On the Stabilization of a Memory-Type Porous Thermoelastic System

Abstract

In this work, we consider a one-dimensional porous thermoelastic system with memory effects. We prove a general decay result, for which exponential and polynomial decay results are special cases, depending only on the kernel of the memory effects. Our result is established irrespective of the wave speeds of the system. The result obtained is new and improves previous results in the literature.

1 Introduction

Elastic materials with voids have stimulated a lot on interest in recent years and many results have been published, most notably in the area of petroleum industry, material science, soil mechanics, foundation engineering, powder technology, and biology. It is widely known that an extension of the classical elasticity theory to porous media was established by Goodman and Cowin [1]. They introduced the concept of a continuum theory of granular materials with interstitial voids. In addition, Nunziato and Cowin [2] introduced the concept that the materials with voids possess a microstructure with the property that the mass at each point is obtained as the product of the mass density of the material matrix by the volume fraction. Later, Ieşan [3, 5], and Ieşan and Quintanilla [6] added the temperature as well as the microtemperature elements to the theory. For extensive discussion on these materials, we refer interested reader to [7]–[10] and the references therein.

In this work, we are concerned with the asymptotic behavior of the solution of porous thermoelastic system with memory effects

$$\begin{aligned}&\rho u_{tt}- \mu u_{xx}-b\phi _x+\beta \theta _{x}=0, \nonumber \\&J\phi _{tt}-\delta \phi _{xx}+ b u_{x}+\xi \phi -m\theta +\int _{0}^{t}g(t-s)\phi _{xx}(x, s)\mathrm{d}s=0, \nonumber \\&c\theta _{t}-\kappa \theta _{xx}+ \beta u_{tx}+m\phi _t=0, \nonumber \\&u(x,0)= u_0(x),\ u_{t}(x, 0)= u_1(x), \nonumber \\&\phi (x,0)=\phi _0(x),\ \phi _{t}(x, 0)=\phi _1(x),\ \theta (x, 0)=\theta _0(x), \nonumber \\&u_x(0,t)= u_x(1, t)=\phi (0,t)=\phi (1, t)=\theta (0,t)=\theta (1,t)=0, \end{aligned}$$

(1.1)

where $(x, t)\in (0, 1)\times [0, +\infty ), u$ is the longitudinal displacement, $\phi$ is the volume fraction of the solid elastic material, $\theta$ is the temperature difference, $u_0, u_1, \phi _0, \phi _1, \theta _0$ are given initial data, and $\rho , \mu , J, \delta , \xi , m, c, \kappa , \beta$ are constitutive constants which are positive. Furthermore, the constants $\mu$ and $\xi$ satisfy $\mu \xi > b^{2}$, where $b\ne 0$ is a real number. The integral represents the memory effect and g is the relaxation function satisfying the following:

• (H1) g: $\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}$ is a $C^{1}$ decreasing function satisfying

  $$\begin{aligned} g(0)> 0, \delta -\int _{0}^{\infty }g(s)\mathrm{d}s=l > 0. \end{aligned}$$

• (H2) There exists a nonincreasing differentiable function $\zeta : \mathbb {R}_{+}\rightarrow \mathbb {R}_{+}$ satisfying

  $$\begin{aligned} g^\prime (t) \le -\zeta (t) g(t),\ \qquad t\ge 0. \end{aligned}$$

The basic evolution together with the constitutive equations, for one-dimensional theories of porous materials, with memory effect is

$$\begin{aligned} \rho u_{tt}=T_x,\quad J\phi _{tt}=H_x+G,\quad \rho T_0\eta _t=q_x, \end{aligned}$$

(1.2)

and

$$\begin{aligned}&T=\mu u_x+b\phi -\beta \theta ,\quad H=\delta \phi _x-\int _{0}^{t}g(t-s)\phi _x \mathrm{d}s, \nonumber \\&\eta =c\theta +\beta u_x+m\phi ,\quad q=k\theta _x,\quad G=-bu_x-\xi \phi +m\theta , \end{aligned}$$

(1.3)

respectively. Here, $\eta$ is the entropy, T is the stress tensor, H is the equilibrated stress vector, G is the equilibrated body force, q is the heat flux vector, and $T_0$ is the absolute temperature in the reference configuration. By substituting (1.3) into (1.2), we obtain the first three equations in (1.1).

The asymptotic behavior of system (1.1) has been considered in the literature with various types of dissipative mechanisms. We first mention the case where the memory term in (1.1) is replaced with a porous dissipation. That is

$$\begin{aligned} \rho u_{tt}=\mu u_{xx}+b\phi _{x}-\beta \theta _x,&\qquad x\in (0, L),\; t> 0, \nonumber \\ J\phi _{tt}=\delta \phi _{xx}-b u_{x}-\xi \phi +m\theta -\tau \phi _{t},&\qquad x\in (0, L),\; t> 0, \nonumber \\ c\theta _{t}=\kappa \theta _{xx}-\beta u_{xt}-m\phi _{t},&\qquad x\in (0, L),\; t > 0. \end{aligned}$$

(1.4)

Casas and Quintanilla [11] considered (1.4) and used the semigroup theory together with the method developed by Liu and Zheng [12] to establish the exponential decay of the solutions. Whereas in absence of porous dissipation ($\tau =0$), the same authors showed in [13] that the heat effect alone is not strong enough to exponentially stabilize the system. However, the heat effect together with microtemperature produced an exponential decay result. Similarly, when $\tau =0$ and $\gamma u_{xxt}$ is added to the first equation in (1.4), Pamplona et al. [14] proved that the system lacks exponential stability, however, by taking some regular initial data, a polynomial stability is obtained. Also, for $\tau =0,$ Soufyane et al. [15] considered (1.4) with some boundary controls and obtained a general decay result, from which the usual exponential and polynomial decay rates are just special cases.

In the absence of the heat effect, (1.4) becomes

$$\begin{aligned} \rho u_{tt}=\mu u_{xx}+b\phi _{x},&\qquad x\in (0, L),\; t> 0, \nonumber \\ J\phi _{tt}=\delta \phi _{xx}-b u_{x}-\xi \phi -\tau \phi _{t},&\qquad x\in (0, L),\; t > 0. \end{aligned}$$

(1.5)

Quintanilla [16] considered (1.5) and proved that the porous dissipation is not strong enough to bring about an exponential decay. However, Apalara [17] considered the same system and proved that the system is exponentially stable provided the wave speeds of the two systems are equal. Equivalently, Apalara [18] replaced the porous dissipation in (1.5) with a memory term of the form $\displaystyle \int _{0}^{t}g(t-s)\phi _{xx}(x, s)\mathrm{d}s$ and obtained a general decay result depending on the kernel of the memory term and the wave speeds of the system. We refer reader to [19]–[23] and the references therein for more results.

Obviously, when $\mu =b=\xi$ and $m=\beta$ then (1.1) is equivalent to the following Timoshenko system

$$\begin{aligned}&\rho u_{tt}- \mu {\left( u+\phi _{x}\right) }_{x}+\beta \theta _{x}=0, \nonumber \\&J\phi _{tt}-\delta \phi _{xx}+ \mu \left( u_{x}+\phi \right) -\beta \theta +\int _0^{t}g(t-s)\phi _{xx}( x,s)\mathrm{d}s =0, \nonumber \\&\rho _{3}\theta _{t}-\kappa \theta _{xx}+\beta (u_{x}+\phi )_t=0. \end{aligned}$$

(1.6)

In the absence of memory term ($g=0$), Almeida Júnior et al [24] considered (1.6) and proved that the system is exponentially stable if and only if

$$\begin{aligned} \dfrac{\mu }{\rho }=\dfrac{\delta }{J} \end{aligned}$$

(1.7)

holds. Prior to the results in [24], Messaoudi and Fareh [25, 26] considered (1.6) with initial data and fully Dirichlet boundary conditions and established some general decay results depending on (1.7) and the kernel g of the memory term. In other words, the viscoelastic dissipation given by the memory term is not strong enough to neutralize the condition of equal wave speeds required to obtain an exponential decay result as established in [24]. However, Apalara [27] recently proved that the memory term together with the heat effect is strong enough to uniformly stabilize system (1.6) without imposing condition (1.7).

The main question which can be asked here is the following: Is the memory term together with the heat effect strong enough to exponentially stabilize system (1.1) irrespective of the wave speeds as in [27] for Timoshenko system? The aim of the present work is to give a positive answer to the question by considering (1.1) and establish a general stability result without imposing (1.7). Our result depends only on the kernel g of the memory term. Meanwhile, from (1.1)$_{1}$ and the boundary conditions, it follows that

$$\begin{aligned} \dfrac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\int _{0}^{1} u(x, t) \mathrm{d}x = 0 . \end{aligned}$$

(1.8)

So, by solving (1.8) and using the initial data of u,  we obtain

$$\begin{aligned} \int _{0}^{1} u(x, t)\mathrm{d}x=t\int _{0}^{1} u_{1}(x)\mathrm{d}x+\int _{0}^{1} u_{0}(x)\mathrm{d}x. \end{aligned}$$

Consequently, if we let

$$\begin{aligned} \overline{ u}(x, t)= u(x, t)-t\int _{0}^{1} u_{1}(x)\mathrm{d}x-\int _{0}^{1} u_{0}(x)\mathrm{d}x, \end{aligned}$$

we obtain

$$\begin{aligned} \int _{0}^{1}\overline{ u}(x, t)\mathrm{d}x=0, \quad \forall t\ge 0. \end{aligned}$$

Consequently, the use of Poincaré’s inequality for $\overline{ u}$ is justified. Furthermore, simple substitution shows that $(\overline{ u}, \phi , \theta )$ satisfies system (1.1) with initial data for $\overline{ u}$ given as

$$\begin{aligned} \overline{ u}_{0}(x)= u_{0}(x)-\int _{0}^{1} u_{0}(x)\mathrm{d}x\ \text { and }\ \overline{ u}_{1}(x)= u_{1}(x)-\int _{0}^{1} u_{1}(x)\mathrm{d}x. \end{aligned}$$

Henceforth, we work with $\overline{ u}$ instead of u but write u for simplicity of notation.

For the well-posedness result, we consider the following space

$$\begin{aligned} H_{*}^1(0,1)= H^1(0,1)\cap L_{*}^2(0,1), \text { where } L_{*}^2(0,1)=\displaystyle \left\{ u\in L^2(0,1) \ \vert \ \int _{0}^{1}u(x)\mathrm{d}x=0\right\} \end{aligned}$$

and state without proof the following result.

Proposition 1.1

Let $(u_0, \phi _0, \theta _0)\in H_{*}^1(0,1)\times \left( H_{0}^1(0,1)\right) ^2$ and $(u_1, \phi _1)\in \left( L^2(0,1)\right) ^2$ be given. Assume that (H1) and (H2) are satisfied, then problem (1.1) has a unique global solution $(u, \phi , \theta )$ which satisfies

$$\begin{aligned}&u\in C(\mathbb {R}_{+}, H_{*}^1(0,1))\cap C^1(\mathbb {R}_{+},L^2(0,1)),\ \phi \in C(\mathbb {R}_{+}, H_{0}^1(0,1))\\&\quad \cap C^1(\mathbb {R}_{+},L^2(0,1)),\theta \in C(\mathbb {R}_{+}, H_{0}^1(0,1)). \end{aligned}$$

Moreover, if $(u_0, \phi _0, \theta _0)\in H^2(0,1)\cap H_{*}^1(0,1)\times \left( H^2(0,1)\cap H_{0}^1(0,1)\right) ^2$ and $(u_1, \phi _1)\in H_{*}^1(0,1)\times H_{0}^1(0,1)$ then the solution satisfies

$$\begin{aligned}&u\in C(\mathbb {R}_{+}, H^2(0,1)\cap H_{*}^1(0, 1))\cap C^1(\mathbb {R}_{+}, H_{*}^1(0,1))\cap C^2(\mathbb {R}_{+},L^2(0,1)),\\&\phi \in C(\mathbb {R}_{+}, H^2(0,1)\cap H_0^1(0, 1))\cap C^1(\mathbb {R}_{+}, H_0^1(0,1))\cap C^2(\mathbb {R}_{+},L^2(0,1)),\\&\theta \in C(\mathbb {R}_{+}, H^2(0,1)\cap H_0^1(0, 1))\cap C^1(\mathbb {R}_{+}, H_0^1(0,1)). \end{aligned}$$

Remark 1.2

The proof can be established using the Galerkin method.

The rest of our paper is organized as follows. In Sect. 2, we state and prove some technical lemmas. In Sect. 3, we state and prove our stability result. We use $c_1$ throughout this paper to denote a generic positive constant.

2 Technical Lemmas

In this section, we state and prove some technical lemmas needed in the proof of our stability result.

Lemma 2.1

Under assumptions (H1) and (H2), the energy functional E,  defined by

$$\begin{aligned} E(t) =&\frac{1}{2}\int _{0}^{1}\left[ \rho u_{t}^{2}+\mu u_x^2+J\phi _{t}^{2}+c\theta ^{2} + \left( \delta -\int _{0}^{t}g(s)\mathrm{d}s\right) \phi _{x}^{2}+\xi \phi ^{2}+2bu_x\phi \right] \mathrm{d}x \nonumber \\&+\dfrac{1}{2}g \circ \phi _x, \end{aligned}$$

(2.1)

satisfies

$$\begin{aligned} E^{\prime }(t)=-\kappa \int _{0}^{1} \theta _{x}^{2} \mathrm{d}x+\dfrac{1}{2}g'\circ \phi _x-\dfrac{1}{2}g(t)\int _{0}^{1}\phi _x^2\mathrm{d}x\le 0, \end{aligned}$$

(2.2)

where

$$\begin{aligned} (g\circ \phi _x)(t)=\int _{0}^{1}\int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))^2\mathrm{d}s\mathrm{d}x. \end{aligned}$$

Proof

Multiplying the first three equations of (1.1) by $u_t, \phi _{t},$ and $\theta ,$ respectively, integrating by parts over (0, 1),  and using the boundary conditions, we obtain

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int _{0}^{1}\left[ \rho u_{t}^{2}+\mu u_x^2+J\phi _{t}^{2}+ c\theta ^{2} +\delta \phi _{x}^{2}+\xi \phi ^{2}+2bu_x\phi \right] \mathrm{d}x \nonumber \\&\qquad -\int _{0}^{1}\phi _{xt}\int _{0}^{t}g(t-s)\phi _x(s)\mathrm{d}s\mathrm{d}x \nonumber \\&\quad = -\kappa \int _{0}^{1}\theta _x^2\mathrm{d}x. \end{aligned}$$

(2.3)

The last term in the left hand side of (2.3) is estimated as follows.

$$\begin{aligned}&-\int _{0}^{1}\phi _{xt}\int _{0}^{t}g(t-s)\phi _x(s)\mathrm{d}s\mathrm{d}x = \int _{0}^{1}\phi _{xt} \int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\mathrm{d}x \nonumber \\&\qquad -\int _{0}^{t}g(s)\mathrm{d}s\int _{0}^{1}\phi _x\phi _{tx}\mathrm{d}x \nonumber \\&\quad = \dfrac{1}{2}\dfrac{\mathrm{d}}{\mathrm{d}t} g\circ \phi _x-\dfrac{1}{2}\dfrac{\mathrm{d}}{\mathrm{d}t}\int _{0}^{t}g(s)\mathrm{d}s \int _{0}^{1}\phi _x^2\mathrm{d}x-\dfrac{1}{2}g'\circ \phi _x+\dfrac{1}{2}g(t)\int _{0}^{1}\phi _x^2\mathrm{d}x.\qquad \quad \end{aligned}$$

(2.4)

The substitution of (2.4) into (2.3), bearing in mind (2.1), yields (2.2).$\square$

Remark 2.2

The energy functional E(t) defined by (2.1) is nonnegative. In fact, it can easily be verified that

$$\begin{aligned} \mu u_x^2+2bu_x\phi +\xi \phi ^2 =&\dfrac{1}{2}\left[ \mu \left( u_{x}+\dfrac{b}{\mu }\phi \right) ^2+\xi \left( \phi +\dfrac{b}{\xi }u_{x}\right) ^2 \right. \\&\quad \left. +\left( \mu -\dfrac{b^2}{\xi }\right) u_x^2+\left( \xi -\dfrac{b^2}{\mu }\right) \phi ^2\right] . \end{aligned}$$

So, using the fact that $\mu \xi > b^{2},$ we obtain

$$\begin{aligned} \mu u_x^2+2bu_x\phi +\xi \phi ^2>\dfrac{1}{2}\left[ \left( \mu -\dfrac{b^2}{\xi }\right) u_x^2+\left( \xi -\dfrac{b^2}{\mu }\right) \phi ^2\right] >0. \end{aligned}$$

Consequently,

$$\begin{aligned} E(t) > \frac{1}{2}\int _{0}^{1}\left[ \rho u_{t}^{2}+\mu _1 u_x^2+J\phi _{t}^{2}+c\theta ^{2} + \left( \delta -\int _{0}^{t}g(s)\mathrm{d}s\right) \phi _{x}^{2}+\xi _1\phi ^{2}\right] \mathrm{d}x+\dfrac{1}{2}g \circ \phi _x,\nonumber \\ \end{aligned}$$

(2.5)

where $2\mu _1=\mu -\dfrac{b^2}{\xi }>0$ and $2\xi _1=\xi -\dfrac{b^2}{\mu }>0.$

Lemma 2.3

The functional

$$\begin{aligned} F_{1}(t):=-c\int _{0}^{1}\theta \left( \int _{0}^{x} u_t(y,t)\mathrm{d}y\right) \mathrm{d}x \end{aligned}$$

satisfies, along the solution of (1.1), for any $\varepsilon _1>0$, the estimate

$$\begin{aligned} F_{1}^\prime (t) \le -\dfrac{\beta }{2}\int _{0}^{1} u_{t}^{2} \mathrm{d}x+\varepsilon _1\int _{0}^{1}(u_{x}^{2}+\phi ^{2})\mathrm{d}x+c_1\int _{0}^{1}\phi _t^2\mathrm{d}x+c_1\left( 1+\dfrac{1}{\varepsilon _1}\right) \int _{0}^{1}\theta _{x}^{2}\mathrm{d}x.\nonumber \\ \end{aligned}$$

(2.6)

Proof

Direct computations using (1.1) yield

$$\begin{aligned} F_{1}^\prime (t)&= -\beta \int _{0}^{1} u_{t}^{2} \mathrm{d}x +\kappa \int _{0}^{1}\theta _x u_{t} \mathrm{d}x+\dfrac{c\beta }{\rho }\int _{0}^{1}\theta ^{2}\mathrm{d}x-\dfrac{c\mu }{\rho }\int _{0}^{1}\theta u_{x}\mathrm{d}x \nonumber \\&\quad -\dfrac{cb}{\rho }\int _{0}^{1}\theta \phi \mathrm{d}x+m\int _{0}^{1}\phi _t\left( \int _{0}^{x} u_t(y,t)\mathrm{d}y\right) \mathrm{d}x. \end{aligned}$$

(2.7)

By Young’s inequality, for any $\varepsilon _1 > 0,$ we obtain

$$\begin{aligned} \kappa \int _{0}^{1}\theta _x u_{t} \mathrm{d}x \le \dfrac{\beta }{4}\int _{0}^{1} u_t^{2}\mathrm{d}x+\dfrac{\kappa ^2}{\beta }\int _{0}^{1} \theta _x^{2}\mathrm{d}x \end{aligned}$$

(2.8)

$$\begin{aligned} -\dfrac{c\mu }{\rho }\int _{0}^{1}\theta u_{x}\mathrm{d}x \le \varepsilon _1\int _{0}^{1} u_x^{2}\mathrm{d}x+\dfrac{c^2\mu ^2}{4\rho ^2\varepsilon _1}\int _{0}^{1} \theta ^{2}\mathrm{d}x \end{aligned}$$

(2.9)

$$\begin{aligned} -\dfrac{cb}{\rho }\int _{0}^{1}\theta \phi \mathrm{d}x \le \varepsilon _1\int _{0}^{1} \phi ^{2}\mathrm{d}x+\dfrac{c^2b^2}{4\rho ^2\varepsilon _1}\int _{0}^{1} \theta ^{2}\mathrm{d}x \end{aligned}$$

(2.10)

$$\begin{aligned} m\int _{0}^{1}\phi _t\left( \int _{0}^{x} u_t(y,t)\mathrm{d}y\right) \mathrm{d}x \le \dfrac{\beta }{4}\int _{0}^{1}\left( \int _{0}^{x} u_t(y,t)\mathrm{d}y\right) ^2\mathrm{d}x+\dfrac{m^2}{\beta }\int _{0}^{1} \phi _t^{2}\mathrm{d}x.\nonumber \\ \end{aligned}$$

(2.11)

The combination of (2.7)–(2.11) yields

$$\begin{aligned} F_{1}^\prime (t)\le & {} -\dfrac{3\beta }{4} \int _{0}^{1} u_{t}^{2} \mathrm{d}x+\dfrac{\kappa ^2}{\beta }\int _{0}^{1}\theta _{x}^{2}\mathrm{d}x+\varepsilon _1\int _{0}^{1}u_{x}^{2}\mathrm{d}x+c_1\left( 1+\dfrac{1}{\varepsilon _1}\right) \int _{0}^{1}\theta ^{2}\mathrm{d}x \nonumber \\&+\varepsilon _1\int _{0}^{1}\phi ^2\mathrm{d}x+\dfrac{m^2}{\beta }\int _{0}^{1}\phi _t^2\mathrm{d}x+\dfrac{\beta }{4}\int _{0}^{1}\left( \int _{0}^{x} u_t(y,t)\mathrm{d}y\right) ^2\mathrm{d}x. \end{aligned}$$

(2.12)

By Cauchy-Schwarz inequality, it is clear that

$$\begin{aligned} \left( \int _{0}^{x} u_t(y,t)\mathrm{d}y\right) ^2\le \left( \int _{0}^{1} u_t\mathrm{d}x\right) ^2\le \int _{0}^{1} u_t^2 \mathrm{d}x. \end{aligned}$$

(2.13)

By substituting (2.13) into (2.12), and using Poincaré’s inequality, we obtain (2.6).$\square$

Lemma 2.4

The functional

$$\begin{aligned} F_{2}(t):=\rho \int _{0}^{1} u u_t\mathrm{d}x \end{aligned}$$

satisfies, along the solution of (1.1),

$$\begin{aligned} F_{2}^\prime (t)&\le -\dfrac{ \mu }{2} \int _{0}^{1}u_{x}^{2}\mathrm{d}x+\rho \int _{0}^{1} u_{t}^{2}\mathrm{d}x \nonumber \\&\quad +c_1\int _{0}^{1}\phi ^2\mathrm{d}x+c_1\int _{0}^{1}\theta _{x}^{2}\mathrm{d}x. \end{aligned}$$

(2.14)

Proof

By taking a derivative of $F_2,$ using (1.1), and then integrating by parts, we obtain

$$\begin{aligned} F_{2}^\prime (t) =-\mu \int _{0}^{1}u_{x}^{2}\mathrm{d}x+\rho \int _{0}^{1} u_{t}^{2}\mathrm{d}x-b\int _{0}^{1}u_x\phi \mathrm{d}x+\beta \int _{0}^{1}u_{x}\theta \mathrm{d}x. \end{aligned}$$

Using Young’s and Poincaré’s inequalities as in the proof of Lemma 2.3, we obtain (2.14).$\square$

Lemma 2.5

The functional

$$\begin{aligned} F_3(t):=-J\int _{0}^{1}\phi _t\int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s \mathrm{d}x \end{aligned}$$

satisfies, for some fixed $t_0>0$ and for any $\varepsilon _2>0,$ the estimate

$$\begin{aligned} F_3'(t)&\le -\dfrac{ Jg_0}{2} \int _{0}^{1}\phi _{t}^{2}\mathrm{d}x+\varepsilon _2\int _{0}^{1} (u_x^2+\phi ^2+\phi _x^2)\mathrm{d}x+c_1\int _{0}^{1}\theta _x^2\mathrm{d}x \nonumber \\&\quad +\,c_1\left( 1+\dfrac{1}{\varepsilon _2}\right) g\circ \phi _x-c_1g'\circ \phi _x, \end{aligned}$$

(2.15)

where $g_0=\displaystyle \int _{0}^{t_0}g(s)\mathrm{d}s$.

Proof

Differentiating $F_3$, taking into account (1.1), and using integrating by parts together with the boundary conditions, we obtain

$$\begin{aligned} F_3' (t) =&- J \int _{0}^{t}g(s)\mathrm{d}s \int _{0}^{1}\phi _{t}^{2}\mathrm{d}x- J\int _{0}^{1}\phi _t\int _{0}^{t}g'(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x \nonumber \\&+\,\delta \int _{0}^{1}\phi _x\int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\mathrm{d}x \nonumber \\&-\,m\int _{0}^{1}\theta \int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x \nonumber \\&+\,b\int _{0}^{1} u_x\int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x \nonumber \\&+\,\xi \int _{0}^{1}\phi \int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x \nonumber \\&-\,\int _{0}^{1}\int _{0}^{t}g(t-s)\phi _x(s)\mathrm{d}s\int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\mathrm{d}x. \end{aligned}$$

(2.16)

Now, we estimate the terms in the right-hand side of (2.16), using Young’s, Cauchy-Schwarz, and Poincaré’s inequalities, and the fact that

$$\begin{aligned} \int _{0}^{t}g(s)\mathrm{d}s\le \delta -l>0. \end{aligned}$$

(2.17)

So, for any $\delta _1, \varepsilon _2>0$, we obtain

$$\begin{aligned} I_{1}= & {} - J\int _{0}^{1}\phi _t\int _{0}^{t}g'(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x \nonumber \\\le & {} J\delta _{1}\int _{0}^{1} \phi ^2_{t}\mathrm{d}x+\frac{ J}{4\delta _{1}} \int _{0}^{1}\left( \int _0^{t}g'(t-s) (\phi (t)-\phi (s))\mathrm{d}s\right) ^{2} \mathrm{d}x \nonumber \\\le & {} J\delta _{1}\int _{0}^{1} \phi ^2_{t}\mathrm{d}x+\frac{ J}{4\delta _{1}} \int _{0}^{1}\left( \int _0^{t}-g^{\prime } (s)\mathrm{d}s\right) \int _0^{t}-g' (t-s) (\phi (t)-\phi (s))^{2} \mathrm{d}s\mathrm{d}x \nonumber \\\le & {} J\delta _{1}\int _{0}^{1} \phi ^2_{t}\mathrm{d}x-\frac{c_1}{\delta _{1}}g' \circ \phi _x, \end{aligned}$$

(2.18)

$$\begin{aligned} I_2= & {} \delta \int _{0}^{1}\phi _x\int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\mathrm{d}x \nonumber \\\le & {} \dfrac{\varepsilon _2}{2}\int _{0}^{1}\phi _x ^{2}\mathrm{d}x+\dfrac{\delta ^2}{2\varepsilon _2}\int _{0}^{1} \left( \int _0^{t}g(t-s)(\phi _x(t)-\phi _x(s)) \mathrm{d}s\right) ^{2}\mathrm{d}x \nonumber \\\le & {} \dfrac{\varepsilon _2}{2}\int _{0}^{1}\phi _x^{2}\mathrm{d}x+\dfrac{\delta ^2}{2\varepsilon _2}\int _0^{t}g(s)\mathrm{d}s\int _{0}^{1} \int _0^{t}g(t-s)(\phi _x(t)-\phi _x(s))^{2}\mathrm{d}s\mathrm{d}x \nonumber \\\le & {} \dfrac{\varepsilon _2}{2}\int _{0}^{1}\phi _x^{2}\mathrm{d}x+\dfrac{c_1}{\varepsilon _2}g\circ \phi _x, \end{aligned}$$

(2.19)

Similar to $I_2,$ we have

$$\begin{aligned} I_3 = b\int _{0}^{1} u_x\int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x\le \varepsilon _2\int _{0}^{1} u_x^{2}\mathrm{d}x+\dfrac{c_1}{\varepsilon _2}g \circ \phi _x,\qquad \end{aligned}$$

(2.20)

$$\begin{aligned} I_4 = \xi \int _{0}^{1}\phi \int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x\le \varepsilon _2\int _{0}^{1}\phi ^{2}\mathrm{d}x+\dfrac{c_1}{\varepsilon _2}g \circ \phi _x,\qquad \end{aligned}$$

(2.21)

$$\begin{aligned} I_5= & {} -m\int _{0}^{1}\theta \int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s\mathrm{d}x\le \dfrac{m}{2}\int _{0}^{1}\theta ^{2}\mathrm{d}x+\dfrac{m}{2}g \circ \phi , \nonumber \\\le & {} c_1\int _{0}^{1}\theta _x^{2}\mathrm{d}x+c_1g \circ \phi _x, \end{aligned}$$

(2.22)

$$\begin{aligned} I_6 =&-\int _{0}^{1}\int _{0}^{t}g(t-s)\phi _x(s)\mathrm{d}s\int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\mathrm{d}x \nonumber \\ =&-\int _{0}^{t}g(s)\mathrm{d}s\int _{0}^{1}\phi _x\int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\mathrm{d}x \nonumber \\&\quad +\int _{0}^{1}\left( \int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\right) ^2\mathrm{d}x \nonumber \\&\le \dfrac{\varepsilon _2}{2}\int _{0}^{1}\phi _x^{2}\mathrm{d}x+\dfrac{1}{2\varepsilon _2}\left( \int _{0}^{t}g(s)\mathrm{d}s\right) ^2\int _{0}^{1} \left( \int _0^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\right) ^{2}\mathrm{d}x \nonumber \\&\quad +\int _{0}^{1}\left( \int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\right) ^2\mathrm{d}x \nonumber \\&\le \dfrac{\varepsilon _2}{2}\int _{0}^{1}\phi _x^{2}\mathrm{d}x+\dfrac{c_1}{\varepsilon _2}g\circ \phi _x. \end{aligned}$$

(2.23)

Since the function g is positive, continuous and $g(0) > 0$, then, for any $t \ge t_0 > 0$, we have

$$\begin{aligned} \int _0^{t}g(s)\mathrm{d}s \ge \int _0^{t_0}g(s)\mathrm{d}s=g_0. \end{aligned}$$

(2.24)

By substituting (2.18)−(2.23) into (2.16), and bearing in mind (2.24), we obtain

$$\begin{aligned} F_3'(t)&\le - J\left[ g_0-\delta _1\right] \int _{0}^{1}\phi _{t}^{2}\mathrm{d}x+\varepsilon _2\int _{0}^{1}( u_x^2+\phi ^2+\phi ^2_x)\mathrm{d}x+c_1\int _{0}^{1}\theta _x^{2}\mathrm{d}x \nonumber \\&\quad +\,c\left( 1+\dfrac{1}{\varepsilon _2}\right) g\circ \phi _x-\dfrac{c}{\delta _1}g'\circ \phi _x, \end{aligned}$$

for all $t\ge t_0$. By letting $\delta _1=\dfrac{g_0}{2}$, we obtain (2.15).$\square$

Lemma 2.6

Let $( u, \phi , \theta )$ be the solution of (1.1). Then the functional

$$\begin{aligned} F_4(t):= J\int _{0}^{1}\phi _{t}\phi \mathrm{d}x+\dfrac{b\rho }{\mu }\int _{0}^{1}\phi \int _{0}^{x}u_t(y)\mathrm{d}y\mathrm{d}x \end{aligned}$$

satisfies, for any $\varepsilon _3>0,$ the estimate

$$\begin{aligned} F'_4(t) \le&-\dfrac{l}{2}\int _{0}^{1}\phi _{x} ^{2}\mathrm{d}x-\xi _1\int _{0}^{1}\phi ^2\mathrm{d}x+\varepsilon _3\int _{0}^{1}u_{t}^{2}\mathrm{d}x+c_1\left( 1+\dfrac{1}{\varepsilon _3}\right) \int _{0}^{1}\phi _{t}^{2}\mathrm{d}x \nonumber \\&+ c_1 \int _{0}^{1}\theta _x^2\mathrm{d}x+ c_1g\circ \phi _x. \end{aligned}$$

(2.25)

Proof

Direct differentiation of $F_4,$ using (1.1) and integration by parts, yields

$$\begin{aligned} F'_4 (t)= & {} -\delta \int _{0}^{1} \phi _{x}^2\mathrm{d}x+J\int _{0}^{1}\phi _{t} ^{2}\mathrm{d}x- \left( \xi -\frac{b^2}{\mu }\right) \int _{0}^{1}\phi ^2\mathrm{d}x+\dfrac{b\rho }{\mu }\int _{0}^{1}\phi _t\int _{0}^{x}u_t(y)\mathrm{d}y\mathrm{d}x \nonumber \\&+\left( m-\dfrac{\beta b}{\mu }\right) \int _{0}^{1}\phi \theta \mathrm{d}x+\int _{0}^{1}\phi _x\int _{0}^{t}g(t-s)\phi _x(s)\mathrm{d}s\mathrm{d}x. \end{aligned}$$

(2.26)

In what follows, we use Cauchy-Schwarz and Young’s inequalities, for $\delta _3>0.$

$$\begin{aligned}&\left( m-\dfrac{\beta b}{\mu }\right) \int _{0}^{1}\phi \theta \mathrm{d}x\le \delta _2\int _{0}^{1}\phi ^{2}\mathrm{d}x+ \left( m-\dfrac{\beta b}{\mu }\right) ^2\dfrac{1}{4\delta _2}\int _{0}^{1}\theta ^{2}\mathrm{d}x, \nonumber \\&\dfrac{b\rho }{\mu }\int _{0}^{1}\phi _t\int _{0}^{x}u_t(y)\mathrm{d}y\mathrm{d}x\le \varepsilon _3\int _{0}^{1}\left( \int _{0}^{x} u_t(y,t)\mathrm{d}y\right) ^2\mathrm{d}x+\left( \dfrac{b\rho }{\mu }\right) ^2\dfrac{1}{4\varepsilon _3}\int _{0}^{1}\phi _t^2\mathrm{d}x,\nonumber \\ \end{aligned}$$

(2.27)

by using (2.13), we obtain

$$\begin{aligned} \dfrac{b\rho }{\mu }\int _{0}^{1}\phi _t\int _{0}^{x}u_t(y)\mathrm{d}y\mathrm{d}x\le \varepsilon _3\int _{0}^{1}u^2_t\mathrm{d}x+\dfrac{c_1}{\varepsilon _3}\int _{0}^{1}\phi _t^2\mathrm{d}x, \end{aligned}$$

(2.28)

$$\begin{aligned}&\int _{0}^{1}\phi _x\int _{0}^{t}g(t-s)\phi _x(s)\mathrm{d}s\mathrm{d}x\nonumber \\&\quad = \int _{0}^{t}g(s)\mathrm{d}s\int _{0}^{1}\phi ^2_x\mathrm{d}x \nonumber \\&\qquad -\int _{0}^{1}\phi _x\int _{0}^{t}g(t-s)(\phi _x(t)-\phi _x(s))\mathrm{d}s\mathrm{d}x \nonumber \\&\quad \le \left( \delta _3+\int _{0}^{t}g(s)\mathrm{d}s\right) \int _{0}^{1}\phi ^2_x\mathrm{d}x+\dfrac{1}{4\delta _3}\left( \int _{0}^{t}g(s)\mathrm{d}s\right) g\circ \phi _x. \end{aligned}$$

(2.29)

By substituting (2.27)–(2.29) into (2.26), and using Young’s and Poincaré’s inequalities together with (2.17) and the fact that $2\xi _1=\xi -\frac{b^2}{\mu }>0$ since $\mu \xi >b^2$ and $\mu >0$, we arrive at

$$\begin{aligned} F'_4 (t)\le&- \left( l-\delta _3\right) \int _{0}^{1}\phi ^2_x\mathrm{d}x-\left( 2\xi _1-\delta _2\right) \int _{0}^{1}\phi ^2\mathrm{d}x+\varepsilon _3\int _{0}^{1}u^2_t\mathrm{d}x \nonumber \\&+c\left( 1+\dfrac{1}{\varepsilon _3}\right) \int _{0}^{1}\phi _{t} ^{2}\mathrm{d}x+\dfrac{c_1}{\delta _2}\int _{0}^{1}\theta _x^2\mathrm{d}x+\dfrac{c_1}{\delta _3}g\circ \phi _x. \end{aligned}$$

(2.30)

By taking $\delta _3=\dfrac{l}{2}$ and $\delta _2=\xi _1,$ we obtain estimate (2.25).$\square$

3 General Stability Result

In this section, we state and prove our result. To achieve this, we define a Lyapunov functional $\mathcal {L}$ and show that it is equivalent to the energy functional E.

Lemma 3.1

For N sufficiently large, the functional defined by

$$\begin{aligned} \mathcal {L}(t):=NE(t)+ \dfrac{4\rho }{\beta }F_{1}(t)+ F_{2}(t)+ N_1F_3(t)+ N_2F_4(t), \end{aligned}$$

(3.1)

where $N_{1}$ and $N_{2}$ are positive real numbers to be chosen appropriately later, satisfies

$$\begin{aligned} c_{3}E(t) \le \mathcal {L}(t) \le c_{4}E(t), \qquad \forall t \ge 0, \end{aligned}$$

(3.2)

for two positive constants $c_{3}$ and $c_{4}.$

Proof

It follows that

$$\begin{aligned} \vert \mathcal {L}(t)-NE(t)\vert \le&\dfrac{4c\rho }{\beta }\int _{0}^{1}\left| \theta \int _{0}^{x} u_t(y,t)\mathrm{d}y\right| \mathrm{d}x+ \rho \int _{0}^{1}\left| u_x\int _{0}^{x} u_t(y,t)\mathrm{d}y\right| \mathrm{d}x \nonumber \\&+ JN_1\int _{0}^{1}\left| \phi _t\int _{0}^{t}g(t-s)(\phi (t)-\phi (s))\mathrm{d}s\right| \mathrm{d}x \nonumber \\&+JN_2\int _{0}^{1}\vert \phi _{t}\phi \vert \mathrm{d}x+\dfrac{b\rho }{\mu }N_2\int _{0}^{1}\left| \phi \int _{0}^{x}u_t(y)\mathrm{d}y\right| \mathrm{d}x. \end{aligned}$$

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

$$\begin{aligned} \vert \mathcal {L}(t)-NE(t)\vert \le c_1 \int _{0}^{1}\left( u_{t}^{2} + \phi _{t}^{2} +\phi _{x}^{2}+ u^2_{x}+\phi ^{2}+\theta ^{2}\right) \mathrm{d}x+c_1g\circ \phi _x. \end{aligned}$$

Using (2.5), we obtain

$$\begin{aligned} \vert \mathcal {L}(t)-NE(t)\vert \le c_1E(t), \end{aligned}$$

that is

$$\begin{aligned} (N-c_1)E(t)\le \mathcal {L}(t)\le (N+c_1)E(t). \end{aligned}$$

By choosing N large enough, (3.2) follows.$\square$

Next, we state and prove the main result of this section.

Theorem 3.2

Let $(u_0, \phi _0, \theta _0)\in H_{*}^1(0,1)\times \left( H_{0}^1(0,1)\right) ^2$ and $(u_1, \phi _1)\in \left( L^2(0,1)\right) ^2$ be given. Assume (H1) and (H2) hold. Then, there exist positive constants $\lambda _1$ and $\lambda _2$ such that the energy functional given by (2.1) satisfies

$$\begin{aligned} E(t)\le \lambda _1e^{-\lambda _2\int _{0}^{t}\zeta (s)\mathrm{d}s}, \quad \forall t\ge 0. \end{aligned}$$

(3.3)

Proof

By differentiating (3.1), recalling (2.2), (2.6), (2.14), (2.15), (2.25), and letting

$$\begin{aligned} \varepsilon _1=\dfrac{\beta \mu }{16\rho },\quad \varepsilon _3=\dfrac{\rho }{2N_2} \end{aligned}$$

we obtain

$$\begin{aligned} \mathcal {L}'(t)\le&-\left[ \kappa N-c_1N_1-c_1N_2-c_1\right] \int _{0}^{1}\theta _{x}^{2}\mathrm{d}x-\dfrac{\rho }{2}\int _{0}^{1} u_{t}^{2}\mathrm{d}x \nonumber \\&-\left[ \dfrac{ \mu }{4}-\varepsilon _2N_1\right] \int _{0}^{1} u_x^{2}\mathrm{d}x-\left[ \dfrac{ Jg_0}{2}N_1-c_1N_2\left( 1+N_2\right) -c_1\right] \int _{0}^{1}\phi _{t}^{2}\mathrm{d}x \nonumber \\&-\left[ \dfrac{l}{2}N_2-\varepsilon _2N_1\right] \int _{0}^{1}\phi _x^2\mathrm{d}x-\left[ \xi _1N_2-\varepsilon _2N_1-c_1\right] \int _{0}^{1}\phi ^{2}\mathrm{d}x\\&+\left[ \dfrac{N}{2}-c_1N_1\right] g'\circ \phi _x+c_1\left[ N_2+N_1\left( 1+\dfrac{1}{\varepsilon _2}\right) \right] g\circ \phi _x. \end{aligned}$$

We choose $N_{2}$ large enough such that

$$\begin{aligned} \alpha _1=\xi _1N_2-c_1 > 0, \end{aligned}$$

then we choose $N_1$ large enough such that

$$\begin{aligned} \alpha _2=\dfrac{ Jg_0}{2}N_1-c_1N_2\left( 1+N_2\right) -c_1 > 0. \end{aligned}$$

Next, we pick $\varepsilon _2$ small enough such that

$$\begin{aligned} \varepsilon _2<\min \left( \dfrac{\alpha _1}{N_1},\ \dfrac{\mu }{4N_1},\ \dfrac{lN_2}{2N_1}\right) , \end{aligned}$$

consequently, we obtain

$$\begin{aligned} \alpha _3=\dfrac{l}{2}N_2-\varepsilon _2N_1> 0,\quad \alpha _4=\dfrac{ \mu }{4}-\varepsilon _2N_1>0,\quad \alpha _5=\alpha _1-\varepsilon _2N_1>0. \end{aligned}$$

Finally, we choose N large enough such that (3.2) remains valid and

$$\begin{aligned} \alpha _6=\kappa N-c_1N_1-c_1N_2-c_1> 0, \quad \alpha _7=\dfrac{N}{2}-c_1N_2>0. \end{aligned}$$

So, we arrive at

$$\begin{aligned} \mathcal {L}'(t)\le&-\alpha _6\int _{0}^{1}\theta _{x}^{2}\mathrm{d}x-\dfrac{\rho }{2}\int _{0}^{1} u_{t}^{2}\mathrm{d}x-\alpha _4\int _{0}^{1} u_x^{2}\mathrm{d}x-\alpha _2\int _{0}^{1}\phi ^2_t\mathrm{d}x-\alpha _3\int _{0}^{1}\phi _{x}^{2}\mathrm{d}x \nonumber \\&-\alpha _5\int _{0}^{1}\phi ^2\mathrm{d}x+\alpha _7g'\circ \phi _{x}+c_1g\circ \phi _{x}. \end{aligned}$$

(3.4)

On the other hand, from (2.1), using Young’s and Poincaré’s inequalities, we obtain

$$\begin{aligned} E(t) \le&\frac{1}{2}\int _{0}^{1}\left[ c_1\theta ^2_{x}+\rho u_{t}^{2}+(\mu +b) u_x^2+\left( \delta -\int _{0}^{t}g(s)\mathrm{d}s\right) \phi _{x}^{2}+(\xi +b)\phi ^{2}+J\phi _{t}^{2}\right] \mathrm{d}x +\dfrac{1}{2}g \circ \phi _x \nonumber \\ \le&c_1\int _{0}^{1}\left[ \theta ^2_{x}+u_{t}^{2}+u_x^2+\phi _{x}^{2}+\phi ^{2}+\phi _{t}^{2}\right] \mathrm{d}x+\dfrac{1}{2}g \circ \phi _x \end{aligned}$$

which implies that

$$\begin{aligned} -\int _{0}^{1}\left[ \theta ^2_{x}+u_{t}^{2}+u_x^2+\phi _{x}^{2}+\phi ^{2}+\phi _{t}^{2}\right] \mathrm{d}x\le -c'E(t)+c''g \circ \phi _x. \end{aligned}$$

(3.5)

The combination of (3.4) and (3.5) gives

$$\begin{aligned} \mathcal {L}'(t)\le -k_1E(t)+k_2(g \circ \phi _x)(t), \quad \forall t \ge t_0, \end{aligned}$$

(3.6)

for some positive constants $k_1$ and $k_2$. By multiplying (3.6) by $\zeta (t)$ and using (H2) and (2.2), we arrive at

$$\begin{aligned} \zeta (t)\mathcal {L}'(t)\le - k_1\zeta (t) E(t)-2k_2 E'(t), \quad \forall t \ge t_0, \end{aligned}$$

which can be rewritten as

$$\begin{aligned} \left( \zeta (t)\mathcal {L}(t)+2k_2 E(t)\right) '-\zeta '(t)\mathcal {L}(t) \le - k_1\zeta (t)E(t), \; \forall t\ge t_0. \end{aligned}$$

Using the fact that $\zeta '(t) \le 0, \forall t \ge 0,$ we have

$$\begin{aligned} \left( \zeta (t)\mathcal {L}(t)+2k_2 E(t)\right) ' \le - k_1\zeta (t)E(t), \quad \forall t\ge t_0. \end{aligned}$$

By exploiting (3.2), it can easily be shown that

$$\begin{aligned} \mathcal {R}(t) =\zeta (t) \mathcal {L}(t)+2k_2E(t)\sim E(t). \end{aligned}$$

(3.7)

Consequently, for some positive constant $\lambda _2$, we obtain

$$\begin{aligned} \mathcal {R}'(t)\le -\lambda _2\zeta (t)\mathcal {R}(t), \quad \forall t\ge t_0. \end{aligned}$$

(3.8)

A simple integration of (3.8) over $(t_0,t)$ leads to

$$\begin{aligned} \mathcal {R}(t)\le \mathcal {R}(t_0)e^{-\lambda _2\int _{t_0}^{t}\zeta (s)\mathrm{d}s}, \quad \forall t\ge t_0. \end{aligned}$$

(3.9)

Using (3.7) for some positive constant $\tilde{\lambda }_{1},$ we obtain,

$$\begin{aligned} E(t)\le \tilde{\lambda }_{1}e^{-\lambda _2\int _{t_0}^{t}\zeta (s)\mathrm{d}s}, \quad \forall t\ge t_0. \end{aligned}$$

(3.10)

Consequently, (3.3) is established by virtue of the continuity and boundedness of E and $\zeta$. In other words, since $E(t)\le E(t_{0})\le E(0), \ \forall t\ge t_{0}>0,$ we get

$$\begin{aligned} E(t)\le \tilde{\lambda }_{1}E(0)e^{\lambda _{2}\int _{0}^{t_{0}}\zeta (s)\mathrm{d}s}e^{-\lambda _{2}\int _{0}^{t}\zeta (s)\mathrm{d}s}, \quad \forall t\ge t_{0}>0. \end{aligned}$$

Consequently, by taking $\lambda _{1}=\tilde{\lambda }_{1}E(0)e^{\lambda _{2}\int _{0}^{t_{0}}\zeta (s)\mathrm{d}s}$ we obtain (3.3).$\square$

Remark 3.3

The result given by Theorem 3.2 shows that the memory term together with the heat effect is strong enough to uniformly stabilize the system without imposing the condition of equal wave of speeds. This same result was obtained by Apalara [27] for equivalent Timoshenko system.

3.1 Example

Now, we give two examples to illustrate the energy decay rates obtained by Theorem 3.2. Given that $\tau , \gamma >0$ with $\tau <\gamma \delta$.

• (1) If $g(t) = \tau e^{-\gamma t}$, then

  $$\begin{aligned} E(t) \le c_0e^{-\gamma c_1t}, \quad \forall t \ge 0. \end{aligned}$$

• (2) If $g(t) = \frac{\tau }{(1+t)^{\gamma +1}}$, then

  $$\begin{aligned} E(t) \le \frac{c_0}{(1+t)^{(\gamma +1)c_1}}, \quad \forall t \ge 0. \end{aligned}$$