On the stabilization of a vibrating equation

Nonlinear Analysis 39 (2000) 537–558www.elsevier.nl/locate/na

On the stabilization of a vibrating equation

B. Chentouf, C.Z. Xu, G. Sallet ∗

INRIA-Lorraine (CONGE project) & URA CNRS 399 (MMAS). ISGMP, Bat. A, Universit�e de Metz,Ile de Saulcy 57045 Metz cedex 01, France

Received 22 May 1997; accepted 8 April 1998

Keywords: Boundary velocity feedback; Nonlinear contractions semigroup; Asymptotic stability;LaSalle’s invariance principle; Decay rate estimate; Multiplier

1. Introduction and main results

We consider the following modelization of a exible torque arm [10], controlled bythe two feedback laws U1 and U2 to be determined:

ytt(x; t)− (ayx)x (x; t) + �yt(x; t) + �y(x; t) = 0; 0¡ x ¡ 1; t ¿ 0;

(ayx) (0; t) = �1U1(t); t ¿ 0;

(ayx) (1; t) = �2U2(t); t ¿ 0;

(1.1)

where

� ≥ 0; � ¿ 0; �1; �2 ≥ 0; �1 + �2 6= 0;a ∈ W 1;∞(0; 1); a(x) ≥ a0 ¿ 0 for all x ∈ [0; 1]:

(1.2)

In Eq. (1.1), U1 and U2 will be determined such that the closed-loop system is asymp-totically stable, i.e, (y(·; t); yt(·; t))→ 0 as t → +∞ in some functional space. In thispaper, we prove that for all � ¿ 0 and for all � ≥ 0, system (1.1) is asymptoticallystabilized by the nonlinear feedback law depending only on the boundary velocities:

U1(t) = f(yt(0; t));

U2(t) = g(yt(1; t));(F)

∗ Corresponding author. Tel.: 0033387 54 72 67; fax: 0033 387 54 72 77; e-mail: chentouf,xu,[email protected]

0362-546X/99/$ - see front matter ? 1999 Elsevier Science Ltd. All rights reserved.PII: S0362 -546X(98)00220 -X

538 B. Chentouf et al. / Nonlinear Analysis 39 (2000) 537–558

Fig. 1. Illustration of nonlinear function f.

where f and g are suitable nonlinear functions. The proof is based on LaSalle’s prin-ciple. Moreover, we show that the closed-loop system is exponentially stable undersome conditions on f, g and a(x). Actually, we have a great degree of freedom todetermine the feedback functions f and g (see Fig. 1). Decay rate estimates of theenergy are given for the di�erent feedback laws proposed. The multiplier method issuccessfully used. Furthermore, the exponential or strong stability that we obtain hereis global.In the case where � = 0, the boundary stabilization of (1.1) has been studied by

many authors. Rao [17] has proved that the feedback law

U1(t) = y(0; t) + F(yt(0; t));

U2(t) = −Mytt(1; t); M; ¿ 0;

stabilizes asymptotically the system under a suitable choice of F . A stabilization resulthas also been obtained by Conrad and Rao [7] with the following feedback:

U1(t) = �y(0; t) + F(yt(0; t));

U2(t) = −�y(1; t)− F(yt(1; t)); � ¿ 0:

In [1], for a(x) = a1x + a0 the authors have established asymptotic stabilization with

U1(t) = 0;

U2(t) = −k1y(1; t)− h(yt(1; t)); k1 ¿ 0;

where h is an appropriate function. A stabilization result has been obtained in [10, 4]by the feedback

U1(t) = kpy(0; t) + kvyt(0; t) +∫ 1

0G(x)y(x; t) dx +

kvkp

∫ 1

0G(x)yt(x; t) dx;

U2(t) = 0;

with kp; kv ¿ 0 and G(x) is a function in L2(0; 1). The stabilization problem of aquasilinear string equation has been extensively studied (see [14,18] and the referencestherein). For instance, in Slemrod’s work [18], at the di�erence of our model the

B. Chentouf et al. / Nonlinear Analysis 39 (2000) 537–558 539

nonlinear string is supposed to be �xed at one end and controlled at the other end.Slemrod has proposed a stabilizing feedback law which contains the boundary velocityand proved exponential stability of the controlled system. However, he has obtainedthe exponential stabilization only for small and smooth initial conditions. In all thereferences cited above, the stabilizing feedback laws for U1 or U2 contain not onlyboundary velocities but also a boundary position term.The stabilization problem for system (1.1) with � 6= 0 has been considered in Haraux

[9] and Komornik [13]. Since our objective is to stabilize the system by means offeedback depending only on boundary velocity, we have to justify why we choose� ¿ 0. If � = 0 and the feedback law only depends on velocities, we may encounterthe situation where the closed-loop system is not well-posed in terms of the semigroupsin the Hilbert space H 1

0 (0; 1)× L2(0; 1), where

Hm0 (0; 1) =

{f ∈ Hm(0; 1);

∫ 1

0f(x) dx = 0

}for m = 1; 2; : : : :

As a matter of fact, let U1(t) = kyt(0; t); U2(t) = −kyt(1; t) (k ¿ 0) and consider thecontrolled system in the linear case:

ytt(x; t)− yxx(x; t) = 0; 0¡ x ¡ 1; t ¿ 0;

yx(0; t) = kyt(0; t); t ¿ 0;

yx(1; t) = −kyt(1; t); t ¿ 0:

From this system, de�ne the operator A by

D(A) ={(y; z) ∈ H 2

0 (0; 1)× H 10 (0; 1); yx(0) = kz(0); yx(1) = −kz(1)

};

and

A(y; z) = (z; yxx) ; ∀(y; z) ∈ D(A):

It is easy to show that the operator A is dissipative in the Hilbert space H 10 (0; 1)× L2

(0; 1) equipped with the the inner product 〈(f1; g1); (f2; g2)〉 =∫ 10 (f1xf2x + g1g2) dx.

We claim that A is not a generator of a C0-semigroup on H 10 (0; 1)× L2(0; 1). Suppose

the contrary to be true. Necessarily, the semigroup is a contraction one. This impliesthat R(I − A) = H 1

0 (0; 1) × L2(0; 1) (see [16], p. 14). However there is no (y; z) inD(A) such that (I − A)(y; z) = (f; g) with

f(x) = 6x2 − 6x + 1;g(x) = 1− f(x);

so, the claim is proved. On the other hand, if � = 0 and � ¿ 0, the system (1.1)is conservative without control. Because of these reasons, we suppose that � ¿ 0throughout the paper.


With the feedback law in (F), we obtain the closed-loop system:

ytt(x; t)− (ayx)x (x; t) + �yt(x; t) + �y(x; t) = 0; 0¡ x ¡ 1; t ¿ 0;

(ayx) (0; t) = �1f(yt(0; t)); t ¿ 0;

(ayx) (1; t) = �2g(yt(1; t)); t ¿ 0:

(1.3)

Introduce the energy associated to Eq. (1.3) as follows:

E(t) =12

∫ 1

0

(�y2(x; t) + y2t (x; t) + a(x)y2x (x; t)

)dx:

An easy formal computation shows that

E(t) = �2yt(1; t)g(yt(1; t))− �1yt(0; t)f(yt(0; t))− �∫ 1

0y2t (x; t) dx: (1.4)

To make the energy decreasing, we assume in (1.4) that f and g satisfy the condition:f(�)� ≥ 0 and g(�)� ≤ 0 for all � ∈ R: Let the Hilbert spaceH=H 1(0; 1)×L2(0; 1)be equipped with the inner product

〈(f1; g1); (f2; g2)〉H =∫ 1

0(af1xf2x + �f1f2 + g1g2 ) dx:

De�ne a nonlinear operator A by

D(A) =

{(y; z);y ∈ H 2(0; 1); z ∈ H 1(0; 1);

(ayx)(0) = �1f(z(0))

(ayx)(1) = �2g(z(1))

}; (1.5)

and for all (y; z) ∈ D(A)

A(y; z)(x) = (−z;−(ayx)x + �y + �z) : (1.6)

With the initial data � = (�1; �2), the closed-loop system (1.3) can be formulated inthe form of the evolution equation over H

�(t) + A�(t) = 0;

�(0) = �;(�)

where �(t) = (y(: ; t); yt(: ; t)). Throughout this paper, f and g are assumed to satisfythe hypothesis (H)

f ∈ C0(R) is increasing such that f(0) = 0; f(�)� ¿ 0 ∀� 6= 0;g ∈ C0(R) is decreasing such that g(0) = 0; g(�)� ¡ 0 ∀� 6= 0:

(H)

Our main results are stated as follows:


Theorem 1. For any initial data (�1; �2) ∈ H; the energy E(t) of (�) tends asymp-totically to zero in H as t → +∞:

Theorem 2. We assume that � = 0 (without damping) and that a(x) is a monotonefunction. Let (y; z) be the solution of (�) stemmed from (�1; �2) ∈ H.(i) If there exist positive constants �1; �2; �3 and �4 such that for all x ∈ R (see

Fig. 1),

�1 | x |≤| f(x) |≤ �2 | x |; �3 | x |≤| g(x) |≤ �4 | x |; (H1)

there exists a constant ! ¿ 0 such that

E(t) ≤ exp(1− t

!

)E(0); ∀t ≥ 0:

(ii) If there exist positive constants �1; �2; �3 and �4 and p¿1 such that for allx ∈ R;

�1 min{| x |; | x |p} ≤| f(x) |≤ �2 | x |;

�3 min{| x |; | x |p} ≤| g(x) |≤ �4 | x |;

(H2)

then, given any M ¿ 1 there exists a constant ! ¿ 0 depending on E(0) such that

E(t) ≤ M 2=(p+1)E(0)(1 + !t)−2=(p−1); ∀t ≥ 0:

(iii) If there exist positive constants �1; �2; �3 and �4 and 0¡ p ¡ 1 such that forall x ∈ R;

�1 | x |≤| f(x) |≤ �2 max{| x |; | x |p} ;

�3 | x |≤| g(x) |≤ �4 max{| x |; | x |p} ;

(H3)

then, given any M ¿ 1 there exists a constant ! ¿ 0 depending on E(0) such that

E(t) ≤ M 2p=(p+1)E(0)(1 + !t)−2p=(1−p):

Theorem 3. Assume that � ¿ 0 (with damping) and let (y; z) be the solution of (�)stemmed from (�1; �2) ∈ H. Then, all the assertions (i)–(iii) of Theorem 2 are alsotrue without assuming a(x) to be monotone.

The paper is organized as follows. The next section is devoted to the proof ofour main results. It is divided into three subsections. In the �rst one, we prove thatour closed-loop system is well-posed in terms of the semigroup theory. The secondsubsection contains the proof of Theorem 1. The third subsection gives a proof ofTheorems 2 and 3. A short conclusion ends our paper.


2. Proof of the main results

2.1. Well-posedness of the system (�)

Lemma 1. The operator A de�ned by Eqs. (1.5) and (1.6) is maximal monotone onH with the domain D(A) dense in H. So, −A is the generator of a continuoussemigroup of contractions S(t) on H.

Proof of Lemma 1. Let (y; z); (y; z) ∈ D(A). A straightforward computation showsthat

〈 A(y; z)− A(y; z); (y; z)− (y; z) 〉H= −�2 (z(1)− z(1) ) ( g(z(1))− g(z(1)))

+�1 (z(0)− z(0)) ( f(z(0))− f(z(0)))

+�∫ 1

0(z − z)2 dx:

By virtue of Eq. (1.2) and (H), we see that these three terms are nonnegative. Thisproves the monotonicity of the operator A.

The maximality of A remains to be shown, i.e, for any given (U; V ) ∈ H, thereexists (y; z) ∈ D(A) such that (I + A)(y; z) = (U; V ). Equivalently, we seek y and zsatisfying

y − z = U;

z − (ayx)x + �z + �y = V;

(ayx) (0) = �1f(z(0));

(ayx) (1) = �2g(z(1));

y ∈ H 2(0; 1); z ∈ H 1(0; 1):

(2.1)

Eliminating the unknown z in Eq. (2.1), one obtains

(�+ � + 1)y − (ayx)x = (�+ 1)U + V;

(ayx) (0) = �1f(y(0)− U (0));

(ayx) (1) = �2g(y(1)− U (1));

y ∈ H 2(0; 1):

(2.2)

Now let us de�ne the function J (·) on H 1(0; 1) by

J ( ) =12

∫ 1

0[a 2x + (�+ � + 1) 2] dx −

∫ 1

0[(�+ 1)U + V ] dx

+F( (0)− U (0))− G( (1)− U (1));


where

F(x) = �1

∫ x

0f(�) d� and G(x) = �2

∫ x

0g(�) d�; ∀x ∈ R: (2.3)

From Eq. (1.2) and (H), we deduce that F and −G are convex, non negative and theyare in C1(R). Then, it is easy to check that J (·) is convex and coercive. Moreover,one can verify that J (·) is continuous in H 1(0; 1). Hence by a minimization theorem(Proposition 38.15, pp.155, [19]), there exists a function y ∈ H 1(0; 1) such that

J (y) = inf ∈H 1(0;1)

J ( ):

This implies that the function � : � → �(�) = J (y+ � ) admits a minimum at � = 0and thus

dd�(J (y + � )) |�=0 = 0; ∀ ∈ H 1(0; 1):

This means that for any ∈ H 1(0; 1), we have

(�+ � + 1)∫ 1

0y dx +

∫ 1

0ayx x dx −

∫ 1

0[(�+ 1)U + V ] dx

+�1 (0) f(y(0)− U (0))− �2 (1) g(y(1)− U (1)) = 0:

(2.4)

In particular, for any ∈ C∞0 (0; 1)

−∫ 1

0ayx x dx =

∫ 1

0[(�+ � + 1)y − (�+ 1)U − V ] dx: (2.5)

This implies that

y ∈ H 2(0; 1);

(�+ � + 1)y − (ayx)x = (�+ 1)U + V:(2.6)

Integrating Eq. (2.4) by parts and using Eq. (2.6), we show that

(ayx) (0) = �1f(y(0)− U (0));

(ayx) (1) = �2g(y(1)− U (1)):(2.7)

Combining Eqs. (2.6) and (2.7), we deduce that y is a solution of system (2.2). Now,we de�ne an element (y; z) by

y; solution of (2:2);

z = y − U;

which satis�es clearly system (2.1) and thus the maximality of the operator A is proved.


The density of D(A) in H is a direct consequence of monotonicity, maximality ofA and the assumptions f(0) = g(0) = 0.

The following lemma is a direct consequence of Lemma 1 and the method ofBrezis [3].

Lemma 2. (i) For any initial data (�1; �2) ∈ D(A); the equation (�) admits a uniquesolution (y; z) ∈ D(A) such that

ddt(y; z) ∈ L∞(R+;H):

The solution (y; z) is given by (y; z) = S(t)(�1; �2); for all t ≥ 0. Moreover; thefunction t 7−→ ‖A(y; z)‖H is decreasing.(ii) For any initial data (�1; �2) ∈ H; the equation (�) admits a unique weak

solution (y; z) = S(t)(�1; �2) such that (y; z) ∈ C0(R+;H):

2.2. Proof of Theorem 1

Note that by a standard argument of density of D(A) in H and the contraction of thesemigroup S(t), it su�ces to prove Theorem 1 for any initial data (�1; �2) ∈ D(A).Let (�1; �2) ∈ D(A). By virtue of i) of Lemma 2, we know that the trajectory ofsolution {(y; z)}t≥0 is a bounded set for the graph norm. Furthermore, one can showthat the injection i : (D(A); ‖ ‖D(A))→ H is compact. This implies that the trajectoryis precompact in H. Applying LaSalle’s invariance principle [8, 12], we deduce thatthe !-limit set

! (� 1; � 2) ={( 1; 2) ∈ H; ( 1; 2)

= limn→∞ S(tn) (�1; �2) with tn → ∞ as n → ∞

}is nonempty, compact, invariant under the semigroup S(t) and

S(t) (�1; �2)→ ! (�1; �2) as t → +∞:

Moreover, we deduce from the maximality of A that ! (�1; �2)⊂D(A) (see [3]). Inorder to prove the asymptotic stability, it is su�cient to show that the !-limit set! (�1; �2) reduces to {0}. For this, let (�1; �2) ∈ ! (�1; �2) and

(y; z) = S(t)(�1; �2)⊂! (�1; �2)⊂D(A):

On the other hand, ‖(y; z)‖H = ‖S(t)(�1; �2)‖H is a constant [9]. Thus,

ddt

‖(y; z)‖2H = 0:

This, together with (i) of Lemma 2, implies that

〈 A(y; z); (y; z) 〉H = 0 (2.8)


Now, from the proof of Lemma 2, we obtain

〈 A(y; z); (y; z) 〉H = �1z(0) · f (z(0))−�2z(1) · g (z(1)) + �‖ y ‖2L2(0;1): (2.9)

Without loss of generality, we assume that �2 6= 0. The other cases can be studied inthe same way. We deduce from Eqs. (2.8), (2.9) and (H) that y veri�es the followingsystem:

ytt(x; t)−(ayx

)x (x; t) + �y(x; t) = 0; 0¡ x ¡ 1; t ¿ 0;

yx(0; t) = yx(1; t) = 0; t ¿ 0;

yt(1; t) = 0; t ¿ 0;

�‖ y ‖2L2(0;1) = 0;(y(0); yt(0)) = (�1; �2) ∈ ! (�1; �2) :

(2.10)

If � 6= 0, then it is obvious that y = yt = 0. Otherwise, let us consider as in [5] thefollowing linear operator:

D(A0) ={(y; z);y ∈ H 2(0; 1); z ∈ H 1(0; 1); yx(0) = 0; yx(1) = 0

};

and for all (y; z) ∈ D(A0),

A0(y; z)(x) = (−z;−(ayx)x + �y) :

A complex number � is an eigenvalue of A0 if and only if there exists a non trivial(y; z) ∈ D(A0) such that (� − A0)(y; z) = 0, i.e,

�y = z;

(ayx)x − �y = �2y;

yx(0) = yx(1) = 0:

(2.11)

Thanks to a classical result of Mikha��lov [15], the spectral system

(ayx)x − �y = �y;

yx(0) = yx(1) = 0;

admits an in�nite number of real eigenvalues 0¿�1¿�2¿: : : , such that (�n) → −∞as n → +∞ and the associated eigenfunctions v1; v2; : : : form an orthonormal basis ofL2(0; 1): Thus, the eigenvalues �k and the associated eigenfunctions V ∗

k of the operatorA0 are given by

�k = ±i√−�k ; k = 1; 2; : : : ;

V ∗k =

(vk ; ±i

√−�k vk ;

); k = 1; 2; : : : :


We note that for all k = 1; 2; : : :

‖ V ∗k ‖2H = −2�k :

We set

�k = i√−�k ; k = 1; 2; : : : ;

�−k = −i√−�k ; k = 1; 2; : : :

and

Vk =1√−2�k

(vk ; −i

√−�k vk

); k = 1; 2; : : : ;

V−k =1√−2�k

(vk ; i

√−�k vk

); k = 1; 2; : : :

which form an orthonormal basis of H. Then, the solution of (2.10) is given by(y; yt

)(t) =

∑k∈Z

Cke−�k tVk ; (2.12)

where Ck = 〈 (�1; �2); Vk 〉H (for the complexed scalar product), for any k ∈ Z: One�nds that

Ck = ak + ibk ; k = 1; 2; : : : ;

C−k = ak − ibk ; k = 1; 2; : : :(2.13)

with

ak =1√−2�k

∫ 1

0

(a�1xvk x + ��1vk

)dx and bk = − 1√

2

∫ 1

0�2vk dx: (2.14)

After an easy computation, we get from Eqs. (2.12) and (2.14)

y(t) =∞∑n=1

(an cos(

√−�nt)− bn sin(

√−�nt)

) √2√−�n

vn;

yt(t) =∞∑n=1

(an sin(

√−�nt) + bn cos(

√−�nt)

)√2vn:

(2.15)

Following the method used by Conrad and Pierre [6], we will prove that an = bn =0 for any n = 1; 2; : : : and thus (y(t); yt(t)) = (0; 0): Indeed, we know from thetheory of linear di�erential equations that vn(1) 6= 0 for n = 1; 2; : : : . Moreover, since(y(0); yt(0)) ∈ D(A0), i.e,

∞∑n=1

an

√−�nvn ∈ L2(0; 1);

yt(0) = �2 =√2

∞∑n=1

bnvn ∈ H 1(0; 1);(2.16)


and since (vn)n≥1 (resp.(vn=

√−�n)n≥1 is an orthonormal basis for L2(0; 1) (resp.

H 1(0; 1), one can verify that the series de�ning yt(t) in Eq. (2.15) converges inH 1(0; 1) uniformly in t. By continuity of the trace operator u 7→ u(1) in H 1(0; 1),the equation yt(1; t) = 0 reads as

yt(1; t) =∞∑n=1

(an sin(

√−�nt) + bn cos(

√−�nt)

)vn(1)

=12i

∑k∈Z

Cke�k tvk(1) = 0;

(2.17)

where

v−n = −vn; n = 1; 2; : : :

On the other hand, the distribution of eigenvalues√−�n is given in [2] (Theorem 9,

p. 303) by

√−�n = n�+

O(1)n

;

which implies that

limn→+∞

(√−�n+1 −

√−�n

)= � ¿ 0:

Now, let

SN (t) =n=N∑

n=−N

Cne�ntvn(1); t ¿ 0:

We know from Eq. (2.17) that

limN→+∞

SN (t) = 0 uniformly in t ∈ [−T; T ]:

Then, using Ingham’s inequality [11], we deduce that there exists a constant ¿ 0such that

n=N∑n=−N

| Cnvn(1) |2≤ ∫ T

−T| SN (t) |2 dt:

This implies, when N → +∞, that∑k∈Z

| CkVk(1) |2≤ 0:

This means that Ck = 0 for any k ∈ Z and thus (y; yt) = 0. Therefore, the proof ofTheorem 1 is complete.


2.3. Proof of Theorem 2 and Theorem 3

Proof of Theorem 2. (i) First, we assume that a(x) is monotone nonincreasing. Let usmultiply Eq. (1.1) by 2x(C0 + �)yx + C0y (where C0; � ¿ 0 are given later in Eqs.(2.24) and (2.27)) and integrate with respect to x and t. A straightforward computationshows that∫ T2

T1

∫ 1

0

[�y2t +

(C0 + (�+ C0)(1− x

ax

a))ay2x − ��y2

]dx dt

+(C0 + �)∫ T2

T1

[−y2t (1; t)−

�2a(1)

g2 (yt(1; t)) + �y2(1; t)]dt

= −∫ 1

0[2(C0 + �)xytyx + C0yty]

t=T2t=T1 dx

+C0

∫ T2

T1

[�2g (yt(1; t))y(1; t)− �1f (yt(0; t))y(0; t)] dt: (2.18)

We need the following important inequality in H 1(0; 1) whose proof is postponed tothe Appendix. For any � ¿ 1 and u ∈ H 1(0; 1), we have∫ 1

0u2(x) dx ≤

(�2

�2 − 1)

u2(1) +(

�4

�2 − 1)∫ 1

0u2x(x) dx: (2.19)

It follows that

− ��∫ 1

0y2(x; t) dx ≥ −

(��2

�2 − 1)

y2(1; t)− ��4

a0(�2 − 1)∫ 1

0ay2x (x; t) dx: (2.20)

By means of Cauchy–Schwartz’s inequality and the monotonic decrease of E(t), weobtain

−∫ 1

0[2(C0 + �)xytyx + C0yty]

t=T2t=T1 dx ≤ M1E(T1); (2.21)

where

M1 = max

(2C0 + �√

a0;C0√�

):

Moreover, since a(x) is a positive monotone nonincreasing function, we have

C0 + (�+ C0)(1− x

ax

a

)≥ �+ 2C0 ∀x ∈ (0; 1): (2.22)

Substituting Eqs. (2.20), (2.21) and Eqs. (2.22) into (2.18), we get∫ T2

T1

∫ 1

0

[�y2t +

((�+ 2C0)− ��4

a0(�2 − 1))

ay2x

]dx dt

+(C0 + �− ��2

�2 − 1)∫ T2

T1�y2(1; t) dt


≤ M1E(T1) + (C0 + �)∫ T2

T1

[y2t (1; t) +

�2a(1)

g2 (yt(1; t))]dt

+C0

∫ T2

T1

[�2g (yt(1; t))y(1; t)− �1f (yt(0; t))y(0; t)] dt: (2.23)

We choose C0 as

C0 = max{

��4

2a0(�2 − 1) ; 3 +�

�2 − 1}

; (2.24)

so that

(�+ 2C0)− ��4

a0(�2 − 1) ≥ � and C0 + �− ��2

�2 − 1 ≥ 3:

Then, we can rewrite (2.23) as∫ T2

T1

∫ 1

0

(�y2t + �ay2x

)dx dt + 3�

∫ T2

T1y2(1; t) dt

≤ M1E(T1) + (C0 + �)∫ T2

T1

[y2t (1; t) +

�2a(1)

g2 (yt(1; t))]dt

+C0

∫ T2

T1

[�2g (yt(1; t))y(1; t)− �1f (yt(0; t))y(0; t)] dt: (2.25)

Assume that �1�2 6= 0. (Otherwise, it is su�cient to consider one term). ApplyingCauchy–Schwartz’s inequality, we have for any C3; C4 ¿ 0

C0�2g (yt(1; t))y(1; t) ≤ 12

[C20 �

22C

23g2 (yt(1; t)) +

1C23

y2(1; t)];

−C0�1f (yt(0; t))y(0; t) ≤ 12

[C20 �

21C

24f

2 (yt(0; t)) +1C24

y2(0; t)]:

(2.26)

On the other hand, a direct computation gives

y2(0; t) ≤ y2(1; t) +∫ 1

0

[y2(x; t) +

1a0

ay2x (x; t)]dx;

which implies, together with Eq. (2.19) that

y2(0; t) ≤(1 +

�2

�2 − 1)

y2(1; t) +1a0

(1 +

�4

�2 − 1)∫ 1

0ay2x (x; t) dx:

Using this inequality and Eq. (2.26), we �nd that

C0

∫ T2

T1

[�2g (yt(1; t))y(1; t)− �1f (yt(0; t))y(0; t)] dt

≤ 12a0C24

(1 +

�4

�2 − 1)∫ T2

T1

∫ 1

0ay2x dx


+12

(1C23+1C24

(1 +

�2

�2 − 1))∫ T2

T1y2(1; t) dt

+C202

∫ T2

T1

[�22C

23g2 (yt(1; t)) + �21C

24f

2 (yt(0; t))]dt:

Inserting the above inequality into Eq. (2.25), and choosing �; C3 and C4 such that

� =4�a0

(1 +

�4

�2 − 1)(

1 +�2

�2 − 1)−1

; C24 ≥1

a0�

(1 +

�4

�2 − 1)

;

C23 ≥ 1=�;(2.27)

we get∫ T2

T1

∫ 1

0

(�y2t +

�2ay2x

)dx dt +

�2

∫ T2

T1y2(1; t) dt

≤ M1E(T1) +∫ T2

T1

[(C0 + �)y2t (1; t)

+((C0 + �)�2

a(1)+

C20 �22C

23

2

)g2 (yt(1; t))

]dt

+C20 �

21C

24

2

∫ T2

T1f2 (yt(0; t)) dt:

(2.28)

Since f and g satisfy the hypothesis (H1) in (i) of Theorem 2, we deduce from Eq.(2.28) that∫ T2

T1

∫ 1

0

(�y2t +

�2ay2x

)dx dt +

�2

∫ T2

T1y2(1; t) dt

≤ M1E(T1) +∫ T2

T1

[M2yt(0; t)f(yt(0; t))−M3yt(1; t)g(yt(1; t))] dt;

with

M2 =C20 �

21C

24�22

2�1; M3 =

(C0 + �)�3

+�24�3

((C0 + �)�2

a(1)+

C20 �22C

23

2

):

Using once again Eq. (2.19), it is easy to see that

�y2(1; t) + �∫ 1

0ay2x dx ≥ M4

∫ 1

0

(�y2 + ay2x

)dx;

where � is given by Eq. (2.27) and M4 =(�2=(�2 − 1))−1 min (�; �=‖a‖∞) :

Then, combinating the two last inequalities with Eq. (1.4) gives∫ T2

T1E(t) dt ≤ M

CE(T1);


with

C = min(2�;M4) and M = M1 + max(M2

�1;M3

�2

):

Finally, thanks to a classic result (Theorem 8.1, p. 103, [13]), we obtain

E(t) ≤ exp(1− Ct

M

)E(0); ∀t ≥ 0:

We have thus proved the �rst part of Theorem 1 with ! = M=C:Now, if a(x) is monotone nondecreasing, using the multiplier 2(x− 1) (C0 + �)yx+

C0y we prove exponential stability in the same way.(ii) For the sake of clarity, we assume that g = −f, without loss of generality.

Moreover, we assume as in (i) that a(x) is monotone nonincreasing and �1�2 6= 0: Wede�ne a functional

�(t) = 2∫ 1

0

(C0 + �

)xytyx dx + 2C0

∫ 1

0yty dx; (2.29)

where C0 and � are positive constants. As in the proof of (i), we can show that thereexist positive constants K0; K1; K2; K3 and K4 such that for any t ≥ 0,

| �(t) |≤ K0E(t); (2.30)

�(t) ≤ −K1E(t) + K2y2t (1; t) + K3f2 (yt(1; t)) + K4f2 (yt(0; t)) : (2.31)

Given � ¿ 0, we introduce (see [7, 17]) the perturbed energy by

E�(t) = E(t) + ��(t) (E(t))(p−1)=2 : (2.32)

This together with the monotonic decrease of E(t) implies that for any M ¿ 1,

M− 12 (E�(t))

(p+1)=2 ≤ (E(t))(p+1)=2 ≤ M12 (E�(t))

(p+1)=2 ; (2.33)

with � ≤ K−10 (E(0))(1−p)=2 (1−M−1=(p+1)) :

On the other hand, from Eq. (2.31) and the hypothesis (H2) of Theorem 2 (ii), we get

�(t) ≤ −K1E(t) + K5y2t (1; t) + K4�22 (yt(0; t))2 ; (2.34)

where

K5 = K2 + K3�22:

In addition, from Eq. (2.32) we have

E�(t) = E(t) + ��(t) (E(t))(p−1)=2 + �(p− 12

)�(t)E(t) (E(t))(p−3)=2 : (2.35)

Plugging Eqs. (1.4), (2.30) and (2.34) into Eq. (2.35), one obtains

E�(t)≤ (1− �p− 1=2K1 (E(0))(p−1)=2)(−�1yt(0; t)f(yt(0; t))− �2yt(1; t)f(yt(1; t)))

+ � (E(t))(p−1)=2(−K1E(t) + K5y2t (1; t) + K4�22y

2t (0; t)

): (2.36)


If y2t (1; t)¿ 1, it follows from hypothesis (H) and (H2) that

�K5 (E(t))(p−1)=2 y2t (1; t) ≤

�K5�1(E(0))(p−1)=2 yt(1; t)f(yt(1; t)): (2.37)

However, while y2t (1; t) ≤ 1, using Young’s inequality we have for any constant B ¿ 0

�K5 (E(t))(p−1)=2 y2t (1; t) ≤

2�p+ 1

(K5B)(p+1)=2 | yt(1; t) |(p+1)=2

+�(p− 1p+ 1

)B−(p+1)=(p−1) (E(t))(p+1)=2 :

This implies thanks to (H2) that

�K5 (E(t))(p−1)=2 y2t (1; t)≤

2�p+ 1

(K5B)(p+1)=2

�1yt(1; t)f(yt(1; t))

+�(p− 1p+ 1

)B−(p+1)=(p−1) (E(t))(p+1)=2 : (2.38)

Combining Eqs. (2.37) and (2.38), one has

�K5 (E(t))(p−1)=2 y2t (1; t) ≤ �

(p− 1p+ 1

)B−(p+1)=(p−1) (E(t))(p+1)=2

+K�6yt(1; t)f(yt(1; t));

(2.39)

where

K�6 =

�K5�1(E(0))(p−1)=2 +

2�p+ 1

(K5B)(p+1)=2

�1:

Similarly, one can prove that

�K4�22 (E(t))(p−1)=2 y2t (0; t) ≤ �

(p− 1p+ 1

)B−(p+1)=(p−1) (E(t))(p+1)=2

+K�7yt(0; t)f(yt(0; t));

(2.40)

with

K�7 =

�K4�22�1

(E(0))(p−1)=2 +2�

p+ 1

(K4�22B

)(p+1)=2�1

:

Inserting Eqs. (2.39) and (2.40) into Eq. (2.36), we obtain

E�(t) ≤ �(2(p− 1p+ 1

)B−(p+1)=(p−1) − K1

)(E(t))(p+1)=2

+(−�1 + ��1

p− 12

K1 (E(0))(p−1)=2 + K�

7

)yt(0; t)f(yt(0; t))


+(−�2 + ��2

p− 12

K1 (E(0))(p−1)=2 + K�

6

)yt(1; t)f(yt(1; t)):

This implies that

E�(t) ≤ −�� (E(t))(p+1)=2 ; (2.41)

provided that B is chosen such that for some � ¿ 0,

2(p− 1p+ 1

)B−(p+1)=(p−1) − K1 ≤ −� ¡ 0;

and � is chosen as follows

−�1 + ��1p− 12

K1 (E(0))(p−1)=2 + K�

7 ≤ 0;

−�2 + ��2p− 12

K1 (E(0))(p−1)=2 + K�

6 ≤ 0:Finally, we deduce from Eqs. (2.33) and (2.41) that

E�(t) ≤ −��M−1=2 (E�(t))(p+1)=2 : (2.42)

Moreover, we claim that

if E(0)¿ 0; then E�(t)¿ 0 for any �nite time t ¿ 0: (P)

(For the proof, see the appendix.) Solving the above di�erential inequality (where wecan divide by E�(t) thanks to (P)) and using Eq. (2.33), we get

E(t) ≤ M 2=(p+1) (E(0)) (1 + !t)−2=(p−1) ; ∀t ≥ 0;with

! =p− 12

M−1=(p+1)�� (E(0))(p−1)=2 :

This proves the second part (ii) of Theorem 2.(iii) We just give a sketch of the proof, for it is analogous to that one of (ii).

First, we introduce for � ¿ 0 the perturbed energy by

E�(t) = E(t) + ��(t) (E(t))(1−p)=2 :

Choosing � su�ciently small, we can obtain

M− 12 (E�(t))

(p+1)=2 ≤ (E(t))(p+1)=2 ≤ (E�(t))(p+1)=2 :

Now a straightforward computation similar to (ii) shows that

E(t) ≤ M 2p=(p+1) (E(0)) (1 + !t)−2p=(1−p) ; ∀t ≥ 0;with

! =1− p2p

M−p=(p+1)�� (E(0))(1−p)=2p :

The proof of Theorem 2 is thus complete.


Proof of Theorem 3. We will show brie y the �rst assertion (i). The others can beproved in the same way as in the proof of Theorem 2.We consider the function

’(x) =∫ x

0exp

(∫ x

�

∣∣∣ax

a

∣∣∣ d�) d�; ∀x ∈ [0; 1]:

It is easy to see that

’(0) = 0; ’(x)¿ 0 ∀x ∈ (0; 1];

’x ≥ 1; a(’a

)x≥ 1:

Performing similar computations as in Eqs. (2.18)–(2.28) with the multiplier ’yx +C∗0 y, we show that there exists a constant K

∗ ¿ 0 such that∫ T2

T1E(t) dt ≤ K∗E(T1) +

∫ T2

T1

∫ 1

0

[�2

(’x − C∗

0

)+ 2]y2

+(−C∗

0 +2a0+

�’(1)2a0

)ay2x dx dt:

One can verify that

K∗ = sup(12�3

[’(1)�2

(1 +

�24a(1)

)+ C∗

0 �2�24

];C∗0 �2�

22

2�1;C∗0

�+

’(1)2

)

+2 sup(C∗0 +

’(1)√a0

;C∗0

�(1 + �)

):

Then, by choosing C∗0 as follows:

�2

(’x − C∗

0

)+ 2 ≤ 0; −C∗

0 +2a0+

�’(1)2a0

≤ 0;

we deduce that∫ T2

T1E(t) dt ≤ K∗E(T1):

The exponential stability is now obvious.

Remark 1. In applications; we may meet mass densities a(x) which are monotone byparts in (0; 1). For example; the mass density given by

a(x) =√

p2 + (x − 1=2)2; ∀x ∈ [0; 1];

is monotone in each interval (0; 12 ) and (12 ; 1). It is interesting to treat this case.

Actually; we can extend our results to this case. Assume that there exists x0 ∈ (0; 1)such that a(x) is monotone nonincreasing (resp. nondecreasing) on (0; x0) and


monotone nondecreasing (resp. nonincreasing) on (x0; 1). We de�ne on [0,1] a function (resp. ) by

(x) = 2(x0 − x) (C + �) (resp. = 2(x − x0) (C + �));

where C and � are two positive constants. It is obvious that (resp. ) is positive(resp. negative) on (0; x0) and negative (resp. positive) on (x0; 1)). With suitableconstants C and �; we show that the multiplier yx + C1y (resp. yx + C1y) allowsus to obtain Theorem 2 with an analogous proof.

3. Conclusion

In this paper, we have shown that a nonlinear boundary velocity feedback is su�cientto stabilize asymptotically the considered vibrating equation. Moreover, under growthconditions on the feedback, the uniform and the rational decay rate of the energy isalso estimated.

Appendix

Proof of Eq. (2.19). For any u ∈ H 1(0; 1), one can verify that

∫ 1

0u2(x) dx =

∫ 1

0

(∫ x

1(u2(�))� d�+ u2(1)

)dx

= u2(1) + 2∫ 1

0

∫ x

1u(�) u�(�) d� dx

= u2(1) + 2∫ 1

0

∫ x

1�u(�)

u�(�)�

d� dx; ∀� ∈ R:

Then, applying Cauchy–Schwartz’s inequality, we have∫ 1

0u2(x) dx ≤ u2(1) + �2

∫ 1

0u2�(�) d�+

1�2

∫ 1

0u2(�) d�;

which implies that(1− 1

�2

)∫ 1

0u2(x) dx ≤ u2(1) + �2

∫ 1

0u2x(x) dx:

Then, for any � ¿ 1∫ 1

0u2(x) dx ≤

(�2

�2 − 1)

u2(1) +(

�4

�2 − 1)∫ 1

0u2x(x) dx:

We have thus proved (2.19).


Proof of the claim (P). Performing a similar computation as in Eq. (2.42), we showthat there exist constants (depending on �) c�1; c�2 ¿ 0 such that

ddt(E�(t) + c�1E(t)) ≥ −c�2 (E(t))

(p+1)=2 : (2.43)

Indeed, Eq. (2.35) implies that

E�(t) ≥ E(t)− � | �(t) | (E(t))(p−1)=2

+ �(p− 12

)| �(t) | E(t) (E(t))(p−3)=2 :

By the use of Eqs. (2.30)–(2.34), the above inequality becomes

E�(t) ≥ E(t)(1 + �

(p− 12

)K0 (E(t))

(p−1)=2)

−�K1 (E(t))(p+1)=2 − � (E(t))(p−1)=2

(K5y2t (1; t) + K4�22y

2t (0; t)

);

and thus by Eqs. (2.39)–(2.40) with B = 1,

E�(t) ≥ E(t)(1 + �

(p− 12

)K0 (E(t))

(p−1)=2)

−�(K1 + 2

(p− 1p+ 1

))(E(t))(p+1)=2

−K�7yt(0; t)f(yt(0; t))− K�

6yt(1; t)f(yt(1; t)):

This together with Eq. (1.4) implies that

E�(t) ≥ E(t)(1 + �

(p− 12

)K0 (E(0))

(p−1)=2 +K�6

�2+

K�7

�1

)

−�(K1 + 2

(p− 1p+ 1

))(E(t))(p+1)=2 : (2.44)

The di�erential inequality (2.43) follows from (2.44) with

c�1 = 1 + �((p− 1)2

)K0 (E(0))

(p−1)=2 +K�6

�2+

K�7

�1; c�2 = �

(K1 + 2

(p− 1p+ 1

)):

On the other hand, using Eq. (2.33) we get

E(t) ≤ M 1=(p+1) (E�(t) + c�1E(t)) ;

This implies together with (2.43) that

ddt

E�(t) ≥ −c�4(E�(t)

)(p+1)=2; (2.45)


where

E�(t) = E�(t) + c�1E(t);

c�4 = c�2M1=2:

(2.46)

Since E(t) and E�(t) are non-increasing (see Eqs. (1.4) and (2.42)), then it is the samefor E�(t). Moreover, it is easy to show that Eq. (2.45) leads us to

ddt

[E�(t) exp

(∫ t

0c�4(E�(�)

)(p−1)=2d�)]

≥ 0:

Since E(0)¿ 0, we deduce from Eqs. (2.33) and (2.46) that E�(0)¿ 0 and

E�(t) exp(∫ t

0c�4(E�(�)

)(p−1)=2d�)

≥ E�(0)¿ 0:

Using the monotonic decrease of E�(t), the above inequality gives

E�(t) ≥ exp(−c�4(E�(0)

)(p−1)=2t)E�(0)¿ 0; ∀t ≥ 0:

Then the claim (P) follows from Eqs. (2.33) and (2.46).

Reference

[1] B. D’Andrea-Novel, F. Boustany, B. Rao, Feedback stabilisation of a hybrid PDE-ODE system:application to an overhead crane, MCSS 7 (1994) 1–22.

[2] G. Birkho�, G.C. Rota, Ordinary Di�erential Equations, Waltham, Massachusetts, Toronto, London,1969.

[3] H. Brezis, Op�erateurs Maximaux Monotones et Semi-groupes de Contractions dans les Espaces deHilbert, North-Holland, Amsterdam, London, 1973.

[4] M. Cherkaoui, Sur la stabilisation d’une poutre d�eformable en torsion ou en �exion par une classe decontrole fronti�ere, Th�ese, Universit�e de Nancy1, 1994.

[5] B. Chentouf, C.Z. Xu, G. Sallet, Stabilisation d’une �equation de vibrations, Rapport de recherche INRIA3085, 1997.

[6] F. Conrad, M. Pierre, Stabilisation of Euler–Bernoulli beam by nonlinear boundary feedback, Rapportde recherche INRIA 1235, 1990.

[7] F. Conrad, B. Rao, Decay of solutions of the wave equation in a star-shaped domain with nonlinearboundary feedback, Asymptotic Analysis 7 (1993) 159–177.

[8] C.M. Dafermos, M. Slemrod, Asymptotic behaviour of non linear contractions semi-groups, J. Func.Anal. 14 (1973) 97–106.

[9] A. Haraux, Syst�emes Dynamiques Dissipatifs et Applications, Collection RMA (17) (1991).[10] S. Icart, J. Leblond, C. Samson, Some results on feedback stabilisation of a one-link exible arm,

Rapport de recherche INRIA-Sophia Antipolis 1682, 1992.[11] A.E. Ingham, Some trigonometrical inequalities with applications to the theory of series, Math.

Zeitschrift 41 (1936) 367–379.[12] V. Jurdjevic, J.P. Quin, Controllability and stability, J. Di�. Equations 28(3) (1978) 381–389.[13] V. Komornik, Exact Controllability and Stabilisation. The Multiplier Method, Wiley, Masson, 1994.[14] T.S. Li, Global classical solutions for quasilinear hyperbolic systems, Wiley, Masson, 1994.[15] V. Mikha�ilov, Equations aux D�eriv�ees Partielles, Mir, Moscou, 1980.[16] A. Pazy, Semigroups of Linear Operators and Applications to Partial Di�erential Equations, Springer,

New York, 1983.


[17] B. Rao, Decay estimate of solution for hybrid system of exible structures, Euro. J. Appl. Math. 4(1993) 303–319.

[18] M. Slemrod, Boundary feedback stabilization for a quasilinear wave equation, Control theory fordistributed parameter systems and applications, Lecture Notes in Control and Information Sciences,54, Springer, New York, 1983, pp. 221–237.

[19] A. Zeidler, Non Linear Functional Analysis and its Applications, vol. 2, Springer, New York, 1986.

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On the stabilization of a vibrating equation

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