BLOW-UP FOR THE EULER-BERNOULLI BEAM PROBLEM WITH A ...

Dynamic Systems and Applications 17 (2008) 109-120

BLOW-UP FOR THE EULER-BERNOULLI BEAM PROBLEM WITH A FRACTIONAL BOUNDARY DISSIPATION SORAYA LABIDI AND NASSER-EDDINE TATAR Laboratoire de Mathématiques Appliquées, Université de Versailles, 45, Avenue des Etats-Unis 78035, Versailles Cedex France ([email protected]) King Fahd University of Petroleum and Minerals, Mathematical Sciences Department, Dhahran 31261, Saudi Arabia ([email protected]) ABSTRACT. We consider a beam problem with a polynomial source and a boundary damping of order between 0 and 1. Sufficient conditions on the initial data are established to have blow up of solutions in finite time. Key words and phrases: Blow up, fractional derivative, boundary feedback, Euler-Bernoulli beam AMS (MOS) Subject Classification: 45K05, 35B40, 35B45, 35B37, 93B52, 26A33

1. INTRODUCTION In this paper we investigate the collapse in finite time of solutions to the following initial boundary value problem known as the Euler-Bernoulli beam problem   utt + ∆2 u = a |u|p−1 u, x ∈ Ω, t > 0    ∂u    u = ∂υ = 0, x ∈ Γ1 , t > 0 (1) ∆u = 0, x ∈ Γ0 , t > 0  Rt  ∂∆u b  = (t − s)β−1 us (s)ds, x ∈ Γ0 , t > 0  ∂υ Γ(β) 0    u (x, 0) = u0 (x) , ut (x, 0) = u1 (x) , x ∈ Ω

where Ω is a bounded domain in Rn with smooth boundary Γ such that Γ = Γ0 ∪ Γ1 and Γ0 ∩ Γ1 =. The constants a and b are positive. The exponent p is greater than 1. The initial data u0 (x) and u1 (x) are given functions, ∂/∂υ denotes the outward normal derivative and Γ(β) is the usual Euler gamma function. The power β in the integral term is such that 0 < β < 1. This problem describes the damping of transversal vibrations of a beam. Controls are forces or torques applied on a portion of the boundary of the beam. This feedback is intended to reduce the effect of reflected waves. Received July 09, 2006

c 1056-2176 $15.00 Dynamic Publishers, Inc.

110

S. LABIDI AND N. TATAR

Note that the integral term in (1) is the fractional derivative (in the sense of Caputo) of order 1 − β (see [16]) defined by ∂t1−β w (t) = I β

d w (t) , dt

0 0     ∂u   u= = 0, x ∈ Γ1 , t > 0   ∂υ  ∆u = ut , x ∈ Γ0 , t > 0   ∂∆u ∂utt b Rt ∂ut  p−1  − h + a |u| u = , x ∈ Γ0 , t > 0 (t − s)β−1 ut ds +    ∂υ ∂υ Γ(β) 0 ∂υ    u (x, 0) = u (x) , u (x, 0) = u (x) x ∈ Ω. 0 t 1 It is clear that here the situation is somehow favorable. Indeed, as both the source and the damping are acting on the same location, one does not have the difficulties of passing from one part of the boundary or the domain to another. In the present work we have to find a way to go from part of the boundary to the whole domain or vice-versa. In addition, in [8] an exponential growth result is proved in an infinite time interval. That is when time goes to infinity, whereas in the present work we prove a stronger result. We prove that solutions blow up in finite time.

For the wave equation without the source term (that is a = 0) a similar problem has been studied by Mbodje and Montseny [12] and recently by Mbodje [11]. In both papers the problem is first converted into a coupled wave-diffusion system called ˙ an augmented system which may be put in the operator theoretical form X(t) = AX(t). Then, they apply the well-known techniques in the semigroup theory to prove existence and uniqueness and LaSalle’s invariance principle to derive an asymptotic decay of solutions. In [11], the same problem but with an exponentially modified kernel tβ−1 e−δt /Γ(β), 0 < β < 1, δ > 0 (see [17]) has been reconsidered. In addition to the well-posedness, the asymptotic convergence towards zero and a decay rate of convergence, the author proves that solutions cannot decay uniformly exponentially to zero. An interesting problem is also studied in [14].

EULER-BERNOULLI BEAM PROBLEM

111

A closely related problem to ours is when the damping acts in the whole domain rather than on the boundary or part of the boundary. This problem has been investigated by Matignon et al. [10] using the same method mentioned above. An asymptotic decay result has been proven but without a precise decay rate. The problem with a strong fractional derivative (and a = 0) has been studied by the second author (Tatar) in [19]. Solutions are proved to be asymptotically convergent to the stationary solution zero in an exponential manner. The case a 6= 0 is considered in [13]. The same problem as in [10] (with an internal fractional damping and a polynomial source) has been treated in a series of papers [6, 18–20, 1]. In these papers the authors established several results on exponential growth and blow up in finite time. Finally, we mention here that the case Ω = Rn and with h(t, x) |u|p as nonlinear source has been discussed in [7]. In this work we prove that solutions of the problem (1) with fractional boundary feedback may blow up in finite time. In fact, it is shown that for any fixed T > 0 we can find initial data whose associated solution blows up in a finite time T∗ < T . The standard existing methods and techniques do not apply to this problem because of the singularity present in the definition of the fractional derivative. The remaining part of the paper is organized as follows: In the next section we prepare some material which will be needed in the sequel. Section 3 is devoted to the statement and proof of our result on blow up in finite time. 2. PRELIMINARIES In this section we prepare some definitions and some lemmas which will be needed in the proof of our result. We assume, without loss of generality, that a = 1 (in fact we could do the same with b). Let us define the classical energy associated to problem (1) by Z 1 2 1 1 2 p+1 dx. E(t) := u + |∆u| − |u| 2 t 2 p+1 Ω It is easily seen that

dE(t) b =− dt Γ(β)

Z

ut Γ0

Z

t 0

(t − s)β−1 ut ds dσ.

This shows that there is no guarantee that the system is dissipative with respect to the classical energy. Replacing t by s and s by z then integrating from 0 to t, we get Z s Z tZ b (s − z)β−1 uz (z)dz dσ ds. ut E(t) − E(0) = − Γ(β) 0 Γ0 0 Hence, E(t) ≤ E (0) for all t ≥ 0, because tβ−1 , 0 < β < 1, is a positive definite function. We will also need the following inequality.

112


Lemma 2.1 (Young inequality, see [2]). Let f ∈ Lp (R) and g ∈ Lq (R) with 1 ≤ p, q ≤ ∞ and 1r = 1p + 1q − 1 ≥ 0. Then f ∗ g ∈ Lr (R) and kf ∗ gkLr ≤ kf kLp kgkLq . Lemma 2.2 (See [9]). Let Ω be a bounded open set in Rn with smooth boundary, then θ s kuk s2 kukH (1−θ)s1 +θs2 (Ω) ≤ c (θ, s1 , s2 ) kuk1−θ H H 1 0

0

0

for all si > 0, 0 < θ < 1. In particular (when the Poincaré inequality is valid), we have kukH θ (Ω) ≤ c (θ) kuk1−θ k∇ukθ2 2 0

for 0 < θ < 1. Lemma 2.3 (See [9]). Let Ω be a bounded open set in Rn with smooth boundary Γ0 , then kukl,Γ0 ≤ C(l, s, Ω) kukH s (Ω) for all l ≥ 2 and s =

n 2

−

n−1 l

> 0.

Remark 2.4. Notice that by Hölder inequality and Lemma 2, we have kukm,Γ0 ≤ . Moreover, a combination of the C(l, s, Ω) kukH s (Ω) for 0 < s < 1 and s ≥ n2 − n−1 m 1−θ previous two lemmas yield kukm,Γ0 ≤ C kuk2 k∇ukθ2 , for 0 < θ < 1. 3. BLOW UP IN FINITE TIME This section contains the statement and proof of our result. To deal with the nonlinear source we need the embedding H01 ,→ Lp and therefore we assume that 2 ≤ p < 2n/(n − 2) if n ≥ 3 and p ≥ 2 if n = 1, 2. Concerning the existence and uniqueness of a local solution, (u, ut ) ∈ H 2 (Ω) × L2 (Ω) and u ∈ C 0 ((0, Tm ), H02 (Ω)) ∩ C 1 ((0, Tm ), L2 (Ω)) (where H02 (Ω) := {v ∈ H 2 (Ω) : v = ∂v/∂υ = 0 on Γ1 }), we refer the reader to the papers [11,12] and also to [3]. Theorem 3.1. Let u be a solution of (1) with 0 < β < 1 and p > 3. Then, for every fixed T > 0 there exist sufficiently large initial data u0 , u1 and 0 < T ∗ < T for which u blows up at T ∗ . Proof. Let us define Z Z tZ 1 2 1 1 p+1 2 |u| − ut − |∆u| dx ds + (kt + l) u20 dx, H(t) = p+1 2 2 Ω 0 Ω where k and l are positive constants to be determined later. Clearly, o R R n 1 0 |u|p+1 − 21 u2t − 12 |∆u|2 dx + k Ω u20 dx H (t) = Ω p+1 R R = k Ω u20 dx − E(t) ≥ k Ω u20 dx − E(0).


113

We can already choose k > E(0)/ Ω u20 dx so that Z 0 (2) k u20 dx − E(0) = H (0) > 0. R

Ω

0

In this way H (t) > 0 and (3) b H (0) − H (t) = E(t) − E(0) = − Γ(β) 0

0

Z tZ 0

us Γ0

Z

s 0

(s − z)β−1 uz (z)dz dσ ds ≤ 0.

The main idea of the proof is to come up with a functional Φ (t) which satisfies an 0 inequality of the form Φ (t) ≥ CΦω (t) with ω > 1. This implies blow up in finite time of the solution. This argument has been used in [3] for the wave equation with an internal dissipation of the form |ut |m−1 ut . Here we shall use a different functional which allows us to include a larger class of initial data. In doing so, several difficulties arouse when dealing with the nonlocal term given by the fractional damping and which contains a singular kernel. The functional we suggest is the following Z Z ε 2 2 1−γ u dx − u0 dx (4) Φ (t) = H (t) + 2 Ω Ω where ε > 0 and 0 < γ = (5)

p−1 2(p+1)

< 1. At t = 0, we have

Φ (0) = H

1−γ

Z 1−γ 2 (0) = l u0 dx Ω

and the derivative of Φ (t) is 0

(6)

Φ (t) = (1 − γ) H

−γ

0

(t) H (t) + ε

Z

uut dx. Ω

We perform a differentiation and then an integration on (6) with respect to t to arrive at (7) Z Z Z Z Z 0

t

0

Φ (t) = (1 − γ) H −γ (t) H (t) + ε

t

u0 u1 dx + ε

Ω

0

Ω

u2s dx ds + ε

uussdx ds.

0

Ω

The objective behind this is to be able to use the properties of positive definite functions. We can have an idea on the last term in (7) by a multiplication of the equation in (1) by u. It appears that (8) RtR RtR Rs RtR b u 0 (s − z)β−1 uz (z)dz dσ ds uutt dx ds = − 0 Ω |∆u|2 dx ds − Γ(β) 0 Ω 0 Γ0 RtR + 0 Ω |u|p+1 dx ds.

Let us denote

tβ−1 Γ(β)

by kβ (t), and for fixed t, we define the extensions by 0 as follows ( w (τ ) , if τ ∈ [0, t] Lw (τ ) = 0, if τ ∈ R\ [0, t]

114


(this extension depends of course on t but this dependence is not mentioned for notational convenience), and ( kβ (τ ) , if τ > 0 Lkβ (τ ) = 0, if τ ≤ 0. It is easily seen that

=

Rs Rt 1 u (s) 0 (s − z)β−1 uz (z) dz ds Γ(β) 0 R +∞ R +∞ Lu (s) −∞ Lkβ (s − z) (Luz ) (z) dz −∞

ds.

Moreover, by Parseval theorem (see [2,5]), we get R +∞ R +∞ Lu (s) −∞ Lkβ (s − z) (Luz ) (z) dz ds −∞ R +∞ = −∞ F (Lu) (σ)F (Lkβ ∗ Luz )(σ)dσ,

where F (f ) denotes the usual Fourier transform of f . Since the Fourier transform of the convolution gives the product of the Fourier transforms and the fact that (see [17]) kγ+η (t) = (kγ ∗ kη )(t) for all 0 < γ, η < 1, we can split the right hand side of the previous identity into two terms and apply the Cauchy-Schwarz inequality and the Young inequality to obtain R +∞ R +∞ Lu (s) −∞ Lkβ (s − z) Lut dz ds −∞ R 2 21 2 21 R +∞ +∞ F (Lkβ/2 )F (Lut ) dσ ≤ −∞ F (Lkβ/2 )F (Lu) dσ −∞ 2 R R 2 +∞ +∞ ≤ 1 F (Lkβ/2 )F (Lu) dσ + δ F (Lkβ/2 )F (Lut ) dσ, 4δ

−∞

−∞

for some δ > 0. Further, by Theorem 16.5.1 in [4], we find (9) h R R +∞ R +∞ +∞ 1 Lk (s − z) Lu dz ds ≤ Lu (s) δ −∞ Lut (s)(Lkβ ∗ Lut )(s)ds β t −∞ −∞ cos(βπ/2) i R +∞ + 4δ1 −∞ Lu(s)(Lkβ ∗ Lu)(s)ds .

Gathering (7), (8) and (9), we get

R RtR 0 0 Φ (t) ≥ (1 − γ) H −γ (t) H (t) + ε Ω u0 u1 dx + ε 0 Ω u2s dx ds RtR RtR bε −ε 0 Ω |∆u|2 dx ds + ε 0 Ω |u|p+1 dx ds − cos(βπ/2) i h R R R R +∞ +∞ 1 × δ Γ0 −∞ Lus (s)(Lkβ ∗ Luz )(s)ds dσ + 4δ Γ0 −∞ Lu(s)(Lkβ ∗ Lu)(s)ds dσ .

Thanks to (3), we can write that R 0 0 0 εδ εδ Φ (t) ≥ (1 − γ) H −γ (t) − cos βπ H (t) + cos(βπ/2) H (0) + ε Ω u0 u1 dx ( ) R t R2 RtR RtR 2 (10) +ε 0 Ω ut dx ds − ε 0 Ω |∆u|2 dx ds + ε 0 Ω |u|p+1 dx ds R R +∞ bε − 4δ cos(βπ/2) Lu(s)(Lkβ ∗ Lu)(s)ds dσ. Γ0 −∞


115

For the last term in the right hand side of (10), the Cauchy-Schwarz inequality yields R R +∞ I = Γ0 −∞ Lu(s)(Lkβ ∗ Lu)(s)ds dσ 21 21 R R R +∞ +∞ 2 dσ. (Lk ∗ Lu) (s)ds ≤ Γ0 −∞ |Lu|2 ds β −∞ Furthermore, appealing to Lemma 1 (Young inequality), we see that R R R R R +∞ +∞ t 2 |Lu| ds dσ I = Γ0 −∞ Lu(s)(Lkβ ∗ Lu)(s)ds dσ ≤ 0 (t − s)β−1 ds Γ0 −∞ R R +∞ 2 tβ ≤ β Γ0 −∞ |Lu| ds dσ Back to (10), we find

i h 0 0 0 εδ εδ H (t) + cos(βπ/2) H (0) Φ (t) ≥ (1 − γ) H −γ (t) − cos(βπ/2) R RtR RtR RtR +ε Ω u0 u1 dx + ε 0 Ω u2t dx ds − ε 0 Ω |∆u|2 dx ds + ε 0 Ω |u|p+1 dx ds R Rt 2 bεtβ |u| ds dσ. − 4δβ cos(βπ/2) Γ0 0

As H (t) 6= 0 for all t ≥ 0, then taking δ = M cos (βπ/2) H −γ (t), we get R 0 0 0 Φ (t) ≥ [(1 − γ) − M ε] H −γ (t) H (t) + εM H −γ (t) H (0) + ε Ω u0 u1 dx RtR RtR RtR (11) +ε 0 Ω u2t dx ds − ε 0 Ω |∆u|2 dx ds + ε 0 Ω |u|p+1 dx ds R Rt 2 bεtβ γ − 4M β cos |u| ds dσ. 2 (βπ/2) H (t) Γ 0 0

Now comes the crucial step of estimating the last term in (11) by existing terms in the inequality. By the definition of H (t), we have R Rt J = H γ (t) Γ0 0 |u|2 ds dσ h R 2 iγ R R t 2 RtR (12) p+1 1 ≤ p+1 0 Ω |u| dx ds + (kt + l) Ω u0 dx |u| ds dσ. Γ0 0

From Lemma 2 and Lemma 3 with θ = 12 (see also Remark 1) we derive 12 Z Z t 21 Z Z t Z Z t 2 2 2 |∇u| ds dx . |u| ds dσ ≤ C1 |u| ds dx Γ0

(13)

0

Ω

0

Ω

0

R 2 p+1 |u| dx ds + (kT + l) u dx 0 Ω Ω 0 p+1 p+1 R R p+3 R Rt p+3 t 2 2 p+3 + 2(p+1) Ω 0 |u| ds dx |∇u| ds dx Ω 0

J≤

C2 (p−1) 2(p+1)

1 p+1

RtR

2(p+1)

where C2 := C1 p−1 . The Hölder inequality allows us to write p+1 p+3 p+1 2 R R p+1 R R p+3 p−1 p−1 t t p+1 2 p+1 ≤ |Ω| T p+1 0 Ω |u| dx ds |u| ds dσ Ω 0 2 R R p+3 t ≤ A1 0 Ω |u|p+1 dx ds p−1

with A1 = (|Ω| T ) p+3 . Then (13) becomes h RtR R 2 i p+1 1 2 |u| dx ds + (kT + l) u dx J ≤ (p−1)C 2(p+1) p+1 0 Ω Ω 0 2 R R p+3 p+1 R Rt p+3 t p+1 2 1 |u| dx ds + (p+3)A . |∇u| ds dx 2(p+1) 0 Ω Ω 0

116


Notice now that p+3 and p+3 are conjugate exponents. Therefore, applying once again 2 p+1 the Young inequality, we obtain h RtR R 2 i p+1 1 2 J ≤ (p−1)C |u| dx ds + (kT + l) u dx 2(p+1) p+1 0 Ω Ω 0 h i R R R R t t p+1 2 (p+3)A1 p+1 2 + 2(p+1) p+3 0 Ω |u| dx ds + p+3 Ω 0 |∇u| ds dx .

With the help of Poincaré inequality, we deduce that Z tZ Z tZ p+1 (14) J ≤ A2 |u| dx ds + A3 |∆u|2 dx ds + A4 0

where A2 = (p−1)C2 2(p+1)

Ω

0

Ω

(p−1)C2 A1 + p+1 , A3 = Cp2A1 (Cp is the Poincaré constant) and A4 = 2(p+1)2 R l) Ω u20 dx. From the relations (14), (11) and the fact that H (t) and

(kT + H (0) are positive, we infer that 0

(15)

R 0 0 Φ (t) ≥ [(1 − γ) − M ε] H −γ (t) H (t) + ε Ω u0 u1 dx RtR RtR RtR +ε 0 Ω u2t dx ds − ε 0 Ω |∆u|2 dx ds + ε 0 Ω |u|p+1 dx ds R R t p+1 R Rt 1 2 3 − εB |u| ds dx − εB |∆u|2 ds dx − εB M Ω 0 M Ω 0 M β

β

β

bT A2 bT A3 bT A4 where B1 = 4β cos 2 (βπ/2) , B2 = 4β cos2 (βπ/2) and B3 = 4β cos2 (βπ/2) . Choosing ε such that , (15) implies that 0 < ε ≤ 1−γ M Rt R R RtR 0 Φ (t) ≥ ε Ω u0 u1 dx + ε 0 Ω u2t dx ds − ε 1 + BM2 0 Ω |∆u|2 dx ds R Rt (16) 3 . +ε 1 − BM1 Ω 0 |u|p+1 ds dx − εB M

Let us add and substract 4εH(t) to the right hand side of (16), we find R RtR 0 Φ (t) ≥ 4εH (t) + ε Ωu0 u1 dx + 3ε 0 Ω u2t dx ds R Rt 4 ε Ω 0 |u|p+1 ds dx + 1 − BM1 − p+1 RtR R 3 . + 1 − BM2 ε 0 Ω |∆u|2 dx ds − 4ε (kT + l) Ω u20 dx − εB M

Select the initial data u0 and u1 such that  R R 2  u u dx − 4 (kT + l) u dx − 0 1  Ω Ω 0 p−3 B1 ˜ (17) − M ≥b>0 p+1   1 − BM2 ≥ 0

B3 M

≥0

for some ˜b > 0. It suffices to select first u0 and u1 such that Z Z (18) u0 u1 dx − 4 (kT + l) u20 dx > 0 Ω

Ω

and then choose M large enough so that Z Z B3 >0 u0 u1 dx − 4 (kT + l) u20 dx ≥ M Ω Ω

and also the second and third conditions in (17) are satisfied. Consequently, (19)

0

Φ (t) ≥ 4εH (t) + 3ε

Z tZ 0

Ω

u2t dx ds

+ εb1

Z Z Ω

t 0

|u|p+1 ds dx.


117

On the other hand, we have (20)

Φ

1 1−γ

(t) ≤ 2

1 1−γ

"

H (t) + ε

1 1−γ

Z t Z 0

uut dx ds Ω

1 # 1−γ

.

It is easy to see that, by the Cauchy-Schwarz inequality and Hölder inequality, we have 12 21 R RtR Rt R 2 2 |u | dx ds uu dx ds ≤ |u| dx t t Ω 0 Ω 0 Ω 1 p+1 21 Rt R R p+1 2 ≤ C3 |u| dx |ut | dx ds 1/2 ΩR R 1/2 R R 0 Ω 2 t t 2 p+1 p+1 ds |ut | dx ds ≤ C3 0 Ω |u| dx . 0 Ω 2 (p+1)(1−2γ)

= 1, it follows that   1 1−2γ  1 R R 1−γ R R R R 2 t t t uut dx ds |ut |2 dx ds + 0 Ω |u|p+1 dx p+1 ds ≤ C4 0 Ω 0 Ω   2 R R (p+1)(1−2γ) RtR t ≤ C4 0 Ω |ut |2 dx ds + T µ 0 Ω |u|p+1 dx ds ,

Moreover, because

where µ = Φ

1 1−γ

p−1 . (p+1)(1−2γ)

This estimate, when used in (20), yields Z t Z Z tZ 1 1 1 2 p+1 µ 1−γ 1−γ 1−γ (t) ≤ 2 H (t) + 2 ε C4 |ut | dx ds + T |u| dx ds 0

Ω

0

Ω

and therefore with the help of (19) we see that (21)

Φ

1 1−γ

0

(t) ≤ KΦ (t)

for some sufficiently large K (depending on T ). Integrating (21) over (0, t), we find Φ

γ 1−γ

(t) ≥

1 Φ

γ − 1−γ

(0) −

γ t K(1−γ)

Consequently, Φ (t) blows up at some time T ∗ ≤

inequality holds if Φ (0) is chosen so that Φ (0) R γ that lγ > K (1 − γ) /γT Ω u20 dx .

γ 1−γ

γ K(1−γ) − 1−γ Φ (0) γ K(1−γ) > γT that is

< T . The last if l is chosen so

proposition 3.2. The set of initial data u0 and u1 satisfying (2) and (18) is not empty. Proof. First, we show that we can find u0 such that Z Z Z Z 1 1 2 2 2 (22) 16 (kT + l) u0 dx + |∆u0 | dx < k u0 dx + |u0 |p+1 dx. 2 p + 1 Ω Ω Ω Ω

Suppose for contradiction that we always have Z Z Z Z 1 1 p+1 2 2 k u0 dx + |u0 | dx ≤ 16 (kT + l) u0 dx + |∆u0 |2 dx p+1 Ω 2 Ω Ω Ω

118


Let u0 = δv0 for an arbitrary δ > 0, then Z Z Z Z δ p+1 δ2 p+1 2 2 2 2 v0 dx + kδ |v0 | dx ≤ 16δ (kT + l) v0 dx + |∆v0 |2 dx. p + 1 2 Ω Ω Ω Ω

Simplifying by δ 2 , we get Z Z Z δ p−1 1 p+1 2 |v0 | dx ≤ |∆v0 | dx + 16 (kT + l) v02 dx. p+1 Ω 2 Ω Ω

Observe that the right hand side is independent on δ. This is impossible and hence √ there exists u0 such that (22) holds. Select now u1 > 4 2 (kT + l) u0 such that Z Z Z Z Z 1 1 1 p+1 2 2 2 u1 dx < k u0 dx + |u0 | dx − |∆u0 |2 dx. 16 (kT + l) u0 dx < 2 Ω p+1 Ω 2 Ω Ω Ω The second inequality means that (2) is satisfied and Z Z Z √ 2 u0 u1 dx > 4 2 (kT + l) u0 dx > 4 (kT + l) u20 dx Ω

Ω

Ω

implies that (18) holds. Acknowledgments: The second author is very grateful to King Fahd University of Petroleum and Minerals for its continuous support. REFERENCES [1] M. R. Alaimia and N.-e. Tatar, Blow up for the wave equation with a fractional damping, J. Appl. Anal. Vol. 11 No. 1 (2005) 133–144. [2] H. Brezis, Analyse Fonctionnelle: Théorie et Applications, Masson, Paris, New York, Barcelone, Milan, Mexico, Sao Paulo, 1983. [3] V. Georgiev and G. Todorova, Existence of a solution of the wave equation with nonlinear damping and source terms, J. Diff. Eqs. 109 (1994) 295–308. [4] G. Gripenberg, S. O. Londen and O. Staffans, Volterra integral and functional equations, Cambridge: Univ. Press, Cambridge 1990. [5] L. H¨ ormander, The Analysis of Linear Partial Differential Operator, Vol. I. Berlin et al. SpringerVerlag 1983. [6] M. Kirane and N.-e. Tatar, Exponential growth for a fractionally damped wave equation, Zeit. Anal. Anw. (J. Anal. Appl.) Vol. 22 No. 1 (2003) 167–177. [7] M. Kirane and N.-e. Tatar, Nonexistence of solutions to a hyperbolic equation with a time fractional damping, Zeit. Anal. Anw. (Journal for Analysis and its Applications), 25 No. 2 (2006), 131–142. [8] S. Labidi and N.-e. Tatar, Unboundedness for the Euler-Bernoulli beam equation with fractional boundary dissipation, Appl. Math. Comp. 161 (2005) 697–706. [9] J. L. Lions and E. Magenes, Problèmes aux limites non homogènes et applications, Dunod, 1968. [10] D. Matignon, J. Audounet and G. Montseny, Energy decay for wave equations with damping of fractional order, in Proc. Fourth International Conference on Mathematical and Numerical Aspects of Wave Propagation Phenomena, pages 638–640. INRIA-SIAM, Golden, Colorado, June 1998.


119

[11] B. Mbodje, Wave energy decay under fractional derivative controls, IMA J. Math. Control and Information, (2005) 1–21. [12] B. Mbodje and G. Montseny, Boundary fractional derivative control of the wave equation, IEEE Transactions on Automatic Control, Vol. 40, No. 2 (1995) 378–382. [13] S. Messaoudi, B. Said-Houari and N.-e. Tatar, Global existence and asymptotic behavior for a fractional differential equation, Appl. Math. Comp. 188 (2007), 1955–1962. [14] G. Montseny, J. Audounet and D. Matignon, Fractional integrodifferential boundary control of the Euler-Bernoulli beam, Publication du Département Traitement du Signal et de l’Image, Ecole Normale Superieur de Télécommunication, 98C003. October 1998. [15] K. B. Oldham and J. Spanier, The fractional calculus, New York: Acad. Press 1974. [16] I. Podlubny, Fractional differential equation, (Math. in Sci. and Eng.: Vol. 198), New York, London: Acad. Press 1999. [17] S. G. Samko, A. A. Kilbas and O. I. Marichev, Fractional integrals and derivatives, Amsterdam: Gordon and Breach 1993 [Engl. Trans. from the Russian 1987]. [18] N.-e. Tatar, A wave equation with fractional damping, Zeit. Anal. Anw. (J. Anal. Appl.), Vol. 22 No. 3 (2003) 1–10. [19] N.-e. Tatar, The decay rate for a fractional differential equations, J. Math. Anal. Appl. 295 (2004) 303–314. [20] N.-e. Tatar, A blow up result for a fractionally damped wave equation, Nonl. Diff. Eqs. Appl. (NoDEA) Vol. 12, No. 2 (2005) 215–226.