International Journal of Analysis and Applications

Volume 19, Number 1 (2021), 77-90

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-19-2021-77

EXPONENTIAL STABILITY FOR A NONLINEAR TIMOSHENKO SYSTEM WITH

DISTRIBUTED DELAY

LAMINE BOUZETTOUTA∗, FAHIMA HEBHOUB, KARIMA GHENNAM, SABRINA

BENFERDI

University of 20 August 1955, Skikda, Algeria

∗Corresponding author: lami-750000@yahoo.fr

Abstract. This paper is concerned with a nonlinear Timoshenko system modeling clamped thin elastic

beams with distributed delay time. The distributed delay is defined on feedback term associated to the

equation for rotation angle. Under suitable assumptions on the data, we establish the exponential stability

of the system under the usual equal wave speeds assumption.

1. Introduction

In this work, we consider the following non linear Timoshenko system with distributed delay,

(1.1)



ρ1ϕtt −k(ϕx + ψ)x = 0,

ρ2ψtt − bψxx + k(ϕx + ψ) + µ1ψt

+
∫ τ2
τ1
µ2(s)ψt(x,t−s)ds + f(ψ) = 0,

where t denotes the time variable and x the space variable along a beam of length 1 in its equilibrium

configuration. Here, ϕ = ϕ(x,t) and ψ = ψ(x,t) denotes the transverse displacement of the beam and the

rotation angle of its filament, respectively. The term µ1ψt represents a frictional damping and f(ψ) is a

forcing term. The coefficients, ρ1,ρ2,k are positive constants represent the density, the polar momentum

Received October 13th, 2020; accepted November 5th, 2020; published December 4th, 2020.

2010 Mathematics Subject Classification. 93D15, 93D20, 35B37, 35B40, 74D05.

Key words and phrases. Timoshenko system, distributed delay time, exponential stability, Lyapunov functional.

©2021 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

77

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-19-2021-77


Int. J. Anal. Appl. 19 (1) (2021) 78

of inertia of a cross section, shear modulus respectively, and b = EI where E is the young’s modulus of

elasticity, I is the moment of inertia cross-section.

System (1.1) is supplemented with the following initial conditions

(1.2)
ϕ(x, 0) = ϕ0, ϕt(x, 0) = ϕt, ψ(x, 0) = ψ0, ψt(x, 0) = ψ1

ψt(x,−t) = f0(x,t),

and Dirichlet boundary conditions

(1.3) ϕ(0, t) = ϕ(1, t) = ψ(0, t) = ψ(1, t),

where x ∈ (0, 1), t ∈ (τ1,τ2). The initial data (ϕ0,ψ0,ϕ1,ψ1,f0) belongs to a suitable functional spacial.

This type of problems (without delay), has been considered, first in [15] where µ1 = µ2 = f = 0. The stability

of this problems has received much attention in last years, we can find in the literature many results about

different stability of Timoshenko systems depending, in particular, on the weights µ1 and µ2 (see [14])

Recently also a great consideration ha been addressed to time delay effects. On such problems, it was showed

that a small delay acted on a boundary control, or internal can destabilize a system which is uniformly

asymptotically stable in the absence of delays. See for instance ( [5]) .

In [13] S. Nicaise and C. Pignotti examined a system of wave equation with initial feedback

(1.4)




utt −uxx + µ0ut +
∫ τ2
τ1
a(x)µ(s)ut(t−s)ds

u = 0 on Γ0(0,α)

∂u

∂ν
= 0 on Γ1(0,α)

u(x, 0) = u0(x) and ut(x, 0) = u1(x) in Ω

ut(x,−t) = f0(x,−t) in Ω(0,τ2)

where a ∈ L2(Ω) is a function chosen with some assumptions. They proved that the above system is

exponentially stable under the condition

µ0 > ||a||α
∫ τ2
τ1
µ(s)ds

Similarly result was obtained by the authors when the distributed delay acted on the part of boundary.

In [11] Mustapha considered a Timoshenko system of thermoelasticity of type III with distributed delay

and establish the stability for the case of equal and non equal speeds of wave propagation .Appalara [1]



Int. J. Anal. Appl. 19 (1) (2021) 79

investigated a thermo-elastic system of Timoshenko type with second sound and distributed delay

(1.5)




ρ1ϕtt −k(ϕx + ψ)x + γ1ϕt +
∫ τ2
τ1
γ2(s)ϕt(x,t−s) = 0

ρ2ψtt − bψxx + k(ϕx + ψ) + δθx = 0

ρ3θt + qx + δψtx = 0

τqt + βq + θx = 0.

in (0, 1)(0,α), this system is exponentially stable regardless the speeds of wave propagation. The same

author studied in [2] a one dimensional Timoshenko system with linear frictional damping and a distributed

delay acting on the displacement equation,he showed that dissipation through the frictional damping is

strong enough to uniformly stabilize the system. for other results about different types of time delay

(discrete and continuos delay) we refer the reader to see ( [1–4, 8, 10]).

B. Feng and H. L. Pelier [6] considered a following non linear Timoshenko system with constant delay

and forcing term:

(1.6)



ρ1ϕtt −k(ϕx + ψ)x = 0,

ρ2ψtt − bψxx + k(ϕx + ψ) + µ1ψt + µ2(s)ψt(x,t− τ) + f(ψ) = 0,

and obtained an exponential stability under equal wave speeds.

Recently S. A. Messaoudi, B. Said-Houari [10] established the stability of a thermoelastic Timoshenko system

of type III with past history and distributed delay for the cases of equal and non equal speeds of wave

propagation respectively.

In the present work, we extend the result of Feng and Pelier, [6] where constant delay is replaced by distributed

delay.

2. Preliminaries

In this section we present the some assumptions needed later to prove our results. As in [12], we introduce

the following new dependent variable

z (x,ρ,s,t) = ψt (x,t−ρs) , x ∈ (0, 1) , ρ ∈ (0, 1) , t,s ∈ (τ1,τ2).

Then, the above variable z satisfies

szt (x,ρ,s,t) + zρ (x,ρ,s,t) = 0, (x,ρ,s,t) ∈ (0, 1) × (0, 1) × (τ1,τ2) × (0, +∞) .



Int. J. Anal. Appl. 19 (1) (2021) 80

Therefore, the problem (1.1) is equivalent to

(2.1)




ρ1ϕtt −k(ϕx + ψ)x = 0, x ∈ (0, 1) , t > 0,

ρ2ψtt − bψxx + k(ϕx + ψ) + µ1ψt

+
∫ τ2
τ1
µ2(s)z(x, 1,s,t)ds + f(ψ) = 0, x ∈ (0, 1) , t > 0,

szt (x,ρ,s,t) + zρ (x,ρ,s,t) = 0, ρ ∈ (0, 1) , s ∈ (τ1,τ2) , t > 0,

with the following initial and boundary conditions

(2.2)




ϕ(x, 0) = ϕ0, ϕt(x, 0) = ϕ1, x ∈ (0, 1) ,

ψ(x, 0) = ψ0, ψt(x, 0) = ψ1, x ∈ (0, 1) ,

z(x,ρ,s, 0) = f0(x,ρs), x ∈ (0, 1), ρ ∈ (0, 1) , s ∈ (0,τ2),

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

Concerning the weight of the delay, we only assume that

(2.3)

∫ τ2
τ1

|µ2 (s)|ds < µ1.

In addition, we give some hypothesis on the forcing term f(ψ(x,t)). We assume that f : IR → IR satisfies

the following condition

(2.4)
∣∣f(ψ1) −f(ψ2)∣∣ ≤ k0 (∣∣ψ1∣∣θ − ∣∣ψ2∣∣θ)∣∣ψ1 −ψ2∣∣

for all ψ1,ψ2 ∈ IR, where k0 > 0, θ > 0. Also

(2.5) 0 ≤ f̃(ψ) ≤ f(ψ)ψ, for all ψ ∈ IR,

with

f̃(y) =

∫ y
0

f(s)ds.

We introduce the Hilbert space,

H = H10 (0, 1) ×L
2 (0, 1) ×H10 (0, 1) ×L

2 (0, 1) ×L2 ((0, 1) × (0, 1) × (τ1,τ2))

For U = (ϕ,u,ψ,v,z)
T
,
(
ϕ̃, ũ, ψ̃, ṽ, z̃

)T
equipped with the scalar product

〈u,ũ〉H =
∫ 1
0

[
ρ1uũ + ρ2vṽ + k (ϕx + ψ)

(
ϕ̃x + ψ̃

)
+ bψxψ̃x

]
dx

+

∫ 1
0

∫ τ2
τ1

s |µ2 (s)|
∫ 1
0

z (x,ρ,s,t) z̃ (x,ρ,s,t) dρdsdx.

We introduce two new dependent variables ϕt = u and ψt = v, then the system (2.1)-(2.2) can be written

as

(2.6)




∂U

∂t
= AU + F, t > 0

U (x, 0) = U0 (x) =
(
ϕ0,ϕ1,ψ0,ψ1,f0

)T
,



Int. J. Anal. Appl. 19 (1) (2021) 81

and

(2.7) AU =




u

k
ρ1

(ϕxx + ψx)

v

b
ρ2
ψxx − kρ2 (ϕx + ψ) −

µ1
ρ2
v − µ1

ρ2

∫ τ2
τ1
µ2 (s) z (x,ρ,s,t) ds

−1
τ
zρ (x,ρ,s,t)



,

F =




0

0

0

−1
ρ2
f(ψ)

0




with the domain

D (A) =
{

(ϕ,u,ψ,v,z)
T ∈ H : v = z (x, 0,s,t) in (0, 1)

}
,

where

H =
(
H2 (0, 1) ∩H10 (0, 1)

)
×H10 (0, 1) ×

(
H2 (0, 1) ∩H10 (0, 1)

)
×H10 (0, 1) ×L

2 ((0, 1) × (0, 1) × (τ1,τ2)) .

Clearly, D(A) is dense in H, we have the following existence and uniqueness result (see [6]).

Theorem 2.1. Let U0 ∈ H and assume that (2.4)-(2.5) and µ2 < µ1 hold. Then, there exists a unique

solution U ∈ C (R+,H) of problem (2.1). Moreover, if U0 ∈ D(A), then U ∈ C (R+,D(A)) ∩C (R+,H) .

3. Stability result

In this section, we use the energy method to show that the solution of problem (2.1)–(2.2) decays expo-

nentially, below we shall give the stability result.

Theorem 3.1. Assume that (2.4)-(2.5) and µ2 < µ1 hold. Assume that
ρ1
ρ2

= k
b

also holds. Then, with

respect to mild solutions, there exist $1 > 0 and $2 > 0 such that

(3.1) E (t) ≤ $1e−$2t, t ≥ 0.

To achieve our goal we state and prove the following lemmas.

Lemma 3.1. The energy functional E (t) of problem (2.1)–(2.2), defined by

E (t) =
1

2

∫ 1
0

(
ρ1ϕ

2
t + ρ2ψ

2
t

)
dx +

1

2

∫ 1
0

{
K (ϕx + ψ)

2
+ bψ2x

}
dx

+

∫ 1
0

∫ 1
0

∫ τ2
τ1

s |µ2 (s)|z (x,ρ,s,t) dsdρdx +
∫ 1
0

f̃(ψ)dx(3.2)



Int. J. Anal. Appl. 19 (1) (2021) 82

satisfies

(3.3)
dE (t)

dt
≤−m1

∫ 1
0

ψ2t dx ≤ 0,

where m1 = µ1 −
∫ τ2
τ1
|µ2 (s)|ds.

Proof. Multiplying the first equation in (2.1) by ϕt, the second equation by ψt, integrating over (0, 1) and

summing them up we get

1

2

d

dt

∫ 1
0

(
ρ1ϕ

2
t + ρ2ψ

2
t

)
dx +

1

2

d

dt

∫ 1
0

{
K (ϕx + ψ)

2
+ bψ2x

}
dx

= −µ1
∫ 1
0

ψ2t dx−µ1
∫ 1
0

f (ψ) ψtdx−
∫ 1
0

∫ τ2
τ1

ψtµ2 (s) z (x, 1,s,t) dsdx.(3.4)

Multiplying the third equation of (2.1) by |µ2 (s)|z (x,ρ,s,t) and integrating over (0, 1) × (0, 1) × (τ1,τ2),

we obtain

1

2

d

dt

∫ 1
0

∫ 1
0

∫ τ2
τ1

s |µ2 (s)|z2 (x,ρ,s,t) dsdρdx

+
1

2

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 1,s,t) dsdx

−
1

2

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 0,s,t) dsdx = 0,(3.5)

by summing (3.5), (3.4) and using the fact that z (x, 0,s,t) = ϕt (x,t) , we have

dE (t)

dt
= −

(
µ1 −

1

2

∫ τ2
τ1

|µ2 (s)|ds
)∫ 1

0

ψ2t dx

−
1

2

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 1,s,t) dsdx

−
∫ 1
0

ψt

∫ τ2
τ1

µ2 (s) z(x, 1,s,t)dsdx.(3.6)

Now, using Young’s inequality, we arrive at

−
∫ 1
0

ψt

∫ τ2
τ1

µ2 (s) z (x, 1,s,t) dsdx

≤
1

2

∫ τ2
τ1

|µ2 (s)|ds
∫ 1
0

ψ2t dx +
1

2

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 1,s,t) dsdx.(3.7)

Inserting (3.7) in (3.6) and using (2.3), we have (3.2) and (3.3). The proof is complete. �

Lemma 3.2. Let (ϕ,ψ,z) be the solution of (2.1)–(2.2). Then, the functional

(3.8) I1 (t) := −
∫ 1
0

(ρ1ϕϕt + ρ2ψψt)dx−
µ1
2

∫ 1
0

ψ2dx.



Int. J. Anal. Appl. 19 (1) (2021) 83

satisfies

dI1 (t)

dt
≤−

∫ 1
0

(ρ1ϕ
2
t + ρ2ψ

2
t )dx + c0

∫ 1
0

ψ2xdx + k

∫ 1
0

(ϕx + ψ)
2dx

+
µ1
4

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2(x, 1,s,t)dsdx,(3.9)

Proof. Differentiating I1 (t), we obtain

dI1 (t)

dt
= −ρ1

∫ 1
0

ϕ2tdx−ρ1
∫ 1
0

ϕϕttdx−ρ2
∫ 1
0

ψ2t dx

−ρ2
∫ 1
0

ψψttdx−µ1
∫ 1
0

ψψtdx,

and using (2.1)1, (2.1)2, we get

dI1 (t)

dt
= −ρ1

∫ 1
0

ϕ2tdx−ρ2
∫ 1
0

ψ2t dx + b

∫ 1
0

ψ2xdx + k

∫ 1
0

(ϕx + ψ)
2dx

+

∫ 1
0

f(ψ)ψdx +

∫ 1
0

ψ

∫ τ2
τ1

µ2 (s) z(x, 1,s,t)dsdx.(3.10)

Applying Young’s and Poincaré inequalities, we have∫ 1
0

ψ

∫ τ2
τ1

|µ2 (s)|z (x, 1,s,t) dsdx

≤ µ1
∫ 1
0

ψ2xdx +
µ1
4

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 1,s,t) dsdx,(3.11)

∫ 1
0

|f(ψ)ψ|dx ≤
∫ 1
0

|ψ|θ |ψ| |ψ|dx

≤‖ψ‖θ2(θ+1) ‖ψ‖2(θ+1) ‖ψ‖

≤ c1
∫ 1
0

ψ2xdx.(3.12)

By substituting (3.11), (3.12) in (3.10), we obtain (3.9). �

Now, let w be the solution of

(3.13) −wxx = ψx, w (0) = w (1) = 0,

then we get

w (x,t) = −
∫ x
0

ψ (y,t) dy + x

(∫ 1
0

ψ (y,t) dy

)
.

We have the following inequalities.

Lemma 3.3. The solution of (3.13) satisfies∫ 1
0

w2xdx ≤
∫ 1
0

ψ2dx and

∫ 1
0

w2tdx ≤
∫ 1
0

ψ2t dx.



Int. J. Anal. Appl. 19 (1) (2021) 84

Proof. We multiply equation (3.13) by w, integrate by parts and use the Cauchy–Schwarz inequality to

obtain

(3.14)

∫ 1
0

w2xdx ≤
∫ 1
0

ψ2dx

Next, we differentiate (3.13) with respect to t and by the same procedure as above, we obtain

(3.15)

∫ 1
0

w2tdx ≤
∫ 1
0

ψ2t dx.

This completes the proof of Lemma (3.3). �

Lemma 3.4. Let (ϕ,ψ,z) be the solution of (2.1)–(2.2). Then, for any ε2 > 0, the functional

(3.16) I2 (t) :=

∫ 1
0

(
ρ2ψtψ + ρ1ϕtw +

µ1
2
ψ2
)
dx,

satisfies

dI2 (t)

dt
≤−

b

2

∫ 1
0

ψ2xdx +

(
ρ1
4ε2

+ ρ2

)∫ 1
0

ψ2t dx + ρ1ε2

∫ 1
0

ϕ2tdx(3.17)

+
µ1
4ε2

∫ 1
0

(∫ τ2
τ1

|µ2 (s)|z2(x, 1,s,t)ds
)
dx−

∫ 1
0

f̃(ψ)dx

Proof. By differentiation I2 (t) , we obtain and by using (2.1)1, (2.1)2, we have

dI2 (t)

dt
= ρ2

∫ 1
0

ψ2t dx− b
∫ 1
0

ψ2xdx + ρ1

∫ 1
0

ϕtwtdx−k
∫ 1
0

ψ2dx + k

∫ 1
0

w2xdx

−
∫ 1
0

f(ψ)ψdx−
∫ 1
0

ψ

(∫ τ2
τ1

µ2 (s) z(x, 1,s,t)ds

)
dx(3.18)

Using Young’s inequality and (3.15), we have

ρ1

∫ 1
0

ϕtwtdx ≤ ρ1ε2
∫ 1
0

ϕ2tdx +
ρ1
4ε2

∫ 1
0

w2tdx

≤ ρ1ε2
∫ 1
0

ϕ2tdx +
ρ1
4ε2

∫ 1
0

ψ2t dx(3.19)

Using Young’s, Cauchy-Schwarz, Poincaré inequalities, we get

−
∫ 1
0

ψ

(∫ τ2
τ1

|µ2 (s) z(x, 1,s,t)|ds
)
dx

≤ δ1
∫ 1
0

ψ2xdx +
µ1
4δ1

∫ 1
0

(∫ τ2
τ1

|µ2 (s)|z2(x, 1,s,t)ds
)
dx(3.20)

Cauchy-Schwarz and Poincaré’s inequalities, give∫ 1
0

|f(ψ)ψ|dx ≤
∫ 1
0

|ψ|θ |ψ| |ψ|dx

≤‖ψ‖θ2(θ+1) ‖ψ‖2(θ+1) ‖ψ‖

≤ c1
∫ 1
0

ψ2xdx.(3.21)



Int. J. Anal. Appl. 19 (1) (2021) 85

By substituting (3.19), (3.20), (3.21) in (3.18), recalling (3.14), (3.15), (2.5) and letting δ1 =
b

2
, we obtain

(3.17). The proof is now complete. �

Lemma 3.5. Let (ϕ,ψ,z) be the solution of (2.1)–(2.2). Then, the functional

(3.22) I3(t) := ρ2

∫ 1
0

ψt(ϕx + ψ) + ρ2

∫ 1
0

ψxϕtdx,

satisfies

dI3 (t)

dt
≤ b [ψxϕx]

1
0 dx +

(
ρ2 +

µ21
k

)∫ 1
0

ψ2t dx−
k

4

∫ 1
0

(ϕx + ψ)
2dx

+ c1

∫ 1
0

ψ2xdx +
µ1
k

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2(x, 1,s,t)dsdx−
∫ 1
0

f̃(ψ)dx

+

(
ρ2k −ρ1b

ρ1

)∫ 1
0

ψx (ϕx + ψ)x dx,(3.23)

where c1 is a positive constant.

Proof. By differentiation I3 (t) and using (2.1)1, (2.1)2, we obtain

dI3 (t)

dt
= b [ψxϕx]

1
0 + ρ2

∫ 1
0

ψ2t dx−k
∫ 1
0

(ϕx + ψ)
2dx−µ1

∫ 1
0

ψt(ϕx + ψ)dx

−
∫ 1
0

∫ τ2
τ1

µ2 (s) (ϕx + ψ)z(x, 1,s,t)dsdx−
∫ 1
0

f(ψ)(ϕx + ψ)dx,(3.24)

By using Young’s inequality, we have

(3.25) µ1

∫ 1
0

|ψt(ϕx + ψ)|dx ≤
k

4

∫ 1
0

(ϕx + ψ)
2dx +

µ21
k

∫ 1
0

ψ2t dx.

Using Young’s and Cauchy Schwarz inequalities, we get∫ 1
0

(ϕx + ψ)

∫ τ2
τ1

|µ2 (s) z(x, 1,s,t)|dsdx

≤
k

4

∫ 1
0

(ϕx + ψ)
2dx +

µ21
k

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2(x, 1,s,t)dsdx.(3.26)

Young’s, Cauchy Schwarz and Poincaré inequalities lead to∫ 1
0

f(ψ)ϕxdx ≤‖ϕx‖‖ψ‖
θ
2(θ+1) ‖ψ‖2(θ+1)

≤
δ0
2b2

∫ 1
0

ϕ2xdx +
b2

2δ0λ1

∫ 1
0

ψ2xdx

≤
δ0
2b2

∫ 1
0

(ϕx + ψ)
2dx +

δ0
2b2

∫ 1
0

ψ2dx +
b2

2δ0λ1

∫ 1
0

ψ2xdx

≤
δ0
2b2

∫ 1
0

(ϕx + ψ)
2dx +

(
δ0

2λ1b2
+

b2

2δ0λ1

)∫ 1
0

ψ2xdx.(3.27)

Inserting (3.25)-(3.27) in (3.24) and letting δ0 =
1

2
kb2, we obtain (3.23). �



Int. J. Anal. Appl. 19 (1) (2021) 86

Next, in order to handle the boundary terms, appearing in (3.23) , we define the function

q(x) = −4x + 2, x ∈ (0, 1)

So, we have the following result.

Lemma 3.6. Let (ϕ,ψ,z) be the solution of (2.1)–(2.2), then for any ε1 > 0, the following estimate holds

b [ψxϕx]
1
0 ≤−

bρ2
4ε1

d

dt

∫ 1
0

qψtψxdx−
ρ1ε1
k

d

dt

∫ 1
0

qϕtϕxdx + 3ε1

∫ 1
0

ϕ2xdx

+ (
2ρ1ε1
k

+
bρ2
2ε1

)

∫ 1
0

ψ2t dx + (
k2ε21

4
+

ε1
4b2

)

∫ 1
0

(ϕx + ψ)
2dx

+
b

4ε1

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 1,s,t) dx

+ (
b2

2ε21
+

1

4λ1b2
+

b2

8ε21λ1
+
µ1b

4ε1
+

b2

4ε31
+ ε1)

∫ 1
0

ψ2xdx(3.28)

Proof. By using Young’s inequality, we easily see that, for ε1 > 0,

(3.29) b [ψxϕx]
1
0 ≤ ε1

[
ϕ2x (1) + ϕ

2
x (0)

]
+

b2

4ε1

[
ψ2x (1) + ψ

2
x (0)

]
,

we need the following fact

d

dt

∫ 1
0

bρ2qψtψxdx = bρ2

∫ 1
0

qψttψxdx + bρ2

∫ 1
0

qψtψxtdx.

On the other hand

bρ2

∫ 1
0

qψttψxdx = b
2

∫ 1
0

qψxxψxdx−kb
∫ 1
0

q(ϕt + ψ)ψxdx

− b
∫ 1
0

∫ τ2
τ1

qψxµ2 (s) z (x, 1,s,t) dsdx− b
∫ 1
0

qf(ψ)ϕxdx

≤−b2
[
ψ2x (1) + ψ

2
x (0)

]
+ 2b2

∫ 1
0

ψ2xdx + (k
2ε2 +

ε

b2
)

∫ 1
0

(ϕx + ψ)
2dx

+ (
b2

ε2
+

ε

2λ1b2
+

b2

2ελ1
+ µ1b)

∫ 1
0

ψ2xdx

+ b

∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 1,s,t) dsdx.(3.30)

Therefore

bρ2

∫ 1
0

qψtψxtdx = 2ρ2b

∫ 1
0

ψ2t dx

Similarly

d

dt

∫ 1
0

ρ1qϕtϕxdx =

∫ 1
0

q(ϕt + ψ)ϕxdx +

∫ 1
0

ρ1qϕtϕxtdx

≤−k
[
ϕ2x (1) + ϕ

2
x (0)

]
+ 3k

∫ 1
0

ϕ2xdx

+ k

∫ 1
0

ψ2xdx + 2ρ1

∫ 1
0

ψ2t dx



Int. J. Anal. Appl. 19 (1) (2021) 87

which, along with (3.29)-(3.30), gives us (3.28). The proof is now complete. �

Lemma 3.7. Let (ϕ,ψ,z) be the solution of (2.1)–(2.2). Then, for η1 > 0, the functional

(3.31) F4 (t) =

∫ 1
0

∫ 1
0

∫ τ2
τ1

se−sρ |µ2 (s)|z2 (x,ρ,s,t) dsdρdx,

satisfies

F ′4 (t) ≤−η1
∫ 1
0

∫ 1
0

∫ τ2
τ1

s |µ2(s)|z2(x,ρ,s,t)dsdρdx

−η1
∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2 (x, 1,s,t) dsdx + µ1
∫ 1
0

ψ2t dx.(3.32)

Proof. Differentiating F4 (t) and using (2.1)3, we obtain

F ′4(t) = −2
∫ 1
0

∫ 1
0

∫ τ2
τ1

e−sρ |µ2(s)|z(x,ρ,s,t)zρ(x,ρ,s,t)dsdρdx

= −
∫ 1
0

∫ 1
0

∫ τ2
τ1

e−sρ |µ2(s)|
∂

∂ρ
[z2(x,ρ,s,t)]dsdρdx

Integration by parts gives and using the fact that z(x, 0,s,t) = ψt and e
−s ≤ e−sρ ≤ 1, we get for all ρ ∈ [0, 1]

F ′4(t) ≤−
∫ 1
0

∫ τ2
τ1

e−s |µ2(s)|z2(x, 1,s,t)dsdx +
∫ τ2
τ1

|µ2(s)|ds
∫ 1
0

ψ2t dx

−
∫ 1
0

∫ 1
0

∫ τ2
τ1

se−s |µ2(s)|z2(x,ρ,s,t)dsdρdx.

Since −e−s is an increasing function, we have −e−s ≤ −e−τ2 for all s ∈ [τ1,τ2] . Finally, setting η1 =

e−τ2 and recalling (2.3), we obtain (3.32). �

Now, we define the Lyapunov functional L(t) by

(3.33) L (t) := NE (t) +
1

8
I1 (t) + N1I2 (t) + I3 (t) + N2I4 (t) ,

where N1,N2 and N are positive constants.

Lemma 3.8. Let (ϕ,ψ,z) be the solution of (2.1)–(2.2). Then, there exists two positive constants β1 and

β2 such that the Lyapunov functional L(t) satisfies

(3.34) β1E (t) ≤ L (t) ≤ β2E (t) , ∀t ≥ 0,

and

(3.35) L′ (t) ≤−λ1E (t) +
(
ρ2k −ρ1b

ρ1

)∫ 1
0

ψx (ϕx + ψ)x dx.



Int. J. Anal. Appl. 19 (1) (2021) 88

Proof. Let

L (t) := NE (t) +
1

8
I1 (t) + N1I2 (t) + I3 (t) + N2I4 (t) ,

then

|L (t) −NE (t)| ≤
ρ1
8

∫ 1
0

|ϕϕt|dx +
ρ2
8

∫ 1
0

|ψψt|dx +
µ1
16

∫ 1
0

ψ2dx

+ N1ρ2

∫ 1
0

|ψtψ|dx + N1ρ1
∫ 1
0

|ϕtw|dx + N1
µ1
2

∫ 1
0

ψ2dx

+ ρ2

∫ 1
0

|ψt(ϕx + ψ)|dx + ρ2
∫ 1
0

|ψxϕt|dx

+ N2

∫ 1
0

∫ 1
0

∫ τ2
τ1

se−sρ |µ2 (s)|z2 (x,ρ,s,t) dsdρdx.

Exploiting Young’s, Poincaré and Cauchy–Schwarz inequalities, we obtain

|L (t) −NE (t)| ≤ C
∫ 1
0

(
ψ2x + ψ

2
t + ϕ

2
t + (ϕx + ψ)

2
+

∫ 1
0

∫ τ2
τ1

s |µ2 (s)|z2 (x, 1,s,t) dsdρ
)
dx

+

∫ 1
0

f̃(ψ)dx

≤ CE (t) ,

Now, combining (3.3), (3.9), (3.17), (3.23) and (3.32), we get

By differentiating L (t), exploiting (3.3), (3.9), (3.17), (3.23), (3.28), (3.32) and setting ε2 =
ρ1

16N1
, we get

dL (t)

dt
= −

(
Nm1 −N1 (4N1 + ρ2) −N2µ1 −

(
ρ2 +

µ21
k

)
− (

2ρ1ε1
k

+
bρ2
2ε1

) + ρ2

)∫ 1
0

ψ2t dx

−
(
b

2
N1 −

c0
8
− (

b2

2ε21
+

1

4b2
+

b2

8ε21
+
µ1b

4ε1
+

b2

4ε31
+ ε1)

)∫ 1
0

ψ2xdx

−
(
k

8
−ε1

(
k2ε1 +

1

b2

))∫ 1
0

(ϕx + ψ)
2dx

−N2β
∫ 1
0

∫ 1
0

∫ τ2
τ1

s |µ2 (s)|z2 (x,ρ,s,t) dsdρdx

−
(
N2β − 4N21 µ1 −

µ1
32
−
µ1
k
−

b

4ε1

)∫ 1
0

∫ τ2
τ1

|µ2 (s)|z2(x, 1,s,t)dsdx

−
ρ1
16

∫ 1
0

ϕ2tdx− (N1 + 1)
∫ 1
0

f̃(ψ)dx, +

(
ρ2k −ρ1b

ρ1

)∫ 1
0

ψx (ϕx + ψ)x dx.

First, we choose ε1 small enough such that

k

8
−ε1

(
k2ε1 +

1

b2

)
> 0.

After, we take N1 large so that

b

2
N1 −

c0
8
− (

b2

2ε21
+

1

4b2
+

b2

8ε21
+
µ1b

4ε1
+

b2

4ε31
+ ε1 > 0.



Int. J. Anal. Appl. 19 (1) (2021) 89

Then, we select N2 large to satisfies

N2β − 4N21 µ1 −
µ1
32
−
µ1
k
−

b

4ε1
> 0.

By finally choose N large enough (even larger so that 1.1 remains valid) such that

Nm1 −N1
(
ρ1
4ε2

+ ρ2

)
−N2µ1 −

(
ρ2 +

µ21
k

)
− (

2ρ1ε1
k

+
bρ2
2ε1

) + ρ2 > 0,

we obtain (3.35). The proof is complete. �

Acknowledgments: The authors wish to thank deeply the anonymous referee for his/her useful remarks

and his/her careful reading of the proofs presented in this paper.

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication

of this paper.

References

[1] T.A. Apalara, Well-posedness and exponential stability for a linear damped Timoshenko system with second sound and

internal distributed delay. Electron. J. Differ. Equ. 2014 (2014), 254.

[2] T.A. Apalara, Uniform decay in weakly dissipative timoshenko system with internal distributed delay feedbacks, Acta

Math. Sci. 36 (2016), 815–830.

[3] L. Bouzettouta, D. Abdelhak, Exponential stabilization of the full von Kármán beam by a thermal effect and a frictional

damping and distributed delay, J. Math. Phys. 60 (2019), 041506.

[4] L. Bouzettouta, S. Zitouni, Kh. Zennir. and H. Sissaoui, Stability of Bresse system with internal distributed delay. J. Math.

Comput. Sci. 7(1) (2017), 92–118.

[5] R. Datko, J. Lagnese and M. P. Polis, An example on the effect of time delays in boundary feedback stabilization of wave

equations. SIAM J. Control Optim. 24(1) (1986), 152–156.

[6] B.W. Feng and M. L. Pelicer, Global existence and exponential stability for a nonlinear Timoshenko system with delay.

Bound. Value Probl. 2015 (2015), Article ID 206.

[7] H.E. Khochemane, A. Djebabla, S. Zitouni, L. Bouzettouta, Well-posedness and general decay of a nonlinear damping

porous-elastic system with infinite memory, J. Math. Phys. 61 (2020), 021505.

[8] H.E. Khochemane, L. Bouzettouta, A. Guerouah, Exponential decay and well-posedness for a one-dimensional porous-

elastic system with distributed delay, Appl. Anal. (2019), 1–15. https://doi.org/10.1080/00036811.2019.1703958.

[9] H. E. Khochemane , S. Zitouni and L. Bouzettouta, Stability result for a nonlinear damping porous-elastic system with

delay term, Nonlinear studies, 27(2) (2020), 1–17.

[10] S. A. Messaoudi and B. Said-Houari, Energy decay in a Timoshenko-type system for thermoelasticity of type III with

distributed delay and past history, Electron. J. Differ. Equ. 2018 (2018), 75.

[11] M.I. Mustafa, A uniform stability result for thermoelasticity of type III with boundary distributed delay, J. Math. Anal.

Appl. 415 (2014), 148–158.

[12] S. Nicaise and C. Pignotti, Stability and instability results of the wave equation with a delay term in the boundary or

internal feedbacks. SIAM J. Control Optim. 45(5) (2006), 1561–1585.

[13] S. Nicaise and C. Pignotti, Interior feedback stabilization of wave equations with time dependent delay. Electron. J. Differ.

Equ. 2011 (2011), 41.



Int. J. Anal. Appl. 19 (1) (2021) 90

[14] C. Pignotti, A note on stabilization of locally damped wave equations with time delay, Syst. Control Lett. 61 (2012), 92–97.

[15] S. P. Timoshenko, On the correction for shear of the differential equation for transverse vibrations of prismatic bars. Phil.

Mag. Ser. 6(41) (1921), 744–746.

[16] S. Zitouni, L. Bouzettouta, Kh. Zennir and D. Ouchenane, Exponential decay of thermo-elastic Bresse system with dis-

tributed delay term, Hacettepe J. Math. Stat. 47(5) (2018), 1216–1230.


	1. Introduction
	2. Preliminaries
	3. Stability result
	References