Polynomial Decay Rate for a Coupled Lamé System with Viscoelastic Damping and Distributed Delay Terms

Abstract

In this paper, we prove a general energy decay results of a coupled Lamé system with distributed time delay. By assuming a more general of relaxation functions and using some properties of convex functions, we establish the general energy decay results to the system by using an appropriate Lyapunov functional.

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1. Introduction

In this work, we shall be concerned with studying the general decay rate of the following Lamé system in $\Omega \times {ℝ}^{+}$: $$\begin{array}{}\text{(1)}& \begin{array}{c}{u}_{tt}-{\Delta }_{e}u+{\int }_0^t{g}_{1}t-s\Delta usds-k_1\Delta u_t-{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \Delta u_{t}x,t-\rho d\rho =f_{1}u,v,\hfill \\ v_{tt}-{\Delta }_{e}v+{\int }_0^t{g}_{2}t-s\Delta vsds-k_2\Delta v_t-{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho \Delta v_{t}x,t-\rho d\rho =f_{2}u,v.\hfill \end{array}\end{array}$$

Equations (1) are associated with the following boundary and initial conditions $$\begin{array}{}\text{(2)}& \begin{array}{c}ux,t=vx,t=0,on\partial \Omega \times {ℝ}^{+},\hfill \\ ux,0=u_{0}x,vx,0=v_{0}x,u_{t}x,0=u_{1}x,v_{t}x,0=v_{1}x,x\in \Omega ,\hfill \\ u_{t}x,-t,v_{t}x,-t=f_{0}x,t,g_{0}x,t,\mathit{\text{in}}\Omega \times 0,{\tau }_2,\hfill \end{array}\end{array}$$where $\Omega$ is a bounded domain in ${ℝ}^n$$n=1,2,3$, with smooth boundary $\partial \Omega$. The elasticity differential operator ${\Delta }_e$ is given by $$\begin{array}{}\text{(3)}& {\Delta }_{e}u=\mu \Delta u+\mu +\lambda \nabla divu,\end{array}$$and the Lamé constants $\mu$ and $\lambda$ are satisfying the following conditions $$\begin{array}{}\text{(4)}& \mu >0,\mu +\lambda >0.\end{array}$$

The parameters $k_1,k_2,{\tau }_1$, and ${\tau }_2$ are positive constants, with ${\tau }_1<{\tau }_2$. The functions ${\mu }_1,{\mu }_2:{\tau }_1,{\tau }_2\to ℝ$ are bounded. The functions $f_{1}u,v$ and $f_{2}u,v$ which represent the source terms will be specified later.

After several authors have studied the problems of coupled systems and hyperbolic systems, their stability is associated with velocities and is proven under conditions imposed on the subgroup [1]. The researchers also studied behavior of the energy in a limited field with nonlinear damping and external force and a varying delay of time to find solutions to the Lame system [1–9].

Recently, problems that contain viscoelasticity have been addressed, and many results have been found regarding the global existence and stability of solutions (see [2, 9]), under conditions on the relaxation function, whether exponential or polynomial decay. In addition, in [10], Boulaaras obtained the stability result of the global solution to the Lamé system with the flexible viscous term by adding logarithmic nonlinearity, even though the kernel is not necessarily decreasing in contrast to what he studied [2].

Introducing a distributed delay term makes our problem different from those considered so far in the literature.

The importance of this term appears in many works, and this is due to the fact that many phenomena depends on their past. Also, it is influence on the asymptotic behavior of the solution for the different types of problems such that Timoshenko system [3, 11–13], transmission problem [14], wave equation [15], and thermoelastic system [16, 17].

In the present work, we extend the general decay result obtained by Feng in [18] to the case of distributed term delay, namely, we will make sure that the result is achieved if the distributed delay term exists.

This paper is organized as follows. In the second section, we give some preliminaries related to problem (1). In Section 3, we prove our main result.

2. Preliminaries

In this section, we provide some materials and necessary assumptions which we need in the prove of our results. We use the standard Lebesgue and Sobolev spaces with their scaler products and norms. For simplicity, we would write $.$ instead of ${.}_2$. Throughout this work, we used a generic positive constant $c$.

For the relaxation functions $g_1$ and $g_2$, we assume, for $i=1,2$,

(A1) $g_{i}t\text{:}{ℝ}^{+}\to {ℝ}^{+}$ are nonincreasing $C^1$ functions satisfying $$\begin{array}{}\text{(5)}& g_{i}0>0\text{and}\mu -{\int }_0^{\infty }{g}_{i}sds=l_i>0.\end{array}$$

We assume further that for $i=1,2:$

(A2) There exist two $C^1$ functions $G_i:{ℝ}^{+}\to {ℝ}^{+}$, with $G_{i}0=G_i^{\prime }0=0$, which are linear or are strictly increasing and strictly convex functions of class $C^2{ℝ}^{+}$ on $0,r,r\le g_{i}0$, such that $$\begin{array}{}\text{(6)}& g_i^{\prime }t\le -{\xi }_{i}t{G}_i{g}_{i}t,\forall t\ge 0,\end{array}$$where ${\xi }_{i}t$ are $C^1$ functions satisfying $$\begin{array}{}\text{(7)}& {\xi }_{i}t>0,{\xi }_i^{\prime }t\le 0,\forall t\ge 0.\end{array}$$

(A3) For the source terms $f_1$ and $f_2$, we take $$\begin{array}{c}\text{(8)}& f_{1}u,v=\alpha {u+v}^{p-1}u+v+\beta u^{p-3/2}u{v}^{p+1/2},\hspace{1em}\forall u,v\in {ℝ}^n{}^2,f_{2}u,v=\alpha {u+v}^{p-1}u+v+\beta v^{p-3/2}v{u}^{p+1/2},\hspace{1em}\forall u,v\in {ℝ}^n{}^2,\end{array}$$with $\alpha ,\beta >0$. Clearly, $$\begin{array}{}\text{(9)}& u{f}_{1}u,v+v{f}_{2}u,v=p+1Fu,v,\forall u,v\in {ℝ}^n{}^2,\end{array}$$where $$\begin{array}{c}\text{(10)}& Fu,v=\frac{1}{p+1}\alpha {u+v}^{p+1}+2\beta {uv}^{p+1/2},\hspace{1em}\forall u,v\in {ℝ}^n{}^2,f_{1}u,v=\frac{\partial F}{\partial u},f_{2}u,v=\frac{\partial F}{\partial v}.\end{array}$$Further, we assume that there is $C>0$, such that $$\begin{array}{}\text{(11)}& \frac{\partial f_i}{\partial u}u,v+\frac{\partial f_i}{\partial v}u,v\le C{u}^{p-1}+v^{p-1},i=1,2\text{where}1\le p\le 6.\end{array}$$

(A4) $$\begin{array}{}\text{(12)}& \text{if}n=1,2;p\ge 3,\text{if}n=3;p=3.\end{array}$$

So, we have the embedding $$\begin{array}{c}\text{(13)}& H_0^1{\Omega }^{°}{L}^q\Omega \text{for}2\le q\le \frac{2n}{n-2}\text{if}n\ge 3\text{or}q\ge 2\text{if}n=1,2,L^{r°}{L}^q\text{for}q<r.\end{array}$$

Let $c_s$ the same embedding constant, so we have $$\begin{array}{}\text{(14)}& {\nu }_q\le c_s{\nabla \nu }_2,{\nu }_q\le c_s{\nu }_r\text{for}\nu \in H_0^1\Omega .\end{array}$$

Remark 1.

There exist two constants ${\Lambda }_1>0$ and ${\Lambda }_2>0$ such that $$\begin{array}{}\text{(15)}& {\int }_{\Omega }{f_{i}u,v}^{2}dx\le {\Lambda }_i{l_1{\nabla u}^2+l_2{\nabla v}^2}^p,i=1,2.\end{array}$$

As in many papers, we introduce the following new variables $$\begin{array}{}\text{(16)}& \begin{array}{c}zx,\rho ,\rho ,t=u_{t}x,t-\rho \rho ,\hfill \\ yx,\rho ,\rho ,t=v_{t}x,t-\rho \rho ,\hfill \end{array}\end{array}$$then, we obtain $$\begin{array}{}\text{(17)}& \begin{array}{c}\rho z_{t}x,\rho ,\rho ,t+z_{\rho }x,\rho ,\rho ,t=0,\hfill \\ zx,0,\rho ,t=u_{t}x,t,\hfill \end{array}\text{(18)}& \begin{array}{c}\rho y_{t}x,\rho ,\rho ,t+y_{\rho }x,\rho ,\rho ,t=0,\hfill \\ yx,0,\rho ,t=v_{t}x,t.\hfill \end{array}\end{array}$$

Consequently, the problem (1) is equivalent to $$\begin{array}{}\text{(19)}& \begin{array}{c}{u}_{tt}-{\Delta }_{e}u+{\int }_0^t{g}_{1}t-s\Delta usds-k_1\Delta u_t-{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \Delta zx,1,\rho ,td\rho =f_{1}u,v,\hfill \\ v_{tt}-{\Delta }_{e}v+{\int }_0^t{g}_{2}t-s\Delta vsds-k_2\Delta v_t-{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho \Delta yx,1,\rho ,td\rho =f_{2}u,v,\hfill \\ \rho z_{t}x,\rho ,\rho ,t+z_{\rho }x,\rho ,\rho ,t=0,\hfill \\ \rho y_{t}x,\rho ,\rho ,t+y_{\rho }x,\rho ,\rho ,t=0,\hfill \end{array}\end{array}$$with the initial data and boundary conditions $$\begin{array}{}\text{(20)}& \begin{array}{c}ux,0,vx,0=u_{0}x,v_{0}x,\text{in}\Omega ,\hfill \\ u_{t}x,0,v_{t}x,0=u_{1}x,v_{1}x,\text{in}\Omega ,\hfill \\ u_{t}x,-t,v_{t}x,-t=f_{0}x,t,g_{0}x,t,\text{in}\Omega \times 0,{\tau }_2,\hfill \\ ux,t=vx,t=0,\text{in}\partial \Omega \times 0,\infty ,\hfill \\ zx,\rho ,\rho ,0,yx,\rho ,\rho ,0=f_{0}x,\rho \rho ,g_{0}x,\rho \rho ,\text{in}\Omega \times 0,1\times 0,{\tau }_2,\hfill \end{array}\end{array}$$where $$\begin{array}{}\text{(21)}& x,\rho ,\rho ,t\in \Omega \times 0,1\times {\tau }_1,{\tau }_2\times 0,\infty .\end{array}$$

We recall the following notations $$\begin{array}{c}\text{(22)}& h\ast \phi ={\int }_0^{t}ht-s\phi sdsdx,h\circ \phi t={\int }_0^{t}ht-s{\phi t-\phi s}^{2}ds.\end{array}$$

Thus, we have the following important property $$\begin{array}{}\text{(23)}& {\int }_{\Omega }h\ast \phi {\phi }_{t}dx=-\frac{1}{2}ht{\phi t}^2+\frac{1}{2}{h}^{\prime }\circ \phi t-\frac{1}{2}\frac{d}{dt}h\circ \phi t-{\int }_0^{t}hsds{\phi t}^2.\end{array}$$

The energy modified associated to the problem (19) is defined by $$\begin{array}{}\text{(24)}& Et=\frac{1}{2}{u_t}^2+\mu -{\int }_0^t{g}_{1}sds{\nabla u}^2+\lambda +\mu {divu}^2+\frac{1}{2}{g}_1\circ \nabla ut+\eta {\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^{2}d\rho d\rho dx+\frac{1}{2}{v_t}^2+\mu -{\int }_0^t{g}_{2}sds{\nabla v}^2+\lambda +\mu {divv}^2+\frac{1}{2}{g}_2\circ \nabla vt+\eta {\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}d\rho d\rho dx-{\int }_{\Omega }Fu,vdx.\end{array}$$

First, we prove in the following theorem the result of energy identity.

Lemma 2.

Assume that $$\begin{array}{}\text{(25)}& {\int }_{{\tau }_1}^{{\tau }_2}{\mu }_i\rho d\rho <k_i,i=1,2.\end{array}$$

Then, the energy modified defined by (24) satisfies, along the solution $u,v,z,y$ of (19), the estimate $$\begin{array}{}\text{(26)}& \frac{d}{dt}Et\le -\frac{1}{2}{g}_{1}t{\nabla u}^2+\frac{1}{2}{g}_1^{\prime }\circ \nabla ut-\frac{1}{2}{g}_{2}t{\nabla v}^2+\frac{1}{2}{g}_2^{\prime }\circ \nabla vt-k_1-\frac{\eta +1}{2}{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho d\rho {\nabla u_t}^2-\frac{\eta -1}{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla zx,1,\rho ,t}^{2}d\rho dx-k_2-\frac{\eta +1}{2}{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho d\rho {\nabla v_t}^2-\frac{\eta -1}{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla yx,1,\rho ,t}^{2}d\rho dx\le 0,\end{array}$$for $$\begin{array}{}\text{(27)}& 1<\eta <\min \frac{2k_1}{{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho d\rho },\frac{2{\text{k}}_2}{{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho d\rho }-1.\end{array}$$

Proof.

First multiplying the equation ${0.14}_1$ by $u_t$ and integrating by parts over $\Omega ,$ we obtain $$\begin{array}{}\text{(28)}& \frac{1}{2}\frac{d}{dt}{u_t}^2+\mu {\nabla u}^2+\lambda +\mu {divu}^2-{\int }_{\Omega }\nabla u_t{\int }_0^t{g}_{1}t-s\nabla usds+k_1{\nabla u_t}^2+{\int }_{\Omega }\nabla u_t{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \nabla zx,1,\rho ,td\rho dx={\int }_{\Omega }{f}_{1}u,v.udx,\end{array}$$by using (23), we obtain $$\begin{array}{}\text{(29)}& \frac{1}{2}\frac{d}{dt}{u_t}^2+\mu -{\int }_0^t{g}_{1}sds{\nabla u}^2+\lambda +\mu {divu}^2+g_1\circ \nabla ut=-\frac{1}{2}{g}_1{\nabla u}^2+\frac{1}{2}{g}_1^{\prime }\circ \nabla ut+{\int }_{\Omega }{\text{u}}_t{f}_{1}u,vdx-k_1{\nabla u_t}^2+{\int }_{\Omega }\nabla u_t{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,1,\rho ,td\rho dx.\end{array}$$

Similarly, multiplying the equation (19) by $v_t$ and integrating over $\Omega ,$ we obtain $$\begin{array}{}\text{(30)}& \frac{1}{2}\frac{d}{dt}{v_t}^2+\mu -{\int }_0^t{g}_{2}sds{\nabla v}^2+\lambda +\mu {divv}^2+g_2\circ \nabla vt=-\frac{1}{2}{g}_2{\nabla v}^2+\frac{1}{2}{g}_2^{\prime }\circ \nabla vt+{\int }_{\Omega }{v}_t{f}_{2}u,vdx-k_2{\nabla v_t}^2+{\int }_{\Omega }\nabla v_t{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho \nabla yx,1,\rho ,td\rho dx.\end{array}$$

Multiplying the equation (19) by $-\eta {\mu }_1\rho \Delta zx,\rho ,\rho ,t$ and integrating by parts over $\Omega \times 0,1\times {\tau }_1,{\tau }_2,$ we obtain $$\begin{array}{}\text{(31)}& \eta {\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_1\rho \nabla zx,\rho ,\rho ,t\nabla z_{t}x,\rho ,\rho ,td\rho d\rho dx=-\eta {\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,\rho ,\rho ,t\nabla z_{\rho }x,\rho ,\rho ,td\rho d\rho dx,\end{array}$$therefore $$\begin{array}{}\text{(32)}& \frac{d}{dt}\frac{\eta }{2}{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^{2}d\rho d\rho dx=-\frac{\eta }{2}{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \frac{d}{d\rho }{\nabla zx,\rho ,\rho ,t}^{2}d\rho d\rho dx=-\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla zx,1,\rho ,t}^{2}d\rho dx+\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla u_{t}x,t}^{2}d\rho dx.\end{array}$$

Multiplying the fourth equation of (19) by $-\eta {\mu }_2\rho \Delta yx,\rho ,\rho ,t$ and integrating over $\Omega \times 0,1\times {\tau }_1,{\tau }_2,$ we obtain $$\begin{array}{}\text{(33)}& \frac{d}{dt}\frac{\eta }{2}{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}d\rho d\rho dx=-\frac{\eta }{2}{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho \frac{d}{d\rho }{\nabla yx,\rho ,\rho ,t}^{2}d\rho d\rho dx=-\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla yx,1,\rho ,t}^{2}d\rho dx+\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla v_{t}x,t}^{2}d\rho dx.\end{array}$$

For the source term, we have $$\begin{array}{}\text{(34)}& {\int }_{\Omega }{u}_t{f}_{1}u,vdx+{\int }_{\Omega }{v}_t{f}_{2}u,vdx={\int }_{\Omega }{u}_t\alpha {u+v}^{p-1}u+v+\beta u^{p-3/2}u{v}^{p+1/2}+{\int }_{\Omega }{v}_t\alpha {u+v}^{p-1}u+v+\beta v^{p-3/2}v{u}^{p+1/2}={\int }_{\Omega }\alpha {u+v}^{p-1}u+v{u}_t+v_t+\beta u^{p-3/2}u.u_t{v}^{p+1/2}+\beta v^{p-3/2}v.v_t{u}^{p+1/2}dx=\frac{d}{dt}{\int }_{\Omega }\frac{\alpha }{p+1}{u+v}^{p+1}+\frac{2\beta }{p+1}{uv}^{p+1/2}dx=\frac{d}{dt}{\int }_{\Omega }Fu,vdx.\end{array}$$

By collecting the previous equations (29)–(34), we get $$\begin{array}{}\text{(35)}& \frac{d}{dt}Et=-\frac{1}{2}{g}_1{\nabla u}^2+\frac{1}{2}{g}_1^{\prime }\circ \nabla ut-k_1\nabla u_t+{\int }_{\Omega }\nabla u_t{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,1,\rho ,td\rho dx-\frac{1}{2}{g}_2{\nabla v}^2+\frac{1}{2}{g}_2^{\prime }\circ \nabla vt-k_2\nabla v_t+{\int }_{\Omega }\nabla v_t{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho \nabla yx,1,\rho ,td\rho dx-\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla zx,1,\rho ,t}^{2}d\rho dx+\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla u_{t}x,t}^{2}d\rho dx-\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla yx,1,\rho ,t}^{2}d\rho dx+\frac{\eta }{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla v_{t}x,t}^{2}d\rho dx.\end{array}$$

Using Young’s inequality, we obtain $$\begin{array}{}\text{(36)}& {\int }_{\Omega }\nabla u_t{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,1,\rho ,td\rho dx\le {\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\nabla u_t\sqrt{{\mu }_1\rho }\sqrt{{\mu }_1\rho }\nabla zx,1,\rho ,td\rho dx\le \frac{1}{2}{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho d\rho \nabla u_t+\frac{1}{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla zx,1,\rho ,t}^{2}d\rho dx,\end{array}$$similarly $$\begin{array}{}\text{(37)}& {\int }_{\Omega }\nabla v_t{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho \nabla yx,1,\rho ,td\rho dx\le \frac{1}{2}{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho d\rho \nabla v_t+\frac{1}{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla yx,1,\rho ,t}^{2}d\rho dx.\end{array}$$

This completes the proof.

3. General Decay

In this section we will prove that the solution of problems (19)–(20) decay generally to trivial solution. Using the energy method and suitable Lyapunov functional.

In the following, we will present our main stability result:

Theorem 3 (Decay rates of energy).

Assume that (A1)–(A3) hold. Then, for every $t_0>0$, there exist two positive constants ${\alpha }_1$ and ${\alpha }_2$ such that the energy defined by (24) satisfies the following decay $$\begin{array}{}\text{(38)}& Et\le {\alpha }_2{G}_4^{-1}{\alpha }_1{\int }_{g^{-1}r}^t\xi sds,\forall t\ge g^{-1}r,\end{array}$$where $$\begin{array}{}\text{(39)}& G_{4}t={\int }_t^r\frac{1}{s{G}_{0}s}ds,G_{0}t=\min G_1^{\prime }t,G_2^{\prime }t,\end{array}$$and $\xi t=\min {\xi }_{1}t,{\xi }_{2}t,gt=\max g_{1}t,g_{2}t.$

This theorem will be proved later after providing some remarks.

Remark 4.

(1) In case ${\int }_0^{\infty }{\xi }_{i}tdt=\infty$, Theorem 3 ensures ${\lim }_{t\to \infty }Et=0$.

(2) From (A2), we infer that ${\lim }_{t\to \infty }{g}_{i}t=0$. Then, there exists some $t_1\ge 0$ large enough such that

(a) $$\begin{array}{}\text{(40)}& g_i{t}_1=r\Rightarrow g_{i}t\le r,\forall t\ge t_1.\end{array}$$

As $G_i$ is positive continuous functions, and$g_i$ and ${\xi }_i$ are positive nonincreasing continuous functions, then, for all $0\le t\le t_1,$$$\begin{array}{}\text{(41)}& 0<g_i{t}_1\le g_{i}t\le g_{i}0\text{and}0<{\xi }_i{t}_1\le {\xi }_{i}t\le {\xi }_{i}0,\end{array}$$which implies for some positive constants $a_i$ and $b_i$, $$\begin{array}{}\text{(42)}& a_i\le {\xi }_{i}t{G}_i{g}_{i}t\le b_i.\end{array}$$

Consequently, $$\begin{array}{}\text{(43)}& g_i^{\prime }t\le -{\xi }_{i}t{G}_i{g}_{i}t\le -\frac{a_i}{g_{i}0}{g}_{i}0\le -\frac{a_i}{g_{i}0}{g}_{i}t,\text{for}t\in 0,t_1.\end{array}$$

(3) We also mention Johnson’s inequality, which is very important for proving our result. If $G$ is a convex function on $a,b$, $g:\Omega \to a,b]$, we have

$$\begin{array}{}\text{(44)}& G\frac{1}{k}{\int }_{\Omega }gxhxdx\le \frac{1}{k}{\int }_{\Omega }Ggxhxdx,\end{array}$$where $h$ is a function that satisfies $$\begin{array}{}\text{(45)}& hx\ge 0\text{and}{\int }_{\Omega }hxdx=k>0.\end{array}$$

To prove the desired result, we create a Lyapunov functional equivalent to $E$. For this, we define some functions that allow us to construct this Lyapunov function.

As in Baowei [18] and Mustafa ([19, 20]), we define $$\begin{array}{}\text{(46)}& C_{\zeta ,i}={\int }_0^{\infty }\frac{g_i^{2}s}{\sqrt{\zeta g_{i}s-g_i^{\prime }s}}\text{and}{h}_{i}t=\zeta g_{i}t-g_i^{\prime }t,i=1,2,\end{array}$$for any $0<\zeta <1$.

Lemma 5.

Let $u,v,z,y$ be a solution of the problem (19). Then, the functional $$\begin{array}{}\text{(47)}& \phi t={\int }_{\Omega }ut{u}_{t}tdx+{\int }_{\Omega }vt{v}_{t}tdx,\end{array}$$satisfies the estimate $$\begin{array}{}\text{(48)}& {\phi }^{\prime }t\le -\frac{l_1}{2}{\nabla ut}^2-\frac{l_2}{2}{\nabla vt}^2+{u_{t}t}^2+{v_{t}t}^2-\lambda +\mu {divut}^2-\lambda +\mu {divvt}^2+\frac{3C_{\zeta ,1}}{2l_1}{h}_1\circ \nabla ut+\frac{3k_1^2}{2l_1}{\nabla u_t}^2+\frac{3k_2^2}{2l_2}{\nabla v_t}^2+\frac{3C_{\zeta ,2}}{2l_2}{h}_2\circ \nabla vt+p+1{\int }_{\Omega }Fut,vtdx+\frac{3k_1}{2l_1}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla zx,1,\rho ,t}^{2}d\rho dx+\frac{3k_2}{2l_2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla yx,1,\rho ,t}^{2}d\rho dx.\end{array}$$

Proof.

Taking the derivative of (47), we obtain $$\begin{array}{}\text{(49)}& {\phi }^{\prime }t={\int }_{\Omega }{u_{t}t}^{2}dx+{\int }_{\Omega }ut{u}_{tt}tdx+{\int }_{\Omega }{v_{t}t}^{2}dx+{\int }_{\Omega }vt{v}_{tt}tdx.\end{array}$$

From problem (19) and using integration by parts, we get $$\begin{array}{}\text{(50)}& {\phi }^{\prime }t={u_{t}t}^2+v_t{t}^2+{\int }_{\Omega }ut{\Delta }_{e}u-{\int }_0^t{g}_{1}t-s\Delta usds+k_1\Delta u_t+{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\varrho \Delta zx,1,\varrho ,td\varrho +f_{1}u,vdx+{\int }_{\Omega }vt{\Delta }_{e}v-{\int }_0^t{g}_{2}t-s\Delta vsds+k_2\Delta v_t+{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\varrho \Delta yx,1,\varrho ,td\varrho +f_{2}u,vdx={u_{t}t}^2+{v_{t}t}^2-k_1{\int }_{\Omega }\nabla u\nabla u_{t}dx-k_2{\int }_{\Omega }\nabla v\nabla v_{t}dx-u{\nabla ut}^2-\lambda +\mu {divut}^2+{\int }_{\Omega }\nabla ut{\int }_0^t{g}_{1}t-s\nabla usds-{\int }_{\Omega }\nabla ut{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\varrho \nabla zx,1,\varrho ,td\varrho +{\int }_{\Omega }ut{f}_{1}u,vdx-\mu {\nabla vt}^2-\lambda +\mu {divvt}^2+{\int }_{\Omega }\nabla vt{\int }_0^t{g}_{2}t-s\nabla vsds-{\int }_{\Omega }\nabla vt{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\varrho \nabla yx,1,\varrho ,td\varrho +{\int }_{\Omega }vt{f}_{2}u,vdx={u_{t}t}^2+{v_{t}t}^2-k_1{\int }_{\Omega }\nabla u\nabla u_{t}dx-k_2{\int }_{\Omega }\nabla v\nabla v_{t}dx-\mu -{\int }_0^t{g}_{1}sds{\nabla ut}^2-\lambda +\mu {divut}^2+{\int }_{\Omega }\nabla ut{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx-{\int }_{\Omega }\nabla ut{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\varrho \nabla zx,1,\varrho ,td\varrho dx+{\int }_{\Omega }ut{f}_{1}u,vdx-\mu -{\int }_0^t{g}_{2}sds{\nabla vt}^2-\lambda +\mu {divvt}^2+{\int }_{\Omega }\nabla vt{\int }_0^t{g}_{2}t-s\nabla vs-\nabla vtdsdx-{\int }_{\Omega }\nabla ut{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\varrho \nabla yx,1,\varrho ,td\varrho dx+{\int }_{\Omega }vt{f}_{2}u,vdx.\end{array}$$

By using Hölder and Young’s inequalities, we have $$\begin{array}{}\text{(51)}& {\int }_{\Omega }\nabla ut{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx\le \frac{l_1}{6}{\nabla ut}^2+\frac{3}{2l_1}{\int }_{\Omega }{\int }_0^t{g_{1}t-s\nabla us-\nabla utds}^{2}dx\le \frac{l_1}{6}{\nabla ut}^2+\frac{3}{2l_1}{\int }_{\Omega }{\int }_0^t{\frac{g_{1}t-s}{\sqrt{\zeta g_{1}t-s-g_1^{\prime }t-s}}\sqrt{\zeta g_{1}t-s-g_1^{\prime }t-s}\nabla us-\nabla utds}^{2}dx\le \frac{l_1}{6}{\nabla ut}^2+\frac{3}{2l_1}{\int }_0^t\frac{g_{1}s}{\sqrt{\zeta g_{1}s-g_1^{\prime }s}}ds{\int }_{\Omega }{\int }_0^t\zeta g_{1}t-s-g_1^{\prime }t-s{\nabla us-\nabla ut}^{2}ds\le \frac{l_1}{6}{\nabla ut}^2+\frac{3C_{\zeta ,1}}{2l_1}{h}_1\circ \nabla ut.\end{array}$$

Similarly, we obtain $$\begin{array}{c}\text{(52)}& {\int }_{\Omega }\nabla vt{\int }_0^t{g}_{2}t-s\nabla vs-\nabla vtdsdx\le \frac{l_2}{6}{\nabla vt}^2+\frac{3C_{\zeta ,2}}{2l_2}{h}_2\circ \nabla vt,{\int }_{\Omega }\nabla ut{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,1,\rho ,td\rho dx\le \frac{l_1}{6}{\nabla ut}^2+\frac{3}{2l_1}{\int }_{\Omega }{{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid \nabla zx,1,\rho ,td\rho }^{2}dx\le \frac{l_1}{6}{\nabla ut}^2+\frac{3}{2l_1}{\int }_{\Omega }{{\int }_{{\tau }_1}^{{\tau }_2}\sqrt{\mid {\mu }_1\rho \mid }\sqrt{\mid {\mu }_1\rho \mid }\nabla zx,1,\rho ,td\rho }^2\le \frac{l_1}{6}{\nabla ut}^2+\frac{3}{2l_1}{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid d\rho {\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho \le \frac{l_1}{6}{\nabla ut}^2+\frac{3k_1}{2l_1}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho ,\\ {\int }_{\Omega }\nabla vt{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho \nabla yx,1,\rho ,td\rho dx\le \frac{l_2}{6}{\nabla vt}^2+\frac{3}{2l_2}{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_2\rho \mid d\rho {\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_2\rho \mid {\nabla yx,1,\rho ,t}^{2}d\rho \le \frac{l_2}{6}{\nabla vt}^2+\frac{3k_2}{2l_2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_2\rho \mid {\nabla yx,1,\rho ,t}^{2}d\rho .\end{array}$$

The Young’s inequality gives $$\begin{array}{c}\text{(53)}& k_1{\int }_{\Omega }\nabla ut\nabla u_{t}tdx\le \frac{l_1}{6}{\nabla ut}^2+\frac{3k_1^2}{2l_1}{\nabla u_{t}t}^2,k_2{\int }_{\Omega }\nabla vt\nabla v_{t}tdx\le \frac{l_2}{6}{\nabla vt}^2+\frac{3k_2^2}{2l_2}{\nabla v_{t}t}^2.\end{array}$$

For the source term, we have $$\begin{array}{}\text{(54)}& {\int }_{\Omega }ut{f}_{1}u,vdx+{\int }_{\Omega }vt{f}_{2}u,vdx=p+1{\int }_{\Omega }Fu,vdx.\end{array}$$

Combining the equations (51)–(54), thus, our proof is completed.

Lemma 6.

Let ($u$, $v$, $z$, $y$) be a solution of the problem (19). Then, the functional $$\begin{array}{}\text{(55)}& \psi t={\int }_{\Omega }{u}_{t}t{\int }_0^t{g}_{1}sus-utdsdx+{\int }_{\Omega }{v}_{t}t{\int }_0^t{g}_{2}svs-vtdsdx={\psi }_{1}t+{\psi }_{2}t,\end{array}$$

satisfies for any $\delta >0$ the estimate $$\begin{array}{}\text{(56)}& {\psi }^{\prime }t\le \delta +\delta {\Lambda }_3{l}_1{\nabla ut}^2+\delta {divu}^2+\delta {\Lambda }_3{l}_2{\nabla vt}^2+\delta -{\int }_0^t{g}_{1}sds{u_{t}t}^2+\frac{c{C}_{\zeta ,1}+1}{\delta }{h}_1\circ \nabla ut+\delta k_1{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho dx+\delta k_1{\nabla u_t}^2+\delta +\delta {\Lambda }_4{l}_2{\nabla vt}^2+\delta {divv}^2+\delta {\Lambda }_4{l}_1{\nabla ut}^2+\delta -{\int }_0^t{g}_{2}sds{v_{t}t}^2+\frac{c{C}_{\zeta ,2}+1}{\delta }{h}_2\circ \nabla vt+\delta k_2{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_2\rho \mid {\nabla yx,1,\rho ,t}^{2}d\rho dx+\delta k_1{\nabla v_t}^2,\end{array}$$where ${\Lambda }_3$ and ${\Lambda }_4$ are two positive constants.

Proof.

First, we begin to estimate ${\psi }_1^{\prime }t$$$\begin{array}{}\text{(57)}& {\psi }_1^{\prime }t={\int }_{\Omega }{u}_{tt}t{\int }_0^t{g}_{1}t-sus-utdsdx+{\int }_{\Omega }{u}_{t}t{\int }_0^t{g}_1^{\prime }t-sus-utdsdx-{\int }_0^t{g}_{1}sds{u_{t}t}^2={\int }_{\Omega }{\Delta }_{e}u-{\int }_0^t{g}_{1}t-s\Delta usds+k_1\Delta u+{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \Delta zx,1,\rho ,td\rho +f_{1}u,v\times {\int }_0^t{g}_{1}t-sus-utdsdx+{\int }_{\Omega }{u}_{t}t{\int }_0^t{g}_1^{\prime }t-sus-utdsdx-{\int }_0^t{g}_{1}sds{u_{t}t}^2={\int }_{\Omega }{u}_{t}t{\int }_0^t{g}_1^{\prime }t-sus-utdsdx-{\int }_0^t{g}_{1}sds{u_{t}t}^2-\mu {\int }_{\Omega }\nabla u{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx-\lambda +\mu {\int }_{\Omega }divut{\int }_0^t{g}_{1}t-sdivus-divutdsdx+{\int }_0^t{g}_{1}t-s\nabla usds{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx+{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,1,\rho ,td\rho {\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx+{\int }_{\Omega }{f}_{1}u,v{\int }_0^t{g}_{1}t-sus-utdsdx-k_1{\int }_{\Omega }\nabla u_t{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx.\end{array}$$

Then, we have $$\begin{array}{}\text{(58)}& {\psi }_1^{\prime }t={\int }_{\Omega }{u}_{t}t{\int }_0^t{g}_1^{\prime }t-sus-utdsdx-{\int }_0^t{g}_{1}sds{u_{t}t}^2-\mu -{\int }_0^t{g}_{1}sds{\int }_{\Omega }\nabla u{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx-\lambda +\mu {\int }_{\Omega }divut{\int }_0^t{g}_{1}t-sdivus-di\text{v}utdsdx+{{\int }_{\Omega }{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx}^2-k_1{\int }_{\Omega }\nabla u_t{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx+{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,1,\rho ,td\rho .{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx+{\int }_{\Omega }{\text{f}}_{1}u,v{\int }_0^t{g}_{1}t-sus-utdsdx.\end{array}$$

As in previous proof and by using Young’s inequality, we conclude that for any $\delta >0$, $$\begin{array}{}\text{(59)}& \mu -{\int }_0^t{g}_{1}sds{\int }_{\Omega }\nabla u{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx\le \delta {\nabla u}^2+\frac{c.C_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut.\end{array}$$

Similarly and by using the fact ${divu}^2\le c{\nabla u}^2$, we have $$\begin{array}{}\text{(60)}& \lambda +\mu {\int }_{\Omega }divut{\int }_0^t{g}_{1}t-sdivus-divutdsdx\le \delta {divu}^2+\frac{c.C_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut.\end{array}$$

The same argument for $$\begin{array}{}\text{(61)}& {\int }_{\Omega }{f}_{1}u,v{\int }_0^t{g}_{1}t-sus-utdsdx\le \delta {\int }_{\Omega }{f_{1}u,v}^{2}dx+\frac{1}{4\delta }{\int }_{\Omega }{{\int }_0^t{g}_{1}t-sus-utds}^{2}dx\le \delta {\Lambda }_1{l_1{\nabla u}^2+l_2{\nabla v}^2}^p+\frac{c.C_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut\le \delta {\Lambda }_1{\frac{2p+1}{p-1}E0}^{p-1}{l}_1{\nabla u}^2+l_2{\nabla v}^2+\frac{c.C_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut\le \delta {\Lambda }_3{l}_1{\nabla u}^2+l_2{\nabla v}^2+\frac{c.C_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut,\end{array}$$

where ${\Lambda }_3={2p+1/p-1E0}^{p-1}$.

From (46), we have $$\begin{array}{c}\text{(62)}& {\int }_{\Omega }{u}_{t}t{\int }_0^t{g}_1^{\prime }t-sus-utdsdx={\int }_{\Omega }{u}_{t}t{\int }_0^t{h}_{1}t-sus-utdsdx-{\int }_{\Omega }{u}_{t}t{\int }_0^t\zeta g_{1}t-sus-utdsdx\le \delta {u_{t}t}^2+\frac{c}{\delta }{\int }_{\Omega }{{\int }_0^t\sqrt{h_{1}t-s}\sqrt{h_{1}t-s}us-utds}^{2}dx+\frac{c{\zeta }^2}{\delta }{\int }_{\Omega }{{\int }_0^t{g}_{1}t-sus-utds}^{2}dx\le \delta {u_{t}t}^2+\frac{c}{\delta }{\int }_0^t{h}_{1}sds{h}_1\circ ut+\frac{c{\zeta }^2{C}_{\zeta ,1}}{\delta }{h}_1\circ ut\le \delta {u_{t}t}^2+\frac{c}{\delta }{h}_1\circ \nabla ut+\frac{c{\zeta }^2{C}_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut\le \delta {u_{t}t}^2+\frac{c{C}_{\zeta ,1}+1}{\delta }{h}_1\circ \nabla ut,{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \nabla zx,1,\rho ,td\rho .{\int }_0^t{g}_{1}t-s\nabla us-\nabla utdsdx\le \delta {\int }_{\Omega }{{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid \nabla zx,1,\rho ,td\rho }^{2}dx+\frac{1}{4\delta }{\int }_{\Omega }{{\int }_0^t{g}_{1}t-s\nabla us-\nabla utds}^{2}dx\le \delta {\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid d\rho {\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho dx+\frac{c{C}_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut\le \delta k_1{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho dx+\frac{c{C}_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut.\end{array}$$

Finally, Young’s inequality gives $$\begin{array}{}\text{(63)}& k_1{\int }_{\Omega }\nabla u_t{\int }_0^t{g}_{1}t-sus-utdsdx\le \delta k_1{\nabla u_t}^2+\frac{k_1}{4\delta }{\int }_{\Omega }{{\int }_0^t{g}_{1}t-sus-utds}^{2}dx\le \delta k_1\nabla u_t+\frac{c{C}_{\zeta ,1}}{\delta }{h}_1\circ \nabla ut.\end{array}$$

Then $$\begin{array}{}\text{(64)}& {\psi }_1^{\prime }t\le \delta -{\int }_0^t{g}_{1}sds{u_{t}t}^2+\delta +\delta {\Lambda }_3{l}_1{\nabla u}^2+\delta {divu}^2+\delta {\Lambda }_3{l}_2{\nabla v}^2+\frac{c{C}_{\zeta ,1}+1}{\delta }{h}_1\circ \nabla ut+\delta k_1{\nabla u_t}^2+\delta k_1{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho dx.\end{array}$$

The same steps can be taken to get the next estimate for ${\psi }_2^{\prime }t,$$$\begin{array}{}\text{(65)}& {\psi }_2^{\prime }t\le \delta -{\int }_0^t{g}_{2}sds{v_{t}t}^2+\delta +\delta {\Lambda }_4{l}_2{\nabla v}^2+\delta {divv}^2+\delta {\Lambda }_4{l}_1{\nabla u}^2+\frac{c{C}_{\zeta ,2}+1}{\delta }{h}_2\circ \nabla vt+\delta k_2{\nabla v_t}^2+\delta k_2{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_2\rho \mid {\nabla yx,1,\rho ,t}^{2}d\rho dx,\end{array}$$where ${\Lambda }_4={\Lambda }_2{2p+1/p-1E0}^{p-1}$.

Lemma 7.

Let $u,v,z,y$ be a solution of the problem (19). Then, the functional $$\begin{array}{}\text{(66)}& It={\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho e^{-\rho \rho }{\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^2+{\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}dxd\rho d\rho ,\end{array}$$satisfies the estimate $$\begin{array}{}\text{(67)}& I^{\prime }t\le -e^{-{\tau }_2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho {\nabla zx,1,\rho ,t}^{2}dxd\rho -e^{-{\tau }_2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho {\nabla yx,1,\rho ,t}^{2}dxd\rho +k_1{\nabla u_{t}t}^2+k_2{\nabla v_{t}t}^2-It.\end{array}$$

Proof.

Differentiating (66) with respect to $t$, we get $$\begin{array}{}\text{(68)}& \frac{d}{dt}It=2{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho e^{-\rho \rho }{\mu }_1\rho \nabla zx,\rho ,\rho ,t\nabla z_{t}x,\rho ,\rho ,t+{\mu }_2\rho \nabla yx,\rho ,\rho ,t\nabla y_{t}x,\rho ,\rho ,tdxd\rho d\rho .\end{array}$$

By using (17) and (18), we have $$\begin{array}{}\text{(69)}& \frac{d}{dt}It=-{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{e}^{-\rho \rho }{\mu }_1\rho \frac{\partial }{\partial \rho }{\nabla zx,\rho ,\rho ,t}^2+{\mu }_2\rho \frac{\partial }{\partial \rho }{\nabla yx,\rho ,\rho ,t}^{2}dxd\rho d\rho =-{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho \frac{\partial }{\partial \rho }{e}^{-\rho \rho }\nabla z{x,\rho ,\rho ,t}^2+{\mu }_2\rho \frac{\partial }{\partial \rho }{e}^{-\rho \rho }{\nabla yx,\rho ,\rho ,t}^{2}dxd\rho d\rho -{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho e^{-\rho \rho }{\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^2+{\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}dxd\rho d\rho .\end{array}$$

Thus, $$\begin{array}{}\text{(70)}& \frac{d}{dt}It=-{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{e}^{-\rho }{\mu }_1\rho {\nabla zx,1,\rho ,t}^{2}dxd\rho +{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho d\rho {\nabla u_{t}x,\rho ,t}^2-{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}{e}^{-\rho }{\mu }_2\rho {\nabla yx,1,\rho ,t}^{2}dxd\rho +{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho d\rho {\nabla v_{t}x,\rho ,t}^2-{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho e^{-\rho \rho }{\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^2+{\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}dxd\rho d\rho .\end{array}$$

Since $e^{-\rho }$ is deceasing function over ${\tau }_1,{\tau }_2$, the desired estimate (67) follows immediately from (25).

The following lemmas are needed to prove the general decay when the functions $G_{i}ti=1,2$ are nonlinear. The proof can be found in Mustafa [19].

Lemma 8.

The functional $$\begin{array}{}\text{(71)}& {\theta }_{1}t={\int }_{\Omega }{\int }_0^t{\sigma }_{1}t-s{\nabla us}^{2}dsdx,\end{array}$$where ${\sigma }_{1}t={\int }_t^{\infty }{g}_{1}sds$ satisfies $$\begin{array}{}\text{(72)}& {\theta }_1^{\prime }t\le -\frac{1}{2}{g}_1\circ \nabla ut+3\mu -l_1{\nabla u}^2.\end{array}$$

Lemma 9.

The functional $$\begin{array}{}\text{(73)}& {\theta }_{2}t={\int }_{\Omega }{\int }_0^t{\sigma }_{2}t-s{\nabla vs}^{2}dsdx,\end{array}$$where ${\sigma }_{2}t={\int }_t^{\infty }{g}_{2}sds$ satisfies $$\begin{array}{}\text{(74)}& {\theta }_2^{\prime }t\le -\frac{1}{2}{g}_2\circ \nabla vt+3\mu -l_2{\nabla v}^2.\end{array}$$

Now, we define the following functional $$\begin{array}{}\text{(75)}& \mathcal{F}t=NEt+N_1\varphi t+N_2\psi t+It,\end{array}$$

where $N,N_1$, and $N_2$ are positive constants. It is easy to prove $Ft$ and $Et$ are equivalent, namely, there exist two positive constants ${\kappa }_1$ and ${\kappa }_2$ such that $$\begin{array}{}\text{(76)}& {\kappa }_{1}Et\le \mathcal{F}t\le {\kappa }_{2}Et.\end{array}$$

By Young’s inequality, we get $$\begin{array}{}\text{(77)}& \mathcal{F}t\le \frac{N}{2}+\frac{N_1}{2}+\frac{N_2}{2}{u_t}^2+{v_t}^2+\frac{N}{2}\mu -{\int }_0^t{g}_{1}sds+c\frac{N_1}{2}+c\frac{N_2}{2}{\nabla u}^2+\frac{N}{2}\mu -{\int }_0^t{g}_{2}sds+c\frac{N_1}{2}+c\frac{N_2}{2}{\nabla v}^2+\frac{N}{2}+c\frac{N_2}{2}{\int }_0^t{g}_{1}sds{g}_1\circ \nabla ut+\frac{N}{2}+c\frac{N_2}{2}{\int }_0^t{g}_{2}sds{g}_2\circ \nabla ut+\frac{N}{2}\eta +C{\int }_{\Omega }{\int }_0^t{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^{2}d\rho d\rho dx+\frac{N}{2}\eta +C{\int }_{\Omega }{\int }_0^t{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}d\rho d\rho dx+\frac{N}{2}\lambda +\mu {divu}^2+{divv}^2-N{\int }_{\Omega }Fu,vdx.\end{array}$$

Then, for any $N$, there exists ${\kappa }_1>0$ such that $$\begin{array}{}\text{(78)}& \mathcal{F}\le {\kappa }_{1}Et.\end{array}$$

On the other hand, we can find $$\begin{array}{}\text{(79)}& \mathcal{F}t\ge \frac{N}{2}-\frac{N_1}{2}-\frac{N_2}{2}{u_t}^2+{v_t}^2+\frac{N}{2}\mu -{\int }_0^t{g}_{1}sds-c\frac{N_1}{2}-c\frac{N_2}{2}{\nabla u}^2+\frac{N}{2}\mu -{\int }_0^t{g}_{2}sds-c\frac{N_1}{2}-c\frac{N_2}{2}{\nabla v}^2+\frac{N}{2}-c\frac{N_2}{2}{\int }_0^t{g}_{1}sds{g}_1\circ \nabla ut+\frac{N}{2}-c\frac{N_2}{2}{\int }_0^t{g}_{2}sds{g}_2\circ \nabla ut+\frac{N}{2}\eta +c{\int }_{\Omega }{\int }_0^t{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^{2}d\rho d\rho dx+\frac{N}{2}\eta +c{\int }_{\Omega }{\int }_0^t{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}d\rho d\rho dx+\frac{N}{2}\lambda +\mu {divu}^2+{divv}^2-N{\int }_{\Omega }Fu,vdx.\end{array}$$

We choose $N$ large enough so that $$\begin{array}{c}\text{(80)}& \frac{N}{2}-\frac{N_1}{2}-\frac{N_2}{2}>0,\frac{N}{2}\mu -{\int }_0^t{g}_{i}sds-c\frac{N_1}{2}-c\frac{N_2}{2}>0,\frac{N}{2}-c\frac{N_2}{2}{\int }_0^t{g}_{i}sds>0,i=1,2.\end{array}$$

Then, there exist ${\kappa }_2>0$ such that $$\begin{array}{}\text{(81)}& \mathcal{F}t\ge {\kappa }_{2}Et.\end{array}$$

Lemma 10.

The functional $\mathcal{F}t$ satisfies for any $t\ge t_1$, $$\begin{array}{}\text{(82)}& {\mathcal{F}}^{\prime }t\le -4\mu -l_1{\nabla ut}^2-4\mu -l_2{\nabla vt}^2-{u_{t}t}^2-{v_{t}t}^2-{divut}^2-{divvt}^2+\frac{1}{4}{g}_1\circ \nabla ut+\frac{1}{4}{g}_2\circ \nabla vt+c{\int }_{\Omega }Fut,vtdx-{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^{2}d\rho d\rho dx-{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}d\rho d\rho dx.\end{array}$$

Proof.

Let $$\begin{array}{}\text{(83)}& g_0=\min {\int }_0^{t_1}{g}_{1}sds,{\int }_0^{t_1}{g}_{2}sds.\end{array}$$

From Lemmas 5, 6, and 7, noting that $g_i^{\prime }=\zeta g_i-h_i$ we have for any $t\ge t_1,$$$\begin{array}{}\text{(84)}& {\mathcal{F}}^{\prime }t\le -\frac{l_1}{2}{N}_1-N_2\delta 1+{\Lambda }_3{l}_1-N_2\delta {\Lambda }_4{l}_1{\nabla ut}^2-\frac{l_2}{2}{N}_1-N_2\delta 1+{\Lambda }_4{l}_2-N_2\delta {\Lambda }_3{l}_2{\nabla vt}^2-g_0{N}_2-\delta N_2-N_1{u_{t}t}^2+{v_{t}t}^2+\frac{\zeta N}{2}{g}_1\circ \nabla ut+g_2\circ \nabla vt+N_{1}p+1{\int }_{\Omega }Fut,vtdx-\frac{N}{2}-N_2\frac{c{C}_{\zeta ,1}+1}{\delta }-\frac{3N_1{C}_{\zeta ,1}}{2l_1}{h}_1\circ \nabla ut-\frac{N}{2}-N_2\frac{c{C}_{\zeta ,2}+1}{\delta }-\frac{3N_1{C}_{\zeta ,2}}{2l_2}{h}_2\circ \nabla vt-\lambda +\mu N_1-\delta N_2{divut}^2+{divvt}^2-N\sigma +e^{-{\tau }_2}-N_1\frac{3}{2l_1}{k}_1-\delta N_2{k}_1{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho dx-N\sigma +e^{-{\tau }_2}-N_1\frac{3}{2l_2}{k}_2-\delta N_2{k}_2{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_2\rho \mid {\nabla yx,1,\rho ,t}^{2}d\rho dx-N{\sigma }_1-N_1\frac{3k_1^2}{2l_1}-\delta N_2{k}_1-k_1{\nabla u_{t}t}^2-N{\sigma }_2-N_1\frac{3k_2^2}{2l_2}-\delta N_2{k}_2-k_2{\nabla v_{t}t}^2-It,\end{array}$$where $$\begin{array}{c}\text{(85)}& {\sigma }_1=k_1-\frac{\eta +1}{2}{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_1\rho d\rho ,{\sigma }_2=k_2-\frac{\eta +1}{2}{\int }_{{\tau }_1}^{{\tau }_2}{\mu }_2\rho d\rho ,\sigma =\frac{\eta -1}{2}.\end{array}$$

Taking $\delta =1/2N_2$, we can get $$\begin{array}{}\text{(86)}& {\mathcal{F}}^{\prime }t\le -\frac{l_1}{2}{N}_1-\frac{1}{2}1+{\Lambda }_3{l}_1-\frac{1}{2}{\Lambda }_4{l}_1{\nabla ut}^2-\frac{l_2}{2}{N}_1-\frac{1}{2}1+{\Lambda }_4{l}_2-\frac{1}{2}{\Lambda }_3{l}_2{\nabla vt}^2-g_0{N}_2-\frac{1}{2}-N_1{u_{t}t}^2+{v_{t}t}^2+\frac{\zeta N}{2}{g}_1\circ \nabla ut+g_2\circ \nabla vt+N_{1}p+1{\int }_{\Omega }Fut,vtdx-\frac{N}{2}-2c{N}_2^2-C_{\zeta ,1}2c{N}_2^2+\frac{3N_1}{2l_1}{h}_1\circ \nabla ut-\frac{N}{2}-2c{N}_2^2-C_{\zeta ,2}2c{N}_2^2+\frac{3N_1}{2l_2}{h}_2\circ \nabla vt-\lambda +\mu N_1-\frac{1}{2}{divut}^2+{divvt}^2-N\sigma +{\text{e}}^{-{\tau }_2}-N_1\frac{3}{2l_1}{k}_1-\frac{k_1}{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_1\rho \mid {\nabla zx,1,\rho ,t}^{2}d\rho dx-N\sigma +e^{-{\tau }_2}-N_1\frac{3}{2l_2}{k}_2-\frac{k_2}{2}{\int }_{\Omega }{\int }_{{\tau }_1}^{{\tau }_2}\mid {\mu }_2\rho \mid {\nabla yx,1,\rho ,t}^{2}d\rho dx-N{\sigma }_1-N_1\frac{3k_1^2}{2l_1}-\frac{3k_1}{2}{\nabla u_{t}t}^2-N{\sigma }_2-N_1\frac{3k_2^2}{2l_2}-\frac{3k_2}{2}{\nabla v_{t}t}^2-It.\end{array}$$

First, we take $N_1>0$ large such that $$\begin{array}{c}\text{(87)}& \lambda +\mu N_1-\frac{1}{2}>0,\frac{l_1}{2}{N}_1-\frac{1}{2}1+{\Lambda }_3{l}_1-\frac{1}{2}{\Lambda }_4{l}_1>4\mu -l_1,\frac{l_2}{2}{N}_1-\frac{1}{2}1+{\Lambda }_3{l}_2-\frac{1}{2}{\Lambda }_4{l}_2>4\mu -l_2.\end{array}$$

We choose $N_2>0$ large enough so that $$\begin{array}{}\text{(88)}& g_0{N}_2-\frac{1}{2}-N_1>1.\end{array}$$

Note that $$\begin{array}{}\text{(89)}& 0<\frac{\zeta g_i^{2}s}{\zeta g_{i}s-g_i^{\prime }s}<\frac{\zeta g_i^{2}s}{-g_i^{\prime }s},i=1,2.\end{array}$$

Then, for any $s\in 0,\infty$, we get $$\begin{array}{}\text{(90)}& \underset{\zeta \to 0}{\lim }\frac{\zeta g_i^{2}s}{\zeta g_{i}s-{g_i}^{\prime }s}=0,i=1,2.\end{array}$$

By using the fact $\zeta g_i^{2}s/\zeta g_{i}s-g_i^{\prime }s<g_{i}s,i=1,2$ and using Lebesgue-dominated convergence theorem, we can get $$\begin{array}{}\text{(91)}& \underset{\zeta \to 0}{\lim }\zeta C_{\zeta ,i}=\underset{\zeta \to 0}{\lim }{\int }_0^{\infty }\frac{\zeta g_i^{2}s}{\zeta g_{i}s-g_i^{\prime }s}=0,i=1,2.\end{array}$$

Thus, there exist some ${\zeta }_{0}0<{\zeta }_0<1$ such that if $\zeta <{\zeta }_0$, then $$\begin{array}{}\text{(92)}& \zeta C_{\zeta ,2}<\frac{1}{8N_1/2l_1+2c{N}_2^2}\text{and}\zeta C_{\zeta ,2}<\frac{1}{8N_1/2l_2+2c{N}_2^2}.\end{array}$$

At last, we choose $N$ large enough and choose $\zeta$ satisfying $$\begin{array}{}\text{(93)}& \frac{1}{4}N-2c{N}_2^2>0\text{and}\zeta =\frac{1}{2N}>{\zeta }_0,\end{array}$$

so, we arrive at $$\begin{array}{}\text{(94)}& \frac{N}{2}-2c{N}_2^2-C_{\zeta ,1}2c{N}_2^2+\frac{3N_1}{2l_1}>0\text{and}\frac{N}{2}-2c{N}_2^2-C_{\zeta ,1}2c{N}_2^2+\frac{3N_1}{2l_1}>0.\end{array}$$

Therefore, we choose $N$ even larger (if needed) so that $$\begin{array}{c}\text{(95)}& N{n}_1+e^{-{\tau }_2}-N_1\frac{3}{2l_1}{k}_1-\frac{k_1}{2}>0,N{n}_1+e^{-{\tau }_2}-N_1\frac{3}{2l_2}{k}_2-\frac{k_2}{2}>0,Nm-N_1\frac{3k_1^2}{2l_1}-\frac{3k_1}{2}>0\text{and}Nm-N_1\frac{3k_2^2}{2l_2}-\frac{3k_2}{2}>0.\end{array}$$

Thus, (82) is established.

Proof of Theorem 11.

Taking into account (43) and (26), we obtain that for any $t\ge t_1$, $$\begin{array}{c}\text{(96)}& {\int }_0^{t_1}{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds\le -\frac{g_{1}0}{a_1}{\int }_0^{t_1}{g}_1^{\prime }s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds\le -c{E}^{\prime }t,{\int }_0^{t_1}{g}_{2}s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds\le -\frac{g_{2}0}{a_2}{\int }_0^{t_1}{g}_2^{\prime }s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds\le -c{E}^{\prime }t.\end{array}$$

Noting (82), we shall see that there exists a constant $m>0$ such that for all $t\ge t_1$, $$\begin{array}{}\text{(97)}& {\mathcal{F}}^{\prime }t\le -mEt-c{E}^{\prime }+c{\int }_{t_1}^t{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds+c{\int }_{t_1}^t{g}_{2}s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds.\end{array}$$

Denote $\mathcal{L}t=\mathcal{F}t+cEt$. It is obvious that $\mathcal{L}t$ is equivalent to $Et$. It follows from (97) that $$\begin{array}{}\text{(98)}& {\mathcal{L}}^{\prime }t\le -mEt+c{\int }_{t_1}^t{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds+c{\int }_{t_1}^t{g}_{2}s{\int }_{\Omega }{\nabla v\text{t}-\nabla vt-s}^{2}dxds.\end{array}$$

We consider two cases.

Case 11.

$Gt$ is linear: By multiplying (98) by $\xi t$ and using (A2) and (26), we obtain $$\begin{array}{}\text{(99)}& \xi t{\mathcal{L}}^{\prime }t\le -m\xi tEt+c\xi t{\int }_{t_1}^t{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds+c\xi t{\int }_{t_1}^t{g}_{2}s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds\le -m\xi tEt+c{\int }_{t_1}^t{\xi }_{1}t{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds+c{\int }_{t_1}^t{\xi }_{2}t{g}_{2}s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds\le -m\xi tEt-c{\int }_{t_1}^t{g}_1^{\prime }s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds-c{\int }_{t_1}^t{g}_2^{\prime }s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds\le -m\xi tEt-c{E}^{\prime }t,\end{array}$$

which gives as $\xi t$ is nonincreasing, $$\begin{array}{}\text{(100)}& {\xi t\mathcal{L}t+cEt}^{\prime }\le \xi t{\mathcal{L}}^{\prime }t+c{E}^{\prime }t\le -m\xi tEt\forall t\ge {\text{t}}_1.\end{array}$$

Denote $Kt=\xi t\mathcal{L}t+cEt$. We get $$\begin{array}{}\text{(101)}& K{t}^{\prime }\le -m\xi tEt.\end{array}$$

Hence, using the fact that $Kt~Et$, we easily obtain $$\begin{array}{}\text{(102)}& Et\le c_1\exp -c_2{\int }_{t_1}^t\xi sds.\end{array}$$

Case 12.

$Gt$ is nonlinear: First, we use Lemmas 8 and 9 to deduce that $$\begin{array}{}\text{(103)}& Jt=\mathcal{F}t+{\theta }_{1}t+{\theta }_{2}t,\end{array}$$is nonnegative, and it satisfies for some positive constant $k$ and for any $t\ge t_1$, $$\begin{array}{}\text{(104)}& J^{\prime }t\le -\mu -l_1{\nabla u}^2-\mu -l_2{\nabla v}^2-{u_t}^2-{v_t}^2-{divu}^2-{divv}^2-\frac{1}{4}{g}_1\circ \nabla u-\frac{1}{4}{g}_2\circ \nabla v+c{\int }_{\Omega }Fu,vdx-{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_1\rho {\nabla zx,\rho ,\rho ,t}^{2}d\rho d\rho dx-{\int }_{\Omega }{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\rho {\mu }_2\rho {\nabla yx,\rho ,\rho ,t}^{2}d\rho d\rho dx\le -kEt\le 0.\end{array}$$

Therefore, $$\begin{array}{}\text{(105)}& k{\int }_{t_1}^{t}Esds\le J{t}_1-Jt\le J{t}_1,\end{array}$$this implies that $$\begin{array}{}\text{(106)}& {\int }_0^{\infty }Esds<\infty .\end{array}$$

Now, we define $I_{i}t$ by $$\begin{array}{c}\text{(107)}& I_{1}t≔q{\int }_{t_1}^t{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds,I_{2}t≔q{\int }_{t_1}^t{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds,\end{array}$$where (106) allows for a constant $0<q<1$ chosen so that for all $t\ge t_1$$$\begin{array}{}\text{(108)}& I_{i}t<1,i=1,2.\end{array}$$

We also assume without loss of generality that $I_{i}t>0$ for all $t\ge t_1$; otherwise, (98) yields an exponential decay. Also, we define ${\lambda }_{1}t$ and ${\lambda }_{2}t$ by $$\begin{array}{c}\text{(109)}& {\lambda }_{1}t≔-{\int }_{t_1}^t{g}_1^{\prime }s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds,{\lambda }_{2}t≔-{\int }_{t_1}^t{g}_2^{\prime }s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds.\end{array}$$

We observe that ${\lambda }_{i}t\le -c{E}^{\prime }ti=1,2.$ Noting that $G_{i}t$ is strictly convex on $0,r$ and $G_{i}0=0$, then $$\begin{array}{}\text{(110)}& G_i\nu x\le \nu G_{i}x,i=1,2,\end{array}$$provided $0\le \nu \le 1$ and $x\in 0,r$. By using (A2), (108), and Jensen’s inequality, we can obtain $$\begin{array}{}\text{(111)}& {\lambda }_{1}t=\frac{1}{q{I}_{1}t}{\int }_{t_1}^t{I}_{1}t-g_1^{\prime }s{\int }_{\Omega }q{\nabla ut-\nabla ut-s}^{2}dxds\ge \frac{1}{q{I}_{1}t}{\int }_{t_1}^t{I}_{1}t{\xi }_{1}s{G}_1{g}_{1}s{\int }_{\Omega }q{\nabla ut-\nabla ut-s}^{2}dxds,\ge \frac{{\xi }_{1}t}{q{I}_{1}t}{\int }_{t_1}^t{G}_1{I}_{1}t{g}_{1}s{\int }_{\Omega }q{\nabla ut-\nabla ut-s}^{2}dxds,\ge \frac{{\xi }_{1}t}{q}{G}_1\frac{1}{I_{1}t}{\int }_{t_1}^t{I}_{1}t{g}_{1}s{\int }_{\Omega }q{\nabla ut-\nabla ut-s}^{2}dxds=\frac{{\xi }_{1}t}{q}{G}_{1}q{\int }_{t_1}^t{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds=\frac{{\xi }_{1}t}{q}\overline{G_1}q{\int }_{t_1}^t{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds,\end{array}$$

where $\overline{G_1}$ is an extension of $G_1$ such that $\overline{G_1}$ is strictly increasing and strictly convex $C^2$ function on $0,+\infty$, see [19]. We have from (111) $$\begin{array}{}\text{(112)}& {\int }_{t_1}^t{g}_{1}s{\int }_{\Omega }{\nabla ut-\nabla ut-s}^{2}dxds\le \frac{1}{q}{\overline{G_1}}^{-1}\frac{q{\lambda }_{1}t}{{\xi }_{1}t}.\end{array}$$

Similarly, we have $$\begin{array}{}\text{(113)}& {\int }_{t_1}^t{g}_{2}s{\int }_{\Omega }{\nabla vt-\nabla vt-s}^{2}dxds\le \frac{1}{q}{\overline{G_2}}^{-1}\frac{q{\lambda }_{2}t}{{\xi }_{2}t}.\end{array}$$

We infer from (98), (112), and (113) that for any $t\ge t_1$$$\begin{array}{}\text{(114)}& {\mathcal{L}}^{\prime }t\le -mEt+c{\overline{G_1}}^{-1}\frac{q{\lambda }_{1}t}{{\xi }_{1}t}+c{\overline{G_2}}^{-1}\frac{q{\lambda }_{2}t}{{\xi }_{2}t}.\end{array}$$

Let us denote $$\begin{array}{}\text{(115)}& G_{0}t=\min {\overline{G_1}}^{-1},{\overline{G_2}}^{-1}.\end{array}$$

For ${\epsilon }_0<r$, the functional ${\mathcal{K}}_{1}t$ defined by: $$\begin{array}{}\text{(116)}& {\mathcal{K}}_{1}t=G_0{\epsilon }_0\frac{Et}{\text{E}0}\mathcal{L}t+Et,\end{array}$$is equivalent to $E$, and by using the fact that $E^{\prime }\le 0,{\overline{G_i}}^{\prime }>0,$ and ${\overline{G_i}}^{\prime \prime }>0$, we infer from (114) that $$\begin{array}{}\text{(117)}& {\mathcal{K}}_1^{\prime }t={\epsilon }_0\frac{E^{\prime }t}{E0}{G}_0^{\prime }{\epsilon }_0\frac{Et}{E0}\mathcal{L}t+G_0{\epsilon }_0\frac{Et}{E0}{\mathcal{L}}^{\prime }t+E^{\prime }t\le -mEt{G}_0{\epsilon }_0\frac{Et}{E0}+c{G}_0{\epsilon }_0\frac{Et}{E0}{\overline{G_1}}^{-1}\frac{q{\lambda }_{1}t}{{\xi }_{1}t}+c{G}_0{\epsilon }_0\frac{Et}{E0}{\overline{G_2}}^{-1}\frac{q{\lambda }_{2}t}{{\xi }_{2}t}.\end{array}$$

Let ${\overline{G}}^{\ast }$ be the convex conjugate of $\overline{G}$ in the sense of Young (see Arnold [21]), then $$\begin{array}{}\text{(118)}& {\overline{G_i}}^{\ast }s=s{{\overline{G_i}}^{\prime }}^{-1}s-\overline{G_i}{{\overline{G_i}}^{\prime }}^{-1}s,i=1,2,\end{array}$$and ${\overline{G}}^{\ast }$ satisfies the following Young’s inequality $$\begin{array}{}\text{(119)}& M{D}_i\le {\overline{G_i}}^{\ast }M+\overline{G_i}{D}_i,i=1,2.\end{array}$$

With $M=G_0{\epsilon }_{0}Et/E0$ and $D_i={\overline{G_i}}^{-1}q{\lambda }_{i}t/{\xi }_{i}t$ and noting ${\overline{G_i}}^{\ast }t\le t{{\overline{G_i}}^{\prime }}^{-1}t$ and (117), we conclude $$\begin{array}{}\text{(120)}& {\mathcal{K}}_1^{\prime }t\le -mEt{G}_0{\epsilon }_0\frac{Et}{E0}+c{\overline{G_1}}^{\ast }{G}_0{\epsilon }_0\frac{Et}{E0}+c\frac{q{\lambda }_{1}t}{{\xi }_{1}t}+c{\overline{G_2}}^{\ast }{G}_0{\epsilon }_0\frac{Et}{E0}+c\frac{q{\lambda }_{2}t}{{\xi }_{2}t}\le -mEt{G}_0{\epsilon }_0\frac{Et}{E0}+c{G}_0{\epsilon }_0\frac{Et}{E0}{{\overline{G_1}}^{\prime }}^{-1}{G}_0{\epsilon }_0\frac{Et}{E0}+c\frac{q{\lambda }_{1}t}{{\xi }_{1}t}+c{G}_0{\epsilon }_0\frac{Et}{E0}{{\overline{G_2}}^{\prime }}^{-1}{G}_0{\epsilon }_0\frac{Et}{E0}+c\frac{q{\lambda }_{2}t}{{\xi }_{2}t}\le -mEt{G}_0{\epsilon }_0\frac{Et}{E0}+c{G}_0{\epsilon }_0\frac{Et}{E0}{{\overline{G_1}}^{\prime }}^{-1}{\overline{G_1}}^{\prime }{\epsilon }_0\frac{Et}{E0}+c\frac{q{\lambda }_{1}t}{{\xi }_{1}t}+c{G}_0{\epsilon }_0\frac{Et}{E0}{{\overline{G_2}}^{\prime }}^{-1}{\overline{G_2}}^{\prime }{\epsilon }_0\frac{Et}{E0}+c\frac{q{\lambda }_{2}t}{{\xi }_{2}t}\le -mE0-c{\epsilon }_0\frac{Et}{E0}{G}_0{\epsilon }_0\frac{Et}{E0}+cq\frac{{\lambda }_{1}t}{{\xi }_{1}t}+\frac{{\lambda }_{2}t}{{\xi }_{2}t}.\end{array}$$

Multiplying by $\xi t$, we get $$\begin{array}{}\text{(121)}& \xi t{\mathcal{K}}_1^{\prime }t\le -mE0-c{\epsilon }_0\xi t\frac{Et}{E0}{G}_0{\epsilon }_0\frac{Et}{E0}+cq{\lambda }_{1}t+{\lambda }_{2}t\le -mE0-c{\epsilon }_0\xi t\frac{Et}{E0}{G}_0{\epsilon }_0\frac{Et}{E0}-c{E}^{\prime }t.\end{array}$$

Consequently, with ${\mathcal{K}}_{2}t=\xi t{\mathcal{K}}_{1}t+cEt$, which satisfies for some ${\beta }_1,{\beta }_2>0$, $$\begin{array}{}\text{(122)}& {\beta }_1{\mathcal{K}}_{2}t\le Et\le {\beta }_2{\mathcal{K}}_{2}t.\end{array}$$

Choosing a suitable ${\epsilon }_0$, we can get from (121) that there exists a constant $k>0$, $$\begin{array}{}\text{(123)}& {\mathcal{K}}_2^{\prime }t\le -k\xi t\frac{Et}{E0}{G}_0{\epsilon }_0\frac{Et}{E0}≔-k\xi t{G}_3{\epsilon }_0\frac{Et}{E0},\end{array}$$where $G_{3}t=t{G}_0{\epsilon }_{0}t.$

From $0\le {\epsilon }_{0}Et/E0<r$, we conclude that for any $t>0$, $$\begin{array}{}\text{(124)}& G_0{\epsilon }_0\frac{Et}{E0}=\min {\overline{G_1}}^{\prime }{\epsilon }_0\frac{Et}{E0},{\overline{G_2}}^{\prime }{\epsilon }_0\frac{Et}{E0}=\min {G_1}^{\prime }{\epsilon }_0\frac{\text{E}t}{E0},{G_2}^{\prime }{\epsilon }_0\frac{Et}{E0}.\end{array}$$

Denote $Rt={\beta }_1{\mathcal{K}}_{2}t/E0$. Using (122), then $$\begin{array}{}\text{(125)}& Rt~Et.\end{array}$$

Since $G_3^{\prime }t=G_0{\epsilon }_{0}t+{\epsilon }_{0}t{G}_0^{\prime }{\epsilon }_{0}t$, then, using the strict convexity of $G_0$ on $0,r$, we arrive at $G_3^{\prime }t,G_{3}t>0$ on $0,1$. We infer from (123) that there exists a constant $b_1>0$ such that for all $t\le t_1$, $$\begin{array}{}\text{(126)}& R^{\prime }t\le -b_1\xi t{G}_{3}Rt.\end{array}$$

Then, by integration over $t_1,t$, we have $$\begin{array}{}\text{(127)}& {\int }_{t_1}^t\frac{-R^{\prime }s}{G_{3}Rs}ds\ge b_1{\int }_{t_1}^t\xi sds{\int }_{{\epsilon }_{0}Rt}^{{\epsilon }_{0}R{t}_1}\frac{1}{\text{s}{G}_{0}s}ds\ge b_1{\int }_{t_1}^t\xi sdsRt\le \frac{1}{{\epsilon }_0}{G}_4^{-1}{b}_1{\int }_{t_1}^t\xi sds,\end{array}$$where $G_{4}t={\int }_t^{r}1/s{G}_{0}sds$ is strictly decreasing on $0,r$ and ${\lim }_{t\to 0}{G}_{4}t=+\infty$. A combination of (125) and (127), estimate (38), is established.

4. Conclusion

In this work, we have proved a general energy decay of a coupled Lamé system with distributed time delay. This result is an natural extension of Feng’s work in [18]. In order to complete this work, the study of the global existence and the blow-up of the solutions of (1) and (2) will be the subject of forthcoming works.

Authors’ Contributions

The authors contributed equally in this article. They have all read and approved the final manuscript.

Acknowledgments

The second author would like to thank all the professors of the mathematics department at the University of Annaba in Algeria, especially his Professors/Scientists Pr. Mohamed Haiour, Pr. Ahmed-Salah Chibi, and Pr. Azzedine Benchettah for the important content of masters and PhD courses in pure and applied mathematics which he received during his studies. Moreover, he thanks them for the additional help they provided to him during office hours in their office about the few concepts/difficulties he had encountered, and he appreciates their talent and dedication for their postgraduate students currently and previously. In addition, the idea and research project in this paper was presented by the second author and was carried out by all authors. The authors are grateful to the anonymous referees for the careful reading and their important observations/suggestions for the sake of improving this paper.

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