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Article

Exploring Thermoelastic Effects in Damped Bresse Systems with Distributed Delay

1
Department of Mathematics, Faculty of Exact Sciences, University of El Oued, El Oued 39000, Algeria
2
Laboratory of Pure and Applied Mathematic, University of Laghouat, Laghouat 03000, Algeria
3
Department of Mathematics and Statistics, Imam Mohammad Ibn Saud Islamic University (IMSIU), Riyadh 13318, Saudi Arabia
4
Department of Mathematics, College of Science, Qassim University, Buraydah 51452, Saudi Arabia
5
Department of Mathematics, Faculty of Sciences, University of 20 Août 1955, Skikda 21000, Algeria
*
Author to whom correspondence should be addressed.
Mathematics 2024, 12(6), 857; https://doi.org/10.3390/math12060857
Submission received: 3 February 2024 / Revised: 6 March 2024 / Accepted: 12 March 2024 / Published: 14 March 2024

Abstract

:
In this work, we consider the one-dimensional thermoelastic Bresse system by addressing the aspects of nonlinear damping and distributed delay term acting on the first and the second equations. We prove a stability result without the common assumption regarding wave speeds under Neumann boundary conditions. We discover a new relationship between the decay rate of the solution and the growth of ϖ at infinity. Our results were achieved using the multiplier method and the perturbed modified energy, named Lyapunov functions together with some properties of convex functions.

1. Introduction and Relevance of Subject

Originally, the Bresse system consists of three wave equations where the main variables describe the longitudinal, vertical and shear angle displacements, which can be represented in [1] as
ρ 1 φ t t = Q x + l N + F 1 ρ 2 ψ t t = M x Q + F 2 ρ 1 w t t = N x I Q + F 3 ,
where
N = k 0 ( w x l φ ) , Q = k ( φ x + l w + ψ ) , M = b ψ x .
We use N , Q and M to denote the axial force, the shear force and the bending moment. By w , φ and ψ , we are denoting the longitudinal, vertical and shear angle displacements. Here, ρ 1 = ρ A = ρ I , k 0 = E A , k = k G A and l = R 1 . We use ρ for density, E for the modulus of elasticity, G for the shear modulus, K for the shear factor, A for the cross-sectional area, I for the second moment of area of the cross-section and R for the radius of curvature, and we assume that all these quantities are positives. Also, by F i we are denoting external forces. The Bresse system (1) is more general than the well-known Timoshenko system where the longitudinal displacement w is not considered l = 0 . There are a number of publications concerning the stabilization of Timoshenko system with a different kind of damping; in this regard, we note the next references (see [2,3,4,5,6,7]).
System (1) is an un-damped system and its associated energy remains constant when the time t evolves. To stabilize system (1), many damping terms have been considered by several authors (see [8,9,10,11,12,13,14]).
By considering damping terms as infinite memories acting in the three equations, the system (1) has been recently studied in [11]
ρ 1 φ t t G h ( φ x + l w + ψ ) x E h l ( w x l φ ) + 0 g 1 ( t s ) φ x x ( s ) d s = 0 ρ 2 ψ t t E I ψ x x + G h ( φ x + l w + ψ ) + 0 g 2 ( t s ) ψ x x ( s ) d s = 0 ρ 1 w t t E h ( w x l φ ) x + l G h ( φ x + l w + ψ ) + 0 g 3 ( t s ) w x x ( s ) d s = 0 ,
where ( x , t ) ] 0 , L [ × R + , g i : R + R + , i = 1 , 2 , 3 are given functions. The authors proved, under suitable conditions on the initial data and the memories g i , that the system is well-posed and its energy converges to zero when time goes to infinity, and they provided a connection between the decay rate of energy and the growth of g i at infinity, whereas in our system (3), the nonlinear damping dominates and makes the energy decay following its rate with some assumptions on distributed delay, as seen in (40) since μ 3 ( s ) ϑ ( s ) is the coefficient of nonlinear damping. The proof is based on the semi-groups theory for the well-posedness, and the energy method with the approach introduced in [15] for the stability.
In [8], the authors considered the Bresse system in a bounded domain with delay terms in the internal feedback
ρ 1 φ t t G h ( φ x + l w + ψ ) x E h l ( w x l φ ) + μ 1 φ t + μ 2 φ t ( x , t τ 1 ) = 0 ρ 2 ψ t t E I ψ x x + G h ( φ x + l w + ψ ) + μ 1 ^ ψ t + μ 2 ^ ψ t ( x , t τ 2 ) = 0 ρ 1 w t t E h ( w x l φ ) x + l G h ( φ x + l w + ψ ) + μ 1 ^ ^ ψ t + μ 2 ^ ^ ψ t ( x , t τ 3 ) = 0
where τ i > 0 ( i = 1 , 2 , 3 ) are a time delay, μ 1 , μ 2 , μ 1 ^ , μ 2 ^ , μ 1 ^ ^ , μ 2 ^ ^ are a positive real numbers. This system is subjected to the Dirichlet boundary conditions and to the initial conditions which belong to a suitable Sobolev space. First, the authors proved the global existence of its solutions in Sobolev spaces by means of the semi-group theory under a condition between the weight of the delay terms in the feedback and the weight of the terms without delay. Furthermore, the authors studied the asymptotic behavior of solutions using multiplies methods.
Motivated by the works mentioned above, for x 0 , 1 , t 0 , , we investigate the following Bresse system:
ρ 1 φ t t = k ( φ x + l w + ψ ) x + k 0 l ( w x l φ ) μ 1 φ t τ 1 τ 2 | μ 2 ( s ) | φ t x , t s ρ 2 ψ t t = β ψ x x k ( φ x + l w + ψ ) 0 t ϖ ( t s ) [ ν ( x ) ψ x ( s ) ] x d s μ 3 ( t ) θ ( x ) f ( ψ t ) γ ϱ x ρ 1 w t t = k 0 ( w x l φ ) x k l ( φ x + l w + ψ ) ρ 3 ϱ t = k q x γ ψ t x ρ 4 q t = δ q k ϱ x ,
with boundary conditions
φ x = 0 , t = φ x = 1 , t = 0 ψ x = 0 , t = ψ x = 1 , t = 0 w x = 0 , t = w x = 1 , t = 0 w x ( x = 0 , t ) = w x ( x = 1 , t ) = 0 φ x ( x = 0 , t ) = φ x ( x = 1 , t ) = 0 , t 0 ,
ϱ ( x = 0 , t ) = ϱ ( x = 1 , t ) = 0 q ( x = 1 , t ) = q ( x = 0 , t ) = 0 , t 0 ,
and the initial data
φ x , t = 0 = φ 0 x , φ t x , t = 0 = φ 1 x , ψ x , t = 0 = ψ 0 x , ψ t x , t = 0 = ψ 1 x , w ( x , t = 0 ) = w 0 ( x ) , w t ( x , t = 0 ) = w 1 ( x ) , ϱ ( x , t = 0 ) = ϱ 0 ( x ) , q ( x , t = 0 ) = q 0 ( x ) .
and
ψ t ( x , t ) = f 0 ( x , t ) .
Here ν , θ , f are specific functions, and ρ 1 , ρ 2 , ρ 3 , ρ 4 , δ , β , γ , k , l , k 0 are constitutive constants, while τ 1 , τ 2 are two real numbers with 0 τ 1 τ 2 , μ 1 is positive constant, μ 2 is an L function, μ 3 is a bounded function and ϖ is the relaxation function satisfying:
 Hypothesis 1 (H1). 
ϖ : R + R +  is a  C 1  function satisfying
ϖ 0 > 0 , β ν ( x ) 0 ϖ s d s = r > 0 .
 Hypothesis 2 (H2). 
There exists a positive nonincreasing differentiable function  ϑ : R + R +  satisfying 
ϖ t ϑ t ϖ t , t 0 .
 Hypothesis 3 (H3). 
μ 2 : [ τ 1 , τ 2 ] R  is a bounded function satisfying 
τ 1 τ 2 | μ 2 ( s ) | d s μ 1 .
 Hypothesis 4 (H4). 
The functions  ν , θ  satisfy 
ν C 1 ( [ 0 , 1 ] ) , ν = 0 , o r ν ( 0 ) + ν ( 1 ) > 0 , inf x [ 0 , 1 ] { ν ( x ) + θ ( x ) } > 0 .
 Hypothesis 5 (H5). 
f : R R  is a continuous and non-decreasing function such that there exist positive constants  k 1 , k 2  and  l 1  and  h : R + R +  is a convex, continuous and increasing function of class  C 1 ( R + ) C 2 ( ] 0 , [ )  satisfying:  h ( 0 ) = 0 , and  h = 0  on  [ 0 , l 1 ]  or  h ( 0 ) = 0  and  h > 0  on  [ 0 , l 1 ]  such that 
h ( s 2 + f 2 ( s ) ) f ( s ) s ,   f o r | s | l 1 , k 1 s 2 f ( s ) s k 2 s 2 ,   f o r | s | > l 1 .
 Remark 1. 
Since ϖ is positive and  ϖ ( 0 ) > 0 ; then, for any  t 0 > 0 , we have 
ϖ 1 = 0 ϖ ( s ) d s > 0 t ϖ ( s ) d s 0 t 0 ϖ ( s ) d s = ϖ 0 > 0 .
Using the fact that ν ( 0 ) > 0 and a is continuous, then there exist ε > 0 such that inf x [ 0 , ε ] ν ( x ) 0 . Let us denote
d = m i n { ε , inf x [ 0 , ε ] { ν ( x ) + θ ( x ) } } > 0 ,
and let α C 1 ( [ 0 , 1 ] ) be such that 0 α a and
α ( x ) = 0 , i f ν ( x ) d 4 α ( x ) = ν ( x ) , i f ν ( x ) d 4 .
Under suitable assumptions, we show that, even in the presence of the thermoelastic, we can establish a general energy decay of the solution for (3). We prove our result by using the energy method together with some properties of convex functions. The advantage to propose system (3) is to discover the interaction between the distributed delay term (located in the first equation) and nonlinear damping in its general case (located in the second equation) with the presence of linear memory and their influence on the stability of the system. We found a good interaction between them by outlining minimal conditions to stabilize the system. We consider that these two terms are considered as damping and each term has a special way to stabilize the system.
 Lemma 1 
([16]). The function α is not identically zero and satisfies
inf x [ 0 , 1 ] { α ( x ) + θ ( x ) } d 2 .
Let us denote by h * the conjugate function in the sense of Young of the convex function h as in [17]
h * ( p ) = sup t R + ( p t h ( t ) ) .
Assume that h > 0 ; then, for p 0 a given number, h * is the Legendre transform of h, which is given as in [18] by
h * ( p ) = p [ h ] 1 ( p ) h ( [ h ] 1 ( p ) ) ,
and which satisfies the following inequality (Young’s inequality):
p x h ( x ) + h * ( p ) , p , x 0 .
The relation (9) and the fact that h ( 0 ) = 0 and ( h ) 1 , h are increasing functions yield
h * ( p ) p [ h ] 1 ( p ) , p 0 .
Now, for ε 0 , we define the functions J and K by
J ( t ) = t , i f h = 0 o n [ 0 , l 1 ] t h ( ε 0 t ) , i f h ( 0 ) = 0 a n d h > 0 o n [ 0 , l 1 ] ,
and
K ( t ) = t 1 1 J ( s ) d s .
The following notations will be used:
( ϖ u ) ( t ) = 0 1 α ( x ) 0 t ϖ ( s ) ( u ( t ) u ( s ) ) d s d x , u L 2 ( 0 , 1 )
( ϖ o u ) ( t ) = 0 1 ν ( x ) 0 t ϖ ( t s ) ( u ( t ) u ( s ) ) 2 d s d x , u L 2 ( 0 , 1 ) ,
there exists a positive constant c so that
( ϖ u ) 2 c ϖ o u x , u H 0 1 ( 0 , 1 ) .
We organized our paper as follows: In Section 2, we prepare some Lemmas and present some appropriate functional to state the main Theorem 1. Notably, the research establishes a stability result in Section 3 without the conventional assumptions on wave speeds, particularly under Neumann boundary conditions.

2. Main Result

In this section, we prove our decay result for the energy of the systems (3)–(5) using the multiplier technique. To achieve our goal, as in [19], we use the following new variable:
z ( x , τ , s , t ) = φ t ( x , t s τ ) ,
and then we obtain
s z t ( x , τ , s , t ) + z τ ( x , τ , s , t ) = 0 z ( x , 0 , s , t ) = φ t ( x , t ) .
consequently, the problem is equivalent to
ρ 1 φ t t = k ( φ x + l w + ψ ) x + k 0 l ( w x l φ ) μ 1 φ t τ 1 τ 2 | μ 2 ( s ) | z ( x , 1 , s , t ) d s ρ 2 ψ t t = β ψ x x k ( φ x + l w + ψ ) 0 t ϖ ( t s ) [ ν ( x ) ψ x ( s ) ] x d s μ 3 ( t ) θ ( x ) f ( ψ t ) γ ϱ x ρ 1 w t t = k 0 ( w x l φ ) x k l ( φ x + l w + ψ ) ρ 3 ϱ t = k q x γ ψ t x ρ 4 q t = δ q k ϱ x s z t ( x , τ , s , t ) + z τ ( x , τ , s , t ) = 0 ,
where
( τ , s ) ( 0 , 1 ) × ( τ 1 , τ 2 ) .
We need a several Lemmas.
 Lemma 2. 
The energy functional E is defined by
E t = 1 2 0 1 [ ρ 1 φ t 2 + ρ 2 ψ t 2 + ρ 1 w t 2 + ( β ν ( x ) 0 t ϖ ( s ) d s ) ψ x 2 + ρ 3 ϱ 2 + ρ 4 q 2 + k ( φ x + l w + ψ ) 2 + k 0 ( w x l φ ) 2 ] d x + 1 2 ϖ o ψ x + 1 2 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x ,
satisfies
E t = δ 0 1 q 2 d x + 1 2 ϖ o ψ x 1 2 ϖ ( t ) 0 1 ν ( x ) ψ x 2 d x μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x μ 1 τ 1 τ 2 | μ 2 ( s ) | d s 0 1 φ t 2 d x ,
and
E ( t ) δ 0 1 q 2 d x + 1 2 ϖ o ψ x η 0 0 1 φ t 2 d x μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x 0 ,
where η 0 = μ 1 τ 1 τ 2 | μ 2 ( s ) | d s 0 .
 Proof. 
Multiplying (3)1 by φ t , (3)2 by ψ t , (3)3 by w t , (3)4 by ϱ and (3)5 by q, with integration by parts over 0 , 1 and using (4), we obtain
1 2 d d t 0 1 [ ρ 1 φ t 2 + ρ 2 ψ t 2 + ρ 1 w t 2 + β ψ x 2 + ρ 3 ϱ 2 + ρ 4 q 2 + k ( φ x + l w + ψ ) 2 + k 0 ( w x l φ ) 2 ] d x + 1 2 ϖ o ψ x + δ 0 1 q 2 d x + μ 1 0 1 φ t 2 d x + 0 1 φ t τ 1 τ 2 | μ 2 ( s ) | z x , 1 , s , t d s d x 0 1 ψ x t 0 t ϖ ( t s ) ν ( x ) ψ x ( s ) d s d x + μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x = 0 .
The last term in the LHS of (17) is estimated as follows:
0 1 ψ x t 0 t ϖ ( t s ) ν ( x ) ψ x ( s ) d s d x = 1 2 d d t ϖ o ψ x 1 2 d d t 0 t ϖ ( s ) d s 0 1 ν ( x ) ψ x 2 1 2 ϖ o ψ x + 1 2 ϖ ( t ) 0 1 ν ( x ) ψ x 2 d x .
Now, multiplying the last equation in (13) by z | μ 2 ( s ) | and integrating the result over ( 0 , 1 ) × ( 0 , 1 ) × ( τ 1 , τ 2 )
d d t 1 2 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 ( x , τ , s , t ) d s d τ d x = 0 1 0 1 τ 1 τ 2 | μ 2 ( s ) | z z τ x , τ , s , t d s d τ d x = 1 2 0 1 0 1 τ 1 τ 2 | μ 2 ( s ) | d d τ z 2 x , τ , s , t d s d τ d x = 1 2 0 1 τ 1 τ 2 | μ 2 ( s ) | ( z 2 x , 0 , s , t z 2 ( x , 1 , s , t ) ) d s d x = 1 2 ( τ 1 τ 2 | μ 2 ( s ) | d s ) 0 1 φ t 2 d x 1 2 0 1 τ 1 τ 2 | μ 2 ( s ) | z 2 x , 1 , s , t d s d x .
From (14), (17), (18) and (19), we obtain (15).
The energy functional satisfies
E ( t ) δ 0 1 q 2 d x + 1 2 ϖ o ψ x ( μ 1 τ 1 τ 2 | μ 2 ( s ) | ) 0 1 φ t 2 d x μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x ,
then, by (8), there exists a positive constant η 0 such that
E ( t ) δ 0 1 q 2 d x + 1 2 ϖ o ψ x η 0 0 1 φ t 2 d x μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x ,
then, we obtain E as a non-increasing function. □
 Lemma 3. 
The functional
F 1 t = ρ 2 0 1 α ( x ) ψ t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x + γ ρ 4 k 0 1 α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x ,
satisfies
F 1 t ρ 2 ϖ 0 2 0 1 α ( x ) ψ t 2 d x + 3 c ε 1 0 1 ψ x 2 d x + k 2 ε 2 0 1 ( φ x + l w + ψ ) 2 d x + c 0 1 θ ( x ) f 2 ( ψ t ) d x + c 0 1 q d x + c ( 1 + ε 1 + 1 ε 1 + 1 ε 2 ) ϖ o ψ x + c ϖ o ψ x d s d x .
 Proof. 
We set
F 1 ( t ) = I 1 ( t ) + I 2 ( t ) ,
where
I 1 t = ρ 2 0 1 α ( x ) ψ t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x , I 2 ( t ) = γ ρ 4 k 0 1 α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x .
Differentiation of I 1 gives
I 1 t = ρ 2 0 1 α ( x ) ψ t t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x ρ 2 0 1 α ( x ) ψ t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x ρ 2 0 1 α ( x ) ψ t 2 0 t ϖ ( s ) d s d x .
Using (13), we obtain
ρ 2 0 1 α ( x ) ψ t t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x = β 0 1 α ( x ) ψ x 0 t ϖ ( t s ) ( ψ x ( t ) ψ x ( s ) ) d s d x + k 0 1 α ( x ) ( φ x + l w + ψ ) 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x 0 1 α ( x ) ν ( x ) 0 t ϖ ( t s ) ψ x ( s ) d s 0 t ϖ ( t s ) ( ψ x ( t ) ψ x ( s ) ) d s d x + μ 3 ( t ) 0 1 α ( x ) θ ( x ) f ( ψ t ) 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x + γ 0 1 α ( x ) ϱ x ( 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s ) d x + 0 1 α ( x ) β ψ x ν ( x ) 0 t ϖ ( s ) ψ x ( s ) d s 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x .
Next, by using (12), we have for any δ 1 > 0
ρ 2 0 1 α ( x ) ψ t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x c δ 1 0 1 α ( x ) ψ t 2 d x c δ 1 ϖ o ψ x .
Also
I 2 ( t ) = γ ρ 4 k 0 1 α ( x ) q t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x + γ ρ 4 k 0 1 α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x γ ρ 4 k 0 1 α ( x ) q ψ t ( 0 t ϖ ( s ) d s ) d x .
Using (13), we obtain
I 2 ( t ) = γ δ k 0 1 α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x γ 0 1 α ( x ) ϱ x ( 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s ) d x γ ρ 4 k 0 1 α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x γ ρ 4 k 0 1 α ( x ) q ψ t ( 0 t ϖ ( s ) d s ) d x .
Similarly to (22), we treat the terms in the RHS of (21) as follows:
β 0 1 α ( x ) ψ x 0 t ϖ ( t s ) ( ψ x ( t ) ψ x ( s ) ) d s d x c ε 1 0 1 ψ x 2 d x + c ε 1 ϖ o ψ x .
Also,
k 0 1 α ( x ) ( φ x + l w + ψ ) 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x c ε 2 0 1 ( φ x + l w + ψ ) 2 d x + c ε 2 ϖ o ψ x .
By the same method used in [18], we have these estimates
0 1 α ( x ) ν ( x ) 0 t ϖ ( s ) ψ x ( s ) d s 0 t ϖ ( t s ) ( ψ x ( t ) ψ x ( s ) ) d s d x c ε 1 0 1 ψ x 2 d x + c ( ε 1 + 1 ε 1 ) ϖ o ψ x ,
and
μ 3 ( t ) 0 1 α ( x ) θ ( x ) f ( ψ t ) 0 t ϖ ( t s ) ( ψ x ( t ) ψ x ( s ) ) d s d x c 0 1 θ ( x ) f 2 ( ψ t ) d x + c ϖ o ψ x .
Finally,
0 1 α ( x ) β ψ x ν ( x ) 0 t ϖ ( s ) ψ x ( s ) d s 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x c ε 1 0 1 ψ x 2 d x + c ( ε 1 + 1 ε 1 ) ϖ o ψ x .
As in (22), we easily find that
γ ρ 4 k 0 1 α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x c 0 1 q 2 d x c ϖ o ψ x .
Also, we estimate the first term in the RHS of (23) as follows:
γ δ k 0 1 α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s d x c 0 1 q 2 d x c ϖ o ψ x ,
and
γ ρ 4 k 0 1 α ( x ) q ψ t 0 t ϖ ( s ) d s d x c δ 2 0 1 ψ t 2 d x + c δ 2 0 1 q 2 d x .
By combining the estimates (20)–(31), we complete the proof. □
 Lemma 4. 
Assume that (H1) hold. Then, the functional
F 2 t = ρ 2 0 1 ψ ψ t d x ,
satisfies
F 2 t r 2 0 1 ψ x 2 d x + ρ 2 0 1 ψ t 2 d x + 2 c k 2 r 0 1 ( φ x + l w + ψ ) 2 d x + c 0 1 θ ( x ) f 2 ( ψ t ) d x + c 0 1 ϱ 2 d x + c ϖ o ψ x .
 Proof. 
With direct computations using integration by parts and Young’s inequality, we have
F 2 t = ρ 2 0 1 ψ t 2 d x β 0 1 ψ x 2 d x k 0 1 ψ ( φ x + l w + ψ ) d x + 0 1 ψ x 0 t ϖ ( t s ) ν ( x ) ψ x ( s ) d s d x γ 0 1 ψ ϱ x d x μ 3 ( t ) 0 1 θ ( x ) ψ f ( ψ t ) d x .
Using ( H 1 ) , Young’s, Cauchy–Schwarz and Poincare’s inequalities, we obtain (32).
 Lemma 5. 
The functional
F 3 t = ρ 3 ρ 4 0 1 ϱ 0 x q ( y ) d y d x ,
satisfies
F 3 t k ρ 3 2 0 1 ϱ 2 d x + ε 3 0 1 ψ t 2 d x + c ( 1 + 1 ε 3 ) 0 1 q 2 d x .
 Proof. 
Direct computations give
F 3 t = k ρ 3 0 1 ( φ x + l w + ψ ) ϱ d x + ρ 3 k 0 l 0 1 ( w x l φ ) 0 x ϱ ( y ) d y d x + γ ρ 3 0 1 ϱ 2 d x + ρ 1 0 1 q φ t d x ρ 1 γ 0 1 φ t 2 d x + ρ 1 l 0 1 φ t ϱ x d x .
by using Young’s inequality, we obtain (34). □
In the following Lemma, we suppose that
k = k 0 .
 Lemma 6. 
The functional
F 4 t = ρ 1 0 1 φ t ( w x l φ ) d x ρ 1 0 1 w t ( φ x + l w + ψ ) d x .
satisfies
F 4 t k 0 l 2 0 1 ( w x l φ ) 2 d x ρ 1 l 2 0 1 w t 2 d x + c 0 1 φ t 2 d x + c 0 1 ψ t 2 d x + k l 0 1 ( φ x + l w + ψ ) 2 d x + c 0 1 τ 1 τ 2 | μ 2 ( s ) | z 2 ( x , 1 , s , t ) d s d x ,
 Proof. 
By differentiating F 4 , then using (3), integration by parts and (4), we obtain
F 4 t = k 0 1 ( φ x + l w + ψ ) x ( w x l φ ) d x k 0 l 0 1 ( w x l φ ) 2 d x + μ 1 0 1 φ t ( w x l φ ) d x 0 1 ( w x l φ ) τ 1 τ 2 | μ 2 ( s ) | z ( x , 1 , s , t ) d s d x ρ 1 0 1 φ t ( w t x l φ t ) d x k 0 0 1 ( w x l φ ) x ( φ x + l w + ψ ) d x + k l 0 1 ( φ x + l w + ψ ) 2 d x ρ 1 0 1 w t ( φ x t + l w t + ψ t ) d x .
Using Young’s, Cauchy–Schwarz and Poincare’s inequalities, bearing in mind (35) yields (36). □
In the following Lemma, we suppose that
γ k β ρ 3 ρ 1 γ χ + k 2 β ρ 4 γ χ = 0 ,
where χ = ( ρ 1 k ρ 2 β ) and (35) holds.
 Lemma 7. 
The functional
F 5 t = ρ 2 0 1 ψ t ( φ x + l w + ψ ) d x + β ρ 1 k 0 1 ψ x φ t d x + β ρ 3 γ χ 0 1 ϱ φ t d x k β γ χ 0 1 q ( φ x + l w + ψ ) d x β ρ 2 l 2 k 0 0 1 ψ ψ t d x + β ρ 1 l k 0 0 1 ψ w t d x ρ 1 k 0 1 φ t 0 t ϖ ( t s ) ν ( x ) ψ x ( s ) d s d x + l ρ 1 k 0 t ϖ ( s ) d s 0 1 ν ( x ) ψ x 0 x w t ( y ) d y d ,
satisfies
F 5 t k 2 0 1 ( φ x + l w + ψ ) 2 d x + 5 ε 4 0 1 w t 2 d x + 3 c ε 6 0 1 ( w x l φ ) 2 d x + 8 c ε 5 0 1 ψ x 2 d x + β 2 l 2 k 0 + 3 l 4 ϖ 0 c 2 k 0 1 ψ x 2 d x + c ( 1 + 1 ε 4 ) 0 1 ψ t 2 d x + c ( 1 + 1 ε 5 ) 0 1 φ t 2 d x + c ( 1 + 1 ε 4 ) 0 1 q 2 d x + c ( 1 + 1 ε 5 + 1 ε 6 ) 0 1 ϱ 2 d x + c ( 1 + 1 ε 5 ) 0 1 τ 1 τ 2 | μ 2 ( s ) | z 2 ( x , 1 , s , t ) d s d x + c ( 1 + 1 ε 4 + 1 ε 5 ) ϖ o ψ x c ϖ o ψ x .
 Proof. 
Direct computations give
F 5 t = ρ 2 0 1 ψ t t ( φ x + l w + ψ ) d x + ρ 2 0 1 ψ t ( φ t x + l w t + ψ t ) d x + β ρ 1 k 0 1 ψ t x φ t d x + β ρ 1 k 0 1 ψ x φ t t d x + β ρ 3 γ χ 0 1 ϱ t φ t d x + β ρ 3 γ χ 0 1 ϱ φ t t d x k β γ χ 0 1 q t ( φ x + l w + ψ ) d x k β γ χ 0 1 q ( φ t x + l w t + ψ t ) d x β ρ 2 l 2 k 0 0 1 ψ ψ t t d x β ρ 2 l 2 k 0 0 1 ψ t 2 d x + β ρ 1 l k 0 0 1 ψ t w t d x + β ρ 1 l k 0 0 1 ψ w t t d x ρ 1 k 0 1 φ t t 0 t ϖ ( t s ) ν ( x ) ψ x ( s ) d s d x ρ 1 k 0 1 φ t 0 t ϖ ( t s ) ν ( x ) ψ x ( s ) d s d x + l ρ 1 k 0 t ϖ ( s ) d s 0 1 ν ( x ) ψ t x 0 x w t ( y ) d y d x + l ρ 1 k 0 t ϖ ( s ) d s 0 1 ν ( x ) ψ x 0 x w t t ( y ) d y d x .
Using Young, Cauchy–Schwarz and Poincare’s inequalities, bearing in mind (35) and (37) yields (38). □
 Lemma 8. 
The functional
F 6 t = 0 1 0 1 τ 1 τ 2 s e s τ | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x ,
satisfies
F 6 t η 1 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x + μ 1 0 1 φ t 2 d x η 1 0 1 τ 1 τ 2 | μ 2 ( s ) | z 2 x , 1 , s , t d s d x
where η 1 is a positive constant.
 Proof. 
By differentiating F 6 , with respect to t and using the last equation in (13), we have
F 6 t = 2 0 1 0 1 τ 1 τ 2 e s τ | μ 2 ( s ) | z z τ x , τ , s , t d s d τ d x = 0 1 0 1 τ 1 τ 2 s e s τ | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x 0 1 τ 1 τ 2 | μ 2 ( s ) | [ e s z 2 x , 1 , s , t z 2 x , 0 , s , t ] d s d x .
Using the fact that z ( x , 0 , s , t ) = φ t ( x , t ) , and e s e s τ 1 , for all 0 < τ < 1 , we obtain
F 6 t = η 1 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , ρ , s , t d s d ρ d x 0 1 τ 1 τ 2 e s | μ 2 ( s ) | z 2 x , 1 , s , t d s d x + τ 1 τ 2 | μ 2 ( s ) | d s 0 1 φ t 2 d x .
Due to the fact that e s is a increasing function, we have e s e τ 2 , for all s [ τ 1 , τ 2 ] .
Finally, setting η 1 = e τ 2 and recalling ( H 3 ) , we obtain (39). □
We are now ready to prove the following result.
 Theorem 1. 
Assume (H1)–(H5), and there exist positive constants λ 1 and λ 2 such that the energy functional given by (14) satisfies
E t λ 2 K 1 λ 1 0 t μ 3 ( s ) ϑ ( s ) d s , t 0 ,
where ϑ = 1 if ν 0 . K is defined in (11).

3. Proof of Main Result

We define a Lyapunov functional
L t = N E t + N 1 F 1 ( t ) + N 2 F 2 ( t ) + N 3 F 3 ( t ) + F 4 ( t ) + N 5 F 5 ( t ) + N 6 F 6 ( t ) ,
where N , N 1 , N 2 , N 3 , N 5 and N 6 are positive constants to be selected later.
 Proof. 
(Of Theorem 1) By differentiating (41) and using (16), (19), (32), (34), (36), (38), (39), we have
L t ρ 1 ϖ 0 2 0 1 ( α ( x ) + θ ( x ) ) ψ t 2 d x + ρ 1 ϖ 0 2 0 1 θ ( x ) ψ t 2 d x r N 2 2 2 c ε 1 N 1 + 8 c ε 5 N 5 N 5 ( β 2 l 2 k 0 + 3 l 4 ϖ 0 c 2 k ) 0 1 ψ x 2 d x + ρ 2 N 2 + ε 3 N 3 + c ( 1 + 1 ε 4 ) N 5 + c 0 1 ψ t 2 d x N η 0 c N 5 ( 1 + 1 ε 5 ) μ 1 N 6 c 0 1 φ t 2 d x ρ 1 l 2 5 ε 4 N 5 0 1 w t 2 d x N δ c N 1 c N 3 ( 1 + 1 ε 3 ) c N 5 ( 1 + 1 ε 4 ) 0 1 q 2 d x N 3 k ρ 3 2 c N 2 c N 5 ( 1 + 1 ε 5 + 1 ε 6 ) 0 1 ϱ 2 d x + [ c N 2 + c N 1 ( 1 + ε 2 + 1 ε 2 + 1 ε 3 ) + c N 5 ( 1 + 1 ε 4 + 1 ε 5 ) ] ϖ o ψ x + [ N 2 c N 1 c N 5 ] ϖ o ψ x N 6 η 1 c N 5 ( 1 + 1 ε 5 ) c 0 1 τ 1 τ 2 | μ 2 ( s ) | z 2 x , 1 , s , t d s d x [ N 6 η 1 ] 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x k 2 N 5 k 2 ε 2 N 1 2 c k 2 r N 2 0 1 ( φ x + l w + ψ ) 2 d x k 0 l 2 3 c ε 6 N 5 0 1 ( w x l φ ) 2 d x
c N 1 + c N 2 + c N 5 0 1 θ ( x ) f 2 ( ψ t ) d x N μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x .
By setting
ε 4 = l ρ 1 20 N 5 , ε 6 = k 0 l 12 c N 5 , ε 5 = 1 8 c N 5 , N 2 = r 8 k c N 5 ,
we obtain
L t ρ 1 ϖ 0 2 0 1 ( α ( x ) + θ ( x ) ) ψ t 2 d x + ρ 1 ϖ 0 2 0 1 θ ( x ) ψ t 2 d x r N 2 2 ( 1 4 l 2 m ) 2 c ε 1 N 1 0 1 ψ x 2 d x + ρ 2 N 2 + ε 3 N 3 + c ( 1 + N 5 ) N 5 + c 0 1 ψ t 2 d x N η 0 c N 5 ( 1 + N 5 ) μ 1 N 6 c 0 1 φ t 2 d x ρ 1 l 4 0 1 w t 2 d x N δ c N 1 c N 3 ( 1 + 1 ε 3 ) c N 5 ( 1 + N 5 ) 0 1 q 2 d x N 3 k ρ 3 2 c N 2 c N 5 ( 1 + N 5 ) 0 1 ϱ 2 d x + [ c N 2 + c N 1 ( 1 + ε 2 + 1 ε 2 + 1 ε 3 ) + c N 5 ( 1 + N 5 ) ] ϖ o ψ x + [ N 2 c N 1 c N 5 ] ϖ o ψ x N 6 η 1 c N 5 ( 1 + N 5 ) c 0 1 τ 1 τ 2 | μ 2 ( s ) | z 2 x , 1 , s , t d s d x [ N 6 η 1 ] 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x k 4 N 5 k 2 ε 2 N 1 c 0 1 ( φ x + l w + ψ ) 2 d x k 0 l 4 0 1 ( w x l φ ) 2 d x c N 1 + c N 2 + c N 5 0 1 θ ( x ) f 2 ( ψ t ) d x N μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x .
Next, we carefully choose our constants so that the terms inside the brackets are positive. Let us take ε 1 = ε 2 = 1 N 1 and we choose N 5 large enough such that
α 1 = k N 5 4 ε 2 N 1 c > 0 ,
then, we choose l small enough and we fix N 2 such that
α 2 = r 2 ( 1 4 l 2 m ) N 2 2 ε 2 N 1 c > 0 ,
where
m = ( β 2 c r 2 + 3 l 2 ϖ 0 c 2 r 2 ) ,
then, we take ε 3 = 1 N 3 and we choose N 1 large enough such that
α 3 = d ρ 1 4 N 1 ρ 2 N 2 ε 3 N 3 c N 5 ( 1 + N 5 ) > 0 ,
then, we choose N 3 large enough such that
α 4 = ρ 3 k 2 N 3 c N 1 ( 1 + N 1 ) c N 2 c N 5 ( 1 + N 5 ) > 0 ,
and N 6 large enough such that
α 5 = η 1 N 6 c N 5 ( 1 + N 5 ) > 0 .
Thus, we arrive at
L t α 2 0 1 ψ x 2 d x α 3 0 1 ψ t 2 d x [ N η 0 c ] 0 1 φ t 2 d x α 7 0 1 w t 2 d x α 5 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , 1 , s , t d x α 1 0 1 ( φ x + l w + ψ ) 2 d x α 8 0 1 ( w x l φ ) 2 d x δ N c 0 1 q 2 d x α 4 0 1 ϱ 2 d x + [ N 2 c ] ϖ o ψ x + c ϖ o ψ x α 6 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x + c 0 1 θ ( x ) f 2 ( ψ t ) d x N μ 3 ( t ) 0 1 θ ( x ) ψ t f ( ψ t ) d x ,
where α 6 = η 1 N 6 , α 7 = ρ 1 l 4 , α 8 = k 0 l 4 .
On the other hand, if we let
T t = N 1 F 1 ( t ) + N 2 F 2 ( t ) + N 3 F 3 ( t ) + F 4 ( t ) + N 5 F 5 ( t ) + N 6 F 6 ( t ) ,
then
T t ρ 2 N 1 0 1 | α ( x ) ψ t 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s | d x + γ ρ 4 k N 1 0 1 | α ( x ) q 0 t ϖ ( t s ) ( ψ ( t ) ψ ( s ) ) d s | d x + N 2 0 1 ψ ψ t d x + ρ 3 ρ 4 N 3 0 1 ϱ 0 x q ( y ) d y d x + ρ 1 0 1 φ t ( w x l φ ) d x + ρ 1 0 1 w t ( φ x + l w + ψ ) d x + ρ 2 N 5 0 1 ψ t ( φ x + l w + ψ ) d x + β ρ 1 k N 5 0 1 ψ x φ t d x + β ρ 3 γ χ N 5 0 1 ϱ φ t d x + β k γ χ N 5 0 1 q ( φ x + l w + ψ ) d x + β ρ 2 l 2 k 0 N 5 0 1 ψ ψ t d x + β ρ 1 l k 0 N 5 0 1 ψ w t d x + ρ 1 k N 5 0 1 ν ( x ) φ t 0 t ϖ ( t s ) ψ x ( s ) d s d x + l ρ 1 k 0 t ϖ ( s ) d s N 5 0 1 ν ( x ) ψ x 0 x w t ( y ) d y d x + N 6 F 6 ( t ) .
Exploiting Young, Cauchy-Schwarz and Poincaré’s inequalities, we obtain
T t c 0 1 φ t 2 + ψ t 2 + ψ x 2 + w t 2 + ( φ x + l w + ψ ) 2 + ( w x l φ ) 2 + ϱ 2 + q 2 d x + c ϖ o ψ x + c 0 1 0 1 τ 1 τ 2 s | μ 2 ( s ) | z 2 x , τ , s , t d s d τ d x . c E t .
Consequently, we obtain
T t = L t N E t c E t ,
that is
N c E t L t N + c E t .
Now, by choosing N large enough such that
N c > 0 , N 2 c > 0 , N δ c > 0 , N η 0 c > 0 ,
and exploiting (14), estimates (44) and (45), respectively, give
c 1 E t L t c 2 E t , t 0 ,
for some c 1 , c 2 > 0 .
We have
L ( t ) E ( t ) ,
and
L t c E t + c ϖ o ψ x + c 0 1 θ ( x ) ( ψ t 2 + f 2 ( ψ t ) ) d x , t t 0 .
Let us define the following sets:
Σ ψ = { x ( 0 , 1 ) : | ψ t ( x , t ) | > l 1 } , ϱ ψ = ( 0 , 1 ) Σ ψ .
We estimate the last term in the RHS of (47).
First, note that
0 1 θ ( x ) ( ψ t 2 + f 2 ( ψ t ) ) d x = Σ ψ θ ( x ) ( ψ t 2 + f 2 ( ψ t ) ) d x + ϱ ψ θ ( x ) ( ψ t 2 + f 2 ( ψ t ) ) d x .
Using assumption ( H 5 ) and (16), we easily show that
μ 3 ( t ) Σ ψ θ ( x ) ( ψ t 2 + f 2 ( ψ t ) ) d x ( k 1 1 + k 2 ) Σ ψ μ 3 ( t ) θ ( x ) ψ t f ( ψ t ) d x ( k 1 1 + k 2 ) 0 1 μ 3 ( t ) θ ( x ) ψ t f ( ψ t ) d x c E ( t ) .
If h = 0 on [ 0 , l 1 ] :
This implies that there exist k 1 , k 2 such that k 1 s 2 f ( s ) s k 2 s 2 for all s R + and then (48) is also satisfied for | ψ t ( x , t ) | l 1 , then on all ( 0 , 1 ) . From (47) and (48) and the fact that μ 3 0 , we arrive at
( μ 3 ( t ) L + c E ( t ) ) μ 3 ( t ) J ( E ( t ) ) + c ϖ o ψ x , t t 0 ,
where J is defined in (10).
If h ( 0 ) = 0 and h > 0 on ( 0 , l 1 ] :
Since h is convex and increasing, h 1 is concave and increasing, and by using (H5), the reversed Jensen’s inequality for concave function (see [20]), and (16), we obtain
μ 3 ( t ) ϱ ψ θ ( x ) ( ψ t 2 + f 2 ( ψ t ) ) d x μ 3 ( t ) ϱ ψ θ ( x ) h 1 ( ψ t f ( ψ t ) ) d x μ 3 ( t ) ϱ ψ h 1 ( θ ( x ) ψ t f ( ψ t ) ) d x μ 3 ( t ) | ϱ ψ | h 1 ( ϱ ψ h 1 1 | ϱ ψ | ( θ ( x ) ψ t f ( ψ t ) ) d x ) c μ 3 ( t ) h 1 ( ϱ ψ h 1 ( θ ( x ) ψ t f ( ψ t ) ) d x ) c μ 3 ( t ) h 1 ( 0 1 h 1 ( θ ( x ) ψ t f ( ψ t ) ) d x ) c μ 3 ( t ) h 1 ( c E ( t ) ) .
Therefore, from (47), (48) and (50), we find that
μ 3 L t c μ 3 E t + c μ 3 h 1 ( c E ( t ) ) c E ( t ) + c ϖ o ψ x , t t 0 .
By using Young’s inequality (49) and the fact that
h * ( p ) p [ h ] 1 ( p ) , E 0 , h > 0 a n d μ 3 0 ,
we obtain for ε 0 > 0 small enough and c 0 > 0 large enough,
[ h ( ε 0 E ( t ) ) [ μ 3 ( t ) L ( t ) + c E ( t ) ] + c 0 E ( t ) ] = ε 0 E ( t ) h ( ε 0 E ( t ) ) [ μ 3 L ( t ) + c E ( t ) ] + h ( ε 0 E ( t ) ) [ μ 3 L ( t ) + μ 3 L ( t ) + c E ( t ) ] + c 0 E ( t ) c μ 3 h ( ε 0 E ( t ) ) E ( t ) + c μ 3 h ( ε 0 E ( t ) ) h 1 ( c E ( t ) ) + c 0 E ( t ) + c h ( ε 0 E ( t ) ) ϖ o ψ x c μ 3 h ( ε 0 E ( t ) ) E ( t ) + c μ 3 h * ( h ( ε 0 E ( t ) ) c E ( t ) + c 0 E ( t ) + c h ( ε 0 E ( t ) ) ϖ o ψ x c μ 3 h ( ε 0 E ( t ) ) E ( t ) + c ε 0 μ 3 h ( ε 0 E ( t ) E ( t ) + c ϖ o ψ x c μ 3 h ( ε 0 E ( t ) ) E ( t ) + c ϖ o ψ x = c μ 3 J ( E ( t ) ) + c ϖ o ψ x .
Now, let us define the following functional:
D ( t ) = μ 3 ( t ) L ( t ) c E ( t ) i f h = 0 , o n [ 0 , l 1 ] h ( ε 0 E ( t ) ) [ μ 3 ( t ) L ( t ) c E ( t ) ] + c 0 E ( t ) , i f   h ( 0 ) = 0 a n d h > 0 o n [ 0 , l 1 ] .
Using (46), we have
D ( t ) E ( t ) ,
and exploiting (49) and (51), we easily deduce that
D ( t ) c μ 3 ( t ) J ( E ( t ) ) + c ϖ o ψ x , t t 0 .
By using (16) and (H2), we obtain
( ϑ ( t ) D ( t ) ) = ϑ ( t ) D ( t ) + ϑ ( t ) D ( t ) c μ 3 ( t ) ϑ ( t ) J ( E ( t ) ) + c ϑ ( t ) ϖ o ψ x c μ 3 ( t ) ϑ ( t ) J ( E ( t ) ) + c ( ϑ ( t ) ϖ ) o ψ x c μ 3 ( t ) ϑ ( t ) J ( E ( t ) ) c ϖ o ψ x c μ 3 ( t ) ϑ ( t ) J ( E ( t ) ) c E ( t ) .
Next, let
H = ε ( ϑ ( t ) D ( t ) + c E ( t ) ) ,
where 0 < ε < ε ¯ and ε ¯ is a positive constant satisfying
ϑ ( t ) D ( t ) + c E ( t ) 1 ε ¯ E ( t ) , t 0 .
We also have
H E ,
and for t t 0
H ( t ) c ε μ 3 ( t ) ϑ ( t ) J ( H ) .
Noting that K = 1 J (see (11)), we obtain from (54)
H ( t ) K ( H ( t ) ) c ε μ 3 ( t ) ϑ ( t ) , t t 0 .
A simple integration over ( t 0 , t ) then yields
K ( H ( t ) ) K ( H ( t 0 ) ) + c ε 0 t μ 3 ( s ) ϑ ( s ) d s c ε 0 t 0 μ 3 ( s ) ϑ ( s ) d s .
On the other hand, since lim t 0 + K ( t ) = + and
0 H ( t 0 ) ε ε ¯ E ( t 0 ) ε ε ¯ E ( 0 ) .
We obtain for ε small enough
K ( H ( t 0 ) ) c ε 0 t 0 μ 3 ( s ) ϑ ( s ) d s > 0 .
Then, thanks to the fact that K 1 is decreasing, we infer
H ( t ) K 1 ( K ( H ( t 0 ) ) + c ε 0 t μ 3 ( s ) ϑ ( s ) d s c ε 0 t 0 μ 3 ( s ) ϑ ( s ) d s ) K 1 ( c ε 0 t μ 3 ( s ) ϑ ( s ) d s ) .
From this end inequality and (53), we easily obtain (40). Then, the proof is completed. □

4. Conclusions

Our system concerns a Bresse system along with structural damping, distributed delay and in the presence of both temperatures of second sound type effects introduced in (3). As a main novelty, a general decay result for the solution with few constraints regarding the speeds of wave propagation is obtained. This new result is considered, as far as we know, as an extension of previous results in the literature for such type of system. We mention here that the nonlinear damping in our system creates a good interaction between the distributed delay and the other damping terms of system (3). This type of damping gives more information and qualitative properties on the solution and its impact on stability is also very important, as it is shown in the requirement of Theorem 1. Of course, the other terms (both temperatures and strong damping effects) act as balances in the stability of the system.
Open problem: Open problem: It will be very interesting to analysis the stability of the same system in more complicated cases from the mathematical point of view, but it will be very useful for the application point of view. Namely, one can consider the system
ρ 1 φ t t = k ( φ x + l w + ψ ) x + k 0 l ( w x l φ ) μ 1 φ t ρ 2 ψ t t = β ψ x x k ( φ x + l w + ψ ) μ 2 ( t ) θ ( x ) f ( ψ t ) γ ϱ x ρ 1 w t t = k 0 ( w x l φ ) x k l ( φ x + l w + ψ ) ρ 3 ϱ t = c 0 ϖ ( s ) ϱ x x ( t s ) d s γ ψ t x ,
with a good choice of boundary and initial conditions. The thermoplastic effect is taken in viscoelasticity, and the most interesting question will be asked as follows: Can the system be stabilized? What kind of stability can be found? What is the role of μ 2 ( t ) θ ( x ) f ( ψ t ) ?

Author Contributions

All authors contributed equally to the founding of the paper. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported and funded by the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University (IMSIU) (grant number IMSIU-RG23058).

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The authors declare no conflicts of interest.

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Choucha, A.; Ouchenane, D.; Mirgani, S.M.; Hassan, E.I.; Alfedeel, A.H.A.; Zennir, K. Exploring Thermoelastic Effects in Damped Bresse Systems with Distributed Delay. Mathematics 2024, 12, 857. https://doi.org/10.3390/math12060857

AMA Style

Choucha A, Ouchenane D, Mirgani SM, Hassan EI, Alfedeel AHA, Zennir K. Exploring Thermoelastic Effects in Damped Bresse Systems with Distributed Delay. Mathematics. 2024; 12(6):857. https://doi.org/10.3390/math12060857

Chicago/Turabian Style

Choucha, Abdelbaki, Djamel Ouchenane, Safa M. Mirgani, Eltigan I. Hassan, A. H. A. Alfedeel, and Khaled Zennir. 2024. "Exploring Thermoelastic Effects in Damped Bresse Systems with Distributed Delay" Mathematics 12, no. 6: 857. https://doi.org/10.3390/math12060857

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