In this paper, we study the inertial manifolds for a class of asymmetrically coupled generalized Higher-order Kirchhoff equations:

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)

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)

the boundary conditions:

∂ i u ∂ n i = 0 , i = 0 , 1 , 2 , ⋯ , m − 1 , x ∈ ∂ Ω , t > 0 , (3)

∂ j v ∂ v j = 0 , j = 0 , 1 , 2 , ⋯ , 2 m − 1 , x ∈ ∂ Ω , t > 0 , (4)

the initial conditions:

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)

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 > 1 , β is real number.

Recently, the existence of inertial manifolds for Kirchhoff-type equation has been favored by many scholars. Many scholars have done a lot of research on this kind of problems and obtained good results [1] [2] [3].

Lin Chen, Wei Wang and Guoguang Lin [1] studied higher-order Kirchhoff-type equation with nonlinear strong dissipation in n dimensional space:

u t t + ( − Δ ) m u t + ϕ ( ‖ ∇ u ‖ 2 ) ( − Δ ) m u + g ( u ) = f ( x ) , x ∈ Ω , t > 0 , m > 1 ,

u ( x , t ) = 0 , ∂ i u ∂ v i = 0 , i = 1 , 2 , ⋯ , m − 1 , x ∈ ∂ Ω , t > 0 ,

u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) ,

for the above equation, they made some suitable assumptions about ϕ ( s ) and g ( u ) to get the existence of exponential attractors and inertial manifolds.

Guoguang Lin, Ming Zhang [2] studied the initial boundary value problem for a class of Kirchhoff-type coupled equations:

u t t − M ( ‖ ∇ u ‖ 2 + ‖ ∇ v ‖ 2 ) Δ u − β Δ u t + g 1 ( u , v ) = f 1 ( x ) ,

v t t − M ( ‖ ∇ u ‖ 2 + ‖ ∇ v ‖ 2 ) Δ v − β Δ v t + g 2 ( u , v ) = f 2 ( x ) ,

they used the one order coupled evolution equation which is equivalent to Kirchhoff-type coupled Equations. Then by using the graph norm in X, they get the existence of the inertial manifolds.

Lin Guoguang, Yang Lujiao [3] studied the existence of exponential attractors and a family of inertial manifolds for a class of generalized Kirchhoff-type equation with damping term:

u t t + M ( ‖ ∇ m u ‖ p p ) ( − Δ ) 2 m u + β ( − Δ ) 2 m u t + g ( u ) = f ( x ) ,

by using Hadamard’s graph transformation method, they proved the spectral interval condition is true; then they obtained the existence of a family of the inertial manifolds for the equation.

For more significant research results about the existence of inertial manifolds for Kirchhoff-type equations, please refer to the literature [4] - [17].

This paper is organized as follows. In Section 2, we present the preliminaries and some lemmas. In Section 3, the inertial manifold is obtained.

The following symbols and assumptions are introduced for the convenience of the statement:

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

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:

(H1) g ( u t , v ) , g ( u , v t ) ∈ C 1 ( Ω ) ,

(H2) ε ≤ m 0 ≤ M ( s ) ≤ m 1 .

Definition 1 [6] Assuming S = ( S ( t ) ) t ≥ 0 is a solution semigroup on Banach space E k , subset μ k ⊂ E k is said to be a family of inertial manifolds, if they satisfy the following three properties:

1) μ k is a finite-dimensional Lipschitz manifold;

2) μ k is positively invariant, i.e., S ( t ) μ k ⊆ μ k , t > 0 ;

3) μ k attracts exponentially all orbits of solution , that is, for any x ∈ E k , there are constants η > 0 , C > 0 such that

d i s t ( S ( t ) x , μ k ) ≤ C e − η t , t ≥ 0 , (6)

Definition 2 [6] Let A : X → X be an operator and assume that F ∈ C b ( X , X ) satisfies the Lipschitz condition:

‖ F ( U ) − F ( V ) ‖ X ≤ l F ‖ U − V ‖ X , (7)

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

Λ 1 = sup { Re λ | λ ∈ σ 1 } , Λ 2 = inf { Re λ | λ ∈ σ 2 } , (8)

X i = s p a n { ω j | λ j ∈ σ i } , i = 1 , 2. (9)

Then

Λ 2 − Λ 1 > 4 l F , (10)

and the orthogonal decomposition

X = X 1 ⊕ X 2 , (11)

holds with continuous orthogonal projections P 1 : X → X 1 and P 2 : X → X 2 .

Lemma 1 [6] Let the eigenvalues μ j ± , j ≥ 1 be arranged in nondecreasing order, for all m ∈ N , there is N ≥ m such that μ N − and μ N + 1 − are consecutive.

Equations (1)-(5) are equivalent to the following one-order evolution equation:

U t + A U = F ( U ) , U ∈ E k (12)

where U = ( u , p , v , q ) , p = u t , q = v t , and

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

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

We consider the usual graph norm in E k , as follows

( U , U ) E k = ( M ( s ) ⋅ ∇ m + k u , ∇ m + k a ¯ ) + ( ∇ k p , ∇ k b ¯ )     + ( M ( s ) ⋅ ∇ 2 m + 2 k v , ∇ 2 m + 2 k c ¯ ) + ( ∇ 2 k q , ∇ 2 k d ¯ ) , (13)

where U = ( u , p , v , q ) T , V = ( a , b , c , d ) T , s = ‖ ∇ m u ‖ 2 + ‖ ∇ m v ‖ 2 , a ¯ , b ¯ , c ¯ , d ¯ denote the conjugation of a , b , c , d respectively. Evidently, the operator A is monotone, and we obtain

( A U , U ) E k = − ( M ( s ) ∇ m + k p , ∇ m + k u ¯ ) + ( M ( s ) ∇ m + k u , ∇ m + k p ¯ )     + β ( ∇ m + k p , ∇ m + k p ¯ ) − ( M ( s ) ∇ 2 m + 2 k q , ∇ 2 m + 2 k v ¯ )     + ( M ( s ) ∇ 2 m + 2 k v , ∇ 2 m + 2 k q ¯ ) + β ( ∇ 2 m + 2 k q , ∇ 2 m + 2 k q ¯ ) = β ( ‖ ∇ m + k p ‖ 2 + ‖ ∇ 2 m + 2 k q ‖ 2 ) ≥ 0 , (14)

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

In order to determine the eigenvalues of A, we consider the eigenvalues equation:

A U = λ U , U = ( u , p , v , q ) T ∈ E k , (15)

that is

{ − p = λ u , M ( s ) ( − Δ ) m u + β ( − Δ ) m p = λ p , − q = λ v , M ( s ) ( − Δ ) 2 m v + β ( − Δ ) 2 m q = λ q , (16)

combined with (16), we obtain

{ λ 2 u + M ( s ) ( − Δ ) m u − β λ ( − Δ ) m u = 0 , λ 2 v + M ( s ) ( − Δ ) 2 m v − β λ ( − Δ ) 2 m v = 0. (17)

where u | ∂ Ω = ( − Δ ) m u | ∂ Ω = v | ∂ Ω = ( − Δ ) 2 m v | ∂ Ω = 0 .

Taking ( − Δ ) k u and ( − Δ ) 2 k v inner product with the Equations (17), we have

{ λ 2 ‖ ∇ k u ‖ 2 + M ( s ) ‖ ∇ m + k u ‖ 2 − β λ ‖ ∇ m + k u ‖ 2 = 0 , λ 2 ‖ ∇ 2 k v ‖ 2 + M ( s ) ‖ ∇ 2 m + 2 k v ‖ 2 − β λ ‖ ∇ 2 m + 2 k v ‖ 2 = 0. (18)

adding them together,

λ 2 ( ‖ ∇ k u ‖ 2 + ‖ ∇ 2 k v ‖ 2 ) + M ( s ) ( ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 )   − β λ ( ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 ) = 0 , (19)

(19) is regard as a quadratic equation with one unknown about λ , so we get

λ j ± = β μ j ± β 2 μ j 2 − 4 M ( s ) ⋅ μ j 2 , (20)

where μ j is the eigenvalue of ( Δ m 0 0 Δ 2 m ) , and μ j is non-derogatory, for ∀ j ≥ 1 , we have

‖ ∇ k u j ‖ 2 + ‖ ∇ 2 k v j ‖ 2 = 1 , ‖ ∇ m + k u j ‖ 2 + ‖ ∇ 2 m + 2 k v j ‖ 2 = μ j ,

‖ ∇ − m − k u j ‖ 2 + ‖ ∇ − 2 m − 2 k v j ‖ 2 = 1 μ j .

If μ j ≥ 4 β 2 M ( s ) , we can get the eigenvalues of A are all positive and real numbers. The corresponding eigenfunction is as follows

U j ± = ( u j , − λ j ± u j , v j , − λ j ± v j ) . (21)

Lemma 2 g ( u t , v ) : V k × V 2 m + 2 k → V k × V 2 m + 2 k , g ( u , v t ) : V m + k × V 2 k → V m + k × V 2 k is uniformly bounded and globally Lipschitz continuous.

Proof. ∀ ( u t , v ) , ( u ¯ t , v ¯ ) ∈ V k × V 2 m + 2 k → V k × V 2 m + 2 k , by (H1), we have

‖ g ( u ¯ t , v ¯ ) − g ( u t , v ) ‖ V k × V 2 m + 2 k = ‖ g u t ( u ¯ t + θ ( u ¯ t − u t ) , v ¯ + θ ( v ¯ − v ) ) ( u ¯ t − u )         + g v ( u ¯ t + θ ( u ¯ t − u t ) , v ¯ + θ ( v ¯ − v ) ) ( v ¯ − v ) ‖ V k × V 2 m + 2 k ≤ l ‖ u ¯ t − u t ‖ V k + l ‖ v ¯ − v ‖ V 2 m + 2 k ≤ l ( ‖ p ¯ − p ‖ V k + ‖ v ¯ − v ‖ V 2 m + 2 k ) ; (22)

Similarly, we have ‖ g ( u ¯ , v ¯ t ) − g ( u , v t ) ‖ V m + k × V 2 k ≤ l ( ‖ u ¯ − u ‖ V m + k + ‖ q ¯ − q ‖ V 2 k ) , where θ ∈ ( 0 , 1 ) , l is Lipschitz coefficient of g.

Theorem 1 When μ j ≥ 4 β 2 m 1 , l is Lipschitz constant of g, there is a large enough N 1 ∈ N so that N ≥ N 1 has

β 2 ( μ N + 1 − μ N ) − 2 β 3 μ 1 − 3 β m 1 μ N + 1 − μ N 2 ≥ 4 l 2 β 3 μ 1 − 3 β m 1 + 1 (23)

then operator A satisfies the spectral interval condition of Definition 2.

Proof. when μ k ≥ 4 β 2 m 1 , the eigenvalues of A are all positive and real numbers, meanwhile { λ k − } k ≥ 1 and { λ k + } k ≥ 1 are increasing order.

Next, we divided the whole process of proof into four steps.

Step 1 By Lemma 1, since λ k ± is nondecreasing order, so there exists N, such that λ N − and λ N + 1 − are continuous adjacent values, Then the eigenvalues of A are separate as

σ 1 = { λ i − , λ j + | max ( λ i − , λ j + ) ≤ λ N − } , σ 2 = { λ i − , λ j ± | λ i − ≤ λ N − ≤ min { λ i − , λ j ± } } . (24)

Step 2 The corresponding E k is decomposed into

E k 1 = S p a n { U i − , U j ± | λ i − , λ j ± ∈ σ 1 } , E k 2 = S p a n { U i − , U j ± | λ i − , λ j ± ∈ σ 2 } , (25)

We aim at madding two orthogonal subspaces of E k and verifying the spectral gap condition (11) is true when Λ 1 = λ N − , Λ 2 = λ N + 1 − , Therefore, we further decompose E k 2 = E C ⊕ E R , where

E C = S p a n { U i − | λ i − ≤ λ N − < λ i + } , E R = S p a n { U j ± | λ N − < λ j ± } , (26)

Set E N = E k 1 ⊕ E C , in order to verify the E k 1 and E k 2 are orthogonal, we need to introduce two functions Φ : E N → R , Ψ : E R → R .

Φ ( U , V ) = β ( ∇ m + k u , ∇ m + k a ¯ ) − 3 β M ( s ) ( ∇ k u , ∇ k a ¯ ) + ( ∇ − m − k b ¯ , ∇ m + k u )     + ( ∇ − m − k p ¯ , ∇ m + k a ) + 3 β ( ∇ − m − k p , ∇ − m − k b ¯ ) + β ( ∇ 2 m + 2 k v , ∇ 2 m + 2 k c ¯ )     − 3 β M ( s ) ( ∇ 2 k v , ∇ 2 k c ¯ ) + ( ∇ − 2 m − 2 k d ¯ , ∇ 2 m + 2 k v )     + ( ∇ − 2 m − 2 k q ¯ , ∇ 2 m + 2 k c ) + 3 β ( ∇ − 2 m − 2 k q , ∇ − 2 m − 2 k d ¯ ) , (27)

Ψ ( U , V ) = β ( ∇ m + k u , ∇ m + k a ¯ ) − ( ∇ k c ¯ , ∇ m + k u ) + ( ∇ k p ¯ , ∇ m + k a )     + β μ 1 ( ∇ k p , ∇ k c ¯ ) + β ( ∇ 2 m + 2 k v , ∇ 2 m + 2 k b ¯ ) − ( ∇ − 2 k d ¯ , ∇ 2 m + 2 k v )     + ( ∇ 2 k q ¯ , ∇ 2 m + 2 k b ) + β μ 1 ( ∇ 2 k q , ∇ 2 k d ¯ ) , (28)

where U = ( u , p , v , q ) T , V = ( a , b , c , d ) T ∈ E k are defined before.

Let U = ( u , p , v , q ) T ∈ E N , by (H2), then

Φ ( U , U ) = β ( ∇ m + k u , ∇ m + k u ¯ ) − 3 β M ( s ) ( ∇ k u , ∇ k u ¯ ) + ( ∇ − m − k p ¯ , ∇ m + k u )       + ( ∇ − m − k p ¯ , ∇ m + k u ) + 3 β ( ∇ − m − k p , ∇ − m − k p ¯ ) + β ( ∇ 2 m + 2 k v , ∇ 2 m + 2 k v ¯ )       − 3 β M ( s ) ( ∇ 2 k v , ∇ 2 k v ¯ ) + ( ∇ − 2 m − 2 k q ¯ , ∇ 2 m + 2 k v )       + ( ∇ − 2 m − 2 k q ¯ , ∇ 2 m + 2 k v ) + 3 β ( ∇ − 2 m − 2 k q , ∇ − 2 m − 2 k q ¯ ) ,

≥ β ( ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 ) − 3 β M ( s ) ( ‖ ∇ k u ‖ 2 + ‖ ∇ 2 k v ‖ 2 )       − 3 β ( ‖ ∇ − m − k p ‖ 2 + ‖ ∇ − 2 m − 2 k q ‖ 2 ) − β 3 ( ‖ ∇ m + k u ‖ 2 + ‖ ∇ 2 m + 2 k v ‖ 2 )       + 3 β ( ‖ ∇ − m − k p ‖ 2 + ‖ ∇ − 2 m − 2 k q ‖ 2 ) ≥ ( 2 β 3 μ 1 − 3 β m 1 ) ( ‖ ∇ k u ‖ 2 + ‖ ∇ 2 k v ‖ 2 ) , (29)

since for ∀ j , m 1 ≤ β 2 μ j , we have Φ ( U , U ) ≥ 0 , for ∀ U ∈ E N , then Φ is positive definite.

Similarly, for U ∈ E R , we have

Ψ ( U , U ) = β ( ∇ m + k u , ∇ m + k u ¯ ) − ( ∇ k p ¯ , ∇ m + k u ) + ( ∇ k p ¯ , ∇ m + k u )       + β μ 1 ( ∇ k p , ∇ k p ¯ ) + β ( ∇ 2 m + 2 k v , ∇ 2 m + 2 k v ¯ ) − ( ∇ 2 k q ¯ , ∇ 2 m + 2 k v )       + ( ∇ 2 k q ¯ , ∇ 2 m + 2 k v ) + β μ 1 ( ∇ 2 k q , ∇ 2 k q ¯ ) ≥ β μ 1 ( ‖ ∇ k u ‖ 2 + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k v ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) , (30)

so, for ∀ U ∈ E R , Ψ ( U , U ) ≥ 0 , the Ψ is also positive definite.

Next, we need to define a scale product in E k

〈 〈 U , V 〉 〉 E k = Φ ( P N U , P N V ) + Ψ ( P R U , P R V ) . (31)

where P N and P R are projection E k → E N , E k → E R respectively, for convenience, we rewrite (31) as follows

〈 〈 U , V 〉 〉 E k = Φ ( U , V ) + Ψ ( U , V ) . (32)

We will proof that two subspaces E k 1 and E k 2 in (25) are orthogonal; in fact, we only need to show E N and E C are orthogonal, that is

〈 〈 U j − , U j + 〉 〉 E k = 0 ,   ( U j − ∈ E N , U j + ∈ E C ) . (33)

by (27), (32), we have

〈 〈 U j − , U j + 〉 〉 E k = Φ ( U j − , U j + ) = β ( ∇ m + k u j , ∇ m + k u ¯ j ) − 3 β M ( s ) ( ∇ k u j , ∇ k u ¯ j )       − λ j + ( ∇ − m − k u ¯ j , ∇ m + k u j ) − λ j − ( ∇ − m − k u ¯ j , ∇ m + k u j )       + 3 β λ j − λ j + ( ∇ − m − k u j , ∇ − m − k u ¯ j ) + β ( ∇ 2 m + 2 k v j , ∇ 2 m + 2 k v ¯ j )       − 3 β M ( s ) ( ∇ 2 k v j , ∇ 2 k v ¯ j ) − λ j + ( ∇ − 2 m − 2 k v ¯ j , ∇ 2 m + 2 k v j )

      − λ j − ( ∇ − 2 m − 2 k v ¯ j , ∇ 2 m + 2 k v j ) + 3 β λ j − λ j + ( ∇ − 2 m − 2 k v j , ∇ − 2 m − 2 k v ¯ j ) , = β ( ‖ ∇ m + k u j ‖ 2 + ‖ ∇ 2 m + 2 k v j ‖ 2 ) − 3 β M ( s ) ( ‖ ∇ k u j ‖ 2 + ‖ ∇ 2 k v j ‖ 2 )       − ( λ j − + λ j + ) ( ‖ u j ‖ 2 + ‖ v j ‖ 2 ) + 3 β λ j − λ j + ( ‖ ∇ − m − k u j ‖ 2 + ‖ ∇ − 2 m − 2 k v j ‖ 2 ) = β μ j − 3 β M ( s ) − ( λ j − + λ j + ) + 3 β λ j − λ j + ⋅ 1 μ j . (34)

Through Equation (19), we can get λ j + + λ j − = β μ j , λ j + λ j − = M μ j , therefore

〈 〈 U j − , U j + 〉 〉 E k = 0. (35)

Step 3 Further, we estimate the Lipschitz constant l F of F

F ( U ) = ( 0 f 1 ( x ) − g ( u t , v ) 0 f 2 ( x ) − g ( u , v t ) ) , (36)

from (27), (28), for ∀ U = ( u , p , v , q ) T ∈ E k , we have

‖ U ‖ E k 2 = Φ ( P 1 U , P 1 U ) + Ψ ( P 2 U , P 2 U ) ≥ ( 2 β 3 μ 1 − 3 β m 1 ) ( ‖ ∇ k P 1 u ‖ 2 + ‖ ∇ 2 k P 1 v ‖ 2 )       + β μ 1 ( ‖ ∇ k P 2 u ‖ 2 + ‖ ∇ k P 2 p ‖ 2 + ‖ ∇ 2 k P 2 v ‖ 2 + ‖ ∇ 2 k P 2 q ‖ 2 ) ≥ ( 2 β 3 μ 1 − 3 β m 1 ) ( ‖ ∇ k u ‖ 2 + ‖ ∇ k p ‖ 2 + ‖ ∇ 2 k v ‖ 2 + ‖ ∇ 2 k q ‖ 2 ) . (37)

By lemma 1, where U = ( u , p , v , q ) T , V = ( u ¯ , p ¯ , v ¯ , q ¯ ) T ∈ E k , we can get

‖ F ( U ) − F ( V ) ‖ E k = ‖ g ( u ¯ t , v ¯ ) − g ( u t , v ) ‖ V k × V 2 m + 2 k + ‖ g ( u ¯ , v ¯ t ) − g ( u , v t ) ‖ V m + k × V 2 k ≤ l ( ‖ p ¯ − p ‖ V k + ‖ v ¯ − v ‖ V 2 m + 2 k ) + l ( ‖ u ¯ − u ‖ V m + k + ‖ q ¯ − q ‖ V 2 k ) ≤ l 2 β 3 μ 1 − 3 β m 1 ‖ U − V ‖ E k , (38)

so, we obtain

l F ≤ l 2 β 3 μ 1 − 3 β m 1 . (39)

Step 4 Now, we will show the spectral gap condition (10) holds.

Λ 2 − Λ 1 = λ N + 1 − − λ N − = β 2 ( μ N + 1 − μ N ) + 1 2 ( R ( N ) − R ( N + 1 ) ) , (40)

where R ( N ) = β 2 μ N 2 − 4 M ( s ) μ N .

Let

lim N → ∞ R ( N ) − R ( N + 1 ) + 2 β 3 μ 1 − 3 β m 1 ( μ N + 1 − μ N ) = 0 . (41)

letting

R 1 ( N ) = β 2 μ N − 4 M ( s ) ( 2 β 3 μ 1 − 3 β m 1 ) 2 μ N , (42)

we can compute

R ( N ) − R ( N + 1 ) + 2 β 3 μ 1 − 3 β m 1 ( μ N + 1 − μ N ) = 2 β 3 μ 1 − 3 β m 1 ( μ N + 1 ( 1 − R 1 ( N + 1 ) ) − μ N ( 1 − R 1 ( N ) ) ) , (43)

lim N → ∞ 2 β 3 μ 1 − 3 β m 1 ( μ N + 1 ( 1 − R 1 ( N + 1 ) ) − μ N ( 1 − R 1 ( N ) ) ) = 0 . (44)

then, we can get

Λ 2 − Λ 1 ≥ β 2 ( μ N + 1 − μ N ) − 2 β 3 μ 1 − 3 β m 1 μ N + 1 − μ N 2 − 1 ≥ 4 l 2 β 3 μ 1 − 3 β m 1 ≥ 4 l F . (45)

Theorem 1 is proved.

Theorem 2 Under the condition of Theorem 1, the problem (1)-(5) exist an inertial manifold μ k in E k ,

μ k = g r a p h ( Φ ) = { ξ k + Φ ( ξ k ) | ξ k ∈ E k 1 } , (46)

where Φ : E k 1 → E k 2 is a Lipschitz continuous function.

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

Lin, G.G. and Shao, M. (2022) A Family of Inertial Manifolds for a Class of Asymmetrically Coupled Generalized Higher-Order Kirchhoff Equations. Open Journal of Applied Sciences, 12, 1174-1183. https://doi.org/10.4236/ojapps.2022.127080