Well-Posedness and Stability Results for a Nonlinear Damped Porous–Elastic System with Infinite Memory and Distributed Delay Terms

Abstract

In the present paper, we consider an important problem from the application perspective in science and engineering, namely, one-dimensional porous–elastic systems with nonlinear damping, infinite memory and distributed delay terms. A new minimal conditions, placed on the nonlinear term and the relationship between the weights of the different damping mechanisms, are used to show the well-posedness of the solution using the semigroup theory. The solution energy has an explicit and optimal decay for the cases of equal and nonequal speeds of wave propagation.

1. Introduction

As introduced in [1], the one-dimensional porous–elastic model constitutes a system of two partial differential equations with unknown $(u,\phi )$ given by

$$\begin{array}{c}{\rho }_0{u}_{tt}=\mu u_{xx}+\beta {\phi }_x,\mathrm{in}\left(0,l\right) \times \left(0,L\right),\hfill \\ {\rho }_{0}k{\phi }_{tt}=\alpha {\phi }_{xx}-\beta u_x-\tau {\phi }_t-\xi \phi ,\mathrm{in}\left(0,l\right) \times \left(0,L\right),\hfill \end{array}$$

(1)

where $l,L>0$ the constant $\rho$ is the mass density, $\kappa$ is the equilibrated inertia and the constants $\mu ,\alpha ,\beta ,\tau ,\xi$ are assumed to satisfy the appropriate conditions. This type of problem has been studied by many authors and a lot of results have been shown (please see [1,2,3,4,5,6,7,8,9]). The pioneering contribution was made by [10] for the problem (1). The basic evolution equations for one-dimensional theories of porous materials with memory effect are given by

$$\rho u_{tt}=T_x,J{\varphi }_{tt}=H_x+G,$$

(2)

where T is the stress tensor, H is the equilibrated stress vector and G is the equilibrated body force. The variables u and $\varphi$ are the displacement of the solid elastic material and the volume fraction, respectively. The constitutive equations are

$$T=\mu u_x+b\varphi ,H=\delta {\varphi }_x-{\int }_0^{t}g\left(t-s\right){\varphi }_x\left(s\right)ds,G=-b{u}_x-\xi \varphi .$$

(3)

A porous–elastic system was considered by [11] in the system

$$\left\{\begin{array}{c}\rho u_{tt}-\mu u_{xx}-b{\varphi }_x=0,\mathrm{in}\left(0,1\right)\times \left(0,\infty \right),\hfill \\ J{\varphi }_{tt}-\delta {\varphi }_{xx}+b{u}_x+\xi \varphi +{\int }_0^{t}g\left(t-s\right){\varphi }_{xx}\left(x,s\right)ds=0,\mathrm{in}\left(0,1\right)\times \left(0,\infty \right).\hfill \end{array}\right.$$

(4)

System (4) subjected Neumann–Dirichlet boundary conditions, where g is the relaxation function; the authors obtained a general decay result for the case of equal speeds of wave propagation (See [12,13]). In [14], the authors improved the case of non-equal speed of wave propagation. In [15] the authors considered the following system with memory and distributed delay terms

$$\left\{\begin{array}{c}\rho u_{tt}-\mu u_{xx}-b{\varphi }_x=0\hfill \\ {\displaystyle J{\varphi }_{tt}-\delta {\varphi }_{xx}+b{u}_x+\xi \varphi +{\int }_0^{t}g\left(s\right){\varphi }_{xx}(t-s)ds}\hfill \\ {\displaystyle +{\mu }_1{\varphi }_t+{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|{\varphi }_t(x,t-\varrho )d\varrho =0.}\hfill \end{array}\right.$$

(5)

The exponential stability results of systems with memory and distributed delay terms, for the case of equal speeds of wave propagation under a suitable assumptions, are proved. In [16], the following system was considered

$$\left\{\begin{array}{c}\rho u_{tt}-\mu u_{xx}-b{\varphi }_x=0,\hfill \\ {\displaystyle J{\varphi }_{tt}-\delta {\varphi }_{xx}+b{u}_x+a\varphi +{\int }_0^{\infty }g\left(s\right){\varphi }_{xx}(t-s)ds+\alpha \left(t\right)f\left({\varphi }_t\right)=0.}\hfill \end{array}\right.$$

(6)

The authors proved the global well-posedness and stability results of (6), which has been extended in [17] for the case of nonequal speeds of wave propagation. Very recently, one-dimensional equations of an homogeneous and isotropic porous–elastic solid with an interior time-dependent delay term feedbacks was treated by Borges Filho and M. Santos in [1].

The result in [10] for system (1) was improved by Apalara to exponential stability in [18]. For more papers related to our paper, please see [19,20,21,22].

Motivated by all the above papers, we investigate the well-posedness and stability results with distributed delay for the cases of equal and nonequal speeds of wave propagation, under additional conditions of the following system

$$\left\{\begin{array}{c}\rho u_{tt}-\mu u_{xx}-b{\varphi }_x=0\hfill \\ {\displaystyle J{\varphi }_{tt}-\delta {\varphi }_{xx}+b{u}_x+\xi \varphi +{\int }_0^{\infty }g\left(p\right){\varphi }_{xx}(t-p)dp}\hfill \\ {\displaystyle +{\mu }_1{\varphi }_t+{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|{\varphi }_t(x,t-\varrho )d\varrho +\alpha \left(t\right)f\left({\varphi }_t\right)=0,}\hfill \end{array}\right.$$

(7)

where

$$(x,\varrho ,t)\in (0,1)\times ({\tau }_1,{\tau }_2)\times (0,\infty ),$$

with the Neumann–Dirichlet boundary conditions

$$u_x(0,t)=u_x\left(1,t\right)=\varphi \left(0,t\right)=\varphi \left(1,t\right)=0,t\ge 0,$$

(8)

and the initial data

$$\begin{gathered} \begin{array}{ccc}& & u\left(x,0\right)=u_0\left(x\right),u_t\left(x,0\right)=u_1\left(x\right),x\in (0,1)\hfill \\ \\ & & \varphi \left(x,0\right)={\varphi }_0\left(x\right),{\varphi }_t\left(x,0\right)={\varphi }_1\left(x\right),x\in (0,1)\hfill \\ \\ & & {\varphi }_t(x,-t)=f_0(x,t),(x,t)\in (0,1)\times (0,{\tau }_2).\hfill \end{array} \end{gathered}$$

(9)

Here, $\rho ,\mu ,J,b,\delta ,\xi$ and ${\mu }_1$ are positive constants satisfying $\mu \xi >b^2$, the term $\alpha \left(t\right)f\left({\varphi }_t\right)$, where the functions $\alpha$ and f are specified later, represent the nonlinear damping term. The term ${\displaystyle {\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|{\varphi }_t(x,t-\varrho )d\varrho }$ is a distributed delay that acts only on the porous equation and ${\tau }_1,{\tau }_2$ are two real numbers with $0\le {\tau }_1\le {\tau }_2$, where ${\mu }_2$ is an $L^{\infty }$ function, and the function g is called the relaxation function. We first state the following assumptions:

Hypothesis 1 (H1).

$g\in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ satisfying

$$g\left(0\right)>0,\delta -{\int }_0^{\infty }g\left(p\right)dp=l>0,{\int }_0^{\infty }g\left(p\right)dp=g_0.$$

(10)

Hypothesis 2 (H2).

There exists a non-increasing differentiable function $\alpha ,\eta :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ such that

$$g^{\prime }\left(t\right)\le -\eta \left(t\right)g\left(t\right),t\ge 0,$$

(11)

and

$$\underset{t\to \infty }{lim}\frac{-{\alpha }^{\prime }\left(t\right)}{\alpha \left(t\right)}=0.$$

(12)

Hypothesis 3 (H3).

$f\in C^0(\mathbb{R},\mathbb{R})$ is non-decreasing such that there exist $v_1,v_2,\epsilon >0$ and a strictly increasing function $G\in C^1\left([0,\infty )\right)$, with $G\left(0\right)=0$ and G is a linear or strictly convex $C^2$-function on $(0,\epsilon ]$, such that

$$\begin{gathered} \begin{array}{c}\hfill s^2+f^2\left(s\right)\le G^{-1}\left(sf\left(s\right)\right),\forall \left|s\right|<\epsilon \\ \\ \hfill v_1\left|s\right|\le \left|f\left(s\right)\right|\le v_2\left|s\right|,\forall \left|s\right|\ge \epsilon .\end{array} \end{gathered}$$

(13)

which implies that $sf\left(s\right)>0$ for all $s\ne 0$. The function f satisfies

$$|f\left({\psi }_2\right)-f\left({\psi }_1\right)|\le k_0\left(\right|{\psi }_2{|}^{\beta }+|{\psi }_1{|}^{\beta }\left)\right|{\psi }_2-{\psi }_1|,{\psi }_1,{\psi }_2\in \mathbb{R},$$

(14)

where $k_0,\beta >0$.

Hypothesis 4 (H4).

The bounded function ${\mu }_2:[{\tau }_1,{\tau }_2]\to \mathbb{R}$, satisfying

$${\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|d\varrho <{\mu }_1.$$

(15)

Now, as in [23], taking the following new variable

$$y(x,\rho ,\varrho ,t)={\varphi }_t(x,t-\varrho \rho ),$$

then we obtain

$$\left\{\begin{array}{c}\varrho y_t(x,\rho ,\varrho ,t)+y_{\rho }(x,\rho ,\varrho ,t)=0\hfill \\ y(x,0,\varrho ,t)={\varphi }_t(x,t).\hfill \end{array}\right.$$

As in [24], we introduce the following new variable

$${\eta }^t(x,s)=\varphi (x,t)-\varphi (x,t-s),(x,t,s)\in (0,1)\times {\mathbb{R}}_{+}\times {\mathbb{R}}_{+},$$

where ${\eta }^t$ is the relative history of ϕ satisfies

$${\eta }_t^t+{\eta }_s^t={\varphi }_t(x,t),(x,t,s)\in (0,1)\times (0,1)\times {\mathbb{R}}_{+}\times {\mathbb{R}}_{+}.$$

Consequently, the problem (7) is equivalent to

$$\left\{\begin{array}{c}\rho u_{tt}-\mu u_{xx}-b{\varphi }_x=0\hfill \\ {\displaystyle J{\varphi }_{tt}-l{\varphi }_{xx}+b{u}_x+\xi \varphi +{\int }_0^{\infty }g\left(p\right){\eta }_{xx}^t\left(p\right)dp}\hfill \\ {\displaystyle +{\mu }_1{\varphi }_t+{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y(x,1,\varrho ,t)d\varrho +\alpha \left(t\right)f\left({\varphi }_t\right)=0}\hfill \\ \varrho y_t(x,\rho ,\varrho ,t)+y_{\rho }(x,\rho ,\varrho ,t)=0\hfill \\ {\eta }_t^t+{\eta }_s^t={\varphi }_t(x,t),\hfill \end{array}\right.$$

(16)

where

$$(x,\rho ,\varrho ,t)\in (0,1)\times (0,1)\times ({\tau }_1,{\tau }_2)\times (0,\infty ),$$

with the following boundary and initial conditions

$$u_x\left(0,t\right)=u_x\left(1,t\right)=\varphi \left(0,t\right)=\varphi \left(1,t\right)=0,t\ge 0,$$

(17)

$$\begin{gathered} \begin{array}{ccc}& & u\left(x,0\right)=u_0\left(x\right),u_t\left(x,0\right)=u_1\left(x\right),x\in (0,1)\hfill \\ \\ & & \varphi \left(x,0\right)={\varphi }_0\left(x\right),{\varphi }_t\left(x,0\right)={\varphi }_1\left(x\right),x\in (0,1)\hfill \\ \\ & & y(x,\rho ,\varrho ,0)=f_0(x,\rho \varrho ),x\in (0,1),\rho \in (0,1),\varrho \in (0,{\tau }_2)\hfill \\ \\ & & {\eta }^t(x,0)=0,{\eta }^0(x,s)={\eta }_0(x,s),(x,s)\in (0,1)\times {\mathbb{R}}_{+}.\hfill \end{array} \end{gathered}$$

Meanwhile, from (7)₁ and (9), it follows that

$$\frac{d^2}{d{t}^2}{\int }_0^{1}u\left(x,t\right)dx=0.$$

(18)

Therefire, by solving (18) and using the initial data of u, we get

$${\int }_0^{1}u\left(x,t\right)dx=t{\int }_0^1{u}_1\left(x\right)dx+{\int }_0^1{u}_0\left(x\right)dx.$$

Consequently, if we let

$$\stackrel{-}{u}\left(x,t\right)=u\left(x,t\right)-t{\int }_0^1{u}_1\left(x\right)dx-{\int }_0^1{u}_0\left(x\right)dx,$$

(19)

we get

$${\int }_0^1\stackrel{-}{u}\left(x,t\right)dx=0,\forall t\ge 0.$$

Therefore, the use of Poincaré’s inequality for $\stackrel{-}{u}$ is justified. In addition, a simple substitution shows that $(\stackrel{-}{u},\varphi ,y,{\eta }^t)$ satisfies system (7). Hence, we work with $\stackrel{-}{u}$ instead of u, but write u for simplicity of notation.

By imposing new appropriate conditions (H3), with the help of some special results, we obtain an unusual, weaker decay result using Lyaponov functiona, extending some earlier results known in the existing literature. The main results in this manuscript are as follows: Theorem 1 for the existence and uniqueness of solution and Theorem 2 for the general stability estimates.

2. Well-Posedness

In this section, we prove the existence and uniqueness result of the system (16)–(18) using the semigroup theory. To achieve our goal, we first introduce the vector function

$$U={(u,u_t,\varphi ,{\varphi }_t,y,{\eta }^t)}^T,$$

and the new dependent variables $v=u_t,\psi ={\varphi }_t,\phi ={\eta }^t$; then, the system (16) can be written as follows

$$\left\{\begin{array}{c}{U}_t=\mathcal{A}U+\Gamma \left(U\right)\hfill \\ U\left(0\right)=U_0={(u_0,u_1,{\varphi }_0,{\varphi }_1,f_0,{\eta }_0)}^T,\hfill \end{array}\right.$$

(20)

where $\mathcal{A}:\mathcal{D}\left(\mathcal{A}\right)\subset \mathcal{H}:\to \mathcal{H}$ is the linear operator defined by

$$\mathcal{A}U=\left(\begin{array}{c}v\hfill \\ \frac{\mu }{\rho }{u}_{xx}+\frac{b}{\rho }{\varphi }_x\hfill \\ \psi \hfill \\ {\displaystyle \frac{l}{J}{\psi }_{xx}+\frac{b}{J}{u}_x-\frac{\xi }{J}{\varphi }_x+\frac{1}{J}{\int }_0^{\infty }g\left(p\right){\phi }_{xx}\left(p\right)dp}\hfill \\ {\displaystyle -\frac{{\mu }_1}{J}\psi -\frac{1}{J}{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y(x,1,\varrho ,t)d\varrho }\hfill \\ -\frac{1}{\varrho }{y}_{\rho }\hfill \\ -{\phi }_s+\psi \hfill \end{array}\right),$$

(21)

and

$$\Gamma \left(U\right)=\left(\begin{array}{c}0\hfill \\ 0\hfill \\ 0\hfill \\ -\frac{\alpha \left(t\right)}{J}f\left(\psi \right)\hfill \\ 0\hfill \\ 0\hfill \end{array}\right),$$

(22)

and $\mathcal{H}$ is the energy space given by

$$\begin{array}{ccc}\hfill \mathcal{H}& =& H_{*}^1(0,1)\times L_{*}^2(0,1)\times H_0^1(0,1)\times L^2(0,1)\times L^2((0,1)\times (0,1)\times ({\tau }_1,{\tau }_2))\times L_g(0,1),\hfill \end{array}$$

where

$$\begin{gathered} \begin{array}{ccc}\hfill L_{*}^2(0,1)& =& \{\Phi \in L^2(0,1)/{\int }_0^1\Phi \left(x\right)dx=0\},\hfill \\ \\ \hfill H_{*}^1(0,1)& =& H^1(0,1)\cap L_{*}^2(0,1),\hfill \\ \\ \hfill L_g(0,1)& =& \{\Phi :{\mathbb{R}}_{+}\to H_0^1(0,1),{\int }_0^1{\int }_0^{\infty }g\left(s\right){\Phi }_x^2\left(p\right)dp<\infty \},\hfill \end{array} \end{gathered}$$

where the space $L_g(0,1)$ is endowed with the following inner product

$${\langle {\Phi }_1,{\Phi }_2\rangle }_{L_g(0,1)}={\int }_0^1{\int }_0^{\infty }g\left(p\right){\Phi }_{1x}\left(p\right){\Phi }_{2x}\left(p\right)dp.$$

For any

$$\begin{array}{c}\hfill U={(u,v,\varphi ,\psi ,y,\phi )}^T\in \mathcal{H},\widehat{U}={(\widehat{u},\widehat{v},\widehat{\varphi },\widehat{\psi },\widehat{y},\widehat{\phi })}^T\in \mathcal{H}.\end{array}$$

The space $\mathcal{H}$ equipped with the inner product is defined by

$$\begin{gathered} \begin{array}{cc}\hfill {\langle U,\widehat{U}\rangle }_{\mathcal{H}}& =\rho {\int }_0^{1}v\widehat{v}dx+\mu {\int }_0^1{u}_x{\widehat{u}}_{x}dx+J{\int }_0^1\psi \widehat{\psi }dx\hfill \\ \\ \hfill & +\xi {\int }_0^1\varphi \widehat{\varphi }dx+l{\int }_0^1{\varphi }_x{\widehat{\varphi }}_{x}dx+b{\int }_0^1(u_x\widehat{\varphi }+{\widehat{u}}_x\varphi )dx\hfill \\ \\ \hfill & +{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y\widehat{y}d\varrho d\rho dx+{\langle \phi ,\widehat{\phi }\rangle }_{L_g(0,1)}.\hfill \end{array} \end{gathered}$$

(23)

The domain of $\mathcal{A}$ is given by

$$\mathcal{D}\left(\mathcal{A}\right)=\left\{\begin{array}{c}U\in \mathcal{H}/u\in H_{*}^2\cap H_{*}^1,\varphi \in H^2\cap H_0^1,\hfill \\ v\in H_{*}^1,\psi \in H_0^1(0,1),\phi \in L_g(0,1),\hfill \\ y,y_{\rho }\in L^2((0,1)\times (0,1)\times ({\tau }_1,{\tau }_2)),y(x,0,\varrho ,t)=\psi ,\hfill \end{array}\right\}$$

where

$$H_{*}^2(0,1)=\left\{\Phi \in H^2(0,1)/{\Phi }_x\left(1\right)={\Phi }_x\left(0\right)=0\right\}.$$

Clearly, $\mathcal{D}\left(\mathcal{A}\right)$ is dense in $\mathcal{H}$. Now, we can state and prove the existence result.

Theorem 1.

Let $U_0\in \mathcal{H}$ and assume that (10)–(15) hold. Then, there exists a unique solution $U\in \mathcal{C}({\mathbb{R}}_{+},\mathcal{H})$ of problem (20). Moreover, if $U_0\in \mathcal{D}\left(\mathcal{A}\right)$, then

$$U\in \mathcal{C}({\mathbb{R}}_{+},\mathcal{D}\left(\mathcal{A}\right))\cap {\mathcal{C}}^1({\mathbb{R}}_{+},\mathcal{H}).$$

Proof. 

First, we prove that the operator $\mathcal{A}$ is dissipative. For any $U_0\in \mathcal{D}\left(\mathcal{A}\right)$ and by using (23), we have

$$\begin{gathered} \begin{array}{cc}\hfill {\langle \mathcal{A}U,U\rangle }_{\mathcal{H}}& =-{\mu }_1{\int }_0^1{\psi }^{2}dx-{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|\psi y(x,1,\varrho ,t)d\varrho dx\hfill \\ \\ -{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y_{\rho }yd\varrho d\rho dx-{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_{xp}\left(p\right){\phi }_x\left(p\right)dpdx.\hfill \end{array} \end{gathered}$$

(24)

For the third term of the RHS of (24), we have

$$\begin{gathered} \begin{array}{cc}\hfill -{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y_{\rho }yd\varrho d\rho dx& =-\frac{1}{2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{\int }_0^1\left|{\mu }_2\left(\varrho \right)\right|\frac{d}{d\rho }{y}^{2}d\rho d\varrho dx\hfill \\ \\ =-\frac{1}{2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2(x,1,\varrho ,t)d\varrho dx\hfill \\ \\ +\frac{1}{2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2(x,0,\varrho ,t)d\varrho dx.\hfill \end{array} \end{gathered}$$

(25)

Using Young’s inequality, we obtain

$$\begin{gathered} \begin{array}{cc}\hfill -{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|\psi y(x,1,\varrho ,t)d\varrho dx& \le \frac{1}{2}({\int }_{{\tau }_1}^{{\tau }_2}|{\mu }_2\left(\varrho \right)\left|d\varrho \right){\int }_0^1{\psi }^{2}dx\hfill \\ \\ +\frac{1}{2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2(x,1,\varrho ,t)d\varrho dx.\hfill \end{array} \end{gathered}$$

(26)

By integrating the last term of the right-hand side of (24), we have

$$\begin{array}{c}\hfill -{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_{xp}\left(p\right){\phi }_x\left(p\right)dpdx=\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx.\end{array}$$

(27)

Substituting (25), (26) and (27) into (24), using the fact that $y(x,0,\varrho ,t)=\psi (x,t)$ and (15), we obtained

$$\begin{gathered} \begin{array}{cc}\hfill {\langle \mathcal{A}U,U\rangle }_{\mathcal{H}}& \le -\left({\mu }_1-{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|d\varrho \right){\int }_0^1{\psi }^{2}dx+\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ \le 0.\hfill \end{array} \end{gathered}$$

(28)

Hence, the operator $\mathcal{A}$ is dissipative.

Next, we prove that the operator $\mathcal{A}$ is maximal. This is enough to show that the operator $(\lambda I-\mathcal{A})$ is surjective. Indeed, for any $F={(f_1,f_2,f_3,f_4,f_5,f_6)}^T\in \mathcal{H}$, we prove that there is a unique $V=(u,v,\varphi ,\psi ,y,\phi )\in \mathcal{D}(\mathcal{A}$) such that

$$(\lambda I-\mathcal{A})V=F.$$

(29)

That is

$$\left\{\begin{array}{c}\lambda u-v=f_1\in H_{*}^1(0,1)\hfill \\ \rho \lambda v-\mu u_{xx}-b{\varphi }_x=\rho f_2\in L_{*}^2(0,1)\hfill \\ \lambda \varphi -\psi =f_3\in H_0^1(0,1)\hfill \\ J\lambda \psi -l{\varphi }_{xx}+b{u}_x+\xi \varphi -{\int }_0^{\infty }g\left(p\right){\phi }_{xx}\left(p\right)dp\hfill \\ +{\mu }_1\psi +{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y(x,1,\varrho ,t)d\varrho =J{f}_4\in L^2(0,1)\hfill \\ \lambda \varrho y_t(x,\rho ,\varrho ,t)+y_{\rho }(x,\rho ,\varrho ,t)=\varrho f_5\in L^2((0,1)\times (0,1)\times ({\tau }_1,{\tau }_2))\hfill \\ \lambda \phi +{\phi }_s-\psi =f_6\in L_g(0,1).\hfill \end{array}\right.$$

(30)

We note that the equation (30)₅ with $y(x,0,\varrho ,t)=\psi (x,t)$ has a unique solution, given by

$$y(x,\rho ,\varrho ,t)=e^{-\lambda \rho \varrho }\psi +\varrho e^{\lambda \varrho \rho }{\int }_0^{\rho }{e}^{\lambda \varrho \sigma }{f}_5(x,\sigma ,\varrho ,t)d\sigma ,$$

(31)

then

$$y(x,1,\varrho ,t)=e^{-\lambda \varrho }\psi +\varrho e^{\lambda \varrho }{\int }_0^1{e}^{\lambda \varrho \sigma }{f}_5(x,\sigma ,\varrho ,t)d\sigma ,$$

(32)

and we infer from (30)₆ that

$$\phi =e^{\lambda s}{\int }_0^s{e}^{\tau }(\psi +f_6\left(\tau \right))d\tau ,$$

(33)

and we have

$$v=\lambda u-f_1,\psi =\lambda \varphi -f_3.$$

(34)

Inserting (32), (33) and (34) in (30)₂ and (30)⁴, we get

$$\left\{\begin{array}{c}\rho {\lambda }^{2}u-\mu u_{xx}-b{\varphi }_x=h_1\in L_{*}^2(0,1)\hfill \\ {\mu }_3\varphi -{\mu }_4{\varphi }_{xx}+b{u}_x=h_2\in L^2(0,1),\hfill \end{array}\right.$$

(35)

where

$$\left\{\begin{array}{c}{\mu }_3=J{\lambda }^2+\xi +\lambda {\mu }_1+\lambda {\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|e^{-\lambda \varrho }d\varrho \hfill \\ {\mu }_4=l+{\int }_0^{\infty }g\left(p\right)(1-e^{\lambda p})dp,\hfill \\ h_1=\rho (\lambda f_1+f_2)\hfill \\ h_2=(J\lambda +{\mu }_1+{\int }_{{\tau }_1}^{{\tau }_2}|{\mu }_2\left(\varrho \right)\left|e^{-\lambda \varrho }d\varrho \right)f_3+J{f}_4\hfill \\ -{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|e^{\lambda \varrho }{\int }_0^1{e}^{\lambda \varrho \sigma }{f}_5(x,\sigma ,\varrho ,t)d\sigma d\varrho \hfill \\ +{\int }_0^{\infty }g\left(p\right)e^{\lambda p}{\int }_0^p{e}^{\tau }{(\psi +f_6\left(\tau \right))}_{xx}d\tau dp.\hfill \end{array}\right.$$

(36)

We multiply (35) by $\widehat{u},\widehat{\varphi }$, respectively and integrate their sum over $(0,1)$ to obtain the following variational formulation

$$B((u,\varphi ),(\widehat{u},\widehat{\varphi }))={\rm Y}(\widehat{u},\widehat{\varphi }),$$

(37)

where

$$B:{(H_{*}^1(0,1)\times H_0^1(0,1))}^2\to \mathbb{R},$$

is the bilinear form defined by

$$\begin{gathered} \begin{array}{ccc}\hfill B((u,\varphi ),(\widehat{u},\widehat{\varphi }))& =& {\lambda }^2\rho {\int }_0^{1}u\widehat{u}dx+{\mu }_3{\int }_0^1\varphi \widehat{\varphi }dx+\mu {\int }_0^1{u}_x{\widehat{u}}_{x}dx\hfill \\ \\ & & +{\mu }_4{\int }_0^1{\varphi }_x{\widehat{\varphi }}_{x}dx+b{\int }_0^1(u_x\widehat{\varphi }+\varphi {\widehat{u}}_x)dx,\hfill \end{array} \end{gathered}$$

(38)

and

$${\rm Y}:(H_{*}^1(0,1)\times H_0^1(0,1))\to \mathbb{R},$$

is the linear functional given by

$$\begin{array}{ccc}\hfill {\rm Y}(\widehat{u},\widehat{\varphi })& =& {\int }_0^1{h}_1\widehat{u}dx+{\int }_0^1{h}_2\widehat{\varphi }dx\hfill \end{array}$$

(39)

Now, for $V=H_{*}^1(0,1)\times H_0^1(0,1)$, equipped with the norm

$$\begin{array}{ccc}\hfill {\parallel (u,\varphi )\parallel }_V^2& =& {\parallel u\parallel }_2^2+{\parallel \varphi \parallel }_2^2+\parallel u_x{\parallel }_2^2+{\parallel {\varphi }_x\parallel }_2^2,\hfill \end{array}$$

we have

$$\begin{gathered} \begin{array}{ccc}\hfill B\left(\right(u,\varphi ),(u,\varphi \left)\right)& =& {\lambda }^2\rho {\int }_0^1{u}^{2}dx+{\mu }_3{\int }_0^1{\varphi }^{2}dx+\mu {\int }_0^1{u}_x^{2}dx\hfill \\ \\ & & +2b{\int }_0^1{u}_x\varphi dx+{\mu }_4{\int }_0^1{\varphi }_x^{2}dx.\hfill \end{array} \end{gathered}$$

(40)

On the other hand, we can write

$$\begin{gathered} \begin{array}{ccc}\hfill \mu u_x^2+2b{u}_x\varphi +{\mu }_3{\varphi }^2& =& \frac{1}{2}[\mu {\left(u_x+\frac{b}{\mu }\varphi \right)}^2+{\mu }_3{\left(\varphi +\frac{b}{{\mu }_3}{u}_x\right)}^2\hfill \\ \\ & & +\left(\mu -\frac{b^2}{{\mu }_3}\right)u_x^2+\left({\mu }_3-\frac{b^2}{\mu }\right){\varphi }^2].\hfill \end{array} \end{gathered}$$

Since $\mu \xi >b^2$, we deduce that

$$\mu u_x^2+2b{u}_x\varphi +{\mu }_3{\varphi }^2>\frac{1}{2}\left[\left(\mu -\frac{b^2}{{\mu }_3}\right)u_x^2+\left({\mu }_3-\frac{b^2}{\mu }\right){\varphi }^2\right],$$

then, for some $M_0>0$

$$B((u,\varphi ),(u,\varphi ))\ge M_0{\parallel (u,\varphi )\parallel }_V^2.$$

(41)

Thus, B is coercive, similarly,

$${\rm Y}(\widehat{u},\widehat{\varphi })\ge M_1{\parallel (\widehat{u},\widehat{\varphi })\parallel }_V^2.$$

(42)

Consequently, using Lax–Milgram theorem, we conclude that (16) has a unique solution

$$(u,\varphi )\in H_{*}^1(0,1\times H_0^1(0,1).$$

Substituting $u,\varphi$ into (32), (33) and (34), respectively, we have

$$\begin{gathered} \begin{array}{ccc}& & v\in H_{*}^1(0,1),\psi \in H_0^1(0,1),\phi \in L_g(0,1)\hfill \\ \\ & & y,y_{\rho }\in L^2((0,1)\times (0,1)\times ({\tau }_1,{\tau }_2)).\hfill \end{array} \end{gathered}$$

(43)

Moreover, if we take $\widehat{u}=0\in H_{*}^1(0,1)$ in (37) to obtain

$$\begin{array}{c}\hfill {\mu }_3{\int }_0^1\varphi \widehat{\varphi }dx+b{\int }_0^1{u}_x\widehat{\varphi }dx+{\mu }_4{\int }_0^1{\varphi }_x{\widehat{\varphi }}_{x}dx={\int }_0^1{h}_2\widehat{\varphi }d,\end{array}$$

(44)

we get

$${\mu }_4{\int }_0^1{\varphi }_x{\widehat{\varphi }}_{x}dx={\int }_0^1(h_2-{\mu }_3\varphi -b{u}_x)\widehat{\varphi }dx,\forall \widehat{\varphi }\in H_0^1(0,1),$$

(45)

which yields

$$\begin{array}{c}\hfill {\mu }_4{\varphi }_{xx}=(h_2-{\mu }_3\varphi -b{u}_x)\in L^2(0,1).\end{array}$$

(46)

Thus,

$$\varphi \in H^2(0,1)\cap H_0^1(0,1).$$

(47)

Consequently, (45) takes the following form

$${\int }_0^1(-{\mu }_4{\varphi }_{xx}-h_2+{\mu }_3\varphi +b{u}_x)\widehat{\varphi }dx=0,\forall \widehat{\varphi }\in H_0^1(0,1).$$

Hence, we get

$$-{\mu }_4{\varphi }_{xx}+{\mu }_3\varphi +b{u}_x=h_2.$$

This give (35)₂. Similarly, if we take $\widehat{\varphi }=0\in H_0^1(0,1)$ in (36) to obtain

$$\begin{array}{c}\hfill \mu {\int }_0^1{u}_x{\widehat{u}}_{x}dx+b{\int }_0^1\varphi {\widehat{u}}_{x}dx+{\lambda }^2\rho {\int }_0^{1}u\widehat{u}dx={\int }_0^1{h}_1\widehat{u}dx,\end{array}$$

we obtain

$$\mu {\int }_0^1{u}_x{\widehat{u}}_{x}dx={\int }_0^1(h_1+b{\varphi }_x-{\lambda }^2\rho u)\widehat{u}dx,\forall \widehat{u}\in H_{*}^1(0,1),$$

(48)

which yields

$$\begin{array}{c}\hfill -\mu u_{xx}=(h_1+b{\varphi }_x-{\lambda }^2\rho u)\in L_{*}^2(0,1).\end{array}$$

(49)

Consequently, (48) takes the following form

$${\int }_0^1(-\mu u_{xx}-h_1-b{\varphi }_x+{\lambda }^2\rho u)\widehat{u}dx=0,\forall \widehat{u}\in H_{*}^1(0,1).$$

Hence, we obtain

$$-\mu u_{xx}-b{\varphi }_x+{\lambda }^2\rho u=h_1.$$

This give (35)₁.

Moreover, (48) also holds for any $\Phi \in C^1\left([0,1]\right)$. Then, by using integration by parts, we obtain

$$\mu {\int }_0^1{u}_x{\Phi }_{x}dx+{\int }_0^1(-h_1-b{\varphi }_x+{\lambda }^2\rho u)\Phi dx=0,\forall \Phi \in C^1\left([0,1]\right).$$

(50)

Then, we obtain for any $\Phi \in C^1\left([0,1]\right)$

$$u_x\left(1\right)\Phi \left(1\right)-u_x\left(0\right)\Phi \left(0\right)=0.$$

(51)

Since $\Phi$ is arbitrary, we obtain that $u_x\left(0\right)=u_x\left(1\right)=0$. Hence, $u\in H_{*}^2(0,1)\cap H_{*}^1(0,1)$. Therefore, the application of regularity theory for the linear elliptic equations guarantees the existence of unique $U\in \mathcal{D}\left(\mathcal{A}\right)$ such that (29) is satisfied. Consequently, we conclude that $\mathcal{A}$ is a maximal dissipative operator. Now, we prove that the operator $\Gamma$ defined in

(22) is locally Lipschitz in $\mathcal{H}$. Let

$$U={(u,v,\varphi ,\psi ,y,\phi )}^T\in \mathcal{H},\widehat{U}={(\widehat{u},\widehat{v},\widehat{\varphi },\widehat{\psi },\widehat{y},\widehat{\phi })}^T\in \mathcal{H}.$$

Then, we have

$$\parallel \Gamma \left(U\right)-\Gamma \left(\widehat{U}\right){\parallel }_{\mathcal{H}}\le M_3{\parallel f\left(\psi \right)-f\left(\widehat{\psi }\right)\parallel }_{L^2\left(01\right)}.$$

By using (14) and Holder and Poincaré’s inequalities, we can obtain

$$\begin{gathered} \begin{array}{ccc}\hfill \parallel f\left(\psi \right)-f\left(\widehat{\psi }\right){\parallel }_{L^2\left(01\right)}& \le & k_0{(\parallel \psi \parallel }_{2\beta }^{\beta }+\parallel \widehat{\psi }{\parallel }_{2\beta }^{\beta })\parallel \psi -\widehat{\psi }\parallel \hfill \\ \\ & \le & k_1{\parallel {\psi }_x-{\widehat{\psi }}_x\parallel }_{L^2\left(01\right)},\hfill \end{array} \end{gathered}$$

which gives us

$$\parallel \Gamma \left(U\right)-\Gamma \left(\widehat{U}\right){\parallel }_{\mathcal{H}}\le M_4{\parallel U-\widehat{U}\parallel }_{\mathcal{H}}.$$

Then, the operator $\Gamma$ is locally Lipschitz in $\mathcal{H}$. Consequently, the well-posedness result follows from the Hille–Yosida theorem. The proof is completed. □

3. Stability Result

In this section, we state and prove our decay result for the energy of the system

(16)–(18) using the multiplier technique. We need the following Lemmas.

Lemma 1.

The energy functional $\mathcal{E}$, defined by

$$\begin{gathered} \begin{array}{ccc}\hfill \mathcal{E}\left(t\right)& =& \frac{1}{2}{\int }_0^1\left[\rho u_t^2+\mu u_x^2+J{\varphi }_t^2+l{\varphi }_x^2+\xi {\varphi }^2+2b{u}_x\varphi \right]dx\hfill \\ \\ & & +\frac{1}{2}{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +\frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx,\hfill \end{array} \end{gathered}$$

(52)

satisfies

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{E}}^{\prime }\left(t\right)& \le & -{\eta }_0{\int }_0^1{\varphi }_t^{2}dx+\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx+\alpha \left(t\right){\int }_0^1{\varphi }_{t}f\left({\varphi }_t\right)dx\hfill \\ \\ & \le & 0,\hfill \end{array} \end{gathered}$$

(53)

where ${\eta }_0={\mu }_1-{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|d\varrho >0$ and $\phi \left(s\right)={\eta }^t=\varphi (x,t)-\varphi (x,t-p)$.

Proof. 

Multiplying (16)${}_1$ by $u_t$ and (16)${}_2$ by ${\varphi }_t$, then integration by parts over $\left(0,1\right)$ and using (17), we get

$$\begin{gathered} \begin{array}{ccc}& & \frac{1}{2}\frac{d}{dt}{\int }_0^1\left[\rho u_t^2+\mu u_x^2+J{\varphi }_t^2+\delta {\varphi }_x^2+\xi {\varphi }^2+2b{u}_x\varphi \right]dx\hfill \\ \\ & & -{\int }_0^1{\varphi }_{xt}{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx+{\mu }_1{\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & +{\int }_0^1{\varphi }_t{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y\left(x,1,\varrho ,t\right)d\varrho dx+\alpha \left(t\right){\int }_0^1{\varphi }_{t}f\left({\varphi }_t\right)dx=0.\hfill \end{array} \end{gathered}$$

(54)

The last term in the LHS of (54) is estimated as follows

$$\begin{gathered} \begin{array}{ccc}\hfill {\int }_0^1{\varphi }_t{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y\left(x,1,\varrho ,t\right)d\varrho dx& \le & \frac{1}{2}\left({\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|d\varrho \right){\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & +\frac{1}{2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,1,\varrho ,t\right)d\varrho dx,\hfill \end{array} \end{gathered}$$

(55)

and

$$\begin{gathered} \begin{array}{ccc}\hfill -{\int }_0^1{\varphi }_{xt}{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx& \le & \frac{1}{2}\frac{d}{dt}{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & -\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx.\hfill \end{array} \end{gathered}$$

(56)

Now, multiplying the equation (16)₃ by $y|{\mu }_2\left(\varrho \right)|$ and integrating the result over $(0,1)\times (0,1)\times ({\tau }_1,{\tau }_2)$

$$\begin{gathered} \begin{array}{ccc}& {\displaystyle \frac{d}{dt}}& \frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y^2(x,\rho ,\varrho ,t)d\varrho d\rho dx\hfill \\ \\ & =& -{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y{y}_{\rho }\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & =& -\frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|\frac{d}{d\rho }{y}^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & =& \frac{1}{2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|(y^2\left(x,0,\varrho ,t\right)-y^2(x,1,\varrho ,t))d\varrho dx\hfill \\ \\ & =& \frac{1}{2}({\int }_{{\tau }_1}^{{\tau }_2}|{\mu }_2\left(\varrho \right)\left|d\varrho \right){\int }_0^1{\varphi }_t^{2}dx-\frac{1}{2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,1,\varrho ,t\right)d\varrho dx.\hfill \end{array} \end{gathered}$$

(57)

Now, using (54), (55), (56) and (57), we have

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{E}}^{\prime }\left(t\right)& \le & -\left({\mu }_1-{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|d\varrho \right){\int }_0^1{\varphi }_t^{2}dx+\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & -\alpha \left(t\right){\int }_0^1{\varphi }_{t}f\left({\varphi }_t\right)dx,\hfill \end{array} \end{gathered}$$

(58)

then, by (10), there exists a positive constant ${\eta }_0$, such that

$${\mathcal{E}}^{\prime }\left(t\right)\le -{\eta }_0{\int }_0^1{\varphi }_t^{2}dx+\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx-\alpha \left(t\right){\int }_0^1{\varphi }_{t}f\left({\varphi }_t\right)dx,$$

(59)

hence, by (11)–(15) we obtain $\mathcal{E}$ is a non-increasing function. □

Remark 1.

Using $(\mu \xi >b^2)$, we conclude that the energy $\mathcal{E}\left(t\right)$ definie by (51) satisfies

$$\begin{gathered} \begin{array}{ccc}\hfill \mathcal{E}\left(t\right)& >& \frac{1}{2}{\int }_0^1\left[\rho u_t^2+\widehat{\mu }{u}_x^2+J{\varphi }_t^2+l{\varphi }_x^2+\widehat{\xi }{\varphi }^2\right]dx\hfill \\ \\ & & +\frac{1}{2}{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +\frac{1}{2}{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx,\hfill \end{array} \end{gathered}$$

(60)

where

$$\widehat{\mu }=\frac{1}{2}(\mu -\frac{b^2}{\xi })>0,\widehat{\xi }=\frac{1}{2}(\xi -\frac{b^2}{\mu })>0,$$

then $\mathcal{E}\left(t\right)$ is positive function.

Lemma 2.

The functional

$$D_1\left(t\right):=J{\int }_0^1{\varphi }_t\varphi dx+\frac{b\rho }{\mu }{\int }_0^1\varphi {\int }_0^x{u}_t\left(y\right)dydx+\frac{{\mu }_1}{2}{\int }_0^1{\varphi }^{2}dx,$$

(61)

satisfies

$$\begin{gathered} \begin{array}{ccc}\hfill D_1^{\prime }\left(t\right)& \le & -\frac{l}{2}{\int }_0^1{\varphi }_x^{2}dx-\widehat{\mu }{\int }_0^1{\varphi }^{2}dx+{\epsilon }_1{\int }_0^1{u}_t^{2}dx+c(1+{\frac{1}{\epsilon }}_1){\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & +c{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx+c{\int }_0^1{f}^2\left({\varphi }_t\right)dx\hfill \\ \\ & & +c{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,1,\varrho ,t\right)d\varrho dx,\hfill \end{array} \end{gathered}$$

(62)

where $\widehat{\mu }=\xi -\frac{b^2}{\mu }>0$.

Proof. 

Direct computation using integration by parts and Young’s inequality, for ${\epsilon }_1>0$, yields

$$\begin{gathered} \begin{array}{ccc}\hfill D_1^{\prime }\left(t\right)& =& -l{\int }_0^1{\varphi }_x^{2}dx-\left(\xi -\frac{b^2}{\mu }\right){\int }_0^1{\varphi }^{2}dx+\frac{b\rho }{\mu }{\int }_0^1{\varphi }_t{\int }_0^x{u}_t\left(y\right)dydx\hfill \\ \\ & & +{\int }_0^1{\varphi }_x{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx+\alpha \left(t\right){\int }_0^1\varphi f\left({\varphi }_t\right)dx\hfill \\ \\ & & +J{\int }_0^1{\varphi }_t^{2}dx-{\int }_0^1\varphi {\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y(x,1,\varrho ,t)d\varrho dx\hfill \\ \\ & \le & -l{\int }_0^1{\varphi }_x^{2}dx-\left(\xi -\frac{b^2}{\mu }\right){\int }_0^1{\varphi }^{2}dx+c\left(1+\frac{1}{{\epsilon }_1}\right){\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & +{\epsilon }_1{\int }_0^1{\left({\int }_0^x{u}_t\left(y\right)dy\right)}^{2}dx+{\int }_0^1{\varphi }_x{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx\hfill \\ \\ & & -{\int }_0^1\varphi {\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y(x,1,\varrho ,t)d\varrho dx+\alpha \left(t\right){\int }_0^1\varphi f\left({\varphi }_t\right)dx.\hfill \end{array} \end{gathered}$$

(63)

According to Cauchy–Schwartz inequality, it is clear that

$${\int }_0^1{\left({\int }_0^x{u}_t\left(y\right)dy\right)}^{2}dx\le {\int }_0^1{\left({\int }_0^1{u}_{t}dx\right)}^{2}dx\le {\int }_0^1{u}_t{}^{2}dx.$$

Therefore, estimate (63) becomes

$$\begin{gathered} \begin{array}{ccc}\hfill D_1^{\prime }\left(t\right)& \le & -\delta {\int }_0^1{\varphi }_x^{2}dx-\left(\xi -\frac{b^2}{\mu }\right){\int }_0^1{\varphi }^{2}dx+c\left(1+\frac{1}{{\epsilon }_1}\right){\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & +{\epsilon }_1{\int }_0^1{u}_t{}^{2}dx-{\int }_0^1\varphi {\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y(x,1,\varrho ,t)d\varrho dx\hfill \\ \\ & & +{\int }_0^1{\varphi }_x{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx+\alpha \left(t\right){\int }_0^1\varphi f\left({\varphi }_t\right)dx.\hfill \end{array} \end{gathered}$$

(64)

The last term in the RHS of (64) is estimated as follows

$$\begin{array}{ccc}\hfill {\int }_0^1{\varphi }_x{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx& \le & c{\delta }_1{\int }_0^1{\varphi }_x^{2}dx+\frac{c}{4{\delta }_1}{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx,\hfill \\ \hfill \end{array}$$

(65)

where we used Cauchy–Schwartz, Young and Poincaré’s inequalities, for ${\delta }_1,{\delta }_2,{\delta }_3>0$.

By substituting (65) into (63), we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill D_1^{\prime }\left(t\right)& \le & -\left(l-c{\delta }_1-{\mu }_{1}c{\delta }_2-c{\delta }_3\right){\int }_0^1{\varphi }_x^{2}dx-\left(\xi -\frac{b^2}{\mu }\right){\int }_0^1{\varphi }^{2}dx\hfill \\ \\ & & +{\epsilon }_1{\int }_0^1{u}_t^{2}dx+c(1+\frac{1}{{\epsilon }_1}){\int }_0^1{\varphi }_t^{2}dx+\frac{c}{4{\delta }_1}{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +\frac{1}{4{\delta }_2}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2(x,1,\varrho ,t)d\varrho dx+\frac{1}{4{\delta }_3}{\int }_0^1{f}^2\left({\varphi }_t\right)dx.\hfill \end{array} \end{gathered}$$

(66)

Bearing in mind that $\mu \xi >b^2$ and letting ${\delta }_1=\frac{l}{6c}$, ${\delta }_2=\frac{l}{6c{\mu }_1}$ and ${\delta }_3=\frac{l}{6c}$, we obtain estimate (62). □

Lemma 3.

Then, for any ${\epsilon }_2>0$ the functional

$$D_2\left(t\right):={\int }_0^1{\varphi }_x{u}_{t}dx+{\int }_0^1{\varphi }_t{u}_{x}dx-\frac{\rho }{\mu J}{\int }_0^1{u}_t{\int }_0^{\infty }g\left(p\right){\varphi }_x\left(t-p\right)dpdx,$$

satisfies

$$\begin{gathered} \begin{array}{ccc}\hfill D_2^{\prime }\left(t\right)& \le & -\frac{b}{2J}{\int }_0^1{u}_x^{2}dx+c{\int }_0^1{\varphi }_x^{2}dx+c{\epsilon }_2{\int }_0^1{u}_t^{2}dx+c{\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & +c{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx-\frac{c}{{\epsilon }_2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +c{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2(x,1,\varrho ,t)d\varrho dx+c{\int }_0^1{f}^2\left({\varphi }_t\right)dx\hfill \\ \\ & & +\left(\frac{\delta }{J}-\frac{\mu }{\rho }\right){\int }_0^1{u}_x{\varphi }_{xx}dx.\hfill \end{array} \end{gathered}$$

(67)

Proof. 

By differentiating $D_2$, then using (16), integration by parts and (17) we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill D_2^{\prime }\left(t\right)& =& -\frac{b}{J}{\int }_0^1{u}_x^{2}dx+\left(\frac{l+g_0}{J}-\frac{\mu }{\rho }\right){\int }_0^1{u}_x{\varphi }_{xx}dx+(\frac{b}{\rho }-\frac{b{g}_0}{\mu J}){\int }_0^1{\varphi }_x^{2}dx\hfill \\ \\ & & -\frac{\xi }{J}{\int }_0^1{u}_x\varphi dx-\frac{b}{\mu J}{\int }_0^1{\varphi }_x{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx\hfill \\ \\ & & -\frac{\rho }{\mu J}{\int }_0^1{u}_t{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x\left(p\right)dpdx-\frac{\alpha \left(t\right)}{\mu J}{\int }_0^1{u}_{x}f\left({\varphi }_t\right)dx\hfill \\ \\ & & -\frac{{\mu }_1}{J}{\int }_0^1{\varphi }_t{u}_{x}dx-\frac{1}{J}{\int }_0^1{u}_x{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2(x,1,\varrho ,t)d\varrho dx.\hfill \end{array} \end{gathered}$$

(68)

In what follows, we estimate the last six terms in the RHS of (68), using Young, Cauchy–Schwartz and Poincaré’s inequalities. For ${\delta }_4,{\delta }_5,{\epsilon }_2>0$, we have

$$-\frac{\xi }{J}{\int }_0^1{u}_x\varphi dx\le \frac{\xi }{J}{\delta }_4{\int }_0^1{u}_x^{2}dx+\frac{\xi }{4J{\delta }_4}{\int }_0^1{\varphi }^{2}dx.$$

By letting ${\delta }_4=\frac{b}{6\xi }$, using Poincaré’s inequality, we get

$$-\frac{\xi }{J}{\int }_0^1{u}_x\varphi dx\le \frac{b}{6J}{\int }_0^1{u}_x^{2}dx+c{\int }_0^1{\varphi }_x^{2}dx,$$

(69)

and by Young and Chauchy–Schawrz’s inequalities, we get

$$\begin{array}{c}\hfill -\frac{b}{\mu J}{\int }_0^1{\varphi }_x{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx\le c{\delta }_5{\int }_0^1{\varphi }_x^{2}dx+\frac{c}{4{\delta }_5}{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx.\end{array}$$

By letting ${\delta }_5=\frac{b}{6cJ}$, we obtain

$$-\frac{b}{\mu J}{\int }_0^1{\varphi }_x{\int }_0^{\infty }g\left(p\right){\phi }_x\left(p\right)dpdx\le \frac{b}{6J}{\int }_0^1{\varphi }_x^{2}dx+c{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx.$$

(70)

Similarly, $\forall {\epsilon }_2>0$ we have

$$\begin{array}{c}\hfill \frac{\rho }{\mu J}{\int }_0^1{u}_t{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x\left(p\right)dpdx\le c{\epsilon }_2{\int }_0^1{u}_t^{2}dx+\frac{c}{{\epsilon }_2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx,\end{array}$$

(71)

and

$$-\frac{{\mu }_1}{J}{\int }_0^1{\varphi }_t{u}_{x}dx\le \frac{{\mu }_1{\delta }_6}{2J}{\int }_0^1{u}_x^{2}dx+\frac{{\mu }_1}{2J{\delta }_6}{\int }_0^1{\varphi }_t^{2}dx,$$

(72)

and

$$\begin{gathered} \begin{array}{ccc}\hfill \frac{1}{J}{\int }_0^1{u}_x{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y(x,1,\varrho ,t)d\varrho dx& \le & \frac{{\delta }_7{\mu }_1}{2J}{\int }_0^1{u}_x^{2}dx\hfill \\ \\ & & +\frac{1}{2J{\delta }_7}{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2(x,1,\varrho ,t)d\varrho ,\hfill \end{array} \end{gathered}$$

(73)

and

$$-\frac{\alpha \left(t\right)}{J}{\int }_0^1{u}_{x}f\left({\varphi }_t\right)dx\le \frac{\alpha \left(0\right){\delta }_8}{2J}{\int }_0^1{u}_x^{2}dx+\frac{\alpha \left(0\right)}{2J{\delta }_8}{\int }_0^1{f}^2\left({\varphi }_t\right)dx.$$

(74)

Replacing (69)–(74) into (68) and letting ${\delta }_6={\delta }_7=\frac{b}{6{\mu }_1}$ and ${\delta }_8=\frac{b}{6\alpha \left(0\right)}$, yields (67). □

Lemma 4.

The functional

$$D_3\left(t\right):=-\rho {\int }_0^1{u}_{t}udx,$$

satisfies

$$D_3^{\prime }\left(t\right)\le -\rho {\int }_0^1{u}_t^{2}dx+\frac{3\mu }{2}{\int }_0^1{u}_x^{2}dx+c{\int }_0^1{\varphi }_x^{2}dx.$$

(75)

Proof. 

Direct computations give

$$D_3^{\prime }\left(t\right)=-\rho {\int }_0^1{u}_t^{2}dx+\mu {\int }_0^1{u}_x^{2}dx+b{\int }_0^1{u}_x\varphi dx.$$

The estimate (75) easily follows using Young and Poincaré inequalities.

$$\begin{gathered} \begin{array}{ccc}\hfill D_3^{\prime }\left(t\right)& \le & -\rho {\int }_0^1{u}_t^{2}dx+\mu {\int }_0^1{u}_x^{2}dx+b\epsilon {\int }_0^1{u}_x^{2}dx+\frac{b}{4\epsilon }{\int }_0^1{\varphi }^{2}dx\hfill \\ \\ & \le & -\rho {\int }_0^1{u}_t^{2}dx+\mu {\int }_0^1{u}_x^{2}dx+b\epsilon {\int }_0^1{u}_x^{2}dx+\frac{bc}{4\epsilon }{\int }_0^1{\varphi }_x^{2}dx,\hfill \end{array} \end{gathered}$$

by taking $\epsilon =\frac{\mu }{2b}$, we obtain (75). □

Lemma 5.

The functional

$$D_4\left(t\right):={\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho e^{-\varrho \rho }\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx,$$

satisfies

$$\begin{gathered} \begin{array}{ccc}\hfill D_4^{\prime }\left(t\right)& \le & -{\eta }_1{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx+{\mu }_1{\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & -{\eta }_1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,1,\varrho ,t\right)d\varrho dx,\hfill \end{array} \end{gathered}$$

(76)

where ${\eta }_1$ is a positive constant.

Proof. 

By differentiating $D_4$, with respect to t, using the Equation (16)₃, we have

$$\begin{gathered} \begin{array}{ccc}\hfill D_4^{\prime }\left(t\right)& =& -2{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{e}^{-\varrho \rho }\left|{\mu }_2\left(\varrho \right)\right|y{y}_{\rho }\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & =& -{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho e^{-\varrho \rho }\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & & -{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|[e^{-\varrho }{y}^2\left(x,1,\varrho ,t\right)-y^2\left(x,0,\varrho ,t\right)]d\varrho dx.\hfill \end{array} \end{gathered}$$

Using the fact that $y(x,0,\varrho ,t)={\varphi }_t(x,t)$ and $e^{-\varrho }\le e^{-\varrho \rho }\le 1$, for all $0<\rho <1$, we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill D_4^{\prime }\left(t\right)& =& -{\eta }_1{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & & -{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}{e}^{-\varrho }|{\mu }_2\left(\varrho \right)|y^2\left(x,1,\varrho ,t\right)d\varrho dx+{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|d\varrho {\int }_0^1{\varphi }_t^{2}dx.\hfill \end{array} \end{gathered}$$

□

Since $-e^{-\varrho }$ is an increasing function, we have $-e^{-\varrho }\le -e^{-{\tau }_2}$, for all $\varrho \in [{\tau }_1,{\tau }_2]$.

Finally, setting ${\eta }_1=e^{-{\tau }_2}$ and recalling (15), we obtain (76). We are now ready to prove the main result.

Theorem 2.

Assume (10)–(15) hold. Let $h\left(t\right)=\alpha \left(t\right).\eta \left(t\right)$ be a positive non-increasing function. Then, for any $U_0\in \mathcal{D}\left(\mathcal{A}\right)$, satisfying, for some $c_0>0$

$$max\left\{{\int }_0^1{\varphi }_{0x}^2(x,s)dx,{\int }_0^1{\varphi }_{0sx}^2(x,s)dx\right\}\le c_0,\forall s>0,$$

(77)

there are positive constants ${\beta }_1,{\beta }_2$ and ${\beta }_3$ such that the energy functional given by (52) satisfies

$$\mathcal{E}\left(t\right)\le {\beta }_1{G}_0^{-1}\left(\frac{{\beta }_2+{\beta }_3{\int }_0^{t}h\left(p\right)\varpi \left(p\right)dp}{{\int }_0^{t}h\left(p\right)dp}\right),$$

(78)

where

$$G_0\left(t\right)=t{G}^{\prime }\left({\epsilon }_{0}t\right),\forall {\epsilon }_0\ge 0,and\varpi \left(s\right)={\int }_s^{\infty }g\left(\sigma \right)d\sigma .$$

(79)

Proof. 

We define a Lyapunov functional

$$\mathcal{L}\left(t\right):=N\mathcal{E}\left(t\right)+N_1{D}_1\left(t\right)+N_2{D}_2\left(t\right)+D_3\left(t\right)+N_4{D}_4\left(t\right),$$

(80)

where $N,$$N_1,$$N_2,$ and $N_4$ are positive constants, to be chosen later. By differentiating (80) and using (53), (62), (67), (75), (76), we have

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}^{\prime }\left(t\right)& \le & -\left[\frac{l{N}_1}{2}-c{N}_2-c\right]{\int }_0^1{\varphi }_x^{2}dx-\left[\rho -N_1{\epsilon }_1-N_{2}c{\epsilon }_2\right]{\int }_0^1{u}_t^{2}dx\hfill \\ \\ & & -\left[\frac{b{N}_2}{2J}-\frac{3\mu }{2}\right]{\int }_0^1{u}_x^{2}dx+c\left[N_1+N_2\right]{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & -\left[{\eta }_{0}N-c{N}_1(1+\frac{1}{{\epsilon }_1})-N_{2}c-{\mu }_1{N}_4\right]{\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & -\left[N_4{\eta }_1-c{N}_1-c{N}_2\right]{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,1,\varrho ,t\right)d\varrho dx\hfill \\ \\ & & -N_1\widehat{\mu }{\int }_0^1{\varphi }^{2}dx+\left[\frac{N}{2}-\frac{c{N}_2}{{\epsilon }_2}\right]{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & -N_4{\eta }_1{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & & +c\left[N_1+N_2\right]{\int }_0^1{f}^2\left({\varphi }_t\right)dx+N_2\chi {\int }_0^1{u}_{xx}{\varphi }_{x}dx,\hfill \end{array} \end{gathered}$$

where $\chi =(\frac{\mu }{\rho }-\frac{\delta }{J})$ and by setting

$${\epsilon }_1=\frac{\rho }{4N_1},{\epsilon }_2=\frac{\rho }{4c{N}_2},$$

we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}^{\prime }\left(t\right)& \le & -\left[\frac{l{N}_1}{2}-c{N}_2(1+N_2)-c\right]{\int }_0^1{\varphi }_x^{2}dx-\frac{\rho }{2}{\int }_0^1{u}_t^{2}dx\hfill \\ \\ & & -\left[\frac{b{N}_2}{2J}-\frac{3\mu }{2}\right]{\int }_0^1{u}_x^{2}dx+c\left[N_1+N_2\right]{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & -\left[{\eta }_{0}N-c{N}_1(1+N_1)-c{N}_2-{\mu }_1{N}_4\right]{\int }_0^1{\varphi }_t^{2}dx\hfill \\ \\ & & -\left[N_4{\eta }_1-c{N}_1-c{N}_2\right]{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,1,\varrho ,t\right)d\varrho dx\hfill \\ \\ & & -N_1\widehat{\mu }{\int }_0^1{\varphi }^{2}dx+\left[\frac{N}{2}-c{N}_2^2\right]{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & -N_4{\eta }_1{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(s\right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & & +c\left[N_1+N_2\right]{\int }_0^1{f}^2\left({\varphi }_t\right)dx+N_2\chi {\int }_0^1{u}_{xx}{\varphi }_{x}dx.\hfill \end{array} \end{gathered}$$

Next, we carefully choose our constants so that the terms inside the brackets are positive. We choose a $N_2$ that is large enough that

$${\alpha }_1=\frac{b{N}_2}{2J}-\frac{3\mu }{2}>0,$$

then, we choose a large enough $N_1$ that

$${\alpha }_2=\frac{l{N}_1}{4}-c{N}_2\left(1+N_2\right)-c>0,$$

then we choose a large enough $N_4$ that

$${\alpha }_3=N_4{\eta }_1-c{N}_1-c{N}_2>0,$$

thus, we arrive at

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}^{\prime }\left(t\right)& \le & -{\alpha }_2{\int }_0^1{\varphi }_x^{2}dx-{\alpha }_0{\int }_0^1{\varphi }^{2}dx-\frac{\rho }{2}{\int }_0^1{u}_t^{2}dx-{\alpha }_1{\int }_0^1{u}_x^{2}dx\hfill \\ \\ & & -[{\eta }_{0}N-c]{\int }_0^1{\varphi }_t^{2}dx+\left[\frac{N}{2}-c\right]{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +c{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx-{\alpha }_3{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,1,\varrho ,t\right)d\varrho dx\hfill \\ \\ & & -{\alpha }_4{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx\hfill \\ \\ & & +c{\int }_0^1{f}^2\left({\varphi }_t\right)dx+{\alpha }_5{\int }_0^1{u}_{xx}{\varphi }_{x}dx.\hfill \end{array} \end{gathered}$$

(81)

where ${\alpha }_0=\widehat{\mu }{N}_1=\left(\xi -\frac{b^2}{\mu }\right)N_1,and{\alpha }_5=N_2\chi =N_2(\frac{\mu }{\rho }-\frac{\delta }{J})$. On the other hand, if we let

$$\mathfrak{L}\left(t\right)=N_1{D}_1\left(t\right)+N_2{D}_2\left(t\right)+D_3\left(t\right)+N_4{D}_4\left(t\right),$$

then

$$\begin{gathered} \begin{array}{ccc}\hfill \left|\mathfrak{L}\left(t\right)\right|& \le & J{N}_1{\int }_0^1\left|\varphi {\varphi }_t\right|dx+\frac{b\rho N_1}{\mu }{\int }_0^1\left|\varphi {\int }_0^x{u}_t\left(y\right)dy\right|dx\hfill \\ \\ & & +N_2{\int }_0^1\left|{\varphi }_x{u}_t+u_x{\varphi }_t-\frac{\rho }{\mu J}{u}_t{\int }_0^{\infty }g\left(p\right){\varphi }_x\left(t-p\right)dp\right|dx\hfill \\ \\ & & +\rho {\int }_0^1\left|u_{t}u\right|dx+N_4{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho e^{-\varrho \rho }\left|{\mu }_2\left(\varrho \right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho dx.\hfill \end{array} \end{gathered}$$

Exploiting Young, Cauchy–Schwartz and Poincaré inequalities, we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill \left|\mathfrak{L}\left(t\right)\right|& \le & c{\int }_0^1\left(u_t^2+{\varphi }_t^2+{\varphi }_x^2+u_x^2+{\varphi }^2\right)dx+c{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +c{\int }_0^1{\int }_0^1{\int }_{{\tau }_1}^{{\tau }_2}\varrho \left|{\mu }_2\left(s\right)\right|y^2\left(x,\rho ,\varrho ,t\right)d\varrho d\rho \hfill \\ \\ & \le & c\mathcal{E}\left(t\right).\hfill \end{array} \end{gathered}$$

Consequently, we obtain

$$\left|\mathfrak{L}\left(t\right)\right|=\left|\mathcal{L}\left(t\right)-N\mathcal{E}\left(t\right)\right|\le c\mathcal{E}\left(t\right),$$

that is

$$\left(N-c\right)\mathcal{E}\left(t\right)\le \mathcal{L}\left(t\right)\le \left(N+c\right)\mathcal{E}\left(t\right).$$

(82)

Now, by choosing a large enough N that

$$\frac{N}{2}-c>0,N-c>0,N{\eta }_0-c>0,$$

and exploiting (52), estimates (81) and (82), respectively, we obtain

$$c_2\mathcal{E}\left(t\right)\le \mathcal{L}\left(t\right)\le c_3\mathcal{E}\left(t\right),\forall t\ge 0,$$

(83)

and

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}^{\prime }\left(t\right)& \le & -k_1\mathcal{E}\left(t\right)+k_2{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +k_3{\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx+{\alpha }_5{\int }_0^1{u}_{xx}{\varphi }_{x}dx,\hfill \end{array} \end{gathered}$$

(84)

for some $k_1,k_2,k_3,c_2,c_3>0.$

Case 1.

If $\chi =({\displaystyle \frac{\mu }{\rho }}-{\displaystyle \frac{\delta }{J}})=0$, in this case, (84) takes the form

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}^{\prime }\left(t\right)& \le & -k_1\mathcal{E}\left(t\right)+k_2{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +k_3{\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx.\hfill \end{array} \end{gathered}$$

(85)

By multiplying (85) by $h\left(t\right)=\alpha \left(t\right).\eta \left(t\right)$, we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill h\left(t\right){\mathcal{L}}^{\prime }\left(t\right)& \le & -k_{1}h\left(t\right)\mathcal{E}\left(t\right)+k_{2}h\left(t\right){\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +k_{3}h\left(t\right){\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx.\hfill \end{array} \end{gathered}$$

(86)

We distinguish two cases

• G is linear on $[0,\epsilon ]$. In this case, using the assumption (13)₁ and (53), we can write

  $$k_{3}h\left(t\right){\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx\le k_{3}h\left(t\right){\int }_0^1{\varphi }_{t}f\left({\varphi }_t\right))dx\le -k_3\eta \left(t\right){\mathcal{E}}^{\prime }\left(t\right),$$

  (87)

  and by (11) we have

  $$\begin{gathered} \begin{array}{ccc}\hfill h\left(t\right){\int }_0^1{\int }_0^{t}g\left(p\right){\phi }_x^2\left(p\right)dpdx& =& \alpha \left(t\right){\int }_0^1{\int }_0^t\eta \left(s\right)g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & \le & -\alpha \left(t\right){\int }_0^1{\int }_0^t{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & \le & -\alpha \left(t\right){\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & \le & -2\alpha \left(t\right){\mathcal{E}}^{\prime }\left(t\right),\hfill \end{array} \end{gathered}$$

  (88)

  and by (77) we obtain

  $$\begin{gathered} \begin{array}{ccc}\hfill {\int }_0^1{\phi }_x^2\left(s\right)dx& =& 2{\int }_0^1{\varphi }_x^2(x,t)dx+2{\int }_0^1{\varphi }_x^2(x,t-s)dx\hfill \\ \\ & \le & 4\underset{s>0}{sup}{\int }_0^1{\varphi }_x^2(x,s)dx+2\underset{\tau >0}{sup}{\int }_0^1{\varphi }_{0x}^2(x,\tau )dx\hfill \\ \\ & \le & \frac{8\mathcal{E}\left(0\right)}{l}+2c_0,\hfill \end{array} \end{gathered}$$

  (89)

  then, we obtain

  $$h\left(t\right){\int }_0^1{\int }_t^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\le (\frac{8\mathcal{E}\left(0\right)}{l}+2c_0)h\left(t\right){\int }_t^{\infty }g\left(p\right)dp.$$

  (90)
  Hence,

  $$\begin{gathered} \begin{array}{ccc}\hfill h\left(t\right){\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx& \le & -2\alpha \left(t\right){\mathcal{E}}^{\prime }\left(t\right)\hfill \\ \\ & & +(\frac{8\mathcal{E}\left(0\right)}{l}+2c_0)h\left(t\right)\varpi \left(t\right).\hfill \end{array} \end{gathered}$$

  (91)
  Inserting (87) and (91) in (86). Since $h^{\prime }\left(t\right)\le 0,{\alpha }^{\prime }\left(t\right)\le 0,{\eta }^{\prime }\left(t\right)\le 0$. Then, we have

  $$\begin{array}{c}\hfill {\mathcal{L}}_1^{\prime }\left(t\right)\le -k_{1}h\left(t\right)\mathcal{E}\left(t\right)+\gamma h\left(t\right)\varpi \left(t\right),\end{array}$$

  (92)

  and

  $$m_1\mathcal{E}\left(t\right)\le {\mathcal{L}}_1\left(t\right)\le m_2\mathcal{E}\left(t\right),$$

  (93)

  with

  $$m_1={\tau }_1,m_2=c_{2}h\left(0\right)+k_3\eta \left(0\right)+2k_2\alpha \left(0\right)+{\tau }_1,$$

  where

  $$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}_1\left(t\right)& =& h\left(t\right)\mathcal{L}\left(t\right)+(k_3\eta \left(t\right)+2k_2\alpha \left(t\right)+{\tau }_1)\mathcal{E}\left(t\right)\sim \mathcal{E}\left(t\right),\hfill \\ \\ \hfill \gamma & =& (\frac{8\mathcal{E}\left(0\right)}{l}+2c_0),{\tau }_1>0and\varpi \left(t\right)={\int }_t^{\infty }g\left(p\right)dp.\hfill \end{array} \end{gathered}$$

  (94)
  Since ${\mathcal{E}}^{\prime }\left(t\right)\le 0,\forall t\ge 0$. By using (92), we have

  $$\mathcal{E}\left(T\right){\int }_0^{T}h\left(t\right)dt\le \left(\frac{{\mathcal{L}}_1\left(0\right)}{k_1}+\frac{\gamma }{k_1}{\int }_0^{T}h\left(t\right)\varpi \left(t\right)dt\right).$$

  (95)
  Using the fact that $G_0^{-1}$ is linear. Then,

  $$\mathcal{E}\left(T\right)\le \zeta G_0^{-1}\left(\frac{\frac{{\mathcal{L}}_1\left(0\right)}{k_1}+\frac{\gamma }{k_1}{\int }_0^{T}h\left(t\right)\varpi \left(t\right)dt}{{\int }_0^{T}h\left(t\right)dt}\right).$$

  (96)

  with ${\beta }_1=\zeta ,{\beta }_2=\frac{{\mathcal{L}}_1\left(0\right)}{k_1},{\beta }_3=\frac{\gamma }{k_1}$. This completes the proof.
• G is nonlinear on $[0,\epsilon ]$, we choose $0\le {\epsilon }_1\le \epsilon$ and we consider

  $$I_1\left(t\right)=\{x\in (0,1),|{\varphi }_t|\le {\epsilon }_1\},I_2=\{x\in (0,1),|{\varphi }_t|>{\epsilon }_1\},$$

  we define

  $$I={\int }_{I_1}{\varphi }_{t}f\left({\varphi }_t\right)dt.$$
  Using Jensen’s inequality and the assumption (13)₁, we have

  $$\begin{gathered} \begin{array}{ccc}\hfill k_{3}h\left(t\right){\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx& \le & k_{3}h\left(t\right){\int }_0^1{\varphi }_{t}f\left({\varphi }_t\right))dx\hfill \\ \\ & \le & k_3^{\prime }h\left(t\right)G^{-1}\left(I\left(t\right)\right)-k_3^{\prime }\eta \left(t\right){\mathcal{E}}^{\prime }\left(t\right).\hfill \end{array} \end{gathered}$$

  (97)
  Inserting (97) in (86), since ${\alpha }^{\prime }\left(t\right)\le 0,{\eta }^{\prime }\left(t\right)\le 0$ and ${\mathcal{E}}^{\prime }\left(t\right)\le 0$, we obtain

  $$\begin{array}{c}\hfill {\mathcal{L}}_2^{\prime }\left(t\right)\le -k_{1}h\left(t\right)\mathcal{E}\left(t\right)+\gamma h\left(t\right)\varpi \left(t\right)+k_3^{\prime }h\left(t\right)G^{-1}\left(I\left(t\right)\right).\end{array}$$

  (98)

  and

  $$m_3\mathcal{E}\left(t\right)\le {\mathcal{L}}_2\left(t\right)\le m_4\mathcal{E}\left(t\right),$$

  (99)

  with

  $$m_3={\tau }_1,m_4=c_{2}h\left(0\right)+k_3^{\prime }\eta \left(0\right)+2k_2\alpha \left(0\right)+{\tau }_1,$$

  where

  $$\begin{array}{ccc}\hfill {\mathcal{L}}_2\left(t\right)& =& h\left(t\right)\mathcal{L}\left(t\right)+(k_3^{\prime }\eta \left(t\right)+2k_2\alpha \left(t\right)+{\tau }_1)\mathcal{E}\left(t\right)\sim \mathcal{E}\left(t\right).\hfill \end{array}$$
  Now, for ${\epsilon }_0<{\epsilon }_1$ and by using ${\mathcal{E}}^{\prime }\left(t\right)\le 0,G^{\prime }>0$ and $G^{\prime \prime }>0$ on $(0,\epsilon ]$, we define the functional ${\mathcal{L}}_3\left(t\right)$ by,

  $$\begin{array}{ccc}\hfill {\mathcal{L}}_3\left(t\right)& =& G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right){\mathcal{L}}_2\left(t\right)+{\tau }_2\mathcal{E}\left(t\right)\sim \mathcal{E}\left(t\right),{\tau }_2>0,\hfill \end{array}$$

  satisfies

  $$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}_3^{\prime }\left(t\right)& =& {\mathcal{E}}^{\prime }\left(t\right)({\epsilon }_0{G}^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right){\mathcal{L}}_2\left(t\right)+{\tau }_2)+{\mathcal{L}}_2^{\prime }\left(t\right)G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)\hfill \\ \\ & \le & -k_{1}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)h\left(t\right)\varpi \left(t\right)\hfill \\ \\ & & +k_3^{\prime }h\left(t\right)G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)G^{-1}\left(I\left(t\right)\right).\hfill \end{array} \end{gathered}$$

  (100)
  To estimate the last term of (92), using the general Young’s inequality

  $$\begin{array}{ccc}\hfill AB& \le & G^{*}\left(A\right)+G\left(B\right),ifA\in (0,G^{\prime }\left(\epsilon \right)),B\in (0,\epsilon ),\hfill \end{array}$$

  (101)

  where

  $$\begin{array}{ccc}\hfill G^{*}\left(A\right)& =& s{\left(G^{\prime }\right)}^{-1}\left(s\right)-G\left({\left(G^{\prime }\right)}^{-1}\left(s\right)\right),ifs\in (0,G^{\prime }\left(\epsilon \right)),\hfill \end{array}$$

  satisfies

  $$\begin{array}{ccc}\hfill k_3^{\prime }h\left(t\right)G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)G^{-1}\left(I\left(t\right)\right)& \le & k_3^{\prime }{\epsilon }_{0}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)-k_3^{\prime }\eta \left(t\right){\mathcal{E}}^{\prime }\left(t\right).\hfill \end{array}$$

  (102)
  Inserting (102) in (92) and letting ${\epsilon }_0=\frac{k_1}{2k_3^{\prime }}$ we get

  $$\begin{array}{ccc}\hfill {\mathcal{L}}_3^{\prime }\left(t\right)+k_3^{\prime }\eta \left(t\right){\mathcal{E}}^{\prime }\left(t\right)& \le & -k_{1}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)h\left(t\right)\varpi \left(t\right).\hfill \end{array}$$

  (103)
  Since ${\eta }^{\prime }\left(t\right)\le 0$, then

  $$\begin{array}{ccc}\hfill {\mathcal{L}}_4^{\prime }\left(t\right)& \le & -k_{1}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)h\left(t\right)\varpi \left(t\right),\hfill \end{array}$$

  where

  $$\begin{array}{ccc}\hfill {\mathcal{L}}_4\left(t\right)& =& {\mathcal{L}}_3\left(t\right)+k_3^{\prime }\eta \left(t\right)\mathcal{E}\left(t\right)\sim \mathcal{E}\left(t\right).\hfill \end{array}$$
  Since $\alpha \left(t\right),G_0\left(\mathcal{E}\left(t\right)\right),G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)$ are non-increasing functions,
  then, for any $T>0$

  $$\begin{gathered} \begin{array}{ccc}\hfill k_1{G}_0\left(\mathcal{E}\left(T\right)\right){\int }_0^{T}h\left(t\right)dt& \le & k_1{\int }_0^{T}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)dt\hfill \\ \\ & \le & {\mathcal{L}}_4\left(0\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(0\right)\right){\int }_0^{T}h\left(t\right)\varpi \left(t\right)dt,\hfill \end{array} \end{gathered}$$

  which gives (78) with ${\beta }_1=1,{\beta }_2={\displaystyle \frac{{\mathcal{L}}_4\left(0\right)}{k_1}}$ and ${\beta }_3={\displaystyle \frac{\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(0\right)\right)}{k_1}}$.
  The proof is now completed.

Case 2.

If $\chi =(\frac{\mu }{\rho }-\frac{\delta }{J})\ne 0$ and

$$\left\{\begin{array}{c}\hfill \left|\chi \right|<{\displaystyle \frac{k_1{\mu }^{2}l}{2N_2(l\rho +b\mu )}}if\chi <0\\ \hfill \left|\chi \right|<{\displaystyle \frac{k_1{\mu }^2}{2N_2\rho }}if\chi >0.\end{array}\right.$$

This case is more important from the physical perspective, where waves are not necessarily of equal speeds. Let

$$\mathcal{E}\left(t\right)=\mathcal{E}(u,\varphi ,y,\phi )={\mathcal{E}}_1\left(t\right).$$

Denotes the first-order energy defined in (52) and

$${\mathcal{E}}_2\left(t\right)=\mathcal{E}(u_t,{\varphi }_t,y_t,{\phi }_t).$$

Denotes the second-order energy; then, we have

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{E}}_2^{\prime }\left(t\right)& \le & -{\eta }_0{\int }_0^1{\varphi }_{tt}^{2}dx+\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_{tx}^2\left(p\right)dp\hfill \\ \\ & & -{\alpha }^{\prime }\left(t\right){\int }_0^1{\varphi }_{tt}f\left({\varphi }_t\right)dx-\alpha \left(t\right){\int }_0^1{\varphi }_{tt}^2{f}^{\prime }\left({\varphi }_t\right)dx\hfill \\ \\ & =& -{\eta }_0{\int }_0^1{\varphi }_{tt}^{2}dx+\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(p\right){\phi }_{tx}^2\left(p\right)dp\hfill \\ \\ & & +\alpha \left(t\right)\left(\frac{-{\alpha }^{\prime }\left(t\right)}{\alpha \left(t\right)}{\int }_0^1{\varphi }_{tt}f\left({\varphi }_t\right)dx-{\int }_0^1{\varphi }_{tt}^2{f}^{\prime }\left({\varphi }_t\right)dx\right).\hfill \end{array} \end{gathered}$$

(104)

Since $f,g$ are non-decreasing functions, $\alpha \left(t\right)$ is a positive function and ${lim}_{t\to \infty }\frac{-{\alpha }^{\prime }\left(t\right)}{\alpha \left(t\right)}=0$, we deduce that

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{E}}_2^{\prime }\left(t\right)& \le & -{\eta }_0{\int }_0^1{\varphi }_{tt}^{2}dx+\frac{1}{2}{\int }_0^1{\int }_0^{\infty }{g}^{\prime }\left(s\right){\phi }_{tx}^2\hfill \\ \\ & \le & -{\eta }_0{\int }_0^1{\varphi }_{tt}^{2}dx,\hfill \end{array} \end{gathered}$$

(105)

where ${\eta }_0={\mu }_1-{\int }_{{\tau }_1}^{{\tau }_2}\left|{\mu }_2\left(\varrho \right)\right|d\varrho >0$.

The last term in (84), by using (16)₁, Young’s inequality and by setting $K=\frac{\chi N_2\rho }{\mu }=\frac{{\alpha }_5\rho }{\mu }$ and ${\alpha }_5=\chi N_2$ as follows

$$\begin{gathered} \begin{array}{ccc}\hfill {\alpha }_5{\int }_0^1{u}_{xx}{\varphi }_{x}dx& =& \frac{{\alpha }_5\rho }{\mu }{\int }_0^1{\varphi }_x{u}_{tt}dx-\frac{b{\alpha }_5}{\mu }{\int }_0^1{\varphi }_x^{2}dx\hfill \\ \\ & =& K\left(\frac{d}{dt}\left[{\int }_0^1{\varphi }_t{u}_{x}dx+{\int }_0^1{\varphi }_x{u}_{t}dx\right]\right)\hfill \\ \\ & & -K{\int }_0^1{u}_x{\varphi }_{tt}^{2}dx-\frac{b{\alpha }_5}{\mu }{\int }_0^1{\varphi }_x^{2}dx\hfill \\ \\ & \le & K\left(\frac{d}{dt}\left[{\int }_0^1{\varphi }_t{u}_{x}dx+{\int }_0^1{\varphi }_x{u}_{t}dx\right]\right)\hfill \\ \\ & & +\frac{\left|K\right|}{4}{\int }_0^1{\varphi }_{tt}^{2}dx+\left|K\right|{\int }_0^1{u}_x^{2}dx.\hfill \end{array} \end{gathered}$$

(106)

Let

$$\mathcal{N}\left(t\right)=\left({\int }_0^1{\varphi }_t{u}_{x}dx+{\int }_0^1{\varphi }_x{u}_{t}dx\right),$$

then (84)

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{L}}^{\prime }\left(t\right)+K{\mathcal{N}}^{\prime }\left(t\right)& \le & -k_1{\mathcal{E}}_1\left(t\right)+k_2{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_p^{2}dpdx+\frac{\left|K\right|}{4}{\int }_0^1{\varphi }_{tt}^{2}dx\hfill \\ \\ & & +\left|K\right|{\int }_0^1{u}_x^{2}dx+k_3{\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx\hfill \\ \\ & \le & -k_4{\mathcal{E}}_1\left(t\right)+k_2{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_p^{2}dpdx\hfill \\ \\ & & +\frac{\left|K\right|}{4}{\int }_0^1{\varphi }_{tt}^{2}dx+k_3{\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx,\hfill \end{array} \end{gathered}$$

(107)

where

$$k_4=k_1-2\frac{\left|K\right|}{\mu }>0.$$

Let

$$\mathcal{R}\left(t\right)=\mathcal{L}\left(t\right)+K\mathcal{N}\left(t\right)+N_5({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)).$$

(108)

Indeed, by using Young’s inequality, we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill \left|\mathcal{N}\right(t\left)\right|& =& |{\int }_0^1\varphi u_{xt}dx|+|{\int }_0^1{\varphi }_t{u}_{x}dx|\hfill \\ \\ & \le & \frac{1}{2}{\int }_0^1{u}_t^{2}dx+\frac{1}{2}{\int }_0^1{\varphi }_t^{2}dx+\frac{1}{2}{\int }_0^1{\varphi }_x^{2}dx+\frac{1}{2}{\int }_0^1{u}_x^{2}dx\hfill \\ \\ & \le & C_0{\mathcal{E}}_1\left(t\right),\hfill \end{array} \end{gathered}$$

(109)

where $C_0=max\{\frac{1}{J},\frac{1}{\xi },\frac{1}{\rho },\frac{1}{\mu }\}$.

By (83) and (109), we obtain

$$|\mathcal{R}\left(t\right)-N_5({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right))|\le (c_3+C_0){\mathcal{E}}_1\left(t\right)\le c({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)),$$

(110)

and

$$(N_5-c)({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right))\le \mathcal{R}\left(t\right)\le (N_5+c)({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)),$$

(111)

and by using (105), (107) and (3), we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{R}}^{\prime }\left(t\right)& =& {\mathcal{L}}^{\prime }\left(t\right)+K{\mathcal{N}}^{\prime }\left(t\right)+N_5({\mathcal{E}}_1^{\prime }\left(t\right)+{\mathcal{E}}_2^{\prime }\left(t\right))\hfill \\ \\ & \le & -k_4{\mathcal{E}}_1\left(t\right)+k_2{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_p^{2}dpdx\hfill \\ \\ & & +k_3{\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx-({\eta }_0{N}_5-\frac{\left|K\right|}{4}){\int }_0^1{\varphi }_{tt}^{2}dx.\hfill \end{array} \end{gathered}$$

(112)

We choose a large enough $N_5$ that

$${\eta }_0{N}_5-\frac{\left|K\right|}{4}>0,N_5-c>0,$$

we obtain

$$\mathcal{R}\left(t\right)\sim ({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)),$$

(113)

and

$$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{R}}^{\prime }\left(t\right)& \le & -k_4{\mathcal{E}}_1\left(t\right)+k_2{\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_p^{2}dpdx\hfill \\ \\ & & +k_3{\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx.\hfill \end{array} \end{gathered}$$

(114)

By multiplying (114) by $h\left(t\right)=\alpha \left(t\right).\eta \left(t\right)$, we obtain

$$\begin{gathered} \begin{array}{ccc}\hfill h\left(t\right){\mathcal{R}}^{\prime }\left(t\right)& \le & -k_{4}h\left(t\right)\mathcal{E}\left(t\right)+k_{2}h\left(t\right){\int }_0^1{\int }_0^{\infty }g\left(p\right){\phi }_x^2\left(p\right)dpdx\hfill \\ \\ & & +k_{3}h\left(t\right){\int }_0^1({\varphi }_t^2+f^2\left({\varphi }_t\right))dx.\hfill \end{array} \end{gathered}$$

(115)

We distinguish two cases

• G is linear on $[0,\epsilon ]$. In the same way as in the previous case, we obtain

  $$\begin{array}{c}\hfill {\mathcal{R}}_1^{\prime }\left(t\right)\le -k_{4}h\left(t\right)\mathcal{E}\left(t\right)+\gamma h\left(t\right)\varpi \left(t\right),\end{array}$$

  (116)

  and

  $$m_1({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right))\le {\mathcal{R}}_1\left(t\right)\le m_2({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)),$$

  (117)

  with

  $$m_1={\tau }_1,m_2=c_{2}h\left(0\right)+k_3\eta \left(0\right)+2k_2\alpha \left(0\right)+{\tau }_1,$$

  where

  $$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{R}}_1\left(t\right)& =& h\left(t\right)\mathcal{R}\left(t\right)+(k_3\eta \left(t\right)+2k_2\alpha \left(t\right)+{\tau }_1)\mathcal{E}\left(t\right)\sim ({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right))\hfill \\ \\ \hfill \gamma & =& (\frac{8\mathcal{E}\left(0\right)}{l}+2c_0),{\tau }_1>0and\varpi \left(t\right)={\int }_t^{\infty }g\left(p\right)dp.\hfill \end{array} \end{gathered}$$
  Since ${\mathcal{E}}^{\prime }\left(t\right)\le 0,\forall t\ge 0$. By using (116), we have

  $$\mathcal{E}\left(T\right){\int }_0^{T}h\left(t\right)dt\le \left(\frac{{\mathcal{R}}_1\left(0\right)}{k_4}+\frac{\gamma }{k_4}{\int }_0^{T}h\left(t\right)\varpi \left(t\right)dt\right).$$

  (118)
  Using the fact that $G_0^{-1}$ is linear. Then,

  $$\mathcal{E}\left(T\right)\le \zeta G_0^{-1}\left(\frac{\frac{{\mathcal{R}}_1\left(0\right)}{k_4}+\frac{\gamma }{k_4}{\int }_0^{T}h\left(t\right)\varpi \left(t\right)dt}{{\int }_0^{T}h\left(t\right)dt}\right),$$

  (119)

  with ${\beta }_1=\zeta ,{\beta }_2=\frac{{\mathcal{R}}_1\left(0\right)}{k_4},{\beta }_3=\frac{\gamma }{k_4}$. This completes the proof.
• G is nonlinear on $[0,\epsilon ]$, we choose $0\le {\epsilon }_1\le \epsilon$. In a similar way to that in the previous case, we have

  $$\begin{array}{c}\hfill {\mathcal{R}}_2^{\prime }\left(t\right)\le -k_{1}h\left(t\right)\mathcal{E}\left(t\right)+\gamma h\left(t\right)\varpi \left(t\right)+k_3^{\prime }h\left(t\right)G^{-1}\left(I\left(t\right)\right),\end{array}$$

  (120)

  and

  $$m_3({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right))\le {\mathcal{R}}_2\left(t\right)\le m_4({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)),$$

  (121)

  with

  $$m_3={\tau }_1,m_4=c_{2}h\left(0\right)+k_3^{\prime }\eta \left(0\right)+2k_2\alpha \left(0\right)+{\tau }_1,$$

  where

  $$\begin{array}{ccc}\hfill {\mathcal{R}}_2\left(t\right)& =& h\left(t\right)\mathcal{R}\left(t\right)+(k_3^{\prime }\eta \left(t\right)+2k_2\alpha \left(t\right)+{\tau }_1)\mathcal{E}\left(t\right)\sim ({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)).\hfill \end{array}$$
  Now, for ${\epsilon }_0<{\epsilon }_1$, and by using ${\mathcal{E}}^{\prime }\left(t\right)\le 0,G^{\prime }>0$ and $G^{\prime \prime }>0$ on $(0,\epsilon ]$, we define the functional ${\mathcal{L}}_3\left(t\right)$ by,

  $$\begin{array}{ccc}\hfill {\mathcal{R}}_3\left(t\right)& =& G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right){\mathcal{R}}_2\left(t\right)+{\tau }_2\mathcal{E}\left(t\right)\sim ({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)),{\tau }_2>0,\hfill \end{array}$$

  satisfies

  $$\begin{gathered} \begin{array}{ccc}\hfill {\mathcal{R}}_3^{\prime }\left(t\right)& =& {\mathcal{E}}^{\prime }\left(t\right)({\epsilon }_0{G}^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right){\mathcal{R}}_2\left(t\right)+{\tau }_2)+{\mathcal{R}}_2^{\prime }\left(t\right)G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)\hfill \\ \\ & \le & -k_{4}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)h\left(t\right)\varpi \left(t\right)\hfill \\ \\ & & +k_3^{\prime }h\left(t\right)G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)G^{-1}\left(I\left(t\right)\right).\hfill \end{array} \end{gathered}$$

  (122)
  To estimate the last term of (122), again using the general Young’s inequality (101).
  Inserting (122) in (121) and letting ${\epsilon }_0=\frac{k_1}{2k_3^{\prime }}$, we get

  $$\begin{array}{ccc}\hfill {\mathcal{R}}_3^{\prime }\left(t\right)+k_3^{\prime }\eta \left(t\right){\mathcal{E}}^{\prime }\left(t\right)& \le & -k_{4}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)h\left(t\right)\varpi \left(t\right).\hfill \end{array}$$

  (123)
  Since ${\eta }^{\prime }\left(t\right)\le 0$, then

  $$\begin{array}{ccc}\hfill {\mathcal{R}}_4^{\prime }\left(t\right)& \le & -k_{4}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)h\left(t\right)\varpi \left(t\right),\hfill \end{array}$$

  where

  $$\begin{array}{ccc}\hfill {\mathcal{R}}_4\left(t\right)& =& {\mathcal{R}}_3\left(t\right)+k_3^{\prime }\eta \left(t\right)\mathcal{E}\left(t\right)\sim ({\mathcal{E}}_1\left(t\right)+{\mathcal{E}}_2\left(t\right)).\hfill \end{array}$$
  Since $\alpha \left(t\right),G_0\left(\mathcal{E}\left(t\right)\right),G^{\prime }\left({\epsilon }_0\mathcal{E}\left(t\right)\right)$ are non-increasing functions, then, for any $T>0$

  $$\begin{array}{ccc}\hfill k_4{G}_0\left(E\left(T\right)\right){\int }_0^{T}h\left(t\right)dt& \le & k_4{\int }_0^{T}h\left(t\right)G_0\left(\mathcal{E}\left(t\right)\right)dt\hfill \\ & \le & {\mathcal{R}}_4\left(0\right)+\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(0\right)\right){\int }_0^{T}h\left(t\right)\varpi \left(t\right)dt,\hfill \end{array}$$

  which gives (78) with ${\beta }_1=1,{\beta }_2=\frac{{\mathcal{R}}_4\left(0\right)}{k_4}$ and ${\beta }_3=\frac{\gamma G^{\prime }\left({\epsilon }_0\mathcal{E}\left(0\right)\right)}{k_4}$.
  The proof is completed.

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