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Stability result for Lord Shulman swelling porous thermo-elastic soils with distributed delay term

  • Abdelbaki Choucha , Salah Mahmoud Boulaaras EMAIL logo and Rashid Jan
Published/Copyright: December 8, 2023

Abstract

The Lord Shulman swelling porous thermo-elastic soil system with the presence of a distributed delay term is studied in this work. We will establish the well-posedness of the system and the exponential stability of the system is derived.

MSC 2010: 35B40; 35L70; 74D05; 93D20

1 Introduction and preliminaries

A concept in which a mixture of viscous liquid and particles combined with gas was first introduced by Eringen [1]. We go to the field equations after examining this heat-resistant combination [2]. Expansive (swelling) soils are categorized under the porous media theory, which investigates this kind of issue. That is why there have been several research studies to reduce the damage caused by swelling soil, particularly in architecture and civil engineering; for more information, see [3]–[10]. The linear theory of swelling porous elastic soil fundamental field equations are stated as follows:

(1.1) y y t t = P 1 x + G 1 + K 1 , t t = P 2 x + G 2 + K 2 ,

where elastic solid material and fluid displacement are denoted by , y , and , y > 0 are their respective densities. The partial tension, internal body forces, and eternal forces, which are all acting on the displacement are ( P 1 , G 1 , K 1 ), respectively. Acting similarly to ( P 2 , G 2 , K 2 ) but on the elastic solid. The constitutive equations for partial tensions are also provided by

(1.2) P 1 P 2 = a 1 , a 2 a 2 , a 3 A . y x x ,

where a 1 , a 3 > 0 , and a 2 0 is a real number. A is matrix positive definite with a 1 a 3 > a 2 2 .

Quintanilla [11] studied (1.1) by assuming

G 1 = G 2 = ξ ( y t t ) , K 1 = a 3 y x x t , K 2 = 0 .

They discovered that stability occurs exponentially in cases when ξ > 0 . Similar to this, in [12] the writers examined (1.1) under different conditions

G 1 = G 2 = 0 , K 1 = y γ ( x ) y t , K 2 = 0 ,

where γ ( x ) is an internal viscous damping function with positive mean. They discovered the exponential stability finding utilizing the spectral approach, for more information, refer [13]–[19].

The majority of natural phenomena and industrial equipment involve time delays, which are significant because time lag causes instability and should be taken into consideration. Additionally, other studies have examined this type of problem, including [20]–[26].

In recent times, there has been a substantial surge of interest among scientists in Lord Shulman’s thermoelasticity, leading to an extensive collection of contributions aimed at elucidating this theory. This theoretical framework encompasses the examination of a system comprising four hyperbolic equations coupled with heat transfer dynamics. Moreover, Lord-Shulman thermoelastic theory was introduced to usher in a more robust heat conduction law as it concerns thermoelastic materials exhibiting elastic vibrations. Notably, the heat equation within this context is itself hyperbolic and parallels the equation initially formulated by Fourier’s law. To delve deeper into the specifics and gain a comprehensive understanding of this theory, it is recommended that the reader consult the following papers: [27,28].

The fundamental equations of evolution for one-dimensional theories of swelling porous thermoelasticity with microtemperature and temperature [29]–[34] are provided by

(1.3) y y t t = T x , φ t t = K x + G , T 0 η t = q x , E t = P x * + q Q ,

where K , T 0 , G , T , q , P * , η , E , Q represent the equilibrated stress, the reference temperature, the equilibrated body force, the stress, the heat flux vector, the first heat flux moment, the entropy, the first moment of energy, and the mean heat flux. To make the calculations easier, let T 0 = 1 .

In this study, we take into account the natural counterpart for the microtemperatures of the Lord-Shulman theory. The constitutive equations must be modified in this case to the following form:

(1.4) T = P 1 + G 1 + K 1 , P * = k 2 ω x , K = P 2 + P 3 , η = γ 0 y x + γ 1 + β 1 ( κ t + ) , G = G 2 + K 2 , Q = ( k 1 k 3 ) ω + ( k k 1 ) x , q = k x + k 1 ω , E = β 2 ( κ ω t + ω ) γ 2 x ,

where ω is the microtemperature vector, κ > 0 is the relaxation parameter, and y , , a 1 , a 2 , a 3 , β 1 , β 2 > 0 . The coefficients γ 1 , k , γ 0 denote, the coupling between the volume fraction and the temperature, the thermal conductivity, the coupling between temperature and displacement, respectively.

As coupling is considered, a 2 0 , the coefficients k 1 , k 2 , k 3 , γ 2 > 0 satisfy the following inequalities:

(1.5) a = a 3 a 2 2 a 1 > 0 ,

(1.6) k 1 2 < k k 3 .

The focus of this study is on thermal impacts, thus we assume that the heat capacity β 1 > 0 . Additionally, we do not assume the influence of microtemperature, and β 2 = k 1 = k 2 = k 3 = γ 2 = 0 , and by including the distributed delay term, we create a new problem that is distinct from earlier works. We establish the system’s well-posedness under the proper assumptions, and we use the energy approach to demonstrate the outcome of the exponential stability.

In this study, we take into account:

(1.7) G 1 = G 2 = P 3 = 0 , K 1 = γ 0 ( κ t + ) , K 2 = γ 1 ( κ t + ) ϑ 1 t ν 1 ν 2 ϑ 2 ( s ) t ( x , t s ) d s .

Now, by substituting (1.4) and (1.7) into (1.3), we arrive at the following problem:

y y t t a 1 y x x a 2 x x + γ 0 ( κ t + ) x = 0 , t t a 3 x x a 2 y x x γ 1 ( κ t + ) + ϑ 1 t + ν 1 ν 2 ϑ 2 ( s ) t ( x , t s ) d s = 0 , β ( κ t + ) t + γ 0 y t x + γ 1 t k x x = 0 ,

where

( x , s , t ) H = ( 0 , 1 ) × ( ν 1 , ν 2 ) × ( 0 , ) ,

with initial and boundary conditions

(1.8) y ( x , 0 ) = y 0 ( x ) , y t ( x , 0 ) = y 1 ( x ) , ( x , 0 ) = 0 ( x ) , ( x , 0 ) = 0 ( x ) , t ( x , 0 ) = 1 ( x ) , t ( x , 0 ) = 1 ( x ) , x ( 0 , 1 ) , t ( x , t ) = f 0 ( x , t ) , ( x , t ) ( 0 , 1 ) × ( 0 , ν 2 ) , y ( 0 , t ) = y ( 1 , t ) = ( 0 , t ) = ( 1 , t ) = ( 0 , t ) = ( 1 , t ) = 0 , t 0 .

First, as in [35], we introduce the new variable

Y ( x , , s , t ) = t ( x , t s ) ,

then we obtain

s Y t ( x , , s , t ) + Y ( x , , s , t ) = 0 , Y ( x , 0 , s , t ) = t ( x , t ) .

We can also write our problem in the following form:

(1.9) y y t t a 1 y x x a 2 x x + γ 0 ( κ t + ) x = 0 , t t a 3 x x a 2 y x x γ 1 ( κ t + ) + ϑ 1 t + ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s = 0 , β ( κ t + ) t + γ 0 y t x + γ 1 t k x x = 0 , s Y t ( x , , s , t ) + Y ( x , , s , t ) = 0 ,

where

( x , , s , t ) ( 0 , 1 ) × H ,

initial conditions are

(1.10) y ( x , 0 ) = y 0 ( x ) , y t ( x , 0 ) = y 1 ( x ) , ( x , 0 ) = 0 ( x ) , ( x , 0 ) = 0 ( x ) , t ( x , 0 ) = 1 ( x ) , t ( x , 0 ) = 1 ( x ) , x ( 0 , 1 ) , Y ( x , , s , 0 ) = f 0 ( x , s ) , ( x , , s ) ( 0 , 1 ) × ( 0 , 1 ) × ( 0 , ν 2 ) ,

and boundary conditions are

(1.11) y ( 0 , t ) = y ( 1 , t ) = ( 0 , t ) = ( 1 , t ) = ( 0 , t ) = ( 1 , t ) = 0 , t 0 .

The integral denote the distributed delay terms with ν 1 , ν 2 > 0 are a time delay, ϑ 1 > 0 , and ϑ 2 is an L function satisfying:

(H1) ϑ 2 : [ ν 1 , ν 2 ] R is a bounded function satisfying

(1.12) ν 1 ν 2 ϑ 2 ( s ) d s < ϑ 1 .

Let the solution of system (1.9)–(1.11) be ( y , , , Y ) and the organization of this work is as follows: The well-posedness is demonstrated in Section 2, and the exponential stability is demonstrated in Section 3. We declare that c > 0 in each of the statements that follow.

2 Well-posedness

We demonstrated the well-posedness of the system (1.9)–(1.11) in this section.

Introducing the vector function first

U = ( y , y t , , t , , t , Y ) T ,

and variables v = y t , φ = t , χ = t , system (1.9) can also be written as:

(2.1) U t = J U U ( 0 ) = U 0 = ( y 0 , y 1 , 0 , 1 , 0 , 1 , f 0 ) T ,

where J : D ( J ) Z : Z is the linear operator stated as

(2.2) J U = v 1 y [ a 1 y x x + a 2 x x γ 0 ( κ χ + ) x ] φ 1 a 3 x x + a 2 y x x + γ 1 ( κ χ + ) ϑ 1 φ ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s χ 1 κ β [ γ 0 v x + γ 1 φ k x x + β χ ] 1 s Y ,

and energy space Z is stated as follows:

Z = K 0 1 ( 0 , 1 ) × L 2 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) × L 2 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) × L 2 ( 0 , 1 ) × L 2 ( ( 0 , 1 ) × ( 0 , 1 ) × ( ν 1 , ν 2 ) ) .

For any

U = ( y , v , , φ , , χ , Y ) T Z , U ^ = ( y ^ , v ^ , ^ , φ ^ , ^ , χ ^ , Y ^ ) T Z ,

we equip Z with the inner product stated as follows:

(2.3) U , U ^ Z = y 0 1 v v ^ d x + a 1 0 1 y x y ^ x d x + 0 1 φ φ ^ d x + a 3 0 1 x ^ x d x + a 2 0 1 ( y x ^ x + y ^ x x ) d x + β 0 1 ( κ χ + ) ( κ χ ^ + ^ ) d x + k κ 0 1 x ^ x d x + 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y Y ^ d s d d x .

The domain of J is given by

D ( J ) = U Z y , , K 2 ( 0 , 1 ) K 0 1 ( 0 , 1 ) , v , φ , χ K 0 1 ( 0 , 1 ) , Y , Y L 2 ( ( 0 , 1 ) × ( 0 , 1 ) × ( ν 1 , ν 2 ) ) , Y ( x , 0 , s , t ) = φ .

Therefore, D ( J ) is dense in Z .

Theorem 2.1

Assume that (1.5) and (1.12) hold. Let U 0 Z , then there exists a unique solution U C ( R + , Z ) of problem (2.1).

Furthermore, if U 0 D ( J ) , then U C ( R + , D ( J ) ) C 1 ( R + , Z ) .

Proof

We will show that the operator J is dissipative. Let for any U 0 D ( J ) and by utilizing (2.3), we have

(2.4) J U , U Z = ϑ 1 0 1 φ 2 d x 0 1 ν 1 ν 2 ϑ 2 ( s ) φ Y ( x , 1 , s , t ) d s d x 0 1 0 1 ν 1 ν 2 ϑ 2 ( s ) Y Y d s d d x k 0 1 x 2 d x .

For the third term of the right-hand side (RHS) of (2.4), we have

(2.5) 0 1 0 1 ν 1 ν 2 ϑ 2 ( s ) Y Y d s d d x = 1 2 0 1 ν 1 ν 2 0 1 ϑ 2 ( s ) d d Y 2 d d s d x = 1 2 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x + 1 2 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 0 , s , t ) d s d x .

By utilizing Young’s inequality, we obtain

(2.6) 0 1 ν 1 ν 2 ϑ 2 ( s ) φ Y ( x , 1 , s , t ) d s d x 1 2 ν 1 ν 2 ϑ 2 ( s ) d s 0 1 φ 2 d x + 1 2 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x .

Substituting (2.5) and (2.6) into (2.4), utilizing Y ( x , 0 , s , t ) = φ ( x , t ) and (1.12), we find

(2.7) J U , U Z η 0 0 1 φ 2 d x k 0 1 x 2 d x 0 ,

where η 0 = ( ϑ 1 ν 1 ν 2 ϑ 2 ( s ) d s ) > 0 and J is a dissipative operator.

We then establish that J is a maximal operator. It is sufficient to demonstrate that ( λ I J ) is a surjective operator.

In fact, we demonstrate that for any F = ( f 1 , f 2 , f 3 , f 4 , f 5 , f 6 , f 7 ) T Z , there exists a unique U = ( y , v , , φ , , χ , Y ) T D ( J ) such that

(2.8) ( λ I J ) U = F .

That is

(2.9) λ y v = f 1 K 0 1 ( 0 , 1 ) y λ v a 1 y x x a 2 x x + γ 0 ( κ χ + ) x = y f 2 L 2 ( 0 , 1 ) λ φ = f 3 K 0 1 ( 0 , 1 ) λ φ a 3 x x a 2 y x x γ 1 ( κ χ + ) + ϑ 1 φ + ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s = f 4 L 2 λ χ = f 5 K 0 1 ( 0 , 1 ) β κ λ χ + γ 0 v x + γ 1 φ k x x + β χ = β κ f 6 L 2 ( 0 , 1 ) s λ Y t ( x , , s , t ) + Y ( x , , s , t ) = s f 7 L 2 ( ( 0 , 1 ) × ( 0 , 1 ) × ( ν 1 , ν 2 ) ) .

We observe that equation (2.9)7 with Y ( x , 0 , s , t ) = φ ( x , t ) has a unique solution defined by

(2.10) Y ( x , , s , t ) = e λ s φ + s e s λ 0 e λ s ϱ f 7 ( x , ϱ , s , t ) d ϱ ,

then

(2.11) Y ( x , 1 , s , t ) = e λ s φ + s e λ s 0 1 e λ s ϱ f 7 ( x , ϱ , s , t ) d ϱ ,

and we have

(2.12) v = λ y f 1 , φ = λ f 3 , χ = λ f 5 .

Inserting (2.11) and (2.12) in (2.9)2, (2.9)4, and (2.9)6, we obtain

(2.13) y λ 2 y a 1 y x x a 2 x x + γ 0 ( κ λ + 1 ) x = h 1 ϑ 3 a 3 x x a 2 y x x γ 1 ( κ λ + 1 ) = h 2 ϑ 4 + γ 0 λ y x + γ 1 λ k x x = h 3 ,

where

(2.14) h 1 = y ( λ f 1 + f 2 ) + γ 0 κ f 5 x , h 2 = f 4 + ( λ + ϑ 1 + ν 1 ν 2 ϑ 2 ( s ) e s λ d s ) f 3 γ 1 κ f 5 , ν 1 ν 2 s ϑ 2 ( s ) e s λ 0 1 e λ s ϱ f 7 ( x , ϱ , s , t ) d ϱ d s , h 3 = κ β f 6 + ( γ 0 f 1 x + γ 1 f 3 + β ( 1 + κ λ ) f 5 ) , ϑ 3 = λ 2 + ϑ 1 λ + λ ν 1 ν 2 ϑ 2 ( s ) e λ s d s , ϑ 4 = β λ ( κ λ + 1 ) .

We multiply (2.13) by y ^ , ^ , ^ , respectively, and integrating their sum over ( 0 , 1 ) , we obtain that

(2.15) B ( ( y , , ) , ( y ^ , ^ , ^ ) ) = Γ ( y ^ , ^ , ^ ) ,

where

B : ( K 0 1 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) ) 2 R

is the bilinear form given by

(2.16) B ( ( y , , ) , ( y ^ , ^ , ^ ) ) = y λ 2 0 1 y y ^ d x + a 1 0 1 y x y ^ x d x + a 2 0 1 x y ^ x d x + γ 0 ( κ λ + 1 ) 0 1 x y ^ d x + ϑ 3 0 1 ^ d x + a 3 0 1 x ^ x d x + a 2 0 1 y x ^ x d x γ 1 ( κ λ + 1 ) 0 1 ^ d x + ϑ 4 ( κ λ + 1 ) λ 0 1 ^ d x + γ 0 ( κ λ + 1 ) 0 1 y x ^ d x + γ 1 ( κ λ + 1 ) 0 1 ^ d x + k ( κ λ + 1 ) λ 0 1 x ^ x d x

and

Γ : ( K 0 1 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) ) R

is the linear functional defined by

(2.17) Γ ( y ^ , ^ , ^ ) = 0 1 h 1 y ^ d x + 0 1 h 2 ^ d x + ( κ λ + 1 ) λ 0 1 h 3 ^ d x .

For V = K 0 1 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) × K 0 1 ( 0 , 1 ) , with the norm

( y , , ) V 2 = y 2 2 + y x 2 2 + 2 2 + x 2 2 + 2 2 + x 2 2 ,

then, we have

(2.18) B ( ( y , , ) , ( y , , ) ) = y λ 2 0 1 y 2 d x + a 1 0 1 y x 2 d x + ϑ 3 0 1 2 d x + a 3 0 1 x 2 d x + ϑ 4 ( κ λ + 1 ) λ 0 1 2 d x + 2 a 2 0 1 y x x d x + k ( κ λ + 1 ) λ 0 1 x 2 d x .

On the other hand, we can write

a 1 y x 2 + 2 a 2 y x x + a 3 x 2 = 1 2 a 1 y x + a 2 a 3 x 2 + a 3 x + a 2 a 1 y x 2 + y x 2 a 1 a 2 2 a 3 + x 2 a 3 a 2 2 a 1 .

Since (1.5), we deduce

(2.19) a 1 y x 2 + 2 a 2 y x x + a 3 x 2 > 1 2 y x 2 a 1 a 2 2 a 3 + x 2 a 3 a 2 2 a 1 ,

then, for some M 0 > 0

(2.20) B ( ( y , , ) , ( y , , ) ) M 0 ( y , , ) V 2 .

B is therefore coercive. Therefore, we deduce from the Lax-Milgram theorem that (2.15) has a unique solution.

(2.21) y , , K 0 1 ( 0 , 1 ) .

Substituting y , , and into (2.9)1,3,5, we obtain

v , φ , χ K 0 1 ( 0 , 1 ) .

Also, the compensation of φ in (2.10) with (2.9)7, yields

Y , Y L 2 ( ( 0 , 1 ) × ( 0 , 1 ) × ( ν 1 , ν 2 ) ) .

In addition, if we assume y ^ = ^ = 0 in (2.16), we obtain

a 3 0 1 x ^ x d x + ϑ 3 0 1 ^ d x + a 2 0 1 y x ^ x d x γ 1 ( κ λ + 1 ) 0 1 ^ d x = 0 1 h 2 ^ d x , ^ K 0 1 ( 0 , 1 ) ,

which implies

a 3 0 1 x ^ x d x = 0 1 ( h 2 ϑ 3 + a 2 y x x + γ 1 ( κ λ + 1 ) ) ^ d x , ^ K 0 1 ( 0 , 1 ) ,

that is

(2.22) a 3 x x + a 2 y x x = ϑ 3 γ 1 ( κ λ + 1 ) h 2 L 2 ( 0 , 1 ) .

Similarly, if we take ^ = ^ = 0 in (2.16), we obtain

(2.23) a 2 x x + a 1 y x x = y λ 2 y + γ 0 ( κ λ + 1 ) x h 1 L 2 ( 0 , 1 ) .

Combining (2.22) and (2.23) and by using (1.5) yields that

(2.24) y , K 2 ( 0 , 1 ) K 0 1 ( 0 , 1 ) .

In the same way, if we let y ^ = ^ = 0 in (2.16), we obtain

ϑ 4 ( κ λ + 1 ) λ 0 1 ^ d x + k ( κ λ + 1 ) λ 0 1 x ^ x d x + γ 0 ( κ λ + 1 ) 0 1 y x ^ d x + γ 1 ( κ λ + 1 ) 0 1 ^ d x = ( κ λ + 1 ) λ 0 1 h 3 ^ d x , ^ K 0 1 ( 0 , 1 ) ,

which implies

k x x = ϑ 4 + γ 0 λ y x + γ 1 λ h 3 L 2 ( 0 , 1 ) .

Consequently,

K 2 ( 0 , 1 ) K 0 1 ( 0 , 1 ) .

Finally, the existence of a unique U D ( J ) such that (2.8) holds is guaranteed by the use of regularity theory for the linear elliptic equations.

As a result, we draw the conclusion that J is a maximal dissipative operator. As a result, we have the well-posedness result according to the Lumer-Philips theorem (see [36]).□

3 Exponential decay

We will demonstrate the system (1.9)–(1.11) stability in this section. For the required result we will discuss the following lemmas.

Lemma 3.1

The energy functional E, stated by

(3.1) E ( t ) = 1 2 0 1 [ y y t 2 + a 1 y x 2 + t 2 + a 3 x 2 + 2 a 2 y x x + β ( κ t + ) 2 + k κ x 2 ] d x + 1 2 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x

satisfies

(3.2) E ( t ) k 0 1 x 2 d x η 0 0 1 t 2 d x 0 ,

where η 0 = ϑ 1 ν 1 ν 2 ϑ 2 ( s ) d s > 0 .

Proof

Multiplying the equations (1.9)1,2,3 by y t , t , and ( κ t + ) , integrating by parts over ( 0 , 1 ) , and utilizing (1.11), we obtain

(3.3) 1 2 d d t 0 1 [ y y t 2 + a 1 y x 2 + t 2 + a 3 x 2 + 2 a 2 y x x + β ( κ t + ) 2 + k κ x 2 ] d x + ϑ 1 0 1 t 2 d x + 0 1 t ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s d x + k 0 1 x 2 d x = 0 .

Multiplying the equation (1.9)4 by Y ϑ 2 ( s ) , and integrating the result over ( 0 , 1 ) × ( 0 , 1 ) × ( ν 1 , ν 2 )

(3.4) d d t 1 2 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x = 0 1 0 1 ν 1 ν 2 ϑ 2 ( s ) Y Y ( x , , s , t ) d s d d x = 1 2 0 1 0 1 ν 1 ν 2 ϑ 2 ( s ) d d Y 2 ( x , , s , t ) d s d d x = 1 2 0 1 ν 1 ν 2 ϑ 2 ( s ) ( Y 2 ( x , 0 , s , t ) Y 2 ( x , 1 , s , t ) ) d s d x = 1 2 ν 1 ν 2 ϑ 2 ( s ) d s 0 1 t 2 d x 1 2 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x .

Now, by putting (3.4) in (3.3), and utilizing Young’s inequality, we obtain

E ( t ) k 0 1 x 2 d x ϑ 1 ν 1 ν 2 ϑ 2 ( s ) d s 0 1 t 2 d x ,

then, by (1.12), η 0 > 0 so that

(3.5) E ( t ) k 0 1 x 2 d x η 0 0 1 t 2 d x ,

then we obtain (3.2) ( E is a non-increasing function).□

Remark 3.2

Using (1.5) and (2.19), we conclude that E ( t ) satisfies

E ( t ) > 1 2 0 1 [ y y t 2 + a 4 y x 2 + t 2 + a 5 x 2 + β ( κ t + ) 2 + k κ x 2 ] d x + 1 2 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x ,

where

(3.6) a 4 = 1 2 a 1 a 2 2 a 3 > 0 , a 5 = 1 2 a 3 a 2 2 a 1 > 0 .

Then, the function E ( t ) is non-negative.

Lemma 3.3

The functional

(3.7) D 1 ( t ) 0 1 t d x a 2 a 1 y 0 1 y t d x + ϑ 1 2 0 1 2 d x

satisfies, for any ε 1 > 0

(3.8) D 1 ( t ) a 2 0 1 x 2 d x + ε 1 0 1 y t 2 d x + c 1 + 1 ε 1 0 1 t 2 d x + c 0 1 ( κ t + ) 2 d x + c 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x .

Proof

Direct computation utilizing Young’s inequality and integration by parts produce

(3.9) D 1 ( t ) = a 3 0 1 x 2 d x + 0 1 t 2 d x + a 2 2 a 1 0 1 x 2 d x a 2 a 1 y 0 1 t y t d x + γ 1 0 1 ( κ t + ) d x + a 2 γ 0 a 1 0 1 ( κ t + ) x d x 0 1 ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s d x = a 3 a 2 2 a 1 0 1 x 2 d x + 0 1 t 2 d x a 2 a 1 y 0 1 t y t d x + γ 1 0 1 ( κ t + ) d x a 2 γ 0 a 1 0 1 ( κ t + ) x d x 0 1 ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s d x ,

by using Poincare’s and Young’s inequalities, for δ 1 , δ 2 , δ 3 , ε 1 > 0 , we obtain

(3.10) D 1 ( t ) a 3 a 2 2 a 1 ( ϑ 1 c δ 1 + c δ 2 + δ 3 ) 0 1 x 2 d x + ε 1 0 1 y t 2 d x + c 1 + 1 ε 1 0 1 t 2 d x + c δ 2 + c δ 3 0 1 ( κ t + ) 2 d x + c δ 1 0 t ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x .

Bearing in mind (1.5), and letting δ 1 = a 6 ϑ 1 c , δ 2 = a 6 c , δ 3 = a 6 , we obtain the estimate (3.8).□

Lemma 3.4

The functional

D 2 ( t ) a 2 0 1 t y d x 0 1 y t d x

satisfies

(3.11) D 2 ( t ) a 2 2 2 0 1 y x 2 d x + c 0 1 x 2 d x + c 0 1 ( κ t + ) 2 d x + c 0 1 t 2 d x + c 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x .

Proof

By differentiating D 2 , then utilizing (1.9), integration by parts, and using (1.11) we obtain

(3.12) D 2 ( t ) = a 2 2 0 1 y x 2 d x + a 2 2 y 0 1 x 2 d x a 2 a 3 a 1 a 2 y 0 1 x y x d x a 2 γ 0 y 0 1 ( κ t + ) x d x + a 2 γ 1 0 1 ( κ t + ) y d x a 2 ϑ 1 0 1 y t d x a 2 0 1 y ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s d x .

Now, utilizing Poincare’s and Young’s inequality, we evaluate the final six terms in the RHS of (3.12). For δ 4 , δ 5 > 0 , we have

a 2 a 3 a 1 a 2 y 0 1 x y x d x δ 4 0 1 y x 2 d x + c δ 4 0 1 2 d x , a 2 ϑ 1 0 1 y t d x c δ 5 0 1 y x 2 d x + c δ 5 0 1 t 2 d x , a 2 γ 1 0 1 ( κ t + ) y d x c δ 5 0 1 y x 2 d x + c δ 5 0 1 ( κ t + ) 2 d x

and

a 2 0 1 y ν 1 ν 2 ϑ 2 ( s ) Y ( x , 1 , s , t ) d s d x c δ 5 0 1 y x 2 d x + c δ 5 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x .

By letting δ 4 = a 2 4 , δ 5 = a 2 12 c , and substituting in (3.12), we obtain (3.11).□

Lemma 3.5

The functional

D 3 ( t ) y 0 1 y t y d x

satisfies

(3.13) D 3 ( t ) y 0 1 y t 2 d x + 3 a 1 0 1 y x 2 d x + a 3 4 0 1 x 2 d x + γ 0 2 4 a 1 0 1 ( κ t + ) 2 d x .

Proof

Direct computations give

D 3 ( t ) = y 0 1 y t 2 d x + a 1 0 1 y x 2 d x + a 2 0 1 y x x d x γ 0 0 1 y x ( κ t + ) d x .

Estimate (3.13) easily follows by utilizing Young’s inequality.□

Lemma 3.6

The functional

D 4 ( t ) β κ 2 0 1 t d x β κ 2 0 1 2 d x

satisfies, for any ε 2 > 0

(3.14) D 4 ( t ) β 2 0 1 ( κ t + ) 2 d x + ε 2 0 1 y t 2 d x + c 0 1 t 2 d x + c 1 + 1 ε 2 0 1 x 2 d x .

Proof

Direct computations give

D 4 ( t ) = β κ 0 1 ( κ t + ) t d x β κ 2 0 1 t 2 d x = γ 0 κ 0 1 y t x d x + γ 1 κ 0 1 t d x + k κ 0 1 x 2 d x β κ 2 0 1 t 2 d x .

Estimate (3.14) easily followed by utilizing Young’s inequality for ε 2 > 0 and the fact that

(3.15)□ 0 1 ( κ t ) 2 d x 1 2 0 1 ( κ t + ) 2 d x + 0 1 2 d x .

Now introducing the functional given by

Lemma 3.7

The functional

D 5 ( t ) 0 1 0 1 ν 1 ν 2 s e s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x ,

satisfies

(3.16) D 5 ( t ) η 1 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x + ϑ 1 0 1 t 2 d x η 1 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x ,

where η 1 > 0 .

Proof

By differentiating D 5 , with respect to t and utilizing the last equation in (1.9), we obtain

D 5 ( t ) = 2 0 1 0 1 ν 1 ν 2 e s ϑ 2 ( s ) Y Y ( x , , s , t ) d s d d x = 0 1 0 1 ν 1 ν 2 s e s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x 0 1 ν 1 ν 2 ϑ 2 ( s ) [ e s Y 2 ( x , 1 , s , t ) Y 2 ( x , 0 , s , t ) ] d s d x .

Utilizing that Y ( x , 0 , s , t ) = t ( x , t ) , and e s e s 1 , 0 < < 1 , we obtain

D 5 ( t ) 0 1 0 1 ν 1 ν 2 s e s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x 0 1 ν 1 ν 2 e s ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x + ν 1 ν 2 ϑ 2 ( s ) d s 0 1 t 2 d x .

We have e s e ν 2 , s [ ν 1 , ν 2 ] since e s is an increasing function. Setting η 1 = e ν 2 and remembering (1.12), we find (3.16).

We are now prepared to demonstrate the major finding.

Theorem 3.8

Consider that (1.5) and (1.12) holds. Then, there exist ζ 1 , ζ 2 > 0 such that the energy functional given by (3.1) holds the following:

(3.17) E ( t ) ζ 1 e ζ 2 t , t 0 .

Proof

We define the functional of Lyapunov

(3.18) ( t ) N E ( t ) + N 1 D 1 ( t ) + N 2 D 2 ( t ) + D 3 ( t ) + N 4 D 4 ( t ) + N 5 D 5 ( t ) ,

where N , N 1 , N 2 , N 4 , N 5 > 0 we will assign them later.

By differentiating (3.18) and utilizing (3.2), (3.8), (3.11), (3.13), (3.14), and (3.16), we have

( t ) a N 1 2 c N 2 a 3 4 0 1 x 2 d x [ y ε 1 N 1 ε 2 N 4 ] 0 1 y t 2 d x a 2 2 N 2 2 3 a 1 0 1 y x 2 d x k N c 1 + 1 ε 2 N 4 0 1 x 2 d x η 0 N c N 1 1 + 1 ε 1 c N 2 c N 4 ϑ 1 N 5 0 1 t 2 d x β 2 N 4 c N 1 c N 2 γ 0 4 a 1 0 1 ( κ t + ) 2 d x [ N 5 η 1 c N 1 c N 2 ] 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x N 5 η 1 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x .

By setting

ε 1 = y 4 N 1 , ε 2 = y 4 N 4 ,

we obtain

( t ) a N 1 2 c N 2 a 3 4 0 1 x 2 d x y 2 0 1 y t 2 d x a 2 2 N 2 2 3 a 1 0 1 y x 2 d x [ k N c N 4 ( 1 + N 4 ) ] 0 1 x 2 d x [ η 0 N c N 1 ( 1 + N 1 ) N 2 c N 4 c ϑ 1 N 5 ] 0 1 t 2 d x β 2 N 4 c N 1 c N 2 γ 0 2 4 a 1 0 1 ( κ t + ) 2 d x [ N 5 η 1 c N 1 c N 2 ] 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x N 5 η 1 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x .

Now, choosing our constants.

We select N 2 large enough in such a way that

α 1 = a 2 2 N 2 2 3 a 1 > 0 ,

then we pick N 1 large enough in such a way that

α 2 = a N 1 2 c N 2 a 3 4 > 0 ,

then we pick N 4 and N 5 large enough in such a way that

α 3 = β 2 N 4 c N 1 c N 2 γ 0 2 4 a 1 > 0 α 4 = N 5 η 1 c N 1 c N 2 > 0 .

Thus, we obtain

(3.19) ( t ) α 2 0 1 x 2 d x y 2 0 1 y t 2 d x α 1 0 1 y x 2 d x [ η 0 N c ] 0 1 t 2 d x [ k N c ] 0 1 x 2 d x α 3 0 1 ( κ t + ) 2 d x α 4 0 1 ν 1 ν 2 ϑ 2 ( s ) Y 2 ( x , 1 , s , t ) d s d x α 5 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x ,

where α 5 = η 1 N 5 > 0 .

If we take

L ( t ) = N 1 D 1 ( t ) + N 2 D 2 ( t ) + D 3 ( t ) + N 4 D 4 ( t ) + N 5 D 5 ( t ) ,

then

L ( t ) N 1 0 1 t d x + N 1 a 2 a 1 y 0 1 y t d x + N 1 ϑ 1 2 0 1 2 d x + N 2 a 2 0 1 y t y t d x + y 0 1 y t y d x + N 4 β κ 2 0 1 t d x + N 4 β κ 2 0 1 2 d x + N 5 0 1 0 1 ν 1 ν 2 s e s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d d x .

According to Poincaré, Young’s, and Cauchy-Schwartz inequalities, we find

L ( t ) c 0 1 ( y t 2 + t 2 + x 2 + y x 2 + ( κ t + ) 2 + x 2 ) d x + c 0 1 0 1 ν 1 ν 2 s ϑ 2 ( s ) Y 2 ( x , , s , t ) d s d .

On the other hand, by (2.19) we have

a 1 y x 2 + 2 a 2 x y x + a 3 x 2 > 1 2 a 1 a 2 2 a 3 y x 2 + a 3 a 2 2 a 1 x 2 .

Hence, we obtain

L ( t ) = ( t ) N E ( t ) c E ( t ) ,

that is

(3.20) ( N c ) E ( t ) ( t ) ( N + c ) E ( t ) .

At this point, we pick N large enough in such a way that

N c > 0 , N η 0 c > 0 , N k c > 0 > 0

and exploiting (3.1), the estimates (3.19) and (3.20), respectively, give

(3.21) c 1 E ( t ) ( t ) c 2 E ( t ) , t 0 ,

and

(3.22) ( t ) d 1 E ( t ) , t 0 ,

for some d 1 , c 1 , c 2 > 0 .

Consequently, for some ζ 2 > 0 , we find

(3.23) ( t ) ζ 2 ( t ) , t 0 .

Integration of (3.23) over ( 0 , t ) yields

(3.24) ( t ) ( 0 ) e ζ 2 t , t 0 .

Therefore, (3.17) is achieved by virtue of (3.21) and (3.24).□

4 Conclusion

This work studies a swelling-porous elastic system coupled with thermoelasticity of the Lord-Shulman type and distributed delay which is more general than classical thermoelasticity. Furthermore, the problem circumvented the absurd situation of the infinite propagation of the effect of a thermal or mechanical disturbance in the medium. We established the well-posedness of our problem using the semigroup method. Additionally, we used the energy method to prove the stability result for the system. It is intriguing to know that the result was obtained independently of the wave velocities of the system or any form of interactions between coefficients of the system other than hypotheses (1.5) and (1.12), which guarantees the positivity of the energy of the system. The present result contributes significantly to the existing literature on swelling porous elastic problems. In the future studies to so what happens if the system contains some dampings and sources terms.

Acknowledgment

The authors would like to thank the Deanship of Scientific Research, Qassim University for funding the publication of this project.

  1. Funding information: The authors would like to thank the Deanship of Scientific Research, Qassim University for funding the publication of this project.

  2. Author contributions: The authors contributed equally in this work.

  3. Conflict of interest: The authors state no conflicts of interest.

  4. Data availability statement: No data were used to support the study.

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Received: 2023-09-13
Revised: 2023-11-17
Accepted: 2023-11-27
Published Online: 2023-12-08

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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