1 Introduction

In the recent decades, the study of problems related to elastic solids with voids has attracted the attention of many researchers due to the extensive practical applications of such materials in different fields, such as petroleum industry, foundation engineering, soil mechanics, power technology, biology, material science and so on. Elastic solids with voids are one of the simple extensions of the theory of the classical elasticity. It allows the treatment of porous solids in which the matrix material is elastic and the interstices are void of material.

In 1972, Godman and Cowin [12] proposed an extension of the classical elasticity theory to porous media. They introduced the concept of a continuum theory of granular materials with interstitial voids into the theory of elastic solids with voids. In addition to their usual elastic effects, these materials have 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. This latter idea was introduced by Nunziato and Cowin [22] when they developed a nonlinear theory of elastic materials with voids. We refer the reader to [8, 9, 18, 24] and the references therein for more details.

It is well known that the classical thermoelasticity using Fourier’s Law of heat conduction, which states that the heat flux is proportional to the gradient of temperature (\(q=-\delta \nabla \theta \)), leads to the physical paradox of infinite speed of heat propagation. In other words, any thermal disturbance at one point will be instantaneously transferred to the other parts of the body. This is practically unrealistic. To overcome this physical paradox but still keeping the essentials of heat conduction process, many theories have subsequently emerged. One of such theories was proposed by Green and Naghdi in 1990s [14,15,16]. They used an analogy between the concepts and equations of the purely thermal and the purely mechanical theories and arrived at three types of constitutive equations for heat flow in a stationary rigid solid labeled as type I, II, and III. Consequently, using these constitutive equations, they obtained three models, called thermoelasticity of type I, thermoelasticity of type II, and thermoelasticity of type III. The linear version of the first one coincides with the classical theory based on Fourier’s law, the second one is known as thermoelasticity without energy dissipation because the heat equation is not a dissipative process and the third one is the most general and it contains the former two as limit cases. For a further historical review on these models, we refer the reader to [6, 7, 13,14,15,16].

The basic evolution equations for one-dimensional theories of porous materials with temperature are given by:

$$\begin{aligned} \rho \omega _{tt}-T_x=0, \ \ J\varphi _{tt}-H_x-G=0,\ \ \alpha \theta _{t}+q_x+\beta \varphi _{tx}=0 \end{aligned}$$
(1.1)

where T is the stress tensor, H is the equilibrated stress vector, G is the equilibrated body force, and q is the heat flux vector. The variables \(\omega \), \(\varphi \), and \(\theta \) are the displacement of the solid elastic material, the volume fraction, and the difference temperature, respectively. The positive parameters \(\rho \), J, and \(\beta \) are the mass density, product of the mass density by the equilibrated inertia, and the coupling constant, respectively. Taking into account Green and Naghdi’s theory, the constitutive equations are:

$$\begin{aligned} \begin{array}{l} T=\mu \omega _{x}+b\varphi ,\ \ \ H=\delta \varphi _x-\beta \theta \\ G=-b\omega _{x}-\xi \varphi ,\ \ \ q=-\delta \Theta _x-k\Theta _{tx} \end{array} \end{aligned}$$
(1.2)

where \(\delta ,k\) denote the thermal conductivity, \(\Theta \) is the so-called thermal displacement whose time derivative is the empirical temperature \(\theta \), that is, \(\Theta _t=\theta \), and \(\mu ,\xi \) are constitutive constants which satisfy

$$\begin{aligned} \mu>0,\ \ \xi>0,\ \ \mu \xi >b^2. \end{aligned}$$
(1.3)

To keep the coupling, the constant b must be different from zero. We substitute (1.2) into (1.1) to obtain the following system

$$\begin{aligned} \left\{ \begin{array}{ll} \rho \omega _{tt}-\mu \omega _{xx}-b\varphi _x=0, &{} ~~~~~ \text{ in } ~(0,1)\times (0,+\infty )\\ J \varphi _{tt}-\delta \varphi _{xx}+b\omega _{x}+\xi \varphi +\beta \theta _x=0,&{} ~~~~~\text{ in }~(0,1) \times (0,+\infty )\\ \alpha \theta _{t}-\delta \Theta _{xx}+\beta \varphi _{tx}-k\Theta _{txx}=0, &{} ~~~~~\text{ in }~(0,1) \times (0,+\infty ). \end{array} \right. \end{aligned}$$
(1.4)

For the asymptotic behaviors of the solutions for porous-elastic systems. Quintanilla [25] considered the one-dimensional porous dissipation elasticity:

$$\begin{aligned} \left\{ \begin{array}{ll} \rho \omega _{tt}-\mu \omega _{xx}-b\varphi _x=0, &{} ~~~~~ \text{ in } ~(0,L)\times (0,+\infty )\\ J \varphi _{tt}-\delta \varphi _{xx}+b\omega _{x}+\xi \varphi +\tau \varphi _t=0,&{} ~~~~~\text{ in }~(0,L) \times (0,+\infty ) \end{array} \right. \end{aligned}$$
(1.5)

with initial and boundary conditions. He used Hurtwitz theorem to prove that the damping through porous-viscosity (\(\tau \varphi _t\)) is not strong enough to obtain an exponential decay but only a slow (nonexponential) decay. However, Apalara [2, 3] considered the same system and proved the exponential stability provided \(\dfrac{\mu }{\rho }=\dfrac{\delta }{J}\). For various other damping mechanisms used and more results on porous elasticity, we refer the reader to [5, 26,27,28] and the references therein. Recently, Apalara [4] considered the following porous-elastic system with microtemperature:

$$\begin{aligned} \left\{ \begin{array}{ll} \rho \omega _{tt}-\mu \omega _{xx}-b\varphi _x=0, &{} ~~~~~ \text{ in } ~(0,1)\times (0,+\infty )\\ J \varphi _{tt}-\delta \varphi _{xx}+b\omega _{x}+\xi \varphi +\beta \theta _x=0,&{} ~~~~~\text{ in }~(0,1) \times (0,+\infty )\\ \alpha \theta _{t}-\kappa \theta _{xx}+\beta \varphi _{tx}+k \theta =0, &{} ~~~~~\text{ in }~(0,1) \times (0,+\infty ). \end{array} \right. \end{aligned}$$
(1.6)

with Dirichlet-Neumann-Dirichlet boundary conditions. He showed that the unique dissipation given by microtemperatures is strong enough to produce exponential stability if and only if:

$$\begin{aligned} \chi =\dfrac{\mu }{\rho }-\dfrac{\delta }{J}=0 \end{aligned}$$
(1.7)

and showed that the system is polynomially stable if \(\chi \ne 0\).

In the present work, we consider the system (1.4) which can be written as follows:

$$\begin{aligned} \left\{ \begin{array}{ll} \rho \omega _{tt}-\mu \omega _{xx}-b\varphi _x=0, &{} ~~~~~ \text{ in } ~(0,1)\times (0,+\infty )\\ J \varphi _{tt}-\delta \varphi _{xx}+b\omega _{x}+\xi \varphi +\beta \theta _x=0,&{} ~~~~~\text{ in }~(0,1) \times (0,+\infty )\\ \alpha \theta _{tt}-\delta \theta _{xx}+\beta \varphi _{ttx}-k\theta _{txx}=0, &{} ~~~~~\text{ in }~(0,1) \times (0,+\infty ) \end{array} \right. \end{aligned}$$
(1.8)

with the following boundary conditions

$$\begin{aligned} \omega (0,t)=\omega (1,t)=\varphi _x(0,t)=\varphi _x(1,t)=\theta (0,t)=\theta (1,t)=0,~~\forall ~t\ge 0 \end{aligned}$$
(1.9)

and initial conditions

$$\begin{aligned} \begin{aligned} \left\{ \begin{array}{ll} \omega (x,0)=\omega _0(x),~~\omega _t(x,0)=\omega _1(x), &{}{}~~~~~x \in (0,1) \\ \varphi (x,0)=\varphi _0(x),~~\varphi _t(x,0)=\varphi _1(x), &{}{}~~~~~x\in (0,1)\\ \theta (x,0)=\theta _0(x),~~\theta _t(x,0)=\theta _1(x), &{}{}~~~~~x \in (0,1) \end{array} \right. \end{aligned} \end{aligned}$$
(1.10)

We study the well-posedness and the asymptotic stability of (1.8)–(1.10). Using the semigroup theory, we prove the existence and uniqueness of the solution. We then exploit the energy method to obtain the exponential decay result for the case of equal wave speeds. When (1.7) does not hold, we prove a polynomial decay result.

The paper is organized as follows: in Sect. 2, we state the problem. In Sect. 3, we establish the well-posedness of the system. In Sect. 4, we show that the system is exponentially stable under condition (1.7). The polynomial stability when the wave-propagation speeds are different, is given in Sect. 5. In Sect. 6, we give some numerical illustrations.

2 Statement of the problem

To obtain the dissipative nature of System (1.8), we introduce the new variables: \(u=\omega _t\) and \(\phi =\varphi _t\). So, System (1.8) takes the form

$$\begin{aligned} \left\{ \begin{array}{ll} \rho u_{tt}-\mu u_{xx}-b\phi _x=0, &{} ~~~~~ \text{ in } ~(0,1)\times (0,+\infty )\\ J \phi _{tt}-\delta \phi _{xx}+bu_{x}+\xi \phi +\beta \theta _{tx}=0,&{} ~~~~~\text{ in }~(0,1) \times (0,+\infty )\\ \alpha \theta _{tt}-\delta \theta _{xx}+\beta \phi _{tx}-k\theta _{txx}=0, &{} ~~~~~\text{ in }~(0,1) \times (0,+\infty ) \end{array} \right. \end{aligned}$$
(2.1)

with the following boundary conditions

$$\begin{aligned} u(0,t)=u(1,t)=\phi _x(0,t)=\phi _x(1,t)=\theta (0,t)=\theta (1,t)=0,~~\forall ~t\ge 0 \end{aligned}$$
(2.2)

and initial conditions

$$\begin{aligned} \begin{aligned} \left\{ \begin{array}{ll} u(x,0)=u_0(x),~~u_t(x,0)=u_1(x), &{}{}~~~~~x \in (0,1) \\ \phi (x,0)=\phi _0(x),~~\phi _t(x,0)=\phi _1(x), &{}{}~~~~~x\in (0,1)\\ \theta (x,0)=\theta _0(x),~~\theta _t(x,0)=\theta _1(x), &{}{}~~~~~x \in (0,1). \end{array} \right. \end{aligned} \end{aligned}$$
(2.3)

Since the boundary conditions on \(\phi \) are of Newmann type, we introduce some transformation that allows the use of Poincaré’s inequality on \(\phi \). From the second equation in (2.1) and the boundary conditions (2.2), it follows that

$$\begin{aligned} \dfrac{\mathrm{{d}}^2}{\mathrm{{d}}t^2}\int _{0}^{1}\phi (x,t) \mathrm{{d}}x + \dfrac{\xi }{J}\int _{0}^{1} \phi (x,t) \mathrm{{d}}x=0. \end{aligned}$$
(2.4)

So, by solving (2.4) and using the initial data of \(\phi \), we obtain

$$\begin{aligned} \int _{0}^{1}\phi (x,t) \mathrm{{d}}x=\left( \int _{0}^{1}\phi _0(x) \mathrm{{d}}x\right) \mathrm{{cos}}\left( \sqrt{\frac{\xi }{J}}t\right) +\sqrt{\dfrac{J}{\xi }}\left( \int _{0}^{1}\phi _1(x) \mathrm{{d}}x\right) \mathrm{{sin}}\left( \sqrt{\dfrac{\xi }{J}}t\right) . \end{aligned}$$
(2.5)

Consequently, if we let

$$\begin{aligned} \bar{\phi }(x,t)=\phi (x,t)-\left( \int _{0}^{1}\phi _0(x) \mathrm{{d}}x\right) \mathrm{{cos}}\left( \sqrt{\frac{\xi }{J}}t\right) -\sqrt{\dfrac{J}{\xi }}\left( \int _{0}^{1}\phi _1(x) \mathrm{{d}}x\right) \mathrm{{sin}}\left( \sqrt{\dfrac{\xi }{J}}t\right) ,\end{aligned}$$

we get

$$\begin{aligned} \int _{0}^{1}\bar{\phi }(x,t)\mathrm{{d}}x=0,\ \forall t\ge 0,\end{aligned}$$

which allows the use of Poincaré’s inequality on \(\bar{\phi }\). So, \((u,\bar{\phi },\theta )\) satisfies (2.1), (2.2) and similar initial conditions (2.3). Therefore, we work with \((u,\bar{\phi },\theta )\) but we write \((u,{\phi },\theta )\) for simplicity.

3 The well-posedness of the problem

In this section, we prove the existence, uniqueness and smoothness of solutions for the system (2.1)–(2.3) using the semigroup theory. Introducing the vector function \(U=(u,v,\phi ,\psi ,\theta ,q)^T\), where \(v=u_t\), \(\psi =\phi _{t}\) and \(q=\theta _t\). System (2.1)–(2.3) can be written as

$$\begin{aligned} \left\{ \begin{array}{ll} U'(t)=\mathcal {A}U(t), &{} t>0\\ U(0)=U_0, \end{array} \right. \end{aligned}$$
(3.1)

where \(U_0=(u_0,u_1,\phi _0,\phi _1,\theta _0,\theta _1)^\mathrm{{T}}\) and the operator \(\mathcal {A}\) is defined by

$$\begin{aligned} \mathcal {A}U=\left( \begin{array}{l} ~~~~~v\\ \dfrac{1}{\rho }(\mu u_{xx}+b\phi _x)\\ ~~~~~\psi \\ \dfrac{1}{J}(\delta \phi _{xx}-bu_{x}-\xi \phi -\beta q_{x})\\ ~~~~~q\\ \dfrac{1}{\alpha }(\delta \theta _{xx}-\beta \psi _{x}+kq_{xx}) \end{array} \right) \end{aligned}$$
(3.2)

We consider the energy space

$$\begin{aligned} \mathcal {H}=H_0^1(0,1)\times L^2(0,1)\times H_*^1(0,1)\times L_*^2(0,1)\times H_0^1(0,1)\times L^2(0,1), \end{aligned}$$

where

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

\(\mathcal {H}\) is a Hilbert space with respect to the following inner product

$$\begin{aligned} (U,\tilde{U})_\mathcal {H}:=&{} \rho \int _{0}^{1}v \tilde{v}\mathrm {{d}}x+\xi \int _{0}^{1}\phi \tilde{\phi }\mathrm {{d}}x+ J\int _{0}^{1}\psi \tilde{\psi }\mathrm {{d}}x+\alpha \int _{0}^{1}q \tilde{q}\mathrm {{d}}x \nonumber \\ {}&+\mu \int _{0}^{1}u_x \tilde{u}_x\mathrm {{d}}x+\delta \int _{0}^{1}\left( \phi _x \tilde{\phi }_x+\theta _x \tilde{\theta }_x\right) \mathrm {{d}}x +b\int _{0}^{1}(u_x \tilde{\phi }\ +\phi \tilde{u}_x)\mathrm {{d}}x. \end{aligned}$$
(3.3)

Remark 3.1

Under the hypothesis \(\mu \xi > b^2\), it is easy to see that (3.3) defines an inner product. In fact, from (3.3), we have

$$\begin{aligned} \Vert U\Vert ^2_\mathcal {H}= & {} (U,U)_\mathcal {H}=\rho \int _0^1 v^2\mathrm{{d}}x+\left( \xi -\dfrac{b^2}{\mu }\right) \int _0^1 \phi ^2\mathrm{{d}}x+ J\int _0^1 \psi ^2\mathrm{{d}}x+\alpha \int _0^1 q^2\mathrm{{d}}x\\&+\mu \int _0^1\left( u_x+\dfrac{b}{\mu }\phi \right) ^2\mathrm{{d}}x+\delta \int _0^1 \left( \phi _x^2+ \theta _x^2\right) \mathrm{{d}}x.\end{aligned}$$

Hence, since \(\mu \xi > b^2\) we conclude that \((U,\tilde{U})_\mathcal {H}\) defines an inner product on \(\mathcal {H}\) and the associated norm \(\parallel \cdot \parallel _\mathcal {H}\) is equivalent to the usual one.

The domain of \(\mathcal {A}\) is given by

$$\begin{aligned}&D(\mathcal {A})=\left\{ U\in \mathcal {H}:~u\in H^2(0,1)\cap H^1_0(0,1),~~v\in H^1_0(0,1),~~\phi \in H^2_*(0,1)\cap H^1_*(0,1),\right. \\&\quad \left. \psi \in H^1_*(0,1),~~q\in H^1_0(0,1) ,~~(\delta \theta +kq)\in H^2(0,1)\right\} , \end{aligned}$$

where

$$\begin{aligned}H^2_*(0,1)=\{u\in H^2(0,1):~ u_x(0)=u_x(1)=0\}. \end{aligned}$$

We have the following well-posedness result:

Theorem 3.2

Let \(U_0\in \mathcal {H}\). Then, there exists a unique solution \(U\in C(\mathbb {R}_+,\mathcal {H})\) of problem (2.1)–(2.3). Moreover, if \(U_0\in D(\mathcal {A})\), then \(U\in C(\mathbb {R}_+,D(\mathcal {A}))\cap C^1(\mathbb {R}_+,\mathcal {H})\).

Proof

The result follows from Lumer–Phillips theorem (see [19, 23]) provided we prove that \(\mathcal {A}\) is a maximal dissipative operator, that is \(\mathcal {A}\) is dissipative and that \((I-\mathcal {A})\) is surjective. Thus, for any \(U\in D(\mathcal {A})\), we have

$$\begin{aligned} (\mathcal {A}U,U)_\mathcal {H}= & {} \mu \int _{0}^{1}u_{xx} v\mathrm{{d}}x+b\int _0^1\phi _{x} v\mathrm{{d}}x+\xi \int _{0}^{1}\psi \phi \mathrm{{d}}x+\delta \int _{0}^{1}\phi _{xx} \psi \mathrm{{d}}x-b\int _{0}^{1}u_x \psi \mathrm{{d}}x\\&-\xi \int _{0}^{1} \phi \psi \mathrm{{d}}x-\beta \int _{0}^{1}q_x \psi \mathrm{{d}}x +\delta \int _{0}^{1}\theta _{xx} q\mathrm{{d}}x-\beta \int _{0}^{1}\psi _x q\mathrm{{d}}x+k\int _{0}^{1}q_{xx} q\mathrm{{d}}x\\&+\mu \int _{0}^{1}u_x v_x\mathrm{{d}}x+\delta \int _{0}^{1}\psi _{x} \phi _x\mathrm{{d}}x +\delta \int _{0}^{1}\theta _x q_x\mathrm{{d}}x+b\int _{0}^{1}v_x \phi \mathrm{{d}}x + b \int _{0}^{1} \psi u_x \mathrm{{d}}x\\= & {} -k\int _{0}^{1}q^2_x \mathrm{{d}}x\le 0. \end{aligned}$$

So, \(\mathcal {A}\) is dissipative. Next, we prove that the operator \((I-\mathcal {A})\) is surjective.

Let \(F=(f^1,f^2,f^3,f^4,f^5,f^6)^\mathrm{{T}} \in \mathcal {H}\), we prove that there exists a unique \(U\in D(\mathcal {A})\) satisfying

$$\begin{aligned} (I-\mathcal {A})U=F. \end{aligned}$$
(3.4)

That is,

$$\begin{aligned} \left\{ \begin{array}{l} u-v=f^1\\ \rho v-\mu u_{xx}-b\phi _x=\rho f^2\\ \phi -\psi =f^3\\ J\psi -\delta \phi _{xx}+bu_x+\xi \phi +\beta q_x=Jf^4\\ \theta -q=f^5\\ \alpha q-\delta \theta _{xx}+\beta \psi _{x}-kq_{xx}=\alpha f^6. \end{array} \right. \end{aligned}$$
(3.5)

Using Eqs. (3.5)\(_1\), (3.5)\(_3\), (3.5)\(_5\) in (3.5)\(_2\), (3.5)\(_4\), (3.5)\(_6\), respectively, we obtain

$$\begin{aligned} \left\{ \begin{array}{l} \rho u-\mu u_{xx}-b\phi _x=\rho (f^2+f^1)\\ J\phi -\delta \phi _{xx}+bu_x+\xi \phi +\beta \theta _x =J(f^4+f^3)+\beta f^5_x\\ \alpha \theta -(\delta +k)\theta _{xx}+ \beta \phi _x =\alpha (f^6+f^5)+\beta f^3_x-k f^5_{xx}. \end{array} \right. \end{aligned}$$
(3.6)

To solve (3.6), we consider the following variational formulation

$$\begin{aligned} B\left( (u,\phi ,\theta ), (\tilde{u},\tilde{\phi },\tilde{\theta })\right) =L\left( (\tilde{u},\tilde{\phi },\tilde{\theta })\right) ,~~~~~\forall ~(\tilde{u},\tilde{\phi },\tilde{\theta })\in \mathcal {W}, \end{aligned}$$
(3.7)

where \(\mathcal {W}=H^1_0(0,1)\times H^1_*(0,1)\times H^1_0(0,1)\), \(B:~\mathcal {W}\times \mathcal {W}\longrightarrow \mathbb {R}\) is the bilinear form defined by

$$\begin{aligned}&\rho \int _{0}^{1} u \tilde{u}\mathrm{{d}}x+\mu \int _{0}^{1} u_{x} \tilde{u}_x\mathrm{{d}}x -b\int _{0}^{1}\phi _x \tilde{u}\mathrm{{d}}x+ (J+\xi )\int _{0}^{1}\phi \tilde{\phi }\mathrm{{d}}x +\delta \int _{0}^{1}\phi _{x}\tilde{\phi }_x\mathrm{{d}}x +b\int _{0}^{1}u_x \tilde{\phi }\mathrm{{d}}x\\&\qquad +\beta \int _{0}^{1}(\theta _x \tilde{\phi }+\phi _x \tilde{\theta })\mathrm{{d}}x+ \alpha \int _{0}^{1}\theta \tilde{\theta }\mathrm{{d}}x +(\delta +k)\int _{0}^{1}\theta _{x} \tilde{\theta }_x \mathrm{{d}}x \end{aligned}$$

and \(L:~\mathcal {W}\longrightarrow \mathbb {R}\) is the linear form given by

$$\begin{aligned}\rho \int _{0}^{1}(f^2+f^1) \tilde{u}\mathrm{{d}}x+ \int _{0}^{1}\left( J(f^4+f^3)+\beta f^5_x\right) \tilde{\phi }\mathrm{{d}}x +\int _{0}^{1}\left( \alpha (f^6+f^5)+\beta f^3_x\right) \tilde{\theta }\mathrm{{d}}x+k\int _{0}^{1} f^5_{x}\tilde{\theta }_x\mathrm{{d}}x.\end{aligned}$$

It is clear that \(\mathcal {W}\) is a Hilbert space with the usual norm and we can easily show, using Cauchy–Schwarz inequality, that B and L are continuous. On the other hand, using Young’s inequality and the fact \(\mu \xi >b^2\), we have

$$\begin{aligned} B\left( (u,\phi ,\theta ), (u,\phi ,\theta )\right)\ge & {} \rho \Vert u\Vert ^2+\left( \mu -\dfrac{b^2}{\xi }\right) \Vert u_x\Vert ^2+J\Vert \phi \Vert ^2+\delta \Vert \phi _{x}\Vert ^2+ \alpha \Vert \theta \Vert ^2+(k+\delta )\Vert \theta _{x}\Vert ^2\\\ge & {} c \Vert (u,\phi ,\theta )\Vert ^2_\mathcal {W}, \end{aligned}$$

for some \(c>0\). Hence, B is coercive. Consequently, Lax–Milgram lemma guarantees the existence of a unique \((u,\phi ,\theta )\) in \(\mathcal {W}\) satisfying (3.7). Using (3.5), we have

$$\begin{aligned}v=u-f^1\in H^1_0,~ \psi =\phi -f^3\in H^1_*,~ q=\theta -f^5\in H^1_0.\end{aligned}$$

\(\bullet \) If we take \((\tilde{\phi },\tilde{\theta }) =(0,0)\) in (3.7), we get

$$\begin{aligned}\mu \int _{0}^{1}u_x \tilde{u}_x\mathrm{{d}}x=\int _{0}^{1}\left[ \rho (f^1+f^2-u)+b\phi _x\right] \tilde{u}\mathrm{{d}}x,\;\forall \tilde{u}\in H^1_0(0,1).\end{aligned}$$

Thus, the elliptic regularity theory implies that

$$\begin{aligned}u\in H^2(0,1)\end{aligned}$$

and, moreover, we obtain

$$\begin{aligned}\rho u-\mu u_{xx} -b\phi _x= \rho (f^1+f^2).\end{aligned}$$

Since \(f^1=u-v\), then

$$\begin{aligned}\rho v-\mu u_{xx} -b\phi _x= \rho f^2\end{aligned}$$

which solves (3.5)\(_2\).

\(\bullet \) If \((\tilde{u},\tilde{\theta }) =(0,0)\) in (3.7), then we have

$$\begin{aligned} \delta \int _{0}^{1}\phi _x \tilde{\phi }_x\mathrm{{d}}x=\int _{0}^{1}\left[ J(f^3+f^4)+\beta f^5_x-(J+\xi )\phi -bu_x-\beta \theta _x\right] \tilde{\phi }\mathrm{{d}}x,\;\forall \tilde{\phi }\in H^1_*(0,1). \end{aligned}$$
(3.8)

Here, we can’t use the regularity theorem, because \(\tilde{\phi }\in H^1_*(0,1)\). So, we take \(\tilde{\Psi }\in H^1_0(0,1)\) and set

$$\begin{aligned}\tilde{\phi }(x)=\tilde{\Psi }(x)-\int _{0}^1\tilde{\Psi }(x)\mathrm{{d}}x.\end{aligned}$$

It is clear that \(\tilde{\phi }\in H^1_*(0,1)\). Then, a substitution in (3.8) leads to

$$\begin{aligned}\delta \int _{0}^{1}\phi _x \tilde{\Psi }_x\mathrm{{d}}x=\int _{0}^{1}r\tilde{\Psi }\mathrm{{d}}x,~~\forall \tilde{\Psi }\in H^1_0(0,1),\end{aligned}$$

where

$$\begin{aligned}r= J(f^3+f^4)+\beta f^5_x-(J+\xi )\phi -bu_x-\beta \theta _x \in L^2_*(0,1).\end{aligned}$$

So

$$\begin{aligned} \phi \in H^2(0,1)\end{aligned}$$

and

$$\begin{aligned}-\delta \phi _{xx} = J(f^3+f^4)+\beta f^5_x-(J+\xi )\phi -bu_x-\beta \theta _x.\end{aligned}$$

We use \(f^3=\phi -\psi \) and \(f^5=\theta -q\) to obtain

$$\begin{aligned}J\psi -\delta \phi _{xx}+bu_x+\xi \phi +\beta q_x=Jf^4.\end{aligned}$$

This gives (3.5)\(_4\). Since \(-\delta \phi _{xx}=r(x)\), then

$$\begin{aligned}-\delta \int _{0}^{1}\phi _{xx} \Phi \mathrm{{d}}x=\int _{0}^{1}r\Phi \mathrm{{d}}x,~~\forall \Phi \in H^1(0,1).\end{aligned}$$

Namely,

$$\begin{aligned}-\delta \phi _x \Phi |^1_0+\delta \int _{0}^{1}\phi _{x} \Phi _x \mathrm{{d}}x=\int _{0}^{1}r\Phi \mathrm{{d}}x,~~\forall \Phi \in H^1(0,1).\end{aligned}$$

Since \(H^1_*\subset H^1\). Then, we have

$$\begin{aligned}-\delta \phi _x \tilde{\phi }|^1_0+\delta \int _{0}^{1}\phi _{x} \tilde{\phi }_x \mathrm{{d}}x=\int _{0}^{1}r\tilde{\phi } \mathrm{{d}}x,~~\forall \tilde{\phi }\in H^1_*(0,1),\end{aligned}$$

and the other hand, we have (3.8). Thus

$$\begin{aligned}\phi _x(1) \tilde{\phi }(1)-\phi _x(0) \tilde{\phi }(0)=0,~~\forall \tilde{\phi }\in H^1_*(0,1).\end{aligned}$$

Since \(\tilde{\phi }\in H^1_*\) is arbitrary. Then,

$$\begin{aligned}\phi _x(1)=\phi _x(0)=0,\end{aligned}$$

and, hence,

$$\begin{aligned}\phi \in H^2_*(0,1).\end{aligned}$$

\(\bullet \) If \((\tilde{u},\tilde{\phi }) =(0,0)\) in (3.7) we get, for any \(\tilde{\theta }\in H^1_0(0,1)\),

$$\begin{aligned}\alpha \int _{0}^{1}\theta \tilde{\theta }\mathrm{{d}}x +(\delta +k)\int _{0}^{1}\theta _{x} \tilde{\theta }_x\mathrm{{d}}x+ \beta \int _{0}^{1}\phi _x \tilde{\theta }\mathrm{{d}}x -k\int _{0}^{1} f^5_{x}\tilde{\theta }_x\mathrm{{d}}x=\int _{0}^{1}\left[ \alpha (f^6+f^5)+\beta f^3_x\right] \tilde{\theta }\mathrm{{d}}x.\end{aligned}$$

This, in turns, yields

$$\begin{aligned}\int _{0}^{1}\left[ (\delta +k)\theta _{x}-kf^5_x\right] \tilde{\theta }_x\mathrm{{d}}x=\int _{0}^1 R\tilde{\theta }\mathrm{{d}}x,~~\forall \tilde{\theta }\in H^1_0(0,1),\end{aligned}$$

where

$$\begin{aligned}R=\alpha (f^6+f^5)+\beta f^3_x-\alpha \theta -\beta \phi _{x}.\end{aligned}$$

Then,

$$\begin{aligned}\left[ (\delta +k)\theta -kf^5\right] \in H^2(0,1).\end{aligned}$$

Since \(f^5=\theta -q\), then \((\delta \theta +kq)\in H^2(0,1)\) and we have

$$\begin{aligned}\alpha q-\delta \theta _{xx}+\beta \psi _{x}-kq_{xx}=\alpha f^6\end{aligned}$$

which solves (3.5)\(_6\).

Hence, there exists a unique \(U\in D(\mathcal {A})\) satisfies (3.4). Finally, using Lumer-Phillips theorem we deduce that \(\mathcal {A}\) is an infinitesimal generator of a contraction semigroup in \(\mathcal {H}\) and this complete the proof. \(\square \)

4 Exponential stability

In this section, we use the energy method to prove that system (2.1)–(2.3) is exponentially stable in the case of equal wave-speed propagation (1.7). To achieve this goal, we first establish some technical lemmas needed in the proof of exponential stability result. We also use c to be a positive generic constant.

Lemma 4.1

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3). Then the energy functional E, defined by

$$\begin{aligned} E(t)=\dfrac{1}{2}\int _0^1\big [ \rho u_t^2+J\phi _t^2+\alpha \theta _t^2+\mu u_x^2+\delta \phi _x^2\mathrm{{d}}x+\delta \theta _x^2+2bu_x\phi +\xi \phi ^2\big ] \mathrm{{d}}x \end{aligned}$$
(4.1)

satisfies

$$\begin{aligned} E^{'}(t)=-k\int _0^1 \theta _{tx}^{2}\mathrm{{d}}x\le 0. \end{aligned}$$
(4.2)

Proof

Multiplying (2.1) by \(u_t\), \(\phi _t\) and \(\theta _t\) respectively, integrating over (0, 1) and using integration by parts and the boundary conditions, we obtain

$$\begin{aligned} \dfrac{\rho }{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1 u_{t}^{2}\mathrm{{d}}x+\dfrac{\mu }{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1 u_{x}^2\mathrm{{d}}x+b\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1 \phi u_{x}\mathrm{{d}}x -b\int _0^1 \phi _{t} u_{x}\mathrm{{d}}x=0. \end{aligned}$$
(4.3)

The second equation

$$\begin{aligned} \dfrac{J}{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1 \phi _t^2\mathrm{{d}}x +\dfrac{\delta }{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t} \int _0^1 \phi _x^2\mathrm{{d}}x +b\int _0^1 u_x \phi _t\mathrm{{d}}x+\dfrac{\xi }{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1\phi ^2\mathrm{{d}}x+\beta \int ^1_0\theta _{tx}\phi _t\mathrm{{d}}x=0. \end{aligned}$$

The third equation

$$\begin{aligned} \dfrac{\alpha }{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1 \theta _t^2\mathrm{{d}}x+\dfrac{\delta }{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1\theta _x^2\mathrm{{d}}x-\beta \int _0^1\phi _t\theta _{tx}\mathrm{{d}}x+k\int _0^1\theta ^{2}_{tx}\mathrm{{d}}x=0. \end{aligned}$$

Adding up the above identities we arrive at

$$\begin{aligned}\dfrac{1}{2}\dfrac{\mathrm{{d}}}{\mathrm{{d}}t}\int _0^1\big [ \rho u_t^2+J\phi _t^2+\alpha \theta _t^2+\mu u_x^2+\delta \phi _x^2+\delta \theta _x^2+2b\phi u_x+\xi \phi ^2\big ] \mathrm{{d}}x=-k\int _0^1\theta _{tx}^{2}\mathrm{{d}}x.\end{aligned}$$

This is exactly (4.2).\(\square \)

Lemma 4.2

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3). Then the functional

$$\begin{aligned} F_1(t):=J\int _0^1\phi \phi _t\mathrm{{d}}x-\dfrac{\rho b}{\mu }\int _0^1 u_t \bigg (\int _0^x\phi (y)\mathrm{{d}}y\bigg )\mathrm{{d}}x \end{aligned}$$
(4.4)

satisfies, for any \(\varepsilon _1>0\), the estimate

$$\begin{aligned} F^{'}_1(t)\le -\dfrac{\delta }{2}\int _0^1 \phi _x^2\mathrm{{d}}x-\bigg (\xi -\dfrac{ b^2}{\mu }\bigg )\int _0^1\phi ^2\mathrm{{d}}x+\varepsilon _1\int _0^1u_t^2\mathrm{{d}}x+\dfrac{\beta ^2}{2\delta }\int _0^1\theta _t^2\mathrm{{d}}x+{\bigg (J+\dfrac{c}{\varepsilon _1}\bigg )}\int _0^1\phi _t^2\mathrm{{d}}x. \end{aligned}$$
(4.5)

Proof

By taking the derivative of \(F_1\), using (2.1) and integrating by parts, we get

$$\begin{aligned} F^{'}_1(t)= & {} J\int _0^1 \phi _t^2\mathrm{{d}}x+J \int _0^1\phi \phi _{tt}\mathrm{{d}}x-\dfrac{\rho b}{\mu }\int _0^1u_{tt}\bigg (\int _0^x\phi \mathrm{{d}}y\bigg )\mathrm{{d}}x-\dfrac{\rho b}{\mu }\int _0^1u_{t}\bigg (\int _0^x\phi _t\mathrm{{d}}y\bigg )\mathrm{{d}}x\\= & {} J\int _0^1 \phi _t^2\mathrm{{d}}x+\delta \int _0^1\phi \phi _{xx}\mathrm{{d}}x- b\int _0^1\phi u_x\mathrm{{d}}x-\xi \int _0^1\phi ^2\mathrm{{d}}x-\beta \int _0^1\phi \theta _{tx}\mathrm{{d}}x\\&-b\int _0^1u_{xx}\bigg ( \int _0^x \phi \mathrm{{d}}y\bigg )\mathrm{{d}}x-\dfrac{ b^2}{\mu }\int _0^1\phi _x\bigg (\int _0^x\phi (y)\mathrm{{d}}y\bigg )\mathrm{{d}}x -\dfrac{ b\rho }{\mu }\int _0^1u_t\bigg (\int _0^x\phi _t(y)\mathrm{{d}}y\bigg )\mathrm{{d}}x. \end{aligned}$$

We use integration by parts and \(\int _0^1\phi \mathrm{{d}}x=0\) to obtain

$$\begin{aligned} \int _0^1 u_{xx}\bigg (\int _0^x\phi (y)\mathrm{{d}}y\bigg )\mathrm{{d}}x=-\int _0^1 u_x\phi \mathrm{{d}}x \end{aligned}$$
(4.6)

and

$$\begin{aligned}\int _0^1 \phi _x\bigg (\int _0^x\phi (y)\mathrm{{d}}y\bigg )\mathrm{{d}}x=-\int _0^1 \phi ^2 \mathrm{{d}}x.\end{aligned}$$

So,

$$\begin{aligned} F^{'}_1(t)= & {} J\int _0^1 \phi _t^2\mathrm{{d}}x-\delta \int _0^1\phi _{x}^{2}\mathrm{{d}}x- b\int _0^1\phi u_x\mathrm{{d}}x-\xi \int _0^1\phi ^2\mathrm{{d}}x+\beta \int _0^1\phi _x\theta _{t}\mathrm{{d}}x +b\int _0^1 u_{x}\phi \mathrm{{d}}x \nonumber \\&+\dfrac{ b^2}{\mu } \int _0^1\phi ^2\mathrm{{d}}x -\dfrac{b\rho }{\mu }\int _0^1 u_t\bigg (\int _0^x\phi _t(y)\mathrm{{d}}y\bigg )\mathrm{{d}}x. \end{aligned}$$
(4.7)

Using Young’s and Cauchy–Schwarz inequalities, we have,

$$\begin{aligned}\beta \int _0^1\phi _x\theta _t\mathrm{{d}}x\le \dfrac{\delta }{2}\int _0^1\phi ^2_x\mathrm{{d}}x+\dfrac{\beta ^2}{2\delta }\int _0^1\theta _t^2\mathrm{{d}}x\end{aligned}$$

and, for any \(\varepsilon _1>0,\)

$$\begin{aligned} -\dfrac{b\rho }{\mu }\int _0^1u_t\bigg (\int _0^x\phi _t(y)\mathrm{{d}}y\bigg )\mathrm{{d}}x\le & {} \varepsilon _1\int _0^1u_t^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _1}\int _0^1 \bigg (\int _0^x\phi _t(y)\mathrm{{d}}y\bigg )^2\mathrm{{d}}x \nonumber \\\le & {} \varepsilon _1\int _0^1u_t^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _1}\int _0^1\phi _t^2\mathrm{{d}}x. \end{aligned}$$
(4.8)

Then, by substituting the above inequalities into (4.7), we get

$$\begin{aligned} F^{'}_1(t)\le & {} -\dfrac{\delta }{2}\int _0^1 \phi _x^2\mathrm{{d}}x-\left( \xi -\dfrac{b^2}{\mu }\right) \int _0^1 \phi ^2\mathrm{{d}}x+\varepsilon _1\int _0^1u_t^2\mathrm{{d}}x+\dfrac{\beta ^2}{2\delta }\int _0^1\theta _{t}^{2}\mathrm{{d}}x\\&+\bigg (J+\dfrac{c}{\varepsilon _1}\bigg )\int _0^1\phi _{t}^{2}\mathrm{{d}}x.\qquad \square \end{aligned}$$

Lemma 4.3

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3). Then the functional

$$\begin{aligned} F_2(t):=-\alpha \int _0^1\theta _t\left( \int _0^x\phi _t(y)\mathrm{{d}}y\right) \mathrm{{d}}x \end{aligned}$$
(4.9)

satisfies, for any \(\varepsilon _2>0\), the estimate

$$\begin{aligned} F^{'}_2(t)\le -\dfrac{\beta }{2}\int _0^1 \phi _t^2\mathrm{{d}}x+c\varepsilon _2\int _0^1\bigg (\phi _x^2+u_x^2 \bigg )\mathrm{{d}}x+c \bigg (1+\dfrac{1}{\varepsilon _2}\bigg ) \int _0^1\theta _{tx}^2 \mathrm{{d}}x+\dfrac{\delta ^2}{\beta }\int _0^1 \theta _x^2\mathrm{{d}}x. \end{aligned}$$
(4.10)

Proof

The differentiation of \(F_2\), using (2.1), integration by parts, and the boundary conditions (2.2), gives

$$\begin{aligned} F^{'}_2(t)= & {} -\alpha \int _0^1\theta _{tt}\bigg (\int _0^x\phi _t(y)\mathrm{{d}}y\bigg )\mathrm{{d}}x-\alpha \int _0^1\theta _{t}\int _0^x\phi _{tt}(y)\mathrm{{d}}y\mathrm{{d}}x\\= & {} -\delta \int _0^1\theta _{xx}\int _0^x\phi _t(y)\mathrm{{d}}y\mathrm{{d}}x+\beta \int _0^1\phi _{tx}\bigg (\int _0^x\phi _{t}(y)\mathrm{{d}}y\bigg )\mathrm{{d}}x\\&-k\int _0^1\theta _{txx}\bigg (\int _0^x\phi _{t}(y)\mathrm{{d}}y\bigg )\mathrm{{d}}x-\alpha \int _0^1\theta _{t}\bigg (\int _0^x\phi _{tt}(y)\mathrm{{d}}y\bigg )\mathrm{{d}}x.\\= & {} \delta \int _0^1\theta _{x}\phi _t\mathrm{{d}}x-\beta \int _0^1 \phi _t^2\mathrm{{d}}x+k \int _0^1\theta _{tx}\phi _t\mathrm{{d}}x-\alpha \int _0^1\theta _{t}\int _0^x\phi _{tt}(y)\mathrm{{d}}y\mathrm{{d}}x. \end{aligned}$$

Now, we estimate the terms in the right-hand side of the above identity. Using Young’s and Cauchy–Schwarz inequalities, (2.1), and calculations as in (4.8), we find,

$$\begin{aligned} \delta \int _0^1\theta _{x}\phi _{t}\mathrm{{d}}x\le & {} \dfrac{\beta }{4}\int _0^1\phi _{t}^{2}\mathrm{{d}}x+\dfrac{\delta ^2}{\beta }\int _0^1\theta _{x}^{2}\mathrm{{d}}x.\\ k\int _0^1\theta _{tx}\phi _{t}\mathrm{{d}}x\le & {} \dfrac{\beta }{4}\int _0^1\phi _{t}^{2}\mathrm{{d}}x+\dfrac{k^2}{\beta }\int _0^1\theta _{tx}^{2}\mathrm{{d}}x.\end{aligned}$$

and, for any \(\varepsilon _2>0,\)

$$\begin{aligned} -\alpha \int _0^1\theta _{t}\left( \int _0^x\phi _{tt}(y)\mathrm{{d}}y\right) \mathrm{{d}}x= & {} -\alpha \int _0^1\theta _{t}\left( \int _0^x \bigg (\dfrac{\delta }{J}\phi _{yy}-\dfrac{b}{J}u_y-\dfrac{\xi }{J}\phi -\dfrac{\beta }{J}\theta _{ty}\bigg )\mathrm{{d}}y\right) \mathrm{{d}}x \\= & {} -\dfrac{\alpha \delta }{J}\int _0^1\theta _{t}\phi _{x}\mathrm{{d}}x+\dfrac{\alpha b}{J}\int _0^1\theta _{t}u\mathrm{{d}}x +\dfrac{\alpha \xi }{J}\int _0^1\theta _{t}\bigg (\int _0^x\phi \mathrm{{d}}y\bigg )\mathrm{{d}}x+\dfrac{\alpha \beta }{J}\int _0^1\theta _t^2\mathrm{{d}}x \\\le & {} \varepsilon _2\int _0^1\phi _x^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _2}\int _0^1\theta _t^2\mathrm{{d}}x+\varepsilon _2\int _0^1u^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _2}\int _0^1\theta _t^2\mathrm{{d}}x\\&+\varepsilon _2\int _0^1\phi ^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _2}\int _0^1\theta _t^2\mathrm{{d}}x+\dfrac{\alpha \beta }{J}\int _0^1\theta _t^2\mathrm{{d}}x. \end{aligned}$$

So, by Poincaré’s inequality and the above estimate, we arrive at

$$\begin{aligned} F^{'}_2(t)\le & {} \dfrac{-\beta }{2}\int _0^1\phi _t^2\mathrm{{d}}x+\bigg (\dfrac{\alpha \beta }{J}+\dfrac{c}{\varepsilon _2}\bigg )\int _0^1\theta _{t}^{2}\mathrm{{d}}x+\varepsilon _2 \int _0^1\phi ^{2}\mathrm{{d}}x+\varepsilon _2 \int _0^1\phi _{x}^{2}\mathrm{{d}}x+\varepsilon _2 \int _0^1 u^{2}\mathrm{{d}}x \\&+\dfrac{k^2}{\beta }\int _0^1\theta _{tx}^{2}\mathrm{{d}}x+\dfrac{\delta ^2}{\beta }\int _0^1\theta _{x}^{2}\mathrm{{d}}x\\\le & {} \dfrac{-\beta }{2}\int _0^1\phi _t^2\mathrm{{d}}x+c\bigg (1+\dfrac{1}{\varepsilon _2}\bigg )\int _0^1\theta _{tx}^{2}\mathrm{{d}}x+\varepsilon _2c \int _0^1(\phi _{x}^{2}+u_x^{2})\mathrm{{d}}x+\dfrac{\delta ^2}{\beta }\int _0^1\theta _{x}^{2}\mathrm{{d}}x. \qquad \qquad \qquad \qquad \square \end{aligned}$$

Lemma 4.4

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3). Then the functional

$$\begin{aligned} F_3(t):=\dfrac{\mu }{\rho }\int _0^1\phi _tu_x\mathrm{{d}}x+\dfrac{\delta }{J}\int _0^1\phi _xu_t\mathrm{{d}}x \end{aligned}$$
(4.11)

satisfies, for some positive constant \(m_0\), the estimate

$$\begin{aligned} F^{'}_3(t)\le -m_0\int _0^1 u_x^2\mathrm{{d}}x+c\int _0^1\phi ^2\mathrm{{d}}x+c\int _0^1\theta ^2_{tx}\mathrm{{d}}x+ \dfrac{\delta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x. \end{aligned}$$
(4.12)

Proof

Direct computations, exploiting \(\chi =0\) (\(\chi =0\) defined by (2.7)) and using (2.1) and integration by parts, yield

$$\begin{aligned} F^{'}_3(t)= & {} \dfrac{\mu }{\rho }\int _0^1\phi _{tt}u_x\mathrm{{d}}x+\dfrac{\mu }{\rho }\int _0^1\phi _{t}u_{tx}\mathrm{{d}}x+\dfrac{\delta }{J}\int _0^1\phi _{tx}u_t\mathrm{{d}}x +\dfrac{\delta }{J}\int _0^1\phi _xu_{tt}\mathrm{{d}}x \nonumber \\= & {} \dfrac{\mu \delta }{\rho J}\int _0^1\phi _{xx}u_x\mathrm{{d}}x-\dfrac{b\mu }{\rho J}\int _0^1u_x^2\mathrm{{d}}x-\dfrac{\mu \xi }{\rho J}\int _0^1\phi u_{x}\mathrm{{d}}x-\dfrac{\beta \mu }{\rho J}\int _0^1 \theta _{tx} u_x\mathrm{{d}}x-\chi \int _0^1\phi _{tx}u_{t}\mathrm{{d}}x \nonumber \\&+\dfrac{\delta \mu }{J\rho }\int _0^1\phi _{x}u_{xx}\mathrm{{d}}x +\dfrac{\delta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x \nonumber \\= & {} \dfrac{-b\mu }{\rho J}\int _0^1u_x^2\mathrm{{d}}x-\dfrac{\mu \xi }{\rho J}\int _0^1\phi u_{x}\mathrm{{d}}x-\dfrac{\beta \mu }{\rho J}\int _0^1 \theta _{tx} u_x\mathrm{{d}}x +\dfrac{\delta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x-\chi \int _0^1\phi _{tx}u_{t} \nonumber \\= & {} \dfrac{-b\mu }{\rho J}\int _0^1u_x^2\mathrm{{d}}x-\dfrac{\mu \xi }{\rho J}\int _0^1\phi u_{x}\mathrm{{d}}x-\dfrac{\beta \mu }{\rho J}\int _0^1 \theta _{tx} u_x\mathrm{{d}}x +\dfrac{\delta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x. \end{aligned}$$
(4.13)

Using Young’s inequality, we get, for any \(\varepsilon _3>0\),

$$\begin{aligned} F^{'}_3(t)\le & {} \dfrac{-b\mu }{\rho J}\int _0^1u_x^2\mathrm{{d}}x+\dfrac{\varepsilon _3}{2}\int _0^1u_x^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _3}\int _0^1\phi ^2\mathrm{{d}}x\\&+\dfrac{\varepsilon _3}{2}\int _0^1u_x^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _3}\int _0^1\theta _{tx}^2\mathrm{{d}}x +\dfrac{\delta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x\\\le & {} -\left( \dfrac{b\mu }{\rho J}-\varepsilon _3\right) \int _0^1u_x^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _3}\int _0^1\phi ^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _3}\int _0^1\theta _{tx}^2\mathrm{{d}}x+\dfrac{\delta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x. \end{aligned}$$

Thus, by taking \(\varepsilon _3\) small enough such that

$$\begin{aligned}m_0=\bigg (\dfrac{b\mu }{\rho J}-\varepsilon _3\bigg )>0,\end{aligned}$$

we obtain (4.12).\(\square \)

Lemma 4.5

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3). Then the functional

$$\begin{aligned} F_4(t):=-\rho \int _0^1u_tu\mathrm{{d}}x, \end{aligned}$$
(4.14)

satisfies

$$\begin{aligned} F^{'}_4(t)\le -\rho \int _0^1 u_t^2\mathrm{{d}}x+\dfrac{3\mu }{2}\int _0^1u_x^2\mathrm{{d}}x+c\int _0^1\phi ^2_{x}\mathrm{{d}}x. \end{aligned}$$
(4.15)

Proof

A differentiation of \(F_4\), using (2.1) and integrating by parts and (2.2), gives

$$\begin{aligned} F^{'}_4(t)= & {} -\rho \int _0^1u_{tt}u\mathrm{{d}}x-\rho \int _0^1u_t^2\mathrm{{d}}x\\= & {} -\mu \int _0^1u_{xx}u\mathrm{{d}}x-b\int _0^1\phi _x u\mathrm{{d}}x-\rho \int _0^1u_t^2\mathrm{{d}}x\\= & {} \mu \int _0^1u_{x}^{2}\mathrm{{d}}x+b\int _0^1\phi u_x\mathrm{{d}}x-\rho \int _0^1u_t^2\mathrm{{d}}x. \end{aligned}$$

Then use of Young’s and Poincaré’s inequalities leads to

$$\begin{aligned} \qquad \qquad \qquad F^{'}_4(t)\le & {} -\rho \int _0^1 u_t^2\mathrm{{d}}x+\mu \int _0^1 u_x^2\mathrm{{d}}x+\dfrac{\mu }{2} \int _0^1 u_x^2\mathrm{{d}}x+\dfrac{b}{2\mu }\int _0^1\phi ^2\mathrm{{d}}x\\\le & {} -\rho \int _0^1 u_t^2\mathrm{{d}}x+\dfrac{3\mu }{2} \int _0^1 u_x^2\mathrm{{d}}x+c\int _0^1\phi ^2_x\mathrm{{d}}x. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \square \end{aligned}$$

Lemma 4.6

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3). Then the functional

$$\begin{aligned} F_5(t):=\alpha \int _0^1\theta \theta _t\mathrm{{d}}x+\dfrac{k}{2}\int _0^1\theta _x^2\mathrm{{d}}x+\beta \int _0^1\phi _x\theta \mathrm{{d}}x \end{aligned}$$
(4.16)

satisfies, for \(\varepsilon _2>0\),

$$\begin{aligned} F^{'}_5(t)\le -\delta \int _0^1 \theta _x^2\mathrm{{d}}x+\varepsilon _2\int _0^1\phi _x^2\mathrm{{d}}x+\bigg (\alpha +\dfrac{\beta ^2}{4\varepsilon _2}\bigg )\int _0^1 \theta _t^2\mathrm{{d}}x. \end{aligned}$$
(4.17)

Proof

A simple differentiation of \(F_5\), using (2.1), (2.2) and integrating by parts, leads to

$$\begin{aligned} F^{'}_5(t)= & {} \alpha \int _0^1\theta _t^2\mathrm{{d}}x+\alpha \int _0^1\theta \theta _{tt}\mathrm{{d}}x+k\int _0^1\theta _{tx}\theta _x\mathrm{{d}}x+\beta \int _0^1\phi _{tx}\theta \mathrm{{d}}x+\beta \int _0^1\phi _{x}\theta _{t}\mathrm{{d}}x\\= & {} \alpha \int _0^1\theta _t^2\mathrm{{d}}x+ \delta \int _0^1\theta _{xx}\theta \mathrm{{d}}x-\beta \int _0^1\phi _{tx}\theta \mathrm{{d}}x+k\int _0^1\theta _{txx}\theta \mathrm{{d}}x+k \int _0^1\theta _{tx}\theta _x\mathrm{{d}}x\\&+ \beta \int _0^1\phi _{tx}\theta \mathrm{{d}}x +\beta \int _0^1\phi _{x}\theta _{t}\mathrm{{d}}x\\= & {} \alpha \int _0^1\theta _t^2\mathrm{{d}}x-\delta \int _0^1\theta _x^2\mathrm{{d}}x+\beta \int _0^1\phi _{x}\theta _{t}\mathrm{{d}}x. \end{aligned}$$

Next, by Young’s inequality, we arrive at

$$\begin{aligned} \qquad \qquad \qquad \qquad \qquad F^{'}_5(t)\le & {} -\delta \int _0^1\theta _x^2\mathrm{{d}}x+ \alpha \int _0^1\theta _t^2\mathrm{{d}}x+\varepsilon _2\int _0^1\phi _x^2\mathrm{{d}}x +\dfrac{\beta ^2}{4\varepsilon _2}\int _0^1 \theta _t^2\mathrm{{d}}x \\\le & {} -\delta \int _0^1\theta _x^2\mathrm{{d}}x+\varepsilon _2\int _0^1\phi _x^2\mathrm{{d}}x +\bigg (\alpha +\dfrac{\beta ^2}{4\varepsilon _2}\bigg )\int _0^1 \theta _t^2\mathrm{{d}}x. \qquad \qquad \qquad \qquad \qquad \qquad \square \end{aligned}$$

Lemma 4.7

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3). Then, for \(N,N_1,N_2,N_3,N_5>0,\) to be chosen properly, the Lyapunov functional, defined by

$$\begin{aligned} \mathcal {L}(t):=NE(t)+N_1F_1(t)+N_2F_2(t)+N_3F_3(t)+F_4(t)+N_5F_5(t), \end{aligned}$$
(4.18)

satisfies, for N sufficiently large,

$$\begin{aligned} \mathcal {L}\sim E \end{aligned}$$
(4.19)

and the estimate

$$\begin{aligned} \mathcal {L}'(t)\le -\lambda \int _{0}^1(u^2_t+\phi _t^2+\theta _t^2+u_x^2+\phi _x^2+\theta _{x}^2+\phi ^2)\mathrm{{d}}x, \end{aligned}$$
(4.20)

where \(\lambda \) is a positive constant.

Proof

The equivalence (4.19) is a matter of a routine calculations. See, for instance [5].

To prove (4.20), we differentiate \(\mathcal {L}(t)\), and recall (4.2), (4.5), (4.10), (4.12), (4.15) and (4.17). So, we have

$$\begin{aligned} \mathcal {L}'(t)\le & {} -Nk\int _0^1 \theta _{tx}^{2}\mathrm{{d}}x-\dfrac{N_1\delta }{2}\int _0^1 \phi _x^2\mathrm{{d}}x-N_1\bigg (\xi -\dfrac{ b^2}{\mu }\bigg )\int _0^1\phi ^2\mathrm{{d}}x +N_1\varepsilon _1\int _0^1u_t^2\mathrm{{d}}x+N_1c\int _0^1\theta _t^2\mathrm{{d}}x\\&+N_1\bigg (J+\dfrac{c}{\varepsilon _1}\bigg )\int _0^1\phi _t^2\mathrm{{d}}x-N_2\dfrac{\beta }{2}\int _0^1 \phi _t^2\mathrm{{d}}x+N_2\varepsilon _2c\int _0^1\big (\phi _x^2+u_x^2 \big )\mathrm{{d}}x\\&+N_2c\bigg (1+\dfrac{1}{\varepsilon _2}\bigg ) \int _0^1\theta _{tx}^2\mathrm{{d}}x+\dfrac{\delta ^2}{\beta }N_2 \int _0^1\theta _{x}^2\mathrm{{d}}x-N_3m_0\int _0^1 u_x^2\mathrm{{d}}x+N_3c\int _0^1\phi ^2\mathrm{{d}}x\\&+N_3c\int _0^1\theta ^2_{tx}\mathrm{{d}}x+N_3\dfrac{\delta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x-\rho \int _0^1 u_t^2\mathrm{{d}}x+\dfrac{3\mu }{2}\int _0^1u_x^2\mathrm{{d}}x\\&+c\int _0^1\phi ^2_{x}\mathrm{{d}}x-\delta N_5\int _0^1 \theta _x^2\mathrm{{d}}x+N_5\varepsilon _2\int _0^1\phi _x^2\mathrm{{d}}x+\bigg (\alpha +\dfrac{\beta ^2}{4\varepsilon _2}\bigg )N_5\int _0^1 \theta _t^2\mathrm{{d}}x . \end{aligned}$$

We apply Poincaré’s inequality for \(\theta _t\) and take \(N_5=\dfrac{2\delta }{\beta }N_2\), to get

$$\begin{aligned} \mathcal {L}'(t)\le & {} -\left[ Nk-N_1c-N_2c\bigg (1+\dfrac{1}{\varepsilon _2}\bigg )-{N_3c}\right] \int _0^1 \theta _{tx}^{2}\mathrm{{d}}x\\&-\left[ \dfrac{N_1\delta }{2}-N_2c\varepsilon _2-N_3\dfrac{\delta b}{\rho J}-c \right] \int _0^1 \phi _x^2\mathrm{{d}}x-\left[ N_1\bigg (\xi -\dfrac{ b^2}{\mu }\bigg )-N_3c\right] \int _0^1\phi ^2\mathrm{{d}}x\\&-(\rho -N_1\varepsilon _1)\int _0^1 u_t^2\mathrm{{d}}x-\left[ N_2\dfrac{\beta }{2}-N_1\bigg (J+\dfrac{c}{\varepsilon _1}\bigg )\right] \int _0^1 \phi _t^2\mathrm{{d}}x -\dfrac{\delta ^2}{\beta }N_2\int _0^1\theta _{x}^2\mathrm{{d}}x\\&-\bigg (N_3m_0-N_2c\varepsilon _2-\dfrac{3\mu }{2}\bigg )\int _0^1 u_x^2\mathrm{{d}}x. \end{aligned}$$

At this point, we choose the constants carefully. First, let us take \(\varepsilon _1=\dfrac{\rho }{2N_1}\), and choose \(N_3\) large enough such that

$$\begin{aligned}\alpha _1=N_3m_0-\frac{ 3\mu }{2}>0.\end{aligned}$$

We then choose \(N_1\) large enough so that

$$\begin{aligned}\alpha _2=N_1\bigg (\xi -\dfrac{b^2}{\mu }\bigg )-N_3c>0,~~\alpha _3=N_1\dfrac{\delta }{2}-\bigg (N_3\dfrac{\delta b}{J\rho }+c\bigg )>0.\end{aligned}$$

Next, we select \(N_2\) so large that

$$\begin{aligned}\alpha _4=N_2\dfrac{\beta }{2}-N_1\bigg (J+\dfrac{2cN_1}{\rho }\bigg )>0,\end{aligned}$$

then pick \(\varepsilon _2\) small enough so that

$$\begin{aligned}\alpha _5=\alpha _1-N_2c\varepsilon _2>0,~~\alpha _6=\alpha _3-N_2c\varepsilon _2>0.\end{aligned}$$

Finally, we choose N large enough so that (4.19) remains valid and, further,

$$\begin{aligned}\alpha _7=Nk-N_1c-N_2c\bigg (1+\frac{1}{\varepsilon _2}\bigg )-N_3c>0.\end{aligned}$$

Therefore, we arrive at

$$\begin{aligned} \mathcal {L}'(t)\le & {} - \alpha _7\int _0^1\theta _{tx}^2\mathrm{{d}}x- \alpha _6\int _0^1\phi _{x}^2\mathrm{{d}}x-\alpha _2\int _0^1\phi ^2\mathrm{{d}}x- \dfrac{\rho }{2}\int _0^1u_{t}^2\mathrm{{d}}x\\&- \alpha _4\int _0^1\phi _{t}^2\mathrm{{d}}x- \alpha _5\int _0^1u_{x}^2\mathrm{{d}}x- c\int _0^1\theta _{x}^2\mathrm{{d}}x.\end{aligned}$$

We finally use Poincaré’s inequality to substitute \(- \int _0^1\theta _{xt}^2\mathrm{{d}}x\) by \(- \int _0^1\theta _{t}^2\mathrm{{d}}x\) and, hence, (4.20) is established.

\(\square \)

Theorem 4.8

Let \((u,\phi ,\theta )\) be the solution of (2.1)–(2.3) and assume (1.7). Then there exist two positive constants \(k_1\) and \(k_2\) such that the energy functional (4.1) satisfies

$$\begin{aligned} E(t)\le k_1e^{-k_2t}, \quad \forall t\ge 0. \end{aligned}$$
(4.21)

Proof

First, using Young’s inequality, (4.1) becomes

$$\begin{aligned} E(t)\le c\int _0^1\left[ u_t^2+\phi _{t}^2+\theta _{t}^2+u_x^2+\phi _{x}^2+\theta _{x}^2+\phi ^2\right] \mathrm{{d}}x. \end{aligned}$$
(4.22)

The combination of (4.20) and (4.22) gives

$$\begin{aligned}\mathcal {L}'(t)\le -c E(t), ~~\forall t\ge 0.\end{aligned}$$

Using \(\mathcal {L}\sim E\), we get

$$\begin{aligned}\mathcal {L}'(t)\le -k_2 \mathcal {L}(t), ~~\forall t\ge 0.\end{aligned}$$

A simple integration over (0, t) yields

$$\begin{aligned}\mathcal {L}(t)\le \mathcal {L}(0)e^{-k_2t}, ~~\forall t\ge 0.\end{aligned}$$

Consequently, (4.21) is established by recalling \(\mathcal {L}\sim E\). \(\square \)

5 Polynomial stability

In this section, we prove the polynomial decay result for the non-equal speed of propagation case, that is (1.7) does not holds. To establish our result, we work with the strong solution of (2.1)–(2.3) and define the second-order energy functional

$$\begin{aligned} E_2(t)=\dfrac{1}{2}\int _0^1\big [ \rho u_{tt}^2+J\phi _{tt}^2+\alpha \theta _{tt}^2+\mu u_{tx}^2+\delta \phi _{tx}^2+\delta \theta _{tx}^2+2bu_{tx}\phi _t+\xi \phi _t^2\big ]\mathrm{{d}}x. \end{aligned}$$
(5.1)

Similar calculations, as in Lemma 4.1, lead to

$$\begin{aligned} E_2'(t)=-k\int _0^1 \theta _{ttx}^{2}\mathrm{{d}}x\le 0. \end{aligned}$$
(5.2)

Lemma 5.1

Let \((u,\phi ,\theta )\) be the strong solution of (2.1)–(2.3). Then the functional

$$\begin{aligned} \tilde{F}_3(t):=\beta F_3(t)-\chi k\int _0^1 u_x\theta _{tx}\mathrm{{d}}x-\chi \delta \int _0^1u_x\theta _x\mathrm{{d}}x \end{aligned}$$
(5.3)

satisfies, for any \(\varepsilon _7>0\) and for some positive constant \(m_1\), the estimate

$$\begin{aligned} \tilde{F}_3'(t)\le -m_1\int _0^1 u_x^2\mathrm{{d}}x+c\int _0^1\phi ^2\mathrm{{d}}x+c_2\int _0^1\theta ^2_{tx}\mathrm{{d}}x+ \dfrac{\delta \beta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x+c_7\int _0^1\theta ^2_{ttx}\mathrm{{d}}x+\varepsilon _7\int _0^1 u_t^2\mathrm{{d}}x. \end{aligned}$$
(5.4)

Proof

A simple differentiation of (5.3) gives

$$\begin{aligned} \tilde{F}_3'(t)=\beta F_3'(t)-\chi k\int _0^1 u_{tx}\theta _{tx}\mathrm{{d}}x-\chi k\int _0^1 u_{x}\theta _{ttx}\mathrm{{d}}x-\chi \delta \int _0^1u_{tx}\theta _x\mathrm{{d}}x-\chi \delta \int _0^1u_{x}\theta _{tx}\mathrm{{d}}x. \end{aligned}$$
(5.5)

Using integration by parts for the second term in the right-hand of (5.5) and exploiting (2.1)\(_3\), we get

$$\begin{aligned} -\chi k\int _0^1 u_{tx}\theta _{tx}\mathrm{{d}}x=\chi k\int _0^1 u_{t}\theta _{txx}\mathrm{{d}}x=\alpha \chi \int _0^1 u_{t}\theta _{tt}\mathrm{{d}}x-\delta \chi \int _0^1 u_{t}\theta _{xx}\mathrm{{d}}x+\beta \chi \int _0^1 u_{t}\phi _{tx}\mathrm{{d}}x. \end{aligned}$$
(5.6)

Substituting (5.6) and (4.13) into (5.5), we obtain

$$\begin{aligned} \tilde{F}_3'(t)= & {} \dfrac{-b\beta \mu }{\rho J}\int _0^1u_x^2\mathrm{{d}}x-\dfrac{\mu \beta \xi }{\rho J}\int _0^1\phi u_{x}\mathrm{{d}}x-\dfrac{\beta ^2\mu }{\rho J}\int _0^1 \theta _{tx} u_x\mathrm{{d}}x +\dfrac{\delta \beta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x+\alpha \chi \int _0^1 u_{t}\theta _{tt}\mathrm{{d}}x\\&-\chi k\int _0^1 u_{x}\theta _{ttx}\mathrm{{d}}x-\chi \delta \int _0^1 u_{x}\theta _{tx}\mathrm{{d}}x.\end{aligned}$$

Using Young’s and Poincaré’s inequalities, we find

$$\begin{aligned} \tilde{F}_3'(t)\le & {} -\left( \dfrac{b\beta \mu }{\rho J}-\varepsilon _6\right) \int _0^1u_x^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _6}\int _0^1\phi ^2\mathrm{{d}}x+\dfrac{c}{\varepsilon _6}\int _0^1 \theta _{tx}^2\mathrm{{d}}x +\dfrac{\delta \beta b}{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x+\varepsilon _7 \int _0^1 u_{t}^2\mathrm{{d}}x\\&+c \bigg (\dfrac{1}{\varepsilon _6}+\dfrac{1}{\varepsilon _7}\bigg )\int _0^1 \theta _{ttx}^2\mathrm{{d}}x. \end{aligned}$$

Finally, we choose \(\varepsilon _6\) small enough such that

$$\begin{aligned}m_1= \dfrac{b\beta \mu }{\rho J}-\varepsilon _6 >0,\end{aligned}$$

to obtain (5.4).\(\square \)

Lemma 5.2

Let \((u,\phi ,\theta )\) be the strong solution of (2.1)–(2.3). Then the Lyapunov functional defined by

$$\begin{aligned} \tilde{ \mathcal {L}}(t):=N^*\left( E(t)+E_2(t)\right) +N_1^*F_1(t)+N^*_2F_2(t)+N^*_3\tilde{F}_3(t)+F_4(t)+N_5^* F_5(t) \end{aligned}$$
(5.7)

satisfies, for \(N^*\),\(N^*_1\),\(N^*_2\),\(N^*_3\),\(N_5^*>0\) to be chosen properly, and for a positive constant \(\lambda _1\),

$$\begin{aligned} \tilde{\mathcal {L}}'(t)\le -\lambda _1\int _{0}^1(u^2_t+\phi _t^2+\theta _t^2+u_x^2+\phi _x^2+\theta _{x}^2+\phi ^2)\mathrm{{d}}x. \end{aligned}$$
(5.8)

Proof

By exploiting (5.4) and the fact \(\mu \xi >b^2\) we get

$$\begin{aligned} \tilde{\mathcal {L}}'(t)\le & {} -N^*k\int _0^1 \theta _{tx}^{2}\mathrm{{d}}x-N^*k\int _0^1\theta _{ttx}^2\mathrm{{d}}x-\dfrac{N^*_1\delta }{2}\int _0^1 \phi _x^2\mathrm{{d}}x-N^*_1\bigg (\xi -\dfrac{ b^2}{\mu }\bigg )\int _0^1\phi ^2\mathrm{{d}}x\\&+N^*_1\varepsilon _1\int _0^1u_t^2\mathrm{{d}}x+N^*_1c\int _0^1\theta _t^2\mathrm{{d}}x+N^*_1\bigg (J+\dfrac{c}{\varepsilon _1}\bigg )\int _0^1\phi _t^2\mathrm{{d}}x-N^*_2\dfrac{\beta }{2}\int _0^1 \phi _t^2\mathrm{{d}}x+N_2^*\dfrac{\delta ^2}{\beta }\int _0^1 \theta _{x}^2\mathrm{{d}}x\\&+N^*_2c\varepsilon _2\int _0^1\bigg (\phi _x^2+u_x^2 \bigg )\mathrm{{d}}x+N^*_2c\bigg (1+\dfrac{1}{\varepsilon _2}\bigg ) \int _0^1\theta _{tx}^2\mathrm{{d}}x-N^*_3m_1\int _0^1 u_x^2\mathrm{{d}}x+N^*_3c\int _0^1\phi ^2\mathrm{{d}}x\\&+N^*_3c_2\int _0^1\theta ^2_{tx}\mathrm{{d}}x+N^*_3\dfrac{\delta b\beta }{\rho J}\int _0^1\phi _x^2\mathrm{{d}}x+N^*_3c_3\int _0^1\theta _{ttx}^2\mathrm{{d}}x+N^*_3\varepsilon _7\int _0^1 u_{t}^2\mathrm{{d}}x-\rho \int _0^1 u_t^2\mathrm{{d}}x\\&+\dfrac{3\mu }{2}\int _0^1u_x^2\mathrm{{d}}x+c\int _0^1\phi ^2_{x}\mathrm{{d}}x-N_5^*\delta \int _0^1 \theta _x^2\mathrm{{d}}x+N_5^*\varepsilon _2\int _0^1\phi _x^2\mathrm{{d}}x+N_5^*\bigg (\alpha +\dfrac{\beta ^2}{4\varepsilon _2}\bigg )\int _0^1 \theta _t^2\mathrm{{d}}x. \end{aligned}$$

We apply Poincaré’s inequality for \(\theta _t\) to get

$$\begin{aligned} \tilde{\mathcal {L}}'(t)\le & {} -\left[ N^*k-N^*_1c-N^*_2c\bigg (1+\dfrac{1}{\varepsilon _2}\bigg )-N^*_3c_2-N_5^*c\bigg (1+\dfrac{1}{\varepsilon _2}\bigg )\right] \int _0^1 \theta _{tx}^{2}\mathrm{{d}}x\\&-\left[ \dfrac{N^*_1\delta }{2}-N^*_2c\varepsilon _2-N^*_3\dfrac{\delta \beta b}{\rho J}-c-N_5^*\varepsilon _2 \right] \int _0^1 \phi _x^2\mathrm{{d}}x-\left[ N^*_1\bigg (\xi -\dfrac{ b^2}{\mu }\bigg )-N^*_3c\right] \int _0^1\phi ^2\mathrm{{d}}x\\&-(\rho -N^*_1\varepsilon _1-N_3^*\varepsilon _7)\int _0^1 u_t^2\mathrm{{d}}x-\left[ N^*_2\dfrac{\beta }{2}-N^*_1\bigg (J+\dfrac{c}{\varepsilon _1}\bigg )\right] \int _0^1 \phi _t^2\mathrm{{d}}x - \bigg (\delta N_5^*-\dfrac{\delta ^2}{\beta }N_2^*\bigg )\int _0^1\theta _{x}^2\mathrm{{d}}x\\&-\bigg (N^*_3m_1-N^*_2c\varepsilon _2-\dfrac{3\mu }{2}\bigg )\int _0^1 u_x^2\mathrm{{d}}x-(N^*k-N^*_3c_3)\int _0^1\theta _{ttx}^2\mathrm{{d}}x. \end{aligned}$$

Similarly to what we did with \(\mathcal {L}'\), we take \(\varepsilon _1=\dfrac{\rho }{4N_1^*}\), \( N_5^*=\dfrac{2\delta }{\beta }N_2^*\) and \(\varepsilon _7=\dfrac{\rho }{4N_3^*}\) and then choose \(N_3^*\) large enough such that

$$\begin{aligned}\alpha _1^*=N_3^*m_1-\frac{ 3\mu }{2}>0\end{aligned}$$

and select \(N_1^*\) large enough so that

$$\begin{aligned}\alpha _2^*=N_1^*\bigg (\xi -\dfrac{b^2}{\mu }\bigg )-N_3^*c>0\end{aligned}$$

and

$$\begin{aligned}\alpha _3^*=N_1^*\dfrac{\delta }{2}-\bigg (N_3^*\dfrac{\delta b\beta }{J\rho }+c\bigg )>0.\end{aligned}$$

Next we choose \(N_2^*\) so large that

$$\begin{aligned}\alpha _4^*=N_2^*\dfrac{\beta }{2}-N_1^*\bigg (J+\dfrac{4cN_1}{\rho }\bigg )>0,\end{aligned}$$

then pick \(\varepsilon _2\) small enough such that

$$\begin{aligned}\alpha _5^*=\alpha ^*_1-N_2^*c\varepsilon _2>0\end{aligned}$$

and

$$\begin{aligned}\alpha _6^*=\alpha _3^*-N_2^*c\varepsilon _2>0.\end{aligned}$$

Finally, we take \(N^*\) large enough such that

$$\begin{aligned}\alpha _7^*=N^*k-N_1^*c-N_2^*c\bigg (1+\frac{1}{\varepsilon _2}\bigg )-N_3^*c>0,\end{aligned}$$

and

$$\begin{aligned}\alpha _8^*=N^*k-N_3^*c_3>0.\end{aligned}$$

Therefore, using Poincaré’s inequality, we arrive at

$$\begin{aligned} \mathcal {L}'(t)\le & {} - \alpha _7^*c\int _0^1\theta _{t}^2\mathrm{{d}}x- \alpha _6^*\int _0^1\phi _{x}^2\mathrm{{d}}x-\alpha _2^*\int _0^1\phi ^2\mathrm{{d}}x- \dfrac{\rho }{2}\int _0^1u_{t}^2\mathrm{{d}}x- \alpha _4^*\int _0^1\phi _{t}^2\mathrm{{d}}x- \alpha _5^*\int _0^1u_{x}^2\mathrm{{d}}x\\&-c\int _0^1\theta _{x}^2\mathrm{{d}}x-\alpha _8^*\int _0^1\theta _{ttx}^2\mathrm{{d}}x.\end{aligned}$$

So, there exists \(\lambda _1>0\) such that

$$\begin{aligned}\tilde{\mathcal {L}}'(t)\le -\lambda _1\int _{0}^1(u^2_t+\phi _t^2+\theta _t^2+u_x^2+\phi _x^2+\theta _{x}^2+\phi ^2)\mathrm{{d}}x.\end{aligned}$$

Theorem 5.3

Let \((u,\phi ,\theta )\) be the strong solution of (2.1)–(2.3) and assume that (1.7) does not hold. Then there exists a positive constant \(k_2\), independent of t and the initial data, such that

$$\begin{aligned} E(t)\le \dfrac{k_2\left( E(0)+E_2(0)\right) }{t},~~\forall t>0. \end{aligned}$$
(5.9)

Proof

The combination of (4.22) and (5.8) gives

$$\begin{aligned}\tilde{ \mathcal {L}}'(t)\le -\dfrac{\lambda _1}{c} E(t), ~~\forall t> 0.\end{aligned}$$

We integrate the last inequality over (0, t), and recall that \(\tilde{ \mathcal {L}}'\) is non-increasing, we obtain

$$\begin{aligned}&\int _0^t E(s)\mathrm{{d}}s \le -\dfrac{c}{\lambda _1}\int _0^t \tilde{ \mathcal {L}}'(s)\mathrm{{d}}s,\\&\int _0^t E(s)\mathrm{{d}}s \le \dfrac{c}{\lambda _1}\tilde{ \mathcal {L}}(0),~~\forall t>0. \end{aligned}$$

Using the fact \(tE(t)\le \int _0^t E(s)\mathrm{{d}}s \), we find

$$\begin{aligned} E(t) \le \lambda _2\dfrac{\tilde{ \mathcal {L}}(0)}{t},~~\forall t>0.\end{aligned}$$

Consequently, there exists \(k_2\) positive such that

$$\begin{aligned} E(t) \le \dfrac{k_2\left( E(0)+E_2(0)\right) }{t},~~\forall t>0.\end{aligned}$$

This finishes the proof.\(\square \)

Remark 5.4

We note here that these results hold even for \(\mu \xi =b^2\). In this case, we have to redefine the energy as in [10] and adjust our calculations accordingly. In particular, when \(\mu =\xi =b\), our system reduces to Timoshenko system with thermoelasticity type III. This has been discussed and similar stability results have been established in [20, 21].

6 Numerical tests

To illustrate the theoretical results of this work, we present in this section two numerical tests. We solve the system (2.1) under the initial and boundary conditions (2.2), (2.3). The system is discritized using a second order finite difference method in time and space. For more stability, we implement the conservative scheme of Lax–Wendroff. for more details, we refer to our previous works [1, 11, 17]. We examine the following two tests:

  • TEST 1: Based on the result (4.21) of our Theorem, we examine the exponential decay of the energy (4.1) using the equality condition of the parameters \(\chi =0\), given by (1.7). Here, we take all parameters of the system (2.1) equal to 1.

  • TEST 2: In Test 2, we examine the polynomial decay of the energy (4.1) using the parameters condition \(\chi \ne 0\), where the parameters of the system (2.1) are taken as follows \(\mu =5; \rho =1; \delta =0.05; J=1\) and the remaining parameters are equal to 1.

To ensure the numerical stability of the implemented method and the executed code, we use \(\Delta {t}<<0.5\mathrm{{d}}x\) satisfying the stability condition according to the Courant–Friedrichs–Lewy (CFL) inequality, where dt represents the time step and \(\mathrm{{d}}x\) the spatial step. The spatial interval [0, 1] is subdivided into 200 subintervals and the temporal interval \([0,T_{e}]=[0,1]\) is deduced from the stability condition above. We run our code for 10,000 time steps using the following initial conditions:

$$\begin{aligned} \begin{aligned} u (x,0)=2\sin \left( \pi x\right) ;\ \phi (x,0)=2x\sin \left( \pi x\right) ;\ \theta (x,0)= \frac{1}{4}x(1-x)\quad \text{ in }\ [0,1]. \end{aligned} \end{aligned}$$
(6.1)

Under the same initial and boundary conditions mentioned above, we show in Fig. 1 the numerical results of the exponential decay case. Whereas we present in Fig. 2 the results obtained for the polynomial case. We show three cross section cuts for the numerical solution \((u ,\phi ,\theta )\) at \(x=0.25\), \(x=0.5\) and at \(x=0.75\). For all components of the solution, the decay behavior is clearly demonstrated for both experiments, the exponential and the polynomial decays. Moreover, it should be stressed that the graphical presentations are normalized to ensure a clear comparisons. Therefore, we can clearly compare the energy decay obtained in Test 1 and in Test 2. For this, see Fig. 3.

Fig. 1
figure 1

TEST 1: cross section cuts of the solution for the exponential decay

Full size image
Fig. 2
figure 2

TEST 2: cross section cuts of the solution for the polynomial decay

Full size image
Fig. 3
figure 3

Energy function for the exponential and polynomial decays

Full size image

Finally, we noticed that the case \(\chi =0\) ensures an exponential energy decay and, therefore, the decay of all components of the solution \((u ,\phi ,\theta )\). While the case \(\chi \ne 0\) ensures the polynomial decay. But for some special choices of the system parameters generating the damping speed, we could obtain an exponential-like decay of the energy and a damped waves similar to the exponential case.