Exact boundary controllability results for a Rao–Nakra sandwich beam

Systems & Control Letters 56 (2007) 558 – 567 www.elsevier.com/locate/sysconle

Exact boundary controllability results for a Rao–Nakra sandwich beam Rajeev Rajaram ∗ Department of Computer Science, Mathematics and Engineering, Stutzman Slonaker Hall, Shepherd University, Shepherdstown, WV 25443, USA Received 4 July 2006; received in revised form 22 March 2007; accepted 24 March 2007 Available online 10 May 2007

Abstract The Rao–Nakra model of a three layer sandwich beam is analyzed for exact boundary controllability. The damped and undamped cases are considered. The multiplier method is used to obtain required observability inequalities that imply controllability. It is shown that if the control time T is large enough, under certain other parametric restrictions, the system is exactly controllable. © 2007 Elsevier B.V. All rights reserved. Keywords: Exact controllability; Null controllability; Distributed parameter systems; Composite materials

1. Introduction Control theory of layered beam and plate models has gained considerable importance in the recent past due to various industrial applications. Several multi-layered sandwich beam models have been derived in the past (see e.g. [1,3,18,20,21,24], and references therein), but controllability results for those models appear to be scant. A generalization to multi-layered plates is seen in [5]. The moment method has been used to obtain boundary controllability results for the Mead–Markus [17,18] and Rao–Nakra [20] sandwich beam models in [8–10,19]. A boundary stabilization result for a closely related laminated beam can be found in [22]. In this paper, the Rao–Nakra sandwich beam model with boundary forces affecting the shear angle and longitudinal motion is analyzed for exact controllability. The multiplier method is used in conjunction with the Hilbert uniqueness method (HUM) principle (see [12,13,15,16]) to prove controllability results. The classical Rao–Nakra sandwich beam model consists of two outer “face plate” layers (which are assumed to be relatively stiff) which “sandwich” a much more compliant “core layer”. The Rao–Nakra model is derived using Euler–Bernoulli beam assumptions for the face plate layers and a “no-slip” ∗ Tel.: +1 304 876 5407.

E-mail address: [email protected]. 0167-6911/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.sysconle.2007.03.007

assumption for the displacements along the interface. The outer layers include the longitudinal and rotational inertia terms whereas the core layer includes shear stresses but neglects bending stresses. The equations of motion as described in [5] are given below: ˜ 2 xt ) = 0, mw tt − wxxtt + Kw xxxx − N h2 (G2 x + G

(1)

1 ˜ 2 t ) = 0, − (G2  + G h1 1 vtt1 − h1 E1 vxx

(2)

3 ˜ 2 t ) = 0, h3 3 vtt3 − h3 E3 vxx + (G2  + G

(3)

1 3 ( = h−1 2 (−v + v ) + N w x ).

Boundary conditions: w(0, t) = wx (0, t) = w(1, t) = v 1 (0, t) = v 3 (0, t) = 0,

(4)

wx (1, t) = M(t),

(5)

v 1 (1, t) = g1 (t),

v 3 (1, t) = g3 (t).

Initial conditions: w(x, 0) = w 0 ; v 3 (x, 0) = v 30 ;

wt (x, 0) = w 1 ; vt1 (x, 0) = v 11 ;

v 1 (x, 0) = v 10 , vt3 (x, 0) = v 31 .

(6) (7)

In the equations above, w denotes the transverse displacement of the beam,  denotes the shear angle of the core layer, (v 1 , v 3 )T is the vector of longitudinal displacement along the neutral axis of the outer layers. (i = 1, 3 is for the outer layers, i =2 is for the core layer.) The density of the ith layer is denoted

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567

i , the thickness hi , Young’s modulus Ei , the shear modulus of the core layer is G2 , and the corresponding shear damping  ˜ 2 . We let m = 3i=1 hi i denote the modulus is denoted by G mass density per unit length,  = 1 h31 /12 + 3 h33 /12 denote a moment of inertia parameter, and K = E1 h31 /12 + E3 h33 /12 denote the bending stiffness. The shear angle and the longitudinal displacement of the outer layers at the right end point are controlled through M(t), g1 (t), and g3 (t). The following is the main exact controllability theorem in this paper: ˜ 2 , ∃T > 0 such Theorem 1. For sufficiently small G2 , G that system (1)–(7) is exactly controllable in the space C = H01 (0, 1) × L2 (0, 1) × L2 (0, 1) × L2 (0, 1) × H −1 (0, 1) × H −1 (0, 1). More precisely, for any pair ((w 0 , v 10 , v 30 , w1 , v 11 , v 31 ), (wT0 , vT10 , vT30 , wT1 , vT11 , vT31 )) ∈ C

559

˜ 2 , T such that the where C > 0. Thus, any values of G2 , G constant multiplying E(0) in Eq. (10) is positive gives Eq. (12). ˜ 2 sufficiently small, ∃T > 0 which We note that for G2 and G makes this constant positive. The paper is organized as follows. In Section 2, the semigroup formulation of system (1)–(3) is discussed. In Section 3.1, energy conservation/dissipation is discussed. In Section 3.2, a trace regularity result is derived that allows us to define solutions to the controlled problem (1)–(7), and in Section 3.3 the proof of Theorem (3) is given. 2. Semigroup formulation The homogeneous problem with zero controls is given by (1)–(3) with homogeneous boundary conditions

there exist controls M(t), g1 (t), g3 (t) in L2 [0, T ] such that the solution of (1)–(7) satisfies the following terminal conditions:

w(0, t) = wx (0, t) = w(1, t) = v 1 (0, t) = v 3 (0, t) = 0,

(13)

w(x, T ) = wT0 ,

(8)

wx (1, t) = 0,

(14)

(9)

and initial conditions given by (6)–(7). Let Y =(w1 , w2 , w3 , w4 , w5 , w6 )T =(w, v 1 , v 3 , wt , vt1 , vt3 )T . Let D = j/jx and L = mI − D 2 . Then L : H01 (0, 1) → H −1 (0, 1) is an isomorphism. Eqs. (1)–(3), (13)–(14) and (6)–(7) can be rewritten as follows:

wt (x, T ) = wT1 ,

v 1 (x, T ) = vT10 , vt1 (x, T ) = vT11 ,

v 3 (x, T ) = vT30 ,

vt3 (x, T ) = vT31 .

Remark 2. Solutions of (1)–(7) are defined in the sense of transposition (see Section 3.2 Definition 10). Theorem 1 follows from the HUM principle (see e.g. [13,15,16]) once the following observability estimate is proved: Theorem 3. Let T > 0. Then every smooth solution of (1)–(3) with homogeneous spatial boundary conditions satisfies the following observability estimate:  T K 2 h 1 E1 1 h 3 E3 3 wxx (1, t) + (vx (1, t))2 + (vx (1, t))2 2 2 2 0    ˜ 2) 4C1 (G2 + G Fm −  T 2(FM + 2G2 C1 ) Fm   ˜ 2T C1 24 (FM + 2G2 C1 ) 2G − 1− E(0), (10) Fm Fm

dY = AY, dt

FM = max(, h1 1 , h3 3 , h1 E1 , h3 E3 , m, K),    2 1 max N(N h2 + 1), 2 N + . C1 = 2 h2

⎛ ⎜ ⎜ ⎜ A˜ 2 = ⎜ ⎜ ⎝ 

(15)

L−1 (N 2 h2 D 2 )

−L−1 (N D)

N D h1 1

1 h2 h1  1

−

N D h 3 3

03×3

I3×3

A˜ 1

˜ 2 A˜ 2 G

−

1 h 2 h3  3

L−1 (N D) −

1 h2 h1 1

−

1 h2 h3 3

⎞ ⎟ ⎟ ⎟ ⎟, ⎟ ⎠

 ,

where (11)

D(A) = Hw3 × H∗2 × H∗2 × H02 × H01 (0, 1) × H01 (0, 1) and

Remark 4. Actually, one needs to show the following:  T K 2 h 1 E1 1 h 3 E3 3 wxx (1, t) + (vx (1, t))2 + (vx (1, t))2 2 2 2 0 CE(0),

Y (0) = (w 0 , v 10 , v 30 , w1 , v 11 , v 31 )T ,

⎛L−1 (−KD 4 +N 2 h G D 2 ) L−1 (−N G D) L−1 (N G2 D) ⎞ 2 2 2 ⎜ ⎟ E1 2 G2 N G2 G2 ⎜ ⎟ D D − ⎜ ⎟ A˜ 1 = ⎜ ⎟, h1 1 1 h2 h1 1 h2 h1 1 ⎜ ⎟ ⎝ ⎠ N G2 G2 G2 E3 2 − D D − h3 3 h2 h3 3 3 h2 h3 3

A= and

v 3 (1, t) = 0,

where

where Fm = min(, h1 1 , h3 3 , h1 E1 , h3 E3 , m, K),

v 1 (1, t) = 0,

Hw3 = H 3 (0, 1) ∩ H02 (0, 1), (12)

H∗2 = { ∈ H 2 (0, 1) : (0) = (1) = 0},

560

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567

H = H02 (0, 1) × H01 (0, 1) × H01 (0, 1) × H01 (0, 1) × L2 (0, 1) × L2 (0, 1).

we show the upper bound. We need the following calculation:  1 1 3 2 G2 h2 (h−1 2 (−v + v ) + N w x ) dx 0

In the above, A : D(A) → H. We will refer to the Hilbert space (H, ., .H ) as the energy space, where the energy inner product is given by the following: 3

, wt , vt1 , vt3 ), (w, ˆ vˆ 1 , vˆ 3 , wˆ t , vˆ 1t , vˆ 3t )e

0

+ Kw xx wˆ xx + h1 E1 vx1 vˆ 1x + h3 E3 vx3 vˆ 3x

0

Consequently, we have the following expression for the energy norm: E(t) = (w, v 1 , v 3 , wt , vt1 , vt3 ) 2H  1 2 = mw 2t + wxt + h1 1 (vt1 )2 + h3 3 (vt3 )2 

1 + G2 h2 (h−1 2 (−v

1 0

h3 E3 (vx3 )2

+ v ) + N w x ) dx. 3

2

+ vx3 2L2 (0,1) .

(16)

 ˆ defines a norm on the energy space H at time Clearly, E(t) t. The following theorem shows that E(t) also defines a norm ˆ which moreover is equivalent to E(t). ˆ Theorem 5. The energies E(t) and E(t) are equivalent, i.e. ˆ ˆ Fm E(t)E(t)(F M + 2G2 C1 )E(t),

(18)

where the constants Fm , FM , and C1 are defined in (11). Proof. The lower bound is obvious since

0

1 h2

0



1

((v 1 )2 + (v 3 )2 ) dx (N G2 2 )

0

 + 2G2 

2

1 N+ h2



× ( vx1 2L2 (0,1) + vx3 |2L2 (0,1) ).

(17)

+ v ) + N w x ) dx 0. 3

2 2 + N G2 (2 wxx + 2((v 1 )2 + (v 3 )2 )) + N 2 G2 h2 2 wxx  1 2 = (N G2 2 )(N h2 + 1) wxx dx

× (N h2 + 1) wxx 2L2 (0,1)

+ vt1 2L2 (0,1) + vx1 2L2 (0,1) + vt3 2L2 (0,1)

1 G2 h2 (h−1 2 (−v

h2

 ((v 1 )2 + (v 3 )2 )

+ 2G2 N +

ˆ = wxx 2 2 E(t) + wxt 2L2 (0,1) + wt 2L2 (0,1) L (0,1)

1

2



ˆ We define a new energy E(t) in the following way:



G2 (−v 1 + v 3 )2 + 2N G2 (−v 1 + v 3 )wx h2

1  2G



× (h−1 ˆ x ) dx. 2 (−vˆ 1 + vˆ 3 ) + N w

+ Kw 2xx + h1 E1 (vx1 )2 +

0



1 3 + G2 h2 (h−1 2 (−v + v ) + N w x )

0

1

+ N 2 G2 h2 wx2 ,  1 G2 (−v 1 + v 3 )2 + 2N G2 (−v 1 + v 3 )wx h 2 0  2 2 +N G2 h2 wx dx

(w, v , v  1 = mw t wˆ t + wxt wˆ xt + h1 1 vt1 vˆ 1t + h3 3 vt3 vˆ 3t 1

 =

2

ˆ except for The remaining terms in E(t) resemble those of E(t) ˆ multiplicative constants. Thus, we have E(t) Fm E(t). Next

Once again, the rest of the terms in E(t) along with the right ˆ hand side in the above estimate resembles those in E(t) except for multiplicative constants. Thus, by choosing the maximum of those constants, we get the upper bound. This proves the theorem.  Theorem 6. A : D(A) → H is the infinitesimal generator of a strongly continuous semigroup on H. Proof. It can be shown that A∗ = −A in the undamped case ˜ 2 ) denote the dependence of A on G ˜ 2 in the ˜ 2 =0). Let A(G (G ˜ 2 ))∗ =−A(−G ˜ 2) damped case. Then it can be shown that (A(G with D(A∗ ) = D(A). With this, it is not difficult to prove that A and A∗ are dissipative; i.e. Re(AY, Y e ) 0 and Re(A∗ Y, Y e ) 0 using explicit calculations. Then the theorem follows by the application of Lumer–Phillip’s theorem [2].  3. Regularity and observability estimates 3.1. Energy dissipation/conservation We multiply Eq. (1) by wt , Eq. (2) by vt1 and Eq. (3) by vt3 and integrate by parts in space and time to obtain the following energy identity: ˜ 2 h2  1 G

(t )2 dx where E (t) = − 2 0

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567 1 3  = (h−1 2 (−v + v ) + N w x ),

(19)

˜ 2 > 0 and is conserved which implies that energy decays when G ˜ when G2 = 0. Mainly, we are interested in the damped case here. Integrating Eq. (19), we have ˜ 2 h2  T  1 G E(T ) = E(0) − (t )2 dx dt. 2 0 0

Proof. We derive the following identity that holds for smooth solutions of the homogeneous problem for T > 0: 

h 1 E1 2

=

+ h2 N ˜ 2 C2 G  h2



T 0

+ N w 2xt

ˆ dt E(t)

+ N (vt1 )2

where

T

m 2

+

˜ 2 C1  T ˜ 2T C1 G G  E(t) dt  E(0), Fm 0 Fm

T

0

(vx3 (1, t))2

(wxx (1, t))2 T





T

 0

 2

+

1 0

wt2 +



0

1 0



h 3 3 + 2

C2 = max(1, h22 N 2 , h2 N )



0

−

+ N (vt3 )2 ) dx dt

h 3 E3 2

(vx1 (1, t))2 +



K 2

˜ 2  T  1 −1 1 2 G 3 2  (h (vt ) + h−1 2 (vt ) 2 0 0 2 2 wxt

T

0

+

We also have       T 1 G ˜ 2 h2 −1   1 3 2 (h2 (−vt + vt ) + N w xt ) dx dt  −  0 0  2

2

561

T

2 wxt +



0



h 1 1 2



0

h 1 E1 2  T



1 0

T



0 1

1 0

T

3K 2

1 0



T

0

N 2h

(vt3 )2



1 0

2 G2

(vx1 )2 −

G2 2h2

˜2 − NG

Theorem 7. The semigroup generated by A extends to a strongly continuous group on H.

˜2  T  1 G (−vt1 + vt3 )xv 1x h2 0 0 ˜2  T  1 G − (−vt1 + vt3 )xv 3x h2 0 0  T 1 ˜ − N G2 xv 1x wxt

wx2

0



T



0

1 0

(vx3 )2



T



0

1

(−v 1 + v 3 )2

0

˜ 2 (xw xx wxt + wx wxt ) N 2 h2 G

0



T



0



1 0

T



0

1 0

(xw xx vt1 + wx vt1 ) (xw xx vt3 + wx vt3 )

−

Proof. Eq. (20) implies the bounded invertibility of the semigroup T(t)∀t > 0 generated by A. With this, Theorem 7 is easily proved using the standard (positive time) semigroup properties.

0

3.2. Regularity

Theorem 8. Let T > 0. Then ∃C > 0 such that every smooth solution of (1)–(3) satisfies the following trace regularity estimate:   h1 E1 T 1 h 3 E3 T 3 2 (vx (1, t)) + (vx (1, t))2 2 2 0 0  K T 2 + wxx (1, t) CE(0). (21) 2 0

1

(vt1 )2

˜ 2 needs to be small enough for Eq. (20) to be useful. where G

In order to define solutions of the non-homogeneous problem we will need the following trace regularity result. (We omit writing differentials dx and dt inside the integral whenever the context is clear.)



T

0

h 3 E3 + 2

+

˜2 + NG



2

where in the last step we have made use of the fact that E(t) E(0), ∀t > 0 and C2 h2 C1 . Hence, we have that   ˜ 2T C1 G E(t) E(T )  1 − E(0) ∀t ∈ (0, T ], (20) Fm

0

2 wxx

˜2 + NG



0

T 0



1 0

xv 3x wxt + X,

(22)

where 

1

X= 0

(mw t xw x + h1 1 vt1 xv 1x + h3 3 vt3 xv 3x

− wxxt xw x )|t=T t=0 .

(23)

The identity is obtained by multiplying (1) by xw x , (2) by xv 1x and (3) by xv 3x and integrating by parts in (0, T ) × (0, 1). For brevity we defer the calculation to Appendix A. Next, we obtain a bound on |X| which is defined in Eq. (23). We first

562

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567

where in Eq. (24) we have used the fact that E(t) E(0)∀t > 0 and D 2 FM . Next, we have the following:

look at 

1

Z(t) =

mw t xw x + h1 1 vt1 xv 1x + h3 3 vt3 xv 3x

0



1

−

m 2

wxxt xw x ,

1 0

+

0

1

 0



+

0

1

0



1



1

0 1

0

1 0

 h3 3 |vt3 xv 3x |  1

1 0

h1 1 (|vt1 |2

0

1

wxxt xw x = − 

0



0 1

= −



1 0



(vt1 )2 +

1

2 wxx +

h 1 E1 2

 0

T



1 0

(vx1 )2

2T (G2 C1 + FM ) E(0). Fm

(25)

   1   1 1  N G ˜2 (xw xx vt + wx vt ) 

+ |vx1 |2 ),

0

˜ 2 ((1 + 2 ) wxx 2 2 N G + vt1 2L2 (0,1) ), L (0,1)    1   3 3  −N G ˜2 (xw xx vt + wx vt ) 

wxt (xw x )x

0

˜ 2 ((1 + 2 ) wxx 2 2 N G + vt3 2L2 (0,1) ), L (0,1) wxt (wx + xw xx ).

   1   G  ˜2 1 3 1 (−vt + vt )xv x  −   h2 0

0

We also have by Poincare’s inequality on (0, 1) that wx 2L2 (0,1) 2 wxx 2L2 (0,1) . Hence we have   1      w xw xxt x = 

   − 

0

1





1 0

2 2 (wxx + wxt ).



˜2 2G ( vt1 2L2 (0,1) + vt3 2L2 (0,1) + vx3 2L2 (0,1) ), h2

   −N G ˜2 

Using all of the above, we have that ˆ  D E(t) where |Z(t)|D E(t) Fm    D = max m2 , h1 1 , h3 3 , (2 + 1) . 2

   N G ˜2 

Hence we have 2D 22 FM E(t) E(0), Fm Fm

˜2 2G ( vt1 2L2 (0,1) + vt3 2L2 (0,1) + vx1 2L2 (0,1) ), h2

   1 G   ˜2  (−vt1 + vt3 )xv 3x    h2 0 

  wxt (wx + xw xx )

  (2 + 1) 2

|X| 

0

T

T

˜ 2 ((1 + 2 ) wxx 2 2 N 2 h2 G + wxt 2L2 (0,1) ), L (0,1)

h3 3 (|vt3 |2 + |vx3 |2 ), 

0

0



h 1 1 2

T



   1   2 N h2 G ˜2 (xw xx wxt + wx wxt  

m(|wt |2 + |wx |2 ),

 h1 1 |vt1 xv 1x | 





3K 2

0

1

m|wt xwx | 



0

  N 2 h2 G 2 T 1 2 wx 2 0 0 0 0  1   h 3 E3 T 1 3 2 3 2 (vt ) + (vx ) 2 0 0 0

wt2 +

We also have the following:

  wxxt xwx 

0 1



h1 1 |vt1 xv 1x |

   3 3 h3 3 |vt xv x | +  

0

1

ˆ  2T (G2 C1 + FM )E(0) 1

m|wt xw x | +





0

mw t xw x + h1 1 vt1 xv 1x + h3 3 vt3 xv 3x

  1  − wxxt xw x  , 

T

h 3 3 + 2

0

  |Z(t)| = 



(24)

1 0

1 0

 

xv 1x wxt   

xv 3x wxt 

˜ 2 ( vx1 2 2 N G + wxt 2L2 (0,1) , L (0,1)

˜ 2 ( vx3 2 2 N G + wxt 2L2 (0,1) ). L (0,1)

(26)

We can combine all the estimates above along with the ˆ E(t)/Fm (from Theorem 5, Eq. (18)). Affact that E(t) T 1 2 ter dropping the non-positive terms −(/2) 0 0 wxt , and T 1 −(G2 /2h2 ) 0 0 (−v 1 + v 3 )2 , and combining the estimates

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567

by v 3 and formally integrate by parts to obtain the following identity:  1 ((mI − D 2 )w(T )z1 − (mI − D 2 )wt (T )z0 )

(25) and (26), we get the following regularity estimate: h1 E1 2



T

0

K + 2



T

h 2 E3 2

 0

T

(vx3 (1, t))2

0

+ h1 1 (v 1 (T )y 11 − vt1 (T )y 10 ) + h3 3 (v 3 (T )y 31  1 3 30 ˜ − vt (T )y ) − G2 h2 (T )(T ˜ )

(wxx (1, t))2

0

 

(vx1 (1, t))2 +

˜ 2 ) + FM ) + 22 FM 2T (C1 (G2 + 2G Fm

CE(0),  where C =

 E(t)

0



1

=

((mI − D 2 )w 0 zt (0) − (mI − D 2 )w 1 z(0))

0

˜ 2 ) + FM ) + 22 FM 4T (C1 (G2 + G Fm

This proves Theorem 8.

 .

(27)

0



˜ 2 ˜x − G ˜ xt ) = 0, mztt − zxxtt + Kzxxxx − N h2 (G2 

(28) (29)

3 ˜ 2 + (G2  ˜ −G ˜ t ) = 0, h3 3 ytt3 − h3 E3 yxx

(30)

1 3 ( ˜ = h−1 2 (−y + y ) + N zx ).

Boundary conditions: z(0, t) = zx (0, t) = z(1, t) = y 1 (0, t) = y 3 (0, t) = 0,

(31)

zx (1, t) = 0,

(32)

y 1 (1, t) = 0,

y 3 (1, t) = 0,

T

+ 0

Kzxx (1, t)M(t) + h1 E1 yx1 (1, t)g1 (t)

+ h3 E3 yx3 (1, t)g3 (t).

(35)

Definition 10. (w, v 1 , v 3 , wt , vt1 , v13 ) is a transpositional solution of (1)–(7) if (w, v 1 , v 3 , wt , vt1 , v13 ) ∈ C∀(z0 , y 10 , y 30 , z1 , yt11 , y 31 ) ∈ H,

˜ 2 − (G2  ˜ −G ˜ t ) = 0,

1 − h1 E1 yxx

+ h1 1 (v 10 yt1 (0) − v 11 y 1 (0)) + h3 3 (v 30 yt3 (0)  1 31 3 ˜ (0)(0) ˜ − v y (0)) − G2 h2



The solution of the controlled problem is defined by the method of transposition. Let (z, y 1 , y 3 , zt , yt1 , yt3 ) ∈ H be the solution of the homogeneous backward problem described below:

h1 1 ytt1

563

and the (35) is satisfied ∀T > 0. Using the trace regularity result (Theorem 8 applied to system (28)–(34)) and the regularity of the adjoint semigroup (Theorem 9) one sees that the right hand side of Eq. (35) defines a bounded linear functional on H. Hence, the left hand side of Eq. (35) defines through a Reisz isomorphism, variables ((mI − D 2 )w, v 1 , v 3 , (mI − D 2 )wt , vt1 , v13 ) ∈ H dual to (z, y 1 , y 3 , zt , yt1 , yt3 ) ∈ H ∀T > 0. Thus, we have (w, v 1 , v 3 , wt , vt1 , v13 ) ∈ C

∀T > 0.

3.3. Observability Terminal conditions: z(x, T ) = z0 ; yt1 (x, T ) = y 11 ;

zt (x, T ) = z1 ;

y 1 (x, T ) = y 10 ,

y 3 (x, T ) = y 30 ;

yt3 (x, T ) = y 31 .

(33) (34)

The adjoint semigroup (generated by A∗ ) inherits the same regularity properties as the original semigroup. Hence semigroup solutions of system (28)–(34) exist in H and we have the following theorem: Theorem 9. Given (z0 , y 10 , y 30 , z1 , y 11 , y 31 ) ∈ H, there exists a unique solution Y = (z, y 1 , y 3 , zt , yt1 , yt3 ), where Y ∈ C(R; H). In order to define a solution for the controlled problem (1)–(7), we multiply Eq. (28) by w, Eq. (29) by v 1 and Eq. (30)

In this section, we prove the main result of this paper (Theorem 3). Proof of Theorem 3. We first obtain another identity which is useful towards obtaining the observability inequality in Theorem 3. Let T > 0. Then every smooth solution of (1)–(3) with homogeneous spatial boundary conditions satisfies the following identity:   T  T h1 E1 T 1 2 h 3 E3 3 2 K (v (1, t)) + (vx (1, t)) + (wxx (1, t))2 2 0 x 2 0 2 0     3m T 1 2 h1 1 T 1 1 2 = wt + (vt ) 2 0 0 2 0 0     h 3 3 T 1 3 2 K T 1 2 + (vt ) + w 2 2 0 0 xx 0 0

564

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567

    h 3 E3 T 1 3 2 h 1 E1 T 1 1 2 (vx ) + (vx ) 2 2 0 0 0 0  T 1    T 1 2 2 ˜ + w + N h2 G2 xw xx wxt 2 0 0 xt 0 0  T 1 ˜2 + NG (xw xx vt1 + 2wx vt1 ) +

0

˜2 − NG



T



0

1 0

0

−

1 2

T

0



1 0

   1 −  2

T





0

1 0

T



0 T



0

T



1 0

(vx1 )2 +

ˆ  E(t)

2(FM

 2





T

0

1 0

Fm + 2G2 C1 )

2 wxt



T

E(t),

0

˜ 1 min(3m, K, h1 E1 , h1 1 , h3 E3 , h3 3 , ) and in the where C= 2 last step we have used the fact that C˜ Fm /2. We next estimate the term involving the coupling constant G2 :

(xw xx vt3 + 2wx vt3 )

0

h 1 E1 2  ˜ C

0

0

˜2  T  1 G − (−vt1 + vt3 )xv 1x h2 0 0 ˜2  T  1 G − (−vt1 + vt3 )xv 3x h2 0 0  T 1  ˜2 ˜2 − NG xv 1x wxt + N G 

+

1 0

4G2 C1 Fm

1 G2 h2 (h−1 2 (−v



T

  + v ) + N wx )  2

3

E(t).

0

xv 3x wxt Hence we have the following:

1 3 2

G2 h2 (h−1 2 (−v + v ) + N w x ) + X , (36)

1 2

−

where  1 X = (mw t (xw x − w) + h1 1 vt1 xv 1x

 0

−

0

+ h3 3 vt3 xv 3x − wx (xw xxt + wxx ))|t=T t=0 .

(37)

We multiply Eq. (1) by w, (2) by v 1 and (3) by v 3 and integrate by parts with respect to time and space utilizing the homogeneous boundary conditions given by Eqs. (13) and (14) to obtain an intermediate identity. We add this intermediate identity to Eq. (22) to obtain Eq. (36). Once again for brevity, we defer the calculations to Appendix B. Next, we estimate the term X . We note that  1 (mww t + wx wxx )|t=T X = X − t=0 .

T



1 0

1 3 2 G2 h2 (h−1 2 (−v + v ) + N w x )

4G2 C1 Fm



T

E(t),

0

where C1 = (2 /2) max(N (N h2 + 1), 2(N + 1/ h2 )). Next, we estimate the terms involving the damping factor ˜ 2 . We use the estimates following Eq. (26) and we have the G ˜ 2: following estimate for all the terms involving G ˜ 2 C1  T 4G ˜ E(t). |terms involving G2 |  Fm 0 Hence we have the following:

0

We use similar estimating techniques as in Eq. (24) to get the following:

˜ ˜ 2  − 4 G 2 C1 terms involving G Fm



T

E(t).

0

2D˜ 24 (FM + 2G2 C1 ) |X | E(t) E(t), Fm Fm

Finally, from Eq. (20) we have the following:

where D˜ =max(m4 , h1 1 , h3 3 , (/2)(2 +1)) and in the last step we have used the fact that D˜ 4 (FM + 2G2 C1 ). Hence,



using Eq. (20) we have X  −



˜ 2T C1 2G 24 (FM + 2G2 C1 ) 1− Fm Fm



T

 E(t) T E(T ) T

0

˜ 2T C1 2G 1− Fm

 E(0).

E(0).

We also have       3m T 1 2 h1 1 T 1 1 2 h3 3 T 1 3 2 wt + (vt ) + (vt ) 2 0 0 2 2 0 0 0 0     K T 1 2 h 3 E3 T 1 3 2 + wxx + (vx ) 2 0 0 2 0 0

Putting all of the above together, we have the proof of Theorem 3.  Acknowledgments The author wishes to acknowledge Prof. Scott Hansen, Department of Mathematics, Iowa State University, Ames, IA 50011, for various suggestions.

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567



Appendix A. Calculations for Eq. (22)

= N 2 h2 G 2

We first multiply (1) by xwx and integrate by parts utilizing homogeneous spatial boundary conditions. We get the following:  T 1  1  T 1 t=T m wtt xw x = m wt xw x |t=0 − m wt xwxt , 0



0

T

m



0

0

1

 wt xwxt = m

0

0

= −

0



1

x

0



m 2

Hence we get  T 1  m wtt xw x = m 0



T



T

0

1

0

0

0

0



T



K 0

 K

0

1

wxxx wx = −K

0

T



0

1



T

xw xx wxxx = K

0

1 0



0

K = 2

T



T



0

1 0

wxxx (xw x )x wxxx wx

1

0

0

T 0

1 2 x wxx 2

T



0

= − Next we have  − N 2 h2 G 2

T



0

1

 2

1

 x

1

xw xx wxxx .

0

 x

0

 0

T



1 0

wxx xw x

0



T

= N 2 h2 G2 0



1 0

0

1 2 w 2 x

 x

wx2 .

Next we multiply (2) by xv 1x and integrate by parts utilizing homogeneous spatial boundary conditions (13)–(14).  T 1  1 h1 1 vtt1 xv 1x = h1 1 vt1 xv 1x |t=T t=0 0

0

0

 − h 1 1  h1 1



1





 x

N 2 h2 G 2 2 wx − N G2 (−vx1 + vx3 )xw x 2  1 +m wt xw x |t=T (A.1) t=0 .

xw xx wxxx ,

Hence we get  T 1  T   3K T 1 2 2 K wxxxx xwx = wxx − wxx (1). 2 0 0 0 0 0 Next we have  T 1   wxxt xw x = 

0

1

0

wt2 .

2 wxx (1)

0

K − 2

0



+

2 wxx ,

0



1

T

We leave the coupled terms unchanged i.e.  T 1 −N G2 (−vx1 + vx3 )xw x .

0



T

1



0



0

0

 wx2 + N 2 h2 G2

0 T

−K

1 0

0



= N 2 h2 G 2

1

Putting everything together, from (1), we get   T 1 K T 2 m 2 3K 2 wt + w wxx (1, t) = 2 0 2 xx 0 0 2

wt2 .



T



T

T



0



0

m 2



0

= −K



0

x

wt xw x |t=T t=0 +

Next we have,  T 1  K wxxxx xw x = − K 0

1

1 2 w 2 t

0



T

565

wx (xw x )x

1 2 w 2 xt 2 wxt .

 x

T 0



1 0



1 x (vt1 )2 2

 x



0

0

1 0

1 0

h1 1 =− 2

Hence we get  T 1  h1 1 vtt1 xv 1x = h1 1 0

T



vt1 xv 1xt , T

0

vt1 xv 1x +



1 0

(vt1 )2 .

h 1 1 2



T

0



1 0

(vt1 )2 .

Next we have  T 1  h 1 E1 T 1 1 −h1 E1 vxx xv 1x = − (vx (1, t))2 2 0 0 0   h 1 E1 T 1 1 2 + (vx ) . 2 0 0 We leave the coupled terms unchanged, i.e.    T 1 G2 T 1 1 3 1 − (−v + v )xv x − N G2 wx xv 1x . h2 0 0 0 0 In all, from (2) we get   T 1 h1 E1 T 1 h1 1 1 2 h1 E1 1 2 (vt ) + (vx ) (vx (1, t))2 = 2 2 2 0 0 0 G2 (−v 1 + v 3 )xv 1x h2  T 1 − N G2 wx xv 1x −

0

+ h 1 1

0



1 0

vt1 xv 1x |t=T t=0 .

(A.2)

566

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567

Similarly from (3), we get h3 E3 2



T

0

 (vx3 (1, t))2 =

T



0

h 3 3 3 2 h 3 E 3 3 2 (vt ) + (vx ) 2 2

1 0

+

G2 (−v 1 + v 3 )xv 3x h2  T 1 + N G2 wx xv 3x +

0

+

+ h 3 3

1 0





T 0





T

= 0

1

1

 )

0

˜2 =N G





T 0

1 0

˜2 − NG



T



˜2 = −N G

T 0

1 )

0



1 xw x vxt

(xw xx vt1



1 0

3 xw x vxt

0

0



T

m



0



2

0





−

T

0

0

T



1



wt2

T



1 0

(xw xx vt1 + wx vt1 ) (xw xx vt3 + wx vt3 )

0

T

0 1

wt w|t=T t=0 − m



T

wxxtt w = 

0

T

K



0

1

1

T

−N h2 G2



0

2 

T

1



(wxx (1, t))

2

1 0

˜2 −N 2 h2 G



T 0

˜2 −N G



T 0

1



1

 0

 −



0

1 0

0

T

2 wxt ,



0





T

wxx w = N h2 G2 2

0



1

T

wxxx wx = K

˜2 wxxt w = N 2 h2 G



T

1

1 0

2 wxx ,

wx2 , (−v 1 + v 2 )wx ,

0



1 3 ˜2 (−vxt + vxt )w = N G

1 0

0

0 1

0

wt2 ,



(−vx1 + vx3 )w = N G2 

1

wxtt wx

0



0

0 T

T

−N G2

T 0





T 0

wxt wx |t=T t=0

wxxxx w = −K

0

1



0



2





0

 

K (vx3 (1, t))2 + 

1

0

T

0

 wtt w = m

=

 1 3K  2 + w − w2 2 0 0 xx 2 0 0 xt 0 0     N 2 h2 G 2 T 1 2 h 3  3 T 1 3 2 wx + (vt ) + 2 2 0 0 0 0

m = 2



1

0

Adding Eqs. (A.1)–(A.3) we get, h3 E 3 (vx1 (1, t))2 +

0

We first multiply Eq. (1) by w and integrate by parts utilizing homogeneous spatial boundary conditions (13)–(14). We get the following:

(xw xx vt3 + wx vt3 ).

T

1

Appendix B. Calculations for Eq. (36)

We have one last calculation:     G2 T 1 1 x (−v 1 + v 3 )2 2h2 0 0 2 x  T 1 G2 =− (−v 1 + v 3 )2 . 2h2 0 0 



1 where X = 0 (mw t xwx + h1 1 vt1 xv 1x + h3 3 vt3 xv 3x − wxxt xw x )|t=T t=0 . This finishes the calculation to obtain Eq. (22).

+ wx vt1 ),

˜2  T  1 ˜2  T  1 G G − (−vt1 + vt3 )xv 1x , − (−vt1 + vt3 )xv 3x h2 0 0 h2 0 0  T 1  T 1 ˜2 ˜2 −N G xv 1x wxt , N G xv 3x wxt .

h1 E1 2

T

0

˜ 2 unchanged: We leave the other terms involving G

0

˜ 2 (xw xx wxt + wx wxt ) N 2 h2 G

0

xwx wxxt

1

(vx3 )2

˜2  T  1 G − (−vt1 + vt3 )xv 1x h2 0 0  T 1 ˜2  T  1 G ˜2 − (−vt1 + vt3 )xv 3x − N G xv 1x wxt h2 0 0 0 0  T 1 ˜2 + NG xv 3x wxt + X, (A.4)

˜ 2 (xw xx wxt + wx wxt ) N 2 h2 G T

1



˜2 − NG

0



0

  h 1 1 T 1 1 2 + (vt ) 2 0 0 0  1   G2 T 1 (vx1 )2 − (−v 1 + v 3 )2 2h2 0 0 0 1

0

(A.3)

0

˜2 × NG

T



0

˜ 2 are The terms involving G ˜2 − N 2 h2 G

0



˜2 + NG vt3 xv 3x |t=T t=0 .

T



h 1 E1 2  T 0

0





h 3 E3 + 2

T 0



1

wxt wx ,

0 T

0

 

1 0

(−vt1 + vt3 )wx .

R. Rajaram / Systems & Control Letters 56 (2007) 558 – 567

Hence we have  T 1 2 (mw 2t + wxt − Kw 2xx − N 2 h2 G2 wx2 0

0



T

− NG2 (−v + v )wx ) − 1

3

0



1

˜ 2 wxt wx (N 2 h2 G

0

˜ 2 (−vt1 + vt3 )wx ) + NG  1 (mw t w + wxt wx )|t=T − t=0 = 0.

(B.1)

0

Finally, we add Eq. (B.1) with (A.4) to obtain Eq. (36). References [1] R.A. Ditaranto, Theory of vibratory bending for elastic and viscoelastic layered finite-length beams, J. Appl. Mech. 32 (1965) 881–886. [2] K.J. Engel, H. Nagel, One-Parameter Semigroups for Linear Evolution Equations, Springer, New York, 1991. [3] R.H. Fabiano, S.W. Hansen, Modelling and analysis of a three layer damped sandwich beam, Discrete Continuous Dynamical Systems (2001) 143–155. [5] S.W. Hansen, Several related models for multilayer sandwich plates, Math. Models and Meth. Appl. Sci. 14 (8) (2004) 1103–1132. [8] S.W. Hansen, R. Rajaram, Riesz basis property and related results for a Rao–Nakra sandwich beam, Discrete Continuous Dynamical Systems (2005) 365–375.

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[9] S.W. Hansen, R. Rajaram, Exact boundary controllability of a Rao–Nakra sandwich beam, Proceedings of the SPIE, vol. 5757, 2005, pp. 97–107. [10] S.W. Hansen, R. Rajaram, Simultaneous boundary control of a Rao–Nakra sandwich beam, CDC, 2005. [12] V. Komornik, Exact Controllability and Stabilization: The Multiplier Method, Wiley, New York, 1994. [13] J. Lagnese, Modelling, Analysis and Control of Thin Plates, Masson, Paris, 1988. [15] J.L. Lions, Exact controllability, stabilization and perturbations for distributed parameter systems, SIAM Rev. 30(1) (1988) 1–68. [16] J.L. Lions, E. Magnenes, Problémes aux limites non homogénes et applications, Dunod, Paris, 1968 (vols. 1 and 2). [17] D.J. Mead, A comparison of some equations for the flexural vibration of damped sandwich beams, J. Sound Vib. 83 (1982) 363–377. [18] D.J. Mead, S. Markus, The forced vibration of a three-layer, damped sandwich beam with arbitrary boundary conditions, J. Sound Vibr. 10 (1969) 163–175. [19] R. Rajaram, S.W. Hansen, Null controllability of a damped Mead–Markus sandwich beam, Discrete and Continuous Dynamical Systems (Suppl. vol.) (2005) S746–S755. [20] Y.V.K.S. Rao, B.C. Nakra, Vibrations of unsymmetrical sandwich beams and plates with viscoelastic cores, J. Sound Vibr. 34(3) 309–326. [21] C.T. Sun, Y.P. Lu, Vibration Damping of Structural Elements, PrenticeHall, Englewood Cliffs, NJ, 1995. [22] J.-M. Wang, G.-Q. Xu, S.-P. Yung, Exponential stabilization of laminated beams with structural damping and boundary feedback controls, SIAM J. Control Optim. 44 (5) (2005) 1575–1597. [24] M.-J. Yan, E.H. Dowell, Governing equations for vibratory constrainedlayer damping sandwich plates and beams, J. Appl. Mech. 39 (1972) 1041–1046.