Uniform decay in a Timoshenko-type system with past history

J. Math. Anal. Appl. 360 (2009) 459–475

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Uniform decay in a Timoshenko-type system with past history Salim A. Messaoudi a , Belkacem Said-Houari b,c,∗ a b c

Department of Mathematics and Statistics, KFUPM, Dhahran 31261, Saudi Arabia Institut de Mathématiques de Toulouse, MIP, Université Paul Sabatier, 118, route de Narbonne, 31062 Toulouse Cedex 09, France Laboratoire de Mathématiques Appliquées, Université Badji Mokhtar, B.P. 12, Annaba 23000, Algeria

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 9 October 2008 Available online 26 June 2009 Submitted by M. Nakao

Fernández Sare and Rivera [H.D. Fernández Sare, J.E. Muñoz Rivera, Stability of Timoshenko systems with past history, J. Math. Anal. Appl. 339 (1) (2008) 482–502] considered the following Timoshenko-type system

ρ1 ϕtt − K (ϕx + ψ)x = 0,

Keywords: Exponential decay Polynomial decay Timoshenko Past history Relaxation function

∞

ρ2 ψtt − bψxx +

g (t )ψxx (t − s, .) ds + K (ϕx + ψ) = 0, 0

where g is a positive differentiable exponentially decaying function. They established an exponential decay result in the case of equal wave-speed propagation and a polynomial decay result in the case of nonequal wave-speed propagation. In this paper, we study the same system, for g decaying polynomially, and prove polynomial stability results for the equal and nonequal wave-speed propagation. Our results are established under conditions on the relaxation function weaker than those in [H.D. Fernández Sare, J.E. Muñoz Rivera, Stability of Timoshenko systems with past history, J. Math. Anal. Appl. 339 (1) (2008) 482– 502]. © 2009 Elsevier Inc. All rights reserved.

1. Introduction In 1921, Timoshenko [21] gave, as model for a thick beam, the following system of coupled hyperbolic equations





ρ utt = K (u x − ϕ ) x , in (0, L ) × (0, +∞), I ρ ϕtt = ( E I ϕx )x + K (u x − ϕ ),

in (0, L ) × (0, +∞),

(1.1)

where t denotes the time variable and x is the space variable along the beam of length L, in its equilibrium configuration, u is the transverse displacement of the beam and ϕ is the rotation angle of the filament of the beam. The coefficients ρ , I ρ , E, I and K are respectively the density (the mass per unit length), the polar moment of inertia of a cross section, Young’s modulus of elasticity, the moment of inertia of a cross section, and the shear modulus. An important issue of research is to look for a minimum dissipation by which solutions of system (1.1) decay uniformly to the stable state as time goes to infinity. In this regard, several types of dissipative mechanisms have been introduced. Raposo et al. [18] used two linear frictional dampings acting on both equations and established an exponential decay result. Kim and Renardy [8] considered (1.1) together with two boundary controls of the form

*

Corresponding author at: Laboratoire de Mathématiques Appliquées, Université Badji Mokhtar, B.P. 12, Annaba 23000, Algeria. E-mail addresses: [email protected] (S.A. Messaoudi), [email protected], [email protected] (B. Said-Houari).

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.06.064

©

2009 Elsevier Inc. All rights reserved.

460

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

K ϕ ( L , t ) − K u x ( L , t ) = α ut ( L , t ), E I ϕx ( L , t ) = −β ϕt ( L , t ),

∀t  0,

∀t  0

and used the multiplier techniques to establish an exponential decay result for the natural energy of (1.1). They also provided numerical estimates to the eigenvalues of the operator associated with system (1.1). Yan [22] generalized the result of [8] by considering nonlinear boundary conditions of the form

K









ϕ ( L , t ) − u x ( L , t ) = f 1 ut ( L , t ) , ∀t  0,

  − E I ϕx ( L , t ) = f 2 ϕt ( L , t ) ,

∀t  0,

where f 1 , f 2 are functions with polynomial growth near the origin. Ammar-Khodja et al. [3] studied the stability of a nonuniform Timoshenko beam by considering the following problem





α utt = β(u x + ϕ ) x , in (0, L ) × (0, +∞), γ ϕtt = (δ ϕx )x − K (ut + ϕ ), in (0, L ) × (0, +∞), u (0, t ) = u ( L , t ) = 0,

ϕx (0, t ) = c ϕt (0, t ),

ϕx ( L , t ) = −dϕt ( L , t ), t > 0

(1.2)

1

for positive C -functions α (x), β(x), γ (x), δ(x), and proved that the uniform stability of (1.2) holds if and only if the β wave speeds are equal ( γδ = α on (0, L )); otherwise only the asymptotic stability has been proved. See also recent work by Messaoudi and Mustafa [13], where the decay rate has been discussed for several systems and without imposing any growth condition on the damping functions. Stabilization by one damping has been considered by many authors. Soufyane and Wehbe [20] showed that it is possible to stabilize uniformly (1.1) by using a unique locally distributed feedback. They considered





ρ utt = K (u x − ϕ ) x , in (0, L ) × (0, +∞), I ρ ϕtt = ( E I ϕx )x + K (u x − ϕ ) − b(x)ϕt , u (0, t ) = u ( L , t ) = ϕ (0, t ) = ϕ ( L , t ) = 0,

in (0, L ) × (0, +∞),

∀t > 0,

(1.3)

where b(x) is a positive and continuous function satisfying

b(x)  b0 > 0,

∀x ∈ [a0 , a1 ] ⊂ [0, L ].

They proved that the uniform stability of (1.3) holds if and only if the wave speeds are equal ( ρK = EI I ); otherwise only the ρ asymptotic stability holds. This result has been recently improved by Rivera and Racke [17], where an exponential decay of the solution energy of (1.3) has been established, allowing b to be with an indefinite sign. Ammar-Khodja et al. [2] considered a linear Timoshenko-type system with memory of the form

ρ1 ϕtt − K (ϕx + ψ)x = 0, t

ρ2 ψtt − bψxx +

g (t − s)ψxx (s) ds + K (ϕx + ψ) = 0 0

in (0, L ) × (0, +∞), together with homogeneous boundary conditions. They used the multiplier techniques and proved that the system is uniformly stable if and only if the wave speeds are equal ( ρK = ρb ) and g decays uniformly. Precisely, they 1 2 proved an exponential decay if g decays in an exponential rate and polynomial decay if g decays in a polynomial rate. They   also required some extra technical conditions on both g and g to obtain their result. Guesmia and Messaoudi [7] obtained the same uniform decay results without imposing those extra technical conditions on g  and g  . Recently, Messaoudi and Mustafa [14] improved the results of [2,7] by allowing more general relaxation functions. They established a more general decay result, from which the exponential and the polynomial decay results are only special cases. The boundary feedback of memory type has also been used by Santos [19]. He considered a Timoshenko system and showed that the presence of two feedback of memory type at a portion of the boundary stabilizes the system uniformly. He also obtained the rate of decay of the energy, which is exactly the rate of decay of the relaxation functions. This latter result has been pushed to a multidimensional problem by Messaoudi and Soufyane [9]. Also, Rivera and Racke [16] treated a nonlinear Timoshenko-type system of the form

ρ1 ϕtt − σ1 (ϕx , ψ)x = 0, ρ2 ψtt − χ (ψx )x + σ2 (ϕx , ψ) + dψt = 0

(1.4)

in a one-dimensional bounded domain. The dissipation is produced here through a frictional damping which is only present in the equation for the rotation angle. The authors gave an alternative proof for a necessary and sufficient condition for

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

461

exponential stability in the linear case and then proved a polynomial stability in general. Moreover, they investigated the global existence of small smooth solutions and exponential stability in the nonlinear case. Recently, Fernández Sare and Rivera [5] considered a Timoshenko-type system with a past history of the form

ρ1 ϕtt − K (ϕx + ψ)x = 0, ∞

ρ2 ψtt − bψxx +

g (s)ψxx (t − s, .) ds + K (ϕx + ψ) = 0

(1.5)

0

where ρ1 , ρ2 , K , b are positive constants and g is a positive twice differentiable function satisfying, for some constants k0 , k1 , k2 > 0,

−k0 g (t )  g  (t )  −k1 g (t ),

g (t ) > 0, bˆ = b −

| g  |  k2 g (t ),

∀t  0,

∞ g (s) ds > 0,

(1.6)

0

and showed that the dissipation given by the history term is strong enough to stabilize the system exponentially if and only if the wave speeds are equal. They also proved that the solution decays polynomially for the case of different wave speeds. For Timoshenko systems in thermoelasticity, Rivera and Racke [15] considered

ρ1 ϕtt − σ (ϕx , ψ)x = 0, in (0, ∞) × (0, L ), ρ2 ψtt − bψxx + k(ϕx + ψ) + γ θx = 0, in (0, ∞) × (0, L ), ρ3 θt − kθxx + γ ψtx = 0, in (0, ∞) × (0, L ), where ϕ , ψ , and θ are functions of (x, t ) which model the transverse displacement of the beam, the rotation angle of the filament, and the difference temperature respectively. Under appropriate conditions of σ , ρi , b, k, γ , they proved several exponential decay results for the linearized system and a nonexponential stability result for the case of different wave speeds. Messaoudi et al. [10] studied

ρ1 ϕtt − σ (ϕx , ψ)x + μϕt = 0, (x, t ) ∈ (0, L ) × (0, ∞), ρ2 ψtt − bψxx + k(ϕx + ψ) + βθx = 0, (x, t ) ∈ (0, L ) × (0, ∞), ρ3 θt + γ qx + δψtx = 0, (x, t ) ∈ (0, L ) × (0, ∞), τ0 qt + q + κ θx = 0, (x, t ) ∈ (0, L ) × (0, ∞), where ϕ = ϕ (t , x) is the displacement vector, ψ = ψ(t , x) is the rotation angle of the filament, θ = θ(t , x) is the temperature difference, q = q(t , x) is the heat flux vector, ρ1 , ρ2 , ρ3 , b, k, γ , δ , κ , μ, τ0 are positive constants. The nonlinear function σ is assumed to be sufficiently smooth and satisfy

σϕx (0, 0) = σψ (0, 0) = k and

σϕx ϕx (0, 0) = σϕx ψ (0, 0) = σψψ = 0. Several exponential decay results for both linear and nonlinear cases have been established. Also, Messaoudi and Said-Houari [11] considered a Timoshenko-type system of thermoelasticity type III of the form

ρ1 ϕtt − K (ϕx + ψ)x = 0, in (0, 1) × (0, ∞), ρ2 ψtt − bψxx + K (ϕx + ψ) + βθx = 0, in (0, 1) × (0, ∞), ρ3 θtt − δθxx + γ ψttx − kθtxx = 0, in (0, 1) × (0, ∞) and proved an exponential decay similar to the one in [15] and [10]. For more results concerning well-posedness and controllability of Timoshenko systems, we refer the reader to Alabau-Boussouira [1], Fernández Sare and Racke [6], and Messaoudi and Mustafa [12]. In the present work we consider (1.5), together with the initial and boundary conditions

ϕ (0, t ) = ϕ (1, t ) = ψ(0, t ) = ψ(1, t ) = 0, t  0, ϕ (., 0) = ϕ0 ,

ϕt (., 0) = ϕ1 ,

ψ(., 0) = ψ0 ,

ψ1 (., 0) = Ψ1

(1.7)

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S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

where ρ1 , ρ2 , K , b are positive constants and g is a differentiable function satisfying, for a positive constant k0 and 1  p < 3/2, the following conditions

bˆ = b −

g (t ) > 0,

∞

g  (t )  −k0 g p (t ).

g (s) ds > 0,

(1.8)

0

ρ

ρ

ρ

We will show that, for K1 = b2 , the first energy decays exponentially if p = 1 and polynomially if p > 1. However, for K1 = ρ2 , the decay is in the rate of 1/t p provided that the initial data are regular enough. Our result improves and generalizes b the work of Fernandez and Rivera [5]. It is also established without any restriction on g  as in [5]. Following the idea of Dafermos [4], we introduce

ηt (x, s) = ψ(x, t ) − ψ(x, t − s), s  0;

(1.9)

consequently we obtain the following initial and boundary conditions

ηt (x, 0) = 0, ∀t  0, ηt (0, s) = ηt (1, s) = 0, ∀s, t  0, η0 (x, s) = η0 (x, s), ∀s  0.

(1.10)

Clearly, (1.9) gives

ηtt (x, s) + ηts (x, s) = ψt (x, t ).

(1.11)

Remark 1.1. Under condition (1.8), it is easy to verify that

∞ G0 =

∞ g

1/2

(s) ds < ∞,

g 2− p (s) ds < ∞,

Gp =

0

1  p < 3 /2 .

(1.12)

0

Remark 1.2. For well posedness and semigroup formulation of the problem (1.5), (1.7), (1.10), we refer the reader to [5]. 2. Uniform decay

ρ1 K

= ρb2

In this section, we state and prove a stability result in the case of equal wave-speed propagation. By combining (1.5), (1.7), (1.10), (1.11), we obtain the following system

ρ1 ϕ tt − K (ϕx + ψ)x = 0, ρ2 ψtt − bˆ ψxx + t t (x, s) +

η

η

∞ g (s)ηtxx (x, t − s) ds + K (ϕx + ψ) = 0,

0 t s (x, s) − ψt (x, t )

= 0,

(2.1)

in (0, 1) × (0, ∞), together with the following initial and boundary conditions

ϕ (0, t ) = ϕ (1, t ) = ψ(0, t ) = ψ(1, t ) = ηt (0, s) = ηt (1, s) = 0, t  0, ϕ (., 0) = ϕ0 ,

ϕt (., 0) = ϕ1 ,

ψ(., 0) = ψ0 ,

ηt (x, 0) = 0.

ψ1 (., 0) = ψ1 ,

(2.2)

The associated first-order energy is

E (t ) = E 1



t



ϕ , ψ, η =

1

1



2 1 t

2 2 ψt

ρ ϕ +ρ

2

 1 + K |ϕx + ψ| + bˆ ψx2 dx +

1 ∞

2

2

0

0



2

g (s)ηtx (s) ds dx.

(2.3)

0

Theorem 2.1. Suppose that (1.8) and

ρ1 K

=

ρ2

(2.4)

b

hold and let

ϕ0 , ψ0 ∈ H 01 (0, 1),





η0t ∈ L 2g R+ , H 01 (0, 1) ,

ϕ1 , ψ1 ∈ L 2 (0, 1).

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

463

Then, there exist two positive constants C and ξ , such that

E (t )  C e −ξ t , p = 1, C E (t )  , p > 1, (t + 1)1/( p −1)

(2.5) (2.6)

where

L 2g



+

R



, H 01 (0, 1)

 +

= u:R →

 1 ∞

H 01 (0, 1)

0

  2   g (s) u x (x, s) ds dx < ∞ .

0

The proof of our result will be established through several lemmas. Lemma 2.1. Let (ϕ , ψ, ηt ) be a solution of (2.1), (2.2). Then we have

dE (t ) dt

=

1

1 ∞

2 0



2

g  (s)ηtx (s) ds dx  0.

(2.7)

0

ϕt , and (2.1)2 by ψt , integrating over (0, 1), and summing up, using (1.8), we ob-

Proof. Multiplying Eq. (2.1)1 by tain (2.7). 2

Lemma 2.2. Let (ϕ , ψ, ηt ) be a solution of (2.1), (2.2). Then we have, for 1 < p < 3/2,

1 ∞ 0

2p −1 1 ∞  t 2  2   g (s) ηx (s) ds dx  C0 g p (s)ηtx (s) ds dx

0

0

(2.8)

0

for a constant C 0 > 0. Proof. Using Hölder’s inequality, it is straightforward to see that, for any r > 1, we have

1 ∞ 0

 2 g (s)ηtx (s) ds dx =

0

1 ∞ 0

0

1 ∞  0



2

g 2r (s)ηtx (s) r g 1

2r −1 2r

  2r −2 (s)ηtx (s) r ds dx

1r 1 ∞

r −r 1  t 2  t 2 2r −1 g (s)ηx (s) ds dx g 2r −2 (s)ηx (s) ds dx . 1 2

0

0

0

(1.9), (1.12), (2.3), and (2.7) lead to

1 ∞ 0



2

g 1/2 (s)ηtx (s) ds dx  2E (0)

0

∞ g 1/2 (s) ds = 2G 0 E (0). 0

By taking r = (2p − 1)/(2p − 2), (2.8) follows.

2

Lemma 2.3. For 1  p  3/2, we have

1 ∞

2 g (s)η

0

t x (s) ds

1 ∞ dx  G p

0

0



2

g p (s)ηtx (s) ds dx.

(2.9)

0

Proof. Using Cauchy–Schwarz’ inequality, we get

∞

∞ g (s)η

0

t x (s) ds

=

g

p

1− 2

p 2

(s) g (s)η

0

Therefore (2.9) follows by (1.12).

∞ t x (s) ds



g 0

2

2− p

1/2 ∞

 2 g (s)ηtx (s) ds p

(s) 0

1/2 .

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S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

As in [15], let

1 I 1 :=

(ρ2 ψt ψ + ρ1 ϕt ω) dx,

(2.10)

0

where

ω is the solution of −ωxx = ψx ,

ω(0) = ω(1) = 0.

Lemma 2.4. Let (ϕ , ψ, ηt ) be a solution of (2.1), (2.2). Then we have, for any ε1 , λ1 > 0

dI 1 (t ) dt

 (−bˆ + λ1 )

1

1 ψx2 dx +

ϕt2 dx

ε1 ρ1

0

0t



ρ1 + ρ2 + 4ε1

1 ψt2 dx + 0

1 ∞

Gp 4 λ1

0



2

g p (s)ηtx (s) ds dx.

(2.11)

0

Proof. By taking a derivative of (2.10) and using Eqs. (2.1) we conclude

dI 1 (t ) dt

= −bˆ

1

1

1

ψx2 dx + ρ2 0

ψt2 dx − K 0

1

0

1

ωx2 dx + ρ1

+K

ψ 2 dx

0

1

ϕt ωt dx − 0

∞ g (s)ηtx (s) ds dx.

ψx 0

0

By using Young’s inequality and

1

1

ωx2 dx  0

0

1

1

ωt2 dx  0

1 ψ 2 dx 

ψx2 dx, 0

1 2 ωtx dx 

0

ψt2 dx, 0

we find that

dI 1 (t ) dt

 −bˆ

1



1 ψx2 dx +

2 t dx +

ε1 ρ1

0

ϕ 0

ρ1 ρ2 + 4ε1

1

1 ψt2 dx −

0

∞ g (s)ηtx (s) ds dx.

ψx 0

(2.12)

0

Using Young’s inequality and (2.9), the last term in the right-hand side of (2.12) can be estimated as follows

1

∞ g (s)η

ψx 0

t x (s) ds dx



0



1 4 λ1 Gp 4 λ1

1 ∞

2 g (s)η

0

t x (s) ds

1

0

1 ∞

ψx2 dx

dx + λ1 0

 2 g (s)ηtx (s) ds dx + λ1

1

p

0

0

Inserting (2.13) into (2.12), we obtain the desired result.

ψx2 dx,

λ1 > 0 .

(2.13)

0

2

Next, we set

1 I 2 := −ρ2

∞ g (s)ηt (s) ds dx.

ψt (x, t ) 0

0

(2.14)

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

465

Lemma 2.5. Let (ϕ , ψ, ηt ) be a solution of (2.1), (2.2). Then we have, for any ε2 > 0,

dI 2 (t )

−

dt

1

ρ2 g0

ψt2 dx + ε2 bˆ 2

2

1

0

0

 + Gp 1 +

∞

where g 0 =

0

1 ψx2 dx + ε2 K 2

1 ∞

1 2ε2

0

(ϕx + ψ)2 dx 0



2

1 ∞

g (0)

g p (s)ηtx (s) ds dx −

2ρ2

0

0



2

g  (s)ηtx (s) ds dx,

(2.15)

0

g (s) ds.

Proof. Using the second and third equations of (2.1) we get

dI 2 (t )

= bˆ

dt

1

1 ∞

∞ g (s)η

ψx 0

t x (s) ds dx +

g (s)η

0

0

1

1 ψt2 dx + ρ2

− ρ2 g 0 0

2 t x (s) ds

1 dx + K

0

∞ g (s)ηt (s) ds dx

(ϕx + ψ) 0

0

∞ g (s)ηts (s) ds dx.

ψt 0

(2.16)

0

By using (2.9) and Young’s inequality, we obtain the following estimates

bˆ

1

∞ g (s)η

ψx 0

t x (s) ds dx

 ε2 bˆ 2

1

0

0

1

∞

0

1

g (s)η (s) ds dx  ε2 K

2

0

0

4ε2

0



1 g (s)ηts (s) ds dx = −

0

∞ ψt

0

2

g p (s)ηtx (s) ds dx,

0

Gp

2

(ϕx + ψ) dx +

4ε2

0

∞ ψt

1 ∞

Gp

1 t

(ϕx + ψ)

K

ψx2 dx +

g  (s)ηt (s) ds dx 

ρ2 g0

1 ∞ 0

ψt2 dx − 0

By inserting all the above estimates in (2.16), relation (2.15) follows.

2

0

1

2

0



g p (s)ηtx (s) ds dx,

g (0) 2ρ2

1 ∞ 0



2

g  (s)ηtx (s) ds dx.

0

2

Next we introduce the functional

1 J (t ) := ρ2

ψt (ϕx + ψ) dx +

ρ1 bˆ

1 ψx ϕt dx +

K

0

ρ1

1

g (s)ηˆ tx (s) ds dx.

ϕt (t )

K

0

∞

0

(2.17)

0

Lemma 2.6. Let (ϕ , ψ, ηt ) be a solution of (2.1), (2.2). Assume that (2.4) holds. Then, for ε3 > 0, we conclude

d J (t ) dt





x=1

∞

 ϕx b ψx +

g (s)η

t x (x, s) x=0

0

1 ψt2 dx +

+ ρ2

(ϕx + ψ)2 dx

−K

0

1

1

0

1 ∞ 2 t dx −

ε3

ϕ

g (0)C (ε3 )

0

0



2

g  (s)ηtx (s) ds dx.

(2.18)

0

Proof. Differentiating J (t ), we obtain

d J (t ) dt

1 = ρ2

1 ψtt (ϕx + ψ) dx + ρ2

0

+

ψt (ϕx + ψ)t dx +

ρ1 bˆ

K 0

ψtx ϕt dx +

ρ1

0

1 0

ρ1

1

∞ t g (s)ηtx (s) ds dx

ϕt

K 0

0

∞ g (s)ηtx (s) ds dx.

ϕtt

K

ψx ϕtt dx +

K

0

1 ρ1 bˆ

1

0

(2.19)

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S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

By using Eqs. (2.1), we find

d J (t ) dt

1 =

 (ϕx + ψ) bˆ ψxx +



∞

0

g (s)η

t xx (x, s) ds

1

− K (ϕx + ψ) dx + ρ2

0

 +

ρ1 b K

1

0

1 − ρ2

ψt2 dx + bˆ

ρ1

ψtx ϕt dx −

1

∞

K

0

0

0

1 g (s)ηtsx (s) ds dx +

ϕt (t ) 0

ψx (ϕx + ψ)x dx ∞ g (s)ηtx (s) ds dx

(ϕx + ψ)x 0

0

and exploiting (2.4), we conclude

d J (t ) dt

1

1 (ϕx + ψ)2 dx + ρ2

= −K 0

0



ρ1

ψt2 dx +

∞

1

∞

ϕt

K 0

x=1

1 g  (s)ηtx (s) ds dx + [bˆ ϕx ψx ]xx= =0

0

g (s)ηtx (x, s)

+ ϕx (x, t )

x=0

0

Young’s inequality gives (2.18).

2

Next, to handle the boundary terms appearing in (2.18), we make use of the function

q(x) = 2 − 4x,

x ∈ [0, 1].

Consequently, we have the following result: Lemma 2.7. Let (ϕ , ψ, ηt ) be a solution of (2.1), (2.2). Then we have, for any λ3 > 0 and ε3 > 0,



bˆ ψx +

ϕx

x=1

∞ g (s)η

t x (s, x) x=0

0



1

−ε3 d

1

ρ1 q(x)ϕt ϕx dx + 3ε3

K dt 0

−

+

+

1 d 4ε3 dt 1

ε3

ϕ

2 x dx +

2ρ1 ε3



ρ2 q(x)ψt bˆ ψx +

0

0



∞

g (s)ηtx (s) ds dx 0



bˆ 2 bˆ 2 + + ε32 8 λ3

ρ2 (2b + ε3 ) 4ε3

1

ψt2 dx +

1 2 λ3



ϕx bˆ ψx +

x=1

∞ g (s)η 0

t x (x, s)

0



2

g p (s)ηtx (s) ds dx

0

1 ∞ (ϕx + ψ)2 dx − ρ2 g (0)C (ε3 )

ε3

0



0

 ε3 ϕ

2 x (1, t ) +

1 4ε3



2

g  (s)ηtx (s) ds dx.

(2.20)

0

ε3 > 0,

ϕ



2 x (0, t )

+

x=0

+ By exploiting

1 ∞

1

λ3 K 2

0

4+

4ε3

0

1



G0

ψx2 dx +

Proof. By using Young’s inequality, we easily see that, for



ϕt2 dx

K

0

1

1

bˆ ψx (1, t ) +

1 4ε3

bˆ ψx (0, t ) +

g (s)ηtx (1, s) ds 0

g (s)η

t x (0, s) ds

0

2

∞

2

∞

.

(2.21)

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

1

d dt



ρ2 q(x)ψt bˆ ψx +



∞ g (s)η

0

t x (s) ds

1 =

0



ρ2 q(x)ψtt bˆ ψx +

0

+

g (s)η

bˆ ψtx +

ρ2 q(x)ψt



∞ t x (s) ds

0



1 0

467



∞ t tx (s) ds

g (s)η 0

and Eq. (2.1)2 we get



1

d

bˆ ψx +

ρ2 q(x)ψt

dt 0

1 =

q(x) bˆ ψxx +

0

+

g (s)η

t x (s) ds

0



1



∞



∞



g (s)η

t xx (x, s) ds

− K (ϕx + ψ)

0

ρ2 q(x)ψt bˆ ψtx +

0

bˆ ψx +

g (s)η

t x (s) ds

dx

0



∞



∞

g (s)ηtxt (s) ds .

(2.22)

0

Simple calculation shows

1

q(x) bˆ ψxx +

0



∞ g (s)η

bˆ ψx +

t xx (x, s) ds

0

=−

1

1

2



∞ g (s)η 0

q (x) bˆ ψx + 

0



2

∞ g (s)η

t x (x, s) ds

dx +

t x (s) ds

q(x)

dx



2

bˆ ψx +

0

2 x=1

∞ g (s)η

t x (s) ds

.

(2.23)

x=0

0

By using (2.1)3 , the last term in (2.22) can be treated as follows



1

ρ2 q(x)ψt

bˆ ψtx +

0



∞ t tx (s) ds

g (s)η 0

= ρ2 bˆ

1

1 q(x)ψt ψtx dx + ρ2

0

=−

ρ2 bˆ

1 q

2



(x)ψt2 dx +

=−

2

q (x)ψt2 dx + ρ2

0

=−

1 ρ2 bˆ 2

0

0

1

∞ t g (s)ηtx (s) ds dx

q(x)ψt 0

0

1

∞

0

q (x)ψt2 dx + g 0 ρ2





g (s) ψtx (t ) − ηtsx (s) ds dx

q(x)ψt

0

=−

t g (s)ηtx (s) ds dx

q(x)ψt

ρ2

0

1 ρ2 bˆ

∞

0

1

1 q(x)ψt ψtx − ρ2

0

ρ2 (bˆ + g0 )

1

2

0

q (x)ψt2 dx + ρ2

0

∞

1

∞ q(x)ψt

0

g (s)ηtsx (s) ds dx

q(x)ψt 0

g  (s)ηtx (s) ds dx.

(2.24)

0

Inserting (2.23) and (2.24) into (2.22), we arrive at

bˆ ψx (0, t ) +

2

∞ g (s)η 0

t x (0, s) ds

+ bˆ ψx (1, t ) +

2

∞ g (s)η 0

t x (1, s) ds

468

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

=−

d

1

dt





∞

ρ2 q(x)ψt bˆ ψx +

g (s)η

0

1

t x (s) ds

0

q(x)(ϕx + ψ) bˆ ψx +

−K

g (s)η

0



1

bˆ ψx +

+2

0

2

∞ t x (x, s) ds

dx

0



∞

g (s)ηtx (s) ds dx 0

+ 2ρ2 (bˆ + g 0 )

1

1 ψt2 dx +

∞

ρ2

0

g  (s)ηtx (s) ds dx.

q(x)ψt 0

(2.25)

0

We now estimate the terms in the right-hand side of (2.25), exploiting Young’s inequality and (2.9), as follows: The second term

1

bˆ ψx +

2 0

2

∞ g (s)η

dx  4bˆ 2

t x (x, s) ds

0

1

1 ∞ ψx2 dx + 4G p

0

0



2

g p (s)ηtx (s) ds dx.

(2.26)

0

The third term

 1

 ∞     t  K q(x)(ϕx + ψ) bˆ ψx + g (s)ηx (s) ds dx   0

0

 1

 ∞     t ˆ  2K  (ϕx + ψ) bψx + g (s)ηx (s) ds dx   0

0

1 2

 4K λ3

(ϕx + ψ) dx + 0

(ϕx + ψ)2 dx + 0

bˆ ψx +

4 λ3

1  4K 2 λ3

1

1

2

0

g (s)η

Gp

ψx2 dx +

2 λ3

t x (s) ds

dx

0

1

bˆ 2

2

∞

2 λ3

0

1 ∞ 0



2

g p (s)ηtx (s) ds dx.

(2.27)

0

The last term

 1  ∞ 1 1 ∞    2    t 2 g  (s)ηtx (s) ds dx. ρ2 q(x)ψt g (s)ηx (s) ds dx  ρ2 ε3 ψt dx − ρ2 g (0)C (ε3 )   0

0

0

0

(2.28)

0

Combining (2.25)–(2.28), we obtain

bˆ ψx (0, t ) +



2

∞ g (s)η

t x (0, s) ds

+ bˆ ψx (1, t ) +

0

−

d

1

dt

2

∞ g (s)η

t x (1, s) ds

0



ρ2 q(x)ψt bˆ ψx +

0



∞ g (s)ηtx (s) ds 0

 1 + bˆ 2 4 + 2 λ3 

1

+ 2ρ2 bˆ + g 0 +

 ψx2 dx +

Gp 4 +

0

ε3

1 ∞

1 2 λ3

1

2

0

1 ψt2 dx + 4K 2 λ3

2

0



g p (s)ηtx (s) ds dx

0

Similarly by using Eq. (2.1)1 , we arrive at

1 ∞ 2

(ϕx + ψ) dx − ρ2 g (0)C (ε3 ) 0

0

0



2

g  (s)ηtx (s) ds dx.

(2.29)

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

d

1



1



1

ρ1 qϕt ϕx dx  − K ϕx2 (1) + ϕx2 (0) + 3K

dt 0

ϕx2 dx + K 0

469

1 ψx2 dx + 2ρ1

0

Hence the assertion of the lemma follows from (2.21), (2.29) and (2.30).

ϕt2 dx.

(2.30)

0

2

Let’s introduce the functional

1

1

K(t ) := −ρ1

ϕt ϕ dx − ρ2

ψt ψ dx.

0

0

Lemma 2.8. Let (ϕ , ψ, ηt ) be a solution of (2.1), (2.2). Then we have, for any ε3 > 0,

d dt

1

1

ϕt2 dx − ρ2

K(t )  −ρ1 0

ψt2 dx + (bˆ + ε3 )

0

ψx2 dx 0

1 (ϕx + ψ)2 dx +

+K

1

0

Gp 4ε3

1 ∞ 0



2

g p (s)ηtx (s) ds dx.

(2.31)

0

Proof. It easy to check that

d dt

1

1 2 t dx −

K(t ) = −ρ1

ϕ

ρ2

0

ψt2 dx + bˆ

1

0

1 ψx2 dx +

(ϕx + ψ) dx −

K

0

1 2

0

∞ g (s)ηtx (s) ds dx.

ψx 0

(2.32)

0

2

By using (2.9) and Young’s inequality, (2.31) follows.

To finalize the proof of Theorem 2.1, we define the Lyapunov functional L as follows

L(t ) := N E (t ) + N 1 I 1 + N 2 I 2 + J (t ) +

+

1 4ε3

bˆ ψx +

ρ2 q(x)ψt

1

ρ1 qϕt ϕx dx

K 0



1

ε3

0



∞ g (s)η

t x (s) ds

dx + μK(t )

(2.33)

0

for N , N 1 , N 2 , μ, ε3 positive constants to be chosen later. By using (2.7), (2.11), (2.15), (2.18), (2.20), (2.32), and

1

1

ϕx2 dx  2 0

1 (ϕx + ψ)2 dx + 2

ψx2 dx

0

0

we get

d dt

1

1 ψx2 dx + Λ2

L(t )  Λ1

1 2 t dx + Λ3

0

0

1

0

1 ∞ 2

+ Λ4

ψt2 dx

ϕ

(ϕx + ψ) dx + C 2 0

 1 ∞   t 2  2 N g (s)ηx (s) ds dx + − C1 g  (s)ηtx (s) ds dx, p

2

0

0

where C 1 and C 2 are positive constants independent of N and

Λ1 = − N 1 (bˆ − λ1 ) + N 2 ε2 bˆ + C (λ3 , μ, ε3 ), Λ2 = N 1 ε1 ρ1 + ε3 +

2ρ1 ε3 K

− μρ1 ,

0

0

(2.34)

470

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475



ρ1

Λ3 = N 1 ρ2 + Λ4 = − K +

− N2

4ε1

K 2 λ3

ε3

g 0 ρ2 2

+ ρ2 +

1 4ε3

(2ρ2 b + ρ2 ε3 ) − μρ2 ,

+ N 2 ε2 K 2 + 2(3ε3 + μ K ),

where C (λ3 , μ, ε3 ) is a constant depending on λ3 , ε2 , ε3 only. At this point, we have to choose our constants very carefully. First, we take λ3 = ε32 , pick



ε3 = min Once

K

K ρ1

,

4( K 2 + 6) 8( K + 2ρ1 )

μ = 1/4, and λ1 = bˆ /2, then we

.

ε3 and μ (hence λ3 ) are fixed, we then choose N 1 so large that N1

bˆ 4

> C (λ3 , μ, ε3 ).

ε1 small enough so that ε1  1/16N 1 and N 2 large enough so that  g 0 ρ2 ρ1 1 + ρ2 + N2 > N 1 ρ2 + (2ρ2 b + ρ2 ε3 ) − μρ2 .

After that, we select

4ε1

2

4ε3

ε2 so small that   N1 1 . ε2 < min ,

Now, we take

4N 2 4K N 2

Finally, after fixing all constants, taking in account (1.8), we choose N large enough so that



N > 2 C1 +

C2 k0

.

Consequently, there exists

d dt

 1 L(t )  −σ1



σ1 > 0 such that (2.34) takes the form ψx2

+ ψt2

2 t

2

1 ∞



+ ϕ + (ϕx + ψ) dx +

0

  t 2 g (s)ηx (s) ds dx . p

0

(2.35)

0

On the other hand, we can choose N even larger (if needed) so that

L(t ) ∼ E (t ).

(2.36)

We distinguish two cases. Case 1. p = 1. A combination of (2.35) and (2.36) leads to

d dt

L(t )  −ξ L(t ).

(2.37)

A simple integration of (2.37) over (0, t ) and use of (2.36) lead to (2.5). Case 2. p > 1. We use (2.3) and (2.7) to get

E

2p −1



2p −2 (t )  C E (0)

1



2 t

ϕ

+ ψt2

2

+ |ϕx + ψ|

+ ψx2



0

 2p −2  C E (0)

1

 1 ∞ dx + C

x

0





0

1 ∞

ϕt2 + ψt2 + |ϕx + ψ|2 + ψx2 dx + C C 0

0

2p−1  t 2 g (s)η (s) ds dx

0



2

g p (s)ηtx (s) ds dx.

(2.38)

0

A combination of (2.35), (2.36) and (2.38) gives

L (t )  −c E 2p −1 (t )  −c L2p −1 (t ).

(2.39)

A simple integration of (2.39) leads to

L(t ) 

C

(t + 1)1/(2p −2)

.

(2.40)

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

471

To obtain (2.6), we observe that

1  t 0



2

g (s)ηtx (s) ds dx =

0

1  t 0

0

1  t  0

and use (2.3), (2.40), and

 t 1

η

t x (x, s)

 t 2 η (s) dx ds  2

0

 t 1 0

 2t 0



p

ds dx

( p −1)/ p 1  t

1/ p  t 2  t 2 p η (s) ds dx η (s) ds dx g ( s ) x x

0

0

  ψx (x, t )2 dx ds + 2

0

1

 2( p−1)

(2.41)

0

= ψx (x, t ) − ψx (x, t − s) to conclude that

x

0

2 



g (s)ηtx (s) p ηtx (s)

 t 1 0

  ψx (x, t )2 dx + 4 bˆ

0

t E (t − s) ds 0

t

Ct

(t + 1)1/(2p −2)

  ψx (x, t − s)2 dx ds

ds

+C

(t − s + 1)1/(2p −2) 0   C 2p − 2 1 C  + 1 − (t + 1)(3−2p )/(2p −2) 3 − 2p (t + 1)(3−2p)/(2p−2)  Π, p < 3/2, where Π is a constant independent of t. Hence, we get

∞1

 t 2 η (s) dx ds  Π. x

0

0

Consequently, we have from (2.41)

1 ∞ 0

 2 g (s)ηtx (s) ds dx  Π ( p −1)/ p

1 ∞

1/ p  t 2   g (s) ηx (s) ds dx p

0

0

0

or

∞1 0

 2 g (s)ηtx (s) dx ds

p

∞1 C

0

0



2

g p (s)ηtx (s) dx ds.

(2.42)

0

Similarly to (2.38), we obtain

1 p

E (t )  C



2 t

ϕ

+ ψt2

2

+ |ϕx + ψ|

+ ψx2



 1 ∞ dx + C

x

0

1 C

0



2 t

ϕ

+ ψt2

2

+ |ϕx + ψ|

+ ψx2



0

p  t 2 g (s)η (s) ds dx

0

1 ∞ dx + C 0



2

g p (s)ηtx (s) ds dx.

(2.43)

0

A combination of (2.35), (2.36), and (2.43) yields

L (t )  −c L p (t ).

(2.44)

A simple integration of (2.44) gives

L(t ) 

C

(t + 1)1/( p −1)

.

Again, use of (2.36) leads to (2.6). This completes the proof of Theorem 2.1.

(2.45)

472

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

ρ1

3. Polynomial decay

= ρb2

K

ρ

ρ

In this section, we treat the case of different wave-speed propagation ( K1 = b2 ), which is, in fact, more realistic from the view point of physics; see in this regard [5] and [17]. We will show if the initial data are regular enough the solution energy E (t ) decays in the rate of 1/t when the relaxation function g decays exponentially and in the rate of t −1/(2p −1) when the relaxation function decays polynomially. Let’s define the second-order energy by

E 2 (t ) = E 1





ϕt , ψt , ηtt ,

(3.1)

where E 1 is given in (2.3). Theorem 3.1. Suppose that

ρ1 K

=

ρ2

(3.2)

b

and let





η0t ∈ L 2g R+ , H 2 (0, 1) ∩ H 01 (0, 1) ,

ϕ0 , ψ0 ∈ H 2 (0, 1) ∩ H 01 (0, 1), ϕ1 , ψ1 ∈ H 01 (0, 1).

(3.3)

Then there exists a positive constant C , such that, ∀t  0,

E (t )  Ct −1/(2p −1) ,

p  1.

(3.4)

To prove this result, we need two lemmas. Lemma 3.1. Suppose that (3.3) holds and let (ϕ , ψ, ηt ) be a solution of (2.1). Then we have

dE 2 (t ) dt

=−

1 ∞

1 2

0



2

t g  (s)ηtx (s) ds dx  0.

(3.5)

0

Proof. We differentiate Eqs. (2.1) with respect to time and then multiply by over (0, 1) and summing up, as in Lemma 2.2, we obtain (3.5). 2

ϕtt and ψtt , respectively. By integrating

Lemma 3.2. Suppose that (3.2), (3.3) hold and let (ϕ , ψ, ηt ) be a solution of (2.1). Then, for ε3 > 0, we conclude

d J (t ) dt





x=1

∞

 ϕx b ψx +

g (s)η

t x (x, s) x=0

1 ∞ ψt2 dx − g (0)C (ε3 )

0

0

1 0

0

0



2

g  (s)ηtx (s) ds dx

0

1 ∞

ϕt2 dx + C (ε3 )

+ 2ε3

(ϕx + ψ)2 dx

−K

0

1 + ρ2

1



2

t g p (s)ηtx (s) ds dx.

(3.6)

0

Proof. Similarly to the proof of Lemma 2.7, we have

d J (t ) dt

1 =

 (ϕx + ψ) bˆ ψxx +

0



∞ g (s)η

t xx (x, s) ds

− K (ϕx + ψ) dx + ρ2

0

 +

ρ1 b K

1 − ρ2

ψtx ϕt dx −

0

ρ1

+

1 ψx (ϕx + ψ)x dx 0

∞ g (s)ηtxs (s) ds dx

ϕt (t ) 0

ψt2 dx + bˆ

0

∞ g (s)ηtx (s) ds dx.

(ϕx + ψ)x 0

1

K

0

1

1

0

(3.7)

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

Since

ρ1 K

1 = ρb2 then we have to handle the term 0 ψtx ϕt dx. For this, we use the idea of [5] to obtain

1

where g 0 =



g0

∞ 0

ρ1 bˆ K

1 ∞

1

ψtx ϕt dx = 0

1

473

t tx (s) ds

ϕt dx −

g (s)η

g0

0

0

1 ∞

1 g0



g (s)η 0

t x (s) ds

ϕt dx,

(3.8)

0

g (s) ds. Therefore, using Young’s inequality and (2.9), we estimate the terms of (3.8) as follows

1 ∞

t tx (s) ds

− ρ2

ϕt dx 

g (s)η 0

ε3

1

1 ∞ 2 t dx +

ϕ

2

0

C (ε3 )

0

0



2

t g p (s)ηtx (s) ds dx

(3.9)

0

and

1 g0



ρ1 bˆ K

1 ∞ − ρ2



g (s)η 0

t x (s) ds

ϕt dx 

ε3

1

1 ∞ 2 t dx −

ϕ

2

0

g (0)C (ε3 )

0

0



2

g  (s)ηtx (s) ds dx.

(3.10)

0

By inserting (3.9) and (3.10) into (3.8) and taking in account the estimates of Lemma 2.7, the desired result (3.6) follows.

2

Proof of Theorem 3.1. To finalize the proof of Theorem 3.1, we define the Lyapunov functional L as follows

 ε3 L(t ) := N E 1 (t ) + E 2 (t ) + N 1 I 1 + N 2 I 2 + J (t ) + 

1

ρ1 qϕt ϕx dx

K

+

1

1 4ε3



ρ2 q(x)ψt bˆ ψx +

0

0



∞

g (s)ηtx (s) ds dx + μK(t ).

(3.11)

0

Consequently, by taking the time derivative of L(t ), we obtain

d dt

1

1 ψx2 dx + Λ2

L(t )  Λ1 0

0

 1 ∞

N

− C1

2

0

1 ∞ + C2 0

1 ψt2 dx + Λ4

ϕ

 +

1 2 t dx + Λ3 0



0

2

g  (s)ηtx (s) ds dx +

N

1 ∞

2

0

0



(ϕx + ψ)2 dx

2

g p (s)ηtx (s) ds dx + C (ε3 )

0

1 ∞ 0



2

t g  (s)ηtx (s) ds dx

0



2

t g p (s)ηtx (s) ds dx,

0

where

Λ2 = N 1 ε1 ρ1 + 2ε3 +

2ρ1 ε3 K

− μρ1 .

Choosing the constants carefully as in Section 2, it is easy to see that, for some

d dt

 1 L(t )  −σ2



 ψx2 + ψt2 + ϕt2 + (ϕx + ψ)2 dx

0

1 ∞

 2 g (s)ηtx (s) ds dx +

1 ∞

p

+ 0

σ2 > 0, we have

0

  t 2   g (s) ηtx (s) ds dx . p

0

(3.12)

0

Moreover N can be chosen so large that L(t )  0. To this end, we distinguish two cases: Case 1. p = 1. It is not hard to see that

d dt

L(t )  −α E (t ).

(3.13)

474

S.A. Messaoudi, B. Said-Houari / J. Math. Anal. Appl. 360 (2009) 459–475

Direct integration of (3.13) gives

t E (s) ds  L(0) − L(t )  L(0),

α

∀t  0.

(3.14)

0

σ3 > 0, such that  L(0)  σ3 E 1 (0) + E 2 (0) , ∀t  0.

By using (3.11), one can find



(3.15)

Hence (3.14) and (3.15) imply

t





E (s) ds  C E 1 (0) + E 2 (0) ,

∀t  0.

(3.16)

0

By noting that

 d d t E (t ) = E (t ) + t E (t )  E (t ), dt dt a simple integration leads to

t t E (t ) 





E (s) ds  C E 1 (0) + E 2 (0) ,

∀t  0.

0

Consequently, we get

E (t ) 

C t



E 1 (0) + E 2 (0) ,

∀t  0.

Case 2. p > 1. By using (2.39), we obtain

dL dt

 −c E 2p −1 (t )

which implies

t





E 2p −1 (s) ds  c L(0) − L(t )  c L(0),

∀t  0.

(3.17)

0

On the other hand, we have

d  2p −1  d tE (t ) = E 2p −1 (t ) + (2p − 1)t E 2p −2 E (t )  E 2p −1 (t ). dt dt Similar calculations, using (3.15), (3.17), and (3.18), yield (3.4). This completes the proof of Theorem 3.1.

(3.18)

2

Remark 3.1. Note that this result generalizes the result of [5] to the case p > 1. Moreover, our result is established without any condition on g  . Acknowledgment This work has been partially funded by KFUPM under Project #IN080408.

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