Download - Rayleigh Beam Decay Rates
-
7/28/2019 Rayleigh Beam Decay Rates
1/27
Ownership and Copyright( Springer-Verlag Limited
Math. Control Signals Systems (2002) 15: 229255 Mathematics of Control,Signals, and Systems
Decay Rates for a Beam with Pointwise Force andMoment Feedback*
Kais Ammari,y Zhuangyi Liu,z and Marius Tucsnaky
Abstract. We consider the Rayleigh beam equation and the EulerBernoulli
beam equation with pointwise feedback shear force and bending moment at the
position x in a bounded domain0; p
with certain boundary conditions. The en-
ergy decay rate in both cases is investigated. In the case of the Rayleigh beam, we
show that the decay rate is exponential if and only if x=p is a rational numberwith coprime factorization x=p p=q, where q is odd. Moreover, for any otherlocation of the actuator we give explicit polynomial decay estimates valid for reg-
ular initial data. In the case of the EulerBernoulli beam, even for a nonhomo-
geneous material, exponential decay of the energy is proved, independently of the
position of the actuator.
Key words. Pointwise control, Exponential decay, Polynomial decay, Observ-
ability inequality.
1. Introduction
Pointwise stabilization of beam systems is a subject widely studied in the literature(see, for instance, [7][9], [17] and [20]). To our knowledge, the stabilizers used areeither of force feedback type or of moment feedback type acting at the joints. It isknown that the stability properties of such a system depend on the location of the
joints in an unrobust way. For instance, for a beam hinged at both ends and oflength p, in order to have strong stabilization the abscissa of the joint has to beirrational and for exponential stabilizability this abscissa has to satisfy even morerestrictive conditions (see, for instance, [2] for an EulerBernoulli beam). In thispaper we investigate the stabilization eect of applying both force and momentfeedbacks simultaneously at the joint. As far as we know, this problem has not yetbeen tackled in the literature. Most of the results we obtain are robust with respectto the location of the joint.
First we consider the following initial and boundary value problem for a
* Date received: October 30, 2000. Date revised: December 20, 2001.y Institut Elie Cartan, Departement de Mathematiques, Universite de Nancy I, F-54506 Vandoeuvre
les Nancy Cedex, France. {ammari,tucsnak}@iecn.u-nancy.fr.
z Department of Mathematics and Statistics, University of Minnesota, Duluth, Minnesota 55812-2496, U.S.A. [email protected].
229
-
7/28/2019 Rayleigh Beam Decay Rates
2/27
homogenous Rayleigh beam with force and moment damping at x x:q2u
qt2 q
4u
qx2 qt2 q
4u
qx4 qu
qtx; tdx q
2u
qt qxx; t ddx
dx 0; 0 < x < p; t > 0;
1:1
u0; t up; t 0; q
2u
qx20; t q
2u
qx2p; t 0; t > 0; 1:2
ux; 0 u0x; quqt
x; 0 u1x; 0 < x < p: 1:3
We also consider a nonhomogenous EulerBernoulli beam:
rx q2u
qt2 q
2
qx2px q
2u
qx2
! qu
qtx; tdx q
2u
qx qtx; t ddx
dx 0; 0 < x < p; t > 0;
1:4u0; t up; t q
2u
qx20; t 0; q
2u
qx2p; t 0; t > 0; 1:5
ux; 0 u0x; quqt
x; 0 u1x; 0 < x < p: 1:6
Here dx is the Dirac mass concentrated in the point x A 0; p, u is the transversedisplacement of the beam and the coecient functions rx, px A C10; p sat-isfy rxbr0 > 0, pxbp0 > 0.
Remark 1.1. We denote by H20; pXH10 0; p the dual space of H20; pXH10 0; p with respect to the pivot space L20; p. If u is smooth on 0; pnfxg 0;T, then u satisfies (1.1) in the space C0;T; H20; pXH10 0; p if and onlyif
q2u
qt2 q
4u
qx2 qt2 q
4u
qx4 0; x A 0; pnfxg; t > 0;
and
u ; tx 0;qu
qx ; t x 0;
q2u
qx2 ; t
" #x
q2u
qx qtx; t; q
3u
qx3 ; t
" #x
quqt
x; t9>>>>=>>>>;; Et > 0;
where fx denotes the jump of the function f at the point x. A similar assertionholds if u is smooth on 0; pnfxg 0;T and u satisfies (1.4) so that the termsare in the space C0;T; H20; pXH10 0; p. Thus, (1.1) (respectively (1.4))is equivalent to the equations modeling the vibrations of two Rayleigh beams (re-spectively two EulerBernoulli beams) with a dissipative joint at x x, whereboth feedback force and feedback moment are applied. The results stated in thisremark, which can be checked by standard calculations, are not used in the re-maining part of the paper, and we omit their proof.
230 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
3/27
The purpose of this paper is to give a complete characterization of points x forwhich the solutions of (1.1)(1.3) (respectively (1.4)(1.6)) are exponentially stablein the energy space, and give an explicit decay rate for regular initial data whenwe do not have exponential stability in the energy space. A significant improve-
ment of our new stabilization strategy is that we have avoided the well-known prob-lems concerning the existence of nodal points. The method used to study (1.1)(1.3) is based on some trace regularity results which reduce stability to observ-ability inequalities for the corresponding undamped problem. As far as we know,in the case of unbounded feedback, this method is new. Both the regularity andthe observability results are based on Fourier analysis, so we can tackle only theconstant coecients case. In order to study (1.4)(1.6), we use a frequency domainmethod and combine a contradiction argument with the multiplier technique tocarry out a special analysis for the resolvent.
The paper is organized as follows. The statements of the main results are given
in the following section. Sections 3 and 4 are devoted to some trace regularity andobservability results needed in what follows. The proof of the main results is givenin Sections 5 and 6.
2. Statement of the Main Results
We define the energy of a solution u of (1.1)(1.3) and of a solution u of (1.4)(1.6) at the time instant t by
Eut 12 p0 quqt x; t 2
q2u
qt qx x; t 2
q2u
qx2 x; t 2
! dx 2:1and
~EEut 12p
0
px q2u
qx2x; t
2 rx quqt x; t 2
!dx; 2:2
respectively. We can easily check that every suciently smooth solution of (1.1)(1.3) (respectively (1.4)(1.6)) satisfies the energy identity
Eut2 Eut1 t2t1
quqt x; s 2 q2uqt qx x; s
2( ) ds; 2:3respectively
~EEut2 ~EEut1 t2
t1
qu
qtx; s
2 q2uqt qx x; s 2
( )ds; 2:4
for all t2 > t1b 0 and, therefore, the energy is a nonincreasing function of thetime variable t. Denote
V
H2
0; p
XH10
0; p
:
We define the concept of a finite energy solution of (1.1)(1.3) and of (1.4)(1.6).This concept will be frequently used in what follows.
Decay Rates for a Beam with Pointwise Force and Moment Feedback 231
-
7/28/2019 Rayleigh Beam Decay Rates
4/27
Definition 2.1. A function
u A C0; T; VXC10;T; H10 0; pis called a finite energy solution of (1.1)(1.3) ifqu=qxx; A H10;T, u satisfies(1.3) and (1.1) is satisfied in L
2
0; T; V, where V is the dual space of V withrespect to the pivot space L20; p.
Definition 2.2. A function
u A C0;T; VXC10;T; L20; pis called a finite energy solution of (1.4)(1.6) if qu=qxx; A H10;T and (1.4)is satisfied in L20;T; V.
The result below concerns the well-posedness of the solutions of (1.1)(1.3) and
the behavior of Eut when t ! y. The first assertion in the proposition belowwas proved, for instance, in the survey paper [22] by using the properties of a gen-eral class of conservative systems. The second part of the proposition is provedin Section 5.
Proposition 2.3. Let u0
u1
A V H10 0; p. Then the following assertions hold:
1. Problem (1.1)(1.3) admits a unique finite energy solution such that
ku
x;
k2H1
0;T
qu
qx x;
2
H10;TaC
ku0
k2H2
0;p
ku1
k2H1
0;p
;
2:5
where the constant C> 0 depends only on x and T. Moreover, u satisfies theenergy estimate (2.3).
2. We have limt!y Eut 0, for any initial data u0u1
in V H10 0; p.
The next result concerns the well-posedness of (1.4)(1.6).
Proposition 2.4. Let u0
u1
A V L20; p. Then problem (1.4)(1.6) admits a
unique finite energy solution such that
kux; k2H10;T qu
qxx;
2H10;T
aCku0k2H20;p ku1k2L20;p; 2:6
where the constant C> 0 depends only on x and T. Moreover, u satisfies the energyestimate (2.4).
The main result of this paper concerns the precise asymptotic behavior of thesolutions of (1.1)(1.3) and (1.4)(1.6). Denote
Y uv
A V V j uj0;x A H30; x and ujx;p A H3x; p& ' 2:7
232 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
5/27
and
DA uv
A Y;
d2u
dx20 d
2u
dx2p 0; d
2u
dx2
x
dvdx
x; d3u
dx3
x
vx( )
;
2:8
where gx denotes the jump of the function g at the point x. The correspondingoperator A is defined on DA by
Au
v
v
R d4u
dx4
vxRdx dv
dxxR ddx
dx
0@ 1A;where the operator R is defined by
R I d2
dx2 1
: 2:9It is well known that R is an isomorphism from H10; p onto H10 0; p.
If u0
u1
A Y, we denote
u0
u1
2Y
ku0k2H30;x ku0k2H3x;p ku1k2H20;p: 2:10
Concerning the system (1.1)(1.3) our main result is the following.
Theorem 2.5.1. Let u A C0;T; VXC10;T; H10 0; p be a finite energy solution of the sys-
tem (1.1)(1.3) (the existence and the uniqueness of such a solution followsfrom Proposition 2.3). Then u satisfies the estimate
EutaMeotku0k2V ku1k2L20;p; Etb 0; 2:11for some constants M;o > 0 (depending only on x), if and only if x=p is arational number with coprime factorization
x
p p
q ; where q is odd: 2:122. For any x A 0; p, there exists a constant C> 0 (independent ofx) such that
for all tb 0 we have
Euta Ct 1
u0
u1
2Y
; Eu0
u1
A DA: 2:13
For the system (1.4)(1.6) our main result is the following.
Theorem 2.6. Let u A C0;T; V
XC1
0;T; L2
0; p
be a finite energy solution
of(1.4)(1.6) (the existence and the uniqueness of such a solution follows from Prop-osition 2.4). Then, for allx A 0; p, there exist constants M;o > 0 (depending only
Decay Rates for a Beam with Pointwise Force and Moment Feedback 233
-
7/28/2019 Rayleigh Beam Decay Rates
6/27
on x), such that~EEutaMeotku0k2V ku1k2L20;p; Etb 0: 2:14
In other terms, for allx A 0; p, (1.4)(1.6) define an exponentially stable system inV
L2
0; p
.
3. Regularity Inequalities
Consider the open-loop system associated to (1.1)(1.3):
q2c
qt2 q
4c
qx2 qt2 q
4c
qx4 v1tdx v2t ddx
dx; 0 < x < p; t > 0; 3:1
c
0; t
c
p; t
0;
q2c
qx2
0; t
q2c
qx2
p; t
0; t > 0;
3:2
cx; 0 0; qc
qtx; 0 0; 0 < x < p: 3:3
The finite energy solutions of (3.1)(3.3) are defined as follows.
Definition 3.1. Let v1; v2 A L20;T. Then a function
c A C0;T; VXC10;T; H10 0; pXH20;T; L20; p 3:4
is called a finite energy solution of (3.1)(3.3) if c satisfies (3.3) and (3.1) issatisfied in L20;T; V.
The following existence and trace regularity result for the system (3.1)(3.3) isessential for the proof of Theorem 2.5.
Proposition 3.2. Suppose that v1; v2 A L20;T. Then the system (3.1)(3.3) admits
a unique finite energy solution c. Moreover, qc=qxx; A H10;T and thereexists C> 0 (depending only on T) such that
kcx; k2H10;T qcqx x; 2
H10;TaCkv1k2L20;T kv2k2L20;T: 3:5
Before proving Proposition 3.2, we consider the following homogeneous undampedproblem:
q2j
qt2 q
4j
qx2 qt2 q
4j
qx4 0; 0 < x < p; t > 0; 3:6
j0; t jp; t 0; q2j
qx20; t q
2j
qx2p; t 0; t > 0; 3:7
jx; 0 u0x; qjqt
x; 0 u1x; 0 < x < p: 3:8
234 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
7/27
Lemma 3.3. For any initial data u0
u1
A V H10 0; p there exists a unique
finite energy solution of (3.6)(3.8) (in the sense of Definition 3.1). Moreover,qj=qxx; A H10;T and there exists C> 0, depending only on T, such that
kjx; k2H10;T qjqx x; 2
H10;TaCku0k2H20;p ku1k2H10;p: 3:9
Proof. It is easy to see, by the semigroup method [18], that the problem (3.6)(3.8) is well-posed in the energy space V H10 0; p. In order to prove (3.9) weput
u0x Xkb1
ak sinkx; u1x Xkb1
bk sinkx; 3:10
with k2
ak; kbk A l2
R. A simple calculation shows that
jx; t Xkb1
ak cosk2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 k2p t
bkffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 k2p
k2sin
k2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 k2p t
( )sinkx; 3:11
which implies that, for all T > 0, we have
T
0
qj
qtx; t
2
q2j
qt qxx; t
2
( )dtaC
Xkb1k2 a2k
k4
1
k2
b2k
aCXkb1
k2a2kk2 b2k;
where C> 0 is a constant, depending only on T. This clearly yields (3.9). 9
It can be easily checked that c is a finite energy solution of (3.1)(3.3) if andonly
q2
cqt2
R q4
cqx4
! v1tRdx v2tR ddxdx
in C0;T;H10 0; p; 3:12c0; t cp; t 0; t > 0; 3:13
cx; 0 0; qcqt
x; 0 0; 0 < x < p; 3:14
where the operator R is defined by (2.9). To prove (3.5), we need three technicalresults. The first one gives the boundedness of some functions which will be usedin what follows.
Lemma 3.4. Let a > 0, x A 0; p and denote Ca fl A C j Re l ag. Then the
Decay Rates for a Beam with Pointwise Force and Moment Feedback 235
-
7/28/2019 Rayleigh Beam Decay Rates
8/27
functions Hi: Ca ! C, iA f1; 2; 3; 4g, defined by
H1l sinhlxsinhlp
p0
sinhlp yRdxy dy
x0
sinhlx yRdxy dy;
H2l sinhlxsinhlp
p0
sinhlp yR ddxdy
y dy
x
0
sinhlx yR ddxdy
y dy;
H3l l coshlxsinhlp
p0
sinhlp yRdxy dy
l x0
coshlx yRdxy dy;
H4l l coshlxsinhlp
p0
sinhlp yR ddxdy
y dy
lx
0
coshlx yR ddxdy
y dy;
are bounded on Ca.
Proof. The functions Hil, for i 1; 2; 3; 4, are continuous on Ca. Then, toprove that these functions are bounded on Ca, it is sucient to prove that they arebounded when l A Ca, Im l ! y. A simple calculation shows that
Rdx sinhx p sinhx
sinhp ; x A 0; x;
sinhx sinhx psinhp ; x A x; p;
8>>>>>:R
ddx
dx
coshx p sinhxsinhp
; x A0; x
;
coshx sinhx psinhp ; x A x; p:
8>>>>>:3:15
It follows that
H1l sinhlx sinhx psinhlp sinhp
x0
sinhlp y sinhy dy
sinhlx sinhxsinhlp sinhp
px
sinhlp y sinhy p dy
sinhx psinhp
x0
sinhlx y sinhy dy;
236 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
9/27
H2l sinhlx coshx psinhlp sinhp
x0
sinhlp y sinhy dy
sinhlx coshxsinh
lp
sinh
p
p
x
sinhlp y sinhy p dy
coshx psinhp
x0
sinhlx y sinhy dy:
Moreover, by using the monotonicity of the functions sinh, cosh combined withthe fact that, for any z A C, we have jsinhzja jcoshRe zj, we deduce that
supl ACa
jH1lja coshax sinhp x sinhxsinhp sinhap
fx coshap p x coshap x x sinhapg
and
supl ACa
jH2lja x coshax coshp x sinhxsinhp sinhap fsinhap sinhapg
p x coshap x sinhp x coshax coshxsinhp sinhap :
Thus H1 and H2 are bounded on Ca. In order to show that H3 and H4 are alsobounded on Ca we notice that the definition of H3 and H4, combined with (3.15),gives
H3l l coshlx sinhx psinhlp sinhp
x0
sinhlp y sinhy dy
l coshlx sinhxsinhlp sinhp
px
sinhlp y sinhy p dy
l sinhx psinhp x0 coshlx y sinhy dy
and
H4l l coshlx coshx psinhlp sinhp
x0
sinhlp y sinhy dy
l coshlx coshxsinhlp sinhp
px
sinhlp y sinhy p dy
l coshx psinhp
x0
coshlx y sinhy dy:
Decay Rates for a Beam with Pointwise Force and Moment Feedback 237
-
7/28/2019 Rayleigh Beam Decay Rates
10/27
Thus, by integration by parts, we obtain
H3l coshlx sinhx psinhlp sinhp
x0
coshlp y coshy dy
coshlx sinhxsinhlp sinhp px coshlp y coshy p dy sinhx p
sinhpx
0
sinhlx y coshy dyand
H4l coshlx coshx psinhlp sinhp
x0
coshlp y coshy dy
coshlx coshx
coshlp coshp p
x
cosh
l
p
y
cosh
y
p
dy
coshx psinhp
x0
sinhlx y coshy dy coshlx coshlp xsinhlp :
Then
supl ACa
jH3lja x coshax coshx psinhap sinhp coshap coshx
p x coshax coshxsinhap sinhp coshap x coshx p
x coshx psinhp coshax coshx
and
supl ACa
jH4lja x coshax coshx psinhap sinhp coshap coshx
p x coshax coshxsinhap coshp coshap x coshx p
xcosh
x
p
sinhp coshax coshx cosh
ax
cosh
a
p
x
sinhap : 9
The following result gives a regularity property for the string equation. This re-sult, combined with the fact that the operator Rq4=qx4 is similar (in a sensewhich will be made precise later) to the second-order derivative operator, will beused in the proof of Proposition 3.2.
Lemma 3.5. Suppose that v1; v2 A L20;T and that c2 A C0;T; VX
C10;T; L20; p satisfies the conditions
q2c2qt2
q2c2qx2
v1tRdx v2tR ddxdx
; 0 < x < p; t > 0; 3:16
238 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
11/27
c20; t c2p; t 0; t > 0; 3:17
c2x; 0 0;qc2qt
x; 0 0; 0 < x < p; 3:18
where the operatorR
is defined in (2.9). Then qc2=qxx; A H1
0;T and thereexists C> 0, depending only on T, such that
kc2x; k2H10;T qc2qx
x; 2
H10;TaCkv1k2L20;T kv2k2L20;T: 3:19
Proof. As (3.16) is time reversible, after extending v1; v2 by zero for t A Rn0;T,we can solve (3.16)(3.18), for t A R. In this way, it follows that c2 can be extendedto a function, denoted also by c2, such that
c2 A CBR; VXC1BR; H10 0; p;c2t 0; Eta 0;
& 3:20and c2 satisfies (3.16)(3.17) for all t A R (above we denoted by C
kBR;X the
space of Ck functions from R into X, which are bounded on R, together withtheir derivatives up to the order k). Let cc2c2l, where l g ih, g > 0 and h A R,be the Laplace (with respect to t) transform of c2. Since c2 satisfies (3.20), inorder to prove (3.19) it suces to prove that the functions t ! egtc2x; t andegtqc2=qxx; t belong to H1R and that there exist two constants M1;M2 > 0such that
keg:c2x; k2H1y;yaM1kv1k2L2y;y kv2k2L2y;yand
eg:qc2qx
x; 2
H1y;yaM2kv1k2L2y;y kv2k2L2y;y:
Since the function t ! egtqc2=qxx; t is in L2R, by the Parseval identity (seefor instance p. 212 of [11]), it suces to prove that the functions
h ! g ihcc2x; g ih; g ih q^
cc2qx x; g ih
belong to L2R, for some g > 0, and that there exist two constants M3;M4 > 0such thaty
yjg ihcc2x; g ihj2 dhaM3
yy
jvv1g ihj2 jvv2g ihj2 dh 3:21
and
y
y jg ihqcc2
qx x; g ihj2
dha
M4 y
yjvv1g ihj2
jvv2g ihj2
dh:3:22
Decay Rates for a Beam with Pointwise Force and Moment Feedback 239
-
7/28/2019 Rayleigh Beam Decay Rates
12/27
It can be easily checked that, for all x A 0; p, the function cc2c2 satisfiesl2
cc2c2x; l
q2cc2qx2
x; l vv1lRdxx vv2lR ddxdx
x; Re l > 0;
3:23cc20; l cc2p; l 0; Re l > 0: 3:24
Then
cc2x; l sinhlxl sinhlp
p0
sinhlp y vv1lRdx vv2lR ddxdx
dy
1l
x0
sinhlx y vv1lRdx vv2lR ddxdx
dy;
x A
0; p
; Re l > 0;
3:25
where Rdx and Rddx=dx are given by (3.15).
The relations above imply that, for every l A C, Re l > 0, we have
lcc2c2x; l H1lvv1l H2lvv2l; ERe l > 0; 3:26and
lqcc2c2qx
x; l H3lvv1l H4lvv2l; ERe l > 0; 3:27
where H1;H2;H3;H4 are the functions introduced in Lemma 3.4. By applyingLemma 3.4 we obtain the existence of M3;M4 > 0 such that (3.21) and (3.22)hold. 9
Remark 3.6. The inequality
kc2x; k2H10;TaCkv1k2L20;T kv2k2L20;Tcan also be obtained by semigroup techniques combined with a trace theorem.
The next result will be used in order to reduce the proof of Proposition 3.2 to a
regularity property for a string equation.
Lemma 3.7. Let R be the operator defined in (2.9). Then the linear operator
L R q4
qx4
! q
2
qx2
is bounded from V onto V, i.e. we have that L ALV.
Proof. Let u A V. Equivalently u can be written as
ux Xnb1
an sinnx;
240 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
13/27
withP
nb1 n4a2n
-
7/28/2019 Rayleigh Beam Decay Rates
14/27
with j satisfying (3.6)(3.8). From (3.9) and (3.31) we deduce that there exists aconstant C> 0 (depending only on T) such that
T
0
BS
t
u0
u
1 2
R
2
dtaC
ku0; u1
k
2V
H1
0
0;p
; E
u0
u
1 A DA0: 3:32According to Theorem 3.1 on p. 173 in [5], inequality (3.32) implies that (3.29)and (3.30) admit a unique finite energy solution. Moreover, this solution dependscontinuously on v1; v2, i.e. there exsits a constant C> 0 (depending only on T)such that
c
qc
qt
C0;T;VH100;p
a kv1kL20;T kv2kL20;T: 3:33
We have thus proved the conclusion (3.4).We prove now the trace regularity property (3.5). According to the expression
ofR, we have
Rq4
qx4
! q
2
qx2 L;
where L ALV (see Lemma 3.7). We notice that c can be written asc c1 c2; 3:34
where c2 satisfies (3.16)(3.18) and c1 is the solution of
q2c1qt2
q2c1qx2
Lc; 0 < x < p; t > 0; 3:35
c10; t c1p; t 0; t > 0; 3:36
c1x; 0 0;qc1qt
x; 0 0; 0 < x < p: 3:37
Since Lc A C0;T; V, then by the classical theory for evolution equations of hy-perbolic type [13], we obtain that c1 A C0;T; H30; pXC10; T; H20; p andthat there exists a constant C
C
T> 0 such that
c1qc1qt
C0;T;H30;pH20;p
aCTkckC0;T;V:
This inequality, combined with (3.33) and the standard trace theorem, impliesthat
kc1x; k2H10;T qc1qx
x;
2
H10;TaCkv1k2L20;T kv2k2L20;T;
for some constant C depending only on T. The last inequality, combined with(3.19) and (3.34), implies the conclusion (3.5). 9
242 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
15/27
4. Observability Inequalities
The observability inequality concerning the solutions of (3.6)(3.8) is given in thenext proposition.
Proposition 4.1. Let T> 0 be fixed. Then the following assertions hold:
1. There exists a constant Cx > 0 such that all finite energy solutions j of (3.6)(3.8) satisfyT
0
qj
qtx; t
2 q2jqt qx x; t 2
( )dtbCxku0k2H20;p ku1k2H10;p 4:1
if and only ifx=p is a rational number with coprime factorization
x
p p
q ; where q is odd:
2. There exists a constant C> 0 (independent of x) such that all finite energysolutions j of (3.6)(3.8) satisfyT
0
qj
qtx; t
2 q2jqt qx x; t 2
( )dtbCku0k2H10;p ku1k2L20;p:
Proof. First step. From (3.11) and the BallSlemrod generalization of Inghamsinequality (see [4]), we obtain that, for all T > 2p, there exists a constant C
Tsuch
that T0
qj
qtx; t
2 q2jqt qx x; t 2
( )dt
bCTXkb1
a2kk4
1 k2 b2k
k2 cos2kx sin2kx; 4:2
where ak and bk have beeen defined by (3.10). If x satisfies (2.12), then, byLemma 3.4 in [20], there exists a constant kx > 0 such that
jcoskxjb kx; Ekb 1: 4:3Inequalities (4.2) and (4.3) imply thatT
0
qj
qtx; t
2 q2jqt qx x; t 2
( )dtbC
Xkb1
k4a2k k2b2k;
for some constant C> 0. It follows that (4.1) holds for all x satisfying (2.12).On the other hand ifx does not satisfy (2.12), then we can again apply Lemma
3.4 from [20] to get the existence of a sequence
pm
HN
, limm!y pmy such
thatlim
m!ycospmx 0: 4:4
Decay Rates for a Beam with Pointwise Force and Moment Feedback 243
-
7/28/2019 Rayleigh Beam Decay Rates
16/27
If we denote by jm the solution of (3.6)(3.7) with initial data
jmx; 0 sinpmx;qjmqt
x; 0 0; Ex A 0; p;
a simple calculation using (4.4) implies that
limm!y
T0 fjqjm=qtx; tj2 jq2jm=qxqtx; tj2g dt
kjm0k2H20;p kqjm=qt0k2H10;p 0:
So, (4.1) is false for any x not satisfying (2.12).
Second step. In order to prove assertion 2 of Proposition 4.1 it is sucient tonotice that
T0
qj
qtx; t 2 q2jqt qx x; t 2( ) dtbCT
Xkb1
a2kk4
1 k2 b2k
k2 cos2kx sin2kx
bCTXkb1
a2kk4
1 k2 b2k
bCT
Xkb1
a2kk2 b2k: 9
5. Proof of Proposition 2.3 and Theorem 2.5
Proof of Proposition 2.3. As we already mentioned in Section 2, the existenceand uniqueness of finite energy solutions of (1.1)(1.3) was already proved in theliterature (see, for instance, [22]).
In order to prove the estimate (2.3) and the trace regularity property (2.5), it
suces to remark that they hold for regular solutions i.e. uqu=qt
A C0;T;DA)
and to use the density ofDA in V H10 0; p.Finally, since the embedding ofD
A
in V
H10
0; p
is obviously compact, by
LaSalles invariance principle (see, for instance, [10]) the strong stability estimateat the end of Proposition 2.3 holds if the relations
q2v
qt2 q
4v
qx2 qt2 q
4v
qx4 0; 0 < x < p; t > 0;
v0; t vp; t q2v
qx20; t q
2v
qx2p; t 0; t > 0;
qv
qt x; t
q2v
qt qx x; t
0; t > 0;
imply that v1 0.
244 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
17/27
In order to prove this unique continuation result, we put
vx; 0 Xkb1
ak sinkx; qvqt
x; 0 Xkb1
bk sinkx; 5:1
with k2ak; kbk A l2R. We haveqv
qtx; t
Xkb1
ak k2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 k2p sink2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 k2p t
bk cos k2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 k2p t & '
sinkx 5:2
and
q2v
qt qxx; t
Xkb1
kak k2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 k2p sin
k2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 k2p t
kbk cos k
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 k2p t
& 'coskx;
5:3which implies, according to the BallSlemrod generalization of Inghams inequal-ity, that for T > 2p,T
0
qv
qtx; t
2 q2vqt qx x; t 2
( )dtbC
Xkb1
a2kk4
1 k2 b2k
k2 cos2kx sin2kx
bC
Xkb1a2k
k4
1 k2 b2k
: 5:4
Hence, vx; 0 0 and qv=qtx; 0 0 for all x A 0; p which implies that v1 0:Thus we proved the strong stability result. 9
Let u A C0;T; VXC10;T; H10 0; p be the solution of (1.1)(1.3). Then ucan be written as
u u1 u2; 5:5where u1 is the solution of (3.6)(3.8) and u2 satisfies
q
2
u2qt2 q
4
u2qx2 qt2 q
4
u2qx4 quqt x; tdx q
2
uqt qx x; t ddxdx 0 in 0;p 0;T;
5:6
u20; t u2p; t q2u2
qx20; t q
2u2
qx2p; t 0; t A 0;T; 5:7
u2x; 0 qu2qt
x; 0 0; x A 0; p: 5:8
We note by Proposition 2.3 that q2u=qxqtx; A L20;T, so (5.6) makes sense.The main ingredient of the proof of Theorem 2.5 is the following result.
Lemma 5.1. Suppose that u0; u1 A V H10 0; p. Then the solutions u of
Decay Rates for a Beam with Pointwise Force and Moment Feedback 245
-
7/28/2019 Rayleigh Beam Decay Rates
18/27
(1.1)(1.3) and the solution u1 of (3.6)(3.8) satisfy
C1
T0
qu1
qtx; t
2
q2u1
qt qxx; t
2( )
dt
aT
0
qu
qtx; t 2 q2uqt qx x; t 2( ) dt
a 4
T0
qu1
qtx; t
2 q2u1qt qx x; t 2
( )dt; 5:9
where C1 > 0 is a constant independent ofu0
u1
.
Proof. First, we note by (3.9) that q2u1=qtqxx; A L20;T. Relation (5.5)
implies thatT0
qu1
qtx; t
2 q2u1qt qx x; t 2
( )dta 2
T0
qu
qtx; t
2 q2uqt qx x; t 2
( )dt
T
0
qu2
qtx; t
2 q2u2qt qx x; t 2
( )dt
!:
The estimate above combined with inequality (3.5) in Proposition 3.2 implies the
existence of a constant C1 > 0, independent ofu0
u1 , such that
C1T
0
qu1
qtx; t
2 q2u1qt qx x; t 2( )
dtaT
0
qu
qtx; t
2 q2uqt qx x; t 2( )
dt:
5:10On the other hand, (5.6) can be rewritten as
q2u2
qt2 q
4u2
qx2 qt2 q
4u2
qx4 qu2
qtx; tdx q
2u2
qt qxx; t ddx
dx
qu1qt
x; tdx q2u1
qt qxx; t ddx
dxin 0; p 0;T: 5:11
If we formally take the duality product (with respect to the pivot space L20; p)of both sides of (5.11) by qu2=qt (this can be done rigorously by considering aregularizing sequence) and then we integrate from 0 to Twe obtain that
12
qu2
qt ;T
2L20;p
12q2u2
qx qt ;T
2L20;p
12q2u2
qx2 ;T
2L20;p
T
0
qu2
qtx; t
2
q2u2
qt qxx; t
2( )
dt
T0
qu1
qtx; t qu2
qtx; t dt T
0
q2u1
qt qxx; t q2u2
qt qxx; t dt:
246 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
19/27
The last relation implies thatT0
qu2
qtx; t
2
q2u2
qt qxx; t
2( )
dt
a T0
qu1
qtx; t qu2
qtx; t dt T
0
q2u1
qt qxx; t q
2u2
qt qxx; t dt;
which obviously yieldsT0
qu2
qtx; t
2 q2u2qt qx x; t 2
( )dta
T0
qu1
qtx; t
2 q2u1qt qx x; t 2
( )dt:
Combining (5.5) with the inequality above, we obtain
T0
qu
qtx; t 2 q2uqt qx x; t 2( ) dta 4 T0 qu1qt x; t
2
q2u1
qt qxx; t 2( ) dt:
5:12Therefore, inequality (5.9) follows from (5.10) and (5.12). 9
We are now ready to prove the main results.
Proof of Theorem 2.5. 1. Proof of the first assertion. The energy Eut of a fi-nite energy solution of (1.1)(1.3) decays exponentially if and only if there existsT > 0 and C> 0 (depending on T) such that
Eu0 EuTbCEu0; E u0
u1
A V H10 0; p: 5:13
By using (2.3), the relation above can be rewrittenT0
qu
qtx; t
2
q2u
qt qxx; t
2( )
dtbCEu0; E u0
u1
A V H10 0; p:
By considering again the decomposition u u1 u2, introduced in (5.5) and byusing Lemma 5.1, the above inequality is equivalent toT
0
qu1
qtx; t
2 q2u1qt qx x; t 2
( )dtbCEu0; E u
0
u1
A V H10 0; p:
By Proposition 4.1 the inequality above holds if and only if x=p is a rationalnumber with coprime factorization
x
p p
q
;
where q is odd. Thus we have proved the first assertion of Theorem 2.5.
Decay Rates for a Beam with Pointwise Force and Moment Feedback 247
-
7/28/2019 Rayleigh Beam Decay Rates
20/27
2. Proof of the second assertion. By Proposition 4.1 and Lemma 5.1, thesolution u of (1.1)(1.3) satisfies the inequality
T
0
qu
qt x; t
2
q2u
qt qx x; t
2
( ) dtbK1u0
u1
2
H10;pL20;p;
Eu0
u1
A V H10 0; p;
for some constant K1 > 0. By using (2.3) we see that the left-hand side of the lastinequality is equal to Eu0 EuT, so we obtain that there exists a constantK1 such that, for all
u0
u1
A V H10 0; p we have
u
T
u 0T 2
VH100;pa
u0
u1 2
VH10;p K
1
u0
u1 2
H10;pL20;p;
5:14
where we denote by 0 the derivative with respect to time.By using a simple interpolation inequality (see p. 49 of [16]), the fact that,
according to (2.3), the function
t ! utu 0t
2VH1
00;p
is nonincreasing and by relation (5.14) we obtain the existence of a constant K2 > 0such that
uTu 0T
2VH1
00;pa
u0
u1
2VH1
00;p
K2
uTu 0T
4VH1
00;p
u0
u1
2Y
; 5:15
where Y is given by (2.7).We follow now the method used in [21]. The estimate (5.15) remains valid in
successive intervals kT
; k
1T
, so we haveuk 1Tu 0k 1T
2VH1
00;p
aukTu 0kT
2VH1
00;p
K2
uk 1Tu 0k 1T
4VH1
00;p
ukTu 0kT
2
Y
:
Since A generates a semigroup of contractions in DA and the graph norm onDA is equivalent to k kY, the relation above implies the existence of a constant
248 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
21/27
K3 > 0 such that
uk 1Tu 0k 1T
2
VH100;p
aukTu 0kT
2VH1
00;p
K3
uk 1Tu 0k 1T
4VH1
00;p
u0
u1
2Y
; Eu0
u1
A DA:
5:16If we adopt now the notation
Ek
ukTu 0kT
2
VH100;p
u0
u1 2
Y
;
5:17
relation (5.16) givesEk1aEk K3E2k1; Ekb 0: 5:18
By applying Lemma 1.2 from [21] and using relation (5.18) we obtain theexistence of a constant M> 0 such that
ukT
u 0kT 2
VH10 0;pa
Mu0
u1
2
Y
k 1; Ekb 0;
which leads to (2.13). 9
6. Proof of Proposition 2.4 and Theorem 2.6
Proof of Proposition 2.4. Define
H V L20; pequipped with the inner product
y1v1
; y2v2
H
p0
px d2y1dx2
d2y2dx2
rxv1v2 dx: 6:1Furthermore, we define an operator A in H by
DA yv
AH
y; v A V; px d2y
dx2A H20; xXH2x; p
d2y
dx20 d
2y
dx2p 0; d
dxp
d2y
dx2
x
vx
p
d2y
dx2 x dvdx x
8>>>>>>>>>>>>>>>:
9>>>>>>>>=
>>>>>>>>;;
6:2
Decay Rates for a Beam with Pointwise Force and Moment Feedback 249
-
7/28/2019 Rayleigh Beam Decay Rates
22/27
Ay; v v; 1rx
d2
dx2px d
2y
dx2
vxdx dv
dxx ddx
dx
; 6:3
where the derivatives with respect to x are calculated in D 00; p.The existence and uniqueness of finite energy solutions of (1.4)(1.6) can be
obtained by standard semigroup methods.In order to prove the estimate (2.4), it suces to note that it holds for solutions
uqu
qt
!A C0;T;DA
and to use the density ofDA in V L20; p. 9Lemma 6.1. The spectrum ofA contains no point on the imaginary axis.
Proof. Since A has compact resolvent, its spectrum sA only consists of eigen-values ofA. We will show that the equation
Az ibz 6:4with z y
v
A DA and b0 0 has only the trivial solution.
By taking the inner product of (6.4) with z AH and using
RehAz; ziH jvxj2 dv
dxx
2
; 6:5
we obtain that vx dv=dxx 0. Next, we eliminate v in (6.4) to get a fourth-order ordinary dierential equation with variable coecients:
d2
dx2px d
2y
dx2
b2rxy 0;
y0 yp d2y
dx20 d
2y
dx2p 0;
yx dydx
x 0:
8>>>>>>>>>>>>>:
6:6
By Lemma 2.3 in [15] the above system only has a trivial solution. 9
Proof of Theorem 2.6. By a classical result (see [12] and [19]) it suces to showthat A satisfies the following two conditions:
rAI fibjbA Rg1 iR 6:7and
lim supjbj!y
kibA1k
-
7/28/2019 Rayleigh Beam Decay Rates
23/27
numbers bn ! y and a sequence of vectors zn ynvn
A DA with kznkH 1such that
kibnI AznkH ! 0 as n ! y; 6:9i.e.
ibnyn vn1 fn ! 0 in V; 6:10
ibnrxvn d2
dx2px d
2yn
dx2
1 gn ! 0 in L20; p: 6:11
Our goal is to derive from (6.9) that kznkH converges to zero, thus, a contradiction.The proof is divided into four steps:
First step. We notice that from (6.5) we have
kibnI
A
zn
kHb
jReh
ibnI
A
zn; zniH
j jvn
x
j2
dvn
dx x
2
:6:12
Then, by (6.9),
jvnxj ! 0; dvndx
x ! 0: 6:13
This further leads to
jbnynxj ! 0; bndyn
dxx
! 0; 6:14
due to (6.10) and the trace theorem.
Second step. We now express vn as a function of yn from (6.10) and substitute itinto (6.11) to get
b2nryn d2
dx2p
d2yn
dx2
gn ibn fn: 6:15
Next, we take the inner product of (6.15) with qxdyn=dx in L20; x whereqx A C20; x and q0 0. We obtain that
x
0 b2
n
ryn
d2
dx2p
d2yn
dx2 qx dyndx dxx
0
gn ibnfnqxdyn
dxdx
x
0
gnqx dyndx
dx x
0
qdfn
dxbnyn dx
x
0
fndq
dxbnyn dx fnxqxbnynx: 6:16
It is clear that the right-hand side of (6.16) converges to zero since fn; gn convergeto zero in H2 and L2, respectively.
Decay Rates for a Beam with Pointwise Force and Moment Feedback 251
-
7/28/2019 Rayleigh Beam Decay Rates
24/27
By a straightforward calculation,
Re
x0
b2nrynqdyn
dxdx
& ' 12rxqxjbnynxj2 12
x0
d
dxrqjbnynj2 dx 6:17
andRe
x0
d2
dx2p
d2yn
dx2
q
dyn
dxdx
& ' Re q d
dxp
d2yn
dx2
x p dq
dx
d2yn
dx2x
dyn
dxx
12pxqxd2yn
dx2x
2
x0
12 3pdq
dx dp
dx q d2yndx2 2
Re pd2q
dx2d2yn
dx2dyn
dx " # dx:Notice from (6.11) that d2=dx2pd2yn=dx2=bn is bounded in L20; xwhich further implies the boundedness of jd=dxpd2yn=dx2xj=bn andjpd2yn=dx2xj=bn. Sincex
0
dyn
dx
2 dx 1ibnx
0
vnd2yn
dx2dx 1
ibn
x0
ibnyn vnd2yn
dx2dx
b
n
dyn
dx x 1bn ynx ;
then, from the boundedness of vn, ibnyn vn, d2yn=dx2, in L20; x and (6.14), wehave dyn=dx converges to zero in L
20; x. With these facts and the boundary con-ditions of yn and (6.14) at hand, we simplify (6.16), then take its real parts. Thisleads to x
0
rdq
dx dr
dxq
jbnynj2 dx
x0
3pdq
dx dp
dxq
d2yn
dx2
2 dx pxqx
d2yn
dx2 x 2
! 0: 6:18Similarly, we take the inner product of (6.15) with q1xdyn=dx in L2x; p withq1 A C
2x; p and q1p 0, then repeat the above procedure. This will give uspx
rdq1
dx dr
dxq1
jbnynj2 dx
px
3pdq1
dx dp
dxq1
d2yn
dx2
2 dx pxq1x d
2yn
dx2x
2
! 0: 6:19
Third step. Next, we show that both of jd2yn=dx2xj and jd2yn=dx2xjconverge to zero. To proceed, we take the inner product of (6.15) with
252 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
25/27
1=jnejnhx in L20; x, where jn ffiffiffiffiffiffiffiffijbnjp , and
hx x
x
rsps 1=4
ds:
This leads to
x
0
j3n ynrejnh dx
x0
d2
dx2p
d2yn
dx2
1jn
ejnh dx ! 0: 6:20
Performing integration by parts four times to the second term on the left-hand sideof (6.20), we obtainx
0
d2
dx2p
d2yn
dx2
1
jnejnh dx
1
jn
d
dx
pd2yn
dx2 pd2yn
dx2
dh
dx jn
dyn
dx
pdh
dx 2
jbn
jynp
dh
dx 3
!ejnhjx0x
0
pd2h
dx2d2yn
dx2 d
dxp
dh
dx
2 !jn
dyn
dx d
dxp
dh
dx
3 !bnyne
jnh( )
dx
x
0
j3nh 04ynpejnh dx: 6:21
Since ejnh converges to zero in L20; x, the first integral term on the right-handside of (6.21) will also converge to zero. We then substitute (6.21) into (6.20) to
cancel the other inner product term. Among the remaining boundary terms, thoseat x 0 all converge to zero due to the exponential decay term ejnh0 and theboundedness of d2=dx2pd2yn=dx2=bn in L20; x; those containing ynx,dyn=dxx also converge to zero due to (6.14). Thus, we simplify (6.20) to
1
jn
d
dxp
d2yn
dx2
x px rx
px 1=4
d2yn
dx2x ! 0: 6:22
Similarly, we take the inner product of (6.15) with 1=jnejnh1x in L2x; p where
h1
x
x
x
rsps
1=4
ds:
Repeating the above analysis yields
1
jn
d
dxp
d2yn
dx2
x px rx
px 1=4
d2yn
dx2x ! 0: 6:23
Recall thatd
dxp
d2yn
dx2
x d
dx
d2yn
dx2
x vnx
converges to zero. Then the dierence of (6.22) and (6.23) leads to
d2yn
dx2x d
2yn
dx2x ! 0: 6:24
Decay Rates for a Beam with Pointwise Force and Moment Feedback 253
-
7/28/2019 Rayleigh Beam Decay Rates
26/27
Also recall that
pd2yn
dx2
x p d
2yn
dx2
x dvn
dxx
converges to zero. Therefore, we have proved
d2yn
dx2x ! 0; d
2yn
dx2x ! 0: 6:25
In view of (6.25), we simplify (6.18) and (6.19) tox0
rdq
dx dr
dxq
jbnynj2 dx
x0
3pdq
dx dp
dxq
d2yn
dx2
2 dx ! 0 6:26and
px
r
dq1
dx dr
dx q1 jbnynj2 dx px 3p dq1dx dpdx q1 d2yn
dx2 2
dx ! 0: 6:27Fourth step. Finally, we choose qx and q1x so that rdq=dx dr=dxqand 3pdq=dx dp=dxq are strictly positive, and rdq1=dx dr=dxq1 and3pdq1=dx dp=dxq1 are strictly negative. This can be done by taking
qx eax 1; q1x e bpx 1; 6:28with
a max kdr=dxkyr
0
;kdp=dxky
3p0& '; b min
kdr=dxkyr
0
;kdp=dxky
3p0& ':
Therefore, (6.26) and (6.27) imply
kbnynkL20;p ! 0; kynkV ! 0: 6:29In view of (6.10), we also get
kvnkL20;p ! 0; 6:30which clearly contradicts (6.9). 9
Remark 6.2. A dierent method, working only in the case rx px 1, wasproposed in [1]. This method, combining ideas from [20] and [3], uses the permu-tation of one of the inputs with the corresponding output of the system.
References
[1] K. Ammari, Asymptotic behavior of coupled EulerBernoulli beams with pointwise dissipation,
ESAIM Proceedings, 8 (2000), 112.
[2] K. Ammari and M. Tucsnak, Stabilization of BernoulliEuler beam by means of a pointwise
feedback force, SIAM J. Control. Optim., 39 (2000), 11601181.
[3] K. Ammari and M. Tucsnak, Stabilization of second order equations by a class of unbounded
feedbacks, ESAIM COCV, 6 (2001), 361386.[4] J. M. Ball and M. Slemrod, Nonharmonic Fourier series and the stabilization of semilinear
control systems, Comm. Pure Appl. Math., 32 (1979), 555587.
254 K. Ammari, Z. Liu, and M. Tucsnak
-
7/28/2019 Rayleigh Beam Decay Rates
27/27
[5] A. Bensoussan, G. Da Prato, M. C. Delfour and S. K. Mitter, Representation and Control of
Infinite Dimensional Systems, Vol. I, Birkhauser, Basel, 1992.
[6] H. Brezis, Analyse Fonctionnelle, Theorie et Applications, Masson, Paris, 1983.
[7] G. Chen, M. C. Delfour, A. M. Krall and G. Payre, Modeling, stabilization and control of
serially connected beams, SIAM J. Control Optim., 25 (1987), 526546.
[8] G. Chen, S. G. Krantz, D. W. Ma, C. E. Wayne and H. H. West, The EulerBernoulli beamequation with boundary energy dissipation, in Operator Methods for Optimal Control Problems,
Sung J. Lee, ed., Marcel Dekker, New York, 1988, pp. 6796.
[9] G. Chen, S. G. Krantz, D. L. Russell, C. E. Wayne, H. H. West and M. P. Coleman, Analysis,
design and behavior of dissipative joints for coupled beams, SIAM J. Control Optim., 28 (1990),
423437.
[10] C. Dafermos, Asymptotic behavior of solutions of evolution equations, in Nonlinear Evolution
Equations, M. G. Crandall, ed., Academic Press, New York, 1978, pp. 103123.
[11] G. Doetsch, Introduction to the Theory and Application of the Laplace Transformation, Springer-
Verlag, Berlin, 1974.
[12] F. Huang, Characteristic conditions for exponential stability of linear dynamical systems in
Hilbert space, Ann. Dierential Equations, 1 (1985), 4356.
[13] I. Lasiecka, J.-L. Lions and R. Triggiani, Nonhomogeneous boundary value problems for secondorder hyperbolic operators, J. Math. Pures Appl., 65 (1986), 149192.
[14] P. D. Lax and R. S. Philips, Scattering theory for dissipative hyperbolic systems, J. Funct. Anal.,
14 (1973), 172235.
[15] W. Leighton and Z. Nehari, On the oscillation of solutions of self-adjoint linear dierential
equations of the fourth order, Trans. Amer. Math. Soc., 89 (1958), 325377.
[16] J. L. Lions and E. Magenes, Problemes aux Limites non Homogenes et Applications, Vol. 1,
Dunod, Paris, 1968.
[17] K. Liu and Z. Liu, Boundary stabilization of nonhomogeneous beam with rotatory inertia at the
tip, Comp. Appl. Math., 114 (2000), 110.
[18] A. Pazy, Semigroups of Linear Operators and Applications to Partial Dierential Equations,
Applied Mathematical Sciences, Vol. 44, Springer-Verlag, New York, 1983.
[19] J. Pruss, On the spectrum of C0-semigroups, Trans. Amer. Math. Soc., 248 (1984), 847857.[20] R. Rebarber, Exponential stability of beams with dissipative joints: a frequency domain approach,
SIAM J. Control. Optim., 33 (1995), 128.
[21] D. L. Russell, Decay rates for weakly damped systems in Hilbert space obtained with control
theoretic methods, J. Dierential Equations, 19 (1975), 344370.
[22] G. Weiss, O. J. Staans and M. Tucsnak, Well-posed linear systemsa survey with emphasis on
conservative systems, Appl. Math. Comput. Sci., 11 (2001), 101127.
Decay Rates for a Beam with Pointwise Force and Moment Feedback 255