 2 m u ′ μ , u j ) + ( g ( u ( t ) , v ( t ) ) , u j ) = ( f 1 ( x ) , u j ) .

it satisfies all j, and thus for ∀ u ∈ L ∞ ( 0 , T ; H 0 2 m + k ( Ω ) ∩ L 2 ( Ω ) ) ,

( v ″ μ , v j ) + ( M ( ‖ D m u r ( t ) ‖ p p + ‖ D m v r ( t ) ‖ p p ) ( − Δ ) 2 m v μ , v j ) + ( β ( − Δ ) 2 m v ′ μ , v j ) + ( g ( u ( t ) , v ( t ) ) , v j ) = ( f 2 ( x ) , v j ) .

it satisfies all j, and thus for ∀ u ∈ L ∞ ( 0 , T ; H 0 2 m + k ( Ω ) ∩ L 2 ( Ω ) ) .

It is easy to obtain that the system of Equations (1)-(5) exists

{ u t t + M ( s 1 ) ( − Δ ) 2 m u 1 − M ( s 2 ) ( − Δ ) 2 m u 2 + β ( − Δ ) 2 m u t + g 1 ( u 1 , v 1 ) − g 1 ( u 2 , v 2 ) = 0 v t t + M ( s 1 ) ( − Δ ) 2 m v 1 − M ( s 2 ) ( − Δ ) 2 m v 2 + β ( − Δ ) 2 m v t + g 2 ( u 1 , v 1 ) − g 2 ( u 2 , v 2 ) = 0 (12)

where,

M ( s 1 ) = M ( ‖ D m u 1 ‖ p p + ‖ D m v 1 ‖ p p ) , M ( s 2 ) = M ( ‖ D m u 2 ‖ p p + ‖ D m v 2 ‖ p p ) , u = u 1 − u 2 , v = v 1 − v 2 .

Use u t , v t and Equation (12) as the inner product in turn, and get it

1 2 d d t ( ‖ u t ‖ 2 + ‖ v t ‖ 2 ) + β ( ‖ D 2 m u t ‖ 2 + ‖ D 2 m v t ‖ 2 ) + ( M ( s 1 ) ( − Δ ) 2 m u 1 − M ( s 2 ) ( − Δ ) 2 m u 2 , u t ) + ( M ( s 1 ) ( − Δ ) 2 m v 1 − M ( s 2 ) ( − Δ ) 2 m v 2 , v t ) + ( g 1 ( u 1 , v 1 ) − g 1 ( u 2 , v 2 ) , u t ) + ( g 2 ( u 1 , v 1 ) − g 2 ( u 2 , v 2 ) , v t ) = 0. (13)

Using the Young’s inequality and the Sobolev-Poincare’s inequality, it is derived

( M ( s 1 ) ( − Δ ) 2 m u 1 − M ( s 2 ) ( − Δ ) 2 m u 2 , u t ) = ( M ( s 1 ) ( − Δ ) 2 m u + M ( s 1 ) ( − Δ ) 2 m u 2 − M ( s 2 ) ( − Δ ) 2 m u 2 , u t ) ≥ μ 2 d d t ‖ D 2 m u ‖ 2 + ( μ 0 − μ 1 ) ( ( − Δ ) 2 m u 2 , u t ) ≥ μ 2 d d t ‖ D 2 m u ‖ 2 + c 4 ‖ u t ‖ . (14)

In summary

( M ( s 1 ) ( − Δ ) 2 m v 1 − M ( s 2 ) ( − Δ ) 2 m v 2 , v t ) ≥ μ 2 d d t ‖ D 2 m v ‖ 2 + c 4 ‖ v t ‖ (15)

| ( g 1 ( u 1 , v 1 ) − g 1 ( u 2 , v 2 ) , u t ) | = | ( g 1 ( u 1 , v 1 ) − g 1 ( u 1 , v 2 ) + g 1 ( u 1 , v 2 ) − g 1 ( u 2 , v 2 ) , u t ) | ≤ | ( | v 1 | r 1 − | v 2 | r 1 , u t ) | + | ( | u 1 | r 1 − | u 2 | r 1 , u t ) | ≤ | ( c ( 1 + | v 1 | r 1 − 1 + | v 2 | r 1 − 1 ) v , u t ) | + | ( c ( 1 + | u 1 | r 1 − 1 + | u 2 | r 1 − 1 ) u , u t ) | ≤ ‖ c ( 1 + | v 1 | r 1 − 1 + | v 2 | r 1 − 1 ) ‖ ∞ ‖ v ‖ ‖ u t ‖ + ‖ c ( 1 + | u 1 | r 1 − 1 + | u 2 | r 1 − 1 ) ‖ ∞ ‖ u ‖ ‖ u t ‖ ≤ c 5 ( ‖ v ‖ 2 + ‖ u ‖ 2 + ‖ u t ‖ 2 2 ) ≤ c 5 ( c ∗ 2 ‖ D 2 m v ‖ 2 + c ∗ 2 ‖ D 2 m u ‖ 2 + ‖ u t ‖ 2 2 ) . (16)

Similarly,

| ( g 1 ( u 1 , v 1 ) − g 1 ( u 2 , v 2 ) , u t ) | ≤ c 5 ( c ∗ 2 ‖ D 2 m v ‖ 2 + c ∗ 2 ‖ D 2 m u ‖ 2 + ‖ v t ‖ 2 2 ) (17)

Substituting (14)-(17) into Equation (13), combining lemmas 1 and 2, using the Poincare’s inequality, to obtain

d d t ( ‖ u t ‖ 2 + ‖ v t ‖ 2 + μ ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) ) + ( 2 β c ∗ 2 + 2 c 4 + c 5 ) ( ‖ u t ‖ 2 + ‖ v t ‖ 2 )   + 4 c 5 c ∗ 2 ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) ≤ 0.

Order y ″ ( t ) = ‖ u t ‖ 2 + ‖ v t ‖ 2 + μ ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) ,

Then there are d d t y ″ ( t ) + k 4 y ″ ( t ) ≤ 0 , k 4 = min { 2 β c ∗ 2 + 2 c 4 + c 5 , 4 c 5 c ∗ 2 } ,

Using the Gronwall’s inequality, get y ″ ( t ) ≤ y ″ ( 0 ) e − t , t ≥ 0 ,

So y ″ ( t ) = ‖ u t ‖ 2 + ‖ v t ‖ 2 + μ ( ‖ D 2 m u ‖ 2 + ‖ D 2 m v ‖ 2 ) ≡ 0 .

That’s ‖ u t ‖ 2 = ‖ v t ‖ 2 = ‖ D 2 m u ‖ 2 = ‖ D 2 m v ‖ 2 = 0 , hence w ( x , t ) = ( u ( x , t ) , v ( x , t ) ) = ( 0 , 0 ) .

Theorem 1 is proved.

Theorem 2 [7] Assume E is a Banach space, and { S ( t ) } t ≥ 0 is the operator semigroup on E,

S ( t ) : E → E , S ( t + r ) = S ( t ) + S ( r )   ( ∀ t , r > 0 ) , S ( 0 ) = I .

where Ι is the identity operator，if S ( t ) satisfies

1) Semigroup S ( t ) is uniformly bounded in E;

2) There exists a bounded absorbing set B 0 in E;

3) { S ( t ) } t ≥ 0 is completely continuous operator.

That is to say that operator semigroup S ( t ) has compact global attractor A.

Where (1) means ∀ R > 0 , exists a constant C ( R ) such that when ‖ u ‖ E ≤ R , there is ‖ S ( t ) u ‖ E ≤ c ( R ) ( ∀ t ∈ [ 0 , ∞ ) ) , and (2) means for any bounded set B ⊂ E , there exists a constant t 0 > 0 , such that S ( t ) B ⊂ B 0 ( ∀ t ≥ t 0 ) . In theorem 2, if S ( t ) is a solution semigroup generated by the initial boundary value problem (1)-(5), ( u ( t ) , z ( t ) , v ( t ) , q ( t ) ) = S ( t ) ( u 0 , z 0 , v 0 , q 0 ) , and Banach space E is changed into Hilbert space E k , there will be family of global attractors.

Theorem 3 Let S ( t ) is a solution semigroup generated by the initial boundary value problems (1)-(5), under the hypothesis of lemma 1 and lemma 2. Assuming that the existence and uniqueness of solution, then the equation has a global attraction subfamily. That is:

A k ⊂ E k ⊂ E 0 ( k = 1 , 2 , ⋯ , 2 m ) , and A k = ω ( B 0 k ) = ∩ s ≥ 0 ∪ t ≥ s S ( t ) B 0 k ¯ .

where

B 0 k = { ( u , z , v , q ) ∈ E k : ‖ D 2 m + k u ‖ 2 + ‖ D k z ‖ 2 + ‖ D 2 m + k v ‖ 2 + ‖ D k q ‖ 2 ≤ c ( R k ) } ,

1) Invariability: S ( t ) A k = A k ;

2) Attractiveness: A k attracts all bounded sets of E k , that is, any bounded set B 0 k ⊂ E k , d i s t ( S ( t ) B 0 K , A k ) = sup x ∈ B 0 K inf y ∈ A k ‖ S ( t ) x − y ‖ E k → 0   ( t → 0 ) .

Then compact set A k is called family of global attractors of semigroup S ( t ) .

Proof: Verify theorem 2 to prove the existence of family of global attractors, under the condition of theorem 1, and the initial boundary value problems (1)-(5) generate solution semigroups S ( t ) : E k → E k .

1) So for any bounded set B 0 k ⊂ E k , having

‖ S ( t ) ( u 0 , z 0 , v 0 , q 0 ) ‖ E k 2 = ‖ D 2 m + k u ‖ 2 + ‖ D k z ‖ 2 + ‖ D 2 m + k v ‖ 2 + ‖ D k q ‖ 2 ≤ C ( R k ) ,

where t ≥ 0 and ( u 0 , z 0 , v 0 , q 0 ) ∈ B k , shows that { S ( t ) } t ≥ 0 is uniformly bounded in E k ;

2) ∀ ( u 0 , z 0 , v 0 , q 0 ) ∈ E k , when t ≥ max { t 0 , t 0 k } , there is ‖ S ( t ) ( u 0 , z 0 , v 0 , q 0 ) ‖ E k 2 ≤ C ( R k ) , thus B k is a bounded absorption set of semigroup S ( t ) ;

3) E k is compactly embedded in E 0 , i.e., the bounded set in E k is a compact set in E 0 , so the operator semigroup S ( t ) is completely continuous operator.

Theorem 2 is proved.

Since the solution semigroup S ( t ) has a family of global attractors in space E k , the dimensionality estimates of the global attractors subfamily are now made, and the resulting Hausdorff and Fractal dimensions are finite to prove. Linearize the problems (1)-(5) first, as follows:

{ U t t + M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ ( D m U + D m V ) ( − Δ ) 2 m u + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m U + β ( − Δ ) 2 m U t + g 2 u ( u , v ) U + g 2 v ( u , v ) V = 0 , V t t + M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ ( D m U + D m V ) ( − Δ ) 2 m v + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m V + β ( − Δ ) 2 m V t + g 2 u ( u , v ) U + g 2 v ( u , v ) V = 0 , ∂ i U ∂ n i | ∂ Ω = ∂ i V ∂ n i | ∂ Ω = 0 , t ≥ 0 , U ( x , 0 ) = ξ 1 , U t ( x , 0 ) = ξ 2 , V ( x , 0 ) = η 1 , V t ( x , 0 ) = η 2 . (18)

where ( ξ 1 , η 1 , ξ 2 , η 2 ) ∈ E k , ( u 0 , v 0 , u 1 , v 1 ) ∈ A k , ( u , v , u t , v t ) = S ( t ) ( u 0 , v 0 , u 1 , v 1 ) is the solution to the problem (18) obtained by ( u 0 , v 0 , u 1 , v 1 ) ∈ A k , Given ( u 0 , v 0 , u 1 , v 1 ) ∈ A k , S ( t ) : E k → E k , it can be proved that for any ( ξ 1 , η 1 , ξ 2 , η 2 ) ∈ E k , there is a unique solution to the linear initial edge value problem ( U ( t ) , V ( t ) , U t ( t ) , V t ( t ) ) ∈ L ∞ ( ( 0 , + ∞ ) ; E k ) .

Theorem 4 for the arbitrary t > 0 , r > 0 , map S ( t ) : E k → E k is a Fractal differentiable. Φ 0 = ( u 0 , v 0 , u 1 , v 1 ) T the differential is a linear operator on F : ( ξ 1 , η 1 , ξ 2 , η 2 ) T → ( U ( t ) , V ( t ) , U t ( t ) , V t ( t ) ) T , where ( U ( t ) , V ( t ) , U t ( t ) , V t ( t ) ) is the solution to the problem (18).

Proof: Set Φ 0 = ( u 0 , v 0 , u 1 , v 1 ) T ∈ E k , there is Φ ¯ 0 = ( u 0 + ξ 1 , v 0 + η 1 , u 1 + ξ 2 , v 1 + η 2 ) T ∈ E k , so ‖ Φ 0 ‖ E k ≤ r , ‖ Φ ¯ 0 ‖ E k ≤ r .

Thus one obtains the Lipchitz property of S ( t ) on bounded set E k , hence

‖ S ( t ) Φ 0 − S ( t ) Φ ¯ 0 ‖ E k 2 ≤ e c t ‖ ( ξ 1 , η 1 , ξ 2 , η 2 ) ‖ E k 2 .

Let σ = u ¯ − u − U ,   H = v ¯ − v − V be the solution to the problem, then

{ σ t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m σ + β ( − Δ ) 2 m σ t = h 1 = h 11 + h 12 H t t + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m H + β ( − Δ ) 2 m H t = h 2 = h 21 + h 22 (19)

where

h 11 = M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ ( D m U + D m V ) ( − Δ ) 2 m u               + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m u ¯ + M ( ‖ D m u ¯ ‖ p p + ‖ D m v ¯ ‖ p p ) ( − Δ ) 2 m u ¯ ,

h 12 = − g 1 ( u ¯ , v ¯ ) + g 1 ( u , v ) + g 1 u ( u , v ) U + g 1 v ( u , v ) V ,

h 21 = M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ ( D m U + D m V ) ( − Δ ) 2 m v               + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m v ¯ + M ( ‖ D m u ¯ ‖ p p + ‖ D m v ¯ ‖ p p ) ( − Δ ) 2 m v ¯ ,

h 22 = − g 2 ( u ¯ , v ¯ ) + g 2 ( u , v ) + g 2 u ( u , v ) U + g 2 v ( u , v ) V .

Take the inner product with the first equation of (19) and ( − Δ ) k σ t , and get it

( σ t t , ( − Δ ) k σ t ) = 1 2 d d t ‖ D k σ t ‖ 2 ,

( M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( − Δ ) 2 m σ , ( − Δ ) k σ t ) = 1 2 M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) d d t ‖ D 2 m + k σ t ‖ 2 ,

( β ( − Δ ) 2 m σ t , ( − Δ ) k σ t ) = β ‖ D 2 m + k σ t ‖ 2 .

this is,

1 2 d d t ‖ D k σ t ‖ 2 + 1 2 M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) d d t ‖ D 2 m + k σ t ‖ 2 + β ‖ D 2 m + k σ t ‖ 2 = ( h 1 , ( − Δ ) k σ t ) .

Order

s = D m u , w = D m v , s ¯ = D m u ¯ , w ¯ = D m v ¯ , u ˜ = u ¯ − u , v ˜ = v ¯ − v ,

θ 1 = ( 1 − α 1 ) s ¯ + α 1 s , θ 3 = ( 1 − α 3 ) w ¯ + α 3 w ,

θ 2 = ( 1 − α 2 ) s + α 2 θ 1 , θ 4 = ( 1 − α 4 ) w + α 4 θ 3 .

N ( θ ) = M ′ ( ‖ s ‖ p p + ‖ w ‖ p p ) ( ‖ s ‖ p p + ‖ w ‖ p p ) ′ , α 1 , α 2 , α 3 , α 4 ∈ ( 0 , 1 ) .

get

h 11 = − N ( θ 1 , θ 3 ) ( D m u ˜ + D m v ˜ ) ( − Δ ) 2 m u ¯ + N ( θ ) D m u ˜ ( − Δ ) 2 m u     − N ( θ ) D m σ ( − Δ ) 2 m u − N ( θ ) D m v ˜ ( − Δ ) 2 m u − N ( θ ) D m H ( − Δ ) 2 m u = − N ( θ 1 , θ 3 ) D m u ˜ ( − Δ ) 2 m u ˜ − N ′ ( θ 2 , θ 4 ) ( 1 − α 1 ) ( D m u ˜ ) 2 ( − Δ ) 2 m u     − N ( θ ) D m σ ( − Δ ) 2 m u − N ( θ 1 , θ 3 ) D m v ˜ ( − Δ ) 2 m u ˜     − N ′ ( θ 2 , θ 4 ) ( 1 − α 3 ) ( D m v ˜ ) 2 ( − Δ ) 2 m u − N ( θ ) D m H ( − Δ ) 2 m u .

So

| ( h 11 , ( − Δ ) k σ t ) | ≤ ‖ N ( θ 1 , θ 3 ) ‖ ∞ ‖ D 2 m + k u ˜ ‖ ‖ Δ m u ˜ ‖ ‖ D m + k σ t ‖       + c 1 ‖ N ′ ( θ 2 , θ 4 ) ‖ ∞ ‖ D 2 m + k u ˜ ‖ 2 ‖ ( − Δ ) m u ˜ ‖ ‖ σ t ‖       + ‖ N ( θ ) ‖ ∞ ‖ ( − Δ ) m u ‖ ‖ D m + k σ ‖ ‖ D 2 m + k σ t ‖       + ‖ N ( θ 1 , θ 3 ) ‖ ∞ ‖ D 2 m + k v ˜ ‖ ‖ Δ m u ˜ ‖ ‖ D m + k σ t ‖

      + c 1 ‖ N ′ ( θ 2 , θ 4 ) ‖ ∞ ‖ D 2 m + k v ˜ ‖ 2 ‖ ( − Δ ) m u ‖ ‖ σ t ‖       + ‖ N ( θ ) ‖ ∞ ‖ ( − Δ ) m u ‖ ‖ D m + k H ‖ ‖ D 2 m + k σ t ‖ ≤ c 3 λ 1 k ‖ D k σ t ‖ 2 + c 4 2 λ 1 m ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ ) + c 5 ‖ D 2 m + k σ t ‖ 2       + ( c 6 2 + c 7 2 4 c 5 λ 1 m + k ) ( ‖ D 2 m + k u ˜ ‖ 2 + ‖ D 2 m + k v ˜ ‖ 2 ) .

In the same way, the second equation of (19) and ( − Δ ) k H t is used to take the internal product and organize it

1 2 d d t ( ‖ D k σ t ‖ 2 + ‖ D k H t ‖ 2 + M ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ 2 ) + β ( ‖ D 2 m + k σ t ‖ 2 + ‖ D 2 m + k H t ‖ 2 ) ) ≤ c 3 λ 1 k ‖ D k σ t ‖ 2 + c 4 2 λ 1 m ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ ) + c 5 ‖ D 2 m + k σ t ‖ 2 + ( c 6 2 + c 7 2 4 c 5 λ 1 m + k ) ( ‖ D 2 m + k u ˜ ‖ 2 + ‖ D 2 m + k v ˜ ‖ 2 ) + h 12 + h 22 . (20)

At the same time, when U t + ε U = P , V t + ε V = P ∗ , there is

h 12 = − ( g 1 ( u ¯ , v ¯ ) − g 1 ( u , v ) − g 1 u ( u , v ) ( u ¯ − u ) − g 1 v ( u , v ) ( v ¯ − v ) ) ,

h 22 = − ( g 2 ( u ¯ , v ¯ ) − g 2 ( u , v ) − g 2 u ( u , v ) ( u ¯ − u ) − g 2 v ( u , v ) ( v ¯ − v ) ) .

From there, get

h 12 = − ∫ 0 1 [ { g 1 u ( u + P 1 ( u ¯ − u ) , v + P 1 ( v ¯ − v ) ) − g 1 u ( u , v ) } ( u ¯ − u ) + { g 1 v ( u + P 1 ( u ¯ − u ) , v + P 1 ( v ¯ − v ) ) − g 1 v ( u , v ) } ( v ¯ − v ) ] d P 1 ,

h 22 = − ∫ 0 1 [ { g 2 u ( u + P 2 ( u ¯ − u ) , v + P 2 ( v ¯ − v ) ) − g 2 u ( u , v ) } ( u ¯ − u ) + { g 2 v ( u + P 2 ( u ¯ − u ) , v + P 2 ( v ¯ − v ) ) − g 2 v ( u , v ) } ( v ¯ − v ) ] d P 2 ,

Get ∀ P 1 , P 2 ∈ [ 0 , 1 ] ,

‖ g 1 u ( u + P 1 ( u ¯ − u ) , v + P 1 ( v ¯ − v ) ) − g 1 u ( u , v ) ‖ ≤ l k P 1 κ 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 1 ,

‖ g 1 v ( u + P 1 ( u ¯ − u ) , v + P 1 ( v ¯ − v ) ) − g 1 v ( u , v ) ‖ ≤ l k P 1 κ 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 1 ,

‖ g 2 u ( u + P 2 ( u ¯ − u ) , v + P 2 ( v ¯ − v ) ) − g 2 u ( u , v ) ‖ ≤ l ′ k P 2 κ 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 2 ,

‖ g 2 v ( u + P 2 ( u ¯ − u ) , v + P 2 ( v ¯ − v ) ) − g 2 v ( u , v ) ‖ ≤ l ′ k P 2 κ 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 2 .

thereupon

‖ h 12 ‖ ≤ 2 c 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 1 + 1 ,

Similarly,

‖ h 22 ‖ ≤ 2 c ′ 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ V 2 m + k × V 2 m + k κ 2 + 1 .

By assumptions, the Formula (20) can be obtained

1 2 d d t ( ‖ D k σ t ‖ 2 + ‖ D k H t ‖ 2 + μ 0 ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ 2 ) ) ≤ 2 c ′ 3 λ 1 k ( ‖ D k σ t ‖ 2 + ‖ D k H t ‖ 2 ) + c ′ 4 2 λ 1 m ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ 2 )     + 2 ( c ′ 5 − β ) ( ‖ D 2 m + k σ t ‖ 2 + ‖ D 2 m + k H t ‖ 2 )     + ( c ′ 6 + c ′ 7 2 2 c ′ 5 λ 1 m + k ) ( ‖ D 2 m + k u ˜ ‖ 4 + ‖ D 2 m + k v ˜ ‖ 4 )     + 4 c 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ E k κ 1 + 1 + 4 c ′ 1 ‖ ( u ¯ , v ¯ ) − ( u , v ) ‖ E k κ 2 + 1

≤ ( 2 c ′ 3 λ 1 k + 2 ( c ′ 5 − β ) λ 1 4 m ) ( ‖ D k σ t ‖ 2 + ‖ D k H t ‖ 2 ) + c ′ 4 2 λ 1 m ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ 2 )     + ( c ′ 6 + c ′ 7 2 2 c ′ 5 λ 1 m + k + c 8 ) ( ‖ D 2 m + k u ˜ ‖ 4 + ‖ D 2 m + k v ˜ ‖ 4 ) .

Order α = max { 4 c ′ 3 λ 1 k + 4 ( c ′ 5 − β ) λ 1 4 m , c ′ 4 μ 0 λ 1 m } ,

Then there is

d d t ( ‖ D k σ t ‖ 2 + ‖ D k H t ‖ 2 + μ 0 ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ 2 ) ) ≤ α ( ‖ D k σ t ‖ 2 + ‖ D k H t ‖ 2 + μ 0 ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ 2 ) )       + c 10 ( ‖ D 2 m + k u ˜ ‖ 4 + ‖ D 2 m + k v ˜ ‖ 4 ) .

Using the Gronwall’s inequality, there is

d d t ( ‖ D k σ t ‖ 2 + ‖ D k H t ‖ 2 + μ 0 ( ‖ D 2 m + k σ ‖ 2 + ‖ D 2 m + k H ‖ 2 ) ) ≤ c 11 e c 12 t ‖ ( ξ 1 , η 1 , ξ 2 , η 2 ) T ‖ E k 4 .

Then when ‖ ( ξ 1 , η 1 , ξ 2 , η 2 ) T ‖ E k 2 → 0 ,

‖ S ( t ) Φ 0 − S ( t ) Φ ¯ 0 − F ( ( ξ 1 , η 1 , ξ 2 , η 2 ) T ) ‖ E k 2 ‖ ( ξ 1 , η 1 , ξ 2 , η 2 ) T ‖ E k 2 ≤ c 11 e c 12 t ‖ ( ξ 1 , η 1 , ξ 2 , η 2 ) T ‖ E k 2 → 0.

Theorem 4 is proved.

The Hausdorff and Fractal dimensions of the global attractors subfamily are estimated below.

The problem of linearization is reduced to Ψ ′ + P ( φ ) Ψ = Γ 1 ( φ ) Ψ + Γ 2 ( φ ) Ψ .

At this point, Ψ = ( U , P , V , P ∗ ) T ∈ E k , P = U t + ε U , P ∗ = V t + ε V , φ = ( u , z , v , q ) T ∈ E k are the solutions to Equation (21), Ψ ( 0 ) = { ξ , ζ , η , σ } ∈ E k , t > 0 .

P ( φ ) = ( ε I − I 0 0 ( 1 − β ε ) ( − Δ ) 2 m + ε 2 I β ( − Δ ) 2 m − ε I 0 0 0 0 ε I − I 0 0 ( 1 − β ε ) ( − Δ ) 2 m + ε 2 I β ( − Δ ) 2 m − ε I ) ,

Γ 1 ( φ ) = ( 0 0 0 0 − g 1 u ( u , v ) 0 − g 1 v ( u , v ) 0 0 0 0 0 − g 2 u ( u , v ) 0 − g 2 v ( u , v ) 0 ) ,

Γ 2 ( φ ) Ψ = ( 0 ( 1 − M ( s ) ) ( − Δ ) 2 m U − 2 M ′ ( s ) s ′ ( D m U + D m V ) ( − Δ ) 2 m u 0 ( 1 − M ( s ) ) ( − Δ ) 2 m V − 2 M ′ ( s ) s ′ ( D m U + D m V ) ( − Δ ) 2 m v ) .

where s = ‖ D m u ‖ p p + ‖ D m v ‖ p p , Order

D 1 = M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ D m U ( − Δ ) 2 m u ,

D 2 = M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ D m V ( − Δ ) 2 m u ,

D 3 = M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ D m U ( − Δ ) 2 m v ,

D 4 = M ′ ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ( ‖ D m u ‖ p p + ‖ D m v ‖ p p ) ′ D m V ( − Δ ) 2 m v .

Theorem 5 Under the conditions of theorem 4, problems (1)-(5) the global attraction subfamily A k has the Hausdorff dimension and the Fractal dimension, and

d H ( A k ) ≤ min { N | N ∈ N + , 1 N ∑ j N λ j t k − 1 < l 2 ϒ } ,

d F ( A k ) ≤ 2 N , ( k = 1 , 2 , ⋯ , 2 m )

Proof: Let N ∈ N + , the N solutions of the problem (21) are χ 1 , χ 2 , ⋯ , χ N , considering N of them, given time τ , there is

| χ 1 ( s ) ∧ χ 2 ( s ) ∧ ⋯ ∧ χ N ( s ) | ∧ N E k = | χ 1 ( 0 ) ∧ χ 2 ( 0 ) ∧ ⋯ ∧ χ N ( 0 ) | ∧ N E k exp ∫ 0 s T r F ′ ( φ ( τ ) ) ⋅ B N ( τ ) d τ .

where φ ( τ ) = ( u ( τ ) , p ( τ ) , v ( τ ) , q ( τ ) ) , B N ( τ ) = B N ( τ , φ 0 ; χ 1 ( 0 ) , ⋯ , χ N ( 0 ) ) is an orthogonal projection from E k to s p a n { χ 1 ( τ ) , ⋯ , χ N ( τ ) } , and y j ( τ ) = { ξ j , ζ j , η j , σ j } , j = 1 , ⋯ , N is B N ( τ ) E k = s p a n { χ 1 ( τ ) , χ 2 ( τ ) , ⋯ , χ N ( τ ) } standard orthogonal radicals.

Set the corresponding inner product and norm,

( y j , y ¯ j ) E k = ( D 2 m + k ξ j , D 2 m + k ξ ˜ j ) + ( D k ζ j , D k ζ ˜ j )     + ( D 2 m + k η j , D 2 m + k η ˜ j ) + ( D k σ j , D k σ ˜ j ) ,

‖ y j ‖ E k 2 = ( y j , y j ) E k = ‖ D 2 m + k ξ j ‖ 2 + ‖ D k ζ j ‖ 2 + ‖ D 2 m + k η j ‖ 2 + ‖ D k σ j ‖ 2 = 1.

T r F ′ ( φ ( τ ) ) ⋅ B N ( τ ) = ∑ j = 1 + ∞ ( F ′ ( φ ( τ ) ) ⋅ B N ( τ ) y j ( τ ) , y j ( τ ) ) E k = ∑ j = 1 N ( F ′ ( φ ( τ ) ) y j ( τ ) , y j ( τ ) ) E k .

so,

− ( P ( φ ) y j , y j ) = − ( ε D 2 m + k ξ j − D 2 m + k ζ j , D 2 m + k ξ j ) − ( ε D 2 m + k η j − D 2 m + k σ j , D 2 m + k η j )       − ( ( 1 − β ε ) ( − Δ ) 2 m D k ξ j + ε 2 D k ξ j , D k ζ j ) − ( β ( − Δ ) 2 m D k ζ j , D k ζ j )       − ( ε D k ζ j , D k ζ j ) − ( ( 1 − β ε ) ( − Δ ) 2 m D k η j + ε 2 D k η j , D k σ j )       − ( β ( − Δ ) 2 m D k σ j , D k σ j ) − ( ε D k σ j , D k σ j )

≤ β − 2 2 ε ‖ D 2 m + k ξ j ‖ 2 + β ε c 13 λ 1 2 m 2 ‖ D k ζ j ‖ 2 + ε 2 2 ‖ D k ξ j ‖ 2         + β − 2 2 ε ‖ D 2 m + k η j ‖ 2 + ( ε c 14 − 4 2 β λ 1 2 m + 2 ε 2 ) ‖ D k σ j ‖ 2 + ε 2 2 ‖ D k η j ‖ 2 . (22)

( Γ 1 ( φ ) y j , y j ) E k = ( − g 1 u ( u , v ) D k ξ j , D k ζ j ) + ( − g 1 v ( u , v ) D k η j , D k ζ j )         + ( − g 2 u ( u , v ) D k ξ j , D k σ j ) + ( − g 2 v ( u , v ) D k η j , D k σ j ) ≤ ‖ g 1 u ( u , v ) ‖ ∞ ‖ D k ξ j ‖ ‖ D k ζ j ‖ + ‖ g 1 v ( u , v ) ‖ ∞ ‖ D k η j ‖ ‖ D k ζ j ‖         + ‖ g 2 u ( u , v ) ‖ ∞ ‖ D k ξ j ‖ ‖ D k σ j ‖ + ‖ g 2 v ( u , v ) ‖ ∞ ‖ D k η j ‖ ‖ D k σ j ‖ ≤ c 15 + c 17 2 ‖ D k ξ j ‖ 2 + c 15 + c 16 2 ‖ D k ζ j ‖ 2 + c 16 + c 18 2 ‖ D k η j ‖ 2 + c 17 + c 18 2 ‖ D k σ j ‖ 2 . (23)

( Γ 2 ( φ ) y j , y j ) E k = ( 1 − M ( s ) ) ( ( − Δ ) 2 m D k ξ j , D k ζ j ) − 2 ( D k D 1 ξ j , D k ζ j ) − 2 ( D k D 2 ζ j , D k ζ j )       + ( 1 − M ( s ) ) ( ( − Δ ) 2 m D k η j , D k σ j ) − 2 ( D k D 3 ξ j , D k σ j ) − 2 ( D k D 4 ζ j , D k σ j ) = ( 1 − M ( s ) ) ( D 2 m + k ξ j , D 2 m + k ζ j ) − 2 ( D k D 1 ξ j , D k ζ j ) − 2 ( D k D 2 ζ j , D k ζ j )       + ( 1 − M ( s ) ) ( D 2 m + k η j , D 2 m + k σ j ) − 2 ( D k D 3 ξ j , D k σ j ) − 2 ( D k D 4 ζ j , D k σ j )

≤ 1 − μ 0 2 ( ‖ D 2 m + k ξ j ‖ 2 + ‖ D 2 m + k ζ j ‖ 2 ) + c 19 ‖ D k ξ j ‖ 2 + ( c 19 + 2 c 20 + c 22 ) ‖ D k ζ j ‖ 2       + 1 − μ 0 2 ( ‖ D 2 m + k η j ‖ 2 + ‖ D 2 m + k σ j ‖ 2 ) + c 21 ‖ D k ξ j ‖ 2 + c 21 ‖ D k σ j ‖ 2 ≤ 1 − μ 0 2 ‖ D 2 m + k ξ j ‖ 2 + 1 − μ 0 2 ‖ D 2 m + k η j ‖ 2       + ( c 23 ( 1 − μ 0 ) λ 1 2 m 2 + c 19 + 2 c 20 + c 22 ) ‖ D k ζ j ‖ 2       + ( c 24 ( 1 − μ 0 ) λ 1 2 m 2 + c 21 ) ‖ D k σ j ‖ 2 + ( c 19 + c 21 ) ‖ D k ξ j ‖ 2 . (24)

According to Formulas (22), (23) and (24), there is

( F ′ ( φ ) y j , y j ) E k = ( ( − P ( φ ) + Γ 1 ( φ ) + Γ 2 ( φ ) ) h j , h j ) ≤ ( β − 2 2 ε + 1 − μ 0 2 ) ‖ D 2 m + k ξ j ‖ 2 + ( β ε c 13 λ 1 2 m 2 + c 15 + c 16 2 + c 23 ( 1 − μ 0 ) λ 1 2 m 2         + c 19 + 2 c 20 + c 22 ) ‖ D k ζ j ‖ 2 + ( β − 2 2 ε + 1 − μ 0 2 ) ‖ D 2 m + k η j ‖ 2     + ( ε c 14 − 4 2 β λ 1 2 m + 2 ε 2 + c 17 + c 18 2 + c 24 ( 1 − μ 0 ) λ 1 2 m 2 + c 21 ) ‖ D k σ j ‖ 2     + ( ε 2 2 + c 15 + c 17 2 + c 19 + c 21 ) ‖ D k ξ j ‖ 2 + ( ε 2 2 + c 16 + c 18 2 ) ‖ D k η j ‖ 2 (25)

Order

l = min { 2 − β 2 ε + μ 0 − 1 2 , − β ε c 13 λ 1 2 m 2 − c 15 + c 16 2                         + c 23 ( μ 0 − 1 ) λ 1 2 m 2 − c 19 − 2 c 20 − c 22 ,                         4 − ε c 14 2 β λ 1 2 m − 2 ε 2 − c 17 + c 18 2 + c 24 ( μ 0 − 1 ) λ 1 2 m 2 − c 21 } ,

ϒ = max { ε 2 2 + c 15 + c 17 2 + c 19 + c 21 , ε 2 2 + c 16 + c 18 2 } ,

To all time τ , there are 0 < t k = k 2 m + k < 1 , making

∑ j = 1 N ( F ′ ( φ ) y j , y j ) E k = ∑ j = 1 N ( ( − P ( φ ) + Γ 1 ( φ ) + Γ 2 ( φ ) ) y j , y j ) ≤ − N l + ϒ ∑ j N ( ‖ D k ξ j ‖ 2 + ‖ D k η j ‖ 2 ) ≤ − N l + 2 ϒ ∑ j N λ j t k − 1 ,

If 1 N ∑ j N λ j t k − 1 < l 2 ϒ ,

q N ( s ) = sup φ 0 ∈ A k sup χ j ∈ E k { 1 t ∫ 0 t T r F ′ ( φ ( τ ) ) ∘ P N ∗ ( τ ) d τ } ≤ − N ( l − 2 ϒ N ∑ j N λ j t k − 1 ) ,

then q N = lim s → ∞ sup q N ( s ) < 0 .

Theorem 5 is proved.

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

Lin, G.G. and Zhou, J.Y. (2022) The Family of Global Attractors for Kirchhoff-Type Coupled Equations. Journal of Applied Mathematics and Physics, 10, 2040-2060. https://doi.org/10.4236/jamp.2022.106139