Well-Posedness and Stability Results for Lord Shulman Swelling Porous Thermo-Elastic Soils with Microtemperature and Distributed Delay

Abstract

The Lord Shulman swelling porous thermo-elastic soil system with the effects of microtemperature, temperatures and distributed delay terms is considered in this study. The well-posedness result is established by the Lumer–Phillips corollary applied to the Hille–Yosida theorem. The exponential stability result is proven by the energy method under suitable assumptions.

1. Introduction and Preliminaries

The first theory that included a viscous liquid, solid and gas mixture was proposed by Eringen [1]. The field equations were obtained by investigating this heat-resistant combination [2]. The porous media theory, which investigates this type of issue, has also been used to classify expansive (swelling) soils. Due to numerous investigations aimed at mitigating the adverse effects of expansive soils, especially within the fields of architecture and civil engineering, this subject appears promising for further research exploration. For additional information, visit [3,4,5,6,7,8,9]. From the linear theory of swelling porous elastic soils, the fundamental field equations are

$$\begin{array}{lll}\hfill {\rho }_u{u}_{tt}& =& P_{1x}+G_1+{\mathcal{H}}_1,\hfill \\ \hfill {\rho }_{\vartheta }{\vartheta }_{tt}& =& P_{2x}+G_2+{\mathcal{H}}_2,\hfill \end{array}$$

(1)

in which ${\rho }_{\vartheta },{\rho }_u>0$ are the densities of the elastic solid material and fluid, while their respective displacements are denoted by $u,\vartheta$. Furthermore, ($P_1,G_1,{\mathcal{H}}_1$) represent the partial tension, internal forces of body and eternal forces acting on the displacement. ($P_2,G_2,{\mathcal{H}}_2$) are similar, but applied to the elastic solid. Also, the constitutive equations for partial tensions are provided by

$$\left(\begin{array}{c}{P}_1\hfill \\ P_2\hfill \end{array}\right)=\underset{A}{\underbrace{\left(\begin{array}{c}{a}_1,a_2\hfill \\ a_2,a_3\hfill \end{array}\right)}}.\left(\begin{array}{c}{u}_x\hfill \\ {\vartheta }_x\hfill \end{array}\right),$$

(2)

where $a_1,a_3>0$, and $a_2\ne 0$ is a real number. A is matrix positive definite with $a_1{a}_3>a_2^2$.

Quintanilla [9] studied (1) by considering

$$G_1=G_2=\xi (u_t-{\vartheta }_t),{\mathcal{H}}_1=a_3{u}_{xxt},{\mathcal{H}}_2=0,$$

where $\xi >0$; the exponential stability can be achieved. Also, in [10], the researchers considered (1) by taking different conditions:

$$G_1=G_2=0,{\mathcal{H}}_1=-{\rho }_u\gamma \left(x\right)u_t,{\mathcal{H}}_2=0,$$

where the internal viscous damping function $\gamma \left(x\right)$ has a positive mean. They were able to determine the exponential stability using the spectral approach. To discover more, read [9,10,11,12,13,14,15,16]. Time delays are of significant importance in the majority of natural phenomena and industrial systems, as they have the potential to induce instability and should be treated with utmost consideration. Additionally, there are numerous works that have examined this category of issues, including [17,18,19,20,21,22,23].

Numerous researchers worked on similar problems in the literature from different perspectives [24,25,26,27,28]. In recent times, there has been a substantial surge of interest among scientists in Lord Shulman’s thermo-elasticity, 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 thermo-elastic theory was introduced to usher in a more robust heat conduction law, as it concerns thermo-elastic 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: [29,30]. The core evolutionary equations governing one-dimensional models of porous thermo-elasticity, incorporating both microtemperature and temperature effects [31,32,33,34], can be expressed as follows:

$$\begin{array}{lll}\hfill {\rho }_u{u}_{tt}& =& T_x,\hfill \\ \hfill {\rho }_{\phi }{\vartheta }_{tt}& =& {\mathcal{H}}_x+G,\hfill \\ \hfill \rho T_0{\eta }_t& =& q_x,\hfill \\ \hfill \rho E_t& =& P_x^{*}+q-Q.\hfill \end{array}$$

(3)

In the context provided, the symbols T, $T_0$, $\mathcal{H}$, E, $\eta$, q, G, Q and $P^{*}$ denote the stress, reference temperature, equilibrated stress, first energy moment, entropy, heat flux vector, equilibrated body force, mean heat flux and first heat flux moment, respectively. For simplicity in computations, we set $T_0$ to be equal to 1.

This paper addresses the inherent counterpart of microtemperatures within the Lord Shulman theory. In this scenario, it becomes possible to adapt the constitutive equations in the subsequent manner:

$$\begin{array}{lll}& \hfill T=P_1+G_1+{\mathcal{H}}_1\hfill & P^{*}=-k_2{\Re }_x,\hfill \\ & \hfill \mathcal{H}=P_2+P_3\hfill & \rho \eta ={\gamma }_0{u}_x+{\gamma }_1\vartheta +{\beta }_1(\kappa {\theta }_t+\theta ),\hfill \\ & \hfill G=G_2+{\mathcal{H}}_2\hfill & Q=(k_1-k_3)\Re +(k-k_1){\theta }_x,\hfill \\ & \hfill q=k{\theta }_x+k_1\Re \hfill & \rho E=-{\beta }_2(\kappa {\Re }_t+\Re )-{\gamma }_2{\vartheta }_x,\hfill \end{array}$$

(4)

in which the microtemperature vector is indicated by ℜ, $\kappa >0$ is the relaxation parameter and ${\rho }_u,{\rho }_{\vartheta },a_1,a_2,a_3,{\beta }_1,{\beta }_2>0$. The coefficients ${\gamma }_0,k,{\gamma }_1$ denote the coupling between the temperature and displacement, the thermal conductivity, the coupling between the volume fraction and the temperature, respectively.

Taking $a_2\ne 0$ and the coefficients $k_1,k_2,k_3,{\gamma }_2>0$ satisfies the inequalities

$$\begin{array}{lll}& & a=a_3-\frac{a_2^2}{a_1}>0,\hfill \end{array}$$

(5)

$$\begin{array}{lll}& & k_1^2<k{k}_3.\hfill \end{array}$$

(6)

In the current work, we focus on the thermal effects, which is why we make the assumption ${\beta }_1,{\beta }_2>0$ for heat capicity. To add interest to the problem, we also add a distributed delay term to the second equation, creating a new case that differs from earlier research. Under the right assumptions, the system is shown to be well posed, and we use the energy method to demonstrate the result of the exponential stability.

In this work, the following are taken into account:

$$\begin{array}{lll}& & G_1=G_2=0,P_3=-{\gamma }_2(\kappa {\Re }_t+\Re ),\hfill \\ & & {\mathcal{H}}_1=-{\gamma }_0(\kappa {\theta }_t+\theta ),\hfill \\ & & {\displaystyle {\mathcal{H}}_2={\gamma }_1(\kappa {\theta }_t+\theta )-{\varpi }_1{\vartheta }_t-{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right){\vartheta }_t\left(x,t-s\right)ds.}\hfill \end{array}$$

(7)

By substituting (4)–(7) into (3), we have

$$\left\{\begin{array}{c}{\rho }_u{u}_{tt}-a_1{u}_{xx}-a_2{\vartheta }_{xx}+{\gamma }_0{(\kappa {\theta }_t+\theta )}_x=0,\hfill \\ {\rho }_{\vartheta }{\vartheta }_{tt}-a_3{\vartheta }_{xx}-a_2{u}_{xx}-{\gamma }_1(\kappa {\theta }_t+\theta )+{\gamma }_2{(\kappa {\Re }_t+\Re )}_x\hfill \\ {\displaystyle +{\varpi }_1{\vartheta }_t+{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right){\vartheta }_t\left(x,t-s\right)ds=0,}\hfill \\ {\beta }_1{(\kappa {\theta }_t+\theta )}_t+{\gamma }_0{u}_{tx}+{\gamma }_1{\vartheta }_t-k{\theta }_{xx}-k_1{\Re }_x=0,\hfill \\ {\beta }_2{(\kappa {\Re }_t+\Re )}_t-k_2{\Re }_{xx}+{\gamma }_2{\vartheta }_{tx}+k_3\Re +k_1{\theta }_x=0,\hfill \end{array}\right.$$

where

$$(x,s,t)\in \mathbb{H}=(0,1)\times ({\nu }_1,{\nu }_2)\times (0,\infty ),$$

and

$$\begin{array}{l}\hfill \begin{array}{c}u\left(x,0\right)=u_0\left(x\right),u_t\left(x,0\right)=u_1\left(x\right),\theta (x,0)={\theta }_0\left(x\right),\hfill \\ \vartheta \left(x,0\right)={\vartheta }_0\left(x\right),{\vartheta }_t\left(x,0\right)={\vartheta }_1\left(x\right),{\theta }_t(x,0)={\theta }_1\left(x\right),\hfill \\ \Re \left(x,0\right)={\Re }_0\left(x\right),{\Re }_t\left(x,0\right)={\Re }_1\left(x\right),x\in \left(0,1\right),\hfill \\ u\left(0,t\right)=u\left(1,t\right)=\vartheta \left(0,t\right)=\vartheta \left(1,t\right)=0,0\le t,\hfill \\ {\vartheta }_t(x,-t)=f_0(x,t),(x,t)\in (0,1)\times (0,{\nu }_2),\hfill \\ \theta \left(0,t\right)=\theta \left(1,t\right)=\Re \left(0,t\right)=\Re \left(1,t\right)=0,0\le t.\hfill \end{array}\end{array}$$

(8)

Next, we introduce a new variable, as mentioned in [23]:

$$\mathcal{Y}(x,\rho ,s,t)={\vartheta }_t(x,t-s\rho ),$$

thus, the following is obtained:

$$\left\{\begin{array}{c}s{\mathcal{Y}}_t(x,\rho ,s,t)+{\mathcal{Y}}_{\rho }(x,\rho ,s,t)=0,\hfill \\ \mathcal{Y}(x,0,s,t)={\vartheta }_t(x,t).\hfill \end{array}\right.$$

Our problem can be expressed in the following form:

$$\left\{\begin{array}{c}{\rho }_u{u}_{tt}-a_1{u}_{xx}-a_2{\vartheta }_{xx}+{\gamma }_0{(\kappa {\theta }_t+\theta )}_x=0,\hfill \\ {\rho }_{\vartheta }{\vartheta }_{tt}-a_3{\vartheta }_{xx}-a_2{u}_{xx}-{\gamma }_1(\kappa {\theta }_t+\theta )+{\gamma }_2{(\kappa {\Re }_t+\Re )}_x\hfill \\ {\displaystyle +{\varpi }_1{\vartheta }_t+{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}\left(x,1,s,t\right)ds=0,}\hfill \\ {\beta }_1{(\kappa {\theta }_t+\theta )}_t+{\gamma }_0{u}_{tx}+{\gamma }_1{\vartheta }_t-k{\theta }_{xx}-k_1{\Re }_x=0,\hfill \\ {\beta }_2{(\kappa {\Re }_t+\Re )}_t-k_2{\Re }_{xx}+{\gamma }_2{\vartheta }_{tx}+k_3\Re +k_1{\theta }_x=0,\hfill \\ s{\mathcal{Y}}_t(x,\rho ,s,t)+{\mathcal{Y}}_{\rho }(x,\rho ,s,t)=0,\hfill \end{array}\right.$$

(9)

in which

$$(x,\rho ,s,t)\in (0,1)\times \mathbb{H},$$

with

$$\begin{array}{l}\hfill \left\{\begin{array}{c}u\left(x,0\right)=u_0\left(x\right),u_t\left(x,0\right)=u_1\left(x\right),\theta (x,0)={\theta }_0\left(x\right),\hfill \\ \vartheta \left(x,0\right)={\vartheta }_0\left(x\right),{\vartheta }_t\left(x,0\right)={\vartheta }_1\left(x\right),{\theta }_t(x,0)={\theta }_1\left(x\right),\hfill \\ \Re \left(x,0\right)={\Re }_0\left(x\right),{\Re }_t\left(x,0\right)={\Re }_1\left(x\right),x\in \left(0,1\right),\hfill \\ \mathcal{Y}(x,\rho ,s,0)=f_0(x,s\rho ),(x,\rho ,s)\in (0,1)\times (0,1)\times (0,{\nu }_2),\hfill \end{array}\right.\end{array}$$

(10)

and

$$\begin{array}{lll}& & u\left(0,t\right)=u\left(1,t\right)=\vartheta \left(0,t\right)=\vartheta \left(1,t\right)=0,\hfill \\ & & \theta \left(0,t\right)=\theta \left(1,t\right)=\Re \left(0,t\right)=\Re \left(1,t\right)=0,0\le t.\hfill \end{array}$$

(11)

In this context, the integrals denote the presence of distributed delay components, where ${\nu }_1$ and ${\nu }_2$—both greater than zero—represent time delays. The functions ${\varpi }_1$ and ${\varpi }_2$ are $L^{\infty }$ functions and must adhere to the following conditions.

Hypothesis 1.

${\varpi }_2:[{\nu }_1,{\nu }_2]\to \mathbb{R}$ is a bounded function satisfying

$${\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|ds<{\varpi }_1.$$

(12)

In this investigation, we delve into the realm of the Lord Shulman model for swelling porous thermo-elastic soils, incorporating the influence of microtemperature, temperatures and distributed delay components. Our focus lies in demonstrating the system’s well-posedness and examining the outcomes related to its exponential stability. This work is structured as follows: in Section 2, the well-posedness is illustrated, and the exponential stability is demonstrated in Section 3. We state that $c>0$ in each of the sentences that follow.

2. Well-Posedness

Here, we will establish the well-posedness of the system (9)–(11). The following vector function is first introduced:

$$X={(u,u_t,\vartheta ,{\vartheta }_t,\theta ,{\theta }_t,\Re ,{\Re }_t,\mathcal{Y})}^T,$$

where variables $v=u_t,\phi ={\vartheta }_t,\chi ={\theta }_t,\Sigma ={\Re }_t$; then, the system (9) is written as follows:

$$\left\{\begin{array}{c}{X}_t=\mathcal{T}X\hfill \\ X\left(0\right)=X_0={(u_0,u_1,{\vartheta }_0,{\vartheta }_1,{\theta }_0,{\theta }_1,{\Re }_0,{\Re }_1,f_0)}^T,\hfill \end{array}\right.$$

(13)

where $\mathcal{T}:\mathcal{S}\left(\mathcal{T}\right)\subset \mathfrak{V}:\to \mathfrak{V}$ is a linear operator given by

$$\mathcal{T}X=\left(\begin{array}{c}v\hfill \\ \frac{1}{{\rho }_u}[a_1{u}_{xx}+a_2{\vartheta }_{xx}-{\gamma }_0{(\kappa \chi +\theta )}_x]\hfill \\ \phi \hfill \\ \frac{1}{{\rho }_{\vartheta }}[a_3{\vartheta }_{xx}+a_2{u}_{xx}+{\gamma }_1(\kappa \chi +\theta )-{\gamma }_2{(\kappa \Sigma +\Re )}_x\hfill \\ -{\varpi }_1\phi -{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}(x,1,s,t)ds]\hfill \\ \chi \hfill \\ -\frac{1}{\kappa {\beta }_1}[{\gamma }_0{v}_x+{\gamma }_1\phi -k{\theta }_{xx}+{\beta }_1\chi -k_1{\Re }_x]\hfill \\ \Sigma \hfill \\ -\frac{1}{\kappa {\beta }_2}[-k_2{\Re }_{xx}+{\gamma }_2{\phi }_x+k_3\Re +k_1{\theta }_x+{\beta }_2\Sigma ]\hfill \\ -\frac{1}{s}{\mathcal{Y}}_{\rho }\hfill \end{array}\right),$$

(14)

in which energy space is denoted by $\mathfrak{V}$, such that

$$\begin{array}{lll}\hfill \mathfrak{V}& =& {\mathcal{H}}_0^1(0,1)\times L^2(0,1)\times {\mathcal{H}}_0^1(0,1)\times L^2(0,1)\times {\mathcal{H}}_0^1(0,1)\times L^2(0,1)\hfill \\ & & \times {\mathcal{H}}_0^1(0,1)\times L^2(0,1)\times L^2((0,1)\times (0,1)\times ({\nu }_1,{\nu }_2)),\hfill \end{array}$$

for any

$$\begin{array}{lll}\hfill X& =& {(u,v,\vartheta ,\phi ,\theta ,\chi ,\Re ,\Sigma ,\mathcal{Y})}^T\in \mathfrak{V},\hfill \\ \hfill \widehat{X}& =& {(\widehat{u},\widehat{v},\widehat{\vartheta },\widehat{\phi },\widehat{\theta },\widehat{\chi },\widehat{\Re },\widehat{\Sigma },\widehat{\mathcal{Y}})}^T\in \mathfrak{V},\hfill \end{array}$$

with the following inner product:

$$\begin{array}{lll}\hfill <X,\widehat{X}{>}_{\mathfrak{V}}& =& {\rho }_u{\int }_0^{1}v\widehat{v}dx+a_1{\int }_0^1{u}_x{\widehat{u}}_{x}dx+{\rho }_{\vartheta }{\int }_0^1\phi \widehat{\phi }dx+a_3{\int }_0^1{\vartheta }_x{\widehat{\vartheta }}_{x}dx\hfill \\ & & +a_2{\int }_0^1(u_x{\widehat{\vartheta }}_x+{\widehat{u}}_x{\vartheta }_x)dx+k_1\kappa {\int }_0^1({\theta }_x\widehat{\Re }+\Re {\widehat{\theta }}_x)dx\hfill \\ & & +{\beta }_1{\int }_0^1(\kappa \chi +\theta )(\kappa \widehat{\chi }+\widehat{\theta })dx+{\beta }_2{\int }_0^1(\kappa \Sigma +\Re )(\kappa \widehat{\Sigma }+\widehat{\Re })dx\hfill \\ & & +k\kappa {\int }_0^1{\theta }_x{\widehat{\theta }}_{x}dx+k_2\kappa {\int }_0^1{\Re }_x{\widehat{\Re }}_{x}dx+k_3\kappa {\int }_0^1\Re \widehat{\Re }dx\hfill \\ & & +{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|\mathcal{Y}\widehat{\mathcal{Y}}dsd\rho dx.\hfill \end{array}$$

(15)

The domain of $\mathcal{T}$ is given by

$$\mathcal{S}\left(\mathcal{T}\right)=\left\{\begin{array}{c}X\in \mathfrak{V}/u,\vartheta ,\theta ,\Re \in {\mathcal{H}}^2(0,1)\cap {\mathcal{H}}_0^1(0,1),v,\phi ,\chi ,\Sigma \in {\mathcal{H}}_0^1(0,1),\hfill \\ \mathcal{Y},{\mathcal{Y}}_{\rho }\in L^2((0,1)\times (0,1)\times ({\nu }_1,{\nu }_2)),\mathcal{Y}(x,0,s,t)=\phi \hfill \end{array}\right\}.$$

Clearly, $\mathcal{S}\left(\mathcal{T}\right)$ is dense in $\mathfrak{V}$.

Theorem 1.

Let $X_0\in \mathfrak{V}$ and consider that (5), (6) and (12) hold. Then, a unique solution $X\in \mathcal{C}({\mathbb{R}}_{+},\mathfrak{V})$ for the problem (13) exists. Additionally, if $X_0\in \mathcal{S}\left(\mathcal{T}\right)$, then

$$X\in \mathcal{C}({\mathbb{R}}_{+},\mathcal{S}\left(\mathcal{T}\right))\cap {\mathcal{C}}^1({\mathbb{R}}_{+},\mathfrak{V}).$$

Proof. 

First, we show that $\mathcal{T}$ is a dissipative operator. For any $X_0\in \mathcal{S}\left(\mathcal{T}\right)$ and by utilizing (15), we achieve

$$\begin{array}{lll}\hfill <\mathcal{T}X,X{>}_{\mathfrak{V}}& =& -{\varpi }_1{\int }_0^1{\phi }^{2}dx-{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\phi \mathcal{Y}(x,1,s,t)dsdx\hfill \\ & & -k{\int }_0^1{\theta }_x^{2}dx-k_3{\int }_0^1{\Re }^{2}dx-2k_1{\int }_0^1\Re {\theta }_{x}dx\hfill \\ & & -k_2{\int }_0^1{\Re }_x^{2}dx-{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}_{\rho }\mathcal{Y}dsd\rho dx.\hfill \end{array}$$

(16)

For the last term of the RHS of (16), we have

$$\begin{array}{lll}\hfill -{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}_{\rho }\mathcal{Y}dsd\rho dx& =& -\frac{1}{2}{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}{\int }_0^1\left|{\varpi }_2\left(s\right)\right|\frac{d}{d\rho }{\mathcal{Y}}^{2}d\rho dsdx\hfill \\ & =& -\frac{1}{2}{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2(x,1,s,t)dsdx\hfill \\ & & +\frac{1}{2}{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2(x,0,s,t)dsdx.\hfill \end{array}$$

(17)

Applying the inequality of Young, we have

$$\begin{array}{lll}\hfill -{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\phi \mathcal{Y}(x,1,s,t)dsdx& \le & \frac{1}{2}({\int }_{{\nu }_1}^{{\nu }_2}|{\varpi }_2\left(s\right)|ds){\int }_0^1{\phi }^{2}dx\hfill \\ & & +\frac{1}{2}{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2(x,1,s,t)dsdx.\hfill \end{array}$$

(18)

Substituting (17) and (18) into (16) and utilizing $\mathcal{Y}(x,0,s,t)=\phi (x,t)$ and (12), the following is obtained:

$$\begin{array}{lll}\hfill <\mathcal{T}X,X{>}_{\mathfrak{V}}& \le & -k{\int }_0^1{\theta }_x^{2}dx-k_3{\int }_0^1{\Re }^{2}dx-2k_1{\int }_0^1\Re {\theta }_{x}dx\hfill \\ & & -{\eta }_0{\int }_0^1{\phi }^{2}dx-k_2{\int }_0^1{\Re }_x^{2}dx,\hfill \end{array}$$

(19)

where ${\eta }_0=({\varpi }_1-{\int }_{{\nu }_1}^{{\nu }_2}|{\varpi }_2\left(s\right)|ds)>0$. Additionally, by (6), we have

$$\begin{array}{l}\hfill -k{\theta }_x^2-k_3{\Re }^2-2k_1\Re {\theta }_x<0.\end{array}$$

(20)

Therefore, the operator $\mathcal{T}$ is dissipative. Now, we show that the operator $\mathcal{T}$ is a maximal. It is enough to prove that $(\lambda I-\mathcal{T})$ is a surjective operator. In fact, we demonstrate that a unique $X={(u,v,\vartheta ,\phi ,\theta ,\chi ,\Re ,\Sigma ,\mathcal{Y})}^T\in \mathcal{S}(\mathcal{T}$) exists for any $F={(f_1,f_2,f_3,f_4,f_5,f_6,f_7,f_8,f_9)}^T\in \mathfrak{V}$, such that

$$(\lambda I-\mathcal{T})X=F.$$

(21)

That is,

$$\left\{\begin{array}{c}\lambda u-v=f_1\in {\mathcal{H}}_0^1(0,1)\hfill \\ {\rho }_u\lambda v-a_1{u}_{xx}-a_2{\vartheta }_{xx}+{\gamma }_0{(\kappa \chi +\theta )}_x={\rho }_u{f}_2\in L^2(0,1)\hfill \\ \lambda \vartheta -\phi =f_3\in {\mathcal{H}}_0^1(0,1)\hfill \\ {\rho }_{\vartheta }\lambda \phi -a_3{\vartheta }_{xx}-a_2{u}_{xx}-{\gamma }_1(\kappa \chi +\theta )+{\gamma }_2{(\kappa \Sigma +\Re )}_x+{\varpi }_1\phi \hfill \\ +{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}(x,1,s,t)ds={\rho }_{\vartheta }{f}_4\in L^2\hfill \\ \lambda \theta -\chi =f_5\in {\mathcal{H}}_0^1(0,1)\hfill \\ {\beta }_1\kappa \lambda \chi +{\gamma }_0{v}_x+{\gamma }_1\phi -k{\theta }_{xx}-k_1{\Re }_x+{\beta }_1\chi ={\beta }_1\kappa f_6\in L^2(0,1)\hfill \\ \lambda \Re -\Sigma =f_7\in {\mathcal{H}}_0^1(0,1)\hfill \\ {\beta }_2\kappa \lambda \Sigma -k_2{\Re }_{xx}+{\gamma }_2{\phi }_x+k_3\Re +k_1{\theta }_x+{\beta }_2\Sigma ={\beta }_2\kappa f_8\in L^2(0,1)\hfill \\ s\lambda {\mathcal{Y}}_t(x,\rho ,s,t)+{\mathcal{Y}}_{\rho }(x,\rho ,s,t)=s{f}_9\in L^2((0,1)\times (0,1)\times ({\nu }_1,{\nu }_2)).\hfill \end{array}\right.$$

(22)

It can be noticed that there exists a unique solution of (22)₉ with $\mathcal{Y}(x,0,s,t)=\phi (x,t)$, which is given by

$$\mathcal{Y}(x,\rho ,s,t)=e^{-\lambda \rho s}\phi +s{e}^{s\rho \lambda }{\int }_0^{\rho }{e}^{\lambda s\varrho }{f}_9(x,\varrho ,s,t)d\varrho ,$$

(23)

then,

$$\mathcal{Y}(x,1,s,t)=e^{-\lambda s}\phi +s{e}^{\lambda s}{\int }_0^1{e}^{\lambda s\varrho }{f}_9(x,\varrho ,s,t)d\varrho ,$$

(24)

thus, we obtain

$$v=\lambda u-f_1,\phi =\lambda \vartheta -f_3,\chi =\lambda \theta -f_5,\Sigma =\lambda \Re -f_7.$$

(25)

Inserting (24) and (25) into (22)₂, (22)₄, (22)₆ and (22)₈, we obtain

$$\left\{\begin{array}{c}{\rho }_u{\lambda }^{2}u-a_1{u}_{xx}-a_2{\vartheta }_{xx}+{\gamma }_0(\kappa \lambda +1){\theta }_x=h_1\hfill \\ {\varpi }_3\vartheta -a_3{\vartheta }_{xx}-a_2{u}_{xx}-{\gamma }_1(\kappa \lambda +1)\theta +{\gamma }_2(\kappa \lambda +1){\Re }_x=h_2\hfill \\ {\varpi }_4\theta +{\gamma }_0\lambda u_x+{\gamma }_1\lambda \vartheta -k{\theta }_{xx}-k_1{\Re }_x=h_3\hfill \\ {\varpi }_5\Re -k_2{\Re }_{xx}+{\gamma }_2\lambda {\vartheta }_x+k_1{\theta }_x=h_4,\hfill \end{array}\right.$$

(26)

where

$$\left\{\begin{array}{c}{h}_1={\rho }_u(\lambda f_1+f_2)+{\gamma }_0\kappa f_{5x},\hfill \\ h_2={\rho }_{\vartheta }{f}_4+({\rho }_{\vartheta }\lambda +{\varpi }_1+{\int }_{{\nu }_1}^{{\nu }_2}|{\varpi }_2\left(s\right)|e^{-s\lambda }ds)f_3-{\gamma }_1\kappa f_5+{\gamma }_2\kappa f_{7x},\hfill \\ -{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|e^{s\lambda }{\int }_0^1{e}^{\lambda s\varrho }{f}_9(x,\varrho ,s,t)d\varrho ds,\hfill \\ h_3=\kappa {\beta }_1{f}_6+{\gamma }_0{f}_{1x}+{\gamma }_1{f}_3+{\beta }_1(1+\kappa \lambda )f_5,\hfill \\ h_4=\kappa {\beta }_2{f}_8+{\gamma }_2{f}_{3x}+{\beta }_2(1+\kappa \lambda )f_7,\hfill \\ {\varpi }_3={\rho }_{\vartheta }{\lambda }^2+{\varpi }_1\lambda +\lambda {\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|e^{-\lambda s}ds,\hfill \\ {\varpi }_4={\beta }_1\lambda (\kappa \lambda +1),\hfill \\ {\varpi }_5={\beta }_2\lambda (\kappa \lambda +1)+k_3.\hfill \end{array}\right.$$

(27)

Multiplying (26) by $\widehat{u},\widehat{\vartheta },\widehat{\theta },\widehat{\Re }$, and integrating the sum over $(0,1)$, we have

$$B((u,\vartheta ,\theta ,\Re ),(\widehat{u},\widehat{\vartheta },\widehat{\theta },\widehat{\Re }))=\Gamma (\widehat{u},\widehat{\vartheta },\widehat{\theta },\widehat{\Re }),$$

(28)

where

$$B:{({\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1))}^2\to \mathbb{R},$$

is the bilinear form given by

$$\begin{array}{lll}\hfill B((u,\vartheta ,\theta ,\Re ),(\widehat{u},\widehat{\vartheta },\widehat{\theta },\widehat{\Re }))& =& {\rho }_u{\lambda }^2{\int }_0^{1}u\widehat{u}dx+a_1{\int }_0^1{u}_x{\widehat{u}}_{x}dx\hfill \\ & & +a_2{\int }_0^1{\vartheta }_x{\widehat{u}}_{x}dx+{\gamma }_0(\kappa \lambda +1){\int }_0^1{\theta }_x\widehat{u}dx\hfill \\ & & +{\varpi }_3{\int }_0^1\vartheta \widehat{\vartheta }dx+a_3{\int }_0^1{\vartheta }_x{\widehat{\vartheta }}_{x}dx\hfill \\ & & +a_2{\int }_0^1{u}_x{\widehat{\vartheta }}_{x}dx-{\gamma }_1(\kappa \lambda +1){\int }_0^1\theta \widehat{\vartheta }dx\hfill \\ & & +\frac{k_1(\kappa \lambda +1)}{\lambda }{\int }_0^1{\Re }_x\widehat{\theta }dx+{\gamma }_2(\kappa \lambda +1){\int }_0^1{\Re }_x\widehat{\vartheta }dx\hfill \\ & & +\frac{{\varpi }_4(\kappa \lambda +1)}{\lambda }{\int }_0^1\theta \widehat{\theta }dx+{\gamma }_0(\kappa \lambda +1){\int }_0^1{u}_x\widehat{\theta }dx\hfill \\ & & +{\gamma }_1(\kappa \lambda +1){\int }_0^1\vartheta \widehat{\theta }dx+\frac{k(\kappa \lambda +1)}{\lambda }{\int }_0^1{\theta }_x{\widehat{\theta }}_{x}dx,\hfill \\ & & +\frac{{\varpi }_5(\kappa \lambda +1)}{\lambda }{\int }_0^1\Re \widehat{\Re }dx+{\gamma }_2(\kappa \lambda +1){\int }_0^1{\vartheta }_x\widehat{\Re }dx\hfill \\ & & +\frac{k_1(\kappa \lambda +1)}{\lambda }{\int }_0^1{\theta }_x\widehat{\Re }dx+\frac{k_2(\kappa \lambda +1)}{\lambda }{\int }_0^1{\Re }_x{\widehat{\Re }}_{x}dx,\hfill \end{array}$$

(29)

and

$$\Gamma :({\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1),\times {\mathcal{H}}_0^1(0,1))\to \mathbb{R},$$

is the linear functional defined as

$$\begin{array}{lll}\hfill \Gamma (\widehat{u},\widehat{\vartheta },\widehat{\theta },\widehat{\Re })& =& {\int }_0^1{h}_1\widehat{u}dx+{\int }_0^1{h}_2\widehat{\vartheta }dx+\frac{(\kappa \lambda +1)}{\lambda }{\int }_0^1{h}_3\widehat{\theta }dx\hfill \\ & & +\frac{(\kappa \lambda +1)}{\lambda }{\int }_0^1{h}_4\widehat{\Re }dx.\hfill \end{array}$$

(30)

Now, for $V={\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1)\times {\mathcal{H}}_0^1(0,1)$, with the norm

$$\begin{array}{lll}\hfill {\parallel (u,\vartheta ,\theta ,\Re )\parallel }_V^2& =& {\parallel u\parallel }_2^2+{\parallel u_x\parallel }_2^2+{\parallel \vartheta \parallel }_2^2+{\parallel {\vartheta }_x\parallel }_2^2+{\parallel \theta \parallel }_2^2+{\parallel {\theta }_x\parallel }_2^2+{\parallel \Re \parallel }_2^2+{\parallel {\Re }_x\parallel }_2^2,\hfill \end{array}$$

we obtain

$$\begin{array}{lll}\hfill B\left(\right(u,\vartheta ,\theta ,\Re ),(u,\vartheta ,\theta ,\Re \left)\right)& =& {\rho }_u{\lambda }^2{\int }_0^1{u}^{2}dx+a_1{\int }_0^1{u}_x^{2}dx+{\varpi }_3{\int }_0^1{\vartheta }^{2}dx\hfill \\ & & +a_3{\int }_0^1{\vartheta }_x^{2}dx+\frac{{\varpi }_4(\kappa \lambda +1)}{\lambda }{\int }_0^1{\theta }^{2}dx\hfill \\ & & +2a_2{\int }_0^1{u}_x{\vartheta }_{x}dx+\frac{k(\kappa \lambda +1)}{\lambda }{\int }_0^1{\theta }_x^{2}dx\hfill \\ & & +\frac{{\varpi }_5(\kappa \lambda +1)}{\lambda }{\int }_0^1{\Re }^{2}dx+\frac{k_2(\kappa \lambda +1)}{\lambda }{\int }_0^1{\Re }_x^{2}dx.\hfill \end{array}$$

(31)

Also, we can write

$$\begin{array}{lll}\hfill a_1{u}_x^2+2a_2{u}_x{\vartheta }_x+a_3{\vartheta }_x^2& =& \frac{1}{2}[a_1{(u_x+\frac{a_2}{a_3}{\vartheta }_x)}^2+a_3{({\vartheta }_x+\frac{a_2}{a_1}{u}_x)}^2\hfill \\ & & +u_x^2(a_1-\frac{a_2^2}{a_3})+{\vartheta }_x^2(a_3-\frac{a_2^2}{a_1})].\hfill \end{array}$$

Because of (5), we deduce that

$$\begin{array}{lll}\hfill a_1{u}_x^2+2a_2{u}_x{\vartheta }_x+a_3{\vartheta }_x^2& >& \frac{1}{2}\left[u_x^2(a_1-\frac{a_2^2}{a_3})+{\vartheta }_x^2(a_3-\frac{a_2^2}{a_1})\right],\hfill \end{array}$$

(32)

then, for some $M_0>0$,

$$B((u,\vartheta ,\theta ,\Re ),(u,\vartheta ,\theta ,\Re ))\ge M_0{\parallel (u,\vartheta ,\theta ,\Re )\parallel }_V^2.$$

(33)

B is, hence, coercive. As a result, we determine that (28) has a unique solution using the Lax–Milgram theorem:

$$u,\vartheta ,\theta ,\Re \in {\mathcal{H}}_0^1(0,1).$$

(34)

Putting $u,\vartheta ,\theta$, and ℜ in (22)_1,3,5,7, we have

$$\begin{array}{l}\hfill v,\phi ,\chi ,\Sigma \in {\mathcal{H}}_0^1(0,1).\end{array}$$

In the same way, the compensation of $\phi$ in (23) with (22)₉ implies that

$$\begin{array}{l}\hfill \mathcal{Y},{\mathcal{Y}}_{\rho }\in L^2((0,1)\times (0,1)\times ({\nu }_1,{\nu }_2)).\end{array}$$

Additionally, if we assume $\widehat{u}=\widehat{\theta }=\widehat{\Re }=0$ in (29), we obtain

$$\begin{array}{lll}& & a_3{\int }_0^1{\vartheta }_x{\widehat{\vartheta }}_{x}dx+{\varpi }_3{\int }_0^1\vartheta \widehat{\vartheta }dx+a_2{\int }_0^1{u}_x{\widehat{\vartheta }}_{x}dx\hfill \\ & & -{\gamma }_1(\kappa \lambda +1){\int }_0^1\theta \widehat{\vartheta }dx+{\gamma }_2(\kappa \lambda +1){\int }_0^1{\Re }_x\widehat{\vartheta }dx={\int }_0^1{h}_2\widehat{\vartheta }dx,\forall \widehat{\vartheta }\in {\mathcal{H}}_0^1(0,1),\hfill \end{array}$$

which implies

$$\begin{array}{l}\hfill a_3{\int }_0^1{\vartheta }_x{\widehat{\vartheta }}_{x}dx={\int }_0^1\left(h_2-{\varpi }_3\vartheta +a_2{u}_{xx}+(\kappa \lambda +1)({\gamma }_1\theta -{\gamma }_2{\Re }_x)\right)\widehat{\vartheta }dx,\forall \widehat{\vartheta }\in {\mathcal{H}}_0^1,\end{array}$$

that is,

$$\begin{array}{l}\hfill a_3{\vartheta }_{xx}+a_2{u}_{xx}={\varpi }_3\vartheta -(\kappa \lambda +1)({\gamma }_1\theta -{\gamma }_2{\Re }_x)-h_2\in L^2(0,1).\end{array}$$

(35)

Similarly, if we take $\widehat{\vartheta }=\widehat{\theta }=\widehat{\Re }=0$ in (29), we obtain

$$\begin{array}{l}\hfill a_2{\vartheta }_{xx}+a_1{u}_{xx}={\rho }_u{\lambda }^{2}u+{\gamma }_0(\kappa \lambda +1){\theta }_x-h_1\in L^2(0,1).\end{array}$$

(36)

Combining (35) and (36), and by using (5), we achieve that

$$\begin{array}{lll}& & u,\vartheta \in {\mathcal{H}}^2(0,1)\cap {\mathcal{H}}_0^1(0,1).\hfill \end{array}$$

(37)

In the same way, if we let $\widehat{u}=\widehat{\vartheta }=\widehat{\Re }=0$ in (29), we obtain

$$\begin{array}{lll}& & \frac{{\varpi }_4(\kappa \lambda +1)}{\lambda }{\int }_0^1\theta \widehat{\theta }dx+\frac{k(\kappa \lambda +1)}{\lambda }{\int }_0^1{\theta }_x{\widehat{\theta }}_{x}dx+{\gamma }_0(\kappa \lambda +1){\int }_0^1{u}_x\widehat{\theta }dx\hfill \\ & & +{\gamma }_1(\kappa \lambda +1){\int }_0^1\vartheta \widehat{\theta }dx+\frac{k_1(\kappa \lambda +1)}{\lambda }{\int }_0^1{\Re }_x\widehat{\theta }dx=\frac{(\kappa \lambda +1)}{\lambda }{\int }_0^1{h}_3\widehat{\theta }dx,\forall \widehat{\theta }\in {\mathcal{H}}_0^1,\hfill \end{array}$$

which implies

$$\begin{array}{l}\hfill k{\theta }_{xx}={\varpi }_4\theta +{\gamma }_0\lambda u_x+{\gamma }_1\lambda \vartheta +k_1{\Re }_x-h_3\in L^2(0,1).\end{array}$$

Consequently,

$$\theta \in {\mathcal{H}}^2(0,1)\cap {\mathcal{H}}_0^1(0,1).$$

In a similar way, if we take $\widehat{\vartheta }=\widehat{\theta }=\widehat{u}=0$ in (29), we find

$$\begin{array}{l}\hfill k_2{\Re }_{xx}={\varpi }_5\Re +{\gamma }_2{\vartheta }_x+k_1{\theta }_x-h_4\in L^2(0,1).\end{array}$$

Hence,

$$\Re \in {\mathcal{H}}^2(0,1)\cap {\mathcal{H}}_0^1(0,1).$$

Ultimately, leveraging the principles of regularity theory for linear elliptic equations guarantees the presence of a singular $X\in \mathcal{S}\left(\mathcal{T}\right)$, which satisfies Equation (21) uniquely. In light of this, we deduce that $\mathcal{T}$ is a maximal dissipative operator. The well-posedness finding is obtained as a result of the Lumer–Philips theorem [35]. □

3. Exponential Decay

In this part, we demonstrate the system (9)–(11) stability result.

The following lemmas apply to this.

Lemma 1.

The energy functional E, defined as

$$\begin{array}{lll}\hfill E\left(t\right)& =& \frac{1}{2}{\int }_0^1\left[{\rho }_u{u}_t^2+a_1{u}_x^2+{\rho }_{\vartheta }{\vartheta }_t^2+a_3{\vartheta }_x^2+2a_2{u}_x{\vartheta }_x+{\beta }_1{(\kappa {\theta }_t+\theta )}^2\right]dx\hfill \\ & & +\frac{1}{2}{\int }_0^1\left[{\beta }_2{(\kappa {\Re }_t+\Re )}^2+k_2\kappa {\Re }_x^2+k_3\kappa {\Re }^2+k\kappa {\theta }_x^2+2k_1\kappa \Re {\theta }_x\right]dx\hfill \\ & & +\frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx,\hfill \end{array}$$

(38)

satisfies

$$\begin{array}{lll}\hfill E^{\prime }\left(t\right)& \le & -k_4{\int }_0^1{\theta }_x^{2}dx-k_2{\int }_0^1{\Re }_x^{2}dx-k_5{\int }_0^1{\Re }^{2}dx-{\eta }_0{\int }_0^1{\vartheta }_t^{2}dx\le 0,\hfill \end{array}$$

(39)

where ${\eta }_0={\varpi }_1-{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|ds>0,k_4=\frac{1}{2}(k-\frac{k_1^2}{k_3})>0,k_5=\frac{1}{2}(k_3-\frac{k_1^2}{k})>0$.

Proof. 

First, we multiply Equation (9)_1,2,3,4 by $u_t,{\vartheta }_t,(\kappa {\theta }_t+\theta )$ and $(\kappa {\Re }_t+\Re )$. In addition to this, applying (11), we have the following:

$$\begin{array}{lll}& & \frac{1}{2}\frac{d}{dt}{\int }_0^1\left[{\rho }_u{u}_t^2+a_1{u}_x^2+{\rho }_{\vartheta }{\vartheta }_t^2+a_3{\vartheta }_x^2+2a_2{u}_x{\vartheta }_x+{\beta }_1{(\kappa {\theta }_t+\theta )}^2\right]dx\hfill \\ & & \frac{1}{2}\frac{d}{dt}{\int }_0^1\left[{\beta }_2{(\kappa {\Re }_t+\Re )}^2+k_2\kappa {\Re }_x^2+k_3\kappa {\Re }^2+k\kappa {\theta }_x^2+2k_1\kappa \Re {\theta }_x\right]dx\hfill \\ & & +{\varpi }_1{\int }_0^1{\vartheta }_t^{2}dx+{\int }_0^1{\vartheta }_t{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}\left(x,1,s,t\right)dsdx\hfill \\ & & +k{\int }_0^1{\theta }_x^{2}dx+k_3{\int }_0^1{\Re }^{2}dx+k_2{\int }_0^1{\Re }_x^{2}dx+2k_1{\int }_0^1\Re {\theta }_{x}dx=0.\hfill \end{array}$$

(40)

Next, multiplying (9)₅ by $\mathcal{Y}|{\varpi }_2\left(s\right)|$ and integrating, we have

$$\begin{array}{lll}& & \frac{d}{dt}\frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2(x,\rho ,s,t)dsd\rho dx\hfill \\ & =& -{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|\mathcal{Y}{\mathcal{Y}}_{\rho }\left(x,\rho ,s,t\right)dsd\rho dx\hfill \\ & =& -\frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|\frac{d}{d\rho }{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx\hfill \\ & =& \frac{1}{2}{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|({\mathcal{Y}}^2\left(x,0,s,t\right)-{\mathcal{Y}}^2(x,1,s,t))dsdx\hfill \\ & =& \frac{1}{2}{\int }_{{\nu }_1}^{{\nu }_2}|{\varpi }_2\left(s\right)|ds{\int }_0^1{\vartheta }_t^{2}dx-\frac{1}{2}{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,1,s,t\right)dsdx.\hfill \end{array}$$

(41)

Putting (41) into (40) and utilizing Young’s inequality, the following is obtained:

$$\begin{array}{lll}\hfill E^{\prime }\left(t\right)& \le & -k{\int }_0^1{\theta }_x^{2}dx-k_3{\int }_0^1{\Re }^{2}dx-k_2{\int }_0^1{\Re }_x^{2}dx\hfill \\ & & -2k_1{\int }_0^1\Re {\theta }_{x}dx-\left({\varpi }_1-{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|ds\right){\int }_0^1{\vartheta }_t^{2}dx,\hfill \end{array}$$

and we have the following inequality:

$$\begin{array}{l}\hfill k{\theta }_x^2+k_3{\Re }^2+2k_1\Re {\theta }_x>\frac{1}{2}\left[{\theta }_x^2(k-\frac{k_1^2}{k_3})+{\Re }^2(k_3-\frac{k_1^2}{k})\right].\end{array}$$

(42)

By (6), we obtain

$$\begin{array}{l}\hfill k{\theta }_x^2+k_3{\Re }^2+2k_1\Re {\theta }_x>k_4{\theta }_x^2+k_5{\Re }^2,\end{array}$$

(43)

where

$$k_4=\frac{1}{2}(k-\frac{k_1^2}{k_3})>0,k_5=\frac{1}{2}(k_3-\frac{k_1^2}{k})>0;$$

then, by (12), $\exists {\eta }_0={\varpi }_1-{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|ds>0$, so that

$$E^{\prime }\left(t\right)\le -k_4{\int }_0^1{\theta }_x^{2}dx-k_5{\int }_0^1{\Re }^{2}dx-k_2{\int }_0^1{\Re }_x^{2}dx-{\eta }_0{\int }_0^1{\vartheta }_t^{2}dx.$$

(44)

Thus, we achieve (39) (E is a non-increasing function). □

Remark 1.

Using (5), (6) and (32), we deduce that $E\left(t\right)$ fulfills

$$\begin{array}{lll}\hfill E\left(t\right)& >& \frac{1}{2}{\int }_0^1\left[{\rho }_u{u}_t^2+a_4{u}_x^2+{\rho }_{\vartheta }{\vartheta }_t^2+a_5{\vartheta }_x^2+{\beta }_1{(\kappa {\theta }_t+\theta )}^2\right]dx\hfill \\ & & +\frac{1}{2}{\int }_0^1\left[{\beta }_2{(\kappa {\Re }_t+\Re )}^2+k_2\kappa {\Re }_x^2+k_4\kappa {\theta }_x^2+k_5\kappa {\Re }^2\right]dx\hfill \\ & & +\frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx,\hfill \end{array}$$

where

$$\begin{array}{l}\hfill a_4=\frac{1}{2}\left(a_1-\frac{a_2^2}{a_3}\right)>0,a_5=\frac{1}{2}\left(a_3-\frac{a_2^2}{a_1}\right)>0.\end{array}$$

(45)

Thus, the non-negativity of the function $E\left(t\right)$ is obtained.

Lemma 2.

The functional

$$D_1\left(t\right):={\rho }_{\vartheta }{\int }_0^1{\vartheta }_t\vartheta dx-\frac{a_2}{a_1}{\rho }_u{\int }_0^1\vartheta u_{t}dx+\frac{{\varpi }_1}{2}{\int }_0^1{\vartheta }^{2}dx,$$

(46)

satisfies, for any ${\epsilon }_1>0$,

$$\begin{array}{lll}\hfill D_1^{\prime }\left(t\right)& \le & -\frac{a}{2}{\int }_0^1{\vartheta }_x^{2}dx+{\epsilon }_1{\int }_0^1{u}_t^{2}dx+c(1+{\frac{1}{\epsilon }}_1){\int }_0^1{\vartheta }_t^{2}dx\hfill \\ & & +c{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx+c{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx\hfill \\ & & +c{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,1,s,t\right)dsdx.\hfill \end{array}$$

(47)

Proof. 

Via direct calculation, utilizing integration by parts, we obtain that

$$\begin{array}{lll}\hfill D_1^{\prime }\left(t\right)& =& -a_3{\int }_0^1{\vartheta }_x^{2}dx+{\rho }_{\vartheta }{\int }_0^1{\vartheta }_t^{2}dx+\frac{a_2^2}{a_1}{\int }_0^1{\vartheta }_x^{2}dx-\frac{a_2}{a_1}{\rho }_u{\int }_0^1{\vartheta }_t{u}_{t}dx\hfill \\ & & +{\gamma }_1{\int }_0^1(\kappa {\theta }_t+\theta )\vartheta dx+\frac{a_2{\gamma }_0}{a_1}{\int }_0^1{(\kappa {\theta }_t+\theta )}_x\vartheta dx\hfill \\ & & -{\gamma }_2{\int }_0^1{(\kappa {\Re }_t+\Re )}_x\vartheta dx-{\int }_0^1\vartheta {\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}(x,1,s,t)dsdx\hfill \\ & =& -\left(a_3-\frac{a_2^2}{a_1}\right){\int }_0^1{\vartheta }_x^{2}dx+{\rho }_{\vartheta }{\int }_0^1{\vartheta }_t^{2}dx-\frac{a_2}{a_1}{\rho }_u{\int }_0^1{\vartheta }_t{u}_{t}dx\hfill \\ & & +{\gamma }_1{\int }_0^1(\kappa {\theta }_t+\theta )\vartheta dx-\frac{a_2{\gamma }_0}{a_1}{\int }_0^1(\kappa {\theta }_t+\theta ){\vartheta }_{x}dx\hfill \\ & & +{\gamma }_2{\int }_0^1(\kappa {\Re }_t+\Re ){\vartheta }_{x}dx-{\int }_0^1\vartheta {\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}(x,1,s,t)dsdx.\hfill \end{array}$$

(48)

Using Poincaré’s and Young’s inequalities, for ${\delta }_1,{\delta }_2,{\delta }_3,{\epsilon }_1>0$, we obtain that

$$\begin{array}{lll}\hfill D_1^{\prime }\left(t\right)& \le & -\left((a_3-\frac{a_2^2}{a_1})--({\varpi }_{1}c{\delta }_1+c{\delta }_2+2{\delta }_3)\right){\int }_0^1{\vartheta }_x^{2}dx+{\epsilon }_1{\int }_0^1{u}_t^{2}dx\hfill \\ & & +c(1+\frac{1}{{\epsilon }_1}){\int }_0^1{\vartheta }_t^{2}dx+(\frac{c}{{\delta }_2}+\frac{c}{{\delta }_3}){\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx\hfill \\ & & +\frac{c}{{\delta }_3}{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx+\frac{c}{{\delta }_1}{\int }_0^t{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2(x,1,s,t)dsdx.\hfill \end{array}$$

(49)

Bearing in mind (5) and letting ${\delta }_1={\displaystyle \frac{a}{6{\varpi }_{1}c}},{\delta }_2={\displaystyle \frac{a}{6c}},{\delta }_3={\displaystyle \frac{a}{12}}$, we obtain estimate (47). □

Lemma 3.

The functional

$$D_2\left(t\right):=a_2\left({\int }_0^1{\vartheta }_{t}udx-{\int }_0^1\vartheta u_{t}dx\right),$$

satisfies,

$$\begin{array}{lll}\hfill D_2^{\prime }\left(t\right)& \le & -\frac{a_2^2}{2{\rho }_{\vartheta }}{\int }_0^1{u}_x^{2}dx+c{\int }_0^1{\vartheta }_x^{2}dx+c{\int }_0^1{\vartheta }_t^{2}dx+c{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx\hfill \\ & & +c{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx+c{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2(x,1,s,t)dsdx.\hfill \end{array}$$

(50)

Proof. 

By differentiating $D_2$, then utilizing (9), integration by parts and (11), we obtain

$$\begin{array}{lll}\hfill D_2^{\prime }\left(t\right)& =& -\frac{a_2^2}{{\rho }_{\vartheta }}{\int }_0^1{u}_x^{2}dx+\frac{a_2^2}{{\rho }_u}{\int }_0^1{\vartheta }_x^{2}dx-\left(\frac{a_2{a}_3}{{\rho }_{\vartheta }}-\frac{a_1{a}_2}{{\rho }_u}\right){\int }_0^1{\vartheta }_x{u}_{x}dx\hfill \\ & & -\frac{a_2{\gamma }_0}{{\rho }_u}{\int }_0^1(\kappa {\theta }_t+\theta ){\vartheta }_{x}dx+\frac{a_2{\gamma }_1}{{\rho }_{\vartheta }}{\int }_0^1(\kappa {\theta }_t+\theta )udx\hfill \\ & & +\frac{a_2{\gamma }_2}{{\rho }_{\vartheta }}{\int }_0^1(\kappa {\Re }_t+\Re )u_{x}dx-\frac{a_2{\varpi }_1}{{\rho }_{\vartheta }}{\int }_0^{1}u{\vartheta }_{t}dx\hfill \\ & & -\frac{a_2}{{\rho }_{\vartheta }}{\int }_0^{1}u{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}(x,1,s,t)dsdx.\hfill \end{array}$$

(51)

We now evaluate the final six terms in the right hand side of (51), utilizing Poincaré’s and Young’s inequalities. For ${\delta }_4,{\delta }_5>0$, we have

$$-\left(\frac{a_2{a}_3}{{\rho }_{\vartheta }}-\frac{a_1{a}_2}{{\rho }_u}\right){\int }_0^1{\vartheta }_x{u}_{x}dx\le {\delta }_4{\int }_0^1{u}_x^{2}dx+\frac{c}{{\delta }_4}{\int }_0^1{\vartheta }^{2}dx,$$

$$\begin{array}{lll}\hfill -\frac{a_2{\varpi }_1}{{\rho }_{\vartheta }}{\int }_0^{1}u{\vartheta }_{t}dx& \le & c{\delta }_5{\int }_0^1{u}_x^{2}dx+\frac{c}{{\delta }_5}{\int }_0^1{\vartheta }_t^{2}dx,\hfill \end{array}$$

$$\begin{array}{lll}\hfill \frac{a_2{\gamma }_1}{{\rho }_{\vartheta }}{\int }_0^1(\kappa {\theta }_t+\theta )udx& \le & c{\delta }_5{\int }_0^1{u}_x^{2}dx+\frac{c}{{\delta }_5}{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx,\hfill \end{array}$$

$$\begin{array}{lll}\hfill \frac{a_2{\gamma }_2}{{\rho }_{\vartheta }}{\int }_0^1(\kappa {\Re }_t+\Re )u_{x}dx& \le & {\delta }_4{\int }_0^1{u}_x^{2}dx+\frac{c}{{\delta }_4}{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx,\hfill \end{array}$$

and

$$\begin{array}{lll}\hfill -\frac{a_2}{{\rho }_{\vartheta }}{\int }_0^{1}u{\int }_{{\nu }_1}^{{\nu }_2}{\varpi }_2\left(s\right)\mathcal{Y}(x,1,s,t)dsdx& \le & c{\delta }_5{\int }_0^1{u}_x^{2}dx\hfill \\ & & +\frac{c}{{\delta }_5}{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2(x,1,s,t)dsdx.\hfill \end{array}$$

By letting ${\delta }_4=\frac{a_2}{8{\rho }_{\vartheta }},{\delta }_5=\frac{a_2}{12c{\rho }_{\vartheta }}$ and putting these into (51), we obtain (50). □

Lemma 4.

The functional

$$D_3\left(t\right):=-{\rho }_u{\int }_0^1{u}_{t}udx,$$

satisfies

$$\begin{array}{lll}\hfill D_3^{\prime }\left(t\right)& \le & -{\rho }_u{\int }_0^1{u}_t^{2}dx+3a_1{\int }_0^1{u}_x^{2}dx+\frac{a_3}{4}{\int }_0^1{\vartheta }_x^{2}dx\hfill \\ & & +\frac{{\gamma }_0^2}{4a_1}{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx.\hfill \end{array}$$

(52)

Proof. 

Direct computations give

$$D_3^{\prime }\left(t\right)=-{\rho }_u{\int }_0^1{u}_t^{2}dx+a_1{\int }_0^1{u}_x^{2}dx+a_2{\int }_0^1{u}_x{\vartheta }_{x}dx-{\gamma }_0{\int }_0^1{u}_x(\kappa {\theta }_t+\theta )dx.$$

Estimate (52) easily follows by utilizing Young’s inequality. □

Lemma 5.

The functional

$$D_4\left(t\right):=-{\beta }_1{\kappa }^2{\int }_0^1{\theta }_t\theta dx-\frac{{\beta }_1\kappa }{2}{\int }_0^1{\theta }^{2}dx-{\beta }_2{\kappa }^2{\int }_0^1{\Re }_t\Re dx-\frac{{\beta }_2\kappa }{2}{\int }_0^1{\Re }^{2}dx,$$

satisfies, for any ${\epsilon }_2>0$,

$$\begin{array}{lll}\hfill D_4^{\prime }\left(t\right)& \le & -\frac{{\beta }_1}{2}{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx-\frac{{\beta }_2}{2}{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx+{\epsilon }_2{\int }_0^1{u}_t^{2}dx+c{\int }_0^1{\vartheta }_t^{2}dx\hfill \\ & & +c(1+\frac{1}{{\epsilon }_2}){\int }_0^1{\theta }_x^{2}dx+c{\int }_0^1{\Re }_x^{2}dx+c{\int }_0^1{\Re }^{2}dx.\hfill \end{array}$$

(53)

Proof. 

Direct computations give

$$\begin{array}{lll}\hfill D_4^{\prime }\left(t\right)& =& -{\beta }_1\kappa {\int }_0^1{(\kappa {\theta }_t+\theta )}_t\theta dx-{\beta }_1{\kappa }^2{\int }_0^1{\theta }_t^{2}dx\hfill \\ & & -{\beta }_2\kappa {\int }_0^1{(\kappa {\Re }_t+\Re )}_t\Re dx-{\beta }_2{\kappa }^2{\int }_0^1{\Re }_t^{2}dx\hfill \\ & =& {\gamma }_0\kappa {\int }_0^1\theta u_{tx}dx+{\gamma }_1\kappa {\int }_0^1{\vartheta }_t\theta dx+k\kappa {\int }_0^1{\theta }_x^{2}dx-{\beta }_1{\kappa }^2{\int }_0^1{\theta }_t^{2}dx\hfill \\ & & +2k_1\kappa {\int }_0^1\Re {\theta }_{x}dx+k_2\kappa {\int }_0^1{\Re }_x^{2}dx+k_3\kappa {\int }_0^1{\Re }^{2}dx\hfill \\ & & -{\gamma }_2\kappa {\int }_0^1{\vartheta }_t{\Re }_{x}dx-{\beta }_2{\kappa }^2{\int }_0^1{\Re }_t^{2}dx.\hfill \end{array}$$

Further simplification of (53) leads us to

$$\begin{array}{lll}& & -{\int }_0^1{\left(\kappa {\theta }_t\right)}^{2}dx\le -\frac{1}{2}{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx+{\int }_0^1{\theta }^{2}dx,\hfill \\ & & -{\int }_0^1{\left(\kappa {\Re }_t\right)}^{2}dx\le -\frac{1}{2}{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx+{\int }_0^1{\Re }^{2}dx.\hfill \end{array}$$

(54)

 □

Now, we introduce a functional stated in Lemma 6.

Lemma 6.

The functional

$$D_5\left(t\right):={\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s{e}^{-s\rho }\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx,$$

satisfies

$$\begin{array}{lll}\hfill D_5^{\prime }\left(t\right)& \le & -{\eta }_1{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx+{\varpi }_1{\int }_0^1{\vartheta }_t^{2}dx\hfill \\ & & -{\eta }_1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,1,s,t\right)dsdx,\hfill \end{array}$$

(55)

where ${\eta }_1>0$.

Proof. 

By differentiating $D_5$ with respect to t and utilizing the last equation in (9), we obtain

$$\begin{array}{lll}\hfill D_5^{\prime }\left(t\right)& =& -2{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}{e}^{-s\rho }\left|{\varpi }_2\left(s\right)\right|\mathcal{Y}{\mathcal{Y}}_{\rho }\left(x,\rho ,s,t\right)dsd\rho dx\hfill \\ & =& -{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s{e}^{-s\rho }\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx\hfill \\ & & -{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|[e^{-s}{\mathcal{Y}}^2\left(x,1,s,t\right)-{\mathcal{Y}}^2\left(x,0,s,t\right)]dsdx.\hfill \end{array}$$

Utilizing that $\mathcal{Y}(x,0,s,t)={\vartheta }_t(x,t)$ and $e^{-s}\le e^{-s\rho }\le 1$, ∀$0<\rho <1$, we achieve that

$$\begin{array}{lll}\hfill D_5^{\prime }\left(t\right)& \le & -{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s{e}^{-s}|{\varpi }_2\left(s\right)|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx\hfill \\ & & -{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}{e}^{-s}|{\varpi }_2\left(s\right)|{\mathcal{Y}}^2\left(x,1,s,t\right)dsdx+({\int }_{{\nu }_1}^{{\nu }_2}|{\varpi }_2\left(s\right)|ds){\int }_0^1{\vartheta }_t^{2}dx.\hfill \end{array}$$

 □

As $-e^{-s}$ is an increasing function, we have $-e^{-s}\le -e^{-{\nu }_2}$, for all $s\in [{\nu }_1,{\nu }_2]$. By setting ${\eta }_1=e^{-{\nu }_2}$ and remembering (12), we discover (55). We can now proceed to proving the primary finding.

Theorem 2.

Let (5), (6) and (12) hold. Then, there exist ${\zeta }_1,{\zeta }_2>0$ such that the energy functional provide by (38) holds:

$$E\left(t\right)\le {\zeta }_1{e}^{-{\zeta }_{2}t},\forall t\ge 0.$$

(56)

Proof. 

To prove the required result, we introduce the Lyapunov functional as follows:

$$\mathcal{P}\left(t\right):=NE\left(t\right)+N_1{D}_1\left(t\right)+N_2{D}_2\left(t\right)+D_3\left(t\right)+N_4{D}_4\left(t\right)+N_5{D}_5\left(t\right),$$

(57)

where $N,N_1,N_2,N_4,N_5>0$; we assign them later.

By differentiating (57) and using (39), (47), (50), (52), (53) and (55), we have

$$\begin{array}{lll}\hfill {\mathcal{P}}^{\prime }\left(t\right)& \le & -\left[\frac{a{N}_1}{2}-c{N}_2-\frac{a_3}{4}\right]{\int }_0^1{\vartheta }_x^{2}dx-\left[{\rho }_u-{\epsilon }_1{N}_1-{\epsilon }_2{N}_4\right]{\int }_0^1{u}_t^{2}dx\hfill \\ & & -\left[\frac{a_2^2{N}_2}{2{\rho }_{\vartheta }}-3a_1\right]{\int }_0^1{u}_x^{2}dx-\left[k_{4}N-c(1+\frac{1}{{\epsilon }_2})N_4\right]{\int }_0^1{\theta }_x^{2}dx\hfill \\ & & -\left[{\eta }_{0}N-c{N}_1(1+\frac{1}{{\epsilon }_1})-c{N}_2-c{N}_4-{\varpi }_1{N}_5\right]{\int }_0^1{\vartheta }_t^{2}dx\hfill \\ & & -\left[\frac{{\beta }_1}{2}{N}_4-c{N}_1-c{N}_2-\frac{{\gamma }_0}{4a_1}\right]{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx\hfill \\ & & -\left[\frac{{\beta }_2}{2}{N}_4-c{N}_1-c{N}_2\right]{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx\hfill \\ & & -\left[k_{2}N-c{N}_4\right]{\int }_0^1{\Re }_x^{2}dx-\left[k_{5}N-c{N}_4\right]{\int }_0^1{\Re }^{2}dx\hfill \\ & & -\left[N_5{\eta }_1-c{N}_1-c{N}_2\right]{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,1,s,t\right)dsdx\hfill \\ & & -N_5{\eta }_1{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx.\hfill \end{array}$$

by setting

$${\epsilon }_1=\frac{{\rho }_u}{4N_1},{\epsilon }_2=\frac{{\rho }_u}{4N_4},$$

we obtain

$$\begin{array}{lll}\hfill {\mathcal{P}}^{\prime }\left(t\right)& \le & -\left[\frac{a{N}_1}{2}-c{N}_2-\frac{a_3}{4}\right]{\int }_0^1{\vartheta }_x^{2}dx-\left[\frac{{\rho }_u}{2}\right]{\int }_0^1{u}_t^{2}dx\hfill \\ & & -\left[\frac{a_2^2{N}_2}{2{\rho }_{\vartheta }}-3a_1\right]{\int }_0^1{u}_x^{2}dx-\left[k_{4}N-c{N}_4(1+N_4)\right]{\int }_0^1{\theta }_x^{2}dx\hfill \\ & & -\left[{\eta }_{0}N-c{N}_1(1+N_1)-N_{2}c-N_{4}c-{\varpi }_1{N}_5\right]{\int }_0^1{\vartheta }_t^{2}dx\hfill \\ & & -\left[\frac{{\beta }_1}{2}{N}_4-c{N}_1-c{N}_2-\frac{{\gamma }_0^2}{4a_1}\right]{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx\hfill \\ & & -\left[\frac{{\beta }_2}{2}{N}_4-c{N}_1-c{N}_2\right]{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx\hfill \\ & & -\left[k_{2}N-c{N}_4\right]{\int }_0^1{\Re }_x^{2}dx-\left[k_{5}N-c{N}_4\right]{\int }_0^1{\Re }^{2}dx\hfill \\ & & -\left[N_5{\eta }_1-c{N}_1-c{N}_2\right]{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,1,s,t\right)dsdx\hfill \\ & & -N_5{\eta }_1{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx.\hfill \end{array}$$

Now, we choose our constants.

We take $N_2$ large enough, such that

$${\alpha }_1=\frac{a_2^2{N}_2}{2{\rho }_{\vartheta }}-3a_1>0;$$

then, we pick $N_1$ large enough, in such a way that

$${\alpha }_2=\frac{a{N}_1}{2}-c{N}_2-\frac{a_3}{4}>0.$$

Then, we pick $N_4$ and $N_5$ large enough, in such a way that

$$\begin{array}{lll}\hfill {\alpha }_3& =& \frac{{\beta }_1}{2}{N}_4-c{N}_1-c{N}_2-\frac{{\gamma }_0^2}{4a_1}>0,\hfill \\ \hfill {\alpha }_4& =& \frac{{\beta }_2}{2}{N}_4-c{N}_1-c{N}_2>0,\hfill \\ \hfill {\alpha }_5& =& N_5{\eta }_1-c{N}_1-c{N}_2>0.\hfill \end{array}$$

Thus, we obtain that

$$\begin{array}{lll}\hfill {\mathcal{P}}^{\prime }\left(t\right)& \le & -{\alpha }_2{\int }_0^1{\vartheta }_x^{2}dx-\frac{{\rho }_u}{2}{\int }_0^1{u}_t^{2}dx-{\alpha }_1{\int }_0^1{u}_x^{2}dx-[{\eta }_{0}N-c]{\int }_0^1{\vartheta }_t^{2}dx\hfill \\ & & -[k_{4}N-c]{\int }_0^1{\theta }_x^{2}dx-{\alpha }_3{\int }_0^1{(\kappa {\theta }_t+\theta )}^{2}dx-{\alpha }_4{\int }_0^1{(\kappa {\Re }_t+\Re )}^{2}dx\hfill \\ & & -{\alpha }_5{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,1,s,t\right)dsdx-\left[k_{2}N-c\right]{\int }_0^1{\Re }_x^{2}dx\hfill \\ & & -\left[k_{5}N-c\right]{\int }_0^1{\Re }^{2}dx-{\alpha }_6{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx,\hfill \end{array}$$

(58)

where ${\alpha }_6={\eta }_1{N}_5>0$.

Similarly, if we assume

$$\mathfrak{L}\left(t\right)=N_1{D}_1\left(t\right)+N_2{D}_2\left(t\right)+D_3\left(t\right)+N_4{D}_4\left(t\right)+N_5{D}_5\left(t\right),$$

then

$$\begin{array}{lll}\hfill \left|\mathfrak{L}\left(t\right)\right|& \le & N_1{\rho }_{\vartheta }{\int }_0^1\left|\vartheta {\vartheta }_t\right|dx+N_1\frac{a_2}{a_1}{\rho }_u{\int }_0^1\left|\vartheta u_t\right|dx+N_1\frac{{\varpi }_1}{2}{\int }_0^1{\vartheta }^{2}dx\hfill \\ & & +N_2{a}_2{\int }_0^1\left|\vartheta u_t-u{\vartheta }_t\right|dx+{\rho }_u{\int }_0^1\left|u_{t}u\right|dx+N_4{\beta }_1{\kappa }^2{\int }_0^1\left|{\theta }_t\theta \right|dx\hfill \\ & & +N_4\frac{{\beta }_1\kappa }{2}{\int }_0^1{\theta }^{2}dx+N_4{\beta }_2{\kappa }^2{\int }_0^1|{\Re }_t\Re |dx+N_4\frac{{\beta }_2\kappa }{2}{\int }_0^1{\Re }^{2}dx\hfill \\ & & +N_5{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s{e}^{-s\rho }\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho dx.\hfill \end{array}$$

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

$$\begin{array}{lll}\hfill \left|\mathfrak{L}\left(t\right)\right|& \le & c{\int }_0^1\left(u_t^2+{\vartheta }_t^2+{\vartheta }_x^2+u_x^2+{(\kappa {\theta }_t+\theta )}^2+{\theta }_x^2\right)dx\hfill \\ & & +c{\int }_0^1\left({\Re }^2+{\Re }_x^2+{(\kappa {\Re }_t+\Re )}^2\right)dx\hfill \\ & & +c{\int }_0^1{\int }_0^1{\int }_{{\nu }_1}^{{\nu }_2}s\left|{\varpi }_2\left(s\right)\right|{\mathcal{Y}}^2\left(x,\rho ,s,t\right)dsd\rho .\hfill \end{array}$$

On the other hand, by (32), we have

$$\begin{array}{lll}\hfill a_1{u}_x^2+2a_2{\vartheta }_x{u}_x+a_3{\vartheta }_x^2& >& \frac{1}{2}\left[(a_1-\frac{a_2^2}{a_3})u_x^2+(a_3-\frac{a_2^2}{a_1}){\vartheta }_x^2\right].\hfill \end{array}$$

Similarly, by $(k{k}_3>k_1^2)$, we have

$$\begin{array}{lll}\hfill k{\theta }_x^2+2k_1\Re {\theta }_x+k_3{\Re }^2& >& \frac{1}{2}\left[(k-\frac{k_1^2}{k_3}){\theta }_x^2+(k_3-\frac{k_1^2}{k}){\Re }^2\right].\hfill \end{array}$$

Hence, we obtain

$$\left|\mathfrak{L}\left(t\right)\right|=\left|\mathcal{P}\left(t\right)-NE\left(t\right)\right|\le cE\left(t\right),$$

that is,

$$\left(N-c\right)E\left(t\right)\le \mathcal{P}\left(t\right)\le \left(N+c\right)E\left(t\right).$$

(59)

At this point, we choose N large enough, such that

$$N-c>0,N{\eta }_0-c>0,N{k}_4-c>0>0,N{k}_2-c>0,N{k}_5-c>0>0.$$

Simplification of (38), (58) and (59) leads us to

$$c_{1}E\left(t\right)\le \mathcal{P}\left(t\right)\le c_{2}E\left(t\right),\forall t\ge 0,$$

(60)

and

$${\mathcal{P}}^{\prime }\left(t\right)\le -d_{1}E\left(t\right),\forall t\ge 0,$$

(61)

for some $d_1,c_1,c_2>0.$

Consequently, for some ${\zeta }_2>0$, we find

$${\mathcal{P}}^{\prime }\left(t\right)\le -{\zeta }_2\mathcal{P}\left(t\right),\forall t\ge 0.$$

(62)

From further simplification of (62), we have the following:

$$\mathcal{P}\left(t\right)\le \mathcal{P}\left(0\right)e^{-{\zeta }_{2}t},\forall t\ge 0.$$

(63)

Hence, (56) is achieved by (60) and (63). □

4. Conclusions

This work studies a swelling porous elastic system coupled with thermo-elasticity of the Lord Shulman type, microtemperature and distributed delay, an approach which is more general than classical thermo-elasticity. Furthermore, the problem circumvents 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 (5), (6) and (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 future work, we will investigate the system with some damping and source terms.