rayleigh beam decay rates

7/28/2019 Rayleigh Beam Decay Rates

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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


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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


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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


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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



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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



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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



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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



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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;



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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:



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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



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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



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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;



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withP

nb1 n4a2n


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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



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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



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.



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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



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(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:



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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.



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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



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



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


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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.



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



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



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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.

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rayleigh beam decay rates

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