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Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 129, pp. 114.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

ftp ejde.math.txstate.edu

STABILIZATION OF LAMINATED BEAMS WITH

INTERFACIAL SLIP

ASSANE LO, NASSER-EDDINE TATAR

Abstract. We study a laminated beam consisting of two identical beamsof uniform thickness, which is modeled as Timoshenko beams. An adhesive

of small thickness is bonding the two layers and creating a restoring forceproducing a damping. It has been shown that the interfacial slip between

the layers alone is not enough to stabilize the system exponentially to itsequilibrium state. Some boundary control has been used in the literature forthat purpose. In this paper, we show that for viscoelastic material there is no

need for any kind of internal or boundary control.

1. Introduction

Many structures in mechanical engineering, electrical engineering, civil engineer-ing and aerospace engineering are formed by a single beam or a number of beams.We can cite for instance, robot arms, rotor turbine and helicopter blades, turbo-machineries, electronic equipment, antennas, missiles, panels, pipelines, buildings,bridges, etc. There are mainly three important theories. The first one is named after

Euler and Bernoulli and the second one after Rayleigh. To alleviate the shortcom-ings in these two theories, Timoshenko came up with a new theory which is bettersuited for engineering practice and is nowadays widely used for moderately thickbeams. Both, rotatory inertia and the effect of shear forces are taken into account.In his theory, Timoshenko also assumed that the plane cross-sections perpendicularto the beam centerline remain plane but could become oblique after deformation.An additional kinematics variable is added in the displacement assumptions. Inter-nal and external forces like the weight of the beam, heavy loads, wind, earthquakesand interaction with other bodies or materials are examples of some sources causinghigh stresses accompanying unwanted vibration. These stresses not only bring somediscomfort, reduce the fatigue-life of the material and produce annoying noise butalso are harmful to the structure as they may cause significant damage or completedestruction of the machine or equipment. Therefore, some ways and devices capa-

ble of enhancing dynamic stability must accompany these structures. To this endvarious devices and energy dissipation mechanisms have been designed either in thematerial itself such as smart materials (piezoelectric, pietzoceramic, viscoelastic),

2010 Mathematics Subject Classification. 34B05, 34D05, 34H05.Key words and phrases. Exponential stabilization; vibration reduction; Timoshenko system;

slip; boundary control; multiplier technique.c2015 Texas State University - San Marcos.

Submitted February 24, 2015. Published May 7, 2015.

1

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2 A. LO, N.-E. TATAR EJDE-2015/129

on its surface (viscoelastic layers, sandwich plates,.. . ) or at the boundary (or partof the boundary). Some well-known dampers are: friction dampers, sensors andactuators, special loads, viscoelastic dampers, tuned mass dampers, tuned liquiddampers and tuned mass liquid dampers. Sometimes they are classified into active,semi-active and passive control methods. In this paper, we would like to investigatethe case of two identical beams with an adhesive layer in the interface creating arestoring force. It has been already shown that when this restoring force is pro-portional to the amount of slip the created frictional damping is unable by itselfto stabilize the system exponentially. The first investigators have been forced tocontrol the system by an additional boundary feedback. We intend to seek otherways and means, preferably less costly, less demanding and easy to implement, tostabilize the system exponentially.

Statement of the problem. The original structure consists of a two-layeredbeam with an adhesive layer bonding the two adjoining surfaces. The adhesive

layer creates a restoring force which is assumed proportional to the amount of slip.Therefore, we are in the presence of a structural damping due to interfacial slip.Moreover, we assume that the adhesive layer is of negligible thickness and mass sothat the contribution of its mass to the kinetic energy of the structure can be ig-nored. The equations of motion modeling the system are derived using Timoshenkotheory and a third equation is coupled with the first two describing the dynamic ofthe slip and containing the internal frictional (Kelvin-Voigt) damping. Namely, wehave the system

wtt+ G( wx)x= 0,I(3stt tt) G( wx) D(3sxx xx) = 0,3Istt+ 3G( wx) + 4s + 4st 3Dsxx= 0,

supplemented by the initial data(w, , s)(x, 0) = (w0, 0, s0), (wt, t, st)(x, 0) = (w1, 1, s1)

and cantilever boundary conditions.Here w,,,G, I , D , , are transverse displacement, rotation angle, density,

shear stiffness, mass moment of inertia, flexural rigidity, adhesive stiffness, adhesivedamping parameter ands is proportional to the amount of slip along the interface.The expression := 3s is the effective rotation angle.

It has been shown in[31]that the frictional damping created by the interfacialslip alone is not enough to stabilize the system exponentially to its equilibriumstate. Therefore, a natural question that can be asked is: what are the possibleadditional damping that can ensure the exponential stability and other kinds of sta-bility of the system? We suggest investigating the case of an additional viscoelastic

damping that acts on the effective rotation angle without resorting to any bound-ary control. Viscoelastic material is very efficient in case there is no considerablechange of frequency or temperature in the structure[2]. The viscoelastic dampingis (according to the Boltzmann Principle) represented by a memory term in theform of a convolution which arises in the constitutive equation between the stressand the strain t

0

h(t r)(3s )xx(r)dr.

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EJDE-2015/129 STABILIZATION OF LAMINATED BEAMS 3

There are basically three main papers in this subject [3, 10, 31]. In [10], theproblem has been derived in details. The authors assumed that the adhesive layeris of negligible thickness and mass and that the restoring force created by this layeris proportional to the amount of slip at the interface.

In[31], the system is studied assuming that

G/and

D/I are two differentwave speeds. Putting = 3s , they transformed the original system into

wtt+ G(3s wx)x= 0,Itt G(3s wx) Dxx= 0,

3Istt+ 3G(3s wx) + 4s + 4st 3Dsxx= 0where 0< x < 1 andt >0. In addition to the well-posedness, the authors pointedout that the frictional damping is enough to asymptotically stabilize the system.However, it is not possible to have exponential stability. They justified their claimby the fact that the eigenvalues of two branches are very close to the imaginaryaxis as their moduli go to infinity. To achieve exponential decay of solutions theyimplemented an additional boundary control

w(0, t) = (0, t) = s(0, t) = 0,

x(1, t) = u1(t) :=k1t(1, t), sx(1, t) = 0,3s(1, t) (1, t) wx(1, t) = u2(t) := k2wt(1, t)

wheret >0. The same system but with the boundary control

(0, t) wx(0, t) = u1(t) :=k1wt(0, t) w(0, t),3sx(1, t) x(1, t) = u2(t) :=k2t(1, t) (1, t),

has been studied in [3]. The authors proved an exponential stabilization result

in case k1=

/G, k2=

I/D and the dominant part of the system is itselfexponentially stable.

For the case of a single viscoelastic Timoshenko beam (therefore without in-terfacial slip) there exist many papers in the literature. We can cite a few ofthem[1, 8, 11, 15, 16, 19,20,21,22,23,24,28,29,30,32,33].

Here, we shall consider the system

wtt+ G( wx)x= 0,

I(3stt tt) G( wx) (3s )xx+ t

0

h(t r)(3s )xx(r)dr= 0,

Istt+ G( wx) +43

s +4

3st sxx= 0,

(1.1)

where 0< x < 1 and t >0, with the boundary conditions

(0, t) = s(0, t) = 0,

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

wx(0, t) = 0, w(1, t) = 0.

(1.2)

The well-posedness of the system has been addressed in[3,31](see [4,5,7,9,17]forthe viscoelastic term). We have weak solutions in (V1

L2)3 and strong solutions

in (V2 H1)3 where

Vk

=

v: vHk(0, 1) : v(0) = 0, k= 1, 2.

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We shall discuss the case where the relaxation functionh : R+ R+ is a boundeddifferentiable function satisfying the standard conditions (as we shall not be con-cerned about finding the largest class of admissible kernels, see [6,12,13,14,18, 25,26, 27, 28, 29, 30] for this matter)

0hh 1h, (1.3)for some positive constants 0 and1. Moreover we assume that

:= 1

0

h(r)dr >0. (1.4)

ForG we shall use the following assumption

(H1) If < 12 , then G < min{, 32, 22

29

9 }, and if 12 < < 9 then

assumeG 0.We first give some lemmas that will serve as a support for the proof of this

theorem.

Lemma 2.2. Ifk and are two differentiable functions then

(k )(t)(t) = 12

(k)(t) +1

2

d

dt

t0

k(s)ds

2(t) (k)(t)

12

k(t) 2(t), t > 0

where

stands for the usual convolution.

Proof. The statement of the this follows from the identity

d

dt(k)(t) = (k)(t) + 2

t0

k(s)ds

t(t)(t) 2(k )(t)t(t)

= (k)(t) + d

dt

t0

k(s)ds

2(t)

k(t) 2(t) 2(k )(t)t(t), t > 0.

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Lemma 2.3. The energyE(t) given by (2.1) satisfies

d

dtE(t) =h(t)

2 (3s )x2 4st2 +1

2

10

(h(3s )x)dx, t >0.

Proof. Multiplying the first equation of (1.1) bywt and integrating over (0, 1) weobtain

2

d

dt

wt2 + G 1

0

( wx)xwtdx= 0or

2

d

dt

wt2 G 1

0

( wx)wxtdx + [G( wx)wt]10 = 0and by our boundary conditions (1.2)

2

d

dt [wt2

] G 1

0 ( wx)wxtdx= 0, t > 0.Note that

G

10

( wx)wxtdx=G 1

0

( wx)( wx )tdx

=G2

d

dt[ wx2] + G

10

( wx)tdx.

Therefore,

1

2

d

dt[wt2 + G wx2] G

10

( wx)tdx= 0, t > 0. (2.2)

Similarly multiplying the second equation of (1.1) by 3st

t and integrating over

(0, 1) we obtain

I2

d

dt[3st t2] G

10

( wx)(3st t)dx

1

0

(3s )xx(3st t)dx + 1

0

(3st t) t

0

h(t r)(3s )xx(r) drdx= 0

or, using integration by parts and the boundary conditions (1.2)

1

2

d

dt

I3st t2 + 3sx x2

G 1

0

( wx)(3st t)dx

1

0

(3st t)x t

0

h(t r)(3s )x(r) drdx= 0, t > 0.(2.3)

By using Lemma2.3 we see that 10

(3st t)x t

0

h(t r)(3s )x(r) drdx

=1

2(h (3s )x)(t) h(t)

2 3sx x2

+1

2

d

dt

t0

h(s)ds3sx x2 (h(3s )x)(t)

, t > 0.

(2.4)

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Likewise, multiplying the third equation of (1.1) byst and integrating over (0, 1),we obtain

12

ddt

Ist2 +4

3s2 + sx2 + G

1

0( wx)stdx +4

3st2 = 0, (2.5)

fort >0. Now it is clear from (2.2)(2.5) that

E(t) =4st2 h(t)2

(3s )x2 +12

10

(h(3s )x)dx, t >0.

This completes the proof.

As h(t) 0, we see that E(t) 0 for all t > 0. Therefore the energy isnon-increasing and uniformly bounded above by E(0).

Next we shall construct a Lyapunov functional Fsatisfying the inequalities

1E(t)F(t)2E(t) and ddt

F(t) F(t)for some positive constants 1, 2 and. The first two inequalities show thatE(t)and F(t) are equivalent. The second one gives the exponential decay ofF(t) (andtherefore the exponential decay ofE(t) as well). To this end, we define

F(t) = E(t) +5

i=1iGi(t), i> 0, i= 1, . . . , 5, t0,

where

G1(t) = I(st, s), G2(t) =(wt, w), G3(t) = I(3st t, 3s ), t0,G4(t) =4

G (wt, ) 3

G(sx, wt) + 3I(st, wx), t0,

with (x, t) =1x

s(, t)dand

G5(t) =I

3st t, t

0

h(t r)[(3s )(t) (3s )(r)] dr

, t0.

Using the Cauchy-Schwarz inequality and the Poincare inequality, one can easilysee that all the Gi(t),i = 1, . . . , 5 are bounded (above and below) by an expressioncontaining the existing terms in the energy E(t). This leads to the equivalence ofF(t) and E(t).

We shall now prove several lemmas with the purpose of creating negative coun-terparts of the terms that appear in the energy in the estimations of the derivativesof the above functionals.

Lemma 2.4. Along the solutions of (1.1)(1.2), we have

G1(t)

sx

2 +

G

40+

4

3

s

2 + 0G

wx

2 + I+

42

9

st

2,

for allt >0 and some0, > 0.

Proof. Clearly,

G1(t) = Ist2 + I(stt, s), t > 0and by the third equation in (1.1) we obtain that for t >0,

G1(t) = Ist2 sx2 4

3s2 4

3 (st, s) G( wx, s)

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Ist2 sx2 + G

40 4

3

s2 + 0G wx2 + s2 +429

st2

sx2 + G40 + 43s2 + 0G wx2 + I+ 4

2

9st2.

Lemma 2.5. The derivative ofG2(t) along solutions of (1.1)(1.2) satisfies

G2(t) wt2 + (G + 1) wx2 + G

21x 3sx2 + 9G

21sx2,

for allt >0 and some1 > 0.

Proof. Using the first equation in(1.1) and the boundary conditions(1.2), we havethat for t >0,

G2(t) =wt2 (wtt, w)=wt2 + G(( wx)x, w)=wt2 G( wx, wx) + G [( wx)w]10=wt2 + G( wx, wx) G( wx, )

wt2 + G wx2 + 1G wx2 + G41

x2

wt2 + (G + 1) wx2 + G21

x 3sx2 + 9G21

sx2.

Lemma 2.6. The derivative ofG3(t) along solutions of (1.1)(1.2) satisfies

G

3(t)

I

3st

t

2

(

G

42 )

3sx

x

2 + 2G

wx

2

+1

4

10

(h(3sx x))dx, t >0

for2 > 0, > 0.

Proof. Using the second equation in (1.1) we find that

Id

dt(3st t, 3s ) = I3st t2 3sx x2

+ [(3sx x)(3s )]10+ G(( wx), (3s ))

+ t

0

h(t r)(3sx x)(r)dr, 3sx x

, t > 0.

Then

G3(t)I3st t2 3sx x2 + 2G wx2 + G

423sx x2

+ t

0

h(t r)[(3sx x)(r) (3sx x)(t)] dr, 3sx x

+ t

0

h(r)dr

((3sx x, 3sx x)

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for2 > 0, or

G3(t)I3st t2 3sx x2 + 2G wx2

+ G

423sx x2 + 3sx x2 +1

4

1

0

(h(3sx x))dx

+ (1 )(3sx x)2,for >0. Hence

G3(t)I3st t2 ( G

42 )3sx x2 + 2G wx2

+1

4

10

(h(3sx x))dx, t >0.

Lemma 2.7. The derivative ofG4(t) is estimated as follows

G

4(t) (3G 1) wx2

+ 1(1 + )I3st t2

+ 1wt2

+422

1G2 +

42

1+ (9 +

1

+

9

41)I

st2, t > 0,for1, >0 provided thatI =

G

.

Proof. Using the first and third equations in (1.1),

G4(t) =4

G (wtt, ) 4

G (wt, t) 3

G(sxt, wt) 3

G(sx, wtt)

+ 3I(stt, wx) + 3I(st, t wxt) .Then we find that

G4(t) = 4(( wx)x, ) 4

G (wt, t) 3

G(sxt, wt) + 3(sx, ( wx)x)

+ 3(G( wx) 43

s 43

st+ sxx, wx) + 3I(st, t wxt),fort >0. Next, by the definition of and the assumption I=

G

, we obtain

G4(t) =4

G (wt, t) 3G wx2 4(st, wx) + 3I(st, t),

fort >0. Now, clearly

4

G (wt, t)1wt2 +4

22

1G2st2,

4(st, wx)1 wx2 +42

1st2,

3(st, t)

1

t

2 +

9

41

st

2

1(1 + )3st t2 + (9 +1

+ 9

41)st2

lead to

G4(t)1wt2 +422

1G2st2 3G wx2 + 1 wx2 +4

2

1st2

+ 1(1 + )I3st t2 + (9 +1

+ 9

41)Ist2

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or

G4(t)

(3G

1)

wx

2 + 1(1 + )I

3st

t

2 + 1

wt

2

+422

1G2 +

42

1+

9 +1

+

9

41

Ist2, t0.

For the next lemma we need to get away from zero to ensure strict positivity oft0

h(r)dr. So for that tt0 > 0 we havet

0h(r)dr t00 h(r)dr= h0 > 0.

Lemma 2.8. For the functionalG5(t) we have

G5(t)G wx2 + (G + 4 + 2 )1

4

10

(h (3s )x)dx

+ (2 )3sx x2 + I( h0)3st t2

+ Ih(0)4 1

0(|h|(3s )x)dx, tt0> 0

for >0.

Proof. We recall that

G5(t) =I

3st t, t

0

h(t r)[(3s )(t) (3s )(r)]dr

, t > 0

and therefore

G5(t) =I(3stt tt, t

0

h(t r)[(3s )(t) (3s )(r)]dr)

I3st

t, t

0

h(t

r)[(3s

)(t)

(3s

)(r)]dr

I t

0

h(r)dr3st t2, t > 0.

In view of the second equation in (1.1)and the boundary conditions (1.2) we write

G5(t) =

G( wx) + (3s )xx, t

0

h(t r)[(3s )(t) (3s )(r)]dr

+ t

0

h(t r)(3s )xx(r)dr, t

0

h(t r)[(3s )(t) (3s )(r)]dr

I

3st t, t

0

h(t r)[(3s )(t) (3s )(r)]dr

I t

0 h(r)dr3st t

2

, t > 0.(2.6)

It is easy to see that for t >0,

G

wx, t

0

h(t r)[(3s )(t) (3s )(r)] dr

G wx2 + G(1 )4

10

(h (3s )x)dx,(2.7)

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(3s )xx,

t0

h(t r)[(3s )(t) (3s )(r)] dr

=(3s )x, t

0h(t r)[(3s )x(t) (3s )x(r)] dr

3sx x2 +1 4

10

(h(3s )x)dx,

(2.8)

and t0

h(t r)(3s )xx(r)dr, t

0

h(t r)[(3s )(t) (3s )(r)]dr

= t

0

h(t r)[(3s )x(t) (3s )x(r)]dr2

t

0

h(r)dr

(3s )x, t

0

h(t r)[(3s )x(t) (3s )x(r)]dr

t

0

h(t r)[(3s )x(t) (3s )x(r)]dr2 + (1 )

3sx x2

+ 1

4 t

0

h(t r)[(3s )x(t) (3s )x(r)]dr2

(1 + 1 4

)(1 ) 1

0

(h (3s )x)dx + (1 )3sx x2,

(2.9)

fort >0. Further

I(3st t, t

0

h(t r)[(3s )(t) (3s )(r)]dr)

I

3st

t

2 +

Ih(0)

4

1

0

(

|h

|(3s

)x)dx, t >0.

(2.10)

Taking into account estimates (2.7)(2.10), in (2.6) and considering tt0 > 0, weobtain

G5(t)G wx2 +G(1 )

4

10

(h (3s )x)dx + 3sx x2

+1

4

10

(h(3s )x)dx + (1 +1 4

)(1 ) 1

0

(h (3s )x)dx

+ (1 )3sx x2 + I3st t2

+Ih(0)

4

10

(|h|(3s )x)dx Ih03st t2

or, for t

t0 > 0

G5(t)G wx2 + (G + 4 + 2 )1

4

10

(h (3s )x)dx

+ (2 )3sx x2 + I( h0)3st t2

+Ih(0)

4

10

(|h|(3s )x)dx.

The proof is complete.

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Using the previous lemmas we now give the proof of our main result.

Proof of Theorem2.1. Gathering the estimates in the previous lemmas we find that

F(t) = E(t) +5

i=1iG

i(t) 4st2 h(t)

2 (3s )x2

+1

2

10

(h(3s )x)dx 1sx2 + 1( G40

+ 43

)s2

+ 10G wx2 + 1(I+ 42

9 )st2 2wt2

+ 2(G + 1) wx2 + G221

3sx x2 +9G221

sx2

+ 3I3st t2 3( G42

)3sx x2 + 32G wx2

+ 3 1 4

1

0

(h(3sx x))dx 4(3G 1) wx2

+ 41(1 + )I3st t2 + 4[ 422

2G2 +

42

1+ (9 +

1

+

9

41)I]st2

+ 14wt2 + 5G wx2 + 5(2 )3sx x2

+ 5(G + 4 + 2 ) 1 4

10

(h (3s )x)dx

+ 5I( h0)3st t2 + 5 Ih(0)4

10

(|h|(3s )x)dx, tt0 > 0

or

F(t)

4 1(I+ 429

) 4

(9 +1

+ 941

)I+42

1+4

22

2G2

st2

(19G221

)sx2 + 1( G40

+ 43

)s2

[4(3G 1) 10G 2(G + 1) 32G 5G] wx2

3( G42

) G221

5(2 )3sx x2

(2 24)wt2 + I[3+ 41(1 + ) + 5( h0)] 3st t2

1

2 3 1

4 5(G + 4 + 2 ) 1

4

5

0Ih(0)

4

1

0

(h (3s

)x

)dx.

(2.11)Our strategy for selecting the different coefficients and parameters is as follows:

all thei,i = 1, . . . 5 will be determined in terms of only one of them (here 1). This1will be accountable in front of and1 in the coefficients of the first and the lastterm in (2.11). From the beginning, we have managed in our estimations to balancethe largest coefficients (here 1/) on the terms that appear in the derivative of theenergy. This will allow us to ignore at the beginning of the process of selection.

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Let us ignore for the moment the first and the last terms in (2.11). We shall, atthe same time, ignore the terms having coefficients in . The focus will be on

1 9G21

2 > 0, G40

43

0,

3( G42

) G21

2 > 0,

2 24 > 0, 3+ 41 5h0 < 0,or

9G

212 < 1,

G

40

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