Well-Posedness and Exponential Stability of Swelling Porous with Gurtin–Pipkin Thermoelasticity

Abstract

The focus of this work is to investigate the well-posedness and exponential stability of a swelling porous system with the Gurtin–Pipkin thermal effect as the only source of damping. The well-posedness result is achieved using an essential corollary to the Lumer–Phillips Theorem. By constructing a suitable Lyapunov functional, we establish an exponential stability result without the conventional limitation to the system’s parameters (coined a stability number in the literature). Generally, the study demonstrates that the unique dissipation from the Gurtin–Pipkin thermal law is sufficient to stabilize the system exponentially, irrespective of the system’s parameters.

1. Introduction

The stabilization of swelling porous elastic systems has recently stimulated much interest from scientists and researchers, and many results have been duly published. Soil mechanics (geomechanics) are widely used to describe soil’s swelling properties when the water content of the soil changes, which might significantly affect infrastructures and the environment. It is, therefore, essential to look for means to subdue the potential disaster. Moreover, civil engineers and architects should characterize soils before embarking on any construction on or closer to the soils to prevent future disasters to life and properties. The resources [1,2,3,4] give detailed analyses of swelling soils and other related phenomena.

Our objective in the current research work focuses on the investigation of the asymptotic stability of a swelling porous system (see [5,6]) with a weak single source of dissipation emanates from the Gurtin–Pipkin heat conduction [7]. Precisely, for ${\rho }_1,{\rho }_2,{\rho }_3,a_1,a_3,\delta >0,a_2\ne 0$, and $a_1{a}_3>a_2^2,$ we examine the following swelling porous thermoelastic model

$$\left\{\begin{array}{c}\begin{array}{c}{\rho }_1{Z}_{tt}-a_1{Z}_{xx}-a_2{U}_{xx}=0,\hfill \\ {\rho }_2{U}_{tt}-a_3{U}_{xx}-a_2{Z}_{xx}-\delta {\theta }_x=0,\hfill \\ {\rho }_3{\theta }_t+q_x-\delta U_{xt}=0,\hfill \\ Z(x,0)=Z_0\left(x\right),Z_t(x,0)=Z_1\left(x\right),U(x,0)=U_0\left(x\right),\hfill \\ U_t(x,0)=U_1\left(x\right),\theta (x,-t)={\theta }_0(x,t),\hfill \\ Z_x(0,t)=Z_x(1,t)=U_x(0,t)=U_x(1,t)=\theta (0,t)=\theta (1,t)=0,\hfill \end{array}\hfill \end{array}\right.$$

(1)

where the unknown variables $Z,U,\theta :(x,t)\in (0,1)\times [0,\infty )\to \mathbb{R}$ represent the displacement of fluid, the elastic solid material, and the temperature, respectively. The initial conditions $Z_0,Z_1,U_0,U_1,$ and ${\theta }_0$ are fixed data. The heat flux q according to the Gurtin–Pipkin heat conduction law is given as

$$\begin{array}{c}\hfill q\left(t\right)=-{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }\mu \left(s\right){\theta }_x(x,t-s)ds,\end{array}$$

(2)

where $\tau$ is a positive constant called the relaxation time and $\mu :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ is the memory kernel whose detailed properties will be enumerated in the forthcoming section. A similar problem exists in the literature for some other systems. For the Timoshenko system, we mention the system

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\rho }_1{\phi }_{tt}-\kappa {({\phi }_x+\varphi )}_x=0,\hfill \\ {\rho }_2{\varphi }_{tt}-b{\varphi }_{xx}+\kappa ({\phi }_x+\varphi )+\delta {\theta }_x=0,\hfill \\ {\displaystyle {\rho }_3{\theta }_t-{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }\mu \left(s\right){\theta }_{xx}(x,t-s)ds+\delta {\varphi }_{xt}=0,}\hfill \end{array}\right.\end{array}$$

(3)

where $\phi ,\varphi ,\theta$ denote the transversal displacement of the beam from its equilibrium, the rotational angle of its filament, and the temperature difference, respectively. The coefficients ${\rho }_1,{\rho }_2,{\rho }_3,\kappa ,\delta ,\tau$ are positive constants. Dell’Oro and Pata [8] considered system (3) and proved an exponential stability result provided the stability number

$$\begin{array}{c}\hfill \left({\displaystyle \frac{{\rho }_1}{\kappa {\rho }_3}}-{\displaystyle \frac{\tau }{\mu \left(0\right)}}\right)\left({\displaystyle \frac{{\rho }_1}{\kappa }}-{\displaystyle \frac{{\rho }_2}{b}}\right)-{\displaystyle \frac{\tau }{\mu \left(0\right)}}{\displaystyle \frac{{\rho }_1{\delta }^2}{b\kappa {\rho }_3}}\end{array}$$

equals zero. When system (3) is fully damped, Fareh [9] proved that the stability number played no role in the exponential stability of the system. Readers are directed to [10] for the Cauchy version of the system (3). In another version, Hanni et al. [11] considered the Gurtin–Pipkin Timoshenko system where heat conduction is acting on the shear force equation

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\rho }_1{\phi }_{tt}-\kappa {({\phi }_x+\varphi )}_x+\delta {\theta }_x=0,\hfill \\ {\rho }_2{\varphi }_{tt}-b{\varphi }_{xx}+\kappa ({\phi }_x+\varphi )-\delta \theta =0,\hfill \\ {\displaystyle {\rho }_3{\theta }_t-{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }\mu \left(s\right){\theta }_{xx}(x,t-s)ds+\delta {({\phi }_x+\varphi )}_t=0}\hfill \end{array}\right.\end{array}$$

and obtained an exponential and a polynomial stability results subject to the behavior of the memory kennel $\mu$ and a stability number $\chi$ given by

$$\begin{array}{c}\hfill \chi =\left({\rho }_2-{\displaystyle \frac{b{\rho }_1}{\kappa }}\right)\left(b-{\displaystyle \frac{\mu \left(0\right){\rho }_2}{\tau {\rho }_3}}\right)+{\displaystyle \frac{b{\rho }_2{\delta }^2}{\kappa {\rho }_3}}.\end{array}$$

Precisely, setting

$$g\left(s\right)=-{\mu }^{\prime }\left(s\right)$$

with a differential inequality

$$g^{\prime }\left(s\right)\le -\nu g^p\left(s\right),\nu >0,p\in [1,1.5),$$

where $g^{\prime }\left(s\right)={\displaystyle \frac{dg\left(s\right)}{ds}}$ and ${\mu }^{\prime }\left(s\right)={\displaystyle \frac{d\mu \left(s\right)}{ds}}$, they proved an exponential stability result when

$$\chi =0\mathrm{and}p=1.$$

Otherwise, that is,

$$\chi \ne 0\mathrm{and}p\in (1,1.5),$$

they showed that the system is polynomially influenced.

For the Bresse Gurtin–Pipkin model, we mentioned the work of Dell’Oro [12]. The author analyzed the design

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\rho }_1{\phi }_{tt}-\kappa {({\phi }_x+\varphi +l\psi )}_x-lk({\psi }_x-l\phi )+l\delta \vartheta =0,\hfill \\ {\rho }_2{\varphi }_{tt}-b{\varphi }_{xx}+\kappa ({\phi }_x+\varphi +l\psi )+\delta {\theta }_x=0,\hfill \\ {\rho }_1{\psi }_{tt}+l\kappa ({\phi }_x+\varphi +l\psi )-k{({\psi }_x-l\phi )}_x+\delta {\vartheta }_x=0,\hfill \\ {\displaystyle {\rho }_3{\theta }_t-{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }\mu \left(s\right){\theta }_{xx}(x,t-s)ds+\delta {\varphi }_{xt}=0,}\hfill \\ {\displaystyle {\rho }_3{\vartheta }_t-{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }h\left(s\right){\vartheta }_{xx}(x,t-s)ds+\delta {({\psi }_x-l\phi )}_t=0,}\hfill \end{array}\right.\end{array}$$

(4)

where $\phi ,\varphi ,\psi ,\theta ,\vartheta$ represent the vertical displacement, the rotation angle, the horizontal displacement, the temperature along the vertical direction, and the temperature along the horizontal direction, respectively. All the constants ${\rho }_1,{\rho }_2,{\rho }_3,\kappa ,l,k,\delta ,b,\tau$ are strictly positive. In the stability analysis, the author introduced two stability numbers

$$\begin{array}{c}\hfill {\chi }_{\mu }=\left({\displaystyle \frac{\tau {\rho }_3}{\mu \left(0\right)}}-{\displaystyle \frac{{\rho }_1}{\kappa }}\right)\left(b-{\displaystyle \frac{\kappa {\rho }_2}{{\rho }_1}}\right)+{\displaystyle \frac{\tau {\delta }^2}{\mu \left(0\right)}},{\chi }_h=\left({\displaystyle \frac{\tau {\rho }_3}{h\left(0\right)}}-{\displaystyle \frac{{\rho }_1}{\kappa }}\right)\left(k-\kappa \right)+{\displaystyle \frac{\tau {\delta }^2}{h\left(0\right)}}\end{array}$$

and demonstrated that the semigroup associated with model (4) is exponentially stable if and only if ${\chi }_{\mu }{\chi }_h=0.$ However, when ${\chi }_{\mu }{\chi }_h\ne 0$, the author affirmed that the semigroup is polynomially (optimal) stable. Previously, Dell’Oro [13] showed that in the absence of the temperature along the horizontal direction (i.e., neglecting the forth equation in (4)), the model is exponentially stable if and only if

$${\chi }_{\mu }=0\mathrm{and}\kappa =k.$$

Readers are directed to [14] for the Cauchy version of the system (4) in the absence of the temperature along the horizontal direction. For other models with Gurtin–Pipkin composition, we refer the readers to [15,16,17,18,19,20,21] and the related works in the references therein.

The Gurtin–Pipkin thermal law given by (2) makes system (1) entirely hyperbolic, which is very interesting because it solves the problem of infinite speed propagation of thermal signals of the parabolic system. The well-known Maxwell–Cattaneo’s and classical Fourier laws are exceptional cases of the Gurtin–Pipkin thermal law. Illustratively, simple calculations show that when $\mu$ in (2) is replaced with

$$\begin{array}{c}\hfill \mu \left(t\right)=e^{-\frac{\beta }{\tau }t},\beta ,\tau >0,\end{array}$$

we have

$$\begin{array}{c}\hfill \tau q_t+\beta q+{\theta }_x=0,\end{array}$$

(5)

which is the well-known Maxwell–Cattaneo thermal law. Similarly, setting $\mu \left(t\right)=\tau \beta \delta \left(t\right)$, where $\delta$ is a Dirac Delta function, and using the filtering property of a Delta function, Equation (2) gives

$$\begin{array}{c}\hfill q=-\beta {\theta }_x,\end{array}$$

(6)

which is the classical Fourier law and $\beta$ is the thermal conductivity. Apalara et al. [22] considered (1) with the thermal law given by (5) and stabilized the resulting system exponentially without the stability number restrictions. Likewise, Apalara et al. [23] investigated (1) with the classical Fourier law (6) and obtained exponential stability results regardless of the system’s wave velocities. Concerning other forms of damping mechanisms, we refer the readers to the following references (which are by no means exhaustive) [24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43].

In this work, we investigate the asymptotic stability of system (1) when the heat flux is given by (2). Putting Equation (2) into (1), we get

$$\left\{\begin{array}{c}\begin{array}{c}{\rho }_1{Z}_{tt}-a_1{Z}_{xx}-a_2{U}_{xx}=0,\hfill \\ {\rho }_2{U}_{tt}-a_3{U}_{xx}-a_2{Z}_{xx}-\delta {\theta }_x=0,\hfill \\ {\rho }_3{\theta }_t-{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }\mu \left(s\right){\theta }_{xx}(x,t-s)ds-\delta U_{xt}=0,\hfill \\ Z(x,0)=Z_0\left(x\right),Z_t(x,0)=Z_1\left(x\right),U(x,0)=U_0\left(x\right),U_t(x,0)=U_1\left(x\right),\hfill \\ \theta (x,-t)={\theta }_0(x,t),\hfill \\ Z_x(0,t)=Z_x(1,t)=U_x(0,t)=U_x(1,t)=\theta (0,t)=\theta (1,t)=0.\hfill \end{array}\hfill \end{array}\right.$$

(7)

The preceding review of the work in the literature reveals that any system with Gurtin–Pupkin thermal law as the only source of damping needs a stability number to achieve exponential stability results. However, in the present work, we analyze system (7), a swelling porous system with the Gurtin–Pupkin thermal law, and establish an exponential stability result without commitment to any stability number.

The breakdown of the remaining sections of this work is as follows. Section 2 captures some necessary assumptions and transformations required to facilitate the validation of the well-posedness and the analysis of stability results. The well-posedness result is considered in Section 3. We dedicate Section 4 to the investigation of our exponential stability result. Lastly, we give a concluding remark in Section 5.

2. Assumptions and Transformations

This section focuses on the assumptions and transformations necessitated due to the nature of our problem. Starting with the memory kernel $\mu :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$, which is a $C^2\left({\mathbb{R}}_{+}\right)$ convex summable non-increasing function satisfying

• (P₀) $\mu \left(0\right)>0,\underset{s\to \infty }{\lim }\mu \left(s\right)=0,$ and ${\displaystyle {\int }_0^{\infty }\mu \left(s\right)ds=1.}$ Furthermore, there exists a positive real number r such that

  $$\begin{array}{c}\hfill -{\mu }^{″}\left(s\right)\le r{\mu }^{\prime }\left(s\right),\end{array}$$

  (8)

  where ${\mu }^{″}\left(s\right)={\displaystyle \frac{d^2}{d{s}^2}}\mu \left(s\right)$. Due to the nature of the integral, we set $g=-{\mu }^{\prime }$, and thus, the following condition ensues
• (P) ${\displaystyle g:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+},g\in C^1\left({\mathbb{R}}_{+}\right),g\left(0\right)>0,g_0={\int }_0^{\infty }g\left(s\right)ds=\mu \left(0\right)>0,}$ and ${\displaystyle {\int }_0^{\infty }sg\left(s\right)ds=1.}$ Furthermore, (8) implies

  $$\begin{array}{c}\hfill g^{\prime }\left(s\right)\le -rg\left(s\right).\end{array}$$

  (9)

Banking on the genuity of Dafermos [44], we introduce a variable $\vartheta :(0,1)\times {\mathbb{R}}_{+}\times {\mathbb{R}}_{+}\to \mathbb{R}$ defined as

$$\begin{array}{c}\hfill \vartheta (x,t,s)={\int }_{t-s}^t\theta (x,\sigma )d\sigma \end{array}$$

(10)

with initial datum ${\displaystyle \vartheta (x,0,s)={\int }_0^s{\theta }_0(x,\sigma )d\sigma .}$ Simple inspection shows that $\vartheta$ satisfies

$$\begin{array}{cc}\hfill & {\vartheta }_t(x,t,s)+{\vartheta }_s(x,t,s)=\theta (x,t)(0,1)\times {\mathbb{R}}_{+}\times {\mathbb{R}}_{+},\hfill \\ & \vartheta (0,t,s)=\vartheta (1,t,s)=\vartheta (x,t,0)=0(0,1)\times {\mathbb{R}}_{+}\times {\mathbb{R}}_{+},\hfill \end{array}$$

where the subscripts t and s signify the partial derivatives with respect to t and s, respectively. Direct computations show that

$$\begin{array}{cc}\hfill {\int }_0^{\infty }\mu \left(s\right){\theta }_{xx}(x,t-s)ds=& {\overline{)\underset{b\to \infty }{\lim }\mu \left(s\right){\int }_{t-s}^t{\theta }_{xx}(x,\sigma )d\sigma {|}_{s=0}^{s=b}}}^0-{\int }_0^{\infty }{\mu }^{\prime }\left(s\right){\int }_{t-s}^t{\theta }_{xx}(x,\sigma )d\sigma ds\hfill \\ \hfill & ={\int }_0^{\infty }g\left(s\right){\vartheta }_{xx}(x,t,s)ds.\hfill \end{array}$$

Lastly, due to the Neumann boundary conditions on Z and U, it is not evidence that Poincaré’s inequality can be applied on Z and U. Therefore, using the boundary conditions in (7)${}_6$, it is ascertained that (7)${}_1$ and (7)${}_2$ satisfy

$$\begin{array}{c}\hfill {\displaystyle \frac{d^2}{d{t}^2}}{\int }_0^{1}Z(x,t)dx=0\mathrm{and}{\displaystyle \frac{d^2}{d{t}^2}}{\int }_0^{1}U(x,t)dx=0,\end{array}$$

(11)

respectively. Solving (11) and using the intial condition (7)${}_4$, we obtain

$$\begin{array}{cc}\hfill & {\int }_0^{1}Z(x,t)dx={\int }_0^1{Z}_0\left(x\right)dx+t{\int }_0^1{Z}_1\left(x\right)dx\mathrm{and}\hfill \\ & {\int }_0^{1}U(x,t)dx={\int }_0^1{U}_0\left(x\right)dx+t{\int }_0^1{U}_1\left(x\right)dx.\hfill \end{array}$$

(12)

By setting

$$\begin{array}{c}\hfill z(x,t)=Z(x,t)-{\int }_0^{1}Z(x,t)dx\mathrm{and}u(x,t)=U(x,t)-{\int }_0^{1}U(x,t)dx,\end{array}$$

we get

$$\begin{array}{c}\hfill {\int }_0^{1}z(x,t)dx={\int }_0^{1}u(x,t)dx=0.\end{array}$$

(13)

Thus, the application of Poincaré’s inequality is appropriate. Gathering the transformations, system (7) becomes

$$\left\{\begin{array}{c}\begin{array}{c}{\rho }_1{z}_{tt}-a_1{z}_{xx}-a_2{u}_{xx}=0,\hfill \\ {\rho }_2{u}_{tt}-a_3{u}_{xx}-a_2{z}_{xx}-\delta {\theta }_x=0,\hfill \\ {\rho }_3{\theta }_t-{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }g\left(s\right){\vartheta }_{xx}(x,t,s)ds-\delta u_{xt}=0,\hfill \\ {\vartheta }_t(x,t,s)+{\vartheta }_s(x,t,s)=\theta (x,t),\hfill \\ z(x,0)=z_0\left(x\right),z_t(x,0)=z_1\left(x\right),u(x,0)=u_0\left(x\right),u_t(x,0)=u_1\left(x\right),\hfill \\ \theta (x,-t)={\theta }_0(x,t),{\vartheta }_0(x,s)={\int }_0^s{\theta }_0(x,\sigma )d\sigma ,\hfill \\ z_x(0,t)=z_x(1,t)=u_x(0,t)=u_x(1,t)=\theta (0,t)=\theta (1,t)=0,\hfill \\ \vartheta (0,t,s)=\vartheta (1,t,s)=\vartheta (x,t,0)=0.\hfill \end{array}\hfill \end{array}\right.$$

(14)

Moving forward, we consider system (14) and establish the exponential stability result, not minding any stability number or other relationship between the parameters of the system. From now onward, we only indicate the variables $x,t,$ and s when necessary to prevent the possibility of any ambiguity.

3. The Well-Posedness Result

This section employs the Lumer–Philips thoerem to verify the existence and uniqueness result of system (14). We start by equipping the phase space of system (14)

$$\mathcal{H}={\left(H_{*}^1\times L_{*}^2\right)}^2\times L_g$$

with the inner product

$$\begin{array}{ll}{(\mathcal{V},\tilde{\mathcal{V}})}_{\mathcal{H}}=& a_1{\int }_0^1{z}_x{\tilde{z}}_{x}dx+{\rho }_1{\int }_0^1\varphi \tilde{\varphi }dx+a_3{\int }_0^1{u}_x{\tilde{u}}_{x}dx+{\rho }_2{\int }_0^{1}v\tilde{v}dx+{\rho }_3{\int }_0^1\theta \tilde{\theta }dx\\ & +a_2{\int }_0^1\left(z_x{\tilde{u}}_x+u_x{\tilde{z}}_x\right)dx+{\displaystyle \frac{1}{\tau }}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\vartheta }_x{\tilde{\vartheta }}_{x}dsdx,\end{array}$$

(15)

where $\varphi =z_t,v=u_t,$ and

$$\{\begin{array}{c}\mathcal{V}={(z,\varphi ,u,v,\theta ,\vartheta )}^T,\tilde{\mathcal{V}}={(\tilde{z},\tilde{\varphi },\tilde{u},\tilde{v},\tilde{\theta },\tilde{\vartheta })}^T\in \mathcal{H},\hfill \\ L_{*}^2=\left\{{\displaystyle \vartheta \in L^2(0,1):{\int }_0^1\vartheta \left(x\right)dx=0}\right\},H_{*}^1=H^1(0,1)\cap L_{*}^2,\hfill \\ L_g=\left\{{\displaystyle \vartheta :{\mathbb{R}}_{+}\to H_0^1(0,1):{\Vert \vartheta \Vert }_{L_g}^2:={\int }_0^{\infty }g\left(s\right){\Vert {\vartheta }_x\left(s\right)\Vert }^{2}ds<\infty }\right\}.\hfill \end{array}$$

The $L_g$ is the Hilbert space of $H_0^1(0,1)$-valued functions on ${\mathbb{R}}_{+}$ equipped with the norm as defined. The representations $L^2(0,1)$ and $H^1(0,1)$ are the classical Sobolev spaces with the standard inner product. Next, we formulate system (14) in the first-order initial value problem:

$$\left\{\begin{array}{c}{\displaystyle \frac{d}{dt}}\mathcal{V}\left(t\right)=\mathcal{A}\mathcal{V}\left(t\right),t>0,\hfill \\ \mathcal{V}\left(0\right)={\mathcal{V}}_0={(z_0,z_1,u_0,u_1,{\theta }_0,{\vartheta }_0)}^T,\hfill \end{array}\right.$$

(16)

where the operator $\mathcal{A}:\mathcal{D}\left(\mathcal{A}\right)\subset \mathcal{A}⟶\mathcal{H}$ is defined by

$$\mathcal{A}:=\left(\begin{array}{cccccc}0& \mathrm{I}& 0& 0& 0& 0\\ a_1{\rho }_1^{-1}{\partial }_x^2& 0& a_2{\rho }_1^{-1}{\partial }_x^2& 0& 0& 0\\ 0& 0& 0& \mathrm{I}& 0& 0\\ a_2{\rho }_2^{-1}{\partial }_x^2& 0& a_3{\rho }_2^{-1}{\partial }_x^2& 0& \delta {\rho }_2^{-1}{\partial }_x& 0\\ 0& 0& 0& \delta {\rho }_3^{-1}{\partial }_x& 0& {\displaystyle {\left(\tau {\rho }_3\right)}^{-1}{\int }_0^{\infty }g\left(s\right){\partial }_x^2(·,s)ds}\\ 0& 0& 0& 0& \mathrm{I}& -{\partial }_s\end{array}\right)$$

(17)

with domain

$$\mathcal{D}\left(\mathcal{A}\right)=\left\{\begin{array}{cc}\mathcal{V}\in \mathcal{H}|z,u\in H^2(0,1),& \varphi ,v\in H_{\star }^1,\\ \theta \in H_0^1(0,1),\vartheta ,{\vartheta }_s\in L_g,& {\displaystyle {\int }_0^{\infty }g\left(s\right)\vartheta \left(s\right)ds\in H^2(0,1)}\end{array}\right\}$$

which is dense in $\mathcal{H}.$ We have the following well-posedness result:

Theorem 1. 

Assume that g satisfies (P), then for any ${\mathcal{V}}_0\in \mathcal{H}$, there exists a unique solution $\mathcal{V}\in C({\mathbb{R}}_{+},\mathcal{H})$ of problem (16). Furthermore, if ${\mathcal{V}}_0\in D\left(\mathcal{A}\right)$, then $\mathcal{V}\in C({\mathbb{R}}_{+},D\left(\mathcal{A}\right))\cap C^1({\mathbb{R}}_{+},\mathcal{H}).$

Proof. 

We begin by showing that $\mathcal{A}$ is dissipative in the space $\mathcal{H}$. For any $\mathcal{V}\in D\left(\mathcal{A}\right)$, we have

$$\begin{array}{cc}\hfill {(\mathcal{A}\mathcal{V},\mathcal{V})}_{\mathcal{H}}=-{\displaystyle \frac{1}{\tau }}& {\int }_0^{\infty }g\left(s\right)({\vartheta }_{xs}\left(s\right),{\vartheta }_x\left(s\right))ds\hfill \\ \hfill =-{\displaystyle \frac{1}{2\tau }}& {\int }_0^{\infty }g\left(s\right){\displaystyle \frac{\partial }{\partial s}}{\Vert {\vartheta }_x\left(s\right)\Vert }^{2}ds\hfill \\ \hfill =-{\displaystyle \frac{1}{2\tau }}& {\displaystyle \frac{\partial }{\partial s}}{\int }_0^{\infty }g\left(s\right)\Vert {\vartheta }_x{\left(s\right)\Vert }^{2}ds+{\displaystyle \frac{1}{2\tau }}{\int }_0^{\infty }{g}^{\prime }\left(s\right){\Vert {\vartheta }_x\left(s\right)\Vert }^{2}ds\hfill \\ \hfill =-{\displaystyle \frac{1}{2\tau }}& \underset{b\to \infty }{\lim }g\left(s\right)\Vert {\vartheta }_x{\left(s\right)\Vert }^2{|}_{s=0}^{s=b}+{\displaystyle \frac{1}{2\tau }}{\int }_0^{\infty }{g}^{\prime }\left(s\right){\Vert {\vartheta }_x\left(s\right)\Vert }^{2}ds.\hfill \end{array}$$

Using (10), it is clear that ${\vartheta }_x(x,t,0)=0$. Furthermore, by solving (9), we have

$$g\left(s\right)\le g\left(0\right)e^{-rs}⇝g\left(s\right)\Vert {\vartheta }_x{\left(s\right)\Vert }^2\le g\left(0\right)e^{-rs}{\Vert {\vartheta }_x\left(s\right)\Vert }^2$$

which gives $\underset{s\to \infty }{\lim }g\left(s\right){\Vert {\vartheta }_x\left(s\right)\Vert }^2=0.$ Consequently, we have

$$\begin{array}{c}\hfill {(\mathcal{A}\mathcal{V},\mathcal{V})}_{\mathcal{H}}={\displaystyle \frac{1}{2\tau }}{\int }_0^{\infty }{g}^{\prime }\left(s\right){\Vert {\vartheta }_x\left(s\right)\Vert }^{2}ds\le 0.\end{array}$$

Thus, the dissipativeness of $\mathcal{A}$ follows. Next, we want to show that for all

$$K={(k_1,k_2,k_3,k_4,k_5,k_6)}^T\in \mathcal{H},$$

there exists a unique $\mathcal{V}\in \mathcal{D}\left(\mathcal{A}\right)$ such that

$$\mathcal{A}\mathcal{V}=K.$$

(18)

Equation (18) implies

$$\begin{array}{cc}\hfill & \varphi =k_1\mathrm{in}{H}_{\star }^1\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill & a_1{z}_{xx}+a_2{u}_{xx}={\rho }_1{k}_2\mathrm{in}{L}_{\star }^2\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill & v=k_3\mathrm{in}{H}_{\star }^1\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & a_2{z}_{xx}+a_3{u}_{xx}+\delta {\theta }_x={\rho }_2{k}_4\mathrm{in}{L}_{\star }^2\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & \delta v_x+{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }g\left(s\right){\vartheta }_{xx}={\rho }_3{k}_5\mathrm{in}{L}^2(0,1)\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill & \theta -{\vartheta }_s=k_6\mathrm{in}{L}_g.\hfill \end{array}$$

(24)

From (19) and (21), we have

$$\varphi ,v\in H_{\star }^1.$$

By substituting (21) into (23), we get

$${\int }_0^{\infty }g\left(s\right){\vartheta }_{xx}ds=\delta \tau k_{3x}-\tau {\rho }_3{k}_5\in L^2(0,1).$$

Consequently, we have

$${\int }_0^{\infty }g\left(s\right)\vartheta \left(s\right)ds\in H^2(0,1).$$

It is obvious that

$$\vartheta \left(s\right)=s\theta -{\int }_0^s{k}_6\left(y\right)dy$$

(25)

satisfies (24) and the boundary condition $\vartheta \left(0\right)=0.$

Using (25) and the fact that ${\displaystyle {\int }_0^{\infty }sg\left(s\right)ds=1}$, it is not farfetched to see that

$$\theta ={\int }_0^{\infty }g\left(s\right)\vartheta \left(s\right)ds+{\int }_0^{\infty }g\left(s\right){\int }_0^s{k}_6\left(y\right)dyds.$$

(26)

Meanwhile, by applying Cauchy–Schwarz’s inequality and Fubini Theorem, for $0<r_1<r$, we get

$$\begin{array}{cc}\hfill {\int }_0^{\infty }g\left(s\right){∥{\int }_0^s{k}_{6x}\left(y\right)dy∥}^{2}ds=& {\int }_0^{\infty }g\left(s\right){∥{\int }_0^s{e}^{\frac{r_1}{2}y}{e}^{-\frac{r_1}{2}y}{k}_{6x}\left(y\right)dy∥}^{2}ds\hfill \\ \hfill \le & {\int }_0^{\infty }g\left(s\right)\left({\int }_0^s{e}^{r_{1}y}dy\right){\int }_0^s{e}^{-r_{1}y}{∥k_{6x}\left(y\right)∥}^{2}dyds\hfill \\ \hfill \le & {\displaystyle \frac{1}{r_1}}{\int }_0^{\infty }\underset{h\left(y\right)}{\underbrace{\left({\int }_y^{\infty }g\left(s\right)\left(e^{r_{1}s}-1\right)ds\right)}}{e}^{-r_{1}y}{∥k_{6x}\left(y\right)∥}^{2}dy.\hfill \end{array}$$

(27)

We define

$$\begin{array}{c}\hfill H\left(y\right)=:h\left(y\right)-{\displaystyle \frac{e^{r_{1}y}}{r-r_1}}g\left(y\right).\end{array}$$

It is evident that $\underset{y\to \infty }{\lim }h\left(y\right)=0.$ Using the fact that $g\left(y\right)\le g\left(0\right)e^{-ry}$ and bearing in mind that g is non-negative, we have $\underset{y\to \infty }{\lim }{\displaystyle \frac{e^{r_{1}y}}{r-r_1}}g\left(y\right)=0.$ Therefore, it follows that $\underset{y\to \infty }{\lim }H\left(y\right)=0.$ Furthermore, using (9), we see that

$$\begin{array}{cc}\hfill H^{\prime }\left(y\right)=& -g\left(y\right)e^{r_{1}y}+g\left(y\right)-{\displaystyle \frac{r_1{e}^{r_{1}y}}{r-r_1}}g\left(y\right)-{\displaystyle \frac{e^{r_{1}y}}{r-r_1}}{g}^{\prime }\left(y\right)\hfill \\ \hfill \ge & -g\left(y\right)e^{r_{1}y}+g\left(y\right)-{\displaystyle \frac{r_1{e}^{r_{1}y}}{r-r_1}}g\left(y\right)+{\displaystyle \frac{r{e}^{r_{1}y}}{r-r_1}}g\left(y\right)=g\left(y\right)\ge 0,\hfill \end{array}$$

which implies that H is non-decreasing, then $H\left(y\right)\le 0$, thus

$$\begin{array}{c}\hfill e^{-r_{1}y}h\left(y\right)=e^{-r_{1}y}{\int }_y^{\infty }g\left(s\right)\left(e^{r_{1}s}-1\right)ds\le {\displaystyle \frac{1}{r-r_1}}g\left(y\right).\end{array}$$

(28)

The substitution of (28) into (27) yields

$$\begin{array}{c}\hfill {\int }_0^{\infty }g\left(s\right){∥{\int }_0^s{k}_{6x}\left(y\right)dy∥}^{2}ds\le {\displaystyle \frac{1}{r(r_1-r)}}{\int }_0^{\infty }g\left(y\right){∥k_{6x}\left(y\right)∥}^{2}dy={\displaystyle \frac{1}{r(r_1-r)}}{∥k_6∥}_{L_g}^2<\infty \end{array}$$

(29)

which is true since $k_6\in L_g$. Then, from the definition, we have ${\displaystyle {\int }_0^s{k}_6\left(y\right)dy\in L_g}$. Now, using Cauchy–Schwarz’s inequality, we get

$$\begin{array}{cc}\hfill {∥{\int }_0^{\infty }g\left(s\right){\int }_0^s{k}_{6x}\left(y\right)dyds∥}^2\le & {\int }_0^{\infty }g\left(s\right)ds{\int }_0^{\infty }g\left(s\right)∥{\int }_0^s{k}_{6x}\left(y\right)dy∥\hfill \\ \hfill \le & g_0{\int }_0^{\infty }g\left(s\right)∥{\int }_0^s{k}_{6x}\left(y\right)dy∥<\infty .\hfill \end{array}$$

Therefore, we conclude that ${\displaystyle {\int }_0^{\infty }g\left(s\right){\int }_0^s{k}_6\left(y\right)dyds\in H_0^1(0,1)}$. Consequently, from (26) we have

$$\theta \in H_0^1(0,1).$$

On the other hand, we see that

$$\begin{array}{c}\hfill {\int }_0^{\infty }g\left(s\right){∥{\theta }_x∥}^{2}ds=g_0{∥{\theta }_x∥}^2<\infty \Rightarrow \theta \in L_g.\end{array}$$

Furthermore, using (9), integration by parts, the fact that ${\displaystyle {\int }_0^{\infty }sg\left(s\right)=1}$, and L’Hôpital’s rule, we get

$${\displaystyle {\int }_0^{\infty }{s}^{2}g\left(s\right)\le {\displaystyle \frac{2}{r}}<\infty .}$$

Therefore, we have

$$\begin{array}{c}\hfill {\int }_0^{\infty }g\left(s\right){∥s{\theta }_x∥}^{2}ds=\left({\int }_0^{\infty }{s}^{2}g\left(s\right)ds\right){∥{\theta }_x∥}^2<\infty \Rightarrow s\theta \in L_g.\end{array}$$

Consequently, from (24) and (26), we get

$$\vartheta ,{\vartheta }_s\in L_g.$$

By solving (20) and (22) simultaneously, we have

$$\begin{array}{cc}\hfill z_{xx}=& {\displaystyle \frac{1}{a_1{a}_3-a_2^2}}\left(a_3{\rho }_1{k}_2-a_2{\rho }_2{k}_4+a_2\delta {\theta }_x\right)\mathrm{in}{L}_{\star }^2\hfill \\ \hfill u_{xx}=& {\displaystyle \frac{1}{a_1{a}_3-a_2^2}}\left(a_1{\rho }_2{k}_4-a_2{\rho }_1{k}_2-a_1\delta {\theta }_x\right)\mathrm{in}{L}_{\star }^2.\hfill \end{array}$$

(30)

Adopting the standard theory for the linear elliptic equation yields

$$z,u\in H^2(0,1).$$

Moreover, it is explicit that

$$\begin{array}{cc}\hfill z_x=& {\displaystyle \frac{1}{a_1{a}_3-a_2^2}}\left(a_3{\rho }_1{\int }_0^x{k}_2\left(y\right)dy-a_2{\rho }_2{\int }_0^x{k}_4\left(y\right)dy+a_2\delta \theta \right)\hfill \\ \hfill u_x=& {\displaystyle \frac{1}{a_1{a}_3-a_2^2}}\left(a_1{\rho }_2{\int }_0^x{k}_4\left(y\right)dy-a_2{\rho }_1{\int }_0^x{k}_2\left(y\right)dy-a_1\delta \theta \right)\hfill \end{array}$$

(31)

satisfy (30). Since $k_2,k_4\in L_{\star }^2$ and $\theta \in H_0^1(0,1)$, it follows from (31) that u and z satisfy the boundary conditions, that is

$$z_x\left(0\right)=z_x\left(1\right)=u_x\left(0\right)=u_x\left(1\right)=0.$$

Thus, the unique solvability of (18) is guaranteed. Consequently, we have ${\Vert \mathcal{V}\Vert }_{\mathcal{H}}\le \mathcal{C}{\Vert K\Vert }_{\mathcal{H}},$ where the positive constant $\mathcal{C}$ is independent of $\mathcal{V}$, thus $0\in \rho \left(\mathcal{A}\right)$. Finally, since $\mathcal{A}$ is dissipative and $0\in \rho \left(\mathcal{A}\right)$, using Theorem 1.2.4 of [45] (a Collorary of Lumer–Phillips Theorem), we conclude that $\mathcal{A}$ is the infinitesimal generator of a $C_0$-semigroup of contractions on $\mathcal{H}$. Accordingly, the well-posedness of our problem (system (14)) captured in Theorem 1, follows.  ☐

4. Exponential Stability Result

The section shows that system (14) is exponentially stable. As usual, starting with the statements and proof of some technical lemmas, then concluding with the analysis of our exponential stability result. Quite interesting, the result is established irrespective of any stability number.

Lemma 1. 

Assume that g satisfies (P). Then, the energy functional $E,$ defined by

$$E\left(t\right)=\frac{1}{2}{\int }_0^1\left[{\rho }_1{z}_t^2+a_1{z}_x^2+{\rho }_2{u}_t^2+a_3{u}_x^2+2a_2{z}_x{u}_x+{\rho }_3{\theta }^2+{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}ds\right]dx,$$

(32)

satisfies

$$E^{\prime }\left(t\right)={\displaystyle \frac{1}{2\tau }}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx\le 0.$$

(33)

Proof. 

Direct multiplication of (14)${}_1$, (14)${}_2$, and (14)${}_3$ by $z_t,u_t$, and $\theta$, respectively, with some necessary computations, give

$$\begin{array}{c}\hfill \frac{1}{2}\frac{d}{dt}{\int }_0^1\left[{\rho }_1{z}_t^2+a_1{z}_x^2+{\rho }_2{u}_t^2+a_3{u}_x^2+{\rho }_3{\theta }^2+2a_2{z}_x{u}_x\right]dx+{\displaystyle \frac{1}{\tau }}{\int }_0^1{\theta }_x{\int }_0^{\infty }g\left(s\right){\vartheta }_{x}dsdx=0.\end{array}$$

(34)

Using (14)${}_4$, that is ${\vartheta }_t(x,t,s)+{\vartheta }_s(x,t,s)=\theta (x,t)$, simple calculation shows that

$$\begin{array}{cc}\hfill {\int }_0^1{\theta }_x{\int }_0^{\infty }g\left(s\right){\vartheta }_{x}dsdx=& {\int }_0^1{\int }_0^{\infty }g\left(s\right)\left({\vartheta }_{tx}+{\vartheta }_{sx}\right){\vartheta }_{x}dsdx\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\int }_0^{\infty }g\left(s\right)|{\vartheta }_x{|}^{2}dsdx+{\displaystyle \frac{1}{2}}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\displaystyle \frac{\partial }{\partial s}}{|{\vartheta }_x|}^{2}dsdx\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\int }_0^{\infty }g\left(s\right)|{\vartheta }_x{|}^{2}dsdx-{\displaystyle \frac{1}{2}}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx.\hfill \end{array}$$

(35)

Substituting (35) in (34), bearing in mind (32), yields (33).  ☐

Lemma 2. 

The derivative of the functional

$${\Delta }_1\left(t\right):={\displaystyle \frac{{\rho }_2}{a_2}}{\int }_0^1\left(u_{t}z-u{z}_t\right)dx,$$

satisfies

$$\begin{array}{cc}\hfill {\Delta }_1^{\prime }\left(t\right)\le -& {\displaystyle \frac{1}{2}}{\int }_0^1{z}_x^{2}dx+\left({\displaystyle \frac{1}{a_2^2}}{\left({\displaystyle \frac{a_1{\rho }_2}{{\rho }_1}}-a_3\right)}^2+{\displaystyle \frac{{\rho }_2}{{\rho }_1}}\right){\int }_0^1{u}_x^{2}dx+{\displaystyle \frac{{\delta }^2}{a_2^2}}{\int }_0^1{\theta }^{2}dx.\hfill \end{array}$$

(36)

Proof. 

Differentiating ${\Delta }_1$, then using (14)${}_1$ and (14)${}_2$, we obtain

$$\begin{array}{c}\hfill {\Delta }_1^{\prime }\left(t\right)=-{\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{{\rho }_2}{{\rho }_1}}{\int }_0^1{u}_x^{2}dx-{\displaystyle \frac{\delta }{a_2}}{\int }_0^1{z}_x\theta dx+{\displaystyle \frac{1}{a_2}}\left({\displaystyle \frac{a_1{\rho }_2}{{\rho }_1}}-a_3\right){\int }_0^1{z}_x{u}_{x}dx.\end{array}$$

(37)

Young’s inequality yields

$$\begin{array}{cc}\hfill -{\displaystyle \frac{\delta }{a_2}}{\int }_0^1{z}_x\theta dx\le & {\displaystyle \frac{1}{4}}{\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{{\delta }^2}{a_2^2}}{\int }_0^1{\theta }^{2}dx,\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{a_2}}\left({\displaystyle \frac{a_1{\rho }_2}{{\rho }_1}}-a_3\right){\int }_0^1{z}_x{u}_{x}dx\le & {\displaystyle \frac{1}{4}}{\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{1}{a_2^2}}{\left({\displaystyle \frac{a_1{\rho }_2}{{\rho }_1}}-a_3\right)}^2{\int }_0^1{u}_x^{2}dx.\hfill \end{array}$$

(39)

Direct substitution of (38) and (39) into (37) yields (36).  ☐

Lemma 3. 

For any ${\epsilon }_1>0$ small, the derivative of the functional

$${\Delta }_2\left(t\right):={\displaystyle \frac{1}{\gamma }}\left(a_1{\rho }_2{\int }_0^1{u}_{t}udx-a_2{\rho }_1{\int }_0^1{z}_{t}udx\right),$$

satisfies

$${\Delta }_2^{\prime }\left(t\right)\le -{\displaystyle \frac{1}{2}}{\int }_0^1{u}_x^{2}dx+{\epsilon }_1{\int }_0^1{z}_t^{2}dx+\left({\displaystyle \frac{a_1{\rho }_2}{\gamma }}+{\displaystyle \frac{a_2^2{\rho }_1^2}{4{\epsilon }_1{\gamma }^2}}\right){\int }_0^1{u}_t^{2}dx+{\displaystyle \frac{a_1^2{\delta }^2}{{\gamma }^2}}{\int }_0^1{\theta }^{2}dx,$$

(40)

where $\gamma =a_1{a}_3-a_2^2>0.$

Proof. 

Differentiating ${\Delta }_2$, then using (14)${}_1$ and (14)${}_2$, we obtain

$$\begin{array}{cc}\hfill {\Delta }_2^{\prime }\left(t\right)=-{\int }_0^1& u_x^{2}dx+{\displaystyle \frac{a_1{\rho }_2}{\gamma }}{\int }_0^1{u}_t^{2}dx-{\displaystyle \frac{a_2{\rho }_1}{\gamma }}{\int }_0^1{z}_t{u}_{t}dx-{\displaystyle \frac{a_1\delta }{\gamma }}{\int }_0^1{u}_x\theta dx.\hfill \end{array}$$

(41)

Adopting Young’s inequality ensures that, for any ${\epsilon }_1>0$,

$$\begin{array}{cc}\hfill -{\displaystyle \frac{a_2{\rho }_1}{\gamma }}{\int }_0^1{z}_t{u}_{t}dx\le & {\epsilon }_1{\int }_0^1{z}_t^{2}dx+{\displaystyle \frac{a_2^2{\rho }_1^2}{4{\epsilon }_1{\gamma }^2}}{\int }_0^1{u}_t^{2}dx,\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill -{\displaystyle \frac{a_1\delta }{\gamma }}{\int }_0^1{u}_x\theta dx& \le {\displaystyle \frac{1}{2}}{\int }_0^1{u}_x^{2}dx+{\displaystyle \frac{a_1^2{\delta }^2}{{\gamma }^2}}{\int }_0^1{\theta }^{2}dx.\hfill \end{array}$$

(43)

Substituting (42) and (43) into (41), we end up with estimate (40).  ☐

Lemma 4. 

Assume that g satisfies (P) and for any ${\epsilon }_2,{\epsilon }_3>0$ small, the derivative of the functional

$${\Delta }_3\left(t\right):={\displaystyle \frac{{\rho }_3}{\delta }}{\int }_0^1\theta {\int }_0^x{u}_t\left(\tau \right)d\tau dx,$$

satisfies

$$\begin{array}{cc}\hfill {\Delta }_3^{\prime }\left(t\right)\le -{\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^{2}dx& +{\epsilon }_2{\int }_0^1{u}_x^{2}dx+{\epsilon }_3{\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{{\rho }_3^2}{{\rho }_2^2}}\left({\displaystyle \frac{{\rho }_2}{{\rho }_3}}+{\displaystyle \frac{a_3^2}{4{\delta }^2{\epsilon }_2}}+{\displaystyle \frac{a_2^2}{4{\delta }^2{\epsilon }_3}}\right){\int }_0^1{\theta }^{2}dx\hfill \\ \hfill & +{\displaystyle \frac{g_0}{2{\delta }^2{\tau }^2}}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx.\hfill \end{array}$$

(44)

Proof. 

Differentiating ${\Delta }_3$, then using (14)${}_2$ and (14)${}_3$, we end up with

$$\begin{array}{cc}\hfill {\Delta }_3^{\prime }\left(t\right)=-& {\int }_0^1{u}_t^{2}dx+{\displaystyle \frac{{\rho }_3}{{\rho }_2}}{\int }_0^1{\theta }^{2}dx+{\displaystyle \frac{a_3{\rho }_3}{\delta {\rho }_2}}{\int }_0^1{u}_x\theta dx+{\displaystyle \frac{a_2{\rho }_3}{\delta {\rho }_2}}{\int }_0^1{z}_x\theta dx\hfill \\ \hfill & -{\displaystyle \frac{1}{\delta \tau }}{\int }_0^1{u}_t{\int }_0^{\infty }g\left(s\right){\vartheta }_{x}dsdx.\hfill \end{array}$$

(45)

Enforcing Young’s inequality, for any ${\epsilon }_2,{\epsilon }_3>0$, yields

$$\begin{array}{cc}\hfill {\displaystyle \frac{a_3{\rho }_3}{\delta {\rho }_2}}{\int }_0^1{u}_x\theta dx& \le {\epsilon }_2{\int }_0^1{u}_x^{2}dx+{\displaystyle \frac{a_3^2{\rho }_3^2}{4{\delta }^2{\rho }_2^2{\epsilon }_2}}{\int }_0^1{\theta }^{2}dx,\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill {\displaystyle \frac{a_2{\rho }_3}{\delta {\rho }_2}}{\int }_0^1{z}_x\theta dx& \le {\epsilon }_3{\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{a_2^2{\rho }_3^2}{4{\delta }^2{\rho }_2^2{\epsilon }_3}}{\int }_0^1{\theta }^{2}dx.\hfill \end{array}$$

(47)

For the subsequent estimations, we consecutively apply Young’s and Cauchy–Schwarz inequalities and recall that $g>0$ to achieve

$$\begin{array}{cc}\hfill -{\displaystyle \frac{1}{\delta \tau }}{\int }_0^1{u}_t{\int }_0^{\infty }g\left(s\right){\vartheta }_{x}dsdx& \le {\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^{2}dx+{\displaystyle \frac{1}{2{\delta }^2{\tau }^2}}{\int }_0^1{\left({\int }_0^{\infty }\sqrt{g\left(s\right)}\sqrt{g\left(s\right)}{\vartheta }_{x}ds\right)}^{2}dx\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^{2}dx+{\displaystyle \frac{1}{2{\delta }^2{\tau }^2}}\left({\int }_0^{\infty }g\left(s\right)ds\right){\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^{2}dx+{\displaystyle \frac{g_0}{2{\delta }^2{\tau }^2}}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx.\hfill \end{array}$$

(48)

By the virtue of estimates (46)–(48) in line with (45), we achieve (44).  ☐

Lemma 5. 

Assume that g satisfies (P) and for any ${\epsilon }_4>0$ small, the derivative of the functional

$${\Delta }_4\left(t\right):=-{\displaystyle \frac{1}{g_0}}{\int }_0^1\theta {\int }_0^{\infty }g\left(s\right)\vartheta \left(s\right)dsdx,$$

satisfies

$$\begin{array}{cc}\hfill {\Delta }_4^{\prime }\left(t\right)\le -{\displaystyle \frac{1}{2}}{\int }_0^1{\theta }^{2}dx& +{\epsilon }_4{\int }_0^1{u}_t^{2}dx+\left({\displaystyle \frac{1}{{\rho }_3\tau }}+{\displaystyle \frac{{\delta }^2}{4{\rho }_3^2{g}_0{\epsilon }_4}}\right){\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx\hfill \\ \hfill & -{\displaystyle \frac{g\left(0\right)}{2g_0^2}}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx.\hfill \end{array}$$

(49)

Proof. 

Similar to the previous lemmas and focusing on (14)${}_3$ and (14)${}_4$, we get

$$\begin{array}{cc}\hfill {\Delta }_4^{\prime }\left(t\right)=-& {\int }_0^1{\theta }^{2}dx+{\displaystyle \frac{1}{{\rho }_3{g}_0\tau }}{\int }_0^1{\left({\int }_0^{\infty }g\left(s\right){\vartheta }_{x}ds\right)}^{2}dx+{\displaystyle \frac{\delta }{{\rho }_3{g}_0}}{\int }_0^1{u}_t{\int }_0^{\infty }g\left(s\right){\vartheta }_{x}dsdx\hfill \\ \hfill & +{\displaystyle \frac{1}{g_0}}{\int }_0^1\theta {\int }_0^{\infty }g\left(s\right){\vartheta }_{s}dsdx.\hfill \end{array}$$

(50)

Utilizing Young’s and Cauchy–Schwarz inequalities, for any ${\epsilon }_4>0$, we achieve the following estimates

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{{\rho }_3{g}_0\tau }}{\int }_0^1{\left({\int }_0^{\infty }g\left(s\right){\vartheta }_{x}ds\right)}^{2}dx& \le {\displaystyle \frac{1}{{\rho }_3\tau }}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx,\hfill \end{array}$$

(51)

$$\begin{array}{cc}\hfill {\displaystyle \frac{\delta }{{\rho }_3{g}_0}}{\int }_0^1{u}_t{\int }_0^{\infty }g\left(s\right){\vartheta }_{x}dsdx& \le {\epsilon }_4{\int }_0^1{u}_t^{2}dx+{\displaystyle \frac{{\delta }^2}{4{\rho }_3^2{g}_0{\epsilon }_4}}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx,\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{g_0}}{\int }_0^1\theta {\int }_0^{\infty }g\left(s\right){\vartheta }_{s}dsdx& \le {\displaystyle \frac{1}{2}}{\int }_0^1{\theta }^{2}dx-{\displaystyle \frac{g\left(0\right)}{2g_0^2}}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx.\hfill \end{array}$$

(53)

We accomplish (49) courtesy of the substitution of estimates (51)–(53) into (50).  ☐

Lemma 6. 

The functional

$${\Delta }_5\left(t\right):=-{\int }_0^1{z}_{t}zdx$$

satisfies

$${\Delta }_5^{\prime }\left(t\right)\le -{\int }_0^1{z}_t^{2}dx+{\displaystyle \frac{1}{2{\rho }_1}}\left(2a_1+a_2\right){\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{a_2}{2{\rho }_1}}{\int }_0^1{u}_x^{2}dx.$$

(54)

Proof. 

Using (14)${}_1$ and Young’s inequality ensures that

$$\begin{array}{cc}\hfill {\Delta }_5^{\prime }\left(t\right)=& -{\int }_0^1{z}_t^{2}dx+{\displaystyle \frac{a_1}{{\rho }_1}}{\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{a_2}{{\rho }_1}}{\int }_0^1{z}_x{u}_{x}dx\hfill \\ \hfill \le & -{\int }_0^1{z}_t^{2}dx+{\displaystyle \frac{1}{2{\rho }_1}}\left(2a_1+a_2\right){\int }_0^1{z}_x^{2}dx+{\displaystyle \frac{a_2}{2{\rho }_1}}{\int }_0^1{u}_x^{2}dx,\hfill \end{array}$$

which concludes our goal in Lemma 6.  ☐

In the following lemma, we specify a Lyapunov functional and prove its equivalence with the energy functional. This is crucial to the analysis of our stability result.

Lemma 7. 

Assume that g satisfies (P) and let $\mathcal{K},{\mathcal{K}}_i,i=1\cdots 5,$ be positive constants. Then, the functional

$$\begin{array}{c}\hfill \mathcal{L}\left(t\right):=\mathcal{K}E\left(t\right)+\sum _{i=1}^5{\mathcal{K}}_i{\Delta }_i\left(t\right)\end{array}$$

(55)

satisfies, for some constants $a,b>0$, the equivalence

$$\begin{array}{c}\hfill aE\left(t\right)\le \mathcal{L}\left(t\right)\le bE\left(t\right)t\ge 0.\end{array}$$

(56)

Proof. 

Direct computations, bearing in mind the definition of ${\Delta }_i,i=1\cdots 5,$ yield

$$\begin{array}{cc}\hfill \left|\mathcal{L}\right(t)-\mathcal{K}& E\left(t\right)|\le {\displaystyle \frac{{\rho }_2}{|a_2|}}{\mathcal{K}}_1{\int }_0^1\left|u_{t}z-u{z}_t\right|dx+{\displaystyle \frac{{\mathcal{K}}_2}{\gamma }}\left(a_1{\rho }_2{\int }_0^1|u_{t}u|dx+|a_2|{\rho }_1{\int }_0^1\left|z_{t}u\right|dx\right)\hfill \\ \hfill +& {\displaystyle \frac{{\rho }_3}{\delta }}{\mathcal{K}}_3{\int }_0^1\left|\theta {\int }_0^x{u}_t\left(\tau \right)d\tau \right|dx+{\displaystyle \frac{{\mathcal{K}}_4}{g_0}}{\int }_0^1\left|\theta {\int }_0^{\infty }g\left(s\right)\vartheta \left(s\right)ds\right|dx+{\mathcal{K}}_5{\int }_0^1\left|z_{t}z\right|dx.\hfill \end{array}$$

By Young’s inequality and the Cauchy–Schwarz inequality, we obtain

$$\begin{array}{cc}\hfill \left|\mathcal{L}\right(t)-\mathcal{K}E(t\left)\right|\le & \left({\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2a_2^2}}+{\displaystyle \frac{a_1^2{\rho }_2^2{\mathcal{K}}_2}{2\gamma {\rho }_1}}+{\displaystyle \frac{{\rho }_3{\mathcal{K}}_3}{2\delta }}\right){\int }_0^1{u}_t^{2}dx+\left({\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2}}+{\displaystyle \frac{{\rho }_1{\mathcal{K}}_2}{\gamma }}\right){\int }_0^1{u}^{2}dx\hfill \\ \hfill & +\left({\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2a_2^2}}+{\displaystyle \frac{a_2^2{\rho }_1{\mathcal{K}}_2}{2\gamma }}+{\displaystyle \frac{{\mathcal{K}}_5}{2}}\right){\int }_0^1{z}_t^{2}dx+\left({\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2}}+{\displaystyle \frac{{\mathcal{K}}_5}{2}}\right){\int }_0^1{z}^{2}dx\hfill \\ \hfill & +\left({\displaystyle \frac{{\rho }_3{\mathcal{K}}_3}{2\delta }}+{\displaystyle \frac{{\mathcal{K}}_4}{2}}\right){\int }_0^1{\theta }^{2}dx+{\displaystyle \frac{{\mathcal{K}}_4}{2g_0}}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx.\hfill \end{array}$$

Using the Poincaré inequality, it follows that

$$\begin{array}{c}\hfill |\mathcal{L}\left(t\right)-\mathcal{K}E\left(t\right)|\le {\tau }_0{\int }_0^1\left[u_t^2+z_t^2+u_x^2+z_x^2+{\theta }^2+{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}ds\right]dx,\end{array}$$

(57)

where

$$\begin{array}{cc}\hfill {\tau }_0=\max \{{\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2a_2^2}}& +{\displaystyle \frac{a_1^2{\rho }_2^2{\mathcal{K}}_2}{2\gamma {\rho }_1}}+{\displaystyle \frac{{\rho }_3{\mathcal{K}}_3}{2\delta }},{\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2a_2^2}}+{\displaystyle \frac{a_2^2{\rho }_1{\mathcal{K}}_2}{2\gamma }}+{\displaystyle \frac{{\mathcal{K}}_5}{2}},c_p\left({\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2}}+{\displaystyle \frac{{\rho }_1{\mathcal{K}}_2}{\gamma }}\right),\hfill \\ \hfill & c_p\left({\displaystyle \frac{{\rho }_2{\mathcal{K}}_1}{2}}+{\displaystyle \frac{{\mathcal{K}}_5}{2}}\right),{\displaystyle \frac{{\rho }_3{\mathcal{K}}_3}{2\delta }}+{\displaystyle \frac{{\mathcal{K}}_4}{2}},{\displaystyle \frac{{\mathcal{K}}_4}{2g_0}}\}>0\hfill \end{array}$$

and $c_p$ is a Poincaré constant. On the other hand, the energy functional given by (32) can be manipulated as

$$\begin{array}{cc}\hfill E\left(t\right)=\frac{1}{2}& {\int }_0^1[{\rho }_1{z}_t^2+{\rho }_2{u}_t^2+{\displaystyle \frac{a_1}{2}}{\left(z_x+{\displaystyle \frac{a_2}{a_1}}{u}_x\right)}^2+{\displaystyle \frac{a_3}{2}}{\left(u_x+{\displaystyle \frac{a_2}{a_3}}{z}_x\right)}^2\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left(a_1-{\displaystyle \frac{a_2^2}{a_3}}\right)z_x^2+{\displaystyle \frac{1}{2}}\left(a_3-{\displaystyle \frac{a_2^2}{a_1}}\right)u_x^2+{\rho }_3{\theta }^2+{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}ds]dx\hfill \\ \hfill \ge \frac{1}{2}& {\int }_0^1[{\rho }_1{z}_t^2+{\rho }_2{u}_t^2+{\displaystyle \frac{1}{2}}\left(a_1-{\displaystyle \frac{a_2^2}{a_3}}\right)z_x^2+{\displaystyle \frac{1}{2}}\left(a_3-{\displaystyle \frac{a_2^2}{a_1}}\right)u_x^2+{\rho }_3{\theta }^2\hfill \\ \hfill & +{\displaystyle \frac{1}{\tau }}{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}ds]dx.\hfill \end{array}$$

(58)

Recalling that $a_1{a}_3>a_2^2,$ then inequality (58) yields

$$\begin{array}{cc}\hfill E\left(t\right)\ge & {\tau }_1{\int }_0^1\left[u_t^2+z_t^2+u_x^2+z_x^2+{\theta }^2+{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}ds\right]dx,\hfill \end{array}$$

(59)

where

$$\begin{array}{c}\hfill {\tau }_1=\min {\displaystyle \frac{1}{2}}\left\{{\rho }_1,{\rho }_2,{\rho }_3,{\displaystyle \frac{1}{2}}\left(a_1-{\displaystyle \frac{a_2^2}{a_3}}\right),{\displaystyle \frac{1}{2}}\left(a_3-{\displaystyle \frac{a_2^2}{a_1}}\right),{\displaystyle \frac{1}{\tau }}\right\}>0.\end{array}$$

The merging of (57) and (59) induces

$$\begin{array}{c}\hfill |\mathcal{L}\left(t\right)-\mathcal{K}E\left(t\right)|\le {\displaystyle \frac{{\tau }_0}{{\tau }_1}}E\left(t\right)\Rightarrow \left[\mathcal{K}-{\displaystyle \frac{{\tau }_0}{{\tau }_1}}\right]E\left(t\right)\le \mathcal{L}\left(t\right)\le \left[\mathcal{K}+{\displaystyle \frac{{\tau }_0}{{\tau }_1}}\right]E\left(t\right).\end{array}$$

(60)

Taking $\mathcal{K}=a+{\displaystyle \frac{{\tau }_0}{{\tau }_1}}$ yields (56), where $a>0$ and $b=a+{\displaystyle \frac{2{\tau }_0}{{\tau }_1}}$.  ☐

Having concluded the proof of all the necessary technical lemmas, our focus is now shifted to the stability result.

Theorem 2. 

Assume that g satisfies (P). Then, system (14) is exponentially stable. In other words, there exist two positive constants ${\kappa }_1$ and ${\kappa }_2$ such that the energy designated by (32) satisfies

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

(61)

Proof. 

Differentiating (55) and administering (33), (36), (40), (44), (49), (54), and setting ${\mathcal{K}}_5=2,{\epsilon }_1={\displaystyle \frac{1}{{\mathcal{K}}_2}},{\epsilon }_2={\displaystyle \frac{{\mathcal{K}}_2}{2{\mathcal{K}}_3}},{\epsilon }_3={\displaystyle \frac{{\mathcal{K}}_1}{2{\mathcal{K}}_3}},{\epsilon }_4={\displaystyle \frac{{\mathcal{K}}_3}{2{\mathcal{K}}_4}}$, we get

$$\begin{array}{cc}\hfill {\mathcal{L}}^{\prime }\left(t\right)\le & \left[{\displaystyle \frac{\mathcal{K}}{2\tau }}-{\displaystyle \frac{g\left(0\right){\mathcal{K}}_4}{2g_0^2}}\right]{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx-\left[{\mathcal{K}}_1-{\displaystyle \frac{2a_1+a_2}{{\rho }_1}}\right]{\int }_0^1{z}_x^{2}dx\hfill \\ \hfill & -\left[{\mathcal{K}}_2-\left({\displaystyle \frac{1}{a_2^2}}{\left({\displaystyle \frac{a_1{\rho }_2}{{\rho }_1}}-a_3\right)}^2+{\displaystyle \frac{{\rho }_2}{{\rho }_1}}\right){\mathcal{K}}_1-{\displaystyle \frac{a_2}{{\rho }_1}}\right]{\int }_0^1{u}_x^{2}dx\hfill \\ \hfill & -\left[{\mathcal{K}}_3-\left({\displaystyle \frac{a_1{\rho }_2}{\gamma }}+{\displaystyle \frac{a_2^2{\rho }_1^2{\mathcal{K}}_2}{4{\gamma }^2}}\right){\mathcal{K}}_2\right]{\int }_0^1{u}_t^{2}dx\hfill \\ \hfill & -\left[{\displaystyle \frac{{\mathcal{K}}_4}{2}}-{\displaystyle \frac{{\delta }^2{\mathcal{K}}_1}{a_2^2}}-{\displaystyle \frac{a_1^2{\delta }^2{\mathcal{K}}_2}{{\gamma }^2}}-{\displaystyle \frac{{\rho }_3^2{\mathcal{K}}_3}{{\rho }_2^2}}\left({\displaystyle \frac{{\rho }_2}{{\rho }_3}}+{\displaystyle \frac{a_3^2{\mathcal{K}}_3}{2{\delta }^2{\mathcal{K}}_2}}+{\displaystyle \frac{a_2^2{\mathcal{K}}_3}{2{\delta }^2{\mathcal{K}}_1}}\right)\right]{\int }_0^1{\theta }^{2}dx\hfill \\ \hfill & -{\int }_0^1{z}_t^{2}dx+\left[{\displaystyle \frac{g_0{\mathcal{K}}_3}{2{\delta }^2{\tau }^2}}+\left({\displaystyle \frac{1}{{\rho }_3\tau }}+{\displaystyle \frac{{\delta }^2{\mathcal{K}}_4}{2{\rho }_3^2{g}_0{\mathcal{K}}_3}}\right){\mathcal{K}}_4\right]{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx.\hfill \end{array}$$

(62)

Now, we appropriately select the remaining constants. By letting

$${\mathcal{K}}_1:=1+{\displaystyle \frac{2a_1+a_2}{{\rho }_1}},$$

then

$${\mathcal{K}}_2:=1+\left({\displaystyle \frac{1}{a_2^2}}{\left({\displaystyle \frac{a_1{\rho }_2}{{\rho }_1}}-a_3\right)}^2+{\displaystyle \frac{{\rho }_2}{{\rho }_1}}\right){\mathcal{K}}_1+{\displaystyle \frac{a_2}{{\rho }_1}},$$

then

$${\mathcal{K}}_3:=1+\left({\displaystyle \frac{a_1{\rho }_2}{\gamma }}+{\displaystyle \frac{a_2^2{\rho }_1^2{\mathcal{K}}_2}{4{\gamma }^2}}\right){\mathcal{K}}_2,$$

then

$${\mathcal{K}}_4:=2+{\displaystyle \frac{2{\delta }^2{\mathcal{K}}_1}{a_2^2}}+{\displaystyle \frac{2a_1^2{\delta }^2{\mathcal{K}}_2}{{\gamma }^2}}+{\displaystyle \frac{{\rho }_3^2{\mathcal{K}}_3}{{\rho }_2^2}}\left({\displaystyle \frac{2{\rho }_2}{{\rho }_3}}+{\displaystyle \frac{a_3^2{\mathcal{K}}_3}{{\delta }^2{\mathcal{K}}_2}}+{\displaystyle \frac{a_2^2{\mathcal{K}}_3}{{\delta }^2{\mathcal{K}}_1}}\right)$$

and finally we revist $\mathcal{K}$ (if necessary) and set it $\mathcal{K}=k$, where k is defined as

$$\begin{array}{c}\hfill k=\max \left\{a+{\displaystyle \frac{{\tau }_0}{{\tau }_1}},2\tau \left(a+{\displaystyle \frac{g\left(0\right){\mathcal{K}}_4}{2g_0^2}}\right)\right\}.\end{array}$$

Ultimately, (62) turns out to be

$$\begin{array}{cc}\hfill {\mathcal{L}}^{\prime }\left(t\right)\le & -{\int }_0^1\left(z_t^2+z_x^2+u_t^2+u_x^2+{\theta }^2\right)dx+k{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx,\forall t\ge 0.\hfill \end{array}$$

(63)

Meanwhile, by revisiting the energy functional (32) and utilizing Young’s inequality, it turns out that

$$\begin{array}{cc}\hfill E\left(t\right)& =\frac{1}{2}{\int }_0^1\left[{\rho }_1{z}_t^2+a_1{z}_x^2+{\rho }_2{u}_t^2+a_3{u}_x^2+2a_2{z}_x{u}_x+{\rho }_3{\theta }^2\right]dx+{\displaystyle \frac{1}{2\tau }}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx\hfill \\ \hfill \le & \frac{1}{2}{\int }_0^1\left[{\rho }_1{z}_t^2+(a_1+a_2)z_x^2+{\rho }_2{u}_t^2+\left(a_2+a_3\right)u_x^2+{\rho }_3{\theta }^2\right]dx+{\displaystyle \frac{1}{2\tau }}{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx\hfill \\ \hfill \le & k_0{\int }_0^1\left[z_t^2+z_x^2+u_t^2+u_x^2+{\theta }^2\right]dx+k_0{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx,\hfill \end{array}$$

(64)

where $k_0:=\max \left\{{\rho }_1,{\rho }_2,{\rho }_3,(a_1+a_2),(a_3+a_2),{\displaystyle \frac{1}{2\tau }}\right\}.$

It is obvious from (63) and (64) that there exist some positive constants $k_1$ and $k_2$, such that

$${\mathcal{L}}^{\prime }\left(t\right)\le -k_{1}E\left(t\right)+k_2{\int }_0^1{\int }_0^{\infty }g\left(s\right){\left|{\vartheta }_x\right|}^{2}dsdx,\forall t\ge 0.$$

(65)

Multiplying (65) by r, using $\left(P\right)$ and (33), we end up with

$${\mathcal{Y}}^{\prime }\left(t\right)\le -r{k}_{1}E\left(t\right),\forall t\ge 0,$$

(66)

where $\mathcal{Y}\left(t\right)=r\mathcal{L}\left(t\right)+2\tau k_{2}E\left(t\right)$. Using (56), it readily follows, for some $a_0,b_0>0$, that

$$\begin{array}{c}\hfill a_{0}E\left(t\right)\le \mathcal{Y}\left(t\right)\le b_{0}E\left(t\right)t\ge 0.\end{array}$$

(67)

Consequently, inequality (66) becomes

$${\mathcal{Y}}^{\prime }\left(t\right)\le -{\kappa }_2\mathcal{Y}\left(t\right),\forall t\ge 0,$$

(68)

where ${\kappa }_2={\displaystyle \frac{r{k}_1}{b_0}}.$ A simple integration of (68) over $(0,t)$ induces

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

(69)

Accordingly, by merging (67) and (69), we get (61), which leads to the conclusion of our stability result.  ☐

5. Concluding Remarks

To our knowledge, many authors have investigated different systems with the Gurtin–Pipkin thermal law. The exponential stability obtained for those systems depends on some stability numbers, as demonstrated in the introduction. However, in this work, we analyze a swelling porous system with the Gurtin–Pipkin thermal law and establish an exponential stability result avoiding the imposition of any stability number. This astonishing result opens the further study of the model concerning Gurtin–Pipkin’s law when the temperature is acting on the displacement equation (the first equation in (1)). However, we believe that a similar exponential result will be obtained independently of any stability number. An additional exciting model is to consider (1) with Coleman–Gurtin’s law for the heat flux given by

$$\begin{array}{c}\hfill \tau q\left(t\right)+(1-\alpha ){\theta }_x+\alpha {\int }_0^{\infty }\mu \left(s\right){\theta }_x(x,t-s)ds=0,\alpha \in (0,1).\end{array}$$

where the Fourier, Maxwell–Cattaneo’s and the Gurtin–Pipkin’s laws are special cases.