exponential decay for a thermoelastic beam ...im.ufrj.br/~rivera/art_pub/tesesant.pdfexponential...

Journal of Thermal Stresses, 31: 537–556, 2008Copyright © Taylor & Francis Group, LLCISSN: 0149-5739 print/1521-074X onlineDOI: 10.1080/01495730801978208

EXPONENTIAL DECAY FOR A THERMOELASTICBEAM BETWEEN TWO STOPS

Santina Arantes and Jaime RiveraNational Laboratory for Scientific Computation – LNCC, Petrópolis,Rio de Janeiro, Brazil

A one-dimensional initial-boundary value problem for a thermoelastic beam that isclamped at one end and moves freely between two obstacles at the other, is studiedusing a method of functional analysis. The boundary conditions at the free end arerepresented either by classic Signorini’s conditions or a normal compliance condition.The main result is that the energy associated with the system decays exponentially astime goes to infinity. Numerical examples are included.

Keywords: Contact problem; Dynamic vibrations; Normal compliance condition; Penalized problem;Signorini’s conditions; Thermoelastic beam.

INTRODUCTION

The study of dynamic vibrations of mechanical systems caused, for example,by noises and characteristic vibrations of cars and its components became anarea of increasing interest for the automotive industries. These industries haveemployed great efforts in the sense to identify and to reduce these unwanted noisesand vibrations. The dynamic vibrations problem of beams in contact with a rigidobstacle has been studied for some authors, such as: Kuttler et al. [1], Andrewset al. [2], Dumont et al. [3] and Bajkowski et al. [4]. The one-dimensional problemdescribing the viscoelastic or elastic oscillations of a beam clamped at one endto a vibrations device while the other end oscillates between two rigid or flexiblestops, was first studied by Moon and Shaw in [5]. This model was also studiedby Kuttler and Shillor in [6], where the authors consider the material constitutivelaw to be either elastic or viscoelastic of the Kelvin–Voigt type. The contact ismodeled either with the Signorini’s conditions or with the normal compliancecondition. The authors show the existence of weak solutions and the uniquenessis proved only for the viscoelastic case. In reference [7], Dumont and Paoli studythis problem considering the elastic case and analyse the convergence of a fulldiscretized approximation; prove the stability and convergence of model and presentimplementation results. For other numerical results see, for example, [4, 8] andreferences therein.

Received 11 April 2007; accepted 28 January 2008.Address correspondence to Santina de Fatima Arantes, Laboratorio Nacional de Computacao

Cientifica, Rua Getulio Vargas, 333 Quitandinaha, Petropolis-Rio de Janeiro 25651-070, Brazil. E-mail:[email protected]

537

538 S. ARANTES AND J. RIVERA

It is well known that beams that vibrate with high frequency may generateconsiderable heat. For this reason, in this work we take into account the heatgenerated during the process. The model includes dynamic vibrations of the beamand the associated generation of heat; that is, the thermoelastic beam problem.Because of the lack of uniqueness for the Signorini’s problem, the main contributionof this work is to show that there exists one solution of the Signorini’s problem,which decays exponentially to zero as time goes to infinity. We also presentnumerical simulations that verify our theoretical results.

Throughout this article the same letter C0 will denote various positive constantsthat do not depend on x and t, and C� = C0�

−1 where � is any positive constant.

MODEL FORMULATION

We consider a thermoelastic beam problem. The physical description of themodel is given in Figure 1. The beam is clamped at its left end and its right end is free.The motion in the free end is constrained by two obstacles that are called stops.

We assume that the beam is configured in �0� L� ⊂ �. Here g1 and g2, withg1 < 0 < g2, are the positions of the stops. We make the usual Euler–Bernoulliassumption that the transverse dimensions of the beam are small enough comparedto the length of the beam and radius of curvature of the bending that any planecross-section, initially perpendicular to the axis of the beam, remains plane andperpendicular to the neutral surface during bending. The neutral surface is the onerunning through the length of the beam that suffers no extension or contractionduring its bending.

If the beam has a uniform square cross-section with vertical thickness h, thenthe thermal moment at �x� t� is given by

� = ��x� t� = 12h3

∫ h/2

−h/2y��x� y� t�dy

Figure 1 Beam subject to a constraint at the free end L.

THERMOELASTIC BEAM BOUNDARY VALUE PROBLEM 539

where ��x� y� t� is the temperature (measured relative to that of the environment) at�x� t� and vertical location y. In the discussion of the model that follows, we use theterms temperature difference and thermal moment interchangeably. We also denoteby u = u�x� t� the vertical displacement and the subscripts the partial derivatives.The equations that describe thermoelastic beams follow from [9], where the authorderived the equations [10]. Thus, the complete model is given by

hutt +Duxxxx +D�1+ ��xx = f in �0� L�× �0�� (1)

C�t − k�xx −T0E

1− �uxxt = g in �0� L�× �0�� (2)

the constants which appear in the above equations are given in Table 1.We introduce the following change of variables to set the problem in a

dimensionless form:

x = x

L� t = kt

CL2� ��x� t� = ��x� t�− T0

T0

� u�x� t� = u

L

D

CT0L4

This change is to replace the interval �0� L� by the unit interval �0� 1� and toreplace the temperature by � = 0.

Considering all dimensionless variables, the shearing force per unit area of thebeam � = ��x� t� is given by the relation

��x� t� = −uxxx − b�x (3)

and the heat flux is given by

q�x� t� = −�x

So that, the complete model, in dimensionless variables, is given by

autt + uxxxx + b�xx = f in �0� 1�× �0�� (4)

�t − �xx − cuxxt = g in �0� 1�× �0�� (5)

where a, b and c are dimensionless positive constants given by

a = hk2

C2D� b = D�1+ ��

L3C� c = T0EL

3

�1− ��D

We consider the initial conditions

u�x� 0� = u0�x�� ut�x� 0� = u1�x�� x� 0� = �0�x�� 0 ≤ x ≤ 1� (6)

where u0, u1 and �0 represent, respectively, the initial displacement, the velocity andtemperature difference of the beam. We assume that the beam has been thermallyinsulated at the clamped end and kept at a zero temperature at. Under theseconditions it is not difficult to show that the problem for Eqs. (4)–(5) is well posed.


It would be natural to consider a general case where the temperature at the free endis time dependent. However, in the general case a proof of the exponential stabilityis not possible using our techniques. Thus, for the sake of simplicity, we are takingthe temperature null at this end (see [4, 11]), so that the boundary conditions weconsider for � are

�x�0� t� = ��1� t� = 0 t > 0 (7)

We remark that we are assuming that there is not friction when the beam hits theobstacles, therefore there is not an extra generation of heat.

We assume that the beam is clamped at x = 0, then

u�0� t� = ux�0� t� = 0 t > 0 (8)

and that the moment at the end x = 1 is zero; that is

uxx�1� t� = 0 t > 0 (9)

At the free end of the beam, where contact with the obstacles may occur, wetake the classic Signorini’s condition

g1 ≤ u�1� t� ≤ g2 t > 0 (10)

This condition assures that the displacement at x = 1 is constrained between thestops g1 and g2. When the beam hits one of the obstacles, the stress is in the oppositedirection of the displacement. Then, we have the situations

��1� t� > 0 if u�1� t� = g1

��1� t� < 0 if u�1� t� = g2

��1� t� = 0 if g1 < u�1� t� < g2

To ensure that only one of the conditions occurs at the same time, we impose that

��1� t��g2 − u�1� t��+�u�1� t�− g1�+ = 0 (11)

where f+ = max�f� 0� is the positive part of f .To study this problem with Signorini’s conditions, we consider first an

approximation given by the normal compliance condition, where the stops areassumed to be flexible with resistance force proportional to the deflection.Introducing the penalization parameter 1/�, we find the solution of the penalizedproblem, then we pass to the limit when � → 0 to find the solution of the Signorini’sproblem. The normal compliance condition is given by

��1� t� = −k��u�1� t�− g2�+ − �g1 − u�1� t��+� (12)

where k = 1/� is a positive constant. This condition leads to a system of thepartial differential equations depending on the parameter �. Observe that the normal


compliance condition (12) allows that the final contact penetrates the obstacle; thatis, it may occurs that

u�1� t� > g2 or u�1� t� < g1 t > 0

From a mechanical point of view, 1/� can be interpreted as the stiffness of thestops. On the other hand, from numerical point of view, for large values of 1/�,we obtain the rigid stops and the condition (11) can be seen as a limit case of thenormal compliance condition (12) when 1/� tends to the infinity. Thus, the normalcompliance condition (12) is called regularization of Signorini’s conditions and saysthat condition (12) is an approximation of the problem with Signorini’s conditionsthrough the boundary conditions.

We used a beam of alloy steel, for which thermal and mechanics propertiesare given in Table 1.

Table 1 Material properties

Length L = 1 2m

Thickness h = 0 05mDensity = 8000kg/m3

Young’s modulus E = 207GPaPoisson’s ratio � = 0 3Coefficient of thermal expansion = 11× 10−6 1/�CModulus of flexural rigidity D = Eh3

12�1−�2�= 2 4× 10−3 GPam

Heat capacity C = 3 6× 106 kg/m s2 �CThermal conductivity k = 38kgm/s3 �CReference temperature T0 = 20 �CExternal sources f (kg/m s2) and g (kg/m2 s3)

We take the values for from [7] and E, �, , C and k from [14]. The choiceof h have been taken small and the value for T0 is the ambient temperature for thealloy steel.

EXISTENCE OF SOLUTIONS AND VARIATIONAL FORMULATIONS

Here we establish the existence of weak solutions as well as uniqueness for thepenalized problem. We introduce the following spaces

V = �u ∈ H2�0� 1�� u�0� = ux�0� = 0�

and

E = �� ∈ H1�0� 1�� 1� = 0�

To incorporate the Signorini’s conditions (10)–(11), we consider the convex setof all the admissible displacements in the right end of the beam.

K = �u ∈ V g1 ≤ u�1� ≤ g2�


We define H = L2�0� 1� and we identify H with H ′, topological dual of H , sowe have

V ⊂ H = H ′ ⊂ V ′ and E ⊂ H = H ′ ⊂ E′

The classic formulation of the problem of vibrations of a thermoelastic beambetween two obstacles with Signorini’s conditions is: To find pairs of functions�u� ��, such that the system (4)–(11) is satisfied. The classical problem with normalcompliance condition is: To find pairs of functions �u� �� satisfying (4)–(9) and (12).

The definition of weak solution of the Signorini’s problem (4)–(11), which wewill use in this work is given as follows.

Definition 3.1. Assume that

u0 ∈ K� u1� �0 ∈ L2�0� 1� and f� g ∈ H1�0� T� L2�0� 1��

with T any fixed positive real number. We say that the couple �u� �� is a weaksolution to (4)–(11), when

u ∈ W 1��(0� T� L2�0� 1�) ∩ L��0� T�K�

� ∈ L2�0� T� E� ∩ L��0� T� L2�0� 1��

u�x� 0� = u0� ut�x� 0� = u1� ��x� 0� = �0

−a∫ T

0

∫ 1

0ut�vt − ut�dx dt +

∫ T

0

∫ 1

0uxx�vxx − uxx�dx dt − b

∫ T

0

∫ 1

0�x�vx − ux�dx dt

≥ a∫ 1

0u1�v�x� 0�− u0�dx +

∫ T

0

∫ 1

0f�v− u� dx dt�

−∫ T

0

∫ 1

0�wt dx dt +

∫ T

0

∫ 1

0�xwx dx dt + c

∫ T

0

∫ 1

0uxxwt dx dt

=∫ 1

0�0w�x� 0�dx − c

∫ 1

0u0xxw�x� 0�dx +

∫ T

0

∫ 1

0gw dx dt�

∀v ∈ L2�0� T�K�� ∀w ∈ L2�0� T� E�� ∀wt ∈ L2�0� T� L2�0� 1��

w�x� T� = 0� v�x� T� = u�x� T�

The existence of weak solution of the Signorini’s problem is obtained bytaking the limit when � → 0 of the penalized problem

au�tt + u�

xxxx + b��xx = f in �0� 1�× �0� T�

��t − ��xx − cu�xxt = g in �0� 1�× �0� T�

u��x� 0� = u�0�x�� u�

t �x� 0� = u�1�x�� x� 0� = ��0�x� in �0� 1�

u��0� t� = u�x�0� t� = u�

xx�1� t� = 0 in �0� T�

��x�0� t� = ��1� t� = 0 in �0� T�

��1� t� = − 1��u��1� t�− g2�

+

−�g1 − u��1� t��+�− �u�t �1� t� in �0� T�

(13)


The term −�u�t �1� t� was introduced to get the uniqueness of the solution of the

problem 13. In that follows, we omit the index �. The weak formulation of thepenalized problem is given by

u ∈ W 1��0� T� L2�0� 1�� ∩ L��0� T�K�

� ∈ L2�0� T� E� ∩ L��0� T� L2�0� 1��

with

uxxt ∈ L��0� T� L2�0� 1�� and �xt ∈ L2�0� T� L2�0� 1��

such that

− a∫ T

0

∫ 1

0utvt dx dt +

∫ T

0

∫ 1

0uxxvxx dx dt − b

∫ T

0

∫ 1

0�xvx dx dt

=∫ T

0

∫ 1

0fv dx dt + a

∫ 1

0u1v�x� 0�dx (14)

− 1�

∫ T

0��u�1� t�− g2�

+ − �g1 − u�1� t��+�v�1� t�dt − �∫ T

0ut�1� t�v�1� t�dt

−∫ T

0

∫ 1

0�wt dx dt +

∫ T

0

∫ 1

0�xwx dx dt + c

∫ T

0

∫ 1

0uxxwt dx dt

=∫ 1

0�0w�x� 0�dx − c

∫ 1

0u0xxw�x� 0�dx +

∫ T

0

∫ 1

0gw dx dt (15)

∀v ∈ L2�0� T� V��∀w ∈ L2�0� T� E� and ∀wt ∈ L2�0� T� L2�0� 1��

with v�x� T� = w�x� T� = 0

We summarize the well posedness of the penalized problem in the followingtheorem.

Theorem 3.2. Let us take

u0 ∈ K ∩H4�0� 1�� u1 ∈ H3�0� 1�� 0 ∈ E ∩H2�0� 1� and f� g ∈ H1�0� T� L2�0� 1��

Then, for any � > 0, there exists only one strong solution to problem (14)–(15) with theregularity

u ∈ W 2��0� T� L2�0� 1�� ∩W 1��0� T� V� ∩ L��0� T�H4�0� 1� ∩ V�

� ∈ L��0� T�H2�0� 1� ∩ E� ∩W 1��0� T� L2�0� 1��

Proof. The proof is based on the Galerkin method. Following the same proceduresas in [12], we are able to show the existence and uniqueness of solution as indicatedin the above Theorem. So, we omit here the proof. On the other hand, using thesame arguments as in Kuttler and Shillor in [6], we can show that there exists a


subsequence of �u�� u�t � �

��, we still denote in the same way, that converges to theSignorini’s problem; that is

�u�� u�t � �

��∗⇀ �u� ut� �� in L��0� T� V�× L��0� T� L2�0� 1��× L��0� T� L2�0� 1��

(16)

when �u� ut� �� satisfies Definition 3.1.

So we have:

Theorem 3.3. Let us u0 ∈ V� u1� �0 ∈ L2�0� 1� and f� g ∈ H1�0� T� L2�0� 1��. Then,there exists a weak solution for Signorini’s problem (4)–(11).

Remark 3.4. The uniqueness of the solution to Signorini’s problem (4)–(11)remains an open question, as well as the existence result for dimensional spacen ≥ 2.

EXPONENTIAL DECAY

Next, we show that the solution of the Signorini’s problem decaysexponentially to zero as time goes to infinity. To show this, we introduce the energyfunctional associated with the thermoelastic system

E�t� u� �� = E�t� = 12

∫ 1

0

[a�ut�2 + �uxx�2 +

b

c��2

]dx

We denote E��t� = E�t� u�� . The energy of the penalized problem is defined as

��t� = E��t�+1��t� (17)

where the functional ��t� is given by

��t� =12��u��1� t�− g2�

+�2 + 12��g1 − u��1� t��+�2 (18)

First, we show that the energy associated with the penalized problem ��t�decays exponentially to zero as time goes to infinity with rates that do not depend on�, and later we will pass to the limit when � → 0 to obtain the decay of the energyE�t� associated with the Signorini’s problem. The idea is to construct a Lyapunov’sfunctional �, satisfying

�1��t� ≤ ��t� ≤ �2��t� andd

dt��t� ≤ −��t�� ∀t > 0

where �1� �2 and � are positive constants.


Let us consider the penalized problem (13). For simplicity, we take f = g = 0in (14) and (15) and we omit the index �. Using (14) and (15), with v = ut and w = �,respectively, we arrive at

12d

dt

{a∫ 1

0�ut�2 dx +

∫ 1

0�uxx�2 dx +

b

c

∫ 1

0��2 dx

+1��u�1� t�− g2�

+�2 + ��g1 − u�1� t��+�2�}= −

∫ 1

0��x�2 dx − ��ut�1� t��2

From the above identity and (18), we see that

d

dt��t� ≤ −

∫ 1

0��x�2 dx − ��ut�1� t��2 (19)

Let us introduce the functional �1 as

�1�t� = a∫ 1

0uut dx

Under the above notations, we have

Lemma 4.1. For any � > 0 small enough, there exists a positive constant C� > 0,such that

d

dt�1�t� ≤ a

∫ 1

0�ut�2 dx −

2�� t�− �1− �b�

∫ 1

0�uxx�2 dx + C�

∫ 1

0��x�2 dx (20)

Proof. From definition of �1�t�, it follows that

d

dt�1�t� = a

∫ 1

0�ut�2 dx + a

∫ 1

0uutt dx

Using the system (13) and integrating by parts, we get

d

dt�1�t� = a

∫ 1

0�ut�2 dx −

∫ 1

0u�uxxxx + b�xx�dx

= a∫ 1

0�ut�2 dx − u�1� t��uxxx�1� t�+ b�x�1� t��+

∫ 1

0uxuxxx dx + b

∫ 1

0ux�x dx

Using (3), (13), inequality � ≤ �2 + �−1�2, Poincare’s inequality (one-dimensionalcase is � u L2�� ≤ L ux L2��, where L is the Poincare’s constant, remembering thatin our case L = 1) and integrating by parts, we conclude

d

dt�1�t� ≤ a

∫ 1

0�ut�2 dx + u�1� t��1� t�−

∫ 1

0�uxx�2 dx + �b

∫ 1

0�uxx�2 dx + b�−1

∫ 1

0��x�2 dx

≤ a∫ 1

0�ut�2 dx −

1��u�1� t�− g2�

+�2 + ��g1 − u�1� t��+�2�

− �1− �b�∫ 1

0�uxx�2 dx + C�

∫ 1

0��x�2 dx


Using (18), our result follows. �

Let us introduce the functional

�2�t� = −a∫ 1

0qutux dx

Lemma 4.2. Let us take q ∈ C3��0� 1��, then there exists a constant C0 =max� 3

2 � b� > 0, such that

d

dt�2�t� ≤ −q�1�ux�1� t��1� t�+

12qxx�1��ux�1� t��2 −

12

∫ 1

0qxxx�ux�2 dx

+C0

∫ 1

0�1+ qx��uxx�2 dx +

12q�0��uxx�0� t��2

+ b

2

∫ 1

0�qx + q�2��x�2 dx −

a

2q�1��ut�1� t��2 +

a

2

∫ 1

0qx�ut�2 dx

Proof. Note that

d

dt�2�t� = −a

∫ 1

0quttux dx − a

∫ 1

0qutuxt dx

Using the system (13), we obtain

d

dt�2�t� =

∫ 1

0qux�uxxxx + b�xx�dx −

a

2

∫ 1

0qd

dx�ut�2 dx

Integrating by parts and using the boundary conditions, it follows that

d

dt�2�t� = q�1�ux�1� t��uxxx�1� t�+ b�x�1� t��−

∫ 1

0�qxux + quxx�uxxx dx

− b∫ 1

0�qxux + quxx��x dx −

a

2q�1��ut�1� t��2 +

a

2

∫ 1

0qx�ut�2 dx (21)

Integrating by parts, yields

−∫ 1

0�qxux + quxx�uxxx dx

=∫ 1

0�qxxux + qxuxx�uxx dx −

12

∫ 1

0qd

dx�uxx�2 dx (22)

= 12

∫ 1

0qxx

d

dx�ux�2 dx +

∫ 1

0qx�uxx�2 dx +

12q�0��uxx�0� t��2 +

12

∫ 1

0qx�uxx�2 dx

= 12qxx�1��ux�1� t��2 −

12

∫ 1

0qxxx�ux�2 dx +

32

∫ 1

0qx�uxx�2 dx +

12q�0��uxx�0� t��2


Using inequality∫ 10 qxux�x dx ≤ 1

2

∫ 10 �qx�x�2 dx + 1

2

∫ 10 �ux�2 dx and the Poincare’s

inequality once more, we arrive at

−b∫ 1

0�qxux + quxx��x dx

≤ b

2

∫ 1

0�qx�x�2 dx +

b

2

∫ 1

0�uxx�2 dx +

b

2

∫ 1

0�q�x�2 dx +

b

2

∫ 1

0�uxx�2 dx

≤ b

2

∫ 1

0�qx + q�2��x�2 dx + b

∫ 1

0�uxx�2 dx (23)

Substituting (22) and (23) into (21) and using (3), our result follows. �

Now let us introduce

�3�t� =∫ 1

0

( ∫ x

0��s� t�ds

)qut dx

Lemma 4.3. Let us take q ∈ C3��0� 1��. Then, for � > 0 small enough, there exists aconstant C�, such that

d

dt�3�t� ≤ C�

∫ 1

0�1+ q2 + �1+ q�2 + �q�0��x�2 dx −

∫ 1

0

(cqx2

− �

)�ut�2 dx

+ c

2q�1��ut�1� t��2 −

1a

( ∫ x

0��s� t�ds

)q�x��x� t�

∣∣∣∣x=1

+ �

a�q�0��uxx�0� t��2 +

2�a

∫ 1

0�1+ qx�2�uxx�2 dx

− 3b2a

∫ 1

0qx��2 dx −

1a

∫ 1

0

( ∫ x

0��s� t�ds

)qxx�uxx + b��dx

Proof. Integrating (13)2 over �0� x� and using the boundary conditions, we get

∫ x

0�t�s� t�ds = �x + cuxt

Multiplying the above equation by qut and integrating over �0� 1�, we get

∫ 1

0

( ∫ x

0�t�s� t�ds

)qut dx =

∫ 1

0�xqut dx + c

∫ 1

0uxtqut dx (24)

From definition of �3�t�, it follows that

d

dt�3�t� =

∫ 1

0

( ∫ x

0�t�s� t�ds

)qut dx +

∫ 1

0

( ∫ x

0��s� t�ds

)qutt dx


Substituting �13�1 and (24) into the above equation, using inequality � ≤ �2 +�−1�2 and integrating by parts, we see that

d

dt�3�t� =

∫ 1

0�xqut dx + c

∫ 1

0uxtqut dx −

1a

∫ 1

0

( ∫ x

0��s� t�ds

)q�uxxxx + b�xx�dx

≤ �−1∫ 1

0�q�x�2 dx + �

∫ 1

0�ut�2 dx +

c

2q�1��ut�1� t��2 dx

− c

2

∫ 1

0qx�ut�2 dx −

1a

( ∫ x

0��s� t�ds

)q�x��x� t�

∣∣∣∣x=1

+ 1a

∫ 1

0

[�q +

( ∫ x

0��s� t�ds

)qx

]uxxx dx︸︷︷︸

Ia

+ b

a

∫ 1

0

[�q +

( ∫ x

0��s� t�ds

)qx

]�x dx︸︷︷︸

Ib

(25)

Integrating by parts once more, yields

Ia = −1aq�0��0� t�uxx�0� t�−

1a

∫ 1

0��xq + 2�qx +

( ∫ x

0��s� t�ds

)qxx

]uxx dx

From Sobolev’s imbedding Theorem (sup�w�x�� ≤ wx L2�0�1�, using � ≤ �2 +�−1�2 and Poincare’s inequality, we get

Ia ≤ 1a�q�0� uxx�0� t��

( ∫ 1

0��x�2 dx

) 12

+ �−1

a

∫ 1

0�q�x�2 dx +

�

a

∫ 1

0�uxx�2 dx

+ 2�−1

a

∫ 1

0��x�2 dx +

2�a

∫ 1

0�qxuxx�2 dx −

1a

∫ 1

0

( ∫ x

0��s� t�ds

)qxxuxx dx

≤ �

a�q�0� uxx�0� t��2 + C�

∫ 1

0��x�2 dx + C�

∫ 1

0�1+ q�2��x�2 dx

+ 2�a

∫ 1

0�1+ qx�2�uxx�2 dx −

1a

∫ 1

0

( ∫ x

0��s� t�ds

)qxxuxx dx

Integrating by parts, we obtain

Ib ≤ − b

2aq�0��0� t��2 − 3b

2a

∫ 1

0qx��2 dx −

b

a

∫ 1

0

( ∫ x

0��s� t�ds

)qxx� dx

By Sobolev’s imbedding, we get

Ib ≤ C��q�0��∫ 1

0��x�2 dx −

3b2a

∫ 1

0qx��2 dx −

b

a

∫ 1

0

( ∫ x

0��s� t�ds

)qxx� dx

Substituting �Ia� and �Ib� into (25), Lemma follows. �


To estimate the boundary terms in Lemmas 4.2 and 4.3, we take q�x� = x − 1.Thus, we have that there exists constant C0 > 0 and for � > 0 small enough, thereexists C� > 0, such that

d

dt�2�t� ≤ C0

∫ 1

0�uxx�2 dx −

12�uxx�0� t��2 + C0

∫ 1

0��x�2 dx +

a

2

∫ 1

0�ut�2 dx (26)

d

dt�3�t� ≤ C�

∫ 1

0��x�2 dx −

(c

2− �

) ∫ 1

0�ut�2 dx +

�

a�uxx�0� t��2

+ 8�a

∫ 1

0�uxx�2 dx −

3b2a

∫ 1

0��2 dx (27)

We also need the following result.

Lemma 4.4. For any q ∈ C3��0� 1��, the functionals �1, �2 and �3 defined inLemmas 4.1, 4.2 and 4.3, respectively, satisfy

��i�t�� ≤ kiE�t�� where i = 1� 2� 3

Proof. From the definition of �1�t� and by using � ≤ 12

2 + 12�

2 and thePoincare’s inequality, we obtain

��1�t�� =∣∣∣∣a ∫ 1

0uut dx

∣∣∣∣ ≤ a

2

∫ 1

0�ut�2 dx +

a

2

∫ 1

0�uxx�2 dx ≤ k1E�t��

With similar procedure, we can show the same estimates for �2�t� and �3�t�. �

Now, we are able to show exponential stability of Signorini’s problem.

Theorem 4.5. The solution of problem (4)–(11) decays exponentially to zero as timegoes to infinity; that is, there exist constants �0� � > 0, such that the energy of system(4)–(11), satisfies

E�t� ≤ �0E�0�e−�t ∀t ≥ 0

Proof. Denote by

� �t� = c

16�1�t�+ ��2�t�+

a

4�3�t�

where � is the constant chosen as in Lemmas 4.1 and 4.3. Consider the Lyapunov’sfunctional defined by

��t� = N��t�+ � �t�

where N > 0 is constant. Then, from (19), (20), (26) and (27), it follows

d

dt��t� ≤ −

[N −

(cC�

16+ �C0 +

aC�

4

)] ∫ 1

0��x�2 dx


−[c

16− �

(bc

16+ C0 + 2

)] ∫ 1

0�uxx�2 dx −

�

4�uxx�0� t��2 −

c

8�� t�

−[ac

16− �

3a4

] ∫ 1

0�ut�2 dx −

3b8

∫ 1

0��2 dx − N��ut�1� t��2

Choosing N large enough and � small enough, we conclude that there exist positiveconstants ai, such that

d

dt��t� ≤ −a0

∫ 1

0�ut�2 dx − a1

∫ 1

0�uxx�2 dx − a2

∫ 1

0��2 dx − a3

�� t�

Taking � = min�a0� a1� a2� a3� and using (17), we get

d

dt��t� ≤ −��t� (28)

From Lemma 4.4 and N large enough, we get that there exist constants �1� �2 > 0,such that

�1��t� ≤ ��t� ≤ �2��t� (29)

Combining (28) and (29), we arrive at

d

dt��t� ≤ −��t� where � = �

�2

Consequently

��t� ≤ �0��0�e−�t ∀t ≥ 0 where �0 =

�2

�1

(30)

Here, we obtain the exponential decay of the energy associated with thepenalized problem, as time goes to infinity. We will pass to the limit in (30) as� −→ 0, to get the decay of the energy associated with the Signorini’s problem.From (17), we have

E��t�+1��t� ≤ �0

(E��0�+

1��0�

)e−�t ∀t ≥ 0

Taking the initial data, such that ��0� = 0, letting � −→ 0, using (16), the lowersemicontinuity of the energy and the fact that � does not depend on �, we get that

E�t� ≤ �0E�0�e−�t ∀t ≥ 0

Thus, we our conclusion follows. �


NUMERICAL PROBLEM

To solve the numerical problem, we use the penalized problem (13), assumingenough regularity on the solution. We uncouple the problem (13) as follows (weomit index �)

auntt + un

xxxx + b�nxx = 0 in �0� 1�× �0� T�

�nt − �nxx − cunxxt = 0 in �0� 1�× �0� T�

un�x� 0� = un0�x�� un

t �x� 0� = un1�x�� n�x� 0� = �n0�x� in �0� 1�

un�0� t� = unx�0� t� = un

xx�1� t� = 0 in �0� T�

�nx�0� t� = �n�1� t� = 0 in �0� T�

�n�1� t� = −1��un�1� t�− g2�

+

−�g1 − un�1� t��+�− �unt �1� t� in �0� T�

(31)

where n ∈ �. One can show that the solution of the system (31) converge to thesolution of system (13), as n −→ �. We use finite element method combined withiterative methods to get our solution.

Let us introduce the spaces

un ∈ V = �u ∈ H2�0� 1�� u�0� = ux�0� = 0� and �n ∈ E = �� ∈ H1�0� 1�� 1� = 0�

Then, the variational formulations are given by

a�untt� v�+ �un

xx� vxx�− b��nx� vx�− �n�1� t�v�1� t� = 0 (32)

��nt � �)+ ��nx� �x�+ c�un

xt� �x� = 0 (33)

∀v ∈ Vh and ∀� ∈ Eh. With the initial conditions

un�x� 0� = un0�x�� un

t �x� 0� = un1�x�� n�x� 0� = �n0�x� (34)

The approximate problems, semidiscrete and completely discretized by finiteelement are defined in the usual way (see Hughes [13]). We consider a partition ofthe interval �0� 1� into P subintervals �xi−1� xi� of length h, where x0 = 0 and xP = 1.Let us denote � = �0� 1�. For h ∈ �∗

+, we consider Vh�� ⊂ C1�� and Eh�� ⊂C0�� finite dimensional subspaces given by

Vh = �vh ∈ V� veh ∈ P3��e�� ⊂ V and Eh = ��h ∈ E��eh ∈ P1��e�� ⊂ E

where veh and �eh are, respectively, the restrictions of vh and �h to the element “e”,

and P1��e� and P3��e� are, respectively, the sets of linear polynomials and cubicalpolynomials of Hermite defined in �e. For time discretization, we use the Eulerimplicit method of finite differences, to approximate the terms un

t , untt and �nt . More

precisely, the approximations of the terms unt , u

ntt and �nt are given by

unt �x� tj� =

unj − un

j−1

�t� un

tt�x� tj� =unj+1 − 2un

j + unj−1

�t2� �nt �x� tj� =

�nj − �nj−1

�t


The variational problems (32)–(34) completely discretized, consists of findingunh�j ∈ Vh and �nh�j ∈ Eh, such that

a

�t2�un

h�j+1� vh�+ ��uxx�nh�j+1� �vxx�h�

= a

�t2�2un

h�j − unh�j−1� vh�+ b��x�

nh�j� �vx�h�+ �n

P�jvP�j

1�t

��nh�j� �h�+ ��x�nh�j� ��x�h�

= 1�t

��nh�j−1� �h�−c

�t��ux�

nh�j − �ux�

nh�j−1� ��x�h�

∀vh ∈ Vh and ∀�h ∈ Eh. With the initial conditions

unh�0�x� = �u0�

nh�x��

unh�1�x�− un

h�0�x�

�t= �u1�

nh�x�� nh�0�x� = ��0�

nh�x�

Remark 5.1. To get computational results, we use the implemented code inFortran 90. The graphics were developed using Maple.

To solve the numerical problem, we use the following resolution algorithm:

Given unh�0� u

nh�1� �

nh�0

For j = 1� � k

�0h�j = �h� j−1� u0h�j+1 = uh�j� u0

h�j = uh�j−1

For n = 1� � Np� find �nh�j ∈ Eh and unh�j+1 ∈ Vh� such that

��nh�j� �h�+ �t��x�nh�j� ��x�h�

= ��nh�j−1� �h�− c��ux�nh�j − �ux�

nh�j−1� ��x�h� ∀�h ∈ Eh�

a�unh�j+1� vh�+ �t2��uxx�

nh�j+1� �vxx�h�

= a�2unh�j − un

h�j−1� vh�+ �t2b��x�nh�j� �vx�h�+ �t2�n

P�jvP�j ∀vh ∈ Vh

�h�j = �Np

h�j� uh�j+1 = uNp

h�j+1� uh�j = uNp

h�j

COMPUTATIONAL RESULTS AND CONCLUSION

The dimensionless constants used in the graphics are given in Table 1.The penalization parameter is given by 1/� = 1× 109 and we taking the stopsin positions g1 = −6× 10−4 and g2 = 6× 10−4. We use a uniform mesh with 50elements and �t = 2× 10−5, and we choose the following initial conditions for uand �

u�x� 0� = 0� ��x� 0� = x2 − 2x3 + x4� ut�x� 0� = −5[1− 2 x2 + 4

3x3 − 1

3x4]

We consider the evolution of the beam’s solution u�x� t�, the beam’soscillations at the end x = 1: u�1� t�, the energy E�t� and the temperature difference��x� t�, in the time T = 0 01 Thus, we have the graphics:


Figure 2 Evolution of the beam’s solution: u�x� t�.

Figure 3 Beam’s oscillations at the end x = 1 � u�1� t�.


Figure 4 (a) Energy: E�t�. (b) Temperature difference: ��x� t�.

We observe in Figures 2 and 3 that the trajectory of the displacement u

in 1, remains constrained by the obstacles in the positions g1 and g2. We noticethat initially u�1� t� touches the obstacles and presents irregular behavior due tothe inertia effect. However, as time goes by, the solution begins to decay andto regularize itself. In Figure 4(a), we see the decay of the energy that, also,presents some irregularities. In Figure 4(b), we see how the temperature differenceis distributed along the beam.

In the next graphics, we consider the beam’s oscillations at the end x = 1:u�1� t� and the energy E�t�, in times T = 0 05 and T = 0 1, respectively.

Figure 5 (a) Beam’s oscillations at the end x = 1 � u�1� t�. (b) Energy: E�t�.


Figure 6 (a) Beam’s oscillations at the end x = 1 � u�1� t�. (b) Energy: E�t�.

We notice in Figures 5 and 6 same behavior of the previous graphics.Moreover, as time increases, we see better the exponential decay of the solution.

Finally, we present a case where the penalization parameter is given by 1/� =1× 107 and analyse the beam’s oscillations at the free end x = 1 � u�1� t�, in the timeT = 0 05.

Figure 7 Beam’s oscillations at the end x = 1 � u�1� t�.


In Figure 7, note that close to zero, these values of u�1� t� in some intervalof time are larger than g2 and in other interval of time are smaller than g1, this isbecause of the normal compliance condition, which considers the elastic obstacles.As � −→ 0, the elastic body goes to a rigid body (see Figure 5(a)); characterizingthe numerical convergence of the penalized problem for the Signorini’s problem.Through the numerical simulations, we observe that the smaller is the value of �,the bigger is the computational cost to guarantee the convergence of method.

REFERENCES

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2. K. T. Andrews, M. Shillor, and S. Wright, On the Dynamic Vibrations of an ElasticBeam in Frictional Contact with a Rigid Obstacle, J. Elast., vol. 42, pp. 1–30, 1996.

3. Y. Dumont, K. L. Kuttler, and M. Shillor, Analysis and Simulations of Vibrations ofa Beam with a Slider, J. Eng. Math., vol. 47, pp. 61–82, 2003.

4. J. Bajkowski, J. R. Fernández, K. L. Kuttler, and M. Shillor, A ThermoviscoelasticBeam Model for Brakes, Euro. J. Appl. Math., vol. 15, pp. 181–202, 2004.

5. F. C. Moon and S. W. Shaw, Chaotic Vibrations of a Beam with Nonlinear BoundaryConditions, J. Non-Linear Mech., vol. 18, pp. 465–477, 1983.

6. K. L. Kuttler and M. Shillor, Vibrations of a Beam Between Two Stops. Dynamics ofContinuous, Discrete Impulsive Sys. Ser. B: Appl.-Algorithms, vol. 8, pp. 93–110, 2001.

7. Y. Dumont and L. Paoli, Vibrations of a Beam Between Stops: Convergence of a FullyDiscretized Approximation, Math. Model. Numer. Anal., vol. 40, pp. 705–734, 2006.

8. Y. Dumont, Vibrations of a Beam Between Stops: Numerical Simulations andComparison of Several Numerical Schemes, Math. Comput. Simul., vol. 60(1–2),pp. 45–83, 2002.

9. A. N. Norris, Dynamics of Thermoelastic Thin Plates: A Comparison of Four Theories,Journal of Thermal Stresses, vol. 29, pp. 169–195, 2006.

10. J. E. Lagnese and J. L. Lions, Modelling Analysis and Control of Thin Plates, CollectionRMA, p. 175, Masson, Paris, 1989.

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exponential decay for a thermoelastic beam ...im.ufrj.br/~rivera/art_pub/tesesant.pdfexponential...

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