Open Access

Existence of solutions to strongly damped plate or beam equations

Boundary Value Problems20122012:76

DOI: 10.1186/1687-2770-2012-76

Received: 11 April 2012

Accepted: 3 July 2012

Published: 20 July 2012

Abstract

In this paper, we study a strongly damped plate or beam equation. By using spatial sequence techniques and energy estimate methods, we obtain an existence theorem of the solution to abstract strongly damped plate or beam equation and to a nonlinear plate or beam equation.

MSC:35L05, 35L20, 35D30, 35D35.

Keywords

existence solution plate beam strongly damped

1 Introduction

We consider the following nonlinear strongly damped plate or beam equation:
{ 2 u t 2 k u t = f ( x , 2 u ) + g ( x , u , D u , D 2 u , D 3 u ) , k > 0 , u | Ω = u | Ω = 0 , u ( x , 0 ) = φ , u t ( x , 0 ) = ψ ,
(1.1)

where Δ is the Laplacian operator, Ω denotes an open bounded set of R N ( N = 1 , 2 ) with a smooth boundary Ω and u denotes a vertical displacement at ( x , t ) .

It is well known that flexible structures like suspension bridges or overhead power transmission lines can be subjected to oscillations due to various causes. Simple models for such oscillations are described with second- and fourth-order partial differential equations as can be seen for example in [18]. The problem (1.1) can be applied in the mechanics of elastic constructions for the study of equilibrium forms of the plate and beam, which has a long history. The abstract theory of Eq. (1.1) was investigated by several authors [914].

The main objective of this article is to find proper conditions on f and g to ensure the existence of solutions of Eq. (1.1). This article uses the spatial sequence techniques, each side of the equation to be treated in different spaces, which is an important way to get more extensive and wonderful results.

The outline of the paper is as follows. In Section 2 we provide an essential definition and lemma of solutions to abstract equations from [1518]. In Section 3, we give an existence theorem of solutions to abstract strongly damped plate or beam equations. In Section 4.10, we present the main result and its proof.

2 Preliminaries

We introduce two spatial sequences:
{ X H 3 X 2 X 1 H , X 2 H 2 H 1 H ,
(2.1)
where H, H 1 , H 2 , H 3 are Hilbert spaces, X is a linear space, and X 1 , X 2 are Banach spaces. All embeddings of (2.1) are dense. Let
{ L : X X 1 be one-one dense linear operator , L u , v H = u , v H 1 , u , v X .
(2.2)
Furthermore, L has eigenvectors { e k } satisfying
L e k = λ k e k ( k = 1 , 2 , ) ,
(2.3)

and { e k } constitutes a common orthogonal basis of H and H 3 .

We consider the following abstract equation:
{ d 2 u d t 2 + k d d t L u = G ( u ) , k > 0 , u ( 0 ) = φ , u t ( 0 ) = ψ ,
(2.4)
where G : X 2 × R + X 1 is a mapping, R + = [ 0 , ) and L : X 2 X 1 is a bounded linear operator satisfying
L u , L v H = u , v H 2 , u , v X 2 .
(2.5)

Definition 2.1[15]

We say u W loc 1 , ( ( 0 , ) , H 1 ) L loc ( ( 0 , ) , X 2 ) is a global weak solution of Eq. (2.4) provided that ( φ , ψ ) X 2 × H 1
u t , v H + k L u , v H = 0 t G ( u ) , v d t + ψ , v H + k L φ , v H ,
(2.6)

for all v X 1 and 0 t < .

Lemma 2.2[18]

Let u L loc p ( ( , ) , X ) , X be a Banach space. If u h = 1 h h t + h u ( s ) d s ( 0 < | h | < 1 ), then { u h } L loc p ( ( , ) , X ) , satisfying
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equa_HTML.gif

3 Existence theorem of abstract equation

Let G = A + B : X 2 × R + X 1 . Assume:

(A1) There is a C 1 functional F : X 2 R 1 such that
A u , L v = D F ( u ) , v , u , v X .
(3.1)
(A2) Functional F : X 2 R 1 is coercive, i.e.,
F ( u ) , u X 2 .
(3.2)
(A3) B satisfies
| B u , L v | C F ( u ) + k 2 v H 2 2 + g ( t ) , u , v X ,
(3.3)

for g L loc 1 ( 0 , ) .

Theorem 3.1 If G : X 2 × R + X 1 is bounded and continuous, and DF is monotone, i.e.,
D F ( u 1 ) D F ( u 2 ) , u 1 u 2 0 , u 1 , u 2 X 2 ,
(3.4)
then, for all ( φ , ψ ) X 2 × H 1 , the following assertions hold.
  1. (1)
    If G = A satisfies (A 1) and (A 2), then Eq. (2.4) has a global weak solution
    u W 1 , ( ( 0 , ) , H 1 ) W 1 , 2 ( ( 0 , ) , H 2 ) L ( ( 0 , ) , X 2 ) .
    (3.5)
     
  2. (2)
    If G = A + B satisfies (A 1)-(A 3), and u n u 0 in L ( ( 0 , T ) , X 2 ) such that
    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ13_HTML.gif
    (3.6)
     
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ14_HTML.gif
(3.7)
then Eq. (2.4) has a global weak solution
u W loc 1 , ( ( 0 , ) , H 1 ) W loc 1 , 2 ( ( 0 , ) , H 2 ) L loc ( ( 0 , ) , X 2 ) .
(3.8)
  1. (3)
    Furthermore, if G = A + B satisfies
    | G u , v | C F ( u ) + 1 2 v H 2 + g ( t ) ,
    (3.9)
     

for g L 1 ( 0 , T ) , then u W loc 2 , 2 ( ( 0 , ) ; H ) .

Proof Let { e k } X be a common orthogonal basis of H and H 3 , satisfying (2.3). Set
{ X n = { i = 1 n α i e i | α i R 1 } , X ˜ n = { j = 1 n β j ( t ) e j | β j C 2 [ 0 , ) } .
(3.10)

Clearly, L X n = X n , L X ˜ n = X ˜ n .

By using Galerkin method, there exists u n C 2 ( [ 0 , ) , X n ) satisfying
{ d u n d t , v H + k L u n , v H = 0 t G ( u n ) , v d t + ψ n , v H + k L φ n , v H , u n ( 0 ) = φ n , u n ( 0 ) = ψ n ,
(3.11)
for v X n , and
0 t [ d 2 u n d 2 t , v H + k L d u n d t , v H ] d t = 0 t G u n , v d t
(3.12)

for v X ˜ n .

Firstly, we consider G = A . Let v = d d t L u n in (3.12). Taking into account (2.2)and (3.1), it follows that
0 = 0 t [ 1 2 d d t d u n d t , d u n d t H 1 + k d u n d t , d u n d t H 2 + D F ( u n ) , d u n d t ] d t = 1 2 d u n d t H 1 2 1 2 ψ n H 1 2 + k 0 t d u n d t H 2 2 d t + F ( u n ) F ( φ n ) .
We get
1 2 d u n d t H 1 2 + k 0 t d u n d t H 2 2 d t + F ( u n ) = F ( φ n ) + 1 2 ψ n H 1 2 .
(3.13)
Let φ H 3 . From (2.1) and (2.2), it is known that { e n } is an orthogonal basis of H 1 . We find that φ n φ in H 3 , and ψ n ψ in H 1 . From that H 3 X 2 is an imbedding, it follows that
{ φ n φ in  X 2 , ψ n ψ in  H 1 .
(3.14)
From (3.2), (3.13) and (3.14), we obtain
{ u n } W 1 , ( ( 0 , ) , H 1 ) W 1 , 2 ( ( 0 , ) , H 2 ) L ( ( 0 , ) , X 2 )  is bounded .
Let
{ u n u 0 in  W 1 , ( ( 0 , ) , H 1 ) L ( ( 0 , ) , X 2 ) , u n u 0 in  W 1 , 2 ( ( 0 , ) , H 2 ) ,
(3.15)

which implies that u n u 0 in W 1 , 2 ( ( 0 , ) , H ) is uniformly weakly convergent from that H 2 H is a compact imbedding.

According to (2.2), (2.4), (2.5) and (3.4), we obtain that
0 0 t D F ( v ) D F ( u n ) , u n v d τ = 0 t A v , L v L u n d τ + 0 t A u n , L u n L v d τ = 0 t A v , L v L u n d τ + 0 t d 2 u n d t 2 + k d d t L u n , L u n L v H d τ = 0 t A v , L v L u n d τ + 0 t d 2 u n d t 2 , L u n H d τ + k 0 t d d t L u n , L u n H d τ 0 t d 2 u n d t 2 , L v H d τ k 0 t d d t L u n , L v H d τ = 0 t A v , L v L u n d τ + 0 t d 2 u n d t 2 , u n H 1 d τ + k 0 t d u n d t , u n H 2 d τ 0 t d 2 u n d t 2 , v H 1 d τ k 0 t d u n d t , v H 2 d τ = 0 t A v , L v L u n d τ + u n , d u n d t H 1 φ n , ψ n H 1 0 t d u n d t , d u n d t H 1 d τ + k 2 u n , u n H 2 k 2 φ n , φ n H 2 d u n d t , v H 1 + ψ n , v ( 0 ) H 1 + 0 t d u n d t , d v d t H 1 d τ k u n , v H 2 + k φ n , v ( 0 ) H 2 + k 0 t u n , d v d t H 2 d τ .
Let n . From (3.15), we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ23_HTML.gif
(3.16)

Since n = 1 X ˜ n is dense in W 1 , ( ( 0 , ) , H 1 ) W 1 , 2 ( ( 0 , ) , H 2 ) L ( ( 0 , ) , X 2 ) , the above equality (3.16) holds for v W 1 , ( ( 0 , ) , H 1 ) W 1 , 2 ( ( 0 , ) , H 2 ) L ( ( 0 , ) , X 2 ) .

We set v the following variable:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Eque_HTML.gif
where w X 2 , λ is a real, u ˜ 0 = u 0 if t 0 , and u ˜ 0 = 0 if t < 0 . Thus the equality (3.16) is read as
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ24_HTML.gif
(3.17)
and,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ25_HTML.gif
(3.18)
In view of (3.17) and (3.18), we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ26_HTML.gif
(3.19)
We know that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equf_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equg_HTML.gif
Let h 0 + . (3.19) can be read as
0 t A ( u 0 + λ w ) , λ L w d τ d u 0 d t , λ w H 1 k u 0 , λ w H 2 + ψ , λ w H 1 + k φ , λ w H 2 0 .
According to (2.2) and (2.5), we obtain that
0 t A ( u 0 + λ w ) , L w d τ d u 0 d t , L w H k L u 0 , L w H + ψ , L w H 1 + k L φ , L w H 0 .
Let λ 0 + . It follows that
0 t A ( u 0 ) , L w d τ d u 0 d t , L w H k L u 0 , L w H + ψ , L w H 1 + k L φ , L w H 0 .
Since L : X 2 X 1 is dense, the above inequality can be rewritten as
d u 0 d t , v H + k L u 0 , v H = 0 t A ( u 0 ) , v d τ + ψ , v H + k L φ , v H ,

which implies that u 0 W 1 , ( ( 0 , ) , H 1 ) W 1 , 2 ( ( 0 , ) , H 2 ) L ( ( 0 , ) , X 2 ) is a global weak solution of Eq. (2.4).

Secondly, we consider G = A + B . Let v = d d t L u n in (3.12). In view of (2.2) and (3.1), it follows that
1 2 d u n d t H 1 2 + k 0 t d u n d t H 2 2 d t + F ( u n ) = 0 t B ( u n ) , d d t L u n d t + F ( φ n ) + 1 2 ψ n H 1 2 .
From (3.3), we have
1 2 d u n d t H 1 2 + F ( u n ) + k 0 t d u n d t H 2 2 d t C 0 t [ F ( u n ) + 1 2 d u n d t H 1 2 ] d t + f ( t ) ,
(3.20)

where f ( t ) = 0 t g ( τ ) d τ + 1 2 ψ H 1 2 + sup n F ( φ n ) .

By using the Gronwall inequality, it follows that
1 2 d u n d t H 1 2 + F ( u n ) f ( 0 ) e C t + 0 t f ( τ ) e C ( t τ ) d τ ,
(3.21)
which implies that for all 0 < T < ,
{ u n } W 1 , ( ( 0 , T ) , H 1 ) L ( ( 0 , T ) , X 2 )  is bounded .
From (3.20) and (3.21), it follows that
{ u n } W 1 , 2 ( ( 0 , T ) , H 2 )  is bounded .
Let
{ u n u 0 in  W 1 , ( ( 0 , T ) , H 1 ) L ( ( 0 , T ) , X 2 ) , u n u 0 in  W 1 , 2 ( ( 0 , T ) , H 2 ) ,
(3.22)

which implies that u n u 0 in W 1 , 2 ( ( 0 , T ) , H ) is uniformly weakly convergent from that H 2 H is a compact imbedding.

The remaining part of the proof is same as assertion (1).

Lastly, assume (3.9) holds. Let v = d 2 u n d t 2 in (3.12). It follows that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equo_HTML.gif
From (3.21), the above inequality implies
0 t d 2 u n d t 2 H 2 d τ C ( C > 0  is constant ) .
(3.23)

We see that for all 0 < T < , { u n } W 2 , 2 ( ( 0 , T ) , H ) is bounded. Thus u W 2 , 2 ( ( 0 , T ) , H ) . □

4 Main result

Now, we consider the nonlinear strongly damped plate or beam equation (1.1). Set
F ( x , y ) = 0 y f ( x , z ) d z .
(4.1)
We assume
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ32_HTML.gif
(4.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ33_HTML.gif
(4.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ34_HTML.gif
(4.4)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ35_HTML.gif
(4.5)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ36_HTML.gif
(4.6)

where ξ = { ξ α | | α | 3 } , ξ α corresponds to D α u .

Theorem 4.1 Under the assumptions (4.1)-(4.6), if φ satisfies the bounded condition of Eq. (1.1), for ( φ , ψ ) W 4 , p ( Ω ) H 2 × H 0 1 ( Ω ) , then there exists a global strong solution for Eq. (1.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equp_HTML.gif
Proof We introduce spatial sequences
X = { u C ( Ω ) | k u | Ω = 0 , k = 0 , 1 , 2 , } , X 1 = L p ( Ω ) , X 2 = { W 4 , p ( Ω ) | u | Ω = u | Ω = 0 } , H = L 2 ( Ω ) , H 1 = H 2 ( Ω ) H 0 1 ( Ω ) , H 2 = { H 3 ( Ω ) | u | Ω = u | Ω = 0 } , H 3 = { u H 4 m ( Ω ) | u | Ω = = 2 m 1 u | Ω } ,
where the inner products of H 1 , H 2 and H 3 are defined by
u , v H 1 = Ω u v d x , u , v H 2 = Ω u v d x , u , v H 3 = Ω 2 m u 2 m v d x ,

where m 1 such that H 3 X 2 is an embedding.

Linear operator L : X 2 X 1 and L : X 2 X 1 is defined by
L u = u , L u = 2 u .
It is known that L and L satisfy (2.2), (2.3) and (2.5). Define G = A + B : X 2 X 1 by
A u , v = Ω f ( x , 2 u ) v d x , B u , v = Ω g ( x , u , D u , D 2 u , D 3 u ) v d x , for  v X 1 .
Let F 1 ( u ) = Ω F ( x , u ) d x , where F is the same as in (4.2). We get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equu_HTML.gif

which implies conditions (A1), (A2) of Theorem 3.1.

From (4.3), we have
D F 1 ( u 1 ) D F 1 ( u 2 ) , u 1 u 2 0 .

From (4.5) and (4.6), we obtain that B : X 2 X 1 is a compact operator. Then, B satisfies (3.6) and (3.7).

We will show (3.3) as follows. From (4.4) and (4.5), for u , v X , it follows that
| B u , L v | = Ω | g ( x , u , D u , D 2 u , D 3 u ) 2 v | d x = Ω | g ( x , u , D u , D 2 u , D 3 u ) v | d x k 2 Ω | v | 2 d x + 2 k Ω | g ( x , u , D u , D 2 u , D 3 u ) | 2 d x k 2 v H 2 2 + C Ω [ | D x g | 2 + i = 1 n | α | 3 | D ζ α g | 2 | D i D α u | 2 ] d x k 2 v H 2 2 + C Ω [ | α | 4 | D α u | p + 1 ] d x ,
which implies condition (A3) of Theorem 3.1. From Theorem 3.1, Eq. (1.1) has a solution
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ37_HTML.gif
(4.7)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ38_HTML.gif
(4.8)
Lastly, we show that u L p ( Ω × ( 0 , T ) ) . By Definition 2.1, u satisfies
Ω u t ( x , t ) v d x + Ω u ( x , t ) v d x = 0 t Ω [ f ( x , u ) + g ( x , u , D u , D 2 u , D 3 u ) v ] d x d t + Ω ψ v d x + Ω φ v d x , v L p ( Ω ) .
Then, for any h > 0 , it follows that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equ39_HTML.gif
(4.9)
where h t u = 1 h ( u ( t + h ) u ( t ) ) . Let v = | h t u t | p 2 h t u t . From (4.9), we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equy_HTML.gif
Then, it follows that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-76/MediaObjects/13661_2012_Article_194_Equz_HTML.gif
From (4.2) and (4.5), we have
0 T Ω | u t ( x , t + h ) u t ( x , t ) h | p d x d t C 0 T Ω | h t u ( x , t ) | p d x d t + C 0 T t t + h Ω [ | f | p + | g | p ] d x d τ d t C 0 T Ω | u t ( x , t ) | p p 1 d x d t + C 0 T Ω [ ( | u | p 1 + 1 ) p + ( | α | 3 | D α u | p 2 + 1 ) p ] d x d t C 0 T Ω | u t ( x , t ) | p p 1 d x d t + C 0 T Ω [ | u | p + | α | 3 | D α u | p 2 2 ( p 1 ) + 1 ] d x d t .

By using the Sobolev embedding theorem, it follows that from (4.7) and (4.8) the right of the above inequality is bounded. Then, u t t exists almost everywhere in Ω × ( 0 , T ) , and u t t L p ( Ω × ( 0 , T ) ) , 0 < T < . □

Declarations

Acknowledgement

The authors are very grateful to the anonymous referees whose careful reading of the manuscript and valuable comments enhanced the presentation of the manuscript. Supported by the National Natural Science Foundation of China (NO. 11071177), the NSF of Sichuan Science and Technology Department of China (NO. 2010JY0057) and the NSF of Sichuan Education Department of China (NO. 11ZA102).

Authors’ Affiliations

(1)
College of Mathematics and Software Science, Sichuan Normal University
(2)
College of Mathematics, Sichuan University

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