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 OJAppS  Vol.11 No.7 , July 2021
The Family of Global Attractors for a Generalized Kirchhoff Equations
Abstract: This paper deals with the existence and uniqueness of solutions of generalized Kirchhoff equations and the family of global attractors for the equation and its dimension estimation. First, the stress term of Kirchhoff equation is properly assumed. When certain conditions are met between the order m and the degree p of Banach space , the existence and uniqueness of the solution of equation are obtained by a prior estimation and Galerkin’s method; Then, the bounded absorption set is obtained by prior estimation, and it is proved that the solution semigroup generated by the equation has a family of global attractors in phase space by using Rellich-Kondrachov compact embedding theorem. Further, the equation is linearized and rewritten into a first-order variational equation, and it is proved that the solution semigroup is Fréchet differentiable on ; Finally, the upper bound of Hausdorff dimension and Fractal dimension of is estimated, and the Hausdorff dimension and Fractal dimension are finite.

1. Introduction

This paper will study the initial-boundary value problems of the following generalized Kirchhoff equations:

u t t + M ( m u p p ) ( Δ ) 2 m u + β ( Δ ) 2 m u t + | u | ρ ( u t + u ) = f ( x ) , (1)

u ( x , t ) = 0 , i u v i = 0 , i = 1 , 2 , , 2 m 1 , x Ω , t > 0 , (2)

u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u t ( x ) , x Ω R n . (3)

where m N + , Ω R n ( n 1 ) is a bounded domain with smooth boundary Ω , f ( x ) is an external force term, M ( m u p p ) is the stress term of Kirchhoff equation, β > 0 , β ( Δ 2 m ) u t is a strong dissipative term, | u | ρ ( u t + u ) is a nonlinear source term.

Many scholars have studied the existence of global attractor of Kirchhoff equation with strong dissipative term, [1] - [7] can be referred.

In reference [8], scholars considered the following Kirchhoff type wave equation with nonlinear strong damping term

u t t ε 1 Δ u t + α | u t | p 1 u t + β | u | q 1 u φ ( u 2 ) Δ u = f ( x ) , ( x , t ) Ω × R + , (4)

u ( x , 0 ) = u 0 ( x ) ; u t ( x , 0 ) = u 1 ( x ) , x Ω , (5)

u ( x , t ) | Ω = 0 , Δ u ( x , t ) | Ω = 0 , x Ω . (6)

Here, Ω is a bounded domain with smooth boundary Ω in R N , ε 1 , α , β are normal numbers.

TokioMatsuyama and RyoIkehata [9] proved the global solution and attenuation of the solution of Kirchhoff type wave equation with nonlinear damping:

u t t M ( u ( t ) 2 2 ) Δ u + δ | u t | p 1 u t = μ | u | q 1 u . (7)

With compact boundary conditions

u ( x , t ) | Ω = 0 , t 0 . (8)

FucaiLi discussed the higher-order Kirchhoff type equations with nonlinear terms in reference [10]:

u t t + ( Ω | m u | 2 ) q ( Δ ) m u + u t | u t | r = | u | p u , x Ω , t > 0 , (9)

u ( x , t ) = 0 , i u v i = 0 , i = 1 , 2 , , m 1 , x Ω , t > 0 , (10)

u ( x , 0 ) = u 0 , u t = u 1 ( x ) , x Ω . (11)

In a bounded domain, where m > 1 is a positive integer, p , q , r > 0 is a normal number, if p r , the existence of global solution will be obtained, if p > max { r , 2 q } , for any initial value with negative initial energy, the solution explodes in a finite time. For more related research results, please refer to references [11] [12] [13] [14].

In this paper, for the convenience of narration, the following spaces and marks are defined: H = L 2 ( Ω ) , H 0 2 m ( Ω ) = H 2 m ( Ω ) H 0 1 ( Ω ) , H 0 2 m + k ( Ω ) = H 2 m + k ( Ω ) H 0 1 ( Ω ) , E k = H 0 2 m + k ( Ω ) × H 0 k ( Ω ) , ( k = 0 , 1 , 2 , , 2 m ) . Define ( , ) and to represent the inner product and norm of H respectively,

namely

( u , v ) = Ω u ( x ) v ( x ) d x , ( u , u ) = u 2 .

Let A k be the family of global attractor from E 0 to E k , B 0 k be the bounded absorption set in E k , where k = 1 , 2 , , 2 m . C i ( i = 0 , 1 , 2 , ) represents a constant.

(H1) assume that Kirchhoff type stress term M ( s ) C 2 ( [ 0, + ] , R ) satisfies:

1 < μ 0 M ( s ) μ 1 , μ = { μ 0 , d d t 2 m u 2 0 , μ 1 , d d t 2 m u 2 < 0.

where μ is a constant.

(H2) ρ 8 m n

There exist constant ε , 0 < ε < min { 1 + β λ 1 2 m 2 1 , 2 μ 0 , 2 μ 0 1 λ 1 2 m + β } .

2. The Existence and Uniqueness of Global Solution

In this section, under the assumption of Kirchhoff stress term, the existence and uniqueness of global solution are obtained by prior estimation and Galerkin’s method.

Lemma 2.1 Suppose Kirchhoff stress term M ( s ) satisfies the conditions (H1), Assume that (H2) holds, f H , ( u 0 , v 0 ) E 0 , v = u t + ε u , ε > 0 , then the smooth solution of the initial-boundary value problem (1.1)-(1.3) satisfies ( u , v ) E 0 , u L ( 0, + ; H 2 m ( Ω ) ) , v L ( 0, + ; L 2 ( Ω ) ) L 2 ( 0, T ; H 2 m ( Ω ) ) , and satisfy that following inequality

( u , v ) E 0 2 = 2 m u 2 + v 2 ( v 0 2 + ε 2 u 0 2 + μ 2 m u 0 2 + 2 ( ε + 1 ) ρ + 2 u 0 L ρ + 2 ( Ω ) ρ + 2 ) e α 1 t + C 0 α 1 ( 1 e α 1 t ) , (1)

β 2 0 T 2 m v 2 d t C 0 T + y 1 ( 0 ) . (2)

where α 1 = min { 2 ε , a 1 μ , a 2 , ε ( ρ + 2 ) ε + 1 } , y 1 ( 0 ) = v 0 2 + ε 2 u 0 2 + μ 2 m u 0 2 + 2 ( ε + 1 ) ρ + 2 u 0 L ρ + 2 ( Ω ) ρ + 2 .

So there is a non-negative real number C ( R 1 ) and t = t 1 ( Ω ) > 0 , make

( u , v ) E 0 2 2 C 0 α 1 = C ( R 1 ) , ( t > t 1 ) . (3)

Proof. Set v = u t + ε u , take the inner product of both sides of Equation (1.1) with v in H, we obtain

( u t t + M ( m u p p ) ( Δ ) 2 m u + β ( Δ ) 2 m u t + | u | ρ ( u t + u ) , v ) = ( f ( x ) , v ) . (4)

( u t t , v ) = 1 2 d d t v 2 + ε 2 1 2 d d t u 2 ε v 2 + ε 3 u 2 . (5)

According to hypothesis (H1), we can get

( M ( m u p p ) ( Δ ) 2 m u , v ) = ( M ( m u p p ) ( Δ ) 2 m u , u t + ε u ) μ 2 d d t 2 m u 2 + ε μ 0 2 m u 2 . (6)

By Young’s inequality and Poincaré’s inequality

( β ( Δ ) 2 m u t , v ) = β 2 m v 2 ε β ( 2 m u , 2 m v ) β 2 m v 2 β ( ε 2 m u 2 m v ) β 2 2 m v 2 ε 2 2 2 m u 2 . (7)

where λ 1 is the first eigenvalue of Δ with homogeneous Dirichlet boundary condition on Ω .

we can get

( | u | ρ ( u t + u ) , v ) = ( | u | ρ , u t 2 ) + ε ( | u | ρ u t , u ) + ( | u | ρ u , u t ) + ε ( | u | ρ u , u ) ε Ω | u | ρ + 2 d x + ( ε + 1 ) Ω | u | ρ u u t d x ε u L ρ + 2 ( Ω ) ρ + 2 + ε + 1 ρ + 2 d d t u L ρ + 2 ( Ω ) ρ + 2 . (8)

According to Young’s inequality, we get

( f ( x ) , v ) f v ε 2 2 v 2 + 1 2 ε 2 f 2 . (9)

Substitute (2.5)-(2.9) into (2.4),

d d t ( μ 2 m u 2 + v 2 + ε 2 u 2 + 2 ( ε + 1 ) ρ + 2 u L ρ + 2 ( Ω ) ρ + 2 ) + ( 2 ε μ 0 ε 2 ) 2 m u 2 + β 2 2 m v 2 + ( β λ 1 2 m 2 2 ε ε 2 ) v 2 + 2 ε 3 u 2 + 2 ε u L ρ + 2 ( Ω ) ρ + 2 1 ε 2 f 2 = C 0 . (10)

Make a 1 = 2 ε μ 0 ε 2 0 , a 2 = β λ 1 2 m 2 2 ε ε 2 0 .

Take α 1 = min { 2 ε , a 1 μ , a 2 , ( ρ + 2 ) ε ε + 1 , } .

Then (2.10) can be converted into

d d t ( μ 2 m u 2 + v 2 + ε 2 u 2 + 2 ( ε + 1 ) ρ + 2 u L ρ + 2 ( Ω ) ρ + 2 ) + α 1 ( μ 2 m u 2 + v 2 + ε 2 u 2 + 2 ( ε + 1 ) ρ + 2 u L ρ + 2 ( Ω ) ρ + 2 ) + β 2 2 m v 2 C 0 . (11)

By Gronwall’s inequality

2 m u 2 + v 2 ( v 0 2 + ε 2 u 0 2 + μ 2 m u 0 2 + 2 ( ε + 1 ) ρ + 2 u 0 L ρ + 2 ( Ω ) ρ + 2 ) e α 1 t + C 0 α 1 ( 1 e α 1 t ) , (12)

β 2 0 T 2 m v 2 d t C 0 T + y 1 ( 0 ) . (13)

where y 1 = v 2 + ε 2 u 2 + μ 2 m u 2 + 2 ( ε + 1 ) ρ + 2 u L ρ + 2 ( Ω ) ρ + 2 .

So there is a non-negative real number C ( R 1 ) and t = t 1 ( Ω ) > 0 , such as

( u , v ) E 0 2 2 C 0 α 1 = C ( R 1 ) , ( t > t 1 ) . (14)

Lemma 2.1 is proved.

Lemma 2.2 Assume that (H1), (H2) holds, if f ( x ) H , ( u 0 , v 0 ) E k ( k = 1 , 2 , , 2 m ) , v = u t + ε u , ε > 0 , Then the smooth solution ( u , v ) E k , ( k = 1 , 2 , , 2 m ) of the initial-boundary value problem (1.1)-(1.3) satisfies u L ( 0, + ; H 0 2 m + k ( Ω ) ) , v L ( 0, + ; H k ( Ω ) ) L 2 ( 0, T ; H 0 2 m + k ( Ω ) ) and satisfy that following inequality

( u , v ) E k 2 = 2 m + k u 2 + k v 2 ( k v 0 2 + ( μ β ε ) 2 m + k u 0 2 ) e α 2 t + C 3 α 2 ( 1 e α 2 t ) , (15)

β 6 0 T 2 m + k v 2 d t y 2 ( 0 ) + C 3 T . (16)

So there is a non-negative real number C ( R 2 ) and t = t 2 ( Ω ) > 0 , make

( u , v ) E k 2 2 C 3 α 2 = C ( R 2 ) , ( t > t 2 ) . (17)

Proof. Take the inner product of ( Δ ) k v = ( Δ ) k u t + ( Δ ) k ε u and the two sides of Equation (1.1), and get

( u t t + M ( m u p p ) ( Δ ) 2 m u + β ( Δ ) 2 m u t + | u | ρ ( u t + u ) , ( Δ ) k v ) = ( f ( x ) , ( Δ ) k v ) . (18)

By Young’s inequality, Poincaré’s inequality

( u t t , ( Δ ) k v ) 1 2 d d t k v 2 2 ε + ε 2 2 k v 2 ε 2 2 λ 1 2 m 2 m + k u 2 . (19)

According to hypothesis (H1)

( M ( m u p p ) , ( Δ ) k v ) μ 2 d d t 2 m + k u 2 + ε μ 0 2 m + k u 2 . (20)

By Young’s inequality, Poincaré’s inequality

( β ( Δ ) 2 m u t , ( Δ ) k v ) = β 2 m + k v 2 β ε 2 m + k u 2 m + k v β 6 2 m + k v 2 β ε 2 2 2 m + k u 2 + β λ 1 2 m 3 k v 2 . (21)

where λ 1 is the first eigenvalue of Δ with homogeneous Dirichlet boundary condition on Ω .

we can get

( | u | ρ ( u t + u ) , ( Δ ) k v ) = ( | u | ρ v , ( Δ ) k v ) ε ( | u | ρ u , ( Δ ) k v ) + ( | u | ρ u , ( Δ ) k v ) ( 1 ε ) ( | u | ρ u , ( Δ ) k v ) ,

By Young’s inequality, Poincaré’s inequality

( 1 ε ) ( | u | ρ u , ( Δ ) k v ) ( 1 ε ) 2 k v ( Ω | u | 2 ρ + 2 d x ) 1 2 , C 1 2 k v β 12 2 m + k v 2 + C 2 . (22)

By Schwarz’s inequality, Young’s inequality

( f ( x ) , ( Δ ) k v ) ε 2 2 k v 2 + 1 2 ε 2 k f ( x ) 2 . (23)

Substitute (2.19)-(2.23) into (2.18) to get

d d t ( k v 2 + μ 2 m + k u 2 ) + ( 2 β λ 1 2 m 3 2 ε 2 ε 2 ) k v 2 + ( 2 ε μ 0 ε 2 λ 1 2 m β ε 2 ) 2 m + k u 2 + β 6 2 m + k v 2 C 3 . (24)

Take a 3 = 2 β λ 1 2 m 3 2 ε 2 ε 2 0 , a 4 = 2 ε μ 0 ε 2 λ 1 2 m β ε 2 0 .

Let α 2 = min { 2 β λ 1 2 m 3 2 ε 2 ε 2 , 2 ε μ 0 ε 2 λ 1 2 m β ε 2 μ } .

Then (2.24) can be converted into

d d t ( k v 2 + μ 2 m + k u 2 ) + α 2 ( k v 2 + μ 2 m + k u 2 ) + β 6 2 m + k v 2 C 3 . (25)

By Gronwall’s inequality

2 m + k u 2 + k v 2 ( μ 2 m + k u 0 2 + k v 0 2 ) e α 2 t + C 3 α 2 ( 1 e α 2 t ) , (26)

β 6 0 T 2 m + k v 2 d t y 2 ( 0 ) + C 3 T . (27)

where y 2 = k v 2 + μ 2 m + k u 2 .

So there is a non-negative real number C ( R 2 ) and t = t 2 ( Ω ) > 0 , make

( u , v ) E k 2 2 C 3 α 2 = C ( R 2 ) , ( t > t 2 ) . (28)

Lemma 2.2 is proved.

Theorem 2.1 Under the assumption of lemma 2.1 and lemma 2.2, and satisfy the hypothesis (H1), (H2), Then the initial-boundary value problem (1.1)-(1.3) has a unique smooth solution ( u , v ) L ( 0 , + ; E k ) , v L 2 ( 0 , T ; H 0 2 m + k ( Ω ) ) , ( k = 0 , 1 , 2 , , 2 m ) .

Proof. Existence: Galerkin’s method is used to prove the existence of global smooth solution.

Step 1: construct an approximate solution.

Let ( Δ ) 2 m + k w j = λ j w j , k = 0 , 1 , 2 , , 2 m . where λ j is the eigenvalue of Δ with homogeneous Dirichlet boundary on Ω , w j denotes the eigenfunction determined by the corresponding eigenvalue λ j , w 1 , , w n constitute the orthonormal basis of H from the eigenvalue theory.

Let the approximate solution of the problem (1.1)-(1.3) be u s = u s ( t ) = j = 1 s g j s ( t ) w j , where g j s ( t ) is determined by the following equations.

( u s t t + M ( m u s p p ) ( Δ ) 2 m u s + β ( Δ ) 2 m u s t + | u s | ρ ( u s t + u s ) , w j ) = ( f ( x ) , w j ) . (29)

( 1 j s )

The formula (2.29) satisfies the initial condition u s ( 0 ) = u 0 s , u s t ( 0 ) = u 1 s .

When s + , ( u 0 s , u 1 s ) ( u 0 , u 1 ) in E k , according to the basic theory of ordinary differential equations, the approximate solution u s ( t ) exists on ( 0, t s ) .

Step 2: Prior estimation.

v s ( t ) = u s t ( t ) + ε u s ( t ) , multiplying by g j s ( t ) + ε g j s ( t ) and summing over j, we can get

1) k = 0 , by lemma 2.1, there is

( u s , v s ) E 0 2 y 1 ( 0 ) e α 1 t + C 0 α 1 ( 1 e α 1 t ) , (30)

β 2 0 T 2 m v 2 d t C 0 T + y 1 ( 0 ) . (31)

2) k = 1 , 2 , , 2 m , by lemma 2.2, there is

( u s , v s ) E k 2 y 2 ( 0 ) e α 2 t + C 3 α 2 ( 1 e α 2 t ) , (32)

β 6 0 T 2 m + k v 2 d t y 2 ( 0 ) + C 3 T . (33)

From (2.30) and (2.32), ( u s , v s ) is bounded in L ( 0, + ; E k ) , ( u s , v s ) in L ( 0, + ; E 0 ) is bounded.

It can be seen that the formula (2.30)-(2.33) holds a priori estimates for lemma 2.1 and lemma 2.2 respectively.

Step 3: Limit process.

In E k ( k = 0 , 1 , 2 , , 2 m ) space, select the subsequence { u h } from the sequence { u s } ,

Make ( u h , v h ) ( u , v ) in L ( 0, + ; E k ) weak * convergence.

According to Rellich-Kondrachov compact embedding theorem, E k compactly embeds E 0 , Then ( u h , v h ) ( u , v ) converges strongly almost everywhere in E 0 .

Let s = h in (2.29), and take the limit, for fixed j, h j ,

Then from (2.29), make u s u in L ( 0, + ; H 0 2 m + k ( Ω ) ) weak * convergence.

Thus ( u h ( t ) , ( Δ ) k w j ) ( u ( t ) , λ j k w j ) in L ( 0, + ) weak * convergence.

( u h t ( t ) , ( Δ ) k w j ) ( u t ( t ) , λ j k w j ) in L ( 0, + ) weak * convergence.

So ( u h t t , ( Δ ) k w j ) = d d t ( u h t , ( Δ ) k w j ) ( u h t , λ j k w j ) in D [ 0, + ) convergence.

D [ 0, + ) is a conjugate space of D [ 0, + ) infinite differentiable space.

( M ( m u h p p ) ( Δ ) k u h , ( Δ ) k w j ) ( M ( m u h p p ) ( Δ ) 2 m + k 2 u , λ j 2 m + k 2 w j ) in L ( 0, + ) weak * convergence.

( β ( Δ ) 2 m u h t , ( Δ ) k w j ) β ( ( Δ ) k 2 v , λ j 2 m + k 2 w j ) ε β ( ( Δ ) 2 m + k 2 u , λ j 2 m + k 2 w j ) in L ( 0, + ) weak * convergence.

u h t u t in E 0 strong convergence, almost everywhere convergence.

| u h | ρ ( u h t + u h ) w . (34)

In L ( 0, + ; L ρ + 2 ρ + 1 ( Ω ) ) weak * convergence.

Take D = Ω , g h = | u h | ρ ( u h t + u h ) . (2.34) almost everywhere g h | u | ρ ( u t + u ) , g h w in L ρ + 2 ρ + 1 ( Ω ) weak convergence.

Therefore w = g = | u | ρ ( u t + u ) , so ( | u h | ρ ( u h t + u h ) , ( Δ ) k w j ) ( | u | ρ ( u t + u ) , ( Δ ) k w j ) in L ( 0, + ) weak * convergence. In particular, u 0 h u 0 weak convergence in the E k , u 1 h u 1 weak convergence in E k , From all j and h + ,it can be introduced

( u t t + M ( m u p p ) ( Δ ) 2 m u + β ( Δ ) 2 m u t + | u | ρ ( u t + u ) , ( Δ ) k w j ) = ( f ( x ) , ( Δ ) k w j ) .

Because of the density of w 1 , w 2 , , w k , .

( u t t + M ( m u p p ) ( Δ ) 2 m u + β ( Δ ) 2 m u t + | u | ρ ( u t + u ) , v ) = ( f ( x ) , v ) , v H 0 2 m + k ( Ω ) .

Therefore, the existence is proved.

The uniqueness of the solution.

Set u * , v * is equations of two solutions, make w = u * v * , w satisfies

w t t + M ( m u * p p ) ( Δ ) 2 m u * M ( m v * p p ) ( Δ ) 2 m v * + β ( Δ ) 2 m w t = | v * | ρ ( v t * + v * ) | u * | ρ ( u t * + u * ) , (35)

w ( 0 ) = 0 , w ( 0 ) = 0 , x Ω R n .

Take inner product of (2.35) and w t in H is as follows

( w t t + M ( m u * p p ) ( Δ ) 2 m u * M ( m v * p p ) ( Δ ) 2 m v * + β ( Δ ) 2 m w t , w t ) = ( | v * | ρ ( v t * + v * ) | u * | ρ ( u t * + u * ) , w t ) . (36)

Therefore

( w t t , w t ) = 1 2 d d t w t 2 , (37)

( β ( Δ ) 2 m w t , w t ) = β 2 m w t 2 . (38)

By differential mean value theorem, Young’s inequality

( M ( m u * p p ) ( Δ ) 2 m u * M ( m v * p p ) ( Δ ) 2 m v * , w t ) 1 2 M ( m u * p p ) d d t 2 m w 2 ( M ( ξ 1 ) ( m u * p p m v * p p ) ( Δ ) 2 m v * , w t ) μ 2 d d t 2 m w 2 β 2 2 m w t 2 C 4 2 2 β 2 m w 2 . (39)

where ξ 1 = δ u * + ( 1 δ ) v * , 0 < δ < 1 .

By Young’s inequality

( | v * | ρ ( v t * + v * ) | u * | ρ ( u t * + u * ) , w t ) C 5 w t 2 Ω ( | v * | ρ + | u * | ρ ) d x + C 6 w w t Ω ( | v * | ρ + | u * | ρ ) d x C 5 ( | v * | ρ L ( Ω ) + | u * | ρ L ( Ω ) ) w t 2 + C 6 ( | v * | ρ L ( Ω ) + | u * | ρ L ( Ω ) ) 2 w t 2 + C 6 ( | v * | ρ L ( Ω ) + | u * | ρ L ( Ω ) ) 2 w 2

by the interpolation inequality

v * ρ C 7 2 m v * ρ n 4 m ,

in the same way with

u * ρ C 8 2 m u * ρ n 4 m ,

where ρ 8 m n .

By Poincaré’s inequality

( | v * | ρ ( v t * + v * ) | u * | ρ ( u t * + u * ) , w t ) C 9 ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) w t 2 + C 10 ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) w 2 C 9 ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) w t 2 + C 10 λ 1 2 m ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) 2 m w 2 . (40)

By (2.37)-(2.40)

d d t ( w t 2 + μ 2 m w 2 ) 2 C 9 ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) w t 2 + ( C 4 2 β + 2 C 10 ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) λ 1 2 m ) 2 m w 2 . (41)

Take the

α 3 = 2 C 9 ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) + C 4 2 β + 2 C 10 ( 2 m v t * ρ n 4 m + 2 m u t * ρ n 4 m ) λ 1 2 m μ ,

there are

d d t y 3 ( t ) α 3 ( t ) y 3 ( t ) . (42)

where y 3 ( t ) = w t 2 + μ 2 m w 2 .

By Gronwall’s inequality

y 3 ( t ) y 3 ( 0 ) e 0 T α 3 ( t ) d t = 0. (43)

Thus y 3 ( t ) = 0 , or u * = v * , therefore, the uniqueness is proved.

Theorem 2.2 [11] Let E be a Banach space and S ( t ) : E E semigroups satisfy the following conditions.

1) semigroup S ( t ) is uniformly bounded in E, and R > 0 , there is a constant C ( R ) , so that when u E R , there is

S ( t ) u E C ( R ) . ( t [ 0, ) ) .

2) There is a bounded absorption set B 0 in E.

3) S ( t ) ( t > 0 ) is a fully continuous operator

Then a semigroup S ( t ) is said to have a compact global attractor A 0 .

Theorem 2.2 in Banach space E change to the Hilbert space E k , has the following the existence theorem of the family of global attractor.

Theorem 2.3 If the global smooth solution of the problem (1.1)-(1.3) satisfies the assumptions and conditions of lemma 2.1 and lemma 2.2, then the problem (1.1)-(1.3) have a family of global attractor A k , ( k = 1 , 2 , , 2 m ) . That is, there is a compact set A k , which makes:

1) S ( t ) A k = A k , ( t > 0 ) .

2) lim t d i s t ( S ( t ) B 0 k , A k ) = 0 , ( B 0 k E k ) .

where 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 , S ( t ) is the solution semigroup of (1.1)-(1.3).

Proof. It is necessary to verify the hypothesis (1), (2), (3) of theorem 2.2. It is easy to know that the Equation (1.1) has a solution semigroup S ( t ) : E k E k under the hypothesis of theorem 2.3.

1) by lemma 2.1, lemma 2.2, bounded set for B 0 k E k and contained in { ( u , v ) E k R k } .

S ( t ) ( u 0 , v 0 ) E k 2 = u H 0 2 m + k ( Ω ) 2 + v H 0 k ( Ω ) 2 u 0 H 0 2 m + k ( Ω ) 2 + v 0 H 0 k ( Ω ) 2 R k 2 .

where t 0, ( u 0 , v 0 ) B 0 k , this suggests that the { S ( t ) } ( t 0 ) in E k uniformly bounded.

2) by lemma 2.1, lemma 2.2, there are further

S ( t ) ( u 0 , v 0 ) E k 2 = u H 0 2 m + k ( Ω ) 2 + v H 0 k ( Ω ) 2 R k 2 , t t 0 = t 0 ( R k ) .

So B 0 k = { ( u , v ) E k : u H 0 2 m + k ( Ω ) 2 + v H 0 k ( Ω ) 2 R k 2 } is the bounded absorbing set of semigroup S ( t ) .

3) since E k E 0 is embedded, then E k bounded set of compact set of E 0 , so the family of semigroup operators S ( t ) is continuous, so the equation exists a family of global attractor A k = w ( B 0 k ) = τ 0 t τ S ( t ) B 0 k ¯ .

3. Estimation of the Dimension of the Family of Global Attractor

Firstly, we linearize the equation into a first-order variational equation and prove that the solution semigroup S ( t ) is Fréchet differentiable on E k , and further prove the attenuation of the volume element of the linearization problem. Finally, the upper bound of Hausdorff dimension and Fractal dimension of A k is estimated.

The Equations (1.1)-(1.3) is linearized

U t t + M ( m u p p ) ( Δ ) 2 m U + M ( m u p p ) ( m u p p ) m U ( Δ ) 2 m u + β ( Δ ) 2 m U t + ρ | u | ρ 1 u t U + | u | ρ U t + ( ρ + 1 ) | u | ρ U = 0 , (1)

U ( x , t ) = 0 , x Ω , t > 0 , (2)

U ( x , 0 ) = ξ , U t ( x , 0 ) = η . (3)

where ( ξ , η ) E k , ( u , u t ) = S ( t ) ( u 0 , u 1 ) is the solution of the problem (1.1)-(1.3) with ( u 0 , u 1 ) A k . Given ( u 0 , u 1 ) A k , S ( t ) : E k E k , it can be proved that for any ( ξ , η ) E k , there is a unique solution ( U ( t ) , U t ( t ) ) L ( 0, + ; E k ) to the linearized initial-boundary value problem.

Lemma 3.1 if S ( t ) : E k E k , Fréchet differential on η 0 = ( u 0 , u 1 ) is a linear operator F : ( ξ , η ) ( U ( t ) , U t ( t ) ) , let t > 0 , R > 0 , and the mapping S ( t ) : E k E k is Fréchet differentiable on E k , where ( U ( t ) , U t ( t ) ) is the solution of linearized initial-boundary value problem.

Proof.set η 0 = ( u 0 , u 1 ) E k , η ¯ 0 = ( u 0 + ξ , u 1 + η ) E k , and η 0 E k R , η ¯ 0 E k R , make η 1 = S ( t ) η 0 = ( u , v ) , η ¯ 1 = S ( t ) η ¯ 0 = ( u ¯ , v ¯ ) , In which the semigroup S ( t ) is Lipschitz continuous on the bounded set of E k , that is, S ( t ) η 0 S ( t ) η ¯ 0 E k 2 e C 25 t ( ξ , η ) E k 2 , Make θ = U t + ε U . so you can get it.

θ t ε θ + ε 2 U + M ( m u p p ) ( Δ ) 2 m U + M ( m u p p ) ( m u p p ) m U ( Δ ) 2 m u + β ( Δ ) 2 m ( θ ε U ) + ρ | u | ρ 1 u t U + | u | ρ ( θ ε U ) + ( ρ + 1 ) | u | ρ U = 0 , (4)

θ ( 0 ) = 0 , θ t ( 0 ) = 0. (5)

Make ( ϕ , φ ) = ( u ¯ u U , v ¯ v θ ) .

{ u ¯ t t + M ( m u ¯ p p ) ( Δ ) 2 m u ¯ + β ( Δ ) 2 m u ¯ t + | u ¯ | ρ ( u ¯ t + u ¯ ) = f ( x ) , u t t + M ( m u p p ) ( Δ ) 2 m u + β ( Δ ) 2 m u t + | u | ρ ( u t + u ) = f ( x ) , U t t + M ( m u p p ) ( Δ ) 2 m U + M ( m u p p ) ( m u p p ) m U ( Δ ) 2 m u + β ( Δ ) 2 m U t + ρ | u | ρ 1 u t U + | u | ρ U t + ( ρ + 1 ) | u | ρ U = 0. (6)

Subtract these three equations to get:

ϕ t t + M ( m u ¯ p p ) ( Δ ) 2 m u ¯ M ( m u p p ) ( Δ ) 2 m u M ( m u p p ) ( Δ ) 2 m U M ( m u p p ) ( m u p p ) m U ( Δ ) 2 m u + β ( Δ ) 2 m ϕ t + | u ¯ | ρ ( u ¯ t + u ¯ ) | u | ρ ( u t + u ) ρ | u | ρ 1 u t U | u | ρ U t ( ρ + 1 ) | u | ρ U = 0. (7)

where

H = M ( m u ¯ p p ) ( Δ ) 2 m u ¯ M ( m u p p ) ( Δ ) 2 m u M ( m u p p ) ( Δ ) 2 m U M ( m u p p ) ( m u p p ) m U ( Δ ) 2 m u + | u ¯ | ρ ( u ¯ t + u ¯ ) | u | ρ ( u t + u ) ρ | u | ρ 1 u t U | u | ρ U t ( ρ + 1 ) | u | ρ U . (8)

Make H = h 1 + h 2 .

h 1 = M ( m u ¯ p p ) ( Δ ) 2 m u ¯ M ( m u p p ) ( Δ ) 2 m u M ( m u p p ) ( Δ ) 2 m U M ( m u p p ) ( m u p p ) m U ( Δ ) 2 m u = M ( m u ¯ p p ) ( Δ ) 2 m u ¯ M ( m u p p ) ( Δ ) 2 m u ¯ + M ( m u p p ) ( Δ ) 2 m ϕ M ( m u p p ) ( m u p p ) m ( u ¯ u ϕ ) ( Δ ) 2 m u . (9)

By the differential mean value theorem

h 1 = M ( m ζ ¯ p p ) ( m ζ ¯ p p ) m ( u ¯ u ) ( Δ ) 2 m u ¯ M ( m u p p ) ( m u p p ) m ( u ¯ u ) ( Δ ) 2 m u + M ( m u p p ) ( m u p p ) m ϕ ( Δ ) 2 m u + M ( m u p p ) ( Δ ) 2 m ϕ = M ( m ς ¯ p p ) ( m ς ¯ p p ) ( 1 s ) ( m ( u ¯ u ) ) 2 ( Δ ) 2 m u ¯ + M ( m u p p ) ( m u p p ) m ( u ¯ u ) ( Δ ) 2 m ( u ¯ u ) + M ( m u p p ) ( m u p p ) m ϕ ( Δ ) 2 m u + M ( m u p p ) ( Δ ) 2 m ϕ . (10)

where ζ ¯ = s m u ¯ + ( 1 s ) m u , s ( 0 , 1 ) , ς ¯ = s m ζ ¯ + ( 1 s ) m u , s ( 0 , 1 ) .

Take inner product of h 1 and ( Δ ) k ϕ t , there is

| ( M ( m ς ¯ p p ) ( m ς ¯ p p ) ( 1 s ) ( m ( u ¯ u ) ) 2 ( Δ ) 2 m u ¯ , ( Δ ) k ϕ t ) | C 11 | Ω ( m ( u ¯ u ) ) 2 2 m + k u ¯ 2 m + k ϕ t d t | C 12 2 λ 1 2 m 2 k 2 μ 0 2 m + k ( u ¯ u ) 4 + μ 0 2 2 m + k ϕ t 2 . (11)

where C 11 = M ( m ς ¯ p p ) ( m ς ¯ p p ) ( 1 s ) , C 12 = C 11 2 m + k u ¯ .

| ( M ( m u p p ) ( m u p p ) m ( u ¯ u ) ( Δ ) 2 m ( u ¯ u ) , ( Δ ) k ϕ t ) | C 13 | Ω ( m ( u ¯ u ) ) 2 m + k ( u ¯ u ) 2 m + k ϕ t d x | C 13 λ 1 m + k 2 2 m + k ( u ¯ u ) 2 2 m + k ϕ t μ 0 2 2 m + k ϕ t 2 + C 13 2 λ 1 m k 2 μ 0 2 m + k ( u ¯ u ) 4 . (12)

| ( M ( m u p p ) ( m u p p ) m ϕ ( Δ ) 2 m u , ( Δ ) k ϕ t ) | C 14 | Ω m ϕ 2 m + k u 2 m + k ϕ t d x | C 14 m ϕ 2 m + k ϕ t μ 0 2 2 m + k ϕ t 2 + C 15 2 λ 1 m k μ 0 2 m + k ϕ 2 . (13)

| ( M ( m u p p ) ( Δ ) 2 m ϕ , ( Δ ) k ϕ t ) | μ 0 2 d d t 2 m + k ϕ 2 . (14)

The h 2 = | u ¯ | ρ ( u ¯ t + u ¯ ) | u | ρ ( u t + u ) ρ | u | ρ 1 u t U | u | ρ U t ( ρ + 1 ) | u | ρ U .

Take inner product of h 2 and ( Δ ) k ϕ t .

| ( | u ¯ | ρ ( u ¯ t + u ¯ ) | u | ρ ( u t + u ) ρ | u | ρ 1 u t U | u | ρ U t ( ρ + 1 ) | u | ρ U , ( Δ ) k ϕ t ) | C 16 | Ω ϕ t ( Δ ) k ϕ t d x | C 16 λ 1 k 2 k ϕ t 2 . (15)

where

| h 2 | = | ( | u ¯ | ρ u ¯ t | u | ρ u t | u | ρ U t ) + ( | u ¯ | ρ u ¯ | u | ρ u ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) U ) | C 17 ( | u ¯ | ρ + | u | ρ ) w t | u | ρ U t + C 18 ( | u ¯ | ρ + | u | ρ ) w t ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) U C 19 ( C 17 ( | u ¯ | ρ + | u | ρ ) + | u | ρ ) ϕ t + C 20 ( C 18 ( | u ¯ | ρ + | u | ρ ) + ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) ) ϕ t C 21 ϕ t

Combined with (3.11)-(3.15), there are

d d t ( 2 k ϕ t 2 + μ 0 2 m + k ϕ 2 ) C 12 2 λ 1 2 m 2 k + C 13 2 λ 1 m k μ 0 u ¯ u E k 4 + C 22 ( 2 k ϕ t 2 + μ 1 2 m + k ϕ 2 ) . (16)

Through the Gronwall’s inequality, there is

2 k ϕ t 2 + μ 0 2 m + k ϕ 2 C 22 e C 23 t ( ξ , η ) E k 4 . (17)

When ( ξ , η ) E k 4 0

s ( t ) η ¯ s ( t ) η F s ( t ) ( ξ , η ) ( ξ , η ) E k 2 C 24 e C 25 t ( ξ , η ) E k 2 0. (18)

So the lemma 3.1 is proved.

Lemma 3.2 Under the assumption and condition of lemma 3.1, the family of global attractor A k of initial-boundary value problem (1.1)-(1.3) has Hausdoff dimension and Fractal dimension, and d H ( A k ) 1 7 n , d F ( A k ) 8 7 n .

Proof. Make ϕ = R ε φ = ( u , v ) , φ = ( u , u t ) , v = u t + ε u , R ε : { u , u t } { u , u t + ε u } is an isomorphic mapping, let A i ( i = 1 , 2 , , 2 m ) be the global attractor of { S ( t ) } , and A ε i is the global attractor of { S ε ( t ) } , and they have the same dimension, from lemma 3.1, we can know that S ( t ) : E k E k is Fréchet differentiable. The linearized first-order variational Equation (3.1) can be rewritten as

P t + Λ ( Ψ ) P = 0. (19)

P t = F t ( Ψ ) . (20)

where

Λ ( Ψ ) = ( ε I I M ( A m 2 u p p ) A 2 m + ε 2 I ε β A 2 m ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) ε + ε 2 | u | ρ + Φ β A 2 m ε I + | u | ρ ε ) ,

I is an identity operator. Δ = A , P = ( U , θ ) E k . Make Ψ = R ε φ = ( U , V ) , φ = ( U , U t ) , θ = U t + ε U R ε : { u , u t } { u , u t + ε u } is an isomorphic mapping. For a fixed ( u 0 , v 0 ) E k , let r 1 , r 2 , , r n be n element of E k , let U 1 ( t ) , U 2 ( t ) , , U n ( t ) be n solutions of the linear Equation (3.19), its initial value is U 1 ( 0 ) = r 1 , U 2 ( 0 ) = r 2 , , U n ( 0 ) = r n .

so

d d t U 1 ( t ) Λ U 2 ( t ) Λ Λ U n ( t ) Λ E k 2 2 t r F t ( Ψ ( τ ) Q n ( τ ) ) U 1 ( t ) Λ U 2 ( t ) Λ Λ U n ( t ) Λ E k 2 = 0. (21)

Further, by the same Gronwall’s inequality, available:

U 1 ( t ) Λ U 2 ( t ) Λ Λ U n ( t ) Λ E k 2 = r 1 Λ r 2 Λ Λ r n Λ E k exp ( 0 t t r F t ( Ψ ( τ ) Q n ( τ ) ) d τ ) . (22)

where Λ stands for outer product and tr stands for trace. Q n ( τ ) is an orthogonal projection from space E k to s p a n { U 1 ( t ) , U 2 ( t ) , , U n ( t ) } .

Given a certain moment τ ,

set w j ( τ ) = ( ξ j ( τ ) , η j ( τ ) ) , j = 1 , 2 , , n is s p a n { U 1 ( t ) , U 2 ( t ) , U n ( t ) } orthonormal basis.

We define the inner product in E k is

( ( ξ , η ) , ( ξ ¯ , η ¯ ) ) = ( ( 2 m + k ξ , 2 m + k ξ ¯ ) + ( k η , k η ¯ ) ) . (23)

To sum up, it is available

t r F ( Ψ ( τ ) Q n ( τ ) ) = j = 1 n ( F ( Ψ ( τ ) ) Q n ( τ ) w j ( τ ) , w j ( τ ) ) E k = j = 1 n ( F ( Ψ ( τ ) ) w j ( τ ) w j ( τ ) ) E k . (24)

where ( F ( Ψ ( τ ) ) w j ( τ ) w j ( τ ) ) E k = ( Λ ( Ψ ) w j , w j ) .

= ε 2 m + k ξ j 2 + ( 2 m + k ξ j , 2 m + k η j ) M ( A m 2 u p p ) ( 2 m + k ξ j , 2 m + k η j ) ε 2 ( k ξ j , k η j ) + β ε ( 2 m + k ξ j , 2 m + k η j ) + ε ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) | u t | ρ 1 ( k ξ j , k η j ) Φ ( k ξ j , k η j ) β 2 m + k η j 2 + ε k η j 2 ε 2 | u | ρ ( k ξ j , k η j ) ε | u | ρ k η j 2

ε 2 m + k ξ j 2 + μ 0 β ε 1 2 2 m + k ξ j 2 + μ 0 β ε 1 2 2 m + k η j 2 + ε 2 k ξ j 2 + Φ + ε 2 | u | ρ ε ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) 2 λ 1 2 m 2 m + k ξ j 2 + Φ + ε 2 | u | ρ ε ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) + 3 ε + 2 ε | u | ρ 2 β λ 1 2 m 2 k η j 2 C 26 ( 2 m + k ξ j 2 + k η j 2 ) + ε 2 k ξ j 2 . (25)

where

C 26 = min { ( μ 0 β ε 1 2 ε + Φ + ε 2 | u | ρ ε ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) 2 λ 1 2 m ) , ( μ 0 β ε 1 2 λ 1 2 m + Φ + ε 2 | u | ρ ε ( ρ | u | ρ 1 u t + ( ρ + 1 ) | u | ρ ) + 3 ε + 2 ε | u | ρ 2 β λ 1 2 m 2 ) } .

Above all there is

( F t ( Ψ ( τ ) ) w j ( τ ) , w j ( τ ) ) C 26 ( 2 m + k ξ j 2 + k η j 2 ) + ε 2 k ξ j 2 . (26)

Because w j ( τ ) = ( ξ j ( τ ) , η j ( τ ) ) T , j = 1 , 2 , , n is s p a n { U 1 ( t ) , U 2 ( t ) , , U n ( t ) } orthonormal basis. Therefore

2 m + k ξ j 2 + k η j 2 = 1. (27)

j = 1 n ( F ( Ψ ( τ ) ) w j ( τ ) , w j ( τ ) ) E k n C 26 + ε 2 j = 1 n k ξ j 2 . (28)

Almost all the t.

j = 1 n k ξ j 2 j = 1 n λ j s 1 . (29)

The s 1 = 2 m and s [ 0,1 ] , λ j is A 2 m characteristic value, and λ 1 < λ 2 < < λ n , so there is

T r F ( Ψ ( τ ) Q n ( τ ) ) n C 26 + ε j = 1 n λ j s 1 . (30)

Set

q n ( t ) = sup Ψ 0 B 0 k sup η j E k ( 1 t 0 t T r F ( s ( t ) Ψ 0 ) Q n ( τ ) d τ ) . (31)

and

q n = lim t q n ( t ) . (32)

Therefore,

q n n C 26 + ε j = 1 n λ j s 1 . (33)

Therefore, B 0 k Lyapunov index κ 1 , κ 2 , , κ n ( n > 1 ) is uniformly bounded, and

κ 1 + κ 2 + + κ n n C 26 + ε j = 1 n λ j s 1 . (34)

make

( q j ) + n C 26 + ε j = 1 n λ j s 1 ε j = 1 n λ j s 1 n C 26 8 . (35)

q n n C 26 ( 1 ε n C 26 j = 1 n λ j s 1 ) 7 8 n C 26 . (36)

Further,

max 1 j n ( q j ) + | q n | 1 7 . (37)

From this we can get d H ( A k ) 1 7 n , d F ( A k ) 8 7 n , Then the Hausdorff dimension and Fractal dimension of the family of global attractor A k are finite.

Cite this paper: Lin, G. and Liu, X. (2021) The Family of Global Attractors for a Generalized Kirchhoff Equations. Open Journal of Applied Sciences, 11, 789-806. doi: 10.4236/ojapps.2021.117058.
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