1 Introduction

In later a long time, some so numerous mathematical researchers have considered pondering the asymptotic behavior of solutions to the equations proposed to study different flexible materials with voids [1,2,3,4], which have decent physical properties, are utilized broadly in engineering, such as vehicles, airplanes, expansive space structures and so on. Due to their broad applications, many researchers’ interest comes from the need to establish results concerning the existence and stabilization of elasticity problems.

In addition to the conventional elastic effects, materials with voids have a microstructure in which the mass at each place is calculated by multiplying the material matrix mass density by the volume fraction. Nunziato and Cowin [4] pioneered the latter concept in their groundbreaking work on elastic materials with voids. Iesan [5,6,7] and Iesan an Quintanilla [8] expanded the hypothesis by including temperature and microtemperatures [9,10,11].

According to our knowledge, evaluating the temporal decay in one-dimensional porous-elastic substances was pioneered with the aid of Quintanilla [12], where he proved that porous-viscosity becomes not robust sufficient to stabilize the system exponentially. Interestingly, Casas and Quintanilla [13] proved that the mixture of porous-viscosity and temperature additionally lacks exponential stability [14,15,16]. However, the identical authors [17] confirmed that the mixture of porous-viscosity and thermal effects (each temperature and microtemparatures) stabilized the system exponentially. Similarly, Magana and Quintanilla [18] proved that viscoelasticity collectively with microtemperatures produced exponential stability, while viscoelasticity collectively with temperature lacks exponential stability [19, 20].

It is natural to think that a porous-elastic system with dissipation due to only microtemperatures will lack exponential stability. However Apalara [21] establishes the contrary, he proved that a porous system with microtempearture decays exponentially if and only if \(\chi =0\) where \(\chi =\frac{\mu }{\rho }-\frac{\delta }{J}\), otherwise the system is polynomially stable [22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40].

The equations for a one-dimensional porous elastic system with microtempeartureare of the form

$$\left\{\begin{aligned}& \rho {u}_{tt}={T}_{x},\\ & J{\phi }_{tt}={H}_{x}+G,\\ & \rho {E}_{t}={P}_{x}+q-Q,\end{aligned}\right.$$
(1.1)

where \((x,t)\in (0,l)\times (0,\infty )\), \(t\) is the time, \(x\) is the distance along the center line of the beam structure and \(l\) is the length of the beam, \(T\) is the stress, \(H\) is the equilibrated stress, \(G\) is the equilibrated body force, \(q\) is the heat flux vector, \(P\) is the first heat flux moment, \(Q\) is the mean heat flux and \(E\) is the first moment of energy. The functions \(u(x,t)\) and \(\phi (x,t)\) are the displacement of the solid elastic material and the volume fraction. The constitutive equations are given by

$$\left\{\begin{aligned}& T=\mu {u}_{x}+b\phi ,\\ & H=\delta {\phi }_{x}-dw,\\ & G=-b{u}_{x}-\xi \phi ,\\ & \rho E=-\alpha w-d{\phi }_{x},\\ & P=-\kappa {w}_{x},\\ & q={k}_{1}w,\\ & Q={k}_{2}w,\end{aligned}\right.$$
(1.2)

where \(w\) is the microtemperature, \({k}_{1}\), \({k}_{2}\), \(d\), \(\mu \), \(\delta \), \(\alpha \), \(b\), \(\kappa \) and \(\xi \) are positive constants such that \(\mu \xi -{b}^{2}\ge 0\). Substitute system (1.2) in (1.1) we get the porous elastic system with microtemperature

$$\rho {u}_{tt}-\mu {u}_{xx}-b{\phi }_{x}=0,\quad (x,t)\in (0,l)\times (0,\infty ),$$
(1.3)
$$J{\phi }_{tt}-\delta {\phi }_{xx}+b{u}_{x}+\xi \phi +d{w}_{x}=0,\quad (x,t)\in (0,l)\times (0,\infty ),$$
(1.4)
$$\alpha {w}_{t}-\kappa {w}_{xx}+d{\phi }_{tx}+kw=0,\quad (x,t)\in (0,l)\times (0,\infty ),$$
(1.5)

with Neumann boundary conditions

$$u(0,t)=u(l,t)={\phi }_{x}(0,t)={\phi }_{x}(l,t)=w(0,t)=w(l,t)=0, t>0,$$
(1.6)

and the initial conditions are

$$\begin{aligned}& u(x,0)={u}_{0}(x),\quad {u}_{t}(x,0)={u}_{1}(x),\quad \phi (x,0)={\phi }_{0}(x),\,\, {\phi }_{t}(x,0)={\phi }_{1}(x),\\ & w(x,0)={w}_{0}(x),\quad x\in (0,l),\end{aligned}$$
(1.7)

where \(k={k}_{1}-{k}_{2}>0\).

The presence of Neumann boundary conditions for \(\phi \) hinders the application of Poincaré inequality. In order to overcome this obstacle, we introduce modifications to \(\phi \) in the following manner. Using equation (1.4) and the boundary conditions (1.6), we obtain

$$J\frac{{d}^{2}}{d{t}^{2}}{\int}_{\!\!\!0}^{l}\phi (x,t) dx+\xi {\int}_{\!\!\!0}^{l}\phi (x,t) dx=0,$$

which is solved by

$${\int}_{\!\!\!0}^{l}\phi (x,t) dx=\left({\int}_{\!\!\!0}^{l}{\phi }_{0}(x) dx\right){\text{cos}}\left(\frac{\sqrt{\xi }}{\sqrt{J}}t\right)+\left({\int}_{\!\!\!0}^{l}{\phi }_{1}(x) dx\right){\text{sin}}\left(\frac{\sqrt{\xi }}{\sqrt{J}}t\right).$$

Hence, if we define

$$\overline{\phi }(x,t)=\phi (x,t)-\left({\int}_{\!\!\!0}^{l}{\phi }_{0}(x) dx\right){\text{cos}}\left(\frac{\sqrt{\xi }}{\sqrt{J}}t\right)-\frac{\sqrt{\xi }}{\sqrt{J}}\left({\int}_{\!\!\!0}^{l}{\phi }_{1}(x) dx\right){\text{sin}}\left(\frac{\sqrt{\xi }}{\sqrt{J}}t\right),$$

then, \((u,\overline{\phi },w)\) satisfies (1.3)–(1.7), with the following initial condition for \(\phi \)

$${\overline{\phi }}_{0}(x)={\phi }_{0}(x)-\left({\int}_{\!\!\!0}^{l}{\phi }_{0}(x) dx\right).$$

Moreover

$${\int}_{\!\!\!0}^{l}\overline{\phi }(x,t) dx=0,$$

which allows the application of Poincaré inequality for \(\phi \). In the subsequent analysis, we will utilize \((u,\overline{\phi },w)\) for our calculations, but for the sake of convenience, we will represent it as \((u,\phi ,w)\). Note that the stabilization of the Porous system and the Bresse-Timoshenko system was studied by different researchers with different damping mechanisms (for example see [41,42,43,44,45,46]).

Apalara [7] study system (1.3)–(1.7) in the case where \(\mu \xi >{b}^{2}\). This paper will deal with the case where \(\mu \xi ={b}^{2}\).

The rest of the paper is as follows: In Sect. 2 we will establish exponential stability when \(\mu \xi ={b}^{2}\). In Sect. 3 we will study the system with a second spectrum free. First, we study well-posedness using the Faedo Galerkin approximation and then prove the exponential stability without any assumption on parameters. In Sect. 4 we present some numerical results which demonstrate the exponential behavior of the solution.

2 Exponential stabilty

This section aims to show that the energy of system (1.3)–(1.7) decays exponentially when \(\mu \xi ={b}^{2}\) and under the condition of equal speed limits.

First we state the well-posedness theorem that is proved by Apalara [7].

Theorem 1

For \({U}_{0}\in \mathcal{H}\) there exists a unique solution \(U\in C({\mathbb{R}}_{+},\mathcal{H})\) to system (1.3)–(1.7). Moreover if \({U}_{0}\in \mathcal{D}(\mathcal{A})\) then \(U\in C({\mathbb{R}}_{+},\mathcal{D}(\mathcal{A}))\cap {C}^{1}({\mathbb{R}}_{+},\mathcal{H}),\) where \(\mathcal{H}\) is the Hilbert space defined by

$$\mathcal{H}:={H}_{0}^{1}(0,l)\times {L}^{2}(0,l)\times {H}_{*}^{1}(0,l)\times {L}_{*}^{2}(0,l)\times {L}^{2}(0,l),$$

where the space \({L}_{*}^{2}(0,l)\) is defined as:

$${L}_{*}^{2}(0,l):=\{u\in {L}^{2}(0,l)|{\int}_{\!\!\!0}^{l}u(x) dx=0\},$$

and the space \({H}_{*}^{1}(0,l)\) is defined as:

$${H}_{*}^{1}(0,l):={H}^{1}(0,l)\cap {L}_{*}^{2}(0,l).$$

Let us first define the energy of system (1.3)–(1.7). Multiply equations (1.3), (1.4) and (1.6) by \({u}_{t}\), \({\phi }_{t}\) and \(w\) respectively we get

$$\left\{\begin{aligned}& \frac{1}{2}\frac{d}{dt}\left(\rho ||{u}_{t}|{|}^{2}+\mu ||{u}_{x}|{|}^{2}\right)+b{\int}_{\!\!\!0}^{l}\phi {u}_{xt}dx=0,\\ & \frac{1}{2}\frac{d}{dt}\left(J||{\phi }_{t}|{|}^{2}+\delta ||{\phi }_{x}|{|}^{2}+\xi ||\phi |{|}^{2}\right)+b{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx+d{\int}_{\!\!\!0}^{l}{w}_{x}{\phi }_{t}dx=0,\\ & \frac{1}{2}\frac{d}{dt}\left(\alpha ||w|{|}^{2}\right)+\kappa {\int}_{\!\!\!0}^{l}{w}_{x}^{2}+d{\int}_{\!\!\!0}^{l}{\phi }_{tx}w+k{\int}_{\!\!\!0}^{l}{w}^{2}=0.\end{aligned}\right.$$
(2.1)

Sum the equations of system (2.1) we obtain

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\left(\rho ||{u}_{t}|{|}^{2}+\mu ||{u}_{x}|{|}^{2}+J||{\phi }_{t}|{|}^{2}+\delta ||{\phi }_{x}|{|}^{2}+\xi ||\phi |{|}^{2}+\alpha ||w|{|}^{2}+2b({u}_{x},\phi )\right)\\ & \quad =-\kappa {\int}_{\!\!\!0}^{l}{w}_{x}^{2}-k{\int}_{\!\!\!0}^{l}{w}^{2}.\end{aligned}$$

Define

$${\mathbb{E}}(t)=\frac{1}{2}\left(\rho ||{u}_{t}|{|}^{2}+\mu ||{u}_{x}|{|}^{2}+J||{\phi }_{t}|{|}^{2}+\delta ||{\phi }_{x}|{|}^{2}+\xi ||\phi |{|}^{2}+\alpha ||w|{|}^{2}+2b({u}_{x},\phi )\right).$$

Add then subtract \(\frac{{b}^{2}}{2\xi }\parallel {u}_{x}{\parallel }^{2}\) to the right side of the above equation we arrive at

$${\mathbb{E}}(t)=\frac{1}{2}\left(\rho ||{u}_{t}|{|}^{2}+J||{\phi }_{t}|{|}^{2}+\delta ||{\phi }_{x}|{|}^{2}+||\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi |{|}^{2}+\alpha ||w|{|}^{2}\right)$$

where \(||.||\) denotes the \({L}^{2}\)–norm.

The dissipation law is given by

$$\frac{d}{dt}{\mathbb{E}}(t)=-\kappa {\int}_{\!\!\!0}^{l}{w}_{x}^{2}-k{\int}_{\!\!\!0}^{l}{w}^{2}.$$

Now the exponential stability result is stated in the following theorem.

Theorem 2

If \(\chi =0\), the energy \({\mathbb{E}}(t)\) of the system (1.3)–(1.7) decays exponentially as time t tends to infinity. That is, there exist two positive constants \({M}_{1}\) and \({\omega }_{1}\) such that

$${\mathbb{E}}(t)\le {M}_{1}{\mathbb{E}}(0){e}^{-{\omega }_{1}t}, \quad\forall t\ge 0,$$

where

$$\chi =\frac{\mu }{\rho }-\frac{\delta }{J}.$$

The proof of Theorem 2 will be established through the following technical lemmas. First, we set

$${\mathcal{F}}_{1}(t)=-\rho {\int}_{\!\!\!0}^{l}{u}_{t}udx.$$

Lemma 1

Let \((u,\phi ,w)\) be a solution of the system (1.3)–(1.7). Then we have

$$\begin{aligned}\frac{d}{dt}{\mathcal{F}}_{1}(t)& \le -\rho {\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx+\frac{3}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\end{aligned}$$
$$+\xi {c}_{p}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx.$$
(2.2)

Proof

Multiply equation (1.3) by \(u\) and integrate by parts over \((0,l)\) we get

$$\rho {\int}_{\!\!\!0}^{l}{u}_{tt}udx+\mu {\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx+b{\int}_{\!\!\!0}^{l}{u}_{x}\phi dx=0,$$

add then subtract the term \(\frac{{b}^{2}}{\xi }{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}\) from the above equation we obtain

$$\rho {\int}_{\!\!\!0}^{l}{u}_{tt}udx+\frac{b}{\sqrt{\xi }}{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right){u}_{x}dx=0,$$

taking into account that \(\frac{d}{dt}({u}_{t}u)={u}_{tt}u+|{u}_{t}{|}^{2}\) we arrive at

$$\frac{d}{dt}\left(-\rho {\int}_{\!\!\!0}^{l}{u}_{t}udx\right)=-\rho {\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx+\frac{b}{\sqrt{\xi }}{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right){u}_{x}dx.$$
(2.3)

Apply Young’s inequality and note that

$${\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx\le 2{\int}_{\!\!\!0}^{l}{\left|{u}_{x}+\frac{\xi }{b}\phi \right|}^{2}dx+2{c}_{p}{\int}_{\!\!\!0}^{l}{\left|\frac{\xi }{b}{\phi }_{x}\right|}^{2}dx$$
(2.4)

we get the desired result.\(\hfill\square\)

Set

$${\mathcal{F}}_{2}(t)=J{\int}_{\!\!\!0}^{l}{\phi }_{t}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}{u}_{tx}\phi dx.$$

Lemma 2

Let \((u,\phi ,w)\) be a solution of the system (1.3)–(1.7). Then we have

$$\frac{d}{dt}{\mathcal{F}}_{2}(t)\le J\sqrt{\xi }{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}-\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+{K}_{1}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx,$$
(2.5)

where \({K}_{1}=\frac{{d}^{2}}{2\sqrt{\xi }}.\)

Proof

Multiply equation (1.4) by \(\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\) we get:

$$\begin{aligned} & J{\int}_{\!\!\!0}^{l}{\phi }_{tt}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx+\delta {\int}_{\!\!\!0}^{l}{\phi }_{x}{\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}dx\\ & =-\sqrt{\xi }{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx-d{\int}_{\!\!\!0}^{l}{w}_{x}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx. \end{aligned}$$
(2.6)

Using Young’s inequality we obtain

$$\begin{aligned} & J{\int}_{\!\!\!0}^{l}{\phi }_{tt}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx+\delta {\int}_{\!\!\!0}^{l}{\phi }_{x}{\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}dx\\ & \le -\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\frac{{d}^{2}}{2\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(2.7)

Add then subtract the term \(\frac{\mu \xi }{b}{\phi }_{x}\) to equation (1.3) we get

$${\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}=\frac{\rho b}{\mu \sqrt{\xi }}{u}_{tt}.$$
(2.8)

Substitute \({\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}\) in equation (2.7) we arrive at

$$\begin{aligned} & J{\int}_{\!\!\!0}^{l}{\phi }_{tt}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}\phi {u}_{ttx}dx\\ & \le -\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\frac{{d}^{2}}{2\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(2.9)

Taking into account that \({u}_{ttx}\phi =\frac{d}{dt}({u}_{tx}\phi )-{u}_{tx}{\phi }_{t}\) and \({\phi }_{tt}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)=\frac{d}{dt}\left[{\phi }_{t}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\right]-{\phi }_{t}{\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{t}\) we obtain

$$\begin{aligned} & \frac{d}{dt}\left(J{\int}_{\!\!\!0}^{l}{\phi }_{t}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}\phi {u}_{tx}dx\right)\\ &\quad \le J\sqrt{\xi }{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}+\frac{\delta b}{\sqrt{\xi }}\left(\frac{J}{\delta }-\frac{\rho }{\mu }\right){\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{tx}dx\\ &\quad\quad -\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+{}_{{K}_{1}}{{\frac{{d}^{2}}{2\sqrt{\xi }}}}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned} $$
(2.10)

Set

$$\begin{aligned} {\mathcal{F}}_{3}(t)=-J\alpha {\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx. \end{aligned}$$

\(\hfill \square \)

Lemma 3

Let \((u,\phi ,w)\) be a solution of the system (1.3)–(1.7). Then we have

$$\begin{aligned} \frac{d}{dt}{\mathcal{F}}_{3}(t)& \le \frac{-Jd}{4}{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx+(1+{c}_{p}){\epsilon }_{1}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+{\epsilon }_{2}{c}_{p}{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx\\ &\quad +{K}_{2}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx+{K}_{3}{\int}_{\!\!\!0} ^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(2.11)

Proof

Multiply equation (1.5) by \(\begin{aligned} J{\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\end{aligned}\) and integrate by parts by parts over \((0,l)\) we get

$$\begin{aligned}J\alpha {\int}_{\!\!\!0}^{l}{w}_{t}\left({\int}_{\!\!\!0}^{x}{\phi }_{t}\left(y\right)dy\right)dx+J\kappa {\int}_{\!\!\!0}^{l}{w}_{x}{\phi }_{t}dx-Jd{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx\end{aligned}$$
$$\begin{aligned} +Jk{\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx=0.\end{aligned}$$
(2.12)

Knowing that

$$\begin{aligned} \frac{d}{dt}\left[w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)\right]={w}_{t}\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)+w\left({\int}_{\!\!\!0}^{x}{\phi }_{tt}(y)dy\right) \end{aligned}$$

and by substituting \({\phi }_{tt}\) from equation (1.4) we obtain

$$\begin{aligned} \frac{d}{dt}\left[-J\alpha {\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx\right]& =-\delta \alpha {\int}_{\!\!\!0}^{l}{\phi }_{x}wdx+b\alpha {\int}_{\!\!\!0}^{l}uwdx+d\alpha {\int}_{\!\!\!0}^{l}|w{|}^{2}dx+J\kappa {\int}_{\!\!\!0}^{l}{w}_{x}{\phi }_{t}dx\\ &\quad -Jd{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx+\xi \alpha {\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}\phi (y)dy\right)dx\\ & \quad+Jk{\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx.\end{aligned}$$
(2.13)

Apply Young’s and Poincare’s inequalities, we get

$$\begin{aligned} \frac{d}{dt}\left[-J\alpha {\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx\right]& \le \frac{-Jd}{2}{\int}_{\!\!\!0}^{l}\left|{\phi }_{t}{|}^{2}dx+{\epsilon }_{1}{\int}_{\!\!\!0}^{l}\right|{\phi }_{x}{|}^{2}dx\\ &\quad +\left(d\alpha +\frac{{\delta }^{2}{\alpha }^{2}}{{\epsilon }_{1}}+\frac{{b}^{2}{\alpha }^{2}}{{\epsilon }_{2}}+\frac{{\xi }^{2}{\alpha }^{2}}{{\epsilon }_{3}}+\frac{{J}^{2}{k}^{2}}{{\epsilon }_{4}}\right){\int}_{\!\!\!0}^{l}|w{|}^{2}dx\\ &\quad +\frac{J{\kappa }^{2}}{2d}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx+{\epsilon }_{3}{\int}_{\!\!\!0}^{l}{\left({\int}_{\!\!\!0}^{x}\phi (y)dy\right)}^{2}dx\\ & \quad+{\epsilon }_{2}{c}_{p}{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx+{\epsilon }_{4}{\int}_{\!\!\!0}^{l}{\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)}^{2}dx.\end{aligned}$$
(2.14)

Using Cauchy–Shwarz inequality we obtain

$$\begin{aligned}\frac{d}{dt}\left[-J\alpha {\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx\right]& \le \frac{-Jd}{2}{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx+{}_{{K}_{3}}{{\frac{J{\kappa }^{2}}{2d}}}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx+{\epsilon }_{1}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx\\ &\quad +{}_{{K}_{2}}{{\left(d\alpha +\frac{{\delta }^{2}{\alpha }^{2}}{{\epsilon }_{1}}+\frac{{b}^{2}{\alpha }^{2}}{{\epsilon }_{2}}+\frac{{\xi }^{2}{\alpha }^{2}}{{\epsilon }_{1}}+\frac{{J}^{2}{k}^{2}}{{\epsilon }_{3}}\right)}}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx\\ &\quad +{\epsilon }_{2}{c}_{p}{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx+{\epsilon }_{1}{\int}_{\!\!\!0}^{l}|\phi {|}^{2}dx\\ & \quad+{\epsilon }_{3}{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx.\end{aligned}$$
(2.15)

Applying Poincare’s inequality and taking \({\epsilon }_{3}=\frac{Jd}{4}\) we get the desired result.\(\hfill \square \)

Set

$$\begin{aligned}{\mathcal{F}}_{4}(t)=J{\int}_{\!\!\!0}^{l}{\phi }_{t}\phi dx.\end{aligned}$$

Lemma 4

Let \((u,\phi ,w)\) be a solution of the system (1.3)–(1.7). Then we have

$$\begin{aligned}\frac{d}{dt}{\mathcal{F}}_{4}(t)& \le J{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx-\frac{\delta }{4}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+\frac{\xi {c}_{p}}{\delta }{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ & \quad +{K}_{4}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx.\end{aligned}$$
(2.16)

Proof

Now Multiply equation (1.4) by \(\phi \) and integrate by parts over \((0,l)\) we get

$$\begin{aligned}J{\int}_{\!\!\!0}^{l}{\phi }_{tt}\phi dx+\delta {\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+\sqrt{\xi }{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\phi dx-d{\int}_{\!\!\!0}^{l}w{\phi }_{x}dx=0, \end{aligned}$$
(2.17)

taking into account that \(\frac{d}{dt}({\phi }_{t}\phi )={\phi }_{tt}\phi +|{\phi }_{t}{|}^{2}\) we arrive at

$$\begin{aligned}\frac{d}{dt}\left(J{\int}_{\!\!\!0}^{l}{\phi }_{t}\phi dx\right)& =J{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx-\delta {\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx-\sqrt{\xi }{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\phi dx\\ &\quad +d{\int}_{\!\!\!0}^{l}w{\phi }_{x}dx.\end{aligned}$$
(2.18)

Apply Young’s and Poincaré inequalities we arrive at

$$\begin{aligned}\frac{d}{dt}\left(J{\int}_{\!\!\!0}^{l}{\phi }_{t}\phi dx\right)& \le J{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx-\frac{\delta }{4}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+\frac{\xi {c}_{p}}{\delta }{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ &\quad +{}_{{K}_{4}}{{\frac{{d}^{2}}{\delta }}}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx.\end{aligned}$$
(2.19)

\(\hfill\square \)

Set

$$\begin{aligned} {\mathcal{F}}_{5}(t)=\frac{\rho \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{t}{\phi }_{x}dx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx. \end{aligned}$$

Lemma 5

Let \((u,\phi ,w)\) be a solution of the system (1.3)–(1.7). Then we have

$$\frac{d}{dt}{\mathcal{F}}_{5}(t)\le -\frac{\mu }{2}{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx+({K}_{5}+\delta ){\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+{K}_{6}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.$$
(2.20)

Proof

Differentiating \({\mathcal{F}}_{5}(t)\) with respect to \(t\) we get

$$\begin{aligned} & \frac{d}{dt}\left[\frac{\rho \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{t}{\phi }_{x}dx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx\right]\\ &\quad =\frac{\rho \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{tt}{\phi }_{x}dx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{\phi }_{tt}{u}_{x}dx-\frac{J\rho }{b}\left(\frac{\mu }{\rho }-\frac{\delta }{J}\right){\int}_{\!\!\!0}^{l}{u}_{tx}{\phi }_{t}dx.\end{aligned}$$

Substitute \({u}_{tt}\) and \({\phi }_{tt}\) from equations (1.3) and (1.4) respectively we obtain

$$\begin{aligned}\frac{d}{dt}\left[\frac{\rho \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{t}{\phi }_{x}dx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx\right]& =\frac{\mu \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{xx}{\phi }_{x}dx+\delta {\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+\frac{\mu \delta }{b}{\int}_{\!\!\!0}^{l}{\phi }_{xx}{u}_{x}dx\\ & \quad-\mu {\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx-\frac{\mu \xi }{b}{\int}_{\!\!\!0}^{l}{u}_{x}\phi dx-\frac{\mu d}{b}{\int}_{\!\!\!0}^{l}{u}_{x}{w}_{x}dx\\ &\quad -\frac{J\rho }{b}\chi {\int}_{\!\!\!0}^{l}{u}_{tx}{\phi }_{t}dx.\end{aligned}$$
(2.21)

Apply Young’s and Poincaré inequalities we arrive at

$$\begin{aligned}\frac{d}{dt}\left[\frac{\rho \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{t}{\phi }_{x}dx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx\right]& \le -\frac{\mu }{2}{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx+\delta {\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx\\ &\quad +{}_{{K}_{5}}{{\frac{\mu {\xi }^{2}}{b}{c}_{p}}}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+{}_{{K}_{6}}{{\frac{\mu {d}^{2}}{b}}}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(2.22)

\(\hfill \square \)

Let

$${\mathbb{L}}(t)={\mathbf{N}}_{1}{\mathbb{E}}(t)+{\mathcal{F}}_{1}(t)+{\mathbf{N}}_{2}{\mathcal{F}}_{2}(t)+{\mathbf{N}}_{3}{\mathcal{F}}_{3}(t)+{\mathbf{N}}_{4}{\mathcal{F}}_{4}(t)+{\mathcal{F}}_{5}(t),$$

Where \({\mathbf{N}}_{1}\), \({\mathbf{N}}_{2}\), \({\mathbf{N}}_{3}\) and \({\mathbf{N}}_{4}\) are positive constants to be fixed.

Theorem 3

There exists positive constants \({\sigma }_{1}\) and \({\sigma }_{2}\) such that

$${\sigma }_{1}{\mathbb{E}}(t)\le {\mathbb{L}}(t)\le {\sigma }_{2}{\mathbb{E}}(t),\quad \forall t\ge 0.$$

Proof

We have

$$\begin{aligned}{\mathbb{L}}(t)-{\mathbf{N}}_{1}{\mathbb{E}}(t)& =-\rho {\int}_{\!\!\!0}^{l}{u}_{t}udx+{\mathbf{N}}_{2}J{\int}_{\!\!\!0}^{l}{\phi }_{t}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-{\mathbf{N}}_{2}\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}{u}_{tx}\phi dx\\ & \quad -{\mathbf{N}}_{3}J\alpha {\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx+{\mathbf{N}}_{4}J{\int}_{\!\!\!0}^{l}{\phi }_{t}\phi dx+\frac{\rho \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{t}{\phi }_{x}dx\\ & \quad +\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx.\end{aligned}$$

By positivity of the constant \({\mathbf{N}}_{2}\delta \frac{\rho b}{\mu \sqrt{\xi }}\) we obtain

$$\begin{aligned}{\mathbb{L}}(t)-{\mathbf{N}}_{1}{\mathbb{E}}(t)& \le -\rho {\int}_{\!\!\!0}^{l}{u}_{t}udx+{\mathbf{N}}_{2}J{\int}_{\!\!\!0}^{l}{\phi }_{t}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx\\ & \quad -{\mathbf{N}}_{3}J\alpha {\int}_{\!\!\!0}^{l}w\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)dx+{\mathbf{N}}_{4}J{\int}_{\!\!\!0}^{l}{\phi }_{t}\phi dx\\ & \quad +\frac{\rho \delta }{b}{\int}_{\!\!\!0}^{l}{u}_{t}{\phi }_{x}dx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx.\end{aligned}$$

Applying Young’s and Poincaré inequalities, we obtain

$$\begin{aligned}|{\mathbb{L}}(t)-{\mathbf{N}}_{1}{\mathbb{E}}(t)|& \le \left(\frac{\rho }{2}+\frac{\rho \delta }{2b}\right){\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx+\left(\frac{\rho {c}_{p}}{2}+\frac{J\mu }{2b}\right){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx\\ &\quad +\frac{{\mathbf{N}}_{2}J}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\left(\frac{{\mathbf{N}}_{4}J{c}_{p}}{2}+\frac{\rho \delta }{2b}\right){\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx\\ & \quad +\left(\frac{{\mathbf{N}}_{2}J}{2}+\frac{{\mathbf{N}}_{4}J}{2}+\frac{J\mu }{2b}\right){\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx+\frac{{\mathbf{N}}_{3}J\alpha }{2}{\int}_{\!\!\!0}^{l}{\left({\int}_{\!\!\!0}^{x}{\phi }_{t}(y)dy\right)}^{2}dx\\ & \quad +\frac{{\mathbf{N}}_{3}J\alpha }{2}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx.\end{aligned}$$

Knowing that \(\mu =\frac{{b}^{2}}{\xi }\) we get

$$\begin{aligned}\left(\frac{\rho {c}_{p}}{2}+\frac{J\mu }{2b}\right){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx& =\left(\frac{\rho {c}_{p}}{2\mu }+\frac{J}{2b}\right){\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi -\sqrt{\xi }\phi \right|}^{2}dx\\ & \le \left(\frac{\rho {c}_{p}}{\mu }+\frac{J}{b}\right){\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\left(\frac{\rho {c}_{p}}{\mu }+\frac{J}{b}\right){\int}_{\!\!\!0}^{l}|\sqrt{\xi }\phi {|}^{2}dx,\end{aligned}$$
(2.23)

Using (2.23) and Cauchy–Shwarz inequality we get

$$\begin{aligned}|{\mathbb{L}}(t)-{\mathbf{N}}_{1}{\mathbb{E}}(t)|& \le \left(\frac{\rho }{2}+\frac{\rho \delta }{2b}\right){\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx+\left(\frac{{\mathbf{N}}_{2}J}{2}+\frac{\rho {c}_{p}}{\mu }+\frac{J}{b}\right){\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ & \quad +\left(\frac{{\mathbf{N}}_{2}J}{2}+\frac{{\mathbf{N}}_{4}J}{2}+\frac{J\mu }{2b}\right){\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx+\frac{{\mathbf{N}}_{3}J\alpha }{2}{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx\\ & \quad +\left(\frac{{\mathbf{N}}_{4}J{c}_{p}}{2}+\frac{\rho \delta }{2b}+\frac{\xi \rho {c}_{p}^{2}}{\mu }+\frac{\xi J{c}_{p}}{b}\right){\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+\frac{{\mathbf{N}}_{3}J\alpha }{2}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx.\end{aligned}$$

Define

$$\begin{aligned}{\mathbf{N}}_{0}:=&{\text{max}}\left\{\frac{\rho }{2}+\frac{\rho \delta }{2b};\frac{{\mathbf{N}}_{2}J}{2}+\frac{\rho {c}_{p}}{\mu }+\frac{J}{b};\frac{{\mathbf{N}}_{2}J}{2}+\frac{{\mathbf{N}}_{4}J}{2}+\frac{J\mu }{2b};\frac{{\mathbf{N}}_{3}J\alpha }{2};\frac{{\mathbf{N}}_{4}J{c}_{p}}{2}\right.\\ &\left.+\frac{\rho \delta }{2b}+\frac{\xi \rho {c}_{p}^{2}}{\mu }+\frac{\xi J{c}_{p}}{b}\right\}.\end{aligned}$$

Hence

$$|{\mathbb{L}}(t)-{\mathbf{N}}_{1}{\mathbb{E}}(t)|\le {\mathbf{N}}_{0}{\mathbb{E}}(t),$$

which implies that

$${\sigma }_{1}{\mathbb{E}}(t)\le {\mathbb{L}}(t)\le {\sigma }_{2}{\mathbb{E}}(t),$$

where \({\sigma }_{1}={\mathbf{N}}_{1}-{\mathbf{N}}_{0}\) and \({\sigma }_{2}={\mathbf{N}}_{1}+{\mathbf{N}}_{0}\) and \({\mathbf{N}}_{1}>{\mathbf{N}}_{0}\).\(\hfill \square \)

Proof of Theorem 2

It follows from Lemmas 1, 23, 4, 5 that

$$\begin{aligned}\frac{d}{dt}{\mathbb{L}}(t) & \le -\rho {\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx-{}_{{\eta }_{1}}{{\left(\frac{\mu }{2}-{\mathbf{N}}_{3}{\epsilon }_{2}{c}_{p}\right)}}{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx\\ & \quad -{}_{{\eta }_{2}}{{\left({\mathbf{N}}_{3}\frac{Jd}{4}-{\mathbf{N}}_{4}J-{\mathbf{N}}_{2}J\sqrt{\xi }\right)}}{\int}_{\!\!\!0}^{l}|{\phi }_{t}{|}^{2}dx-{}_{{\eta }_{3}}{{\left({\mathbf{N}}_{2}\frac{\sqrt{\xi }}{2}-{\mathbf{N}}_{4}\frac{\xi {c}_{p}}{\delta }-\frac{3}{2}\right)}}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ \quad& -{}_{{\eta }_{4}}{{\left({\mathbf{N}}_{4}\frac{\delta }{4}-{\mathbf{N}}_{3}(1+{c}_{p}){\epsilon }_{1}-(\xi {c}_{p}+\delta +{K}_{5})\right)}}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx\\ &\quad -{}_{{\eta }_{5}}{{({\mathbf{N}}_{1}k-{\mathbf{N}}_{4}{K}_{4}-{\mathbf{N}}_{3}{K}_{2})}}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx-{}_{{\eta }_{6}}{{\left({\mathbf{N}}_{1}\kappa -{\mathbf{N}}_{2}{K}_{1}-{\mathbf{N}}_{3}{K}_{3}-{K}_{6}\right)}}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx\end{aligned}$$
(2.24)

Choose \({\epsilon }_{2}=\frac{{b}^{2}}{2{c}_{p}\xi {\mathbf{N}}_{3}},\) we get that \({\eta }_{1}=0\),

now take

$${N}_{2}>\frac{2\left({\mathbf{N}}_{4}\frac{\xi {c}_{p}}{\delta }+\frac{3}{2}\right)}{\sqrt{\xi }},$$

which implies that \({\eta }_{3}>0\).

Choose

$${\epsilon }_{1}=\frac{1}{{\mathbf{N}}_{3}(1+{c}_{p})},$$

then take

$${N}_{4}>\frac{4(1+\xi {c}_{p}+\delta +{K}_{5})}{\delta },$$

then we obtain that \({\eta }_{4}>0\).

Now take

$${\mathbf{N}}_{3}>\frac{4({\mathbf{N}}_{4}+{\mathbf{N}}_{2}\sqrt{\xi })}{d},$$

hence we obtain that \({\eta }_{2}>0\).

Take

$${\mathbf{N}}_{1}>{\text{max}}\left\{\frac{{\mathbf{N}}_{4}{K}_{4}+{\mathbf{N}}_{3}{K}_{2}}{k};\frac{{\mathbf{N}}_{2}{K}_{1}+{\mathbf{N}}_{3}{k}_{3}+{K}_{6}}{\kappa }\right\},$$

from where we obtain that \({\eta }_{5},{\eta }_{6}>0\). Therefore we can conclude that there exists a positive constant \({\varvec{\upeta}}=2{\text{min}}\{1,{\eta }_{2},{\eta }_{3},{\eta }_{4},{\eta }_{5},{\eta }_{6}\}\) such that

$$\frac{d}{dt}{\mathbb{L}}(t)\le -{\varvec{\upeta}}{\mathbb{E}}(t),$$

by equivalence between \({\mathbb{E}}(t)\) and \({\mathbb{L}}(t)\) according to Theorem 3 we get:

$$\frac{d}{dt}{\mathbb{L}}(t)\le -{\omega }_{1}{\mathbb{L}}(t),$$

where \({\omega }_{1}=\frac{{\varvec{\upeta}}}{{\sigma }_{2}}\). Now integrate the above inequality over \((0,t)\) we obtain

$${\mathbb{L}}(t)\le {\mathbb{L}}(0){e}^{-{\omega }_{1}t},$$

again by equivalence between \({\mathbb{E}}(t)\) and \({\mathbb{L}}(t)\) according to Theorem 3 we arrive at

$${\mathbb{E}}(t)\le {M}_{1}{\mathbb{E}}(0){e}^{-{\omega }_{1}t},$$

where \({M}_{1}=\frac{{\sigma }_{2}}{{\sigma }_{1}}.\)\(\hfill\square \)

3 Second spectrum approach

In this section, we consider the porous system with second spectrum free and microtemperature:

$$\rho {u}_{tt}-\mu {u}_{xx}-b{\phi }_{x}=0,\quad (x,t)\in (0,l)\times (0.\infty ),$$
(3.1)
$$-J{u}_{ttx}-\delta {\phi }_{xx}+b{u}_{x}+\xi \phi +d{w}_{x}=0,\quad (x,t)\in (0,l)\times (0.\infty ),$$
(3.2)
$$\alpha {w}_{t}-\kappa {w}_{xx}+d{\phi }_{tx}+kw=0,\quad (x,t)\in (0,l)\times (0.\infty ),$$
(3.3)

with Dirichlet boundary conditions

$$u(0,t)=u(l,t)=\phi (0,t)=\phi (l,t)=w(0,t)=w(l,t)=0, t>0,$$
(3.4)

and the initial conditions are

$$\begin{aligned}& u(x,0)={u}_{0}(x),\quad {u}_{t}(x,0)={u}_{1}(x),\quad {u}_{tt}(x,0)={u}_{2}(x),\\ & \phi (x,0)={\phi }_{0}(x),\quad w(x,0)={w}_{0}(x), \quad x\in (0,l).\end{aligned}$$
(3.5)

This system (3.1)–(3.5) is obtained by following the procedure of Elishakoff [28], which involves replacing the term \({\phi }_{tt}\) in (1.4) by \(-{u}_{xtt}\) based on d’ Alembert’s principle for dynamic equilibrium. This eliminates the second spectrum of frequency and its damaging consequences for wave propagation speed. This work aims to get exponential decay without assuming any conditions on the physical parameters.

The dissipation of system (3.1)–(3.5) is obtained from the definition of energy. Indeed, multiply equation (3.1) by \({u}_{t}\), integrate by parts over \((0,l)\) and using boundary conditions (3.4) we get

$$\frac{1}{2}\frac{d}{dt}\left(\rho ||{u}_{t}|{|}^{2}+\mu ||{u}_{x}|{|}^{2}\right)+b{\int}_{\!\!\!0}^{l}\phi {u}_{xt}dx=0.$$
(3.6)

Multiply equation (3.2) by \({\phi }_{t}\), integrate by parts over \((0,l)\) and using boundary conditions (3.4) we get

$$\frac{1}{2}\frac{d}{dt}\left(\delta ||{\phi }_{x}|{|}^{2}+\xi ||\phi |{|}^{2}\right)+b{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx+J{\int}_{\!\!\!0}^{l}{u}_{tt}{\phi }_{tx}dx+d{\int}_{\!\!\!0}^{l}{w}_{x}{\phi }_{t}dx=0.$$
(3.7)

From equation (3.1) we obtain that

$${\phi }_{xt}=\frac{\rho }{b}{u}_{ttt}-\frac{\mu }{b}{u}_{xxt},$$

substitute \({\phi }_{xt}\) in equation (3.7) we arrive at

$$\frac{1}{2}\frac{d}{dt}\left(\delta ||{\phi }_{x}|{|}^{2}+\xi ||\phi |{|}^{2}+\frac{J\rho }{b}\parallel {u}_{tt}{\parallel }^{2}+\frac{J\mu }{b}\parallel {u}_{xt}{\parallel }^{2}\right)+b{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{x}dx+d{\int}_{\!\!\!0}^{l}{w}_{x}{\phi }_{t}dx=0.$$
(3.8)

Now multiply equation (3.3) by \(w\), integrate by parts over \((0,l)\) and using boundary conditions (3.4) we get

$$\frac{1}{2}\frac{d}{dt}\left(\alpha ||w|{|}^{2}\right)+\kappa {\int}_{\!\!\!0}^{l}{w}_{x}^{2}+d{\int}_{\!\!\!0}^{l}{\phi }_{tx}w+k{\int}_{\!\!\!0}^{l}{w}^{2}=0.$$
(3.9)

Add equations (3.6), (3.8) and (3.9) we get

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\left(\rho ||{u}_{t}|{|}^{2}+\mu ||{u}_{x}|{|}^{2}+\delta ||{\phi }_{x}|{|}^{2}+\xi ||\phi |{|}^{2}+\frac{J\rho }{b}\parallel {u}_{tt}{\parallel }^{2}+\frac{J\mu }{b}\parallel {u}_{xt}{\parallel }^{2}+\alpha ||w|{|}^{2}+2b({u}_{x},\phi )\right)\\ & \quad =-\kappa {\int}_{\!\!\!0}^{l}{w}_{x}^{2}-k{\int}_{\!\!\!0}^{l}{w}^{2}.\end{aligned}$$

Define

$$E(t)=\frac{1}{2}\left(\rho ||{u}_{t}|{|}^{2}+\mu ||{u}_{x}|{|}^{2}+\delta ||{\phi }_{x}|{|}^{2}+\xi ||\phi |{|}^{2}+\frac{J\rho }{b}\parallel {u}_{tt}{\parallel }^{2}+\frac{J\mu }{b}\parallel {u}_{xt}{\parallel }^{2}+\alpha ||w|{|}^{2}+2b({u}_{x},\phi )\right).$$

Add then subtract \(\frac{{b}^{2}}{2\xi }\parallel {u}_{x}{\parallel }^{2}\) to the right side of the above equation we arrive at

$$E(t)=\frac{1}{2}\left(\rho ||{u}_{t}|{|}^{2}+\frac{J\rho }{b}||{u}_{tt}|{|}^{2}+(\mu -{b}^{2}/\xi )||{u}_{x}|{|}^{2}+\frac{J\mu }{b}||{u}_{xt}|{|}^{2}+\delta ||{\phi }_{x}|{|}^{2}+||\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi |{|}^{2}+\alpha ||w|{|}^{2}\right)$$

where \(||.||\) denotes the \({L}^{2}\)–norm.

The dissipation law is given by

$$\frac{d}{dt}E(t)=-\kappa {\int}_{\!\!\!0}^{l}{w}_{x}^{2}-k{\int}_{\!\!\!0}^{l}{w}^{2}.$$

3.1 Well-Posedness

This section aims to show the existence and uniqueness of the weak solution of system (3.1)–(3.5). Therefore, we will use the classical Faedo–Galerkin approximation and a priori estimates, then pass the limits using compactness arguments.

Define the Hilbert space

$$\mathcal{H}:={H}_{0}^{1}(0,l)\times {H}_{0}^{1}(0,l)\times {L}^{2}(0,l)\times {H}_{0}^{1}(0,l)\times {L}^{2}(0,l).$$

Now multiply the equations (3.1), (3.2) and (3.3) by \(\overline{u }\), \(\overline{\phi }\), \(\overline{w}\in {H }_{0}^{1}(0,l)\) respectively and integrate by parts over \((0,l)\) we get using boundary conditions (3.4)

$$\left\{\begin{aligned}& \rho ({u}_{tt},\overline{u })+\mu ({u}_{x},{\overline{u} }_{x})+b(\phi ,{\overline{u} }_{x})=0,\\ & J({u}_{tt},{\overline{\phi }}_{x})+\delta ({\phi }_{x},{\overline{\phi }}_{x})+b({u}_{x},\overline{\phi })+\xi (\phi ,\overline{\phi })+d({w}_{x},\overline{\phi })=0,\\ & \alpha ({w}_{t},\overline{w })+\kappa ({w}_{x},{\overline{w} }_{x})+d({\phi }_{tx},\overline{w })+k(w,\overline{w })=0.\end{aligned}\right.$$
(3.10)

Definition 1

Let the initial data \(({u}_{0},{u}_{1},{u}_{2},{\phi }_{0},{w}_{0})\in \mathcal{H}\) then a function \(V=(u,{u}_{t},{u}_{tt},\phi ,w)\in C(0,T;\mathcal{H})\) is said to be a weak solution of (3.1)–(3.5) if it is a solution of the weak problem (3.10) for almost \(t\in [0,T].\)

Theorem 4

Suppose that the initial data \(({u}_{0},{u}_{1},{u}_{2},{\phi }_{0},{w}_{0})\in \mathcal{H}\) then system (3.1)–(3.5) have a weak solution satisfying

$$u\in {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right), {u}_{t}\in {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),$$
$${u}_{tt}\in {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right), \phi \in {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),$$
$$w\in {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right)\cap {L}^{2}\left(0,T;{H}_{0}^{1}(0,l)\right),$$

where the solution \(V=(u,{u}_{t},{u}_{tt},\phi ,w)\) depends continuously on the initial data in \(\mathcal{H}\). In particular \(V\) is unique solution of system (3.1)–(3.5).

Proof

We will use the Faedo-Galerkin method to prove the above theorem and proceed in five steps. Step 1. Approximated solution Let \(({u}_{0},{u}_{1},{u}_{2},{\phi }_{0},{w}_{0})\in \mathcal{H}\). Let \(\{{\eta }_{i}{\}}_{i=1}^{\infty }\subset {C}^{\infty }([0,l])\) be basis for \({H}_{0}^{1}(0,l)\), and let \({V}^{m}=span\{{\eta }_{i}{\}}_{i=1}^{m}\). Now we introduce

$${u}^{m}=\sum_{i=0}^{m}{a}_{i}(t){\eta }_{i}(x), {\phi }^{m}=\sum_{i=0}^{m}{b}_{i}(t){\eta }_{i}(x), {w}^{m}=\sum_{i=0}^{m}{c}_{i}(t){\eta }_{i}(x),$$
(3.11)

which solves the following approximated problem for \(\overline{u }\), \(\overline{\phi }\),\(\overline{w}\in {V }^{m}\)

$$\left\{\begin{aligned}& \rho ({u}_{tt}^{m},\overline{u })+\mu ({u}_{x}^{m},{\overline{u} }_{x})+b({\phi }^{m},{\overline{u} }_{x})=0,\\ & J({u}_{tt}^{m},{\overline{\phi }}_{x})+\delta ({\phi }_{x}^{m},{\overline{\phi }}_{x})+b({u}_{x}^{m},\overline{\phi })+\xi ({\phi }^{m},\overline{\phi })+d({w}_{x}^{m},\overline{\phi })=0,\\ & \alpha ({w}_{t}^{m},\overline{w })+\kappa ({w}_{x}^{m},{\overline{w} }_{x})+d({\phi }_{tx}^{m},\overline{w })+k({w}^{m},\overline{w })=0.\end{aligned}\right.$$
(3.12)

with initial conditions

$$\left({u}^{m}(0),{u}_{t}^{m}(0),{u}_{tt}^{m}(0),{\phi }^{m}(0),{w}^{m}(0)\right)=({u}_{0}^{m},{u}_{1}^{m},{u}_{2}^{m},{\phi }_{0}^{m},{w}_{0}^{m}),$$

such that

$$({u}_{0}^{m},{u}_{1}^{m},{u}_{2}^{m},{\phi }_{0}^{m},{w}_{0}^{m})\to ({u}_{0},{u}_{1},{u}_{2},{\phi }_{0},{w}_{0}) strongly in \mathcal{H}.$$

By using the Carathoedory theorem for standard ordinary differential equations theory, system \((3.12)\) has a local solution \(\left({u}^{m}(t),{u}_{t}^{m}(t),{u}_{tt}^{m}(t),{\phi }^{m}(t),{w}^{m}(t)\right)\) on the maximal interval \([0,{t}_{m})\) with \(0<{t}_{m}\le T\) for every \(m\in {\mathbb{N}}\)

Step 2. A priori estimates

Let \(\overline{u }={u}_{t}^{m}\), \(\overline{\phi }={\phi }_{t}^{m}\) and \(\overline{w }={w}^{m}\) and taking into consideration from equation (3.1) that

$${\phi }_{xt}=\frac{\rho }{b}{u}_{ttt}-\frac{\mu }{b}{u}_{xxt},$$

then system (3.12) becomes

$$\left\{\begin{aligned}& \rho ({u}_{tt}^{m},{u}_{t}^{m})+\mu ({u}_{x}^{m},{u}_{tx}^{m})+b({\phi }^{m},{u}_{tx}^{m})=0\\ & \frac{J\rho }{b}\left({u}_{tt}^{m},{u}_{ttt}^{m}\right)+\frac{J\mu }{b}\left({u}_{ttx}^{m},{u}_{tx}^{m}\right)+\delta ({\phi }_{x}^{m},{\phi }_{tx}^{m})+b({u}_{x}^{m},{\phi }_{t}^{m})+\xi ({\phi }^{m},{\phi }_{t}^{m})+d({w}_{x}^{m},{\phi }_{t}^{m})=0\\ & \alpha ({w}_{t}^{m},{w}^{m})+\kappa ({w}_{x}^{m},{w}_{x}^{m})+d({\phi }_{tx}^{m},{w}^{m})+k({w}^{m},{w}^{m})=0,\end{aligned}\right.$$
(3.13)

which is equivalent to

$$\left\{\begin{aligned}& \frac{d}{2dt}\left(\rho ||{u}_{t}^{m}|{|}^{2}\right)+\frac{d}{2dt}\left(\mu ||{u}_{x}^{m}|{|}^{2}\right)+b({\phi }^{m},{u}_{tx}^{m})=0\\ & \frac{d}{2dt}\left(\frac{J\rho }{b}\parallel {u}_{tt}^{m}{\parallel }^{2}\right)+\frac{d}{2dt}\left(\frac{J\mu }{b}\parallel {u}_{tx}^{m}{\parallel }^{2}\right)+\frac{d}{2dt}\left(\delta ||{\phi }_{x}^{m}|{|}^{2}\right)+b({u}_{x}^{m},{\phi }_{t}^{m})\\ & +\frac{d}{2dt}\left(\xi ||{\phi }^{m}|{|}^{2}\right)+d({w}_{x}^{m},{\phi }_{t}^{m})=0\\ & \frac{d}{2dt}\left(\alpha ||{w}^{m}|{|}^{2}\right)+\kappa ||{w}_{x}^{m}|{|}^{2}+d({\phi }_{tx}^{m},{w}^{m})+k||{w}^{m}|{|}^{2}=0,\end{aligned}\right.$$
(3.14)

where \(||.||\) denotes the norm in \({L}^{2}(0,l)\).

Add the above two equations we get

$$\begin{aligned}& \frac{d}{2dt}\left(\rho ||{u}_{t}^{m}|{|}^{2}+\mu ||{u}_{x}^{m}|{|}^{2}+\frac{J\rho }{b}\parallel {u}_{tt}^{m}{\parallel }^{2}+\frac{J\mu }{b}\parallel {u}_{tx}^{m}{\parallel }^{2}+\delta ||{\phi }_{x}^{m}|{|}^{2}\right.\\ & +\xi ||{\phi }^{m}|{|}^{2}+2b({u}_{x}^{m},{\phi }^{m})+\alpha ||{w}^{m}|{|}^{2})+\kappa ||{w}_{x}^{m}|{|}^{2}+k||{w}^{m}|{|}^{2}=0.\end{aligned}$$
(3.15)

Let

$${E}^{m}(t)=\frac{1}{2}\left(\rho ||{u}_{t}^{m}|{|}^{2}+\mu ||{u}_{x}^{m}|{|}^{2}+\frac{J\rho }{b}\parallel {u}_{tt}^{m}{\parallel }^{2}+\frac{J\mu }{b}\parallel {u}_{tx}^{m}{\parallel }^{2}+\delta ||{\phi }_{x}^{m}|{|}^{2}+\xi ||{\phi }^{m}|{|}^{2}+2b({u}_{x}^{m},{\phi }^{m})+\alpha ||{w}^{m}|{|}^{2}\right).$$

Then equation (3.15) becomes

$$\frac{d}{dt}{E}^{m}(t)+\kappa ||{w}_{x}^{m}|{|}^{2}+k||{w}^{m}|{|}^{2}=0.$$
(3.16)

Now integrate (3.16) from \(0\) to \(t<{t}_{m}\), we obtain from the choice of the initial data that for all \(t\in [0,T]\) and for every \(m\in {\mathbb{N}}\) that

$${E}^{m}(t)+\kappa {\int}_{\!\!\!0}^{t}||{w}_{x}^{m}(s)|{|}^{2}ds+k{\int}_{\!\!\!0}^{t}||{w}^{m}(s)|{|}^{2}ds\le {C}_{0},$$
(3.17)

where \({C}_{0}\) is a positive constant depending on the initial data.

Step 3. Passing to the limit. Using (3.17) and by the definition of \({E}^{m}(t)\) we obtain that

$$\left\{\begin{aligned}& \{{u}^{m}\}\,\quad is\, bounded\, in\,\quad {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),\\ & \{{u}_{t}^{m}\}\,\quad is\, bounded\, in\,\quad {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),\\ & \{{u}_{tt}^{m}\}\,\quad is \,bounded \,in\,\quad {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right),\\ & \{{\phi }^{m}\}\quad \,is\, bounded \,in\,\quad {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),\\ & \{{w}^{m}\}\,\quad is\, bounded \,in\,\quad {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right)\cap {L}^{2}\left(0,T;{H}_{0}^{1}(0,l)\right).\end{aligned}\right.$$

Then we can extract a subsequence of \(\{{u}^{m}\}\), \(\{{\phi }^{m}\}\) and \(\{{w}^{m}\}\) and still denoted by \(\{{u}^{m}\}\), \(\{{\phi }^{m}\}\) and \(\{ {w}^{m} \}\), such that

$$\left\{\begin{aligned}& {u}^{m}\to u \,weakly \,star\, in \,{L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),\\ & {u}_{t}^{m}\to {u}_{t} \,\quad weakly \,star\, in\,\quad {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),\\ & {u}_{tt}^{m}\to {u}_{tt}\,\quad weakly \,star\, in\,\quad {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right),\\ & {\phi }^{m}\to \phi \quad\,weakly \,star \,in\,\quad {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),\\ & {w}^{m}\to w\,\quad weakly \,star\, in\,\quad {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right),\\ & {w}^{m}\to w\,\quad weakly\, in\quad \,{L}^{2}\left(0,T;{H}_{0}^{1}(0,l)\right).\end{aligned}\right.$$

Now pass to the limits in the approximate variational problem (3.12) we get a weak solution satisfying

$$u\in {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),\quad {u}_{t}\in {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),$$
$${u}_{tt}\in {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right),\quad \phi \in {L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right),$$
$$w\in {L}^{\infty }\left(0,T;{L}^{2}(0,l)\right)\cap {L}^{2}\left(0,T;{H}_{0}^{1}(0,l)\right).$$

Step 4. Initial data. Knowing that

$${H}_{0}^{1}(0,l)\subset {L}^{2}(0,l)\subset {H}^{-1}(0,l),$$

where \({H}^{-1}(0,l)\) is the dual space of \({H}_{0}^{1}(0,l)\).

By using Aubin-Lions lemma, see [39], we obtain that \({L}^{\infty }\left(0,T;{H}_{0}^{1}(0,l)\right)\) is compactly embedded in \(C\left(0,T;{L}^{2}(0,l)\right)\). This implies that

$${u}^{m}\to \,u \quad\,strongly \,in\quad \,C\left(0,T;{L}^{2}(0,l)\right),$$
$${u}_{t}^{m}\to {u}_{t}\,\quad strongly\, in\,\quad C\left(0,T;{L}^{2}(0,l)\right),$$

Hence,

$$\left(u(0),{u}_{t}(0)\right)=({u}_{0},{u}_{1}).$$

Now differentiate with respect to \(t\) the first equation of system (3.12) we get

$$\rho ({u}_{ttt}^{m},\overline{u })+\mu ({u}_{tx}^{m},{\overline{u} }_{x})+b({\phi }_{t}^{m},{\overline{u} }_{x})=0,$$

for all \(\overline{u}\in {H }_{0}^{1}(0,l)\).

Multiply the above equation by a test function

$$\lambda \in {H}_{0}^{1}(0,T),\,\quad such\, that\, \quad\lambda (0)=1, \lambda (T)=0,$$

and then integrate by parts over \([0,T]\)

$$-\rho ({u}_{2}^{m},\overline{u })-\rho {\int}_{\!\!\!0}^{T}({u}_{tt}^{m},\overline{u }){\lambda }_{t}dt+\mu {\int}_{\!\!\!0}^{T}({u}_{tx}^{m},{\overline{u} }_{x})\lambda dt+b{\int}_{\!\!\!0}^{T}({\phi }_{t}^{m},{\overline{u} }_{x})\lambda dt=0.$$

Take the limit \(m\to \infty \), we arrive at

$$-\rho ({u}_{2},\overline{u })-\rho {\int}_{\!\!\!0}^{T}({u}_{tt},\overline{u }){\lambda }_{t}dt+\mu {\int}_{\!\!\!0}^{T}({u}_{tx},{\overline{u} }_{x})\lambda dt+b{\int}_{\!\!\!0}^{T}({\phi }_{t},{\overline{u} }_{x})\lambda dt=0.$$
(3.18)

Now differentiate the first equation of system (3.10) with respect to time, then multiply the result by \(\lambda \) under the same conditions above and integrate by parts over \([0,T]\) we get

$$-\rho ({u}_{tt}(0),\overline{u })-\rho {\int}_{\!\!\!0}^{T}({u}_{tt},\overline{u }){\lambda }_{t}dt+\mu {\int}_{\!\!\!0}^{T}({u}_{tx},{\overline{u} }_{x})\lambda dt+b{\int}_{\!\!\!0}^{T}({\phi }_{t},{\overline{u} }_{x})\lambda dt=0.$$
(3.19)

Combine the two equations \((3.18)\) and \((3.19)\) we obtain that \({u}_{tt}(0)={u}_{2}.\) In the same way we can get that \((\phi (0),w(0))=({\phi }_{0},{w}_{0})\).

Step 5. Continuous dependence on initial data. Let \({V}_{1}(t)=(u,{u}_{t},{u}_{tt},\phi ,w)\) and \({V}_{2}(t)=(\widetilde{u},{\widetilde{u}}_{t},{\widetilde{u}}_{tt},\widetilde{\phi },\widetilde{w})\) be two solutions of the system (3.1)–(3.4) with initial data \({V}_{1}(0)=({u}_{0},{u}_{1},{u}_{2},{\phi }_{0},{w}_{0})\) and \({V}_{2}(0)=({\widetilde{u}}_{0},{\widetilde{u}}_{1},{\widetilde{u}}_{1},{\widetilde{\phi }}_{0},{\widetilde{w}}_{0})\) such that \({V}_{1}(0),{V}_{2}(0)\in \mathcal{H}\). Then \((U,{U}_{t},{U}_{tt},\Phi ,W)={V}_{1}(t)-{V}_{2}(t)\) satisfies the following equations

$$\rho {U}_{tt}-\mu {U}_{xx}-b{\Phi }_{x}=0, in (0,l)\times (0.\infty ),$$
(3.20)
$$-J{U}_{ttx}-\delta {\Phi }_{xx}+b{U}_{x}+\xi\Phi +d{W}_{x}=0, in (0,l)\times (0.\infty ),$$
(3.21)
$$\alpha {W}_{t}-\kappa {W}_{xx}+d{\Phi }_{tx}+kW=0, in (0,l)\times (0.\infty ),$$
(3.22)

with initial data \(({U}_{0},{U}_{1},{U}_{2},{\Phi }_{0},{W}_{0})={V}_{1}(0)-{V}_{2}(0)\).

Now multiply (3.20) by \({U}_{t}\), (3.21) by \({\Phi }_{t}\) and (3.22) by \(W\) then integrate the result over \((0,l)\) we arrive at

$$\frac{d}{dt}\widehat{E}(t)=-\kappa {\int}_{\!\!\!0}^{l}{W}_{x}^{2}dx-k{\int}_{\!\!\!0}^{l}{W}^{2}dx,$$
(3.23)

where \(\widehat{E}(t)\) is the energy related to \({V}_{1}(t)-{V}_{2}(t)\) and defined by

$$\begin{aligned} \widehat{E}(t) & =\frac{1}{2}\left(\rho ||{U}_{t}|{|}^{2}+\frac{J\rho }{b}||{U}_{tt}|{|}^{2}+(\mu -{b}^{2}/\xi )||{U}_{x}|{|}^{2}+\frac{J\mu }{b}||{U}_{xt}|{|}^{2}+\delta ||{\Phi }_{x}|{|}^{2}+||\frac{b}{\sqrt{\xi }}{U}_{x}\right.\\ & \quad \left. \quad +\sqrt{\xi }\Phi |{|}^{2}+\alpha ||W|{|}^{2}\right). \end{aligned}$$

Integrate (3.23) over \((0,t)\), w get that there exists a positive constant \({C}_{T}\) such that for any \(t\in [0,T]\),

$$\widehat{E}(t)\le {C}_{T}\widehat{E}(0),$$

which implies that the weak solution depends continuously on the initial data. Consequently the weak solution of system (3.1)–(3.5) is unique.\(\hfill \square \)

3.2 Exponential stability

This section will prove the exponential decay of the system’s energy (3.1)–(3.5). We will study two cases, first, the case when \(\mu \xi >{b}^{2}\), and second, the case when \(\mu \xi ={b}^{2}\). The method of proof is based on multipliers techniques.

3.2.1 First case:

We will study the exponential decay when \(\mu \xi >{b}^{2}\), and our result is stated in the following theorem.

Theorem 5

The energy \(E(t)\) of the system (3.1)–(3.5) decays exponentially as time t tends to infinity. That is, there exist two positive constants \({M}_{2}\) and \({\omega }_{2}\) independent of the initial data and independent of any relationship between coefficients such that.

$$E(t)\le {M}_{2}E(0){e}^{-{\omega }_{2}t}, \forall t\ge 0.$$

The proof of Theorem 5will be established through two lemmas. First, we set

$$\mathcal{F}(t)=\rho {\int}_{\!\!\!0}^{l}{u}_{t}udx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{u}_{xt}{u}_{x}dx.$$

Lemma 6

Let \((u,\phi ,w)\) be a solution of the system (3.1–3.5). Then we have

$$\begin{aligned}\frac{d}{dt}\mathcal{F}(t)& \le \left(\frac{J\mu }{b}+2\rho {c}_{p}\right){\int}_{\!\!\!0}^{l}|{u}_{xt}{|}^{2}dx-\rho {\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx-(\mu -{b}^{2}/\xi ){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx-\frac{\delta }{2}{\int}_{\!\!\!0}^{L}|{\phi }_{x}{|}^{2}dx\\ & \quad -\frac{J\rho }{b}{\int}_{\!\!\!0}^{l}|{u}_{tt}{|}^{2}dx-{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\frac{{d}^{2}}{2\delta }{\int}_{\!\!\!0}^{l}|w{|}^{2}dx.\end{aligned}$$
(3.1)

Proof

Multiply equation (3.1) by \(u\) and integrate by parts over \((0,l)\) we get

$$\rho {\int}_{\!\!\!0}^{l}{u}_{tt}udx+\mu {\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx+b{\int}_{\!\!\!0}^{l}{u}_{x}\phi dx=0,$$

add then subtract the term \(\frac{{b}^{2}}{\xi }{\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}\) from the above equation we obtain

$$\rho {\int}_{\!\!\!0}^{l}{u}_{tt}udx+(\mu -{b}^{2}/\xi ){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx+\frac{b}{\sqrt{\xi }}{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right){u}_{x}dx=0,$$

taking into account that \(\frac{d}{dt}({u}_{t}u)={u}_{tt}u+|{u}_{t}{|}^{2}\) we arrive at

$$\frac{d}{dt}\left(\rho {\int}_{\!\!\!0}^{l}{u}_{t}udx\right)=\rho {\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx-(\mu -{b}^{2}/\xi ){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx-\frac{b}{\sqrt{\xi }}{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right){u}_{x}dx.$$
(3.2)

Multiply equation (3.2) by \(\phi \) and integrate by parts over \((0,l)\) we get

$$J{\int}_{\!\!\!0}^{l}{u}_{tt}{\phi }_{x}dx+\delta {\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+\sqrt{\xi }{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\phi dx+d{\int}_{\!\!\!0}^{l}{w}_{x}\phi dx=0$$
(3.3)

From equation (3.1) we get that \({\phi }_{x}=\frac{\rho }{b}{u}_{tt}-\frac{\mu }{b}{u}_{xx},\) then substitute \({\phi }_{x}\) in equation (3.3) and taking into account that \(\frac{d}{dt}({u}_{tx}{u}_{x})={u}_{ttx}{u}_{x}+|{u}_{tx}{|}^{2}\) we obtain

$$\frac{d}{dt}\left(\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{u}_{xt}{u}_{x}dx\right)-\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}|{u}_{tx}{|}^{2}dx+\frac{J\rho }{b}{\int}_{\!\!\!0}^{l}|{u}_{tt}{|}^{2}dx+\delta {\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx$$
$$+\sqrt{\xi }{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\phi dx-d{\int}_{\!\!\!0}^{l}w{\phi }_{x}dx=0,$$

Using Poincaré and Young’s inequality, we get

$$\begin{aligned}\frac{d}{dt}\left(\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{u}_{xt}{u}_{x}dx\right)& \le \frac{J\mu }{b}{\int}_{\!\!\!0}^{l}|{u}_{tx}{|}^{2}dx-\frac{J\rho }{b}{\int}_{\!\!\!0}^{l}|{u}_{tt}{|}^{2}dx-\frac{\delta }{2}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx\\ & \quad -\sqrt{\xi }{\int}_{\!\!\!0}^{l}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\phi dx+\frac{{d}^{2}}{2\delta }{\int}_{\!\!\!0}^{l}|w{|}^{2}dx.\end{aligned}$$
(3.4)

Add the two equations (3.2) and (3.4) we obtain

$$\begin{aligned}\frac{d}{dt}\left(\rho {\int}_{\!\!\!0}^{l}{u}_{t}udx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{u}_{xt}{u}_{x}dx\right)& \le \frac{J\mu }{b}{\int}_{\!\!\!0}^{l}|{u}_{xt}{|}^{2}dx+\rho {\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx\\ &\quad-(\mu -{b}^{2}/\xi ){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx\\ & \quad -\frac{J\rho }{b}{\int}_{\!\!\!0}^{l}\left|{u}_{tt}\right|^{2}dx-{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ &\quad+\frac{{d}^{2}}{2\delta }{\int}_{\!\!\!0}^{l}|w{|}^{2}dx-\frac{\delta }{2}{\int}_{\!\!\!0}^{L}|{\phi }_{x}{|}^{2}dx.\end{aligned} $$

Add and subtract the term \(\rho {\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx\) to the right side of the above inequality, then use Poincaré inequality we get the desired result. \(\hfill \square \)

Set

$$\mathcal{G}(t)=-J{\int}_{\!\!\!0}^{l}{u}_{tx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}{u}_{tx}\phi dx+\frac{\alpha {C}_{1}}{d}{\int}_{\!\!\!0}^{l}w{u}_{t}dx.$$

Lemma 7

Let \((u,\phi ,w)\) be a solution of the system (3.1)–(3.5). Then we have

$$\begin{aligned}\frac{d}{dt}\mathcal{G}(t)& \le -\frac{Jb}{2\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{u}_{xt}{|}^{2}dx+\frac{{\varepsilon }_{3}}{2}{\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx+\frac{{\varepsilon }_{1}}{2}{\int}_{\!\!\!0}^{l}|{u}_{tt}{|}^{2}dx-\delta (\mu -{b}^{2}/\xi )\frac{\sqrt{\xi }}{\mu }{\int}_{\!\!\!0}^{L}|{\phi }_{x}{|}^{2}dx\\ & \quad-\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+{C}_{2}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx+{C}_{3}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(3.5)

where \({C}_{1}\), \({C}_{2}\) and \({C}_{3}\) are positive constant to be determined.

Proof

Multiply equation (3.2) by \(\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\) we get:

$$\begin{aligned}& -J{\int}_{\!\!\!0}^{l}{u}_{ttx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx+\delta {\int}_{\!\!\!0}^{l}{\phi }_{x}{\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}dx\\ & \quad=-\sqrt{\xi }{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx-d{\int}_{\!\!\!0}^{l}{w}_{x}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx.\end{aligned}$$
(3.6)

Using Young’s inequality we obtain

$$\begin{aligned}& -J{\int}_{\!\!\!0}^{l}{u}_{ttx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx+\delta {\int}_{\!\!\!0}^{l}{\phi }_{x}{\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}dx\\ & \quad\le -\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\frac{{d}^{2}}{2\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(3.7)

Add then subtract the term \(\frac{\mu \xi }{b}{\phi }_{x}\) to equation (3.1) we get

$${\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}=\frac{\rho b}{\mu \sqrt{\xi }}{u}_{tt}+(\mu -{b}^{2}/\xi )\frac{\sqrt{\xi }}{\mu }{\phi }_{x}.$$
(3.8)

Substitute \({\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{x}\) in equation (3.7) we arrive at

$$\begin{aligned}& -J{\int}_{\!\!\!0}^{l}{u}_{ttx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}\phi {u}_{ttx}dx\\ &\quad \le -\delta (\mu -{b}^{2}/\xi )\frac{\sqrt{\xi }}{\mu }{\int}_{\!\!\!0}^{L}|{\phi }_{x}{|}^{2}dx-\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\frac{{d}^{2}}{2\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(3.9)

Taking into account that \({u}_{ttx}\phi =\frac{d}{dt}({u}_{tx}\phi )-{u}_{tx}{\phi }_{t}\) and \({u}_{ttx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)=\frac{d}{dt}\left[{u}_{tx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)\right]-{u}_{tx}{\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)}_{t}\) we obtain

$$\begin{aligned}& \frac{d}{dt}\left(-J{\int}_{\!\!\!0}^{l}{u}_{tx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}\phi {u}_{tx}dx\right)\\ & \quad\le -\frac{Jb}{\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{u}_{tx}{|}^{2}-{}_{{C}_{1}}{{\left(J\sqrt{\xi }+\delta \frac{\rho b}{\mu \sqrt{\xi }}\right)}}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{tx}dx-\delta (\mu -{b}^{2}/\xi )\frac{\sqrt{\xi }}{\mu }{\int}_{\!\!\!0}^{L}|{\phi }_{x}{|}^{2}dx\\ & \qquad -\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx+\frac{{d}^{2}}{2\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx.\end{aligned}$$
(3.10)

Now multiply equation (3.3) by \(\frac{{C}_{1}}{d}{u}_{t}\), integrate by parts over \((0,l)\) we get and using boundary conditions (3.4) we have

$$\frac{\alpha {C}_{1}}{d}{\int}_{\!\!\!0}^{l}{w}_{t}{u}_{t}dx+\frac{\kappa {C}_{1}}{d}{\int}_{\!\!\!0}^{l}{w}_{x}{u}_{tx}dx-{C}_{1}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{tx}dx+\frac{k{C}_{1}}{d}{\int}_{\!\!\!0}^{l}w{u}_{t}dx=0.$$

Taking into account that \({w}_{t}{u}_{t}=\frac{d}{dt}(w{u}_{t})-w{u}_{tt}\) we obtain

$$\frac{d}{dt}\left(\frac{\alpha {C}_{1}}{d}{\int}_{\!\!\!0}^{l}w{u}_{t}dx\right)=\frac{\alpha {C}_{1}}{d}{\int}_{\!\!\!0}^{l}w{u}_{tt}dx-\frac{\kappa {C}_{1}}{d}{\int}_{\!\!\!0}^{l}{w}_{x}{u}_{tx}dx+{C}_{1}{\int}_{\!\!\!0}^{l}{\phi }_{t}{u}_{tx}dx-\frac{k{C}_{1}}{d}{\int}_{\!\!\!0}^{l}w{u}_{t}dx.$$
(3.11)

Add the two equations (3.10) and (3.11) and apply Young’s inequality then for all \({\varepsilon }_{1}\), \({\varepsilon }_{2}\) and \({\varepsilon }_{3}>0\) we have

$$\begin{aligned}& \frac{d}{dt}\left(-J{\int}_{\!\!\!0}^{l}{u}_{ttx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx-\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}\phi {u}_{tx}dx+\frac{\alpha {C}_{1}}{d}{\int}_{\!\!\!0}^{l}w{u}_{t}dx\right)\\ & \quad\le -\frac{Jb}{\sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{u}_{tx}{|}^{2}-\delta (\mu -{b}^{2}/\xi )\frac{\sqrt{\xi }}{\mu }{\int}_{\!\!\!0}^{L}|{\phi }_{x}{|}^{2}dx-\frac{\sqrt{\xi }}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ &\qquad {}_{{C}_{2}}{{\left(\frac{{\alpha }^{2}{C}_{1}^{2}}{2{d}^{2}{\varepsilon }_{1}}+\frac{{k}^{2}{C}_{1}^{2}}{2{d}^{2}{\varepsilon }_{3}}\right)}}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx+\frac{{\varepsilon }_{1}}{2}{\int}_{\!\!\!0}^{l}|{u}_{tt}{|}^{2}dx\\ &\qquad+{}_{{C}_{3}}{{\left(\frac{{\kappa }^{2}{C}_{1}^{2}}{2{d}^{2}{\varepsilon }_{2}}+\frac{{d}^{2}}{2\sqrt{\xi }}\right)}}{\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx+\frac{{\varepsilon }_{2}}{2}{\int}_{\!\!\!0}^{l}|{u}_{tx}{|}^{2}dx +\frac{{\varepsilon }_{3}}{2}{\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx.\end{aligned}$$
(3.12)

Take \({\varepsilon }_{2}=\frac{Jb}{2\sqrt{\xi }}\) we get the desired result.\(\square \)

Let

$$\mathcal{L}(t)={N}_{1}E(t)+\mathcal{F}(t)+{N}_{2}\mathcal{G}(t),$$

where \({N}_{1}\) and \({N}_{2}\) are positive constants to be fixed.

Theorem 6

There exists positive constants \({\nu }_{1}\) and \({\nu }_{2}\) such that

$${\nu }_{1}E(t)\le \mathcal{L}(t)\le {\nu }_{2}E(t), \forall t\ge 0.$$

Proof

We have

$$\begin{aligned}|\mathcal{L}(t)-{N}_{1}E(t)|& \le \rho {\int}_{\!\!\!0}^{l}{u}_{t}udx+\frac{J\mu }{b}{\int}_{\!\!\!0}^{l}{u}_{xt}{u}_{x}dx+{N}_{2}J{\int}_{\!\!\!0}^{l}{u}_{tx}\left(\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right)dx\\ & \quad+{N}_{2}\delta \frac{\rho b}{\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}{u}_{tx}\phi dx+{N}_{2}\frac{\alpha {C}_{1}}{d}{\int}_{\!\!\!0}^{l}w{u}_{t}dx.\end{aligned}$$

Apply Young’s and Poincaré inequalities we obtain

$$\begin{aligned}|\mathcal{L}(t)-{N}_{1}E(t)|& \le \left(\frac{\rho }{2}+{N}_{2}\frac{\alpha {C}_{1}}{2d}\right){\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx+\left(\frac{\rho {c}_{p}}{2}+\frac{J\mu }{2b}\right){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx\\ &\quad +\left(\frac{J\mu }{2b}+\frac{{N}_{2}J}{2}+{N}_{2}\delta \frac{\rho b}{2\mu \sqrt{\xi }}\right){\int}_{\!\!\!0}^{l}|{u}_{tx}{|}^{2}dx+\frac{{N}_{2}J}{2}{\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ &\quad +{N}_{2}\delta {c}_{p}\frac{\rho b}{2\mu \sqrt{\xi }}{\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx+{N}_{2}\frac{\alpha {C}_{1}}{2d}{\int}_{\!\!\!0}^{l}|w{|}^{2}dx\end{aligned}$$

Define

$$\begin{aligned}{N}_{0}:=&\,{\text{max}}\left\{\frac{1}{\rho }\left(\rho +{N}_{3}\frac{\alpha {C}_{1}}{d}\right);\frac{1}{\mu -{b}^{2}/\xi }\left(\rho {c}_{p}+\frac{J\mu }{b}\right);\right.\\& \left.\frac{b}{J\mu }\left(\frac{J\mu }{b}+{N}_{2}J+{N}_{2}\delta \frac{\rho b}{\mu \sqrt{\xi }}\right);{N}_{2}J;{N}_{2}{c}_{p}\frac{\rho b}{\mu \sqrt{\xi }};{N}_{2}\frac{{C}_{1}}{d}\right\}\end{aligned}$$

Hence

$$|\mathcal{L}(t)-{N}_{1}E(t)|\le {N}_{0}E(t),$$

which implies that

$${\nu }_{1}E(t)\le \mathcal{L}(t)\le {\nu }_{2}E(t),$$

where \({\nu }_{1}={N}_{1}-{N}_{0}\) and \({\nu }_{2}={N}_{1}+{N}_{0}\) and \({N}_{1}>{N}_{0}.\)

Proof of Theorem 5

It follows from Lemmas 6 and 7 that

$$\begin{aligned}\frac{d}{dt}\mathcal{L}(t) \le & -\left(\rho -\frac{{N}_{2}{\varepsilon }_{3}}{2}\right){\int}_{\!\!\!0}^{l}|{u}_{t}{|}^{2}dx-\left(\frac{J\rho }{b}-\frac{{N}_{2}{\varepsilon }_{1}}{2}\right){\int}_{\!\!\!0}^{l}|{u}_{tt}{|}^{2}dx-(\mu -{b}^{2}/\xi ){\int}_{\!\!\!0}^{l}|{u}_{x}{|}^{2}dx\\ & \quad -\left({N}_{2}\frac{Jb}{2\sqrt{\xi }}-(\frac{J\mu }{b}+2\rho {c}_{p})\right){\int}_{\!\!\!0}^{l}|{u}_{xt}{|}^{2}dx-\left(1+\frac{{N}_{2}\sqrt{\xi }}{2}\right){\int}_{\!\!\!0}^{l}{\left|\frac{b}{\sqrt{\xi }}{u}_{x}+\sqrt{\xi }\phi \right|}^{2}dx\\ & \quad -\left(\frac{\delta }{2}+{N}_{2}\delta (\mu -{b}^{2}/\xi )\frac{\sqrt{\xi }}{\mu }\right){\int}_{\!\!\!0}^{l}|{\phi }_{x}{|}^{2}dx\\ & \quad -({N}_{1}k-{N}_{2}{C}_{2}-\frac{{d}^{2}}{2\delta }){\int}_{\!\!\!0}^{l}|w{|}^{2}dx-({N}_{1}\kappa -{N}_{2}{C}_{3}){\int}_{\!\!\!0}^{l}|{w}_{x}{|}^{2}dx \end{aligned}$$
(3.13)

Choose \({\varepsilon }_{3}=\frac{\rho }{{N}_{2}}\), \({\varepsilon }_{1}=\frac{J\rho }{b{N}_{2}}\), \({N}_{2}>\frac{2\sqrt{\xi }}{Jb}(\frac{J\mu }{b}+2\rho {c}_{p})\) and \({N}_{1}>{\text{max}}\left\{\frac{{N}_{2}{C}_{2}+\frac{{d}^{2}}{2\delta }}{k};\frac{{N}_{2}{C}_{3}}{\kappa }\right\}\), from where we obtain that \({\zeta }_{1}=\rho -\frac{{N}_{2}{\varepsilon }_{3}}{2}>0\), \({\zeta }_{2}=\frac{J\rho }{b}-\frac{{N}_{2}{\varepsilon }_{1}}{2}>0\), \({\zeta }_{3}={N}_{2}\frac{Jb}{2\sqrt{\xi }}-(\frac{J\mu }{b}+2\rho {c}_{p})>0\), \({\zeta }_{4}=1+\frac{{N}_{2}\sqrt{\xi }}{2}>0\), \({\zeta }_{5}=\frac{\delta }{2}+{N}_{2}\delta (\mu -{b}^{2}/\xi )\frac{\sqrt{\xi }}{\mu }>0\), \({\zeta }_{6}={N}_{1}k-{N}_{2}{C}_{2}-\frac{{d}^{2}}{2\delta }>0\) and \({\zeta }_{7}={N}_{1}\kappa -{N}_{2}{C}_{3}>0\) and from where we can conclude that there exists a positive constant \(\beta =2{\text{min}}\{1,{\zeta }_{1},{\zeta }_{2},{\zeta }_{3},{\zeta }_{4},{\zeta }_{5},{\zeta }_{6},{\zeta }_{7}\}\) such that

$$\frac{d}{dt}\mathcal{L}(t)\le -\beta E(t),$$

by equivalence between \(E(t)\) and \(\mathcal{L}(t)\) according to Theorem \(6\) we get:

$$\frac{d}{dt}\mathcal{L}(t)\le -{\omega }_{2}\mathcal{L}(t),$$

where \({\omega }_{2}=\frac{\beta }{{\nu }_{2}}\). Now integrate the above inequality over \((0,t)\) we obtain

$$\mathcal{L}(t)\le \mathcal{L}(0){e}^{-{\omega }_{2}t},$$

again by equivalence between \(E(t)\) and \(\mathcal{L}(t)\) according to Theorem 3 we arrive at

$$E(t)\le {M}_{2}E(0){e}^{-{\omega }_{2}t},$$

where \({M}_{2}=\frac{{\nu }_{2}}{{\nu }_{1}}.\) \(\hfill\square \)

3.2.2 Second case

We will prove exponential stability for \(\mu \xi ={b}^{2}\); our result is stated in the following theorem.

Theorem 7

The energy \(E(t)\) of the system (3.1–3.5) decays exponentially as time t tends to infinity. That is, there exist two positive constants \({M}_{3}\) and \({\omega }_{3}\) independent of the initial data and independent of any relationship between coefficients such that

$$E(t)\le {M}_{3}E(0)\,\,{e}^{-{\omega }_{3}t}, \forall t\ge 0.$$

Proof

We will use the same multipliers used in Theorem 5, and define the Lyapunov functional.

$$\mathbf{L}(t)={N}_{3}E(t)+\mathcal{F}(t)+{N}_{4}\mathcal{G}(t),$$

where \({N}_{3}\) and \({N}_{4}\) are positive constants to be fixed. We proceed similarly as the proof of Theorem 5 to get the desired result.

Fig. 1
figure 1

Graphs-Classical system with equal speed limits

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Fig. 2
figure 2

Energy as a function of time

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Fig. 3
figure 3

\(-log[energy]\) as function of time

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Fig. 4
figure 4

Graphs-Classical system with non equal speed limits

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Fig. 5
figure 5

Energy as a function of time

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Fig. 6
figure 6

\(-log[energy]\) as function of time

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Fig. 7
figure 7

Graphs-Second spectrum free system

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Fig. 8
figure 8

Energy as a function of time

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Fig. 9
figure 9

\(-log[energy]\) as function of time

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4 Numerical simulations

First we denote by \(\widehat{u}={u}_{t},\widehat{\phi }={\phi }_{t}\), \(\widehat{w}={w}_{t}\) and we introduce the following weak form after multiplying the equations (3.1), (3.2), (3.3) by \(\overline{u },\overline{\phi },\overline{w}\in {H }_{0}^{1}(0,l)\)

$$(WP)\left\{\begin{aligned}& \rho ({\widehat{u}}_{t},\overline{u })+\mu ({u}_{x},{\overline{u} }_{x})+b(\phi ,{\overline{u} }_{x})=0\\ & J({\widehat{u}}_{t},{\overline{\phi }}_{x})+\delta ({\phi }_{x},{\overline{\phi }}_{x})+b({u}_{x},\overline{\phi })+\xi (\phi ,\overline{\phi })+d({w}_{x},\overline{\phi })=0\\ & \alpha (\widehat{w},\overline{w })+\kappa ({w}_{x},{\overline{w} }_{x})+d({\widehat{\phi }}_{x},\overline{w })+k(w,\overline{w })=0\end{aligned}\right.$$
(4.1)

Let us partition the interval \((0;l)\) into subintervals \({I}_{j}=({x}_{j-1};{x}_{j})\) of length \(h=\frac{1}{s}\) with \(0={x}_{0}<{x}_{1}<\cdots<{x}_{s}=l\) and define the associated finite element spaces by

$${S}_{h}^{0}=\{u\in {H}_{0}^{1}(0,l);u\in C([0,l]),u{|}_{{I}_{j}}\in {P}_{1}(K)\}.$$

For a given final time \(T\) and a positive integer\(N\), define the time step \(\Delta t=\frac{T}{N}\) and the nodes\({t}_{n}=n\Delta t,n=0,...,N\). By using the Implicit Euler scheme in time and the finite element variational approximation in space, we introduce the following scheme. For \({\overline{u} }_{h},{\overline{\phi }}_{h},{\overline{w} }_{h}\in {S}_{h}^{0}\), find \({u}_{h}^{n},{\phi }_{h}^{n},{w}_{h}^{n}\in {S}_{h}^{0}\) such that,

$$(NP)\left\{\begin{aligned}& \frac{\rho }{\Delta t}({\widehat{u}}_{h}^{n}-{\widehat{u}}_{h}^{n-1},{\overline{u} }_{h})+\mu ({u}_{hx}^{n},{\overline{u} }_{hx})+b({\phi }_{h}^{n},{\overline{u} }_{hx})=0\\ & \frac{J}{\Delta t}({\widehat{u}}_{h}^{n}-{\widehat{u}}_{h}^{n-1},{\overline{\phi }}_{hx})+\delta ({\phi }_{hx}^{n},{\overline{\phi }}_{hx})+b({u}_{hx}^{n},{\overline{\phi }}_{h})+\xi ({\phi }_{h}^{n},{\overline{\phi }}_{h})+d({w}_{hx}^{n},{\overline{\phi }}_{h})=0\\ & \frac{\alpha }{\Delta t}({w}_{h}^{n}-{w}_{h}^{n-1},{\overline{w} }_{h})+\kappa ({w}_{hx}^{n},{\overline{w} }_{hx})+d({\widehat{\phi }}_{hx}^{n},{\overline{w} }_{h})+k({w}_{h}^{n},{\overline{w} }_{h})=0,\end{aligned}\right.$$
(4.2)

where \({\widehat{u}}_{h}^{n}=\frac{{u}_{h}^{n}-{u}_{h}^{n-1}}{\Delta t}\), \({\widehat{\phi }}_{h}^{n}=\frac{{\phi }_{h}^{n}-{\phi }_{h}^{n-1}}{\Delta t}\) and \({\widehat{w}}_{h}^{n}=\frac{{w}_{h}^{n}-{w}_{h}^{n-1}}{\Delta t}\).

Plugging \({\widehat{u}}_{h}^{n}\), \({\widehat{\phi }}_{h}^{n}\) and \({\widehat{w}}_{h}^{n}\) in system (4.2) we get,

$$\left\{\begin{aligned}& \frac{\rho }{{\left(\Delta t\right)}^{2}}\left({u}_{h}^{n}-2{u}_{h}^{n-1}+{u}_{h}^{n-2},{\overline{u} }_{h}\right)+\mu \left({u}_{hx}^{n},{\overline{u} }_{hx}\right)+b\left({\phi }_{h}^{n},{\overline{u} }_{hx}\right)=0,\\ & \frac{J}{{\left(\Delta t\right)}^{2}}\left({\phi }_{h}^{n}-2{\phi }_{h}^{n-1}+{\phi }_{h}^{n-2},{\overline{\phi }}_{h}\right)+\delta \left({\phi }_{hx}^{n},{\overline{\phi }}_{hx}\right)+b\left({u}_{hx}^{n},{\overline{\phi }}_{h}\right)+\xi \left({\phi }_{h}^{n},{\overline{\phi }}_{h}\right)+d\left({w}_{hx}^{n},{\overline{\phi }}_{h}\right)=0,\\ & \frac{\alpha }{\Delta t}\left({w}_{h}^{n}-{w}_{h}^{n-1},{\overline{w} }_{h}\right)+\kappa \left({w}_{hx}^{n},{\overline{w} }_{hx}\right)+\frac{d}{\Delta t}\left({\phi }_{hx}^{n}-{\phi }_{hx}^{n-1},{\overline{w} }_{h}\right)+k\left({w}_{h}^{n},{\overline{w} }_{h}\right)=0,\end{aligned}\right.$$

note that by the finite element theory, \({u}_{h}^{n}={\sum }_{i=1}^{s}{a}_{i}^{n}{\psi }_{i}\), \({\phi }_{h}^{n}={\sum }_{i=1}^{s}{b}_{i}^{n}{\psi }_{i}\) and \({w}_{h}^{n}={\sum }_{i=1}^{s}{c}_{i}^{n}{\psi }_{i}\) where \({\psi }_{i}\) are bases of the finite space \({S}_{h}^{0}\). Taking \({\overline{u} }_{h}={\psi }_{j}\) we get

$$\left\{\begin{aligned}& \frac{\rho }{{\left(\Delta t\right)}^{2}}Z{A}^{n}+\mu T{A}^{n}+bY{B}^{n}=\frac{2\rho }{{\left(\Delta t\right)}^{2}}Z{A}^{n-1}-\frac{\rho }{{\left(\Delta t\right)}^{2}}Z{A}^{n-2},\\ & \frac{J}{{\left(\Delta t\right)}^{2}}Z{A}^{n}+\delta T{B}^{n}+bX{A}^{n}+\xi Z{B}^{n}+dX{C}^{n}=\frac{2J}{{\left(\Delta t\right)}^{2}}Y{A}^{n-1}-\frac{J}{{\left(\Delta t\right)}^{2}}Y{A}^{n-2}\\ & \frac{\alpha }{\Delta t}Z{C}^{n}+\kappa T{C}^{n}+\frac{d}{\Delta t}X{B}^{n}+kZ{C}^{n}=\frac{\alpha }{\Delta t}Z{C}^{n-1}+\frac{d}{\Delta t}X{B}^{n-1}\end{aligned}\right.$$
(4.3)

where the vectors \({A}^{n}\), \({B}^{n}\) and \({C}^{n}\) are given by

$${A}^{n}=\left\{{a}_{i}^{n}\right\}$$
$${B}^{n}=\left\{{b}_{i}^{n}\right\}$$
$${C}^{n}=\left\{{c}_{i}^{n}\right\}$$

and the matrices \(Z\), \(X\), \(Y\), and \(T\) are given by

$$X=\left({\psi }_{ix},{\psi }_{j}\right)$$
$$Y=\left({\psi }_{i},{\psi }_{jx}\right)$$
$$Z=\left({\psi }_{i},{\psi }_{j}\right)$$
$$T=\left({\psi }_{ix},{\psi }_{jx}\right)$$

We solve (4.3) using the following initial and physical data, \(\rho =d=\alpha =b=\xi =j=0.001\), \(k=1\) and \(\mu =0.01\). The space discritization \(\Delta x=\frac{1}{11}\) and the time dicritization \(\Delta t=\frac{1}{22}\) with total time \(T=25\). The initial data \({u}_{h}^{0}={\phi }_{h}^{0}={w}_{h}^{0}={u}_{h}^{1}={\phi }_{h}^{1}=(1-x)x.\)

4.1 Classical system with equal speed limit

See Figs. 1, 2 and 3.

4.2 Classical system with non-equal speed limit

See Figs. 4, 5 and 3.

4.3 Second spectrum free system

See Figs. 7, 8 and 9.

4.4 Graphical analysis

The case of classical system with equal Speed condition: In this case, the energy decays so fast to zero which shows that the decay type is exponential, and what proves this is the graph of \(log(E(t))\) that shows a straight. This is graphical evidence of exponential behavior. The fast decay of \(u,\phi ,\) and \(w\) also show the exponential decay.

The case of classical system with non-equal Speed condition: Starting with the graphs of the functions \(u,\phi ,\) and \(w\), it is clear that they have much more vibratory behavior than the previous case, and this is a direct indication that the decay is slower than the previous case. Regarding the graph of the energy, although the graph shows a decay to zero the graph of \(log(E(t))\) proves that we lost the exponential decay.

The second spectrum free case: In this case also the energy decays so fast to zero which shows that the decay type is exponential, and what proves this is the graph of \(log(E(t))\) that shows a straight. This is graphical evidence of exponential behavior.