Jun 29, 2018 - Inertial Manifold, Hadamard's Graph Transformation Method, Lipschitz ... Nowadays, the study on the complexity of the space-time of high ...
Applied Mathematics, 2018, 9, 730-737 http://www.scirp.org/journal/am ISSN Online: 2152-7393 ISSN Print: 2152-7385
The Inertial Manifold for Class Kirchhoff-Type Equations with Strongly Damped Terms and Source Terms Guoguang Lin, Xiangshuang Xia Department of Mathematics, Yunnan University, Kunming, China
How to cite this paper: Lin, G.G. and Xia, X.S. (2018) The Inertial Manifold for Class Kirchhoff-Type Equations with Strongly Damped Terms and Source Terms. Applied Mathematics, 9, 730-737. https://doi.org/10.4236/am.2018.96050 Received: April 11, 2018 Accepted: June 26, 2018 Published: June 29, 2018 Copyright © 2018 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0). http://creativecommons.org/licenses/by/4.0/
Abstract In this paper, we study the inertial manifolds for a class of the Kirchhoff-type equations with strongly damped terms and source terms. The inertial manifold is a finite dimensional invariant smooth manifold that contains the global attractor, attracting the solution orbits by the exponential rate. Under appropriate assumptions, we firstly exert the Hadamard’s graph transformation method to structure a graph norm of a Lipschitz continuous function, and then we prove the existence of the inertial manifold by showing that the spectral gap condition is true.
Keywords Inertial Manifold, Hadamard’s Graph Transformation Method, Lipschitz Continuous, Spectral Gap Condition
Open Access
1. Introduction In this paper, we concerned the equation:
( (
u − M ∇u 2 + tt vtt + M ∇u 2 + = u ( x, 0 ) u0 ( x ) , = v ( x, 0 ) v0 ( x ) , u = 0, v = ∂Ω ∂Ω
∇mv
2
∇mv
2
) ∆u − β∆u + g (u, v ) =f ( x ) , ) ( −∆ ) v + β ( −∆ ) v + g (u, v ) = f ( x ) , t
m
1
m
t
ut ( x, 0 ) u1 ( x ) , =
vt ( x, 0 ) v1 ( x ) , = 0,
1
2
x ∈ Ω,
2
(1.1)
x ∈ Ω,
∂iv = 0= ( i 1, 2, , m − 1) , ∂µ i ∂Ω
where Ω is a bounded domain in Rn with a smooth boundary ∂Ω , β > 1 is a DOI: 10.4236/am.2018.96050
Jun. 29, 2018
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constant and fi ( x )( i = 1, 2 ) is a given source term. Moreover, 2 2 is a scalar function. Then the assumptions on M and M ∇u + ∇ m v gi ( u , v ) will be specified later. Nowadays, the study on the complexity of the space-time of high dimensional and infinite dimensional dynamical systems has gradually become the focus of nonlinear scientific research. In recent years, the inertial manifold has been found in the researches of the long time behavior of the solution and the attractor structure. The inertial manifold is a tool to describe the interaction between the low frequency components and the high frequency components [1]. When the flow has an inertial manifold, its high frequency description depends on the low frequency, and it contains attractors and exponentially attracts solution of the track, which realizes that the infinite dimensional dynamical system is reduced to a finite dimensional dynamical systems of the finite dimensional invariable Lipschitz manifold. Therefore, the inertial manifold is a powerful tool to study the long-time behavior of nonlinear dissipative systems and expose the real or seemingly chaotic structure of nonlinear dynamics. In addition, the study of inertial manifold is of great significance. The central idea of the methods that people use to solve practical problems such as Galerkin method, Cellular automaton and Coupled map, are to discuss the infinite dimensional problem into a finite dimensional problem. So, the inertial manifold is of great significance to the development of nonlinear science.
)
(
In 1988, the concept of inertial manifold was first proposed in the study of infinite dimensional dynamical system by R. Temam, C. Foias and Sell G.R. [2]. They considered the equation as following: ut + Au + B ( u , u ) + C ( u ) − f = 0.
(1.2)
where Au is a linear unbounded self-adjoint operator on H with domain D ( A ) dense in H. In 2010, Guoguang Lin and Jingzhu Wu [3] studied the existence of the inertial manifold of Boussinesq equation:
utt − α∆ut − ∆u + u 2 k +1 = F ( x, y ) , u ( x, y, 0 ) = u0 ( x, y ) , t ) u ( x + π, y,= t ) u ( x, y + π,= t ) 0, u ( x, y,=
(1.3)
( x, y ) ∈ Ω,
where= Ω ( 0, π ) ∈ R 2 , t > 0, α > 2 . In 2016, Ling Chen, Wei Wang and Guoguang Lin [4] established the expo2
nential attractors and inertial manifolds of the higher-order Kirchhoff-type equation:
(
utt + ( −∆ ) ut + φ ∇ mu m
2
) ( −∆ )
m
u + g (u ) = f ( x ) .
(1.4)
There are many researches on inertial manifolds for nonlinear wave equations (see [5] [6]). Concerning the inertial manifold, many difficulties are solved. So we take advantage of Hadamard’s graph transformation method in this paper. The paper is arranged as follows. In Section 2, some assumptions, notations DOI: 10.4236/am.2018.96050
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and lemmas are stated. In Section 3, the existence of the inertial manifold is established.
2. Preliminaries For convenience, we first introduce the following notations:
X = H 02 ( Ω ) × H 02 m ( Ω ) × H 01 ( Ω ) × H 0m ( Ω ) , X= H 01 ( Ω ) × H 0m ( Ω ) , 0 ci ( i = 1, 2,) denotes different positive constants,
( ⋅, ⋅) and ⋅ are the inner ⋅ is the norm of H − m ( Ω ) . product and norm of L ( Ω ) , −m Next, we give some assumptions and definition needed in the proof of our results. 2
1, 2 ) . ( A1 ) gi ( u, v ) ∈ C1 ( Ω ) , ( i =
( A2 ) ε ≤ m0 ≤ M ( s ) ≤ m1 ≤
( β − 1) µk 4
.
Definition 2.1. [7] Let A : X → X be an operator and assume that F ∈ Cb ( X , X ) satisfies the Lipschitz condition F (U ) − F (V )
X
≤ lF U − V
X
, U ,V ∈ X .
(2.1)
The operator A is called satisfy the spectral gap condition relative to F, if the point spectrum of the operator A can be divided into two parts σ 1 and σ 2 , of which σ 1 is finite, and such that, if = Λ1 sup {Re λ λ ∈= σ 1} , Λ 2 inf {Re λ λ ∈ σ 2 } ,
(2.2)
and
= X i span w j λ j ∈ σ i , = i 1, 2 .
{
}
(2.3)
Λ 2 − Λ1 > 4lF ,
(2.4)
Then and the orthogonal decomposition X = X 1 ⊕ X 2 holds with continuous orthogonal projections P1 : X → X 1 , P2 : X → X 2 . Lemma 2.1. [8] Let the eigenvalues µ ±j , j ≥ 1 be arranged in nondecreasing order. For all ∀m ∈ N , there exists N ≥ m such that µ N− and µ N− +1 are consecutive.
3. The Inertial Manifold Equation (1.1) is equivalent to the following one order evolution equation: U t + AU = F (U ) ,
(3.1)
where U = ( u , v, p, q ) ∈ X , p = ut , q = vt , T
0 0 −I 0 0 0 0 −I = A = , F (U ) −M ( s ) ∆ 0 − β∆ 0 m m β 0 M s −∆ 0 −∆ ( )( ) ( ) DOI: 10.4236/am.2018.96050
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0 0 (3.2) f1 ( x ) − g1 ( u , v ) f 2 ( x ) − g 2 ( u, v ) Applied Mathematics
G. G. Lin, X. S. Xia
{( u, v ) ∈ H ( Ω ) × H ( Ω )} × H ( Ω ) × H ( Ω ) .
D= ( A)
2m 0
2 0
m 0
1 0
We consider the usual graph norm in X, as follows
( M ( s ) ∇u, ∇u ) + ( M ( s ) ∇ v, ∇ v ) + ( p , p ) + ( q , q ) , (3.3)
(U ,V= )X
m
1
, v, p, q ) , V ( u1 , v1 , p1 , q1 ) ( u= T
where U =
T
m
1
1
1
∈ X , u1 , v1 , p1 , q1 respectively represent
the conjugation of u1 , v1 , p1 , q1 . Evidently, the operator A is monotone, for U ∈ D ( A ) , we obtain
( AU ,U ) X = ( − M ( s ) ∇p , ∇u ) + ( − M ( s ) ∇ m q , ∇ m v )
(
+ ( p, − M ( s ) ∆u − β∆p ) + q , M ( s )( −∆ ) v + β ( −∆ ) q m
= ( − M ( s ) ∇p , ∇u ) + ( ∇p , M ( s ) ∇u ) + β ∇p
(
) (
m
)
2
(3.4)
)
+ − M ( s ) ∇ m q, ∇ m v + ∇ m q , M ( s ) ∇ m v + β ∇ m q
2
2
2
= β ∇p + β ∇ m q ≥ 0. So,
( AU ,U ) X is a nonnegative and real number.
In order to determine the eigenvalues of A, we consider the eigenvalues equation:
( u , v, p , q )
= AU λ= U, U
T
∈X .
(3.5)
That is
λu, − p = −q = λ v, − M ( s ) ∆u − β∆p =λ p, M ( s )( −∆ )m v + β ( −∆ )m q = λ q.
(3.6) (3.7) (3.8) (3.9)
Substitute (3.6), (3.7) into (3.8), (3.9), we obtain
λ 2u − M ( s ) ∆u + βλ∆u = 0, 2 m m λ v + M ( s )( −∆ ) v − βλ ( −∆ ) v = 0.
(3.9) (3.10)
Replacing u, v with uk , vk , taking u, v inner product with the Equations (3.9), (3.10), and adding them together, we have
(
λkk uk
2
+ vk
2
) + M ( s ) ( ∇u
2 k
+ ∇ m vk
2
) − λ β ( ∇u k
2 k
+ ∇ m vk
2
) =0 . (3.11)
(3.11) is regard as a quadratic equation with one unknown about λk , so we get
λk± =
βµk ± β 2 µk2 − 4 M ( s ) µk 2
,
(3.12)
for ∀k ≥ 1 , we have uk
2
+ vk
2
= 1 , ∇u k
2
+ ∇ m vk
2
=µk , ∇ −1uk
and µk is non-derogatory. If β ≥
DOI: 10.4236/am.2018.96050
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4M ( s )
µk
2
+ ∇ − m vk
2
=
1
µk
.
(3.13)
+ 1 , because of β > 1 , then
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G. G. Lin, X. S. Xia
β2 ≥
4M ( s )
µk
, we can get the eigenvalues of A are all positive and real numbers.
The corresponding eigenfunction is as follows
( u , v , −λ u , −λ v ) .
U k± =
k
± k k
k
± k k
(3.14)
Lemma 3.1. gi : X 0 → X 0 , ( i = 1, 2 ) is uniformly bounded and globally Lipschitz continuous. Proof. ∀ ( u , v ) , ( u1 , v1 ) ∈ X 0 , by
( A1 ) , we have
gi ( u , v ) − gi ( u1 , v1 ) ≤ giu (ξ , v )( u − u1 ) ≤ ci1 u − u1
(
H 01 ( Ω )
≤ ci3 u − u1
X0 H 01 ( Ω )
+ giv ( u ,η )( v − v1 )
+ ci2 v − v1
H 01 ( Ω )
+ v − v1
H 0m ( Ω )
(3.15)
H 0m ( Ω )
H 0m ( Ω )
),
where
{
}
ci3 =max ci1 , ci2 , ξ =u + (1 − θ1 ) u1 , η =v + (1 − θ 2 ) v1 . Let li = ci3 , then li is Lipschitz coefficient of gi ( u , v ) . Theorem 3.1. li is Lipschitz constant of gi ( u , v ) , when β ≥
4M ( s )
µk
+ 1 , set
N1 ∈ N , such that N ≥ N1 , we obtain
1 1 − 2 2
( µ N +1 − µ N )
( β − 1) µ1 − 4m1 ≥
4l
( β − 1) µ1 − 4m1
+1 ,
(3.16)
where l = max {l1 , l2 } . By ( A2 ) and Lemma 3.1, the operator A satisfies the spectral gap condition of (2.4). 4M ( s ) Proof. when β ≥ + 1 , the eigenvalues of A are all positive and real µk
{λ }
− k k ≥1
numbers, meanwhile
{λ }
+ k k ≥1
and
are increasing order.
Next, we divided the whole process of proof into four steps. Step 1 By Lemma 2.1, since such that λ as
= σ1
and λ
− N
{λ , λ − j
+ k
− N +1
{λ } ± k
is nondecreasing order, so there exists N,
are continuous. Then the eigenvalues of A are separate
{
}
}
− max λ j− , λk+ ≤ λN= ,σ 2
{λ , λ − j
± k
}
λ j− ≤ λN− ≤ min {λ j+ , λk± } . (3.17)
Step 2 The corresponding X is decomposed into
{
}
{
}
= X 1 span U −j , U k+ λ j− , λk+= ∈ σ 1 , X 2 span U −j , U k± λ j− , λk± ∈ σ 2 .
(3.18)
We aim at madding two orthogonal subspaces of X and verifying the spectral gap condition (2.4) is true when Λ= λN− , Λ 2= λN− +1 . Therefore, we further de1 compose X= X c ⊕ X R , where 2
{
}
{
}
+ = X c span U −j λ −j ≤ λN− < λ= span U k± λN− < λk± . j , XR
(3.19)
Set X= X 1 ⊕ X c , in order to verify the X 1 and X 2 are orthogonal, we N DOI: 10.4236/am.2018.96050
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need to introduce two functions Φ : X N → R , Ψ : X R → R .
(
)
) ( ) ( ) ( +4 ( ∇ p, ∇ p ) + β ( ∇ v, ∇ v ) − 4 M ( s ) ( v, v ) + ( ∇ q , ∇ v ) + ( ∇ q , ∇ v ) + 4 ( ∇ q, ∇ q ) , Ψ (U , V ) = 2 β ( ∇u , ∇u ) − β ( ∇ p , ∇u ) − β ( ∇ p, ∇u ) + β ( ∇ p, ∇ p ) +2 β ( ∇ v, ∇ v ) − β ( ∇ q , ∇ v ) − β ( ∇ q , ∇ v ) + β ( ∇ q, ∇ q ) ,
Φ (U , V )= β ∇u , ∇u1 − 4 M ( s ) u , u1 + ∇ −1 p1 , ∇u + ∇ −1 p, ∇u1 −1
−1
−m
m
1
m
1
m
−m
m
−1
1
−1
−1
−m
m
1
1
−m
1
−1
1
m
1
m
−m
1
−m
where U , V ∈ X are defined before. Let U ( u , v, p, q ) ∈ X N , by ( A2 ) , then =
1
1
−m
1
−m
m
1
)
(
(
(3.21)
1
) (
Φ (U , U )= β ( ∇u , ∇u ) − 4 M ( s )( u , u ) + ∇ −1 p, ∇u + ∇ −1 p, ∇u
(
(3.20)
)
)
+4 ∇ −1 p, ∇ −1 p + β ∇ m v, ∇ m v − 4 M ( s )( v, v )
(
) (
) (
+ ∇ − m q , ∇ m v + ∇ − m q , ∇ m v + 4 ∇ − m q, ∇ − m q
(
≥ ( β − 1) ∇u + ∇ m v 2
(
2
) − 4M ( s ) ( u
≥ ( β − 1) µ1 − 4m1 u + v Since for ∀k , m1 ≤
( β − 1) µk
2
2
2
+ v
2
)
(3.22)
)
).
, therefore Φ (U , U ) ≥ 0 , for ∀U ∈ X N , then
4
Φ is positive definite. Similarly, for U ∈ X R , we have
( ) ( ) p ) + 2 β ( ∇ v, ∇ v ) − β ( ∇ q , ∇ v ) v ) + β ( ∇ q, ∇ q )
Ψ (U , U ) = 2 β ( ∇u , ∇u ) − β ∇ −1 p, ∇u − β ∇ −1 p, ∇u
( − β (∇
+ β ∇ p, ∇
(
−1
−1
−m
q , ∇m
2
≥ β ∇u + ∇ m v
m
m
−m
2
) ≥ βµ ( u 1
−m
−m
2
+ v
2
m
(3.23)
).
So, for ∀U ∈ X R , Ψ (U , U ) ≥ 0 , the Ψ is also positive definite. Next, we need to define a scale product in X U ,V
X
= Φ ( PNU , PNV ) + Ψ ( PRU , PRV ) ,
(3.24)
where PN and PR are projection: X → X N , X → X R respectively, for convenience, we rewrite (3.24) as follows U ,V
X
= Φ (U , V ) + Ψ (U , V ) .
(3.25)
We will proof that two subspaces X1 and X2 in (3.18) are orthogonal. In fact, we only need to show XN and XC are orthogonal, that is U −j , U +j
X
= 0
( ∀U
− j
)
∈ X N , U +j ∈ X c .
(3.26)
By (3.20), (3.25), we have DOI: 10.4236/am.2018.96050
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U −j , U +j
X
(
)
= Φ U −j , U +j
(
)
( ) ( ) ( + 4λ λ ( ∇ u , ∇ u ) + β ( ∇ v , ∇ v ) − 4 M ( s ) ( v , v ) − λ ( ∇ v , ∇ v ) − λ ( ∇ v , ∇ v ) + 4λ λ ( ∇ v , ∇
= β ∇u j , ∇u j − 4 M ( s ) u j , u j − λ +j ∇ −1 u j , ∇u j − λ −j ∇ −1 u j , ∇u j − j
(
−m
+ j
− λ +j
−1
j
m
j
m
j
−m
− j
j
m
j
m
j
j
j
− j
j
−m
+ j
) − M (s)( u + v ) + λ ) ( u + v ) + 4λ λ ( u + v ). 2
= β ∇u j
(
−1
+ j
+ ∇mv j
2
2
2
j
− j
j
j
j
−m
vj
)
2
j
2
− j
)
j
+ j
2
2
j −1
j −m
(3.27)
λ −j λ +j M ( s ) µ j , therefore + λ −j βµ j ,= By (3.12), we can get λ +j= U −j , U +j
X
(
)
= Φ U −j , U +j = 0.
(3.28)
Step 3 Further, we estimate the Lipschitz constant lF of F (U ) (3.2). According to Lemma 3.1, gi : X 3 → X 3 is Lipschitz continuous with Lipschitz
constant li . Let Pi : X → X i ( i = 1, 2 ) is orthogonal projection. From (3.22), (3.23) and (3.24), we have U
2 X
= Φ ( PU 1 , PU 1 ) + Ψ ( P2U , P2U )
( ) + βµ ( P u + P v ) ≥ ( β − 1) µ − 4m ( Pu + P u + Pv + P v ) ≥ ( β − 1) µ − 4m ( u + v ) . ≥ ( β − 1) µ1 − 4m1 Pu + Pv 1 1 2
2
1
1
1
1
2
2
1
2
1
2
2
2
2
1
2
2
2
, v, p, q ) , V ( u1 , v1 , p1 , q1 ) ∈ X , we have ( u= F (U ) − F (V ) X = g1 ( u , v ) − g1 ( u1 , v1 ) X + g 2 ( u , v ) − g 2 ( u1 , v1 ) X T
Given U =
(3.29)
2
2
T
3
≤ l1 ( u − u1 + v − v1 ) + l2 ( u − u1 + v − v1 ≤ l ( u − u1 + v − v1 ) l U −V ≤ ( β − 1) − 4m1
X
)
3
(3.30)
,
where l = max {l1 , l2 } . So, we obtain
lF ≤
l
( β − 1) − 4m1
.
(3.31)
Step 4 Now, we will show the spectral gap condition (2.4) holds. Since Λ= λN− , Λ 2= λN− +1 , then 1
Λ 2 −= Λ1 λN− +1 − = λN−
1 1 ( µ N +1 − µ N ) + Q ( N ) − Q ( N + 1) , 2 2
(3.32)
where Q= ( N ) β 2 µ N2 − 4M ( s ) µ N . Let N1 > 0 , for ∀N ≥ N1 , then
4m1 β2 Q1 ( N ) = 1− − , we can obtain 1 4 m − − β µ 1) µ1 − 4m1 − µ β ( ) 1 1 N ( DOI: 10.4236/am.2018.96050
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Q ( N ) − Q ( N + 1) + =
From
( β − 1) µ1 − 4m1 ( µ N +1Q1 ( N + 1) − µ N Q1 ( N ) )
( A2 ) , we can easily obtain lim
N →∞
(
( β − 1) − 4m1 ( µ N +1 − µ N )
Q ( N ) − Q ( N + 1) +
(3.33)
0 . (3.34) ( β − 1) µ1 − 4m1 ( µ N +1 − µ N ) ) =
Then, according to (3.16), (3.31), (3.32) and (3.34), we have 1 1 Λ 2 − Λ1 > ( µ N +1 − µ N ) − ( β − 1) µ1 − 4m1 − 1 2 2 4l ≥ ≥ 4lF . ( β − 1) µ1 − 4m1
(3.35)
Therefore, Theorem 3.1 is true. Theorem 3.2. Under the condition of Theorem 3.1, the problem (1.1)-(1.5) exist an inertial manifold µ in X,
µ= graph ( m ) := {ζ + m (ζ ) : ζ ∈ X1} ,
(3.36)
where X 1 , X 2 defined in (3.18) and m : X 1 → X 2 is a Lipschitz continuous function.
Acknowledgements The authors would like to thank for the anonymous referees for their valuable comments and suggestions sincerely. These contributions increase the value of the paper.
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DOI: 10.4236/am.2018.96050
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