<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">JAMP</journal-id><journal-title-group><journal-title>Journal of Applied Mathematics and Physics</journal-title></journal-title-group><issn pub-type="epub">2327-4352</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/jamp.2022.107150</article-id><article-id pub-id-type="publisher-id">JAMP-118637</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  The Long-Term Dynamic Behavior of Solutions to a Class of Generalized Higher-Order Kirchhoff-Type Coupled Wave Equations
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Guoguang</surname><given-names>Lin</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Min</surname><given-names>Shao</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib></contrib-group><aff id="aff1"><addr-line>Department of Mathematics, Yunnan University, Kunming, China</addr-line></aff><pub-date pub-type="epub"><day>11</day><month>07</month><year>2022</year></pub-date><volume>10</volume><issue>07</issue><fpage>2181</fpage><lpage>2199</lpage><history><date date-type="received"><day>7,</day>	<month>May</month>	<year>2022</year></date><date date-type="rev-recd"><day>18,</day>	<month>July</month>	<year>2022</year>	</date><date date-type="accepted"><day>21,</day>	<month>July</month>	<year>2022</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  In this paper, we study the long-term dynamic behavior of a class of generalized high-order Kirchhoff-type coupled wave equations. Firstly, the existence of uniqueness global solution of this kind of equations in 
  <em>E<sub>k</sub></em> space is proved by prior estimation and Galerkin method; Then, through using Rellich-Kondrachov compact embedding theorem, it is proved that the solution semigroup 
  <em>S</em>(
  <em>t</em>) has the family of the global attractors 
  <em>A<sub>k</sub></em> in space 
  <em>E<sub>k</sub></em>; Finally, through linearization method, proves that the operator semigroup 
  <em>S</em>(
  <em>t</em>) Frechet differentiable and the attenuation of linearization problem volume element. Furthermore, we can obtain the finite Hausdorff dimension and Fractal dimension of the family of the global attractors 
  <em>A<sub>k</sub></em>.
 
</p></abstract><kwd-group><kwd>Kirchhoff Equation</kwd><kwd> Existence and Uniqueness of Solutions</kwd><kwd> Global Attractor Family</kwd><kwd> Dimension Estimation</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>In this paper, we study the long-term dynamic behavior of a class of generalized high-order Kirchhoff-type coupled wave equations:</p><p>u t t + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m u + β ( − Δ ) m u t + g ( u t , v ) = f 1 ( x ) , (1)</p><p>v t t + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) 2 m v + β ( − Δ ) 2 m v t + g ( u , v t ) = f 2 ( x ) , (2)</p><p>the boundary conditions:</p><p>∂ i u ∂ n i = 0 , i = 0 , 1 , 2 , ⋯ , m − 1 , x ∈ ∂ Ω , t &gt; 0 , (3)</p><p>∂ j v ∂ v j = 0 , j = 0 , 1 , 2 , ⋯ , 2 m − 1 , x ∈ ∂ Ω , t &gt; 0 , (4)</p><p>the initial conditions:</p><p>u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , v ( x , 0 ) = v 0 ( x ) , v t ( x , 0 ) = v 1 ( x ) , x ∈ Ω , (5)</p><p>where Ω is a bounded domain in R n with smooth boundary ∂ Ω , u 0 ( x ) , u 1 ( x ) is a known function, g ( u , v ) , f i ( x ) , i = 1 , 2 are nonlinear source term and the external force interference terms, m &gt; 1 , β is real number.</p><p>Recently, the global attractor and its dimension estimation for Kirchhoff type equations have been favored by many scholars. Many scholars have done a lot of research on this kind of problems and obtained good results [<xref ref-type="bibr" rid="scirp.118637-ref1">1</xref>] [<xref ref-type="bibr" rid="scirp.118637-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.118637-ref3">3</xref>].</p><p>Lin Guoguang, Gao Yunlong [<xref ref-type="bibr" rid="scirp.118637-ref1">1</xref>] studied the longtime behavior of solution to initial boundary value problem for a class of strongly damped higher-order Kirchhoff type equation:</p><p>u t t + ( − Δ ) m u t + ( α + β ‖ ∇ m u ‖ 2 ) q ( − Δ ) m u + g ( u ) = f ( x ) ,     ( x , t ) ∈ Ω &#215; [ 0 , + ∞ ) ,</p><p>they got the existence and uniqueness of the solution by the Galerkin method and obtained the existence of the global attractor in H 0 m ( Ω ) &#215; L 2 ( Ω ) according to the attractor theorem, besides, the estimation of the upper bound of Hausdorff dimension for the attractor was established.</p><p>Guoguang Lin, Ming Zhang [<xref ref-type="bibr" rid="scirp.118637-ref2">2</xref>] studied the initial boundary value problem for a class of Kirchhoff-type coupled equations:</p><p>u t t − M ( ‖ ∇ u ‖ 2 + ‖ ∇ v ‖ 2 ) Δ u − β Δ u t + g 1 ( u , v ) = f 1 ( x ) ,</p><p>v t t − M ( ‖ ∇ u ‖ 2 + ‖ ∇ v ‖ 2 ) Δ v − β Δ v t + g 2 ( u , v ) = f 2 ( x ) ,</p><p>they obtained the existence of the global attractor and a precise estimate of upper bound of Hausdorff dimension.</p><p>Lin Guoguang, Yang Lujiao [<xref ref-type="bibr" rid="scirp.118637-ref3">3</xref>] studied the long-time properties of solutions of generalized Kirchhoff-type equation with strongly damped terms:</p><p>u t t + M ( ‖ ∇ m u ‖ p p ) ( − Δ ) 2 m u + β ( − Δ ) 2 m u t + g ( u ) = f ( x ) , <sup> </sup></p><p>by assuming the nonlinear source terms g ( u ) and Kirchhoff stress term M ( s ) , the author verified the appropriateness of the solution and proved the existence of the global attractor, obtained the upper boundary estimation of the Hausdorff dimension and Fractal dimension of a family of the global attractor.<sup> </sup></p><p>For more significant research results about the global attractor and its dimension estimation of Kirchhoff equation, please refer to the literature [<xref ref-type="bibr" rid="scirp.118637-ref4">4</xref>] - [<xref ref-type="bibr" rid="scirp.118637-ref18">18</xref>].</p></sec><sec id="s2"><title>2. Existence and Uniqueness of Solutions</title><p>The following symbols and assumptions are introduced for the convenience of statement:</p><p>V m = H m ( Ω ) ∩ H 0 1 ( Ω ) , V 2 m = H m ( Ω ) ∩ H 0 1 ( Ω ) , V 4 m = H 4 m ( Ω ) ∩ H 0 1 ( Ω ) , ‖   ⋅   ‖ = ‖   ⋅   ‖ L 2 ( Ω ) , E 0 = V m &#215; V 0 &#215; V 2 m &#215; V 0 , E 1 = V 2 m &#215; V 0 &#215; V 4 m &#215; V 0 , E k = V m + k &#215; V k &#215; V 2 m + 2 k &#215; V 2 k , V 0 = L 2 ( Ω ) .</p><p>In order to obtain our results, we consider system (1)-(5) under some assumptions on M ( s ) and g ( u , v ) . Precisely, we state the general assumptions:</p><p>(A1) M ( s ) ∈ C 2 ( [ 0 , + ∞ ) , R ) is not decreasing function and for positive constants δ 0 , δ 1 ,</p><p>(1) ε + 1 ≤ δ 0 ≤ M ( s ) ≤ δ 1 ,</p><p>(2) M ( s ) is a non-negative Lipschitz function, L is associated with the Lipschitz constant M ( s ) , M ( s ) = M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) .</p><p>(A2) For any u , v , p , q ∈ V , g ( u t , v ) , g ( u , v t ) ∈ C 2 ( ℝ ) , there exist α ≥ 0 , ε ≥ 0 , C ( α , ε ) ≥ 0 , such that</p><p>( g ( u t , v ) , p ) + ( g ( u , v t ) , q ) ≥ α ( ‖ p ‖ 2 + ‖ q ‖ 2 ) − ε ( ‖ u ‖ 2 + ‖ v ‖ 2 ) − C ( α , ε ) ( ‖ u t ‖ 2 + ‖ v t ‖ 2 ) .</p><p>Lemma 1 Assuming (A1)-(A2) are true, letting ( u 0 , p 0 , v 0 , q 0 ) ∈ E 0 , f 1 ( x ) , f 2 ( x ) ∈ L 2 ( Ω ) , then there is a solution ( u , p , v , q ) for problem (1)-(5) which has the following properties:</p><p>(i) ( u , p , v , q ) ∈ L ∞ ( ( 0 , + ∞ ) ; E 0 ) ;</p><p>(ii)</p><p>y ( t ) ≤ y ( 0 ) e − 2 ε 0 k 1 t + k 1 2 ε ε 0 ( ‖ f 1 ( x ) ‖ 2 + ‖ f 2 ( x ) ‖ 2 ) , (6)</p><p>where y ( t ) = ‖ ∇ m u ‖ 2 + ‖ p ‖ 2 + ‖ ∇ 2 m v ‖ 2 + ‖ q ‖ 2 .</p><p>(iii) There are normal numbers C ( R 0 ) and t 0 = t 0 ( Ω ) &gt; 0 , such that</p><p>‖ ( u , p , v , q ) ‖ E 0 2 = ‖ ∇ m u ‖ 2 + ‖ p ‖ 2 + ‖ ∇ 2 m v ‖ 2 + ‖ q ‖ 2 ≤ C ( R 0 ) . (7)</p><p>Proof: Let p = u t + ε u inner product with Equation (1),</p><p>( u t t + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m u + β ( − Δ ) m u t + g ( u t , v ) , p ) = ( f 1 ( x ) , p ) , (8)</p><p>according to the hypothesis (A1), using the Young inequality, Holder inequation, Poincare inequality, etc., there are</p><p>( u t t , p ) = 1 2 d d t ‖ p ‖ 2 − ε ‖ p ‖ 2 + ε 2 ( u , p ) , (9)</p><p>( M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m u , p ) ≥ δ 2 d d t ‖ ∇ m u ‖ 2 + ε δ 0 ‖ ∇ m u ‖ 2 , (10)</p><p>( β ( − Δ ) m u t , p ) = β ‖ ∇ m u t ‖ 2 + β ε 2 d d t ‖ ∇ m u ‖ 2 , (11)</p><p>( f 1 ( x ) , p ) ≤ ε 2 ‖ p ‖ 2 + 1 2 ε ‖ f 1 ( x ) ‖ 2 , (12)</p><p>where δ = δ 0 or δ 1 .</p><p>Similarly, letting q = v t + ε v inner product with Equation (2), the treatment of each item is similar to (9)-(11), and the above results are sorted out,</p><p>1 2 d d t ( ( δ + β ε ) ( ‖ ∇ m u ‖ 2 + ‖ ∇ 2 m v ‖ 2 ) + ‖ p ‖ 2 + ‖ q ‖ 2 ) − 3 ε 2 ( ‖ p ‖ 2 + ‖ q ‖ 2 ) + ε 2 [ ( u , p ) + ( v , q ) ] + ε δ 0 ( ‖ ∇ m u ‖ 2 + ‖ ∇ 2 m v ‖ 2 ) + β ( ‖ ∇ m u t ‖ 2 + ‖ ∇ 2 m v t ‖ 2 ) + ( g ( u t , v ) , p ) + ( g ( u , v t ) , q ) = 1 2 ε ( ‖ f 1 ( x ) ‖ 2 + ‖ f 2 ( x ) ‖ 2 ) , (13)</p><p>using the Young inequality, the Poincare inequality, and the assumptions (A2), The individual items in Equation (13) are treated as follows:</p><p>ε 2 [ ( u , p ) + ( v , q ) ] ≥ − ε ( ‖ u ‖ 2 + ‖ v ‖ 2 ) − ε 3 4 ( ‖ p ‖ 2 + ‖ q ‖ 2 ) , (14)</p><p>by the Poincare inequality has</p><p>− ε ( ‖ u ‖ 2 + ‖ v ‖ 2 ) ≥ − ε ( λ 1 − m + λ 1 − 2 m ) ( ‖ ∇ m u ‖ 2 + ‖ ∇ 2 m v ‖ 2 ) , (15)</p><p>where λ 1 is the first eigenvalue with homogeneous Dirichlet boundary conditions of − Δ , in the same way</p><p>− C ( α , ε ) ( ‖ u t ‖ 2 + ‖ v t ‖ 2 ) ≥ − C ( α , ε ) ( λ 1 − m + λ 1 − 2 m ) ( ‖ ∇ m u t ‖ 2 + ‖ ∇ 2 m v t ‖ 2 ) , (16)</p><p>1 2 d d t ( ( δ + β ε ) ( ‖ ∇ m u ‖ 2 + ‖ ∇ 2 m v ‖ 2 ) + ‖ p ‖ 2 + ‖ q ‖ 2 ) + ( α − 3 ε 2 − ε 3 4 ) ( ‖ p ‖ 2 + ‖ q ‖ 2 ) + ε ( δ 0 − 2 ( λ 1 − m + λ 1 − 2 m ) ) ( ‖ ∇ m u ‖ 2 + ‖ ∇ 2 m v ‖ 2 ) + ( β − C ( α , ε ) ( λ 1 − m + λ 1 − 2 m ) ) ( ‖ ∇ m u t ‖ 2 + ‖ ∇ 2 m v t ‖ 2 ) , ≤ 1 2 ε ( ‖ f 1 ( x ) ‖ 2 + ‖ f 2 ( x ) ‖ 2 ) , (17)</p><p>where α &gt; 3 ε 2 + ε 3 4 , δ 0 &gt; 2 ( λ 1 − m + λ 1 − 2 m ) , β &gt; C ( α , ε ) ( λ 1 − m + λ 1 − 2 m ) .</p><p>Let y &#175; ( t ) = ( δ + β ε ) ( ‖ ∇ m u ‖ 2 + ‖ ∇ 2 m v ‖ 2 ) + ‖ p ‖ 2 + ‖ q ‖ 2 ,</p><p>y ( t ) = ‖ ∇ m u ‖ 2 + ‖ ∇ 2 m v ‖ 2 + ‖ p ‖ 2 + ‖ q ‖ 2 , there are normal numbers k 1 = min { 1 , δ + β ε } , such that</p><p>y &#175; ( t ) ≥ k 1 y ( t ) ≥ 0 , (18)</p><p>let ε 0 = min { α − 3 ε 2 − ε 3 4 , ε ( δ 0 − 2 ( λ 1 − m + λ 1 − 2 m ) ) } , there are</p><p>d d t y ( t ) + 2 ε 0 k 1 y ( t ) ≤ 1 ε ( ‖ f 1 ( x ) ‖ 2 + ‖ f 2 ( x ) ‖ 2 ) , (19)</p><p>by Gronwall inequality,</p><p>y ( t ) ≤ y ( 0 ) e − 2 ε 0 k 1 t + k 1 2 ε ε 0 ( ‖ f 1 ( x ) ‖ 2 + ‖ f 2 ( x ) ‖ 2 ) , (20)</p><p>so there are normal numbers C ( R 0 ) and t 0 = t 0 ( Ω ) &gt; 0 , such that</p><p>‖ ( u , p , v , q ) ‖ E 0 2 = ‖ ∇ m u ‖ 2 + ‖ p ‖ 2 + ‖ ∇ 2 m v ‖ 2 + ‖ q ‖ 2 ≤ C ( R 0 ) . (21)</p><p>Lemma 1 is proved.</p><p>Lemma 2 Assuming (A1)-(A2) are true, ( u 0 , p 0 , v 0 , q 0 ) ∈ E k , f 1 ( x ) ∈ H m ( Ω ) , f 2 ( x ) ∈ H 2 m ( Ω ) , then there is a solution ( u , p , v , q ) for problem (1)-(5), which has the following properties:</p><p>(i) ( u , p , v , q ) ∈ L ∞ ( ( 0 , + ∞ ) ; E k ) ;</p><p>(ii)</p><p>y 1 ( t ) ≤ y 1 ( 0 ) e − 2 ε 1 k 2 t + k 2 2 ε ε 1 ( ‖ ∇ k f 1 ( x ) ‖ 2 + ‖ ∇ 2 k f 2 ( x ) ‖ 2 ) , (22)</p><p>where y ( t ) = ‖ ∇ m u ‖ 2 + ‖ p ‖ 2 + ‖ ∇ 2 m v ‖ 2 + ‖ q ‖ 2 .</p><p>(iii) There are normal numbers C ( R k ) and t 0 k , such that</p><p>‖ ( u , p , v , q ) ‖ E k 2 = ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ≤ C ( R k ) . (23)</p><p>Proof: Let ( − Δ ) k p inner product with Equation (1),</p><p>( u t t + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m u + β ( − Δ ) m u t + g ( u t , v ) , ( − Δ ) k p ) = ( f 1 ( x ) , ( − Δ ) k p ) , (24)</p><p>according to the hypothesis (A1), using the Young inequality, Holder inequation, Poincare inequality, etc., there are</p><p>( u t t , ( − Δ ) k p ) ≥ 1 2 d d t ‖ ∇ k p ‖ 2 − ( ε + ε 2 2 ) ‖ ∇ k p ‖ 2 − ε 2 λ 1 − m 2 ‖ ∇ m + k u ‖ 2 , (25)</p><p>( M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m u , ( − Δ ) k p ) ≥ δ 2 d d t ‖ ∇ m + k u ‖ 2 + ε δ 0 ‖ ∇ m + k u ‖ 2 , (26)</p><p>( β ( − Δ ) m u t , ( − Δ ) k p ) = β ‖ ∇ m + k u t ‖ 2 + β ε 2 d d t ‖ ∇ m + k u ‖ 2 , (27)</p><p>where δ = δ 0 or δ 1 .</p><p>Similarly, letting ( − Δ ) 2 m q inner product with Equation (2), the treatment of each item is similar to (25)-(27), using Young inequality, Poincare inequality, and assumption (A2), the individual terms are treated as follows:</p><p>( g ( u t , v ) , ( − Δ ) k p ) + ( g ( u , v t ) , ( − Δ ) 2 k q ) ≤ C 1 2 ε ( λ 1 − m + λ 1 − 2 m ) ( ‖ ∇ m + k u t ‖ 2 + ‖ ∇ 2 m + 2 k v t ‖ 2 ) + ε C 1 2 ( ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) , (28)</p><p>( f 1 ( x ) , ( − Δ ) k p ) + ( f 2 ( x ) , ( − Δ ) 2 k 2 q ) ≤ ε 2 ( ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) + 1 2 ε ( ‖ ∇ k f 1 ( x ) ‖ 2 + ‖ ∇ 2 k f 2 ( x ) ‖ 2 ) . (29)</p><p>Then sort out the result of appeal and other inner product items, get</p><p>1 2 d d t ( ( δ + β ε ) ( ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 ) + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) + ε 2 − ε C 1 − 3 ε 2 ( ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) + ( ε δ 0 − ε 2 2 ( λ 1 − m + λ 1 − 2 m ) ) ( ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 ) + ( β − C 1 2 ε ( λ 1 − m + λ 1 − 2 m ) ) ( ‖ ∇ m + k u t ‖ 2 + ‖ ∇ 2 m + 2 k v t ‖ 2 ) ≤ 1 2 ε ( ‖ ∇ k f 1 ( x ) ‖ 2 + ‖ ∇ 2 k f 2 ( x ) ‖ 2 ) , (30)</p><p>using the Young inequality, Poincare inequality, and assumption (A2), the individual terms in Equation (30) are treated as follows:</p><p>( g ( u t , v ) , ( − Δ ) k p ) + ( g ( u , v t ) , ( − Δ ) 2 k q ) ≤ C 1 2 ε ( λ 1 − m + λ 1 − 2 m ) ( ‖ ∇ m + k u t ‖ 2 + ‖ ∇ 2 m + 2 k v t ‖ 2 ) + ε C 1 2 ( ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) , (31)</p><p>( f 1 ( x ) , ( − Δ ) k p ) + ( f 2 ( x ) , ( − Δ ) 2 k 2 q ) ≤ ε 2 ( ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) + 1 2 ε ( ‖ ∇ k f 1 ( x ) ‖ 2 + ‖ ∇ 2 k f 2 ( x ) ‖ 2 ) , (32)</p><p>where ε − C 1 − 3 &gt; 0 , δ 0 &gt; ε 2 ( λ 1 − m + λ 1 − 2 m ) , β &gt; C 1 2 ε ( λ 1 − m + λ 1 − 2 m ) .</p><p>Let y &#175; 1 ( t ) = ( δ + β ε ) ( ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 ) + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ,</p><p>y 1 ( t ) = ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 , there are normal numbers k 2 = min { 1 , δ + β ε } , such that</p><p>y &#175; 1 ( t ) ≥ k 2 y 1 ( t ) ≥ 0 , (33)</p><p>let ε 1 = min { ε 2 − ε C 1 − 3 ε 2 − ε 3 4 , ε δ 0 − ε 2 2 ( λ 1 − m + λ 1 − 2 m ) } , get</p><p>d d t y 1 ( t ) + 2 ε 1 k 2 y 1 ( t ) ≤ 1 ε ( ‖ ∇ k f 1 ( x ) ‖ 2 + ‖ ∇ 2 k f 2 ( x ) ‖ 2 ) , (34)</p><p>by Gronwall inequality,</p><p>y 1 ( t ) ≤ y 1 ( 0 ) e − 2 ε 1 k 2 t + k 2 2 ε ε 1 ( ‖ ∇ k f 1 ( x ) ‖ 2 + ‖ ∇ 2 k f 2 ( x ) ‖ 2 ) , (35)</p><p>so there are normal numbers C ( R k ) and t 0 k &gt; 0 , such that</p><p>‖ ( u , p , v , q ) ‖ E k 2 = ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k q ‖ 2 ≤ C ( R k ) (36)</p><p>Lemma 1 is proved.</p><p>Theorem 1 Assuming (A1)-(A2) is true, ( u 0 , p 0 , v 0 , q 0 ) ∈ E k , f 1 ( x ) ∈ V k , f 2 ( x ) ∈ V 2 k , then the initial boundary value problem (1)-(5) has a unique solution</p><p>( u ( x , t ) , p ( x , t ) , v ( x , t ) , q ( x , t ) ) ∈ L ∞ ( ( 0 , + ∞ ) ; E k ) . (37)</p><p>Proof: According to literature [<xref ref-type="bibr" rid="scirp.118637-ref9">9</xref>] and Galerkin method, combining with lemma 1 and lemma 2, we can easily obtain the existence of solutions.</p><p>Next, prove the uniqueness of the solution:</p><p>Assuming ( u 1 , p 1 , v 1 , q 1 ) , ( u 2 , p 2 , v 2 , q 2 ) ∈ E k are the two solutions of the problem (1)-(5), letting u &#175; = u 1 − u 2 , v &#175; = v 1 − v 2 , obtain that</p><p>{ u &#175; t t + M ( s 1 ) ( − Δ ) m u 1 − M ( s 2 ) ( − Δ ) m u 2 + β ( − Δ ) m u &#175; t + g ( u 1 t , v 1 ) − g ( u 2 t , v 2 ) = 0 , v &#175; t t + M ( s 1 ) ( − Δ ) 2 m v 1 − M ( s 2 ) ( − Δ ) 2 m v 2 + β ( − Δ ) 2 m v &#175; t + g ( u 1 , v 1 t ) − g ( u 2 , v 2 t ) = 0 , u &#175; ( x , 0 ) = 0 , u &#175; t ( x , 0 ) = 0 , v &#175; ( x , 0 ) = 0 , v &#175; t ( x , 0 ) = 0 , x ∈ Ω , ∂ i u &#175; ∂ n i = 0 , i = 0 , 1 , 2 , ⋯ , m − 1 , x ∈ ∂ Ω , t &gt; 0 , ∂ j v &#175; ∂ v j = 0 , j = 0 , 1 , 2 , ⋯ , 2 m − 1 , x ∈ ∂ Ω , t &gt; 0 , (38)</p><p>where s 1 = ‖ ∇ m u 1 ‖ 2 − ‖ ∇ m v 1 ‖ 2 , s 2 = ‖ ∇ m u 2 ‖ 2 − ‖ ∇ m v 2 ‖ 2 .</p><p>Let u &#175; t , v &#175; t inner product with the first two equations in (38) and obtain,</p><p>{ 1 2 d d t ‖ u &#175; t ‖ 2 + ( M ( s 1 ) ( − Δ ) m u 1 − M ( s 2 ) ( − Δ ) m u 2 , u &#175; t ) + β ‖ ∇ m u &#175; t ‖ 2 = ( g ( u 2 t , v 2 ) − g ( u 1 t , v 1 ) , u &#175; t ) , 1 2 d d t ‖ v &#175; t ‖ 2 + ( M ( s 1 ) ( − Δ ) 2 m v 1 − M ( s 2 ) ( − Δ ) 2 m v 2 , v &#175; t ) + β ‖ ∇ 2 m v &#175; t ‖ 2 = ( g ( u 2 , v 2 t ) − g ( u 1 , v 1 t ) , v &#175; t ) , (39)</p><p>using the Young inequality, Poincare inequality, as well as the assumptions (A2), for processing on the type of individual items as follows:</p><p>( M ( s 1 ) ( − Δ ) m u &#175; + M ( s 1 ) ( − Δ ) m u 2 − M ( s 2 ) ( − Δ ) m u 2 , u &#175; t ) ≥ M ( s ) 2 d d t ‖ ∇ m u ‖ 2 − ( M ( s 1 ) ( − Δ ) m u 2 − M ( s 2 ) ( − Δ ) m u 2 , u &#175; t ) ≥ M ( s ) 2 d d t ‖ ∇ m u ‖ 2 − L ( ( ‖ ∇ m u 1 ‖ + ‖ ∇ m u 2 ‖ ) ‖ ∇ m u &#175; ‖     + L ( ‖ ∇ m v 1 ‖ + ‖ ∇ m v 2 ‖ ) ‖ ∇ m v &#175; ‖ ) ‖ ( − Δ ) m u &#175; 2 ‖ ‖ u &#175; t ‖ ≥ C 2 ( ‖ ∇ m u &#175; ‖ + ‖ ∇ 2 m v &#175; ‖ ) ‖ u &#175; t ‖ ≥ C 2 ( ‖ ∇ m u &#175; ‖ 2 + ‖ ∇ 2 m v &#175; ‖ 2 + ‖ u &#175; t ‖ 2 2 ) , (40)</p><p>similarly, we obtain</p><p>( M ( s 1 ) ( − Δ ) 2 m v 1 − M ( s 2 ) ( − Δ ) 2 m v 2 , v &#175; t ) ≥ C 3 ( ‖ ∇ m u &#175; ‖ 2 + ‖ ∇ 2 m v &#175; ‖ 2 + ‖ v &#175; t ‖ 2 2 ) , (41)</p><p>| ( g ( u 1 t , v 1 ) − g ( u 2 t , v 1 ) + g ( u 2 t , v 1 ) − g ( u 2 t , v 2 ) , u &#175; t ) | ≤ | ( g ′ ( ξ t , v 1 ) | u &#175; t | + g ′ ( u 2 t , η ) | v &#175; | , u &#175; t ) | ≤ ‖ g ′ ( ξ t , v 1 ) ‖ ∞ ‖ u &#175; t ‖ 2 + ‖ g ′ ( u 2 t , η ) ∞ ‖ ‖ | v &#175; | ‖ ‖ u &#175; t ‖ ≤ C 4 ( ‖ v &#175; ‖ 2 2 + 3 ‖ u &#175; t ‖ 2 2 ) , (42)</p><p>| ( g ( u 1 , v 1 t ) − g ( u 2 , v 2 t ) , v &#175; t ) | ≤ C 5 ( 3 ‖ v &#175; t ‖ 2 2 + ‖ u &#175; ‖ 2 2 ) , (43)</p><p>Through the (40)-(43), finally will become</p><p>1 2 d d t ( ‖ u &#175; t ‖ 2 + ‖ v &#175; t ‖ 2 + M ( s 1 ) ( ‖ ∇ m u &#175; ‖ 2 + ‖ ∇ 2 m v &#175; ‖ 2 ) ) + β ( ‖ ∇ m u &#175; t ‖ 2 + ‖ ∇ 2 m v &#175; t ‖ 2 ) ≤ ( C 2 + C 3 ) ( ‖ ∇ m u &#175; ‖ 2 + ‖ ∇ 2 m v &#175; ‖ 2 ) + C 2 + 3 C 4 2 ‖ u &#175; t ‖ 2     + C 3 + 3 C 5 2 ‖ v &#175; t ‖ 2 + C 4 2 ‖ v ‖ 2 + C 5 2 ‖ u ‖ 2 ≤ C 6 ( ‖ ∇ m u &#175; ‖ 2 + ‖ ∇ 2 m v &#175; ‖ 2 ) + C 7 ( ‖ u &#175; t ‖ 2 + ‖ v &#175; t ‖ 2 ) . (44)</p><p>Let y 2 ( t ) = ‖ u &#175; t ‖ 2 + ‖ v &#175; t ‖ 2 + M ( s 1 ) ( ‖ ∇ m u &#175; ‖ 2 + ‖ ∇ 2 m v &#175; ‖ 2 ) , there are normal numbers k 3 = min { C 7 , C 8 } , where C 6 ≤ C 8 M ( s 1 ) , such that</p><p>d d t y 2 ( t ) ≤ k 3 y 2 ( t ) , (45)</p><p>using the Gronwall inequality, we have</p><p>y 2 ( t ) ≤ y 2 ( 0 ) e k 4 t = 0 , (46)</p><p>y 2 ( t ) = ‖ u &#175; t ‖ 2 + ‖ v &#175; t ‖ 2 + M ( s 1 ) ( ‖ ∇ m u &#175; ‖ 2 + ‖ ∇ 2 m v &#175; ‖ 2 ) = 0 , (47)</p><p>so</p><p>u &#175; = v &#175; = 0. (48)</p><p>Theorem 1 is proved.</p></sec><sec id="s3"><title>3. The Family of Global Attractors and Dimension Estimation</title><p>Theorem 2 [<xref ref-type="bibr" rid="scirp.118637-ref9">9</xref>] Assume E 0 is a Banach space, { S ( t ) } t ≥ 0 is the operator semigroup, S ( t ) : E → E , S ( t + τ ) = S ( t ) ⋅ S ( τ ) ( ∀ t , τ ≥ 0 ) , S ( 0 ) = I , where I is the identity operator. If S ( t ) satisfies</p><p>1) Semigroup S ( t ) is uniformly bounded in E, i.e. ∀ R &gt; 0 , exists a constant C ( R ) such that when ‖ u ‖ E ≤ R , there is ‖ S ( t ) u ‖ E ≤ C ( R ) ( ∀ t ∈ [ 0 , + ∞ ) ) ;</p><p>2) There exists a bounded absorbing set B in E, that is, for any bounded set B ⊂ E , there exists a constant t 0 &gt; 0 , such that</p><p>S ( t ) B ⊂ B ( t &gt; t 0 ) , (49)</p><p>3) { S ( t ) } t ≥ 0 is completely continuous operator.</p><p>Then operator semigroup S ( t ) has compact global attractor A.</p><p>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, then the initial boundary value problems (1)-(5) have the family of global attractors. There are compact sets satisfying:</p><p>A k ⊂ E k ⊂ E 0 and A k = ω ( B 0 k ) = ∩ τ ≥ 0 ∪ t ≥ τ S ( t ) B 0 k &#175; , k = 1 , 2 , ⋯ , m .</p><p>where B 0 k = { ( u , p , v , q ) ∈ E k : ‖ ∇ m + k u ‖ 2 + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 + ‖ ∇ 2 k q ‖ 2 ≤ R 0 k } ,</p><p>1) S ( t ) A k = A k , t &gt; 0 ,</p><p>2) A k attracts all bounded sets of E k , that is, any bounded set B 0 k ⊂ E k , having</p><p>lim t → ∞ d i s t ( S ( t ) B k , A k ) = 0 , where d i s t ( S ( t ) B k , A k ) = sup x ∈ B 0 k inf y ∈ A k ‖ S ( t ) − y ‖ E k .</p><p>then compact set A k are called the family of global attractors of semigroup S ( t ) .</p><p>Proof: From lemma 1, lemma 2, for any bounded set B 0 k ⊂ E k and B 0 k ⊂ { ‖ ( u , p , v , q ) ‖ E k ≤ R k } , the equation has solution semigroups S ( t ) : E k → E k , and</p><p>‖ S ( t ) ( u 0 , p 0 , v 0 , q 0 ) ‖ E k 2 = ‖ u ‖ V m + k 2 + ‖ p ‖ V k 2 + ‖ v ‖ V 2 m + 2 k 2 + ‖ q ‖ V 2 k 2 ≤ C ( R k ) , (50)</p><p>where t 0 ≥ 0 , ( u 0 , v 0 ) ∈ B 0 k , shows that S ( t ) t ≥ 0 is uniformly bounded in E k ;</p><p>Further,</p><p>B 0 k = { ( u , p , v , q ) ∈ E k : ‖ ∇ m + k u ‖ 2 + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 + ‖ ∇ 2 k q ‖ 2 ≤ R 0 k } is a</p><p>bounded absorption set of semigroup S ( t ) ; 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. Then there exists a global attractor family of equations</p><p>A k = ω ( B 0 k ) = ∩ τ ≥ 0 ∪ t ≥ τ S ( t ) B 0 k &#175; , k = 1 , 2 , ⋯ , m .</p><p>Theorem 3 is proved.</p><p>After the family of global attractors are obtained, in order to estimate the Hausdroff dimension and Fractal dimension of the family of global attractors, the initial boundary value problem (1)-(5) is linearized and obtain that</p><p>{ U t t + M ′ ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) [ ( ∇ m u , ∇ m U ) + ( ∇ m v , ∇ m V ) ] ( − Δ ) m u + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m U + β ( − Δ ) m U t + ∂ g ( u t , v ) ∂ u t U t + ∂ g ( u t , v ) ∂ v V = 0 , U t t + M ′ ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) [ ( ∇ m u , ∇ m U ) + ( ∇ m v , ∇ m V ) ] ( − Δ ) m u + M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ( − Δ ) m U + β ( − Δ ) m U t + ∂ g ( u t , v ) ∂ u t U t + ∂ g ( u t , v ) ∂ v V = 0 , U ( x , 0 ) = ξ 1 , U t ( x , 0 ) = ξ 2 , V ( x , 0 ) = η 1 , V t ( x , 0 ) = η 2 , U ( x , 0 ) | x ∈ Ω = V ( x , 0 ) | x ∈ Ω = 0 , t &gt; 0. (51)</p><p>where ( ξ 1 , ξ 2 , η 1 , η 2 ) ∈ E 0 , ( u , p , v , q ) = S ( t ) ( u 0 , p 0 , v 0 , q 0 ) is the solution of the initial boundary value problem (51).</p><p>Given ( u 0 , p 0 , v 0 , q 0 ) ∈ A k , S ( t ) : E k → E k , for any ( ξ 1 , ξ 2 , η 1 , η 2 ) ∈ E k , there exists a unique solution ( U ( t ) , P ( t ) , V ( t ) , Q ( t ) ) ∈ L ∞ ( 0 , T ; E k ) to the linear initial boundary value problem (51).</p><p>Lemma 3 For any t &gt; 0 , R &gt; 0 , the mapping S ( t ) : E k → E k is Frechet differentiable. The derivative on ρ 0 = ( u 0 , p 0 , v 0 , q 0 ) T is a linear operator on E k ,</p><p>D S ( t ) ρ 0 : ( ξ , ζ , η , σ ) → ( U , P , V , Q ) ,</p><p>where ( U ( t ) , P ( t ) , V ( t ) , Q ( t ) ) is the solution of the problem (51).</p><p>Proof: suppose ρ 0 = ( u 0 , p 0 , v 0 , q 0 ) ∈ E k , ρ ˜ 0 = ( u 0 + ξ , p 0 + ζ , v 0 + η , q 0 + σ ) ∈ E k with ‖ ρ 0 ‖ E k ≤ R k , ‖ ρ ˜ 0 ‖ E k ≤ R k , we denote</p><p>S ( t ) ρ 0 = ρ = ( u 1 , p 1 , v 1 , q 1 ) , S ( t ) ρ ˜ 0 = ( u 2 , p 2 , v 2 , q 2 ) .</p><p>First, we can prove a Lipschitz property of S ( t ) on the bounded sets on E k , that is</p><p>‖ S ( t ) ρ 0 − S ( t ) ρ ˜ 0 ‖ E k 2 ≤ e c t ‖ ( ξ , ζ , η , σ ) ‖ E k 2 . (52)</p><p>We now consider the difference θ = u 2 − u 1 − U , ω = v 2 − v 1 − V is the solution to problem (53),</p><p>{ θ t t + M ( ‖ ∇ m u 1 ‖ 2 + ‖ ∇ m v 1 ‖ 2 ) ( − Δ ) m θ + β ( − Δ ) m θ t = h 1 , ω t t + M ( ‖ ∇ m u 1 ‖ 2 + ‖ ∇ m v 1 ‖ 2 ) ( − Δ ) 2 m ω + β ( − Δ ) 2 m ω t = h 2 , ω ( 0 ) = ω t ( 0 ) = 0 , θ ( 0 ) = θ t ( 0 ) = 0. (53)</p><p>where</p><p>h 1 = [ M ( s ) − M ( s ˜ ) ] ( − Δ ) m u 2     + 2 M ′ ( s ) [ ( ∇ m u 1 , ∇ m U ) + ( ∇ m v 1 , ∇ m V ) ] ( − Δ ) m u 1     + ∂ g ( u 1 t , v 1 ) ∂ u 1 t U t + ∂ g ( u 1 t , v 1 ) ∂ v 1 V + g ( u 1 t , v 1 ) − g ( u 2 t , v 2 ) , (54)</p><p>h 2 = [ M ( s ) − M ( s ˜ ) ] ( − Δ ) 2 m v 2     + 2 M ′ ( s ) [ ( ∇ m u 1 , ∇ m U ) + ( ∇ m v 1 , ∇ m V ) ] ( − Δ ) 2 m v 1     + ∂ g ( u 1 , v 1 t ) ∂ u 1 U + ∂ g ( u 1 , v 1 t ) ∂ v 1 t V t + g ( u 1 , v 1 t ) − g ( u 2 , v 2 t ) , (55)</p><p>where s = ‖ ∇ m u 1 ‖ 2 + ‖ ∇ m v 1 ‖ 2 , s ˜ = ‖ ∇ m u 2 ‖ 2 + ‖ ∇ m v 2 ‖ 2 .</p><p>Some items are treated as follows</p><p>[ M ( s ) − M ( s ˜ ) ] ( − Δ ) m u 2 + 2 M ′ ( s ) [ ( ∇ m U , ∇ m u 1 ) + ( ∇ m V , ∇ m v 1 ) ] ( − Δ ) m u 1 = ( M ( s ) − M ( s ˜ ) ) ( − Δ ) m u 2 + 2 M ′ ( s ) [ ( ∇ m u 1 , ∇ m U ) + ( ∇ m v 1 , ∇ m V ) ] ( − Δ ) m u 1 = − 2 M ′ ( s ) [ ( − ∇ m U , ∇ m u 1 ) + ( − ∇ m V , ∇ m v 1 ) ] ( − Δ ) m u 1 + M ′ ( α s + ( 1 − α ) s ˜ )     &#215; [ ( ∇ m ( u 1 − u 2 ) , ∇ m ( u 2 + u 1 ) ) + ( ∇ m ( v 1 − v 2 ) , ∇ m ( v 2 + v 1 ) ) ] ( − Δ ) m u 2</p><p>≤ M ″ ( ξ ) [ ( ∇ m ( u 1 − u 2 ) , ∇ m ( u 2 + u 1 ) ) + ( ∇ m ( v 1 − v 2 ) , ∇ m ( v 2 + v 1 ) ) ] 2 ( − Δ ) m u 2     + M ′ ( s ) [ ( ∇ m ( u 1 − u 2 ) , ∇ m ( u 2 − u 1 ) ) + ( ∇ m ( v 1 − v 2 ) , ∇ m ( v 2 − v 1 ) ) ] ( − Δ ) m u 2     + 2 M ′ ( s ) [ ( ∇ m ( u 1 − u 2 ) , ∇ m u 1 ) + ( ∇ m ( v 1 − v 2 ) , ∇ m v 1 ) ] ( − Δ ) m ( u 2 − u 1 )     − 2 M ′ ( s ) [ ( ∇ m θ , ∇ m u 1 ) + ( ∇ m ω , ∇ m v 1 ) ] ( − Δ ) m u 1 = I 1 + I 2 + I 3 + I 4 . (56)</p><p>Let ( − Δ ) k θ t inner product with the first equation in (53), ( − Δ ) 2 k ω t and inner product with the second equation in (53), obtain</p><p>( I 1 , ( − Δ ) k θ t ) ≤ ( 2 M ″ ( ξ ) [ ( ∇ m ( u 1 − u 2 ) , ∇ m ( u 2 + u 1 ) ) 2 + ( ∇ m ( v 1 − v 2 ) , ∇ m ( v 2 + v 1 ) ) 2 ] ∇ m + k u 2 , ∇ m + k θ t ) ≤ 2 c 1 ( 4 ε 2 ‖ ∇ m ( u 2 − u 1 ) ‖ 4 + 4 ε 2 ‖ ∇ m ( v 2 − v 1 ) ‖ 4 + ε 2 ‖ ∇ m + k θ t ‖ 2 8 ) , ≤ 2 c 1 ( 4 ε 2 ( λ 1 − 2 k + λ 1 − 2 m − 4 k ) ( ‖ ∇ m + k ( u 2 − u 1 ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 2 − v 1 ) ‖ 4 ) + ε 2 ‖ ∇ m + k θ t ‖ 2 8 ) ,</p><p>( I 2 , ( − Δ ) k θ t ) = ( M ′ ( s ) [ ( ∇ m ( u 1 − u 2 ) , ∇ m ( u 2 − u 1 ) ) + ( ∇ m ( v 1 − v 2 ) , ∇ m ( v 2 − v 1 ) ) ] ( − Δ ) m + k u 2 , ∇ m + k θ t ) ≤ c 2 ( 4 ε 2 ‖ ∇ m ( u 2 − u 1 ) ‖ 4 + 4 ε 2 ‖ ∇ m ( v 2 − v 1 ) ‖ 4 + ε 2 ‖ ∇ m + k θ t ‖ 2 8 ) ≤ c 2 ( 4 ε 2 ( λ 1 − 2 k + λ 1 − 2 m − 4 k ) ( ‖ ∇ m + k ( u 2 − u 1 ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 2 − v 1 ) ‖ 4 + ) ε 2 ‖ ∇ m + k θ t ‖ 2 8 ) ,</p><p>( I 3 , ( − Δ ) k θ t ) ≤ ( 2 M ′ ( s ) [ ( ∇ m ( u 1 − u 2 ) , ∇ m u 1 ) + ( ∇ m ( v 1 − v 2 ) , ∇ m v 1 ) ] ∇ m + k ( u 2 − u 1 ) , ∇ m + k θ t ) ≤ 2 c 3 ( 4 ε 2 ‖ ∇ m ( u 2 − u 1 ) ‖ 4 + 4 ε 2 ‖ ∇ m ( v 2 − v 1 ) ‖ 4 + 4 ε 2 ‖ ∇ m + k ( u 2 − u 1 ) ‖ 4 + 3 ε 2 ‖ ∇ m + k θ t ‖ 2 16 ) ≤ 2 c 3 ( 4 ε 2 ( λ 1 − 2 k + λ 1 − 2 m − 4 k + 1 ) ( ‖ ∇ m + k ( u 2 − u 1 ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 2 − v 1 ) ‖ 4 ) + 3 ε 2 ‖ ∇ m + k θ t ‖ 2 16 ) ,</p><p>( I 4 , ( − Δ ) k θ t ) = ( − 2 M ′ ( s ) [ ( ∇ m θ , ∇ m u 1 ) + ( ∇ m ω , ∇ m v 1 ) ] ∇ m + k u 1 , ∇ m + k θ t ) ≤ 2 c 4 ( 4 ‖ ∇ m + k θ ‖ 2 ε 2 + 4 ‖ ∇ m + k ω ‖ 2 ε 2 + ε 2 ‖ ∇ m + k θ t ‖ 2 8 ) ≤ 2 c 4 ( 4 ε 2 ( λ 1 − 2 k + λ 1 − 2 m − 4 k ) ( ‖ ∇ m + k θ ‖ 2 + ‖ ∇ 2 m + 2 k ω ‖ 2 ) + ε 2 ‖ ∇ m + k θ t ‖ 2 8 ) ,</p><p>which implies that</p><p>| ( [ M ( s ) − M ( s ˜ ) ] ( − Δ ) m u 2 + 2 M ′ ( s ) [ ( ∇ m U , ∇ m u 1 ) + ( ∇ m V , ∇ m v 1 ) ] ( − Δ ) m u 1 , ( − Δ ) k θ t ) | ≤ c 5 ε 2 ( λ 1 − 2 k + λ 1 − 2 m − 4 k + 1 ) ( ‖ ∇ m + k ( u 2 − u 1 ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 2 − v 1 ) ‖ 4     + ‖ ∇ m + k θ ‖ 2 + ‖ ∇ 2 m + 2 k ω ‖ 2 ) + c 6 ε 2 ‖ ∇ m + k θ t ‖ 2 . (57)</p><p>Analogously,</p><p>| ( [ M ( s ) − M ( s ˜ ) ] ( − Δ ) 2 m v 2 + 2 M ′ ( s ) [ ( ∇ m u 1 , ∇ m U ) + ( ∇ m v 1 , ∇ m V ) ] ( − Δ ) 2 m v 1 , ( − Δ ) 2 k ω t ) | ≤ c 7 ε 2 ( λ 1 − 2 k + λ 1 − 2 m − 4 k + 1 ) ( ‖ ∇ m + k ( u 2 − u 1 ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 2 − v 1 ) ‖ 4     + ‖ ∇ m + k θ ‖ 2 + ‖ ∇ 2 m + 2 k ω ‖ 2 ) + c 8 ε 2 ‖ ∇ 2 m + 2 k ω t ‖ 2 . (58)</p><p>Further,</p><p>( ∂ g ( u 1 t , v 1 ) ∂ u 1 t U t + ∂ g ( u 1 t , v 1 ) ∂ v 1 V + g ( u 1 t , v 1 ) − g ( u 2 t , v 2 ) , ( − Δ ) k θ t ) = ( ∂ g ( ξ 1 , v 1 ) ∂ ξ 1 ( u 1 t − u 2 t ) + ∂ g ( u 2 t , η 1 ) ∂ η 1 ( v 1 − v 2 ) + ∂ g ( u 1 t , v 1 ) ∂ u 1 t U t + ∂ g ( u 1 t , v 1 ) ∂ v 1 V , ( − Δ ) k θ t ) ≤ ( ∂ 2 g ( ξ 2 , v 1 ) ∂ ξ 2 ( u 1 t − u 2 t ) ( ξ 1 − u 1 t ) + ∂ g ( u 1 t , v 1 ) ∂ u 1 t ( − θ t ) + ∂ 2 g ( u 2 t , η 2 ) ∂ , η 2 2 ( v 1 − v 2 ) ( η 1 − v 1 )     + ∂ 2 g ( ξ 3 , v 1 ) ∂ ξ 3 ∂ v ( v 1 − v 2 ) ( u 2 t − u 1 t ) + ∂ g ( u 1 t , v 1 ) ∂ v 1 ( − ω ) , ( − Δ ) k θ t )</p><p>≤ c 9 λ 1 − k 2 − m ‖ ∇ m + k ( u 1 t − u 2 t ) ‖ 2 ‖ ∇ k θ t ‖ + c 10 ‖ ∇ k θ t ‖ 2 + c 11 λ 1 − 3 k 2 − 2 m ‖ ∇ 2 m + 2 k ( v 1 − v 2 ) ‖ 2 ‖ ∇ k θ t ‖     + c 12 λ 1 − k − 3 m 2 ‖ ∇ m + k ( u 1 t − u 2 t ) ‖ ‖ ∇ 2 k + 2 m ( v 1 − v 2 ) ‖ ‖ ∇ k θ t ‖ + c 13 λ 1 − k 2 − m ‖ ∇ 2 m + 2 k ω ‖ ‖ ∇ k θ t ‖ ≤ c 14 2 ( 3 2 ε 2 ( ‖ ∇ m + k ( u 1 t − u 2 t ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 1 − v 2 ) ‖ 4 ) + ε 2 ( ‖ ∇ k θ t ‖ 2 + 5 ‖ ∇ 2 m + 2 k ω ‖ 2 ) ) , (59)</p><p>where,</p><p>ξ 1 = r 1 u 1 t − ( 1 − r 1 ) u 2 t , η 1 = r 2 v 1 − ( 1 − r 2 ) v 2 , ξ 2 = r 3 ξ 1 − ( 1 − r 3 ) u 1 t , η 2 = r 4 η 1 − ( 1 − r 4 ) v 1 , ξ 3 = r 5 u 2 t − ( 1 − r 5 ) u 1 t , c 14 = max { c 9 λ 1 − k 2 − m , 2 c 10 ε 2 , c 11 λ 1 − 3 k 2 − 2 m , c 12 λ 1 − k − 3 m 2 , c 13 λ 1 − k 2 − m } .</p><p>Similarly</p><p>( ∂ g ( u , v t ) ∂ u U + ∂ g ( u , v t ) ∂ v t V t + g ( u , v t ) − g ( u ˜ , v ˜ t ) , ω t ) ≤ c 15 2 ( 3 2 ε 2 ( ‖ ∇ m + k ( u 1 − u 2 ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 1 t − v 2 t ) ‖ 4 )         + ε 2 ( ‖ ∇ m + k θ ‖ 2 + 5 ‖ ∇ 2 k ω t ‖ 2 ) ) . (60)</p><p>Based on the above Equations (57)-(60), it is sorted out that</p><p>1 2 d d t { ‖ ∇ k θ t ‖ 2 + ‖ ∇ 2 k ω t ‖ 2 + μ ( ‖ ∇ m + k θ ‖ 2 + ‖ ∇ 2 m + 2 k ω ‖ 2 ) } + ( β − ( c 6 + c 8 ) ε 2 ) ( ‖ ∇ m + k θ t ‖ 2 + ‖ ∇ 2 m + 2 k ω t ‖ 2 ) ≤ c 16 ( ‖ ∇ m + k ( u 1 − u 2 ) ‖ 4 + ‖ ∇ m + k ( u 1 t − u 2 t ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 1 − v 2 ) ‖ 4     + ‖ ∇ 2 m + 2 k ( v 1 t − v 2 t ) ‖ 4 ) + c 17 ( ‖ ∇ k θ t ‖ 2 + ‖ ∇ 2 k ω t ‖ 2 + ‖ ∇ m + k θ ‖ 2 + ‖ ∇ 2 m + 2 k ω ‖ 2 ) , (61)</p><p>where β ≥ ( c 6 + c 8 ) ε 2 ,</p><p>by Gronwall inequality</p><p>‖ θ t ‖ 2 + ‖ ω t ‖ 2 + μ ( ‖ ∇ m θ ‖ 2 + ‖ ∇ 2 m ω ‖ 2 ) ≤ c 18 e C 19 t ∫ 0 t ( ‖ ∇ m + k ( u 1 − u 2 ) ‖ 4 + ‖ ∇ m + k ( u 1 t − u 2 t ) ‖ 4       + ‖ ∇ 2 m + 2 k ( v 1 − v 2 ) ‖ 4 + ‖ ∇ 2 m + 2 k ( v 1 t − v 2 t ) ‖ 4 ) d τ ≤ c 20 e c 21 t ‖ ( ξ , ζ , η , σ ) ‖ E k 4 , (62)</p><p>so as ‖ ( ξ , ζ , η , σ ) ‖ E k 2 → 0 in E k , there are</p><p>‖ S ( t ) ρ ˜ 0 − S ( t ) ρ 0 − ( D S ( t ) ρ 0 ) ( ξ 1 , ξ 2 , η 1 , η 2 ) ‖ E k 2 ‖ ( ξ , ζ , η , σ ) ‖ E k 2 ≤ c 20 e c 21 t ‖ ( ξ , ζ , η , σ ) ‖ E k 2 → 0.</p><p>The differentiability of S(t) is proved.</p><p>The next step will be used in demonstrating the process of dimension estimation. it seems obvious that the Equations (1)-(5) also can be written as</p><p>φ t + H ( φ ) = F ( φ ) , (63)</p><p>where φ = ( u , p , v , q ) T , p = u t + ε u , q = v t + ε v ,</p><p>H ( φ ) = ( ε u − p ( − Δ ) m u − ε ( β ( − Δ ) m − ε ) u + ( β ( − Δ ) m − ε ) p ε v − q ( − Δ ) 2 m v − ε ( β ( − Δ ) 2 m − ε ) v + ( β ( − Δ ) 2 m − ε ) q ) , (64)</p><p>F ( φ ) = ( 0 f 1 ( x ) − g ( u t , v ) + ( 1 − M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 ) ) ( − Δ ) m u 0 f 2 ( x ) − g ( u , v t ) + ( 1 − M ( ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ ) ) ( − Δ ) 2 m v ) . (65)</p><p>Consider the first variation equation of (63)</p><p>Ψ ′ + P ( φ ) Ψ = Γ 1 ( φ ) Ψ + Γ 2 ( φ ) Ψ , (66)</p><p>where Ψ = ( U , P , V , Q ) T , P = U t + ε U , Q = V t + ε V , and φ = ( u , p , v , q ) T is the solution to problem (63), and</p><p>P ( φ ) = ( ε I − I 0 0 ( 1 − β ε ) ( − Δ ) m + ε 2 I β ( − Δ ) 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 ) , (67)</p><p>Γ 1 ( φ ) = ( 0 0 0 0 ε ∂ g ( u t , v ) ∂ u t − ∂ g ( u t , v ) ∂ u t − ∂ g ( u t , v ) ∂ v 0 0 0 0 0 − ∂ g ( u , v t ) ∂ u 0 ε ∂ g ( u , v t ) ∂ v t − ∂ g ( u , v t ) ∂ v t ) , (68)</p><p>Γ 2 ( φ ) Ψ = ( 0 ( 1 − M ( s ) ) ( − Δ ) m U − 2 M ′ ( s ) [ ( ∇ m u , ∇ m U ) + ( ∇ m v , ∇ m V ) ] ( − Δ ) m u 0 ( 1 − M ( s ) ) ( − Δ ) 2 m V − 2 M ′ ( s ) [ ( ∇ m u , ∇ m U ) + ( ∇ m v , ∇ m V ) ] ( − Δ ) 2 m v ) . (69)</p><p>Theorem 4 Under the condition of Theorem 3, the global attractors of initial boundary value problems (1)-(5) have finite dimensional Hausdroff dimension</p><p>and fractal dimension, and then d H ( A k ) &lt; 2 3 n 0 , d F ( A k ) &lt; 4 3 n 0 .</p><p>Proof: For any fixed ( u 0 , p 0 , v 0 , q 0 ) ∈ E k , assume that χ 1 , χ 2 , ⋯ , χ n are n 0 elements in E k , and ψ 1 ( t ) , ψ 2 ( t ) , ⋯ , ψ n 0 ( t ) are n 0 solutions of the linearized Equation (66) with an initial value ψ 1 ( 0 ) = χ 1 , ψ 2 ( 0 ) = χ 2 , ⋯ , ψ n 0 ( 0 ) = χ n 0 , where n 0 is a natural number. It can be obtained by calculation</p><p>‖ ψ 1 ( t ) ∧ ⋯ ∧ ψ n 0 ( t ) ‖ ∧ n 0 E k ≤ ‖ χ 1 ∧ ⋯ ∧ χ n ‖ ∧ n 0 E k exp ∫ 0 t t r F ′ ( φ ( τ ) ) ∘ Q n ​ 0 ( τ ) d τ . (70)</p><p>where ∧ represents the outer product, tr represents the trace of the operator, Q n 0 ( τ ) = Q n 0 ( τ , φ 0 ; χ 1 , χ 2 , ⋯ , χ n 0 ) represents the orthogonal projection from E k to s p a n { ψ 1 ( t ) , ψ 2 ( t ) , ⋯ , ψ n 0 ( t ) } .</p><p>At a given time τ , let h j ( τ ) = ( ξ j ( τ ) , ζ j ( τ ) , η j ( τ ) , σ j ( τ ) ) T , j = 1 , 2 , ⋯ , n 0</p><p>are the standard orthogonal basis of space s p a n { ψ 1 ( t ) , ψ 2 ( t ) , ⋯ , ψ n 0 ( t ) } , then define the inner product on E k</p><p>( h j , h j &#175; ) = ( ∇ m + k ξ j , ∇ m + k ξ j &#175; ) + ( ∇ k ζ j , ∇ k ζ j &#175; )     + ( ∇ 2 m + 2 k η j , ∇ 2 m + 2 k η j &#175; ) + ( ∇ 2 k σ j , ∇ 2 k σ j &#175; ) ,</p><p>‖ h j ‖ E k 2 = ( h j , h j ) E k = ‖ ∇ m + k ξ j ‖ 2 + ‖ ∇ k ζ j ‖ 2 + ‖ ∇ 2 m + 2 k η j ‖ 2 + ‖ ∇ 2 k σ j ‖ 2 = 1.</p><p>Through the above conditions, can get</p><p>t r F ′ ( φ ( τ ) ) ∘ Q n 0 ( τ ) = ∑ j = 1 + ∞ ( F ′ ( φ ( τ ) ) ∘ Q n 0 ( τ ) h j ( τ ) , h j ( τ ) ) E k = ∑ j = 1 n 0 ( F ′ ( φ ( τ ) ) h j ( τ ) , h j ( τ ) ) E k ,</p><p>by the Holder inequality, Young and Poincare inequality</p><p>( P ( φ ) h j , h j ) = ε ( ∇ m + k ξ j , ∇ m + k ξ j ) − ( ∇ m + k ζ j , ∇ m + k ξ j ) + ( 1 − β ε ) ( ∇ m + k ξ j , ∇ m + k ζ j )       + ε 2 ( ∇ k ξ j , ∇ k ζ j ) + β ( ∇ m + k ζ j , ∇ m + k ζ j ) − ε ( ∇ k ζ j , ∇ k ζ j )       + ε ( ∇ 2 m + 2 k η j , ∇ 2 m + 2 k η j ) − ( ∇ 2 m + 2 k σ j , ∇ 2 m + 2 k η j ) + ε 2 ( ∇ 2 k η j , ∇ 2 k σ j )       + β ( ∇ 2 m + 2 k σ j , ∇ 2 m + 2 k σ j ) − ε ( ∇ 2 k σ j , ∇ 2 k σ j )</p><p>      + ( 1 − β ε ) ( ∇ 2 m + 2 k η j , ∇ 2 m + 2 k σ j ) ≥ ε ‖ ∇ m + k ξ j ‖ 2 − ( 2 − β ε ) ‖ ∇ m + k ζ j ‖ ‖ ∇ m + k ξ j ‖ − ε 2 ‖ ∇ k ξ j ‖ ‖ ∇ k ζ j ‖       + β ‖ ∇ m + k ζ j ‖ 2 − ε ‖ ∇ k ζ j ‖ 2 − ( 2 − β ε ) ‖ ∇ 2 m + 2 k σ j ‖ ‖ ∇ 2 m + 2 k η j ‖       − ε 2 ‖ ∇ 2 k η j ‖ ‖ ∇ 2 k σ j ‖ + β ‖ ∇ 2 m + 2 k σ j ‖ 2 − ε ‖ ∇ 2 k σ j ‖ 2 + ε ‖ ∇ 2 m + 2 k η j ‖ 2</p><p>≥ ( β λ 1 m − ε 2 + 2 ε 2 − c 21 ( 2 − β ε ) λ 1 m 2 ) ‖ ∇ k ζ j ‖ 2 − ε 2 2 ‖ ∇ k ξ j ‖ 2 − ε 2 2 ‖ ∇ 2 k η j ‖ 2     + ( ε − c 21 ( 2 − β ε ) 2 ) ‖ ∇ m + k ξ j ‖ 2 + ( ε − c 22 ( 2 − β ε ) 2 ) ‖ ∇ 2 m + 2 k η j ‖ 2     + ( β λ 1 2 m − ε 2 + 2 ε 2 − c 22 ( 2 − β ε ) λ 1 2 m 2 ) ‖ ∇ 2 k σ j ‖ 2 , (71)</p><p>( Γ 1 ( φ ) h j , h j ) = ε ( ∂ g ( u t , v ) ∂ u t ∇ k ξ j , ∇ k ζ j ) − ( ∂ g ( u t , v ) ∂ u t ∇ k ζ j , ∇ k ζ j )     − ( ∂ g ( u t , v ) ∂ v ∇ k η j , ∇ k ζ j ) − ( ∂ g ( u , v t ) ∂ u ∇ 2 k ξ j , ∇ 2 k σ j )     + ε ( ∂ g ( u , v t ) ∂ v t ∇ 2 k η j , ∇ 2 k σ j ) − ( ∂ g ( u , v t ) ∂ v t ∇ 2 k σ j , ∇ 2 k σ j )</p><p>≤ ε ‖ ∂ g ( u t , v ) ∂ u t ‖ ∞ ‖ ∇ k ξ j ‖ ‖ ∇ k ζ j ‖ − ‖ ∂ g ( u t , v ) ∂ u t ‖ ∞ ‖ ∇ k ζ j ‖ 2     − ‖ ∂ g ( u t , v ) ∂ v ‖ ∞ ‖ ∇ k η j ‖ ‖ ∇ k ζ j ‖ − ‖ ∂ g ( u , v t ) ∂ u ‖ ∞ ‖ ∇ 2 k ξ j ‖ ‖ ∇ 2 k σ j ‖     + ε ‖ ∂ g ( u , v t ) ∂ v t ‖ ∞ ‖ ∇ 2 k η j ‖ ‖ ∇ 2 k σ j ‖ − ‖ ∂ g ( u , v t ) ∂ v t ‖ ∞ ‖ ∇ 2 k σ j ‖ 2 ≤ ε 2 c 23 2 ‖ ∇ k ξ j ‖ 2 + c 23 − c 24 2 ‖ ∇ k ζ j ‖ 2 + ε 2 c 25 2 ‖ ∇ 2 k η j ‖ 2 + c 23 − c 26 2 ‖ ∇ 2 k σ j ‖ 2 , (72)</p><p>( Γ 2 ( φ ) h j , h j ) = ( 1 − M ( s ) ) ( ∇ m + k ξ j , ∇ m + k ζ j ) − 2 M ′ ( s ) ( ∇ m u , ∇ m ξ j ) ( ∇ 2 m + k u , ∇ k ζ j )       − 2 M ′ ( s ) ( ∇ m v , ∇ m η j ) ( ∇ 2 m + k u , ∇ k ζ j ) + ( 1 − M ( s ) ) ( ∇ 2 m + 2 k η j , ∇ 2 m + 2 k σ j )       − 2 M ′ ( s ) ( ∇ m u , ∇ m ξ j ) ( ∇ 4 m + 2 k v , ∇ 2 k σ j )       − 2 M ′ ( s ) ( ∇ m v , ∇ m η j ) ( ∇ 4 m + 2 k v , ∇ 2 k σ j )</p><p>≤ ( 1 − δ 0 ) λ 1 m 2 ( ‖ ∇ m + k ξ j ‖ ‖ ∇ k ζ j ‖ + ‖ ∇ 2 m + 2 k η j ‖ ‖ ∇ 2 k σ j ‖ )       + 2 c 26 ( λ 1 − k 2 + λ 1 − m 2 − k ) ( ‖ ∇ m + k ξ j ‖ ‖ ∇ k ζ j ‖ + ‖ ∇ 2 m + 2 k η j ‖ ‖ ∇ k ζ j ‖ )       + 2 c 27 ( λ 1 − k 2 + λ 1 − m 2 − k ) ( ‖ ∇ m + k ξ j ‖ ‖ ‖ ∇ 2 k σ j ‖ ‖ + ‖ ∇ 2 m + 2 k η j ‖ ‖ ∇ 2 k σ j ‖ )</p><p>≤ 1 − δ 0 2 λ 1 m 2 ( ‖ ∇ m + k ξ j ‖ 2 + ‖ ∇ k ζ j ‖ 2 + ‖ ∇ 2 m + 2 k η j ‖ 2 + ‖ ∇ 2 k σ j ‖ 2 )       + 2 c 28 ( λ 1 − k 2 + λ 1 − m 2 − k ) ( ‖ ∇ m + k ξ j ‖ 2 + ‖ ∇ k ζ j ‖ 2 + ‖ ∇ 2 m + 2 k η j ‖ 2 + ‖ ∇ 2 k σ j ‖ 2 ) ≤ 1 − δ 0 2 λ 1 m 2 + 2 c 28 ( λ 1 − k 2 + λ 1 − m 2 − k ) . (73)</p><p>Based on the above Equations (71)-(73), it is sorted out that</p><p>( F ′ ( φ ( τ ) ) h j ( τ ) , h j ( τ ) ) E k = ( ( − P ( φ ) + Γ 1 ( φ ) + Γ 2 ( φ ) ) h j , h j ) ≤ − ( β λ 1 m − ε 2 + 2 ε 2 − c 21 ( 2 − β ε ) λ 1 m 2 − c 23 − c 24 2 ) ‖ ∇ k ζ j ‖ 2       − ( β λ 1 2 m − ε 2 + 2 ε 2 − c 22 ( 2 − β ε ) λ 1 2 m 2 − c 25 − c 26 2 ) ‖ ∇ 2 k σ j ‖ 2       − ( ε − c 21 ( 2 − β ε ) 2 ) ‖ ∇ m + k ξ j ‖ 2 − ( ε − c 22 ( 2 − β ε ) 2 ) ‖ ∇ 2 m + 2 k η j ‖ 2       + ε 2 ( c 23 − 1 ) 2 ‖ ∇ k ξ j ‖ 2 + ε 2 ( c 25 − 1 ) 2 ‖ ∇ 2 k η j ‖ 2 + 1 − δ 0 2 λ 1 m 2       + 2 c 28 ( λ 1 − k 2 + λ 1 − m 2 − k ) , (74)</p><p>let</p><p>b = min { β λ 1 m − ε 2 + 2 ε 2 − c 21 ( 2 − β ε ) λ 1 m 2 − c 23 − c 24 2 , ε − c 21 ( 2 − β ε ) 2 ,                   ε − c 22 ( 2 − β ε ) 2 , β λ 1 2 m − ε 2 + 2 ε 2 − c 22 ( 2 − β ε ) λ 1 2 m 2 − c 25 − c 26 2 ,                   − 1 − δ 0 2 λ 1 m 2 − 2 c 28 ( λ 1 − k 2 + λ 1 − m 2 − k ) } ,</p><p>a = max { ε 2 ( c 23 − 1 ) 2 , ε 2 ( c 25 − 1 ) 2 } ,</p><p>we can obtain</p><p>∑ j = 1 n 0 ( F ′ ( φ ( τ ) ) h j ( τ ) , h j ( τ ) ) E k ≤ − n 0 b + a ∑ j = 1 n 0 ( ‖ ∇ k ξ j ‖ 2 + ‖ ∇ 2 k η j ‖ 2 )     , (75)</p><p>for almost all times t, there is ∑ i = 1 n 0 ‖ ∇ k ξ i ‖ 2 ≤ ∑ i = 1 n 0 λ i s ′ − 1 , ∑ j = 1 n 0 ‖ ∇ 2 k η j ‖ 2 ≤ ∑ j = 1 n 0 λ j s ″ − 1 , so</p><p>t r F ′ ( φ ( τ ) ) ∘ Q n 0 ( τ ) ≤ − n 0 b + a ( ∑ i = 1 n 0 λ i s ′ − 1 + ∑ j = 1 n 0 λ j s ″ − 1 ) , (76)</p><p>because of</p><p>q n 0 ( t ) = sup φ 0 ∈ B 0 k sup Ψ j ( 0 ) ∈ E k { 1 t ∫ 0 t t r F ′ ( φ ( τ ) ) ∘ Q n 0 ( τ ) d τ } , q n 0 = lim t → ∞ q n 0 ( t ) , (77)</p><p>q n 0 ( t ) ≤ − n 0 b + a ( ∑ i = 1 n 0 λ i s ′ − 1 + ∑ j = 1 n 0 λ j s ″ − 1 ) , q n 0 ≤ − n 0 b + a ( ∑ i = 1 n 0 λ i s ′ − 1 + ∑ j = 1 n 0 λ j s ″ − 1 ) , (78)</p><p>Therefore, the Lyapunov exponent K 1 , K 2 , ⋯ , K n 0 on set B 0 k is uniformly bounded, and</p><p>K 1 + K 2 + ⋯ + K n 0 ≤ − n 0 b + a ( ∑ i = 1 n 0 λ i s ′ − 1 + ∑ j = 1 n 0 λ j s ″ − 1 ) , (79)</p><p>so</p><p>( q i j ) + ≤ − n 0 b + a ( ∑ i = 1 n 0 λ i s ′ − 1 + ∑ j = 1 n 0 λ j s ″ − 1 ) ≤ a ( ∑ i = 1 n 0 λ i s ′ − 1 + ∑ j = 1 n 0 λ j s ″ − 1 ) ≤ 2 5 n 0 b , (80)</p><p>q n 0 ≤ − n 0 b ( 1 − a n 0 b ( ∑ i = 1 n 0 λ i s ′ − 1 + ∑ j = 1 n 0 λ j s ″ − 1 ) ) ≤ − 3 5 n 0 b , (81)</p><p>further</p><p>max 1 ≤ i , j ≤ n 0 ( q i j ) + | q n 0 | ≤ 2 3 . (82)</p><p>Thus, can obtain d H ( A k ) &lt; 2 3 n 0 , d F ( A k ) &lt; 4 3 n 0 .</p></sec><sec id="s4"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s5"><title>Cite this paper</title><p>Lin, G.G. and Shao, M. (2022) The Long-Term Dynamic Behavior of Solutions to a Class of Generalized Higher-Order Kirchhoff-Type Coupled Wave Equations. Journal of Applied Mathematics and Physics, 10, 2181-2199. https://doi.org/10.4236/jamp.2022.107150</p></sec></body><back><ref-list><title>References</title><ref id="scirp.118637-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Lin, G.G. and Gao, Y.L. (2017) The Global and Exponential Attractors for the Higher-Order Kirchhoff-Type Equation with Strong Linear Damping. 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