A Family of the Inertial Manifolds for a Class of Generalized Kirchhoff-Type Coupled Equations

Abstract

The paper considers the long-time behavior for a class of generalized high-order Kirchhoff-type coupled equations, under the corresponding hypothetical conditions, according to the Hadamard graph transformation method, obtain the equivalent norm in space , and we obtain the existence of a family of the inertial manifolds while such equations satisfy the spectral interval condition.

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Lin, G. and Zhou, J. (2022) A Family of the Inertial Manifolds for a Class of Generalized Kirchhoff-Type Coupled Equations. Open Journal of Applied Sciences, 12, 1116-1127. doi: 10.4236/ojapps.2022.127076.

1. Introduction

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 .

In order to overcome the research difficulties, G. Foias, G. R. Sell and R. Temam [1] proposed the concept of inertial manifolds, which greatly promoted the study of infinite-dimensional dynamical systems. Where the inertial manifold is a positive, finite-dimensional Lipschitz manifold, and the existence of an inertial manifold depends on the establishment of a spectral interval condition. Therefore, the research on a family of inertial manifolds is of great significance from both theoretical and practical aspects, and the relevant theoretical achievements can be referred to [2] - [9].

Guoguang Lin, Lingjuan Hu [10] studied a system of coupled wave equations of higher-order Kirchhoff type with strong damping terms

{ 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. Using the Hadamard graph transformation method, the Lipschitz constant l F of F is further estimated, and the inertial manifolds that satisfies the spectral interval condition is obtained.

Lin Guoguang, Liu Xiaomei [11] studied a family of inertial manifolds for a class of generalized higher-order Kirchhoff equations with strong dissipation terms

{ 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 ( x , t ) = 0 , i u v i = 0 , i = 1 , 2 , , 2 m 1 , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω R n .

where m N + , Ω R n ( n 1 ) is a bounded domain with a smooth boundary in Ω , f ( x ) is an external force term, M ( m u p p ) is the stress term of Kirchhoff equation, β ( Δ ) 2 m u t is a strong dissipative term, | u | ρ ( u t + u ) is a nonlinear source term. Based on appropriate assumptions and the Hadamard graph transformation method, the spectral interval condition is verified, and the existence of a family of the inertial manifolds of the equation is obtained.

On the basis of previous research, rigid term strengthening becomes

M ( D m u p p + D m v p p ) ( Δ ) 2 m u and M ( D m u p p + D m v p p ) ( Δ ) 2 m v , and this paper seeks a family of inertial manifolds. When defining the equivalence norm in space E k , by making reasonable assumptions, it is obtained that the equation satisfies the spectral interval condition so that there is a family of inertial manifolds.

2. Preliminaries

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

Set = D . Consider Hilbert space family 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 there are

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:

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

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

3. A Family of Inertial Manifolds

Definition 1 [12] lets S = { S ( t ) } t 0 be the solution semigroup on Banach space E k = H 0 2 m + k ( Ω ) × H 0 k ( Ω ) ( k = 1 , 2 , , 2 m ) , and a subset μ k E k satisfies:

1) μ k is finite-dimensional Lipschitz popular;

2) μ k is positively unchanging, { S ( t ) } t 0 : u 0 μ k , S ( t ) u 0 μ k , t 0 ;

3) μ k attracts the solution orbit exponentially, i.e. for any u E k , the existence constant η > 0 , c > 0 makes d i s t ( S ( t ) u , μ ) c e η t , t 0 .

Then μ k is called E k is a family of inertial manifolds.

In order to describe the spectral interval condition, first consider that the nonlinear term F : E k E k is integrally bounded and continuous, and has a positive Lipschitz constant l F , and its operator A has several eigenvalues and eigenfunctions of the positive real part.

Definition 2 [12] Set operator A : Χ Χ has several eigenvalues of positive real numbers, and F C b ( Χ , Χ ) satisfies the Lipschitz condition:

F ( u ) F ( v ) Χ l F u v Χ , u , v Χ ,

The point spectrum of the operator A can be divided into two parts σ 1 and σ 2 , and σ 1 is finite,

Λ 1 = sup { Re λ | λ σ 1 } , Λ 2 = sup { Re λ | λ σ 2 } , Χ i = s p a n { ω j | λ j σ i } , i = 1 , 2.

and conditions

Λ 2 Λ 1 > 4 l F (6)

are satisfied.

Where the continuous projection P 1 : Χ Χ 1 , P 2 : Χ Χ 2 , there is orthogonal decomposition Χ = Χ 1 Χ 2 , then the operator A satisfies the spectral interval condition.

Lemma 1 g i : V 2 m + k × V 2 m + k V 2 m + k × V 2 m + k ( i = 1 , 2 ) is a uniform bounded and integral Lipschitz continuous function.

Proof: ( u ˜ , v ˜ ) , ( u , v ) V 2 m + k × V 2 m + k ( k = 1 , 2 , , 2 m ) ,

g 1 ( u ˜ , v ˜ ) g 1 ( u , v ) V 2 m + k × V 2 m + k = g 1 u ( u + θ ( u ˜ u ) , v + θ ( v ˜ v ) ) ( u ˜ u ) + g 1 v ( u + θ ( u ˜ u ) , v + θ ( v ˜ v ) ) ( v ˜ v ) V 2 m + k × V 2 m + k g 1 u ( u + θ ( u ˜ u ) , v + θ ( v ˜ v ) ) ( u ˜ u ) V 2 m + k × V 2 m + k + g 1 v ( u + θ ( u ˜ u ) , v + θ ( v ˜ v ) ) ( v ˜ v ) V 2 m + k × V 2 m + k l ( u ˜ u V 2 m + k + v ˜ v V 2 m + k ) .

Similarly, there are

g 2 ( u ˜ , v ˜ ) g 2 ( u , v ) V 2 m + k × V 2 m + k l ( u ˜ u V 2 m + k + v ˜ v V 2 m + k )

where l is the Lipschitz constant of g i , θ ( 0 , 1 ) .

Lemma 2 [12] lets the sequence of eigenvalues { μ j } j 1 is a non-subtractive sequence, then N 0 N + , for N N 0 , μ N and μ N + 1 are consecutive adjacent values.

In order to verify that the operator satisfies the spectral interval condition, so as to draw the conclusion that there is a family of inertial manifolds in questions (1)-(5), the following definitions and assumptions can be made first.

Based on the above relevant conditions, consider the first-order development equation equivalent to Equations (1)-(5), as follows:

U t + A U = F ( U ) (7)

Of which U = ( u , z , v , q ) ,

A = ( 0 I 0 0 M ( s ) ( Δ ) 2 m β ( Δ ) 2 m 0 0 0 0 0 I 0 0 M ( s ) ( Δ ) 2 m β ( Δ ) 2 m ) ,

F ( U ) = ( 0 f 1 ( x ) g 1 ( u , v ) 0 f 2 ( x ) g 2 ( u , v ) ) ,

D ( A ) = { ( u , v ) V 2 m + k × V 2 m + k | ( u , v ) V 0 × V 0 , ( D 2 m + k u , D 2 m + k v ) V 0 × V 0 } × V k × V k ,

s = D m u p p + D m v p p .

In order to determine the eigenvalue of matrix operator A , first consider graph module ( U , V ) E k = ( M ( s ) D 2 m + k u , D 2 m + k u ¯ ) + ( D k z ¯ , D k z ) ) + ( M ( s ) D 2 m + k v , D 2 m + k v ¯ ) + ( D k q ¯ , D k q ) generated by inner product in E k .

Where U = ( u , z , v , q ) , V = ( u , z , v , q ) , and u ¯ , z ¯ , v ¯ , q ¯ represent the conjugation of u , z , v , q respectively. In addition, operator A is monotonic, and for U D ( A ) , there is

( A U , U ) E k = ( M ( s ) D 2 m + k z , D 2 m + k u ¯ ) + ( D k z ¯ , M ( s ) ( Δ ) 2 m u + β ( Δ ) 2 m z ) ( M ( s ) D 2 m + k q , D 2 m + k v ¯ ) + ( q ¯ , M ( s ) ( Δ ) 2 m v + β ( Δ ) 2 m q ) = ( M ( s ) D 2 m + k z , D 2 m + k u ¯ ) + ( D 2 m + k z ¯ , M ( s ) D 2 m + k u ) + ( D 2 m + k z ¯ , β D 2 m + k z ) ( M ( s ) D 2 m + k q , D 2 m + k v ¯ ) + ( D 2 m + k q ¯ , M ( s ) D 2 m + k v ) + ( D 2 m + k q ¯ , β D 2 m + k q ) = β ( D 2 m + k z 2 + D 2 m + k q 2 ) 0 ,

Therefore, ( A U , U ) E k is a nonnegative real number.

In order to further determine the eigenvalue of the matrix operator A , the following characteristic equation can be considered,

A U = λ U , U = ( u , z , v , q ) E k , (8)

That is

{ z = λ u , M ( s ) ( Δ ) 2 m u + β ( Δ ) 2 m z = λ z , q = λ v , M ( s ) ( Δ ) 2 m v + β ( Δ ) 2 m q = λ q .

Thus u , v meet the eigenvalue problem

{ λ 2 u λ β ( Δ ) 2 m u + M ( s ) ( Δ ) 2 m u = 0 , λ 2 v λ β ( Δ ) 2 m v + M ( s ) ( Δ ) 2 m v = 0 , i u n i | Ω = i v n i | Ω = 0 , i = 0 , 1 , 2 , , 2 m 1 ,

Take the inner product of ( Δ ) k u , ( Δ ) k v and Equations (1) and (2) above respectively, with

{ λ 2 D k u 2 λ β D 2 m + k u 2 + M ( s ) D 2 m + k u 2 = 0 , λ 2 D k v 2 λ β D 2 m + k v 2 + M ( s ) D 2 m + k v 2 = 0 ,

That is

λ 2 ( D k u 2 + D k v 2 ) λ β ( D 2 m + k u 2 + D 2 m + k v 2 ) + M ( s ) ( D 2 m + k u 2 + D 2 m + k v 2 ) = 0. (9)

Equation (9) is a univariate quadratic equation about λ . Replace u , v with u j , v j . For each positive integer j, Equation (8) has paired eigenvalues

λ j ± = β μ j ± ( β μ j ) 2 4 M ( s ) μ j 2 ,

where μ j is the characteristic root of ( Δ ) 2 m in V 2 m × V 2 m , then μ j = λ 1 j 2 m n .

If ( β μ j ) 2 4 M ( s ) μ j , then μ j 4 M ( s ) β 2 , the eigenvalues of operator A are all real numbers, and the corresponding eigenfunction form is

U j ± = ( u j , λ j ± u j , v j , λ j ± v j )

For convenience, mark for any j 1 , there are

D 2 m + k u j 2 + D 2 m + k v j 2 = μ j , D k u j 2 + D k v j 2 = 1 , D 2 m k u j 2 + D 2 m k v j 2 = 1 μ j .

Theorem 1: Assumes that l is the Lipschitz constant of g i ( u , v ) ( i = 1 , 2 ) . When N 0 N + is sufficiently large, for N N 0 , the following inequality holds

( μ N + 1 μ N ) ( β β 2 μ 1 4 M ( s ) ) 32 l β 2 μ 1 4 M ( s ) + 1. (10)

Then all operators A satisfy the spectral interval condition (6).

Proof. Because μ j 4 M ( s ) β 2 and the eigenvalues of A are positive real numbers, { λ j } j 1 and { λ j + } j 1 are single increment sequences.

The following four steps are taken to prove theorem 1:

Step 1: Because { λ j } j 1 and { λ j + } j 1 are non subtractive columns, according to lemma 2, there are N 0 N + , for N N 0 , λ N and λ N + 1 are continuous adjacent values.

Therefore, there is N, so that λ N and λ N + 1 are continuous adjacent values, and the eigenvalue of A can be decomposed into

σ 1 = { λ r | 1 r N } , σ 2 = { λ r + , λ j ± | 1 r N j }

Step 2: Consider the corresponding decomposition of E k into

E k 1 = s p a n { U r | λ r σ 1 } , E k 2 = s p a n { U r + , U j ± | λ r + , λ j ± σ 2 }

The equivalent inner product ( ( U , V ) ) E k given below makes E k 1 , E k 2 orthogonal.

Further decompose E k 2 = E H E R , of which

E H = s p a n { U r + | 1 r N } , E R = s p a n { U j ± | j N }

Because E k 1 and E H are finite dimensional subspaces, U N E k 1 , U N + 1 E R , and E k 1 and E R are orthogonal, while E k 1 and E H are not orthogonal, E k 1 and E k 2 are not orthogonal. So we need to redefine the equivalent norm on E k , so that E k 1 and E H are orthogonal. Order E N = E k 1 E H .

Construct two functions Φ : E N R , Ψ : E R R of which,

Φ ( U , V ) = 2 β ( β 1 ) ( D 2 m + k u , D ( 2 m + k ) u ¯ ) + 2 β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + 2 β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + 4 ( D ( 2 m + k ) z ¯ , D ( 2 m + k ) z ) 4 M ( s ) ( D k u , D k u ¯ ) + 2 β ( D 2 m + k u ¯ , D 2 m + k u ) + 2 β ( β 1 ) ( D 2 m + k v , D 2 m + k v ¯ ) + 2 β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + 2 β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + 4 ( D ( 2 m + k ) q ¯ , D ( 2 m + k ) q ) 4 M ( s ) ( D k v , D k v ¯ ) + 2 β ( D 2 m + k v ¯ , D 2 m + k v ) ,

Ψ ( U , V ) = 2 β ( D 2 m + k u ¯ , D 2 m + k u ) + β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + 4 ( D ( 2 m + k ) z ¯ , D ( 2 m + k ) z ) 2 M ( s ) ( D k u , D k u ¯ ) + 2 β ( β 1 ) ( D 2 m + k u , D ( 2 m + k ) u ¯ ) + 2 β ( D 2 m + k v ¯ , D 2 m + k v ) + β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + 4 ( D ( 2 m + k ) q ¯ , D ( 2 m + k ) q ) 2 M ( s ) ( D k v , D k v ¯ ) + 2 β ( β 1 ) ( D 2 m + k v , D 2 m + k v ¯ ) .

Among them U = ( u , z , v , q ) , V = ( u , z , v , q ) E N or E R .

For U = ( u , z , v , q ) E N , then

Φ ( U , U ) = 2 β ( β 1 ) ( D 2 m + k u , D ( 2 m + k ) u ¯ ) + 2 β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + 2 β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + 4 ( D ( 2 m + k ) z ¯ , D ( 2 m + k ) z ) 4 M ( s ) ( D k u , D k u ¯ ) + 2 β ( D 2 m + k u ¯ , D 2 m + k u ) + 2 β ( β 1 ) ( D 2 m + k v , D 2 m + k v ¯ ) + 2 β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + 2 β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + 4 ( D ( 2 m + k ) q ¯ , D ( 2 m + k ) q )

4 M ( s ) ( D k v , D k v ¯ ) + 2 β ( D 2 m + k v ¯ , D 2 m + k v ) 2 β ( β 1 ) ( D 2 m + k u 2 + D 2 m + k v 2 ) 4 β ( D ( 2 m + k ) z D 2 m + k u + D ( 2 m + k ) q D ( 2 m + k ) v ) + 4 ( D ( 2 m + k ) z 2 + D ( 2 m + k ) q 2 ) 4 M ( s ) ( D k u 2 + D k v 2 ) + 2 β ( D 2 m + k u 2 + D 2 m + k v 2 )

2 β ( β 1 ) ( D 2 m + k u 2 + D 2 m + k v 2 ) 4 ( D ( 2 m + k ) z 2 + D ( 2 m + k ) q 2 ) β 2 ( D 2 m + k u 2 + D 2 m + k v 2 ) + 4 ( D ( 2 m + k ) z 2 + D ( 2 m + k ) q 2 ) 4 M ( s ) ( D k u 2 + D k v 2 ) + 2 β ( D 2 m + k u 2 + D 2 m + k v 2 ) = β 2 ( D 2 m + k u 2 + D 2 m + k v 2 ) 4 M ( s ) ( D k u 2 + D k v 2 ) ( β 2 μ 1 4 M ( s ) ) ( D k u 2 + D k v 2 ) .

For any k, there is β 2 μ k 4 M ( μ k ) . According to hypothesis 1 μ 0 M ( s ) μ 1 β 2 μ k 4 , then Φ ( U , U ) 0 , that is, Φ is positive definite.

Similarly, for any U = ( u , z , v , q ) E R , there is

Ψ ( U , U ) = 2 β ( D 2 m + k u ¯ , D 2 m + k u ) + β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + β ( D ( 2 m + k ) z ¯ , D 2 m + k u ) + 4 ( D ( 2 m + k ) z ¯ , D ( 2 m + k ) z ) 2 M ( s ) ( D k u , D k u ¯ ) + 2 β ( β 1 ) ( D 2 m + k u , D ( 2 m + k ) u ¯ ) + 2 β ( D 2 m + k v ¯ , D 2 m + k v ) + β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + β ( D ( 2 m + k ) q ¯ , D 2 m + k v ) + 4 ( D ( 2 m + k ) q ¯ , D ( 2 m + k ) q )

2 M ( s ) ( D k v , D k v ¯ ) + 2 β ( β 1 ) ( D 2 m + k v , D 2 m + k v ¯ ) . 2 β ( D 2 m + k u 2 + D 2 m + k v 2 ) 2 β ( D ( 2 m + k ) z D 2 m + k u + D ( 2 m + k ) q D 2 m + k v ) + 4 ( D ( 2 m + k ) z 2 + D ( 2 m + k ) q 2 ) 2 M ( s ) ( D k u 2 + D k v 2 ) + 2 β ( β 1 ) ( D 2 m + k u 2 + D 2 m + k v 2 )

2 β ( D 2 m + k u 2 + D 2 m + k v 2 ) 4 ( D ( 2 m + k ) z 2 + D ( 2 m + k ) q 2 ) β 2 ( D 2 m + k u 2 + D 2 m + k v 2 ) + 4 ( D ( 2 m + k ) z 2 + D ( 2 m + k ) q 2 ) 2 M ( s ) ( D k u 2 + D k v 2 ) + 2 β ( β 1 ) ( D 2 m + k u 2 + D 2 m + k v 2 ) = β 2 ( D 2 m + k u 2 + D 2 m + k v 2 ) 2 M ( s ) ( D k u 2 + D k v 2 ) ( β 2 μ 1 2 M ( s ) ) ( D k u 2 + D k v 2 ) .

So there are U = ( u , z , v , q ) E R , Ψ ( U , U ) 0 , then Ψ is also positive definite.

Redefine the inner product of E k :

( ( U , V ) ) E k = Φ ( P N U , P N V ) + Ψ ( P R U , P R V ) (11)

where P N and P R are projections of E k E N and E k E R , respectively Here, Equation (11) is written as

( ( U , V ) ) E k = Φ ( U , V ) + Ψ ( U , V )

Under the redefined inner product of E k , to prove that E k 1 and E k 2 are orthogonal, we only need to prove that E k 1 and E H are orthogonal, that is,

( ( U j , U j + ) ) E k = Φ ( U j , U j + ) = 0.

Because there are U j E k 1 , U j + E H , that is

Φ ( U j , U j + ) = 2 β ( β 1 ) ( D 2 m + k u j , D 2 m + k u ¯ j ) 2 β λ j + ( D ( 2 m + k ) u ¯ j , D 2 m + k u j ) 2 β λ j ( D ( 2 m + k ) u ¯ j , D 2 m + k u j ) + 4 λ j λ j + ( D ( 2 m + k ) u ¯ j , D ( 2 m + k ) u j ) 4 M ( s ) ( D k u j , D k u ¯ j ) + 2 β ( D 2 m + k u ¯ j , D 2 m + k u j ) + 2 β ( β 1 ) ( D 2 m + k v j , D 2 m + k v ¯ j ) 2 β λ j + ( D ( 2 m + k ) v ¯ j , D 2 m + k v j ) 2 β λ j ( D ( 2 m + k ) v ¯ j , D 2 m + k v j ) + 4 λ j λ j + ( D ( 2 m + k ) v ¯ j , D 2 m + k v j )

4 M ( s ) ( D k v j , D k v ¯ j ) + 2 β ( D 2 m + k v ¯ j , D 2 m + k v j ) = 2 β ( β 1 ) ( D 2 m + k u j 2 + D 2 m + k v j 2 ) 2 β ( λ j + λ j + ) ( u j 2 + v j 2 ) + 4 λ j λ j + ( D ( 2 m + k ) u j 2 + D ( 2 m + k ) v j 2 ) 4 M ( s ) ( D k u j 2 + D k v j 2 ) + 2 β ( D 2 m + k u j 2 + D 2 m + k v j 2 ) = 4 M ( μ j ) + 2 β 2 μ j 2 β ( λ j + λ j + ) + 4 λ j λ j + 1 μ j . (12)

Because of Equation (9), there are

λ j + + λ j = β μ j , λ j + λ j = M ( μ j ) μ j .

So Φ ( U j , U j + ) = 4 M ( μ j ) + 2 β 2 μ j 2 β ( λ j + + λ j ) + 4 λ j + λ j 1 μ j = 0 .

Step 3: according to the orthogonal decomposition established above, let’s prove that A satisfies the spectral interval condition. First estimate the Lipschitz constant l F of F, where

F ( U ) = ( 0 , f 1 ( x ) g 1 ( u , v ) , 0 , f 2 ( x ) g 2 ( u , v ) ) T

According to lemma 1, g i ( u , v ) : V 2 m + k × V 2 m + k V 2 m + k × V 2 m + k are uniformly bounded and Lipschitz continuous, if U = ( u , z , v , q ) E k , U i = ( u i , z i , v i , q i ) P i U ( i = 1 , 2 ) ,

Then

P 1 u = u 1 , P 1 v = v 1 , P 2 u = u 2 , P 2 v = v 2 .

U E k 2 = Φ ( P 1 U , P 2 U ) + Ψ ( P 1 U , P 2 U ) ( β 2 μ 1 4 M ( s ) ) ( D k P 1 u 2 + D k P 1 v 2 ) + ( β 2 μ 1 2 M ( s ) ) ( D k P 2 u 2 + D k P 2 v 2 ) ( β 2 μ 1 4 M ( s ) ) ( D k u 2 + D k v 2 ) .

Given U = ( u , z , v , q ) , V = ( u ˜ , z ˜ , v ˜ , q ˜ ) E k , we can get

F ( U ) F ( V ) E k = g 1 ( u , v ) g 1 ( u ˜ , v ˜ ) V 2 m + k × V 2 m + k + g 2 ( u , v ) g 2 ( u ˜ , v ˜ ) V 2 m + k × V 2 m + k 2 l ( u u ˜ V 2 m + k + v v ˜ V 2 m + k ) 4 l β 2 μ 1 4 M ( s ) U V E k .

So

l F 4 l β 2 μ 1 4 M ( s ) (13)

From (13), if

Λ 2 Λ 1 = λ N + 1 λ N > 16 l β 2 μ 1 4 M ( s ) (14)

Then the spectral interval condition (6) holds.

Step 4: according to the above paired eigenvalues, there are

Λ 2 Λ 1 = λ N + 1 λ N = β 2 ( μ N + 1 μ N ) + R ( N ) R ( N + 1 ) 2 . (15)

Of which, R ( N ) = β 2 μ N 2 4 M ( s ) μ N .

There are N 0 N + , for N N 0 , let R 0 ( N ) = R ( N ) β 2 μ 1 4 M ( s ) , there are

R ( N ) R ( N + 1 ) + β 2 μ 1 4 M ( s ) ( μ N + 1 μ N ) = β 2 μ 1 4 M ( s ) ( ( μ N + 1 R ( N + 1 ) β 2 μ 1 4 M ( s ) ) ( μ N R ( N ) β 2 μ 1 4 M ( s ) ) ) = β 2 μ 1 4 M ( s ) ( ( μ N + 1 R 0 ( N + 1 ) ) ( μ N R 0 ( N ) ) ) (16)

And because of lim N + ( μ N R 0 ( N ) ) = lim N + ( μ N R ( N ) β 2 μ 1 4 M ( s ) ) = 0 , there are

lim N + R ( N ) R ( N + 1 ) + β 2 μ 1 4 M ( s ) ( μ N + 1 μ N ) = 0. (17)

According to the hypothesis (10) of Theorem 1 and Equations (13)-(17), there are

Λ 2 Λ 1 1 2 ( ( μ N + 1 μ N ) ( β β 2 μ 1 4 M ( s ) ) 1 ) 16 l β 2 μ 1 4 M ( s ) 4 l F . (18)

Theorem 1 is proved.

Theorem 2 [12] Through theorem1, operator A satisfies the spectral interval condition, and problems (1)-(5) have a family of inertial manifolds μ k , and μ k E k . The form is as follows,

μ k = g r a p h ( Γ ) E k : = { ς + Γ ( ς ) : ς E k 1 }

where Γ : E k 1 E k 2 is Lipschitz continuous and has Lipschitz constant l F , and g r a p h ( Γ ) represents the graph of Γ .

Conflicts of Interest

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

References

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