Existence and Stability Results for Thermodiffusion Laminated Beam System with Delay Feedback

Abstract

In this paper, a one-dimensional thermodiffusion laminated beam system with delay feedback is studied. The existence of a solution for our system is discussed within the context of the semigroup approach. In addition, under different boundary conditions, two results of stability properties independent of initial data are investigated.

1. Introduction

The aim of this study is to investigate the following system, for $(x,t)\in (0,1)\times (0,\infty )$:

$$\left\{\begin{array}{c}\rho {\varrho }_{tt}+G{\left(\varpi -{\varrho }_x\right)}_x+\mu {\varrho }_t(x,t-\tau )=0,\hfill \\ I_{\rho }{(3s-\varpi )}_{tt}-D{(3s-\varpi )}_{xx}-G\left(\varpi -{\varrho }_x\right)-{\gamma }_1{\theta }_x-{\gamma }_2{P}_x=0,\hfill \\ I_{\rho }{s}_{tt}-D{s}_{xx}+G\left(\varpi -{\varrho }_x\right)+{\displaystyle \frac{4}{3}}\gamma s+{\displaystyle \frac{4}{3}}\beta s_t=0,\hfill \\ c{\theta }_t+d{P}_t-\kappa {\theta }_{xx}-{\gamma }_1{(3s-\varpi )}_{xt}=0,\hfill \\ d{\theta }_t+r{P}_t-\hslash P_{xx}-{\gamma }_2{(3s-\varpi )}_{xt}=0,\hfill \end{array}\right.$$

(1)

along with the initial conditions ($x\in (0,1))$:

$$\left\{\begin{array}{c}\varrho (x,t=0)={\varrho }_0\left(x\right),\varpi (x,t=0)={\varpi }_0\left(x\right),s(x,t=0)=s_0\left(x\right),\theta (x,t=0)={\theta }_0\left(x\right),\hfill \\ P(x,t=0)=P_0\left(x\right),{\varrho }_t(x,t=0)={\varrho }_1\left(x\right),{\varpi }_t(x,t=0)={\varpi }_1\left(x\right),s_t(x,t=0)=s_1\left(x\right)\hfill \\ {\varrho }_t(x,-t)=f_0(x,t),\hfill \end{array}\right.$$

(2)

where $\tau >0$ is the time delay, and the non-zero real number $\mu >0$ is the weight of delay. With some restrictions on the parameters, we take into account the two boundary conditions listed below:

Boundary I ($t>0$):

$$\left\{\begin{array}{c}{\varrho }_x(x=0,t)=\varpi (x=0,t)=s(x=0,t)=\theta (x=0,t)=P(x=0,t)=0,\hfill \\ \varrho (x=1,t)={\varpi }_x(x=1,t)=s_x(x=1,t)=\theta (x=1,t)=P(x=1,t)=0,\hfill \end{array}\right.$$

(3)

and Boundary II ($t>0$):

$$\left\{\begin{array}{c}\varrho (x=0,t)=\varpi (x=0,t)=s(x=0,t)=\theta (x=0,t)=P(x=0,t)=0,\hfill \\ \varrho (x=1,t)=\varpi (x=1,t)=s(x=1,t)=\theta (x=1,t)=P(x=1,t)=0.\hfill \end{array}\right.$$

(4)

${\gamma }_1,{\gamma }_2,r,c$ and d are positive constants. We assume that the wave speeds are identical:

$${\displaystyle \frac{\rho }{G}}={\displaystyle \frac{I_{\rho }}{D}}.$$

(5)

We suppose also that the symmetric matrix

$$\Gamma =\left(\begin{array}{cc}c& d\\ d& r\end{array}\right)$$

is positive definite. Then, we deduce that

$$\delta =cr-d^2>0.$$

(6)

Thus, for all $\theta$ and P, we have

$$c{\theta }^2+2d\theta P+r{P}^2>0.$$

(7)

It might be assumed that dissipation can only be explained through thermal conductivity in Timoshenko beams. However, research stemming from the development of advanced technologies after World War II has shown that the role of diffusion in solids cannot be ignored. This leads to a key question: what occurs when both diffusion and thermal effects are considered in laminated beams? Diffusion is the random movement of particles from areas of higher concentration to lower concentration. In elastic solids, thermodiffusion results from the interaction between strain, temperature, and mass diffusion fields. Heat and mass diffusion processes are crucial in various engineering applications, such as satellite systems, returning spacecraft, and aircraft landings on water or land. Currently, there is a strong interest in diffusion processes within the production of integrated circuits, resistors, semiconductor substrates, and MOS transistors. Oil companies are also focusing on this phenomenon to optimize oil extraction conditions.

First, we evaluate similar findings in the literature. Apalara [1] studied

$$\left\{\begin{array}{c}\rho {\phi }_{tt}+G{(\psi -{\phi }_x)}_x=0,\mathrm{in}\left(0,1\right)\times \left(0,+\infty \right),\hfill \\ I_{\rho }{(-\psi +3s)}_{tt}-D{(-\psi +3s)}_{xx}-G(\psi -{\phi }_x)=0,\mathrm{in}\left(0,1\right)\times \left(0,+\infty \right),\hfill \\ I_{\rho }{s}_{tt}-D{s}_{xx}+3G(\psi -{\phi }_x)+4\gamma s+\delta {\theta }_x=0,\mathrm{in}\left(0,1\right)\times \left(0,+\infty \right),\hfill \\ \delta s_{xt}-\lambda {\theta }_{xx}+{\rho }_3{\theta }_t=0,\mathrm{in}\left(0,1\right)\times \left(0,+\infty \right),\hfill \end{array}\right.$$

Its soultion is exponential stable when ${\displaystyle \frac{G}{\rho }}={\displaystyle \frac{D}{I_{\rho }}}$. Aouadi et al. [2] investigated the following Timoshenko beam with thermodiffusion effects:

$$\left\{\begin{array}{c}{\rho }_1{\phi }_{tt}-k{\left({\phi }_x+\psi \right)}_x=0,\mathrm{in}\left(0,L\right)\times \left(0,+\infty \right),\hfill \\ {\rho }_2{\psi }_{tt}-\alpha {\psi }_{xx}+k\left({\phi }_x+\psi \right)-{\gamma }_1{\theta }_x-{\gamma }_2{P}_x=0,\mathrm{in}\left(0,L\right)\times \left(0,+\infty \right),\hfill \\ c{\theta }_t+d{P}_t-k{\theta }_{xx}-{\gamma }_1{\psi }_{xt}=0,\mathrm{in}\left(0,L\right)\times \left(0,+\infty \right),\hfill \\ d{\theta }_t+r{P}_t-\mathfrak{h}{P}_{xx}-{\gamma }_2{\psi }_{xt}=0,\mathrm{in}\left(0,L\right)\times \left(0,+\infty \right),\hfill \end{array}\right.$$

They proved the existence and exponential stability result. We refer the reader to [3,4,5,6] and the references cited therein for more related results.

This work is new due to the idea of mass diffusion in laminated beams, which can have important physical consequences in addition to deformations of the body. In micro-beam resonators, for example, the current research has concentrated on the effect of mass diffusion on the damping ratio (see, e.g., [7]). Moreover, in addition to the traditional critical thickness linked to thermoelastic damping, mass diffusion provides a new one. The aforementioned arguments imply that mass diffusion will be essential to understanding laminated beam models’ thermomechanical behavior. According to the literature review, no numerical or theoretical studies investigating the impact of mass diffusion on the thermal vibrations of laminated beams have been carried out. This work therefore aims to explore the impact of mass diffusion in addition to temperature impacts on laminated beam behavior.

The structure of the paper is as follows: Section 2 provides our first main result about well posedness and the procedure used to achieve this result is a semigroup approach. For the two types of boundary conditions, we shall state and prove the exponential stability in Section 3.

2. Well-Posedness Results of the System

In order to show the well-posedness result of system (1), (2), and (4) or problems (1)–(3), we denote $\xi =3s-\varpi$; then, we introduce this new variable as follows:

$$z\left(x,\sigma ,t\right)={\varrho }_t\left(x,t-\tau \sigma \right),x\in \left(0,1\right),\sigma \in \left(0,1\right),t>0.$$

Then, we obtain the following equation:

$$\tau z_t\left(x,\sigma ,t\right)+z_{\sigma }\left(x,\sigma ,t\right)=0,\left(x,\sigma ,t\right)\in \left(0,1\right)\times \left(0,1\right)\times \left(0,+\infty \right).$$

Then, system (1) can be transformed as follows:

$$\left\{\begin{array}{c}\rho {\varrho }_{tt}+G{\left(3s-\xi -{\varrho }_x\right)}_x+\mu z(x,1,t)=0,\hfill \\ I_{\rho }{\xi }_{tt}-D{\xi }_{xx}-G\left(3s-\xi -{\varrho }_x\right)-{\gamma }_1{\theta }_x-{\gamma }_2{P}_x=0,\hfill \\ I_{\rho }{s}_{tt}-D{s}_{xx}+G\left(3s-\xi -{\varrho }_x\right)+{\displaystyle \frac{4}{3}}\gamma s+{\displaystyle \frac{4}{3}}\beta s_t=0,\hfill \\ c{\theta }_t+d{P}_t-\kappa {\theta }_{xx}-{\gamma }_1{\xi }_{xt}=0,\hfill \\ d{\theta }_t+r{P}_t-\hslash P_{xx}-{\gamma }_2{\xi }_{xt}=0,\hfill \\ \tau z_t\left(x,\sigma ,t\right)+z_{\sigma }\left(x,\sigma ,t\right)=0,\hfill \end{array}\right.$$

(8)

With the common initial data ($x\in (0,1)$), have the following:

$$\left\{\begin{array}{c}\varrho (x,t=0)={\varrho }_0\left(x\right),\xi (x,t=0)={\xi }_0\left(x\right),s(x,t=0)=s_0\left(x\right),\theta (x,0)={\theta }_0\left(x\right),\hfill \\ P(x,t=0)=P_0\left(x\right),{\varrho }_t(x,t=0)={\varrho }_1\left(x\right),{\xi }_t(x,t=0)={\xi }_1\left(x\right),s_t(x,0)=s_1\left(x\right),\hfill \\ z(x,0,t)={\varrho }_t(x,t),z(x,\sigma ,0)=f_0(x,\sigma t),(\sigma ,t)\in \left(0,1\right)\times \left(0,+\infty \right).\hfill \end{array}\right.$$

The boundary conditions are taken in two different cases as

Boundary I ($t>0)$:

$$\left\{\begin{array}{c}{\varrho }_x(x=0,t)=\xi (x=0,t)=s(x=0,t)=\theta (x=0,t)=P(x=0,t)=0,\hfill \\ \varrho (x=1,t)={\xi }_x(x=1,t)=s_x(x=1,t)=\theta (x=1,t)=P(x=1,t)=0,\hfill \end{array}\right.$$

and Boundary II ($t>0)$:

$$\left\{\begin{array}{c}\varrho (x=0,t)=\xi (x=0,t)=s(x=0,t)=\theta (x=0,t)=P(x=0,t)=0,\hfill \\ \varrho (x=1,t)=\xi (x=1,t)=s(x=1,t)=\theta (x=1,t)=P(x=1,t)=0,\hfill \end{array}\right.$$

Let $u={\varrho }_t,v={\xi }_t$ and $y=s_t$, and introducing the vector function

$$U\left(t\right)={\left(\varrho ,u,\xi ,v,s,y,\theta ,P,z\right)}^T,$$

we define the spaces

$$\begin{array}{c}\hfill H_a^1(0,1)=\left\{y\in H^1(0,1):y(x=0)=0\right\},\\ \hfill H_b^1(0,1)=\left\{v\in H^1(0,1):v(x=1)=0\right\},\\ \hfill H_b^2(0,1)=\left\{y\in H^2(0,1):y_x(x=0)=0\right\},\\ \hfill H_a^2(0,1)=\left\{v\in H^2(0,1):v_x(x=1)=0\right\},\end{array}$$

as well as the phase spaces ${\mathcal{H}}_1$ and ${\mathcal{H}}_2$ with respect to Boundary I and Boundary II as

$${\mathcal{H}}_1=H_b^1(0,1)\times L^2(0,1)\times {\left[H_a^1(0,1)\times L^2(0,1)\right]}^2\times {\left[L^2(0,1)\right]}^2\times L^2\left((0,1)\times (0,1)\right),$$

and

$${\mathcal{H}}_2={\left[H_0^1(0,1)\times L^2(0,1)\right]}^3\times {\left[L^2(0,1)\right]}^2\times L^2\left((0,1)\times (0,1)\right),$$

Both state spaces are equipped with the same inner product, which is given by the following:

$$\begin{array}{ccc}\hfill {〈U,\tilde{U}〉}_{{\mathcal{H}}_i}& =& {\int }_0^1[\rho u\tilde{u}+G\left(3s-\xi -{\varrho }_x\right)\left(3\tilde{s}-\tilde{\xi }-{\tilde{\varrho }}_x\right)+I_{\rho }v\tilde{v}+D{\xi }_x{\tilde{\xi }}_x\hfill \\ & & +3I_{\rho }y\tilde{y}+3D{s}_x{\tilde{s}}_x+4\gamma s\tilde{s}+c\theta \tilde{\theta }+d\tilde{\theta }P+d\theta \tilde{P}+rP\tilde{P}]dx\hfill \\ & & +\tau \left|\mu \right|\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}z(x,\sigma )\tilde{z}(x,\sigma )d\sigma dx.\hfill \end{array}$$

This is defined for all $U\left(t\right)=(\varrho ,u,\xi ,v,s,y,\theta ,P,z),\tilde{U}\left(t\right)=(\tilde{\varrho },\tilde{u},\tilde{\xi },\tilde{v},\tilde{s},\tilde{y},\tilde{\theta },\tilde{P},\tilde{z})\in {\mathcal{H}}_i,i=1,2$.

Then, system (8) can be transformed into a first-order Cauchy problem:

$$\left\{\begin{array}{c}{\displaystyle \frac{d}{dt}}U\left(t\right)={\mathcal{A}}_{i}U\left(t\right),i=1,2,\hfill \\ U\left(0\right)=U_0={\left({\varrho }_0,{\varrho }_1,{\xi }_0,{\xi }_1,s_0,s_1,{\theta }_0,P_0,f_0\right)}^T,\hfill \end{array}\right.$$

Therein, the operator ${\mathcal{A}}_i,i=1,2$ is given by

$${\mathcal{A}}_{i}U=\left(\begin{array}{c}u\\ \\ -{\displaystyle \frac{1}{\rho }}\left(G{\left(3s-\xi -{\varrho }_x\right)}_x+\mu z(x,1,t)\right)\\ \\ v\\ \\ {\displaystyle \frac{1}{I_{\rho }}}\left(D{\xi }_{xx}+G\left(3s-\xi -{\varrho }_x\right)+{\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\\ \\ y\\ \\ {\displaystyle \frac{1}{I_{\rho }}}\left(D{s}_{xx}-G\left(3s-\xi -{\varrho }_x\right)-{\displaystyle \frac{4}{3}}\gamma s-{\displaystyle \frac{4\beta }{3}}y\right)\\ \\ {\displaystyle \frac{\kappa r}{\delta }}{\theta }_{xx}-{\displaystyle \frac{\hslash d}{\delta }}{P}_{xx}+{\displaystyle \frac{{\gamma }_{1}r-{\gamma }_{2}d}{\delta }}{v}_x\\ \\ {\displaystyle \frac{\hslash c}{\delta }}{P}_{xx}-{\displaystyle \frac{\kappa d}{\delta }}{\theta }_{xx}+{\displaystyle \frac{{\gamma }_{2}c-{\gamma }_{1}d}{\delta }}{v}_x\\ \\ -{\displaystyle \frac{1}{\tau }}{z}_{\sigma }(x,\sigma )\end{array}\right).$$

Therein, we have

$$\begin{array}{ccc}\hfill \mathcal{D}\left({\mathcal{A}}_1\right)& =& \{U\in {\mathcal{H}}_1/\varrho \in H_b^2(0,1)\cap H_b^1(0,1),\xi ,s\in H_a^2(0,1)\cap H_a^1(0,1),u\in H_b^1(0,1)\hfill \\ & & v,y\in H_a^1(0,1),\theta ,P\in H^2(0,1)\cap H_0^1(0,1),z,z_{\sigma }\in L^2\left((0,1)\times (0,1)\right)\}.\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \mathcal{D}\left({\mathcal{A}}_2\right)& =& \{U\in {\mathcal{H}}_2/\varrho ,\xi ,s\in H^2(0,1)\cap H_0^1(0,1),u,v,y\in H_0^1(0,1),\hfill \\ & & \theta ,P\in H^2(0,1)\cap H_0^1(0,1),z,z_{\sigma }\in L^2\left((0,1)\times (0,1)\right)\}.\hfill \end{array}$$

The first result related to the well posedness in both boundary conditions is presented in the following theorem.

Theorem 1.

For $U_0\in {\mathcal{H}}_1$ (resp. $U_0\in {\mathcal{H}}_2$), the problem (1)−(3) (resp. (1)−(2)) and (4) has a unique weak solution:

$$U\left(t\right)\in C\left([0,\infty ),{\mathcal{H}}_1\right)\left(resp.U\left(t\right)\in C\left([0,\infty ),{\mathcal{H}}_2\right)\right).$$

In addition, if $U_0\in \mathcal{D}\left({\mathcal{A}}_1\right)$ (resp. $U_0\in \mathcal{D}\left({\mathcal{A}}_2\right)$), then

$$U\left(t\right)\in C\left([0,\infty ),\mathcal{D}\left({\mathcal{A}}_1\right)\right)\cap C^1\left([0,\infty ),{\mathcal{H}}_1\right)\left(resp.U\left(t\right)\in C\left([0,\infty ),\mathcal{D}\left({\mathcal{A}}_2\right)\right)\cap C^1\left([0,\infty ),{\mathcal{H}}_2\right).\right.$$

Proof. 

In order to prove Theorem 1, we use the semigroup approach. That is, this is done to show that the operator ${\mathcal{A}}_i$ generates a $C_0-$ semigroup in ${\mathcal{H}}_i$. At this stage, we first prove the dissipativity of the operator

$${\mathcal{B}}_i={\mathcal{A}}_i-c_{1}I,i=1,2.$$

For

$$U=(\varrho ,u,\xi ,v,s,y,\theta ,P,z)\in \mathcal{D}\left({\mathcal{A}}_i\right),i=1,2,$$

we have

$$\begin{array}{ccc}& & {〈A_{i}U\left(t\right),U\left(t\right)〉}_{{\mathcal{H}}_i}\hfill \\ & =& {\int }_0^1\rho u\left[-{\displaystyle \frac{G}{\rho }}{\left(3s-\xi -{\varrho }_x\right)}_x-{\displaystyle \frac{\mu }{\rho }}z(x,1)\right]dx+{\int }_0^{1}G\left(3s-\xi -{\varrho }_x\right)\left(3y-v-u_x\right)dx\hfill \\ & +& {\int }_0^1{I}_{\rho }v\left[{\displaystyle \frac{D}{I_{\rho }}}{\xi }_{xx}+{\displaystyle \frac{G}{I_{\rho }}}\left(3s-\xi -{\varrho }_x\right)+{\displaystyle \frac{{\gamma }_1}{I_{\rho }}}{\theta }_x+{\displaystyle \frac{{\gamma }_2}{I_{\rho }}}{P}_x\right]dx+D{\int }_0^1{\xi }_x{v}_{x}dx\hfill \\ & +& 3I_{\rho }{\int }_0^{1}y\left[{\displaystyle \frac{D}{I_{\rho }}}{s}_{xx}-{\displaystyle \frac{G}{I_{\rho }}}\left(3s-\xi -{\varrho }_x\right)-{\displaystyle \frac{4\gamma }{3I_{\rho }}}s-{\displaystyle \frac{4\beta }{3I_{\rho }}}y\right]dx+3D{\int }_0^1{s}_x{y}_{x}dx\hfill \\ & +& 4\gamma {\int }_0^{1}sydx+c{\int }_0^1\theta \left({\displaystyle \frac{\kappa r}{\delta }}{\theta }_{xx}-{\displaystyle \frac{\hslash d}{\delta }}{P}_{xx}+{\displaystyle \frac{{\gamma }_{1}r-{\gamma }_{2}d}{\delta }}{v}_x\right)dx\hfill \\ & +& r{\int }_0^{1}P\left({\displaystyle \frac{\hslash c}{\delta }}{P}_{xx}-{\displaystyle \frac{\kappa d}{\delta }}{\theta }_{xx}+{\displaystyle \frac{{\gamma }_{2}c-{\gamma }_{1}d}{\delta }}{v}_x\right)dx\hfill \\ & +& d{\int }_0^{1}P\left({\displaystyle \frac{\kappa r}{\delta }}{\theta }_{xx}-{\displaystyle \frac{\hslash d}{\delta }}{P}_{xx}+{\displaystyle \frac{{\gamma }_{1}r-{\gamma }_{2}d}{\delta }}{v}_x\right)dx\hfill \\ & +& d{\int }_0^1\theta \left({\displaystyle \frac{\hslash c}{\delta }}{P}_{xx}-{\displaystyle \frac{\kappa d}{\delta }}{\theta }_{xx}+{\displaystyle \frac{{\gamma }_{2}c-{\gamma }_{1}d}{\delta }}{v}_x\right)dx-\left|\mu \right|\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}z(x,\sigma )z_{\sigma }(x,\sigma )d\sigma dx.\hfill \end{array}$$

(9)

Using integration by parts and taking into account the boundary conditions, we obtain

$$\begin{array}{cc}\hfill {〈{\mathcal{A}}_{i}U\left(t\right),U\left(t\right)〉}_{{\mathcal{H}}_i}& \le -\kappa {\int }_0^1{\theta }_x^{2}dx-\hslash {\int }_0^1{P}_x^{2}dx-4\beta {\int }_0^1{y}^{2}dx+\left|\mu \right|{\int }_0^1{u}^{2}dx,\hfill \\ \hfill & \le \left|\mu \right|{\int }_0^1{u}^{2}dx,\hfill \\ \hfill & \le c_1{〈U\left(t\right),U\left(t\right)〉}_{{\mathcal{H}}_i},\hfill \end{array}$$

where $c_1={\displaystyle \frac{max\left(\right|\mu |,1)}{min(\varrho ,G,3I_{\rho },3D,4\delta ,c,2d,r,\tau |\mu \left|\right)}}$, which shows that the operator ${\mathcal{A}}_i-c_{1}I$$(i=1,2)$ is dissipative.

Let

$$F=\left(f_1,\dots ,f_9\right)\in {\mathcal{H}}_i,$$

and we seek a solution $U\in D\left({\mathcal{A}}_i\right)$ of

$$\left(Id-A_i\right)U=F.$$

Then, we obtain

$$\left\{\begin{array}{c}\varrho -u=f_1,\hfill \\ \rho u+G{\left(3s-\xi -{\varrho }_x\right)}_x+\mu z(x,1)=\rho f_2,\hfill \\ \xi -v=f_3,\hfill \\ I_{\rho }v-D{\xi }_{xx}-G\left(3s-\xi -{\varrho }_x\right)-{\gamma }_1{\theta }_x-{\gamma }_2{P}_x=I_{\rho }{f}_4,\hfill \\ s-y=f_5,\hfill \\ \left(4\beta +3I_{\rho }\right)y-3D{s}_{xx}+3G\left(3s-\xi -{\varrho }_x\right)+4\gamma s=3I_{\rho }{f}_6,\hfill \\ \delta \theta -\kappa r{\theta }_{xx}+\hslash d{P}_{xx}-\left({\gamma }_{1}r-{\gamma }_{2}d\right)v_x=\delta f_7,\hfill \\ \delta P-\hslash c{P}_{xx}+\kappa d{\theta }_{xx}-\left({\gamma }_{2}c-{\gamma }_{1}d\right)v_x=\delta f_8\hfill \\ \tau z+z_{\sigma }=\tau f_9.\hfill \end{array}\right.$$

(10)

Suppose that $\varrho$ is found with the appropriate regularity. Therefore, the first equation in (10) yields

$$u=\varrho -f_1.$$

(11)

It is clear that $u\in H_b^1\left(0,1\right)$ if $U\in {\mathcal{H}}_1$ and $u\in H_0^1\left(0,1\right)$ if $U\in {\mathcal{H}}_2$. Furthermore, we can find z as

$$z\left(x,0\right)=u\left(x\right),\mathrm{for}x\in \left(0,1\right).$$

(12)

We obtain using (10)₉

$$z\left(x,\sigma \right)=u\left(x\right)e^{-\sigma \tau }+\tau e^{-\sigma \tau }{\int }_0^{\sigma }{f}_9\left(x,\eta \right)e^{\eta \tau }d\eta .$$

From (11), we have

$$z\left(x,\sigma \right)=\varrho \left(x\right)e^{-\sigma \tau }-f_1{e}^{-\sigma \tau }+\tau e^{-\sigma \tau }{\int }_0^{\sigma }{f}_9\left(x,\eta \right)e^{\eta \tau }d\eta .$$

(13)

From (13), we have

$$z\left(x,1\right)=\varrho \left(x\right)e^{-\tau }+z_0\left(x\right),$$

(14)

where $x\in \left(0,1\right)$, and

$$z_0\left(x\right)=-f_1{e}^{-\tau }+\tau e^{-\tau }{\int }_0^1{f}_9\left(x,\eta \right)e^{\eta \tau }d\eta .$$

(15)

It is clear from the above formula that $z_0$ depends only on $f_i,$$i=1,9.$

By using (14), (12) in (10), and substituting (10)₁ into (10)₂, (10)₃ into (10)₄, (10)₅ into (10)₆, (10)₇ into (10)₈, we obtain

$$\left\{\begin{array}{c}\rho \varrho +G{\left(3s-\xi -{\varrho }_x\right)}_x+\mu \varrho e^{-\tau }=\rho (f_1+f_2)+\mu f_1{e}^{-\tau }-\mu \tau e^{-\tau }{\int }_0^1{f}_9\left(x,\eta \right)e^{\eta \tau }d\eta ,\hfill \\ I_{\rho }\xi -D{\xi }_{xx}-G\left(3s-\xi -{\varrho }_x\right)-{\gamma }_1{\theta }_x-{\gamma }_2{P}_x=I_{\rho }(f_4+f_3),\hfill \\ \left(4\beta +3I_{\rho }\right)s-3D{s}_{xx}+3G\left(3s-\xi -{\varrho }_x\right)+4\gamma s=3I_{\rho }{f}_6+\left(4\beta +3I_{\rho }\right)f_5,\hfill \\ \delta \theta -\kappa r{\theta }_{xx}+\hslash d{P}_{xx}-\left({\gamma }_{1}r-{\gamma }_{2}d\right){\xi }_x=\delta f_7-\left({\gamma }_{1}r-{\gamma }_{2}d\right)f_{3x},\hfill \\ \delta P-\hslash c{P}_{xx}+\kappa d{\theta }_{xx}-\left({\gamma }_{2}c-{\gamma }_{1}d\right){\xi }_x=\delta f_8-\left({\gamma }_{2}c-{\gamma }_{1}d\right)f_{3x}.\hfill \end{array}\right.$$

(16)

Multiplying (16)₁ by $\overline{\varrho }$, (16)₂ by $\overline{\xi }$, (16)₃ by $\overline{s}$, (16)₄ by ${\displaystyle \frac{c}{\delta }}\overline{\theta }$, (16)₅ by ${\displaystyle \frac{r}{\delta }}\overline{P}$, (16)₄ by ${\displaystyle \frac{d}{\delta }}\overline{P}$ and (16)₅ by ${\displaystyle \frac{d}{\delta }}\overline{\theta }$ and integrating their sum over $(0,1)$ to obtain the subsequent variations formulation, we have

$$\mathcal{B}\left((\mathit{\varrho },\xi ,s,\theta ,P),(\overline{\varrho },\overline{\xi },\overline{s},\overline{\theta },\overline{P})\right)=\mathcal{L}(\overline{\varrho },\overline{\xi },\overline{s},\overline{\theta },\overline{P}),$$

(17)

where the bilinear form

$$\mathcal{B}:{\left[H_b^1(0,1)\times H_a^1(0,1)\times H_a^1(0,1)\times L^2(0,1)\times L^2(0,1)\right]}^2\to \mathbb{R},$$

or

$$\mathcal{B}:{\left[H_0^1(0,1)\times H_0^1(0,1)\times H_0^1(0,1)\times L^2(0,1)\times L^2(0,1)\right]}^2\to \mathbb{R},$$

is given by

$$\begin{array}{cc}& \mathcal{B}((\varrho ,\xi ,s,\theta ,P),(\overline{\varrho },\overline{\xi },\overline{s},\overline{\theta },\overline{P}))\hfill \\ \hfill =& (\rho +\mu e^{-\tau }){\int }_0^1\varrho \overline{\varrho }dx+G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)\left(3\overline{s}-\overline{\xi }-{\overline{\varrho }}_x\right)dx+I_{\rho }{\int }_0^1\xi \overline{\xi }dx\hfill \\ & +D{\int }_0^1{\xi }_x{\overline{\xi }}_{x}dx+\left(4\beta +3I_{\rho }+4\gamma \right){\int }_0^{1}s\overline{s}dx+3D{\int }_0^1{s}_x{\overline{s}}_{x}dx\hfill \\ & +{\gamma }_1{\int }_0^1\theta {\overline{\xi }}_{x}dx-{\gamma }_1{\int }_0^1{\xi }_x\overline{\theta }dx+{\gamma }_2{\int }_0^{1}P{\overline{\xi }}_{x}dx-{\gamma }_2{\int }_0^1{\xi }_x\overline{P}dx\hfill \\ & +c{\int }_0^1\theta \overline{\theta }dx+\kappa {\int }_0^1{\theta }_x{\overline{\theta }}_{x}dx+r{\int }_0^{1}P\overline{P}dx\hfill \\ & +\hslash {\int }_0^1{P}_x{\overline{P}}_{x}dx+d{\int }_0^1\theta \overline{P}dx+d{\int }_0^{1}P\overline{\theta }dx,\hfill \end{array}$$

and the linear form

$$\mathcal{L}:H_b^1(0,1)\times H_a^1(0,1)\times H_a^1(0,1)\times L^2(0,1)\times L^2(0,1)\to \mathbb{R},$$

or

$$\mathcal{L}:H_0^1(0,1)\times H_0^1(0,1)\times H_0^1(0,1)\times L^2(0,1)\times L^2(0,1)\to \mathbb{R},$$

is defined by

$$\begin{array}{cc}& \mathcal{L}(\overline{\varrho },\overline{\xi },\overline{s},\overline{\theta },\overline{P})\hfill \\ & =\rho {\int }_0^1\left(f_1+f_2-{\displaystyle \frac{\mu }{\rho }}{z}_0\right)\overline{\varrho }dx+I_{\rho }{\int }_0^1\left(f_3+f_4\right)\overline{\xi }dx+\left(4\beta +3I_{\rho }\right){\int }_0^1{f}_5\overline{s}dx\hfill \\ & +3I_{\rho }{\int }_0^1{f}_6\overline{s}dx+c{\int }_0^1{f}_7\overline{\theta }dx-{\gamma }_1{\int }_0^1{f}_{3x}\overline{\theta }dx+r{\int }_0^1{f}_8\overline{P}dx\hfill \\ & -{\gamma }_2{\int }_0^1{f}_{3x}\overline{P}dx+d{\int }_0^1{f}_7\overline{P}dx+d{\int }_0^1{f}_8\overline{\theta }dx.\hfill \end{array}$$

It is easy to verify that $\mathcal{B}$ is continuous and coercive, and $\mathcal{L}$ is continuous. So, applying the Lax–Milgram theorem, we deduce that there exists a positive constant C such that

$$\mathcal{B}\left((\varrho ,\xi ,s,\theta ,p),(\overline{\varrho ,}\overline{\xi },\overline{s},\overline{\theta },\overline{p})\right)\ge {C\left|\right|(\varrho ,\xi ,s,\theta ,p)\left|\right|}_{{\mathcal{H}}_i}^2,i=1,2.$$

(18)

This shows that $\mathcal{B}$ is coercive on ${\mathcal{H}}_i\times {\mathcal{H}}_i,i=1,2$.

Owing to the Lax–Milgram Theorem, we find that (17) has a one solution:

$$\varrho \in H_b^1(0,1),\xi ,s\in H_a^1(0,1),\theta ,P\in L^2(0,1),$$

or

$$\varrho ,\xi ,s\in H_0^1(0,1),\theta ,P\in L^2(0,1).$$

To achieve more regularity, we should take

$$(\overline{\xi },\overline{s},\overline{\theta },\overline{P})=(0,0,0,0)\in H_a^1(0,1)\times H_a^1(0,1)\times L^2(0,1)\times L^2(0,1).$$

From (17), we have

$$(\rho +\mu e^{-\tau }){\int }_0^1\varrho \overline{\varrho }dx-G{\int }_0^1\left(3s-\xi -{\varrho }_x\right){\overline{\varrho }}_x=\rho {\int }_0^1\left(f_1+f_2-{\displaystyle \frac{\mu }{\rho }}{z}_0\right)\overline{\varrho }dx,\forall \overline{\varrho }\in H_b^1,$$

(19)

which implies that

$$G{\varrho }_{xx}=(\rho +\mu e^{-\tau })\varrho +G{(3s-\xi )}_x-\rho \left(f_1+f_2-{\displaystyle \frac{\mu }{\rho }}{z}_0\right)\in L^2(0,1).$$

Then, we have

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

As opposed to that, for any $\phi \in C^1[0,1]$ with $\phi (x=1)=0$, we find that (19) also holds true. Then,

$$(\rho +\mu e^{-\tau }){\int }_0^1\varrho \phi dx-G{\int }_0^1\left(3s-\xi -{\varrho }_x\right){\phi }_x-\rho {\int }_0^1\left(f_1+f_2-{\displaystyle \frac{\mu }{\rho }}{z}_0\right)\phi dx=0,\forall \phi \in C^1[0,1].$$

(20)

Utilizing integration by parts, we obtain

$$\begin{array}{c}\hfill -G(3s-\xi -{\varrho }_x)\left(1\right)\phi \left(1\right)-G{\varrho }_x\left(0\right)\phi \left(0\right)+(\rho +\mu e^{-\tau }){\int }_0^1\varrho \phi dx\\ \hfill +G{\int }_0^1{(3s-\xi -{\varrho }_x)}_x\phi dx-\rho {\int }_0^1\left(f_1+f_2-{\displaystyle \frac{\mu }{\rho }}{z}_0\right)\phi dx=0.\end{array}$$

Thus,

$$-G{\varrho }_x(x=0)\phi (x=0)=0,\forall \phi \in C^1[0,1].$$

Since $\phi$ is arbitrary, we can obtain ${\varrho }_x(x=0)=0$; thus,

$$\varrho \in H_b^2(0,1)\cap H_b^1(0,1).$$

By using the same approach, we can arrive at

$$\xi ,s\in H_a^2(0,1)\cap H_a^1(0,1),\theta ,P\in H^2(0,1)\cap H_0^1(0,1).$$

So, by applying the Lax–Milgram theorem, our problem has a unique solution. Consequently, $\left(\varrho ,\xi ,s,\theta ,P\right)\in D\left({\mathcal{A}}_1\right)$ and ${\mathcal{A}}_1$ are maximal likewise, so we may acquire $(\varrho ,\xi ,s,\theta ,P)\in$$D\left({\mathcal{A}}_2\right)$ and ${\mathcal{A}}_2$ as maximal. Therefore, the conclusion of the proof follows from the Lumer–Philips Theorem (see [8] ). □

3. Exponential Stability

The stability of problem (1) arises very concretely during the qualitative study, since the presence of a delay most often leads to oscillations or even instabilities. If the mathematical nature of the delay phenomenon is not taken into account, the only thermal diffusion effect is to provide the stability behavior to permanently accelerate dynamic performances. Various methods are available. The most practical method is the Lyapunov approach. We will use it here because it generally serves as a basis for suitable methods of investigating stability. To begin with, we define the energy of the system as

$$\begin{array}{cc}\hfill E\left(t\right)=& {\displaystyle \frac{1}{2}}{\int }_0^1\left[\rho {\varrho }_t^2+G{\left(3s-\xi -{\varrho }_x\right)}^2+I_{\rho }{\xi }_t^2+D{\xi }_x^2+3I_{\rho }{s}_t^2+3D{s}_x^2\right.\hfill \\ & \left.+4\gamma s^2+c{\theta }^2+2d\theta P+r{P}^2\right]dx+{\displaystyle \frac{\tau \left|\mu \right|}{2}}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{z}^2(x,\sigma ,t)d\sigma dx.\hfill \end{array}$$

Then, according to (7), the energy E is positive.

It is not hard to see for the two boundary conditions that

$${\displaystyle \frac{d}{dt}}E\left(t\right)\le -\kappa {\int }_0^1{\theta }_x^{2}dx-\hslash {\int }_0^1{P}_x^{2}dx-4\beta {\int }_0^1{s}_t^{2}dx+\left|\mu \right|\underset{0}{\overset{1}{\int }}{\varrho }_t^{2}dx.$$

The stability result is stated in the following theorem.

Theorem 2.

Let U be a regular solution of problem (1)–(3) or (1), (2), and (4). Let (5) hold. Then, there exist positive constants $\overline{\mu },a_1$ and $a_2$, independent of initial data, such that $\left|\mu \right|<\overline{\mu }$ and

$$E\left(t\right)\le a_{2}E\left(0\right)e^{-a_{1}t},t>0.$$

(21)

To prove Theorem 2, with the two boundary conditions, we shall divide into the following two subsections.

3.1. Exponential Decay with Boundary I

We need the following several lemmas.

Lemma 1.

Let the functional $F_1$ be defined by

$$F_1\left(t\right)=I_{\rho }{\int }_0^1{\xi }_t\phi dx,$$

where

$$-{\gamma }_1{\phi }_x=c\theta +dP,\phi (x=0)=\phi (x=1)=0.$$

Then, the there exist constants $c_1>0$ and $c_2>0$ such that for any ${\epsilon }_1>0$ and ${\epsilon }_2>0$, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_1\left(t\right)& \le & -{\displaystyle \frac{I_{\rho }}{2}}{\int }_0^1{\xi }_t^{2}dx+{\displaystyle \frac{G{\epsilon }_1}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{D{\epsilon }_2}{2}}{\int }_0^1{\xi }_x^{2}dx\hfill \\ & +& c_1{\int }_0^1{\theta }_x^{2}dx+c_2{\int }_0^1{P}_x^{2}dx.\hfill \end{array}$$

(22)

Proof. 

Differentiate $F_1$ and use (8)₂ to obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_1\left(t\right)& =& {\int }_0^1\left[D{\xi }_{xx}+G\left(3s-\xi -{\varrho }_x\right)+{\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right]\phi dx+I_{\rho }{\int }_0^1{\xi }_t{\phi }_{t}dx\hfill \\ & =& G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)\phi dx-D{\int }_0^1{\xi }_x{\phi }_{x}dx+{\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\phi dx\hfill \\ & & +I_{\rho }{\int }_0^1{\xi }_t{\phi }_{t}dx.\hfill \end{array}$$

(23)

Using Cauchy–Schwarz and Poincare’s inequalities, we obtain $\forall {\epsilon }_1>0$

$$\begin{array}{ccc}\hfill G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)\phi dx& \le & {\displaystyle \frac{{\epsilon }_{1}G}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{G}{2{\epsilon }_1}}{\int }_0^1{\phi }_x^{2}dx\hfill \\ & \le & {\displaystyle \frac{{\epsilon }_{1}G}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{G}{2{\epsilon }_1}}{\int }_0^L{\left({\displaystyle \frac{c}{{\gamma }_1}}\theta +{\displaystyle \frac{d}{{\gamma }_1}}P\right)}^{2}dx\hfill \\ & \le & {\displaystyle \frac{{\epsilon }_{1}G}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{G{d}^2{\beta }_2}{{\epsilon }_1{\gamma }_1^2}}{\int }_0^1{P}_x^{2}dx\hfill \end{array}$$

(24)

$$\begin{array}{ccc}& +& {\displaystyle \frac{G{c}^2{\beta }_1}{{\epsilon }_1{\gamma }_1^2}}{\int }_0^1{\theta }_x^{2}dx,\hfill \end{array}$$

(25)

and for any ${\epsilon }_2>0$

$$-D{\int }_0^1{\xi }_x{\phi }_{x}dx\le {\displaystyle \frac{D{\epsilon }_2}{2}}{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{D{c}^2{\beta }_3}{{\epsilon }_2{\gamma }_1^2}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{D{d}^2{\beta }_4}{{\epsilon }_2{\gamma }_1^2}}{\int }_0^1{P}_x^{2}dx,$$

(26)

and

$$\begin{array}{ccc}\hfill {\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\phi dx& \le & {\displaystyle \frac{1}{2}}{\int }_0^1{\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)}^{2}dx+{\displaystyle \frac{1}{2}}{\int }_0^1{\phi }^{2}dx\hfill \\ & \le & \left({\gamma }_1^2+{\displaystyle \frac{c^2{\beta }_5}{{\gamma }_1^2}}\right){\int }_0^1{\theta }_x^{2}dx+\left({\gamma }_2^2+{\displaystyle \frac{{\beta }_6{d}^2}{{\gamma }_1^2}}\right){\int }_0^1{P}_x^{2}dx.\hfill \end{array}$$

(27)

Using (8)₄ and Young’s inequality, we have

$$\begin{array}{ccc}\hfill I_{\rho }{\int }_0^1{\xi }_t{\phi }_{t}dx& =& -{\displaystyle \frac{I_{\rho }}{{\gamma }_1}}{\int }_0^1{\xi }_t{\int }_0^x\left(\kappa {\theta }_{yy}+{\gamma }_1{\xi }_{ty}\right)dydx\hfill \\ & =& -I_{\rho }{\int }_0^1{\xi }_t^{2}dx-{\displaystyle \frac{I_{\rho }\kappa }{{\gamma }_1}}{\int }_0^1{\xi }_t{\theta }_{x}dx\hfill \\ & \le & -{\displaystyle \frac{I_{\rho }}{2}}{\int }_0^1{\xi }_t^{2}dx+{\displaystyle \frac{I_{\rho }{\kappa }^2}{2{\gamma }_1^2}}{\int }_0^1{\theta }_x^{2}dx.\hfill \end{array}$$

(28)

Inserting (24)–(28) into (23), we establish (22) with

$$c_1={\displaystyle \frac{I_{\rho }{\kappa }^2}{2{\gamma }_1^2}}+{\displaystyle \frac{G{c}^2{\beta }_1}{{\epsilon }_1{\gamma }_1^2}}+{\displaystyle \frac{D{c}^2{\beta }_3}{{\epsilon }_2{\gamma }_1^2}}+\left({\gamma }_1^2+{\displaystyle \frac{c^2{\beta }_5}{{\gamma }_1^2}}\right),$$

and

$$c_2={\displaystyle \frac{G{d}^2{\beta }_2}{{\epsilon }_1{\gamma }_1^2}}+{\displaystyle \frac{D{d}^2{\beta }_4}{{\epsilon }_2{\gamma }_1^2}}+\left({\gamma }_2^2+{\displaystyle \frac{{\beta }_6{d}^2}{{\gamma }_1^2}}\right).$$

□

Lemma 2.

The functional $F_2$ defined by

$$F_2\left(t\right)=I_{\rho }{\int }_0^1{\xi }_t\xi dx,$$

satisfies, for any $t>0$,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_2\left(t\right)& \le & -{\displaystyle \frac{D}{2}}{\int }_0^1{\xi }_x^{2}dx+I_{\rho }{\int }_0^1{\xi }_t^{2}dx+{\displaystyle \frac{G{ϵ}_3}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx\hfill \\ & +& {\displaystyle \frac{2{\gamma }_1^2}{D}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{2{\gamma }_2^2}{D}}{\int }_0^1{P}_x^{2}dx.\hfill \end{array}$$

(29)

Proof. 

By using (8)₂, we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_2\left(t\right)& =& I_{\rho }{\int }_0^1{\xi }_t^{2}dx+G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)\xi dx-D{\int }_0^1{\xi }_x^{2}dx\hfill \\ & +& {\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\xi dx.\hfill \end{array}$$

(30)

Making use of Young’s and Poincare’s inequalities, we obtain

$$\begin{array}{ccc}\hfill G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)\xi dx& \le & {\displaystyle \frac{G}{2{ϵ}_3}}{\int }_0^1{\xi }^{2}dx+{\displaystyle \frac{G{ϵ}_3}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx\hfill \\ & \le & {\displaystyle \frac{G{c}_3}{2{ϵ}_3}}{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{G{ϵ}_3}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx,\hfill \end{array}$$

(31)

and

$${\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\xi dx\le {\displaystyle \frac{D}{4}}{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{2{\gamma }_1^2}{D}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{2{\gamma }_2^2}{D}}{\int }_0^1{P}_x^{2}dx.$$

By using (30)–(31) and choosing $c_3={\displaystyle \frac{D{ϵ}_3}{2G}}$, we obtain (29). □

Lemma 3.

Let $F_3$ a functional defined by

$$F_3\left(t\right)=-I_{\rho }{\int }_0^1{\xi }_t\left(3s-\xi -{\varrho }_x\right)dx+{\displaystyle \frac{\rho D}{G}}{\int }_0^1{\xi }_x{\varrho }_{t}dx.$$

Assume that the condition (5) holds; then, $F_3$ satisfies, for any $t>0$,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_3\left(t\right)& \le & -{\displaystyle \frac{G}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+2I_{\rho }{\int }_0^1{\xi }_t^{2}dx+{\displaystyle \frac{3I_{\rho }}{4}}{\int }_0^1{s}_t^{2}dx\hfill \\ & +& {\displaystyle \frac{{\gamma }_1^2}{G}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{{\gamma }_2^2}{G}}{\int }_0^1{P}_x^{2}dx+{\displaystyle \frac{D^2}{2G^2}}{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{{\mu }^2}{2}}{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

(32)

Proof. 

Using the first and the second equations of (8) and taking the derivative of $F_3$, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_3\left(t\right)& =& {\int }_0^1\left[-G\left(3s-\xi -{\varrho }_x\right)-D{\xi }_{xx}-{\gamma }_1{\theta }_x-{\gamma }_2{P}_x\right]\left(3s-\xi -{\varrho }_x\right)dx\hfill \\ & -& I_{\rho }{\int }_0^1{\xi }_t{\left(3s-\xi -{\varrho }_x\right)}_{t}dx+{\displaystyle \frac{\rho D}{G}}{\int }_0^1{\xi }_{xt}{\varrho }_{t}dx\hfill \\ & +& {\displaystyle \frac{D}{G}}{\int }_0^1{\xi }_x\left[-G{\left(3s-\xi -{\varrho }_x\right)}_x-\mu z(x,1,t)\right]dx\hfill \\ & =& -G{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx-{\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\left(3s-\xi -{\varrho }_x\right)dx\hfill \\ & -& I_{\rho }{\int }_0^1{\xi }_t{(3s-\xi )}_{t}dx+\left({\displaystyle \frac{\rho D}{G}}-I_{\rho }\right){\int }_0^1{\xi }_{xt}{\varrho }_{t}dx\hfill \\ & -& {\displaystyle \frac{D\mu }{G}}{\int }_0^1{\xi }_{x}z(x,1,t)dx.\hfill \end{array}$$

(33)

Young’s inequality gives us

$$\begin{array}{cc}\hfill -{\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\left(3s-\xi -{\varrho }_x\right)dx& \le {\displaystyle \frac{G}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{1}{2G}}{\int }_0^1{\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)}^{2}dx\\ & \le {\displaystyle \frac{G}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{{\gamma }_1^2}{G}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{{\gamma }_2^2}{G}}{\int }_0^1{P}_x^{2}dx,\end{array}$$

(34)

and

$$-{\displaystyle \frac{D\mu }{G}}\underset{0}{\overset{1}{\int }}{\xi }_{x}z(x,1,t)dx\le {\displaystyle \frac{D^2}{2G^2}}\underset{0}{\overset{1}{\int }}{{\xi }_x}^{2}dx+{\displaystyle \frac{{\mu }^2}{2}}\underset{0}{\overset{1}{\int }}{z}^2(x,1,t)dx,$$

(35)

and

$$\begin{array}{ccc}\hfill -I_{\rho }\underset{0}{\overset{1}{\int }}{\xi }_t{(3s-\xi )}_t& =& I_{\rho }\underset{0}{\overset{1}{\int }}{\xi }_t^{2}dx-3I_{\rho }\underset{0}{\overset{1}{\int }}{\xi }_t{s}_{t}dx\hfill \\ & \le & 2I_{\rho }\underset{0}{\overset{1}{\int }}{\xi }_t^{2}dx+{\displaystyle \frac{3I_{\rho }}{4}}\underset{0}{\overset{1}{\int }}{\left(s_t\right)}^{2}dx.\hfill \end{array}$$

(36)

Replacing (34)–(36) into (33) and noting the condition (5), we obtain (32). □

Lemma 4.

Let the functional $F_4$ be defined by

$$F_4\left(t\right)=-\rho {\int }_0^1\varrho {\varrho }_{t}dx-\rho {\int }_0^1{\varrho }_t{\int }_x^1(3s-\xi )\left(y\right)dydx.$$

Then, for all $t>0$, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_4\left(t\right)& \le & -{\displaystyle \frac{\rho }{2}}{\int }_0^1{\varrho }_t^{2}dx+\left(G+1\right){\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx\hfill \\ & +& \rho {\int }_0^1{\xi }_t^{2}dx+9\rho {\int }_0^1{s}_t^{2}dx+{\displaystyle \frac{{\mu }^2}{4}}{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

(37)

Proof. 

It follows from (8) that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_4\left(t\right)& =& -\rho {\int }_0^1{\varrho }_t^{2}dx+G{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}_x\varrho dx+\mu {\int }_0^1\left({\int }_x^1(3s-\xi )\left(y\right)dy+\varrho \right)z(x,1,t)dx\hfill \\ & +& G{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}_x{\int }_x^1(3s-\xi )\left(y\right)dydx-\rho {\int }_0^1{\varrho }_t{\int }_x^1{(3s-\xi )}_t\left(y\right)dydx\hfill \\ & =& -\rho {\int }_0^1{\varrho }_t^{2}dx+G{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+\rho {\int }_0^1{\varrho }_t{\int }_x^1{\xi }_t\left(y\right)dydx\hfill \\ & -& 3\rho {\int }_0^1{\varrho }_t{\int }_x^1{s}_t\left(y\right)dydx+\mu {\int }_0^1\left({\int }_x^1(3s-\xi )\left(y\right)dy+\varrho \right)z(x,1,t)dx,\hfill \end{array}$$

thereby yielding the following estimates

$$\rho {\int }_0^1{\varrho }_t{\int }_x^1{\xi }_t\left(y\right)dydx\le {\displaystyle \frac{\rho }{4}}{\int }_0^1{\varrho }_t^{2}dx+\rho {\int }_0^1{\xi }_t^{2}dx,$$

and

$$-3\rho {\int }_0^1{\varrho }_t{\int }_x^1{s}_t\left(y\right)dydx\le {\displaystyle \frac{\rho }{4}}{\int }_0^1{\varrho }_t^{2}dx+9\rho {\int }_0^1{s}_t^{2}dx,$$

and

$$\begin{array}{ccc}\hfill \mu {\int }_0^1\left({\int }_x^1(3s-\xi )\left(y\right)dy+\varrho \right)z(x,1,t)dx& \le & {\int }_0^1{\left({\int }_x^1(3s-\xi )\left(y\right)dy+\varrho \right)}^{2}dx+{\displaystyle \frac{{\mu }^2}{4}}{\int }_0^1{z}^2(x,1,t)dx\hfill \\ & \le & {\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{{\mu }^2}{4}}{\int }_0^1{z}^2(x,1,t)dx,\hfill \end{array}$$

which imply (37). □

Lemma 5.

Let the functional $F_5$ be defined by

$$F_5\left(t\right)=3I_{\rho }{\int }_0^1{s}_{t}sdx+2\beta {\int }_0^1{s}^{2}dx-3\rho {\int }_0^1{\varrho }_t{\int }_x^{1}s\left(y\right)dydx.$$

Then, for any $t>0$, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_5\left(t\right)& \le & -3D{\int }_0^1{s}_x^{2}dx-3\gamma {\int }_0^1{s}^{2}dx+{\displaystyle \frac{3\rho }{2}}{\int }_0^1{\varrho }_t^{2}dx\hfill \\ & +& 3\left(I_{\rho }+{\displaystyle \frac{1}{2}}\rho \right){\int }_0^1{s}_t^{2}dx+{\displaystyle \frac{9{\mu }^2}{4\gamma }}{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

(38)

Proof. 

By taking the derivative of $F_5$ and using (8)₃, we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_5\left(t\right)& =& 3I_{\rho }{\int }_0^1{s}_t^{2}dx-3D{\int }_0^1{s}_x^{2}dx-4\gamma {\int }_0^1{s}^{2}dx-3\rho {\int }_0^1{\varrho }_t{\int }_x^1{s}_t\left(y\right)dydx\hfill \\ & +& 3\mu {\int }_0^{1}z(x,1,t){\int }_x^{1}s\left(y\right)dydx,\hfill \end{array}$$

along with the estimates

$$-3\rho {\int }_0^1{\varrho }_t{\int }_x^1{s}_t\left(y\right)dydx\le {\displaystyle \frac{3\rho }{2}}{\int }_0^1{\varrho }_t^{2}dx+{\displaystyle \frac{3\rho }{2}}{\int }_0^1{s}_t^{2}dx,$$

and

$$3\mu {\int }_0^{1}z(x,1,t){\int }_x^{1}s\left(y\right)dydx\le {\displaystyle \frac{9{\mu }^2}{4\gamma }}{\int }_0^1{z}^2(x,1,t)dx+\gamma {\int }_0^1{s}^{2}dx,$$

so we establish (38). □

Lemma 6.

The functional $F_6$ defined by

$$F_6\left(t\right)=\tau {\int }_0^1{\int }_0^1{e}^{-\sigma \tau }{z}^2(x,\sigma ,t)d\sigma dx,$$

satisfies for $m_1>0$ the estimate

$${\displaystyle \frac{d}{dt}}{F}_6\left(t\right)\le -m_1{\int }_0^1{z}^2(x,1,t)dx-m_1\tau {\int }_0^1{\int }_0^1{z}^2(x,\sigma ,t)d\sigma dx+{\int }_0^1{\varrho }_t^{2}dx.$$

(39)

Proof. 

Differentiate $F_6$ and use the last equation in (8) and $z(x,0,t)={\varrho }_t$ as follows:

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}& F_6\left(t\right)=-2{\int }_0^1{\int }_0^1{e}^{-\tau \sigma }z(x,\sigma ,t)z_{\sigma }(x,\sigma ,t)d\sigma dx\hfill \\ & =-{\int }_0^1{\int }_0^1{\displaystyle \frac{d}{d\sigma }}\left[e^{-\tau \sigma }{z}^2(x,\sigma ,t)\right]d\sigma dx-\tau {\int }_0^1{\int }_0^1{e}^{-\tau \sigma }{z}^2(x,\sigma ,t)d\sigma dx\hfill \\ \hfill & =-{\int }_0^1\left[e^{-\tau }{z}^2(x,1,t)-z^2(x,0,t)\right]dx-\tau {\int }_0^1{\int }_0^1{e}^{-\tau \sigma }{z}^2(x,\sigma ,t)d\sigma dx\hfill \\ & =-{\int }_0^1{e}^{-\tau }{z}^2(x,1,t)dx+{\int }_0^1{\varrho }_t^{2}dx-\tau {\int }_0^1{\int }_0^1{e}^{-\tau \sigma }{z}^2(x,\sigma ,t)d\sigma dx.\hfill \end{array}$$

Observe that, $\forall \sigma \in (0,1)$, the relation $e^{-\tau }\le e^{-\sigma \tau }\le 1$ holds. Therefore, for some $m_1=e^{-\tau }$, we arrive at (39). □

We are now in position to introduce the Lyapunov functional $\mathcal{L}$ in the following:

$$\mathcal{L}\left(t\right)=NE\left(t\right)+N_1{F}_1\left(t\right)+N_2{F}_2\left(t\right)+N_3{F}_3\left(t\right)+F_4\left(t\right)+N_4{F}_5\left(t\right)+N_5{F}_6\left(t\right),$$

where N and $N_i,i=1,\dots ,5$ will be determined later.

Lemma 7.

There exist two positive constants ${\nu }_1$ and ${\nu }_2$ such that

$${\nu }_{1}E\left(t\right)\le \mathcal{L}\left(t\right)\le {\nu }_{2}E\left(t\right).$$

(40)

Proof. 

Owing to the Young’s inequality, there is a constant $N_0>0$ such that

$$|\mathcal{L}\left(t\right)-NE\left(t\right)|\le N_{0}E\left(t\right).$$

Then, we can take $N>0$ large (if needed) such that $N-N_0>0$ to obtain (40) with ${\nu }_1=N-N_0$ and ${\nu }_2=N+N_0$. □

Proof. 

(Of Theorem 2 for Boundary I.) By using Lemma 1–Lemma 6 and taking

$${\epsilon }_1={\displaystyle \frac{N_3}{4N_1}},{\epsilon }_2={\displaystyle \frac{2}{D{N}_1}},{\epsilon }_3={\displaystyle \frac{N_3}{4N_2}},N_5=\left|\mu \right|,N_4={\displaystyle \frac{1}{6}},$$

we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}\mathcal{L}\left(t\right)& \le & -\left(\kappa N-c_1{N}_1-{\displaystyle \frac{2{\gamma }_1^2}{D}}{N}_2-{\displaystyle \frac{{\gamma }_1^2}{G}}{N}_3\right){\int }_0^1{\theta }_x^{2}dx\hfill \\ & -& \left({\displaystyle \frac{D}{2}}{N}_2-{\displaystyle \frac{D^2}{2G^2}}{N}_3-1\right){\int }_0^1{\xi }_x^{2}dx\hfill \\ & -& \left(\hslash N-c_2{N}_1-{\displaystyle \frac{2{\gamma }_2^2}{D}}{N}_2-{\displaystyle \frac{{\gamma }_2^2}{G}}{N}_3\right){\int }_0^L{P}_x^{2}dx-\left({\displaystyle \frac{\rho }{4}}-(N+1)\left|\mu \right|\right){\int }_0^1{\varrho }_t^{2}dx\hfill \\ & -& \left[4\beta N-{\displaystyle \frac{3I_{\rho }}{4}}{N}_3-9\rho -\left({\displaystyle \frac{I_{\rho }}{2}}+{\displaystyle \frac{\rho }{4}}\right)\right]{\int }_0^1{s}_t^{2}dx-{\displaystyle \frac{D}{2}}{\int }_0^1{s}_x^{2}dx\hfill \\ & -& \left({\displaystyle \frac{I_{\rho }}{2}}{N}_1-I_{\rho }{N}_2-2I_{\rho }{N}_3-\rho \right){\int }_0^1{\xi }_t^{2}dx-{\displaystyle \frac{\gamma }{2}}{\int }_0^1{s}^{2}dx\hfill \\ & -& \left({\displaystyle \frac{G}{4}}{N}_3-\left(G+1\right)\right){\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx-m_1\left|\mu \right|\tau {\int }_0^1{\int }_0^1{e}^{-\sigma \tau }{z}^2(x,\sigma ,t)d\sigma dx\hfill \\ & -& \left|\mu \right|\left[m_1-\left|\mu \right|\left({\displaystyle \frac{3}{8\gamma }}+{\displaystyle \frac{1}{4}}+{\displaystyle \frac{N_3}{2}}\right)\right]{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

We first take $N_3>0$ such that

$${\displaystyle \frac{G}{4}}{N}_3-\left(G+1\right)>0.$$

For fixed $N_3$, we take $N_2>0$ such that

$${\displaystyle \frac{D}{2}}{N}_2-{\displaystyle \frac{D^2}{2G^2}}{N}_3-1>0.$$

For fixed $N_2$ and $N_3$, we choose $N_1>0$ large enough such that

$${\displaystyle \frac{I_{\rho }}{2}}{N}_1-I_{\rho }{N}_2-2I_{\rho }{N}_3-\rho >0.$$

Then, we choose N so large such that

$$\kappa N-c_1{N}_1-{\displaystyle \frac{2{\gamma }_1^2}{D}}{N}_2-{\displaystyle \frac{{\gamma }_1^2}{G}}{N}_3>0,\hslash N-c_2{N}_1-{\displaystyle \frac{2{\gamma }_2^2}{D}}{N}_2-{\displaystyle \frac{{\gamma }_2^2}{G}}{N}_3>0,$$

and

$$4\beta N-{\displaystyle \frac{3I_{\rho }}{4}}{N}_3-9\rho -\left({\displaystyle \frac{I_{\rho }}{2}}+{\displaystyle \frac{\rho }{4}}\right)>0.$$

Finally, we pick

$$\overline{\mu }=min\left\{{\displaystyle \frac{m_1}{\frac{3}{8\gamma }+\frac{1}{4}+\frac{N_3}{2}}},{\displaystyle \frac{\rho }{4(N+1)}}\right\}.$$

Therefore, there exists a positive constant ${\nu }_0$ such that

$${\displaystyle \frac{d}{dt}}\mathcal{L}\left(t\right)\le -{\nu }_{0}E\left(t\right).$$

(41)

According to (40) and (41), we see that

$$E\left(t\right)\le {\displaystyle \frac{{\nu }_2}{{\nu }_1}}E\left(0\right)e^{-{\displaystyle \frac{{\nu }_0}{{\nu }_2}}t}.$$

This completes the proof. □

3.2. Exponential Decay with Boundary $II$

The exponential decay of system (1) with Boundary $II$ is established in this subsection. The functional $F_1\left(t\right)$ still fulfills the estimates (29) for Boundary II, and this is simple to confirm. We first establish the following lemmas in order to circumvent the boundary’s difficulty.

Lemma 8.

Let the functional $H_1$ be defined by

$$H_1\left(t\right)=-\rho {\int }_0^1\varrho {\varrho }_{t}dx.$$

Then,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{H}_1\left(t\right)& \le & -\rho {\int }_0^1{\varrho }_t^{2}dx+4{\int }_0^1{\xi }_x^{2}dx+36{\int }_0^1{s}_x^{2}dx\hfill \\ & +& \left(1+G+{\displaystyle \frac{G^2}{4}}\right){\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{{\mu }^2}{2}}{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

(42)

Proof. 

Differentiating $H_1$ with respect to t and using (8)₁, we have

$${\displaystyle \frac{d}{dt}}{H}_1\left(t\right)=-\rho {\int }_0^1{\varrho }_t^{2}dx+G{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx-G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)(3s-\xi )dx+\mu {\int }_0^1\varrho z(x,1,t)dx.$$

(43)

It follows from Young’s and Poincare’s inequalities that

$$-G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)(3s-\xi )dx\le {\int }_0^1{(3s-\xi )}_x^{2}dx+{\displaystyle \frac{G^2}{4}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx,$$

(44)

$$\mu {\int }_0^1\varrho z(x,1,t)dx\le {\displaystyle \frac{{\mu }^2}{2}}{\int }_0^1{z}^2(x,1,t)dx+{\displaystyle \frac{1}{2}}{\int }_0^1{\varrho }^{2}dx,$$

(45)

and

$${\displaystyle \frac{1}{2}}{\int }_0^1{\varrho }^{2}dx\le {\displaystyle \frac{1}{2}}{\int }_0^1{\varrho }_x^{2}dx={\displaystyle \frac{1}{2}}{\int }_0^1{\left(3s-\xi -(3s-\xi )-{\varrho }_x\right)}^{2}dx\le {\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\int }_0^1{(3s-\xi )}_x^{2}dx.$$

On the other hand,

$${\int }_0^1{(3s-\xi )}_x^{2}dx\le 2{\int }_0^1{\xi }_x^{2}dx+18{\int }_0^1{s}_x^{2}dx,$$

Together with (43)–(45), this yields (42). □

Lemma 9.

Let the functional $H_2$ be defined by

$$H_2\left(t\right)=\rho {\int }_0^1{\varrho }_{t}fdx+I_{\rho }{\int }_0^1{\xi }_t\xi dx,$$

where

$$-f_{xx}={\xi }_x,f(x=0)=f(x=1)=0.$$

Then, $H_2$ satisfies, for any ${\epsilon }_3>0$,

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{H}_2\left(t\right)\le & -{\displaystyle \frac{D}{2}}{\int }_0^1{\xi }_x^{2}dx+{\epsilon }_3{\int }_0^1{\varrho }_t^{2}dx+\left(I_{\rho }+{\displaystyle \frac{{\rho }^2}{4{\epsilon }_3}}\right){\int }_0^1{\xi }_t^{2}dx\hfill \\ \hfill & +{\displaystyle \frac{2{\gamma }_1^2}{D}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{2{\gamma }_2^2}{D}}{\int }_0^1{P}_x^{2}dx+{\displaystyle \frac{{\mu }^2}{2}}{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

(46)

Proof. 

It is easy to obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{H}_2\left(t\right)=& G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)f_{x}dx-\mu {\int }_0^{1}fz(x,1,t)dx+\rho {\int }_0^1{\varrho }_t{f}_{t}dx+I_{\rho }{\int }_0^1{\xi }_t^{2}dx\hfill \\ & -D{\int }_0^1{\xi }_x^{2}dx+G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)\xi dx+{\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\xi dx\hfill \\ \hfill =& \rho {\int }_0^1{\varrho }_t{f}_{t}dx+I_{\rho }{\int }_0^1{\xi }_t^{2}dx-D{\int }_0^1{\xi }_x^{2}dx+{\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\xi dx-\mu {\int }_0^{1}fz(x,1,t)dx.\hfill \end{array}$$

(47)

It follows from Young’s inequality that

$${\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\xi dx\le {\displaystyle \frac{D}{4}}{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{2{\gamma }_1^2}{D}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{2{\gamma }_2^2}{D}}{\int }_0^1{P}_x^{2}dx.$$

and

$$-\mu {\int }_0^{1}fz(x,1,t)dx\le {\displaystyle \frac{{\mu }^2}{2}}{\int }_0^1{z}^2(x,1,t)dx+{\displaystyle \frac{1}{2}}{\int }_0^1{f}^{2}dx,$$

and

$$\rho {\int }_0^1{\varrho }_t{f}_{t}dx\le {\epsilon }_3{\int }_0^1{\varrho }_t^{2}dx+{\displaystyle \frac{{\rho }^2}{4{\epsilon }_3}}{\int }_0^1{\xi }_t^{2}dx,\forall {\epsilon }_3>0,$$

where we used the fact that

$${\displaystyle \frac{1}{2}}{\int }_0^1{f}^{2}dx\le {\displaystyle \frac{1}{2}}{\int }_0^1{f}_x^{2}dx\le {\displaystyle \frac{D}{4}}{\int }_0^1{\xi }_x^{2}dx,$$

and

$${\int }_0^1{f}_t^{2}dx\le {\int }_0^1{f}_{xt}^{2}dx\le {\int }_0^1{\xi }_t^{2}dx.$$

The proof is completed. □

The functional $F_3$ in the following lemma is still defined as in Lemma (3), but because the boundary conditions have changed, we now obtain the new estimate shown below.

Lemma 10.

Assume that the condition (5) holds; then, the functional $F_3$ is satisfied for any ${\epsilon }_4>0$ and ${\epsilon }_5>0$

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{F}_3\left(t\right)& \le \hfill & -\left({\displaystyle \frac{G}{2}}-\left(6+{\displaystyle \frac{2}{G}}\right){\epsilon }_4-{\displaystyle \frac{D}{4{\epsilon }_4}}{\epsilon }_5\right){\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+\left({\displaystyle \frac{36}{G}}+126\right){\epsilon }_4{\int }_0^1{s}_x^{2}dx\hfill \\ & +\hfill & {\displaystyle \frac{2\rho }{G}}{\epsilon }_4{\int }_0^1{\varrho }_t^{2}dx+\left(2I_{\rho }+{\displaystyle \frac{I_{\rho }D}{2{\epsilon }_4}}\right){\int }_0^1{\xi }_t^{2}dx+{\displaystyle \frac{3I_{\rho }}{4}}{\int }_0^1{s}_t^{2}dx\hfill \\ & +\hfill & \left[\left(14+{\displaystyle \frac{4}{G}}\right){\epsilon }_4+{\displaystyle \frac{D}{4{\epsilon }_4}}\left(3D+{\displaystyle \frac{G^2}{{\epsilon }_5}}\right)\right]{\int }_0^1{\xi }_x^{2}dx+\left({\displaystyle \frac{{\gamma }_1^2}{G}}+{\displaystyle \frac{{\gamma }_1^2}{2{\epsilon }_4}}\right){\int }_0^1{\theta }_x^{2}dx\hfill \\ & & +{\displaystyle \frac{{\mu }^2}{2}}\left(1+{\displaystyle \frac{2{\epsilon }_4}{G}}\right){\int }_0^1{z}^2(x,1,t)dx+\left({\displaystyle \frac{{\gamma }_2^2}{G}}+{\displaystyle \frac{{\gamma }_2^2}{2{\epsilon }_4}}\right){\int }_0^1{P}_x^{2}dx\hfill \\ & \hfill & -{\displaystyle \frac{{\epsilon }_4}{G}}{\displaystyle \frac{d}{dt}}{\int }_0^1\rho g\left(x\right){\varrho }_t{\varrho }_{x}dx-{\displaystyle \frac{D}{4{\epsilon }_4}}{\displaystyle \frac{d}{dt}}{\int }_0^1{I}_{\rho }g\left(x\right){\xi }_t{\xi }_{x}dx,\hfill \end{array}$$

(48)

where

$$g\left(x\right)=2-4x,x\in [0,1].$$

Proof. 

Given the boundary conditions (4) and the identical arguments in (33), we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{F}_3\left(t\right)=& -G{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+D{\xi }_x{\varrho }_x{|}_{x=0}^{x=1}-{\int }_0^1\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right)\left(3s-\xi -{\varrho }_x\right)dx\hfill \\ & -I_{\rho }{\int }_0^1{\xi }_t{(3s-\xi )}_{t}dx+\left({\displaystyle \frac{\rho D}{G}}-I_{\rho }\right){\int }_0^1{\xi }_{tx}{\varrho }_{t}dx\hfill \\ & -{\displaystyle \frac{\mu D}{G}}{\int }_0^1{\xi }_{x}z(x,1,t)dx.\hfill \end{array}$$

According to (34) and (36), we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{F}_3\left(t\right)\le & -{\displaystyle \frac{G}{2}}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+D{\xi }_x{\varrho }_x{|}_{x=0}^{x=1}+{\displaystyle \frac{{\gamma }_1^2}{G}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{{\gamma }_2^2}{G}}{\int }_0^1{P}_x^{2}dx\hfill \\ & +\left({\displaystyle \frac{\rho D}{G}}-I_{\rho }\right){\int }_0^1{\xi }_{xt}{\varrho }_{t}dx+2I_{\rho }{\int }_0^1{\xi }_t^{2}dx+{\displaystyle \frac{3I_{\rho }}{4}}{\int }_0^1{s}_t^{2}dx\hfill \\ & +{\displaystyle \frac{D^2}{2G^2}}{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{{\mu }^2}{2}}{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

(49)

We will now estimate the term $D{\xi }_x{\varrho }_x{|}_{x=0}^{x=1}$. Using Young’s inequality, we have $\forall {\epsilon }_4>0$

$$D{\xi }_x{\varrho }_x{|}_{x=0}^{x=1}\le {\epsilon }_4\left[{\varrho }_x^2\left(1\right)+{\varrho }_x^2\left(0\right)\right]+{\displaystyle \frac{D^2}{4{\epsilon }_4}}\left[{\xi }_x^2\left(1\right)+{\xi }_x^2\left(0\right)\right].$$

(50)

It follows from (8)₂ that

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\int }_0^1{I}_{\rho }g\left(x\right){\xi }_t{\xi }_{x}dx& ={\int }_0^{1}g\left(x\right)\left[D{\xi }_{xx}+G\left(3s-\xi -{\varrho }_x\right)+{\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right]{\xi }_{x}dx+I_{\rho }{\int }_0^{1}g\left(x\right){\xi }_t{\xi }_{xt}dx\hfill \\ & ={\displaystyle \frac{D}{2}}g\left(x\right){\xi }_x^2{|}_{x=0}^{x=1}-{\displaystyle \frac{D}{2}}{\int }_0^1{g}^{\prime }\left(x\right){\xi }_x^{2}dx+G{\int }_0^{1}g\left(x\right)\left(3s-\xi -{\varrho }_x\right){\xi }_{x}dx\hfill \\ \hfill & +{\int }_0^{1}g\left(x\right)\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right){\xi }_{x}dx-{\displaystyle \frac{I_{\rho }}{2}}{\int }_0^1{g}^{\prime }\left(x\right){\xi }_t^{2}dx,\hfill \end{array}$$

Notice that

$$g\left(0\right)=2,g\left(1\right)=-2,g^{\prime }\left(x\right)=-4,$$

yields

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\int }_0^1{I}_{\rho }g\left(x\right){\xi }_t{\xi }_{x}dx=& -D\left[{\xi }_x^2\left(1\right)+{\xi }_x^2\left(0\right)\right]+2D{\int }_0^1{\xi }_x^{2}dx+G{\int }_0^{1}g\left(x\right)\left(3s-\xi -{\varrho }_x\right){\xi }_x\hfill \\ & +2I_{\rho }{\int }_0^1{\xi }_t^{2}dx+{\int }_0^{1}g\left(x\right)\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right){\xi }_{x}dx.\hfill \end{array}$$

(51)

From Young’s inequality, we obtain for any ${\epsilon }_5>0$ the following:

$$G{\int }_0^{1}g\left(x\right)\left(3s-\xi -{\varrho }_x\right){\xi }_{x}dx\le {\epsilon }_5{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{G^2}{{\epsilon }_5}}{\int }_0^1{\xi }_x^{2}dx,$$

(52)

and

$${\int }_0^{1}g\left(x\right)\left({\gamma }_1{\theta }_x+{\gamma }_2{P}_x\right){\xi }_{x}dx\le D{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{2{\gamma }_1^2}{D}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{2{\gamma }_2^2}{D}}{\int }_0^L{P}_x^{2}dx.$$

(53)

Replacing (52) and (53) in (51), we see that

$$\begin{array}{cc}\hfill D\left[{\xi }_x^2\left(1\right)+{\xi }_x^2\left(0\right)\right]\le & -{\displaystyle \frac{d}{dt}}{\int }_0^1{I}_{\rho }g\left(x\right){\xi }_t{\xi }_{x}dx+{\epsilon }_5{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+\left(3D+{\displaystyle \frac{G^2}{{\epsilon }_5}}\right){\int }_0^1{\xi }_x^{2}dx\hfill \\ & +{\displaystyle \frac{2{\gamma }_1^2}{D}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{2{\gamma }_2^2}{D}}{\int }_0^1{P}_x^{2}dx+2I_{\rho }{\int }_0^1{\xi }_t^{2}dx.\hfill \end{array}$$

(54)

On the other hand,

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\int }_0^1\rho g\left(x\right){\varrho }_t{\varrho }_{x}dx& =G{\int }_0^{1}g\left(x\right){\varrho }_{xx}{\varrho }_{x}dx-G{\int }_0^{1}g\left(x\right){(3s-\xi )}_x{\varrho }_{x}dx+\rho {\int }_0^{1}g\left(x\right){\varrho }_t{\varrho }_{xt}dx\hfill \\ & -\mu {\int }_0^{1}g\left(x\right){\varrho }_{x}z(x,1,t)dx\hfill \\ & =-G\left[{\varrho }_x^2\left(1\right)+{\varrho }_x^2\left(0\right)\right]+2G{\int }_0^1{\varrho }_x^{2}dx-G{\int }_0^{1}g\left(x\right){(3s-\xi )}_x{\varrho }_{x}dx+2\rho {\int }_0^1{\varrho }_t^{2}dx\hfill \\ \hfill & -\mu {\int }_0^{1}g\left(x\right){\varrho }_{x}z(x,1,t)dx.\hfill \end{array}$$

Then, the following estimates

$$-G{\int }_0^{1}g\left(x\right){(3s-\xi )}_x{\varrho }_{x}dx\le G{\int }_0^1{\varrho }_x^{2}dx+G{\int }_0^1{(3s-\xi )}_x^{2}dx,$$

and

$$-\mu {\int }_0^{1}g\left(x\right){\varrho }_{x}z(x,1,t)dx\le {\int }_0^1{\varrho }_x^{2}dx+{\mu }^2{\int }_0^1{z}^2(x,1,t)dx,$$

give us

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\int }_0^1\rho g\left(x\right){\varrho }_t{\varrho }_{x}dx\le & -G\left[{\varrho }_x^2\left(1\right)+{\varrho }_x^2\left(0\right)\right]+3G{\int }_0^1{\varrho }_x^{2}dx+G{\int }_0^1{(3s-\xi )}_x^{2}dx+2\rho {\int }_0^1{\varrho }_t^{2}dx\hfill \\ & +{\int }_0^1{\varrho }_x^{2}dx+{\mu }^2{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

(55)

Notice that

$${\int }_0^1{(3s-\xi )}_x^{2}dx\le 2{\int }_0^1{\xi }_x^{2}dx+18{\int }_0^1{s}_x^{2}dx,$$

and

$$\begin{array}{cc}\hfill {\int }_0^1{\varrho }_x^{2}dx& ={\int }_0^1{\left(3s-\xi -{\varrho }_x-(3s-\xi )\right)}^{2}dx\hfill \\ & \le 2{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+2{\int }_0^1{(3s-\xi )}_x^{2}dx\hfill \\ & \le 2{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+4{\int }_0^1{\xi }_x^{2}dx+36{\int }_0^1{s}_x^{2}dx.\hfill \end{array}$$

Therefore, we conclude from (55) that

$$\begin{array}{cc}\hfill \left[{\varrho }_x^2\left(1\right)+{\varrho }_x^2\left(0\right)\right]\le & -{\displaystyle \frac{1}{G}}{\displaystyle \frac{d}{dt}}{\int }_0^1\rho g\left(x\right){\varrho }_t{\varrho }_{x}dx+\left({\displaystyle \frac{2}{G}}+6\right){\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+{\displaystyle \frac{{\mu }^2}{G}}{\int }_0^1{z}^2(x,1,t)dx\hfill \\ & +\left({\displaystyle \frac{4}{G}}+14\right){\int }_0^1{\xi }_x^{2}dx+\left({\displaystyle \frac{36}{G}}+126\right){\int }_0^1{s}_x^{2}dx+{\displaystyle \frac{2\rho }{G}}{\int }_0^1{\varrho }_t^{2}dx.\hfill \end{array}$$

(56)

Combining (54) and (56) with (50), we have for any ${\epsilon }_4>0$ and ${\epsilon }_5>0$ that

$$\begin{array}{cc}\hfill D{\xi }_x{\varrho }_x{|}_{x=0}^{x=1}\le & \left((6+{\displaystyle \frac{2}{G}}){\epsilon }_4+{\displaystyle \frac{D}{4{\epsilon }_4}}{\epsilon }_5\right){\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx+\left({\displaystyle \frac{36}{G}}+126\right){\epsilon }_4{\int }_0^1{s}_x^{2}dx\hfill \\ \hfill & +{\displaystyle \frac{2\rho }{G}}{\epsilon }_4{\int }_0^1{\varrho }_t^{2}dx+{\displaystyle \frac{I_{\rho }D}{2{\epsilon }_4}}{\int }_0^1{\xi }_t^{2}dx+{\displaystyle \frac{{\gamma }_1^2}{2{\epsilon }_4}}{\int }_0^1{\theta }_x^{2}dx+{\displaystyle \frac{{\gamma }_2^2}{2{\epsilon }_4}}{\int }_0^L{P}_x^{2}dx\hfill \\ \hfill & +\left[\left({\displaystyle \frac{4}{G}}+14\right){\epsilon }_4+{\displaystyle \frac{D}{4{\epsilon }_4}}\left(3D+{\displaystyle \frac{G^2}{{\epsilon }_5}}\right)\right]{\int }_0^1{\xi }_x^{2}dx+{\displaystyle \frac{{\mu }^2{\epsilon }_4}{G}}{\int }_0^1{z}^2(x,1,t)dx\hfill \\ \hfill & -{\displaystyle \frac{{\epsilon }_4}{G}}{\displaystyle \frac{d}{dt}}{\int }_0^1\rho g\left(x\right){\varrho }_t{\varrho }_{x}dx-{\displaystyle \frac{D}{4{\epsilon }_4}}{\displaystyle \frac{d}{dt}}{\int }_0^1{I}_{\rho }g\left(x\right){\xi }_t{\xi }_{x}dx.\hfill \end{array}$$

(57)

Inserting (57) into (49) and using (5), then (48) follows. □

Lemma 11.

Let the functional $H_3$ be defined by

$$H_3\left(t\right)=3I_{\rho }{\int }_0^{1}s{s}_{t}dx+2\beta {\int }_0^1{s}^{2}dx.$$

Then, $H_3$ satisfies

$${\displaystyle \frac{d}{dt}}{H}_3\left(t\right)\le -3D{\int }_0^1{s}_x^{2}dx-3\gamma {\int }_0^1{s}^{2}dx+3I_{\rho }{\int }_0^1{s}_t^{2}dx+{\displaystyle \frac{9G^2}{4\gamma }}{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx.$$

(58)

Proof. 

Clearly,

$${\displaystyle \frac{d}{dt}}{H}_3\left(t\right)=3I_{\rho }{\int }_0^1{s}_t^{2}dx-3G{\int }_0^1\left(3s-\xi -{\varrho }_x\right)sdx-3D{\int }_0^1{s}_x^{2}dx-4\gamma {\int }_0^1{s}^{2}dx.$$

Afterward, using Young’s inequality, we arrive at (58). □

In the following, we define the functional $\widehat{\mathcal{L}}$ by

$$\begin{array}{cc}\hfill \widehat{\mathcal{L}}\left(t\right)=& ME\left(t\right)+M_1{F}_1\left(t\right)+H_1\left(t\right)+M_3{H}_2\left(t\right)+M_4{F}_3\left(t\right)+M_5{H}_3\left(t\right)+M_6{F}_6\hfill \\ \hfill & +{\displaystyle \frac{{\epsilon }_4}{G}}{M}_4{\int }_0^1\rho g\left(x\right){\varrho }_t{\varrho }_{x}dx+{\displaystyle \frac{D}{4{\epsilon }_4}}{M}_4{\int }_0^1{I}_{p}g\left(x\right){\xi }_t{\xi }_{x}dx,\hfill \end{array}$$

where M and $M_i,i=1,3,4,5,6$ will be determined later. As in (40), there exist two positive constants ${\widehat{\nu }}_1$ and ${\widehat{\nu }}_2$ such that

$${\widehat{\nu }}_{1}E\left(t\right)\le \widehat{\mathcal{L}}\left(t\right)\le {\widehat{\nu }}_{2}E\left(t\right).$$

(59)

Proof. 

(Of Theorem 2 for Boundary $II$)

Combining (22), (42), (46), (48), and (58), as well as taking

$${\epsilon }_1={\displaystyle \frac{1}{M_1}},{\epsilon }_2={\displaystyle \frac{1}{D{M}_1}},{\epsilon }_3={\displaystyle \frac{\rho }{4M_3}},{\epsilon }_4={\displaystyle \frac{G}{8M_4}},{\epsilon }_5={\displaystyle \frac{1}{M_4^2}},M_6=\left|\mu \right|,$$

we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}\widehat{\mathcal{L}}\left(t\right)\le & -\left(\kappa M-c_1{M}_1-{\displaystyle \frac{2{\gamma }_1^2}{D}}{M}_3-{\displaystyle \frac{{\gamma }_1^2}{G}}{M}_4-{\displaystyle \frac{4{\gamma }_1^2}{G}}{M}_4^2\right){\int }_0^1{\theta }_x^{2}dx\hfill \\ & -\left(\hslash M-c_2{M}_1-{\displaystyle \frac{2{\gamma }_2^2}{D}}{M}_3-{\displaystyle \frac{{\gamma }_2^2}{G}}{M}_4-{\displaystyle \frac{4{\gamma }_2^2}{G}}{M}_4^2\right){\int }_0^1{P}_x^{2}dx\hfill \\ & -\left(4\beta M-{\displaystyle \frac{3I_{\rho }}{4}}{M}_4-3I_{\rho }{M}_5\right){\int }_0^1{s}_t^{2}dx-\left({\displaystyle \frac{\rho }{2}}-\left|\mu \right|\left(M+1\right)\right){\int }_0^1{\varrho }_t^{2}dx\hfill \\ \hfill & -\left[{\displaystyle \frac{I_{\rho }}{2}}{M}_1-\left(I_{\rho }+\rho M_3\right)M_3-2I_{\rho }{M}_4-{\displaystyle \frac{4D{I}_{\rho }}{G}}{M}_4^2\right]{\int }_0^1{\xi }_t^{2}dx\hfill \\ \hfill & -\left[{\displaystyle \frac{D}{2}}{M}_3-5-{\displaystyle \frac{7G}{4}}-{\displaystyle \frac{2D}{G}}{M}_4^2\left(3D+G^2{M}_4^2\right)\right]{\int }_0^1{\xi }_x^{2}dx\hfill \\ \hfill & -\left[{\displaystyle \frac{G}{2}}{M}_4-{\displaystyle \frac{9G}{4}}-{\displaystyle \frac{5}{4}}-{\displaystyle \frac{2D}{G}}-{\displaystyle \frac{G^2}{4}}-{\displaystyle \frac{9G^2}{4\gamma }}{M}_5\right]{\int }_0^1{\left(3s-\xi -{\varrho }_x\right)}^{2}dx\hfill \\ & -\left(3D{M}_5-{\displaystyle \frac{81}{2}}-{\displaystyle \frac{63G}{4}}\right){\int }_0^1{s}_x^{2}dx-3\gamma M_5{\int }_0^1{s}^{2}dx\hfill \\ & -m_1\left|\mu \right|\tau {\int }_0^1{\int }_0^1{e}^{-\sigma \tau }{z}^2(x,\sigma ,t)d\sigma dx\hfill \\ & -{\displaystyle \frac{\left|\mu \right|}{2}}\left[m_1-\left|\mu \right|\left(M_3+M_4+{\displaystyle \frac{9}{8}}\right)\right]{\int }_0^1{z}^2(x,1,t)dx.\hfill \end{array}$$

We first take $M_5>0$ such that

$$3D{M}_5-{\displaystyle \frac{81}{2}}-{\displaystyle \frac{63G}{4}}>0.$$

For fixed $M_5$, we choose $M_4>0$ large enough such that

$${\displaystyle \frac{G}{2}}{M}_4-{\displaystyle \frac{9G}{4}}-{\displaystyle \frac{5}{4}}-{\displaystyle \frac{2D}{G}}-{\displaystyle \frac{G^2}{4}}-{\displaystyle \frac{9G^2}{4\gamma }}{M}_5>0.$$

For fixed $M_4$, we choose $M_3>0$ large enough such that

$${\displaystyle \frac{D}{2}}{M}_3-5-{\displaystyle \frac{7G}{4}}-{\displaystyle \frac{2D}{G}}{M}_4^2\left(3D+G^2{M}_4^2\right)>0.$$

For fixed $M_4$ and $M_3$, we choose $M_1>0$ large enough such that

$${\displaystyle \frac{I_{\rho }}{2}}{M}_1-\left(I_{\rho }+\rho M_3\right)M_3-2I_{\rho }{M}_4-{\displaystyle \frac{4D{I}_{\rho }}{G}}{M}_4^2>0.$$

For fixed $M_i,i=1,3,4,5$, we at last take M large enough such that

$$\kappa M-c_1{M}_1-{\displaystyle \frac{2{\gamma }_1^2}{D}}{M}_3-{\displaystyle \frac{{\gamma }_1^2}{G}}{M}_4-{\displaystyle \frac{4{\gamma }_1^2}{G}}{M}_4^2>0,$$

$$\hslash M-c_2{M}_1-{\displaystyle \frac{2{\gamma }_2^2}{D}}{M}_3-{\displaystyle \frac{{\gamma }_2^2}{G}}{M}_4-{\displaystyle \frac{4{\gamma }_2^2}{G}}{M}_4^2>0,$$

and

$$4\beta M-{\displaystyle \frac{3I_{\rho }}{4}}{M}_4-3I_{\rho }{M}_5>0.$$

Finally, we pick

$$\overline{\mu }=min\left\{{\displaystyle \frac{m_1}{M_3+M_4+\frac{9}{8}}},{\displaystyle \frac{\rho }{2(M+1)}}\right\}.$$

We find that there exists a positive constant ${\widehat{\nu }}_0$ so that

$${\displaystyle \frac{d}{dt}}\widehat{\mathcal{L}}\left(t\right)\le -{\widehat{\nu }}_{0}E\left(t\right),$$

together with (59), gives

$$E\left(t\right)\le {\displaystyle \frac{{\widehat{\nu }}_2}{{\widehat{\nu }}_1}}E\left(0\right)e^{-{\displaystyle \frac{{\widehat{\nu }}_0}{{\widehat{\nu }}_2}}t}.$$

□

4. Conclusions

In this work, we have investigated the existence of a solution for our system, which is discussed within the context of the semigroup approach. In addition, under different boundary conditions, two results of stability properties independent of initial data are studied (see [3,9,10,11,12,13]). In future work, we will follow the same study but with a time-varying delay term, and the numerical examples will be investigated.