General decay of a nonlinear damping porous-elastic system with past history

Abstract

In this paper, we consider a one-dimensional porous-elastic system with past history and nonlinear damping term. We established the well-posedness using the semigroup theory and we showed that the dissipation given by this complementary controls guarantees the general stability for the case of equal speed of wave propagation.

1 Introduction

In the present work, we consider the following porous elastic system with past history and nonlinear damping term

$$\begin{aligned} \left\{ \begin{array}{l} \rho u_{tt}-\mu u_{xx}-b\phi _{x}=0, x\in \left( 0,1\right) , t>0, \\ J\phi _{tt}-\delta \phi _{xx}+bu_{x}+\xi \phi +\int \nolimits _{0}^{\infty }g\left( s\right) \phi _{xx}\left( t-s\right) ds \\ +\chi \left( t\right) f\left( \phi _{t}\right) =0, x\in \left( 0,1\right) , t>0, \\ u\left( x,0\right) =u_{0}\left( x\right) ,\ u_{t}\left( x,0\right) =u_{1}\left( x\right) , x\in \left( 0,1\right) , \\ \phi \left( x,0\right) =\phi _{0}\left( x\right) ,\ \phi _{t}\left( x,0\right) =\phi _{1}\left( x\right) , x\in \left( 0,1\right) , \\ u_{x}\left( 0,t\right) =u_{x}\left( 1,t\right) =\phi \left( 0,t\right) =\phi \left( 1,t\right) =0, t>0, \end{array} \right. \end{aligned}$$

(1.1)

where the functions u and $\phi$ represent respectively the longitudinal displacement and the volume fraction. The parameter $\rho$ designates the mass density and J equals to the product of the mass density by the equilibrated inertia. The term $\chi \left( t\right) f\left( \phi _{t}\right)$ is the nonlinear damping term where $\chi$ is a positive non-increasing differentiable function and f is specified in the preliminaries, the integral represents the infinite memory term and g is the relaxation function which satisfies

$$\begin{aligned} {g^{\prime }(t)\le -\alpha (t)g(t),\ t\ge 0,} \end{aligned}$$

(1.2)

where ${\alpha }$ is a positive non-increasing differentiable function. The parameters $\mu ,$b,  $\xi ,$$\delta$ are positive constitutive constants such that

$$\begin{aligned} \mu \xi >b^{2}. \end{aligned}$$

(1.3)

The original motivation of this problem was introduced by Goodman and Cowin [7] in 1972 when they proposed the idea of introducing the concept of a continuum theory of granular materials with interstitial voids into the theory of elastic solids with voids. This idea gives the relation between the elasticity theory and the porous media theory, for more details we cite the works of Cowin and Nunziato [5] from 1983 and Cowin [4] from 1985.

The system (1.1) was constructed by considering the following two basic evolution equations of the one-dimensional porous materials theory

$$\begin{aligned} \rho u_{tt}=T_{x}, J\phi _{tt}=H_{x}+D, \end{aligned}$$

(1.4)

where T,  H and D represent respectively the stress tensor, the equilibrated stress vector and the equilibrated body force. Consequently, to get the system (1.1) we take the constitutive equations T,  H and D in this form

$$\begin{aligned} T= & {} \mu u_{x}+b\phi , H=\delta \phi _{x}-\int _{0}^{\infty }g(s)\phi _{x}\left( t-s\right) ds, \nonumber \\ D= & {} -bu_{x}-\xi \phi -\chi \left( t\right) f\left( \phi _{t}\right) , \end{aligned}$$

(1.5)

and by combining (1.5) and (1.4), we obtain (1.1).

In [16], Quintanilla gave a result concerning the slow decay for a one-dimensional porous dissipation elasticity, after Apalara [2] showed an exponential stability of the same system considered in [16] under the hypothesis (1.7).

In [1], Apalara established a general decay result for the energy of the same problem considered in [2, 16] where he replaced the porous dissipation by a non linear damping term as follows

$$\begin{aligned} \left\{ \begin{array}{l} \rho u_{tt}-\mu u_{xx}-b\phi _{x}=0, \\ J\phi _{tt}-\delta \phi _{xx}+bu_{x}+\xi \phi +\chi \left( t\right) f\left( \phi _{t}\right) =0. \end{array} \right. \end{aligned}$$

(1.6)

Note that, if we consider the problem (1.6) with viscoelasticity $\left( -\gamma u_{txx}\right)$ that is acting only on the first equation and $\chi \left( t\right) =0,$ we come across the work of Magańa and Quintanilla [14] where they showed that viscoelasticity is not strong enough to make the solutions decay in an exponential way. If we consider the same problem (1.6) with $\chi \left( t\right) =1,$ we refer to the work of Boussouira [3] in the case of Timoshenko system where the authors established a general semi explicit decay of the system.

The purpose of this paper is to study the well posedness and the asymptotic behavior of the solution of (1.6) with past history term when this last is acting only on the second equation. We prove the general decay of this system for the case of equal speed of wave propagation in both equations of the system, that is

$$\begin{aligned} \frac{\mu }{\rho }=\frac{\delta }{J}. \end{aligned}$$

(1.7)

Introducing a function $\chi \left( t\right)$ in the nonlinear damping term and $\alpha \left( t\right)$ which satisfies (1.2) makes our problem different from those considered so far in the literature.

The importance of the past history term and its influence on the asymptotic behavior of the solution appears in many works for different types of problems. To learn more about this term we refer the readers to [6, 8,9,10,11, 13, 17] in the case of Timoshenko system, thermoelastic Laminated Beam and the transmission problem.

The paper is organized as follows. In Sect. 2, we introduced some transformations and assumptions needed to prove the main result. In Sect. 3, we used the semigroup method to prove the well-posedness of problem (1.1). In Sect. 4, we considered several lemmas that help us to construct the Lyapunov functional. In Sect. 5, we proved our general stability result.

2 Preliminaries

In this section we present the backgrounds mathematics needed later to prove our main result. We shall use the following hypothesis

• (H1) $g: {\mathbb {R}} _{+}\rightarrow {\mathbb {R}} _{+}$ is a $C^{1}$ function satisfying

  $$\begin{aligned} g(0)>0,\quad \delta -\int _{0}^{\infty }g(s)ds=l>0,\quad \int _{0}^{\infty }g(s)ds=g_{0}. \end{aligned}$$

  (2.1)

• (H2) $f: {\mathbb {R}} \rightarrow {\mathbb {R}}$ is a non-decreasing $C^{0}$-function such that there exist the positive constants $\nu _{1},$$\nu _{2,}$$\epsilon$ and a strictly increasing function $G\in C^{1}\left( \left[ 0,\infty \right) \right) ,$ with $G(0)=0.$ Moreover, G is linear or strictly convex $C^{2}-$function on $\left( 0,\epsilon \right]$ such that

  $$\begin{aligned} \left\{ \begin{array}{c} s^{2}+f^{2}(s)\le G^{-1}\left( sf\left( s\right) \right) ,\forall \left| s\right| \le \epsilon , \\ \nu _{1}\left| s\right| \le \left| f(s)\right| \le \nu _{2}\left| s\right| , \forall \left| s\right| \ge \epsilon , \end{array} \right. \end{aligned}$$

  (2.2)

  which implies that $sf(s)>0$, for all $s\ne 0.$

• (H3) The function f satisfies the following property

  $$\begin{aligned} \left| {f(\psi }_{2}{)-f(\psi _{1})}\right| \le {k_{0}(|\psi }_{1}{ |^{\varrho }+|\psi }_{2}{|^{\varrho })|{\psi }_{1}-{\psi }_{2}|,\quad {\psi } _{1},{\psi }_{2}\in {\mathbb {R}} ,} \end{aligned}$$

  (2.3)

  where $k_{0}>0$, $\varrho >0.$

Note that the hypothesis (H2) was first introduced by Lasiecka and Tataru [12] in 1993.

Consider the following inequalities, that will help us in some estimations; we omit their proof.

Lemma 1

The following inequalities hold,

$$\begin{aligned} \int _{0}^{1}\left( \int _{0}^{\infty }g(s)\left( \phi (t)-\phi (t-s)\right) ds\right) ^{2}dx&\le d_{1}\left( g\circ \phi _{x}\right) \left( t\right) , \end{aligned}$$

(2.4)

$$\begin{aligned} \int _{0}^{1}\left( \int _{0}^{\infty }g^{\prime }(s)\left( \phi _{x}(t)-\phi _{x}(t-s)\right) ds\right) ^{2}dx&\le -g\left( 0\right) \left( g^{\prime }\circ \phi _{x}\right) \left( t\right) , \end{aligned}$$

(2.5)

$$\begin{aligned} \int _{0}^{1}\left( \int _{0}^{\infty }g(s)\left( \phi _{x}(t)-\phi _{x}(t-s)\right) ds\right) ^{2}dx&\le g_{0}\left( g\circ \phi _{x}\right) \left( t\right) , \end{aligned}$$

(2.6)

$$\begin{aligned} \int _{0}^{1}\left( \int _{0}^{\infty }g^{\prime }(s)\left( \phi (t)-\phi (t-s)\right) ds\right) ^{2}dx&\le -d_{2}(g^{\prime }\circ \phi _{x})(t), \end{aligned}$$

(2.7)

where $d_{1}$, $d_{2}$are positive constants and

$$\begin{aligned} (g\circ \nu )(t)=\int _{0}^{1}\int _{0}^{\infty }g(s)(\nu (x,t)-\nu (x,t-s))^{2}\,ds\,dx. \end{aligned}$$

Here are some notations that will help us for the computation of energy

$$\begin{aligned} \eta ^{t}(x,s)=\phi (x,t)-\phi (x,t-s),\;(x,t,s)\in (0,1)\times {\mathbb {R}} _{+}\times {\mathbb {R}} _{+}, \end{aligned}$$

which was adopted in articles [6, 15]. Here $\eta ^{t}$ is the relative history of $\phi$ and verifies

$$\begin{aligned}&\eta _{t}^{t}+\eta _{s}^{t}-\phi _{t}=0,\ \left( x,t,s\right) \in (0,1)\times {\mathbb {R}} _{+}\times {\mathbb {R}} _{+}. \nonumber \\&\eta ^{t}\left( 0,s\right) =\eta ^{t}\left( 1,s\right) =0, t,s>0 \nonumber \\&\eta ^{t}(x,0)=0,\ \eta ^{0}(x,s)=\eta _{0}(x,s), x\in \left( 0,1\right) , t>0, \end{aligned}$$

(2.8)

then, the system (1.1) is equivalent to

$$\begin{aligned} \left\{ \begin{array}{l} \rho u_{tt}-\mu u_{xx}-b\phi _{x}=0, x\in \left( 0,1\right) , t>0, \\ J\phi _{tt}-\delta \phi _{xx}+bu_{x}+\xi \phi +\int \nolimits _{0}^{\infty }g\left( s\right) \phi _{xx}\left( t-s\right) ds \\ +\chi \left( t\right) f\left( \phi _{t}\right) =0, x\in \left( 0,1\right) , t>0, \\ \eta _{t}^{t}+\eta _{s}^{t}=\phi _{t},\ s,t>0, \\ u\left( x,0\right) =u_{0}\left( x\right) ,\ u_{t}\left( x,0\right) =u_{1}\left( x\right) , x\in \left( 0,1\right) , \\ \phi \left( x,0\right) =\phi _{0}\left( x\right) ,\ \phi _{t}\left( x,0\right) =\phi _{1}\left( x\right) , x\in \left( 0,1\right) , \\ \eta ^{t}(x,0)=0,\ \eta ^{0}(x,s)=\eta _{0}(x,s), x\in \left( 0,1\right) , s>0 \\ u_{x}\left( 0,t\right) =u_{x}\left( 1,t\right) =\phi \left( 0,t\right) =\phi \left( 1,t\right) =\eta ^{t}\left( 0,s\right) =\eta ^{t}\left( 1,s\right) =0, s,t>0. \end{array} \right. \end{aligned}$$

(2.9)

In order to be able to use Poincaré’s inequality for u , we introduce

$$\begin{aligned} {\bar{u}}\left( x,t\right) =u\left( x,t\right) -t\int _{0}^{1}u_{1}\left( x\right) dx-\int _{0}^{1}u_{0}\left( x\right) dx. \end{aligned}$$

Using (2.9)$_{1},$ we have

$$\begin{aligned} \int _{0}^{1}{\bar{u}}\left( x,t\right) dx=0, \forall t\ge 0. \end{aligned}$$

In what follows, we will work with ${\bar{u}}$ but, for convenience, we write u instead of ${\bar{u}}$.

3 Well-posedness

In this section, we give the existence and uniqueness result for problem (2.9) using the semigroup theory. First, we introduce the vector function

$$\begin{aligned} \Phi =\left( u,u_{t},\phi ,\phi _{t},\eta ^{t}\right) ^{T}, \end{aligned}$$

and the two new dependent variables

$$\begin{aligned} v=u_{t}, \psi =\phi _{t}. \end{aligned}$$

Note that, the second equation of (2.9) can be rewritten as follows

$$\begin{aligned} J\phi _{tt}-l\phi _{xx}+bu_{x}+\xi \phi -\int \nolimits _{0}^{\infty }g\left( s\right) \eta _{xx}^{t}\left( x,s\right) ds+\chi \left( t\right) f\left( \phi _{t}\right) =0, \end{aligned}$$

then the system (2.9) is equivalent to

$$\begin{aligned} \left\{ \begin{array}{l} \dfrac{\partial }{\partial t}\Phi +{\mathcal {A}}\Phi =\Gamma \left( \Phi \right) , \\ \Phi \left( x,0\right) =\Phi _{0}\left( x\right) =\left( u_{0},u_{1},\phi _{0},\phi _{1},\eta _{0}\right) ^{T}, \end{array} \right. \end{aligned}$$

(3.1)

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

$$\begin{aligned} {\mathcal {A}}\Phi= & {} \left( \begin{array}{c} -v \\ -\frac{\mu }{\rho }u_{xx}-\frac{b}{\rho }\phi _{x} \\ -\psi \\ -\frac{l}{J}\phi _{xx}+\frac{b}{J}u_{x}+\frac{\xi }{J}\phi -\frac{1}{J} \int \nolimits _{0}^{\infty }g\left( s\right) \eta _{xx}^{t}\left( x,s\right) ds \\ \eta _{s}^{t}-\psi \end{array} \right) , \\ \Gamma \left( \Phi \right)= & {} \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ -\frac{\chi \left( t\right) }{J}f\left( \psi \right) \\ 0 \end{array} \right) , \end{aligned}$$

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

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

such that

$$\begin{aligned} H_{*}^{1}\left( 0,1\right)&=H^{1}\left( 0,1\right) \cap L_{*}^{2}\left( 0,1\right) , \\ L_{*}^{2}\left( 0,1\right)&=\left\{ \varphi \in L^{2}\left( 0,1\right) :\int _{0}^{1}\varphi \left( x\right) dx=0\right\} , \\ L_{g}&=\left\{ \varphi : {\mathbb {R}} _{+}\longrightarrow H_{0}^{1}\left( 0,1\right) ,\int _{0}^{1}\int \nolimits _{0}^{\infty }g\left( s\right) \varphi _{x}^{2}dsdx<\infty \right\} , \end{aligned}$$

the space $L_{g}$ is endowed with the following inner product

$$\begin{aligned} \left\langle \varphi _{1,}\varphi _{2}\right\rangle _{L_{g}}=\int _{0}^{1}\int \nolimits _{0}^{\infty }g\left( s\right) \varphi _{1x}\left( s\right) \varphi _{2x}\left( s\right) dsdx. \end{aligned}$$

For any $\Phi =\left( u,v,\phi ,\psi ,\eta ^{t}\right) ^{T}\in {\mathcal {H}},$${\tilde{\Phi }}=\left( {\tilde{u}},{\tilde{v}},{\tilde{\phi }},{\tilde{\psi }},\tilde{ \eta }^{t}\right) ^{T}\in {\mathcal {H}},$ we equip ${\mathcal {H}}$ with the inner product defined by

$$\begin{aligned} \left\langle \Phi ,{\tilde{\Phi }}\right\rangle _{{\mathcal {H}}}&=\rho \int _{0}^{1}v{\tilde{v}}dx+\mu \int _{0}^{1}u_{x}{\tilde{u}}_{x}dx+J\int _{0}^{1} \psi {\tilde{\psi }}dx+b\int _{0}^{1}\left( u_{x}{\tilde{\phi }}+{\tilde{u}}_{x}\phi \right) dx \\&\quad +\xi \int _{0}^{1}\phi {\tilde{\phi }}dx+l\int _{0}^{1}\phi _{x}{\tilde{\phi }} _{x}dx+\left\langle \eta ^{t},{\tilde{\eta }}^{t}\right\rangle _{_{L_{g}}}. \end{aligned}$$

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

$$\begin{aligned} D\left( {\mathcal {A}}\right)&=\left\{ \Phi \in {\mathcal {H}}\mid u\in H_{*}^{2}\left( 0,1\right) \cap H_{*}^{1}\left( 0,1\right) ;\ \phi \in H^{2}\left( 0,1\right) \cap H_{0}^{1}\left( 0,1\right) ;\right. \\&\quad \left. v\in H_{*}^{1}\left( 0,1\right) ;\ \psi \in H_{0}^{1}\left( 0,1\right) ;\ \eta ^{t}\in L_{g}\right\} , \end{aligned}$$

where

$$\begin{aligned} H_{*}^{2}\left( 0,1\right) =\left\{ \varphi \in H^{2}\left( 0,1\right) :\varphi _{x}\left( 0\right) =\varphi _{x}\left( 1\right) =0\right\} . \end{aligned}$$

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

Remark 1

Note that, the inner product $\left\langle \Phi ,\Phi \right\rangle _{{\mathcal {H}}}$ is positive. Indeed

$$\begin{aligned} \left\langle \Phi ,\Phi \right\rangle _{{\mathcal {H}}}&=\rho \int _{0}^{1}v^{2}dx+\mu \int _{0}^{1}u_{x}^{2}dx+J\int _{0}^{1}\psi ^{2}dx+2b\int _{0}^{1}u_{x}\phi dx \\&\quad +\xi \int _{0}^{1}\phi ^{2}dx+l\int _{0}^{1}\phi _{x}^{2}dx+\int \limits _{0}^{1}\int \limits _{0}^{\infty }g\left( s\right) \left( \eta _{x}^{t}\left( x,s\right) \right) ^{2}dsdx, \end{aligned}$$

and it can easily be verified that

$$\begin{aligned} \mu u_{x}^{2}+2bu_{x}\phi +\xi \phi ^{2}&=\frac{1}{2}\left[ \mu \left( u_{x}+\frac{b}{\mu }\phi \right) ^{2}+\xi \left( \phi +\frac{b}{\xi } u_{x}\right) ^{2}\right. \\&\quad \left. +\left( \mu -\frac{b^{2}}{\xi }\right) u_{x}^{2}+\left( \xi -\frac{ b^{2}}{\mu }\right) \phi ^{2}\right] , \end{aligned}$$

thanks to (1.3), we conclude that

$$\begin{aligned} \left\langle \Phi ,\Phi \right\rangle _{{\mathcal {H}}}&\ge \rho \int _{0}^{1}v^{2}dx+J\int _{0}^{1}\psi ^{2}dx+\xi _{1}\int _{0}^{1}\phi ^{2}dx+\mu _{1}\int _{0}^{1}u_{x}^{2}dx \\&\quad +l\int _{0}^{1}\phi _{x}^{2}dx+\int \limits _{0}^{1}\int \limits _{0}^{\infty }g\left( s\right) \left( \eta _{x}^{t}\left( x,s\right) \right) ^{2}dsdx, \end{aligned}$$

where $\xi _{1}=\dfrac{1}{2}\left( \xi -\dfrac{b^{2}}{\mu }\right) >0$ and $\mu _{1}=\dfrac{1}{2}\left( \mu -\dfrac{b^{2}}{\xi }\right) >0.$

Theorem 1

Let $\Phi _{0}\in {\mathcal {H}}$ and assume that $\left( H_{1}\right) -$$\left( H_{3}\right)$ hold. Then, there exists a unique solution $\Phi \in C\left( {\mathbb {R}} _{+},{\mathcal {H}}\right)$ of problem (3.1). Moreover, if $\Phi _{0}\in D\left( {\mathcal {A}}\right)$ then

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

Proof

We use the semigroup approach. It is sufficient to show that ${\mathcal {A}}$ is a maximal monotone operator. First, we give the expression of $\left\langle {\mathcal {A}}\Phi ,\Theta \right\rangle _{{\mathcal {H}}}$ for any $\Phi =\left( u,v,\phi ,\psi ,\eta ^{t}\right) ^{T}\in {\mathcal {H}},$$\Theta =\left( \Theta _{1},\Theta _{2},\Theta _{3},\Theta _{4},\Theta _{5}\right) ^{T}\in {\mathcal {H}}$. Then, by a simple calculation using the integration by parts, we have

$$\begin{aligned} \left\langle {\mathcal {A}}\Phi ,\Theta \right\rangle _{{\mathcal {H}}}= & {} \mu \int \limits _{0}^{1}\left( u_{x}\Theta _{2x}-v_{x}\Theta _{1x}\right) dx+b\int \limits _{0}^{1}\left( \phi \Theta _{2x}-v_{x}\Theta _{3}\right) dx\\&+\,l\int \limits _{0}^{1}\left( \psi \Theta _{3xx}-\phi _{xx}\Theta _{4}\right) dx+\xi \int \limits _{0}^{1}\left( \phi \Theta _{4}-\psi \Theta _{3}\right) dx \\&+\,b\int \limits _{0}^{1}\left( u_{x}\Theta _{4}-\psi \Theta _{1x}\right) dx-\int \limits _{0}^{1}\Theta _{4}\int \limits _{0}^{\infty }g\left( s\right) \eta _{xx}^{t}\left( x,s\right) dsdx \\&+\int \limits _{0}^{1}\int \limits _{0}^{\infty }g\left( s\right) \left( \eta _{s}^{t}\left( x,s\right) -\psi \right) _{x}\Theta _{5x}dsdx. \end{aligned}$$

Therefore, using the integration by parts and the boundary conditions, we can conclude that

$$\begin{aligned} \left\langle {\mathcal {A}}\Phi ,\Phi \right\rangle _{{\mathcal {H}}}&=-\int _{0}^{1}\psi \int \nolimits _{0}^{\infty }g\left( s\right) \eta _{xx}^{t}\left( x,s\right) dsdx \\&\quad +\int \limits _{0}^{1}\int \limits _{0}^{\infty }g\left( s\right) \left( \eta _{s}^{t}\left( x,s\right) -\psi \right) _{x}\eta _{x}^{t}\left( x,s\right) dsdx \\&=\int \limits _{0}^{1}\int \limits _{0}^{\infty }g\left( s\right) \eta _{sx}^{t}\left( x,s\right) \eta _{x}^{t}\left( x,s\right) dsdx. \end{aligned}$$

Again, integrating by parts with respect to s and using the fact that $\eta _{x}^{t}\left( x,0\right) =0,$ we obtain (see also Lemma 1)

$$\begin{aligned} \left\langle {\mathcal {A}}\Phi ,\Phi \right\rangle _{{\mathcal {H}}}= & {} \int \limits _{0}^{\infty }g\left( s\right) \int \limits _{0}^{1}\eta _{sx}^{t}\left( x,s\right) \eta _{x}^{t}\left( x,s\right) dxds \\= & {} -\frac{1}{2}\int \limits _{0}^{1}\int \limits _{0}^{\infty }g^{\prime }\left( s\right) \left( \eta _{x}^{t}\left( x,s\right) \right) ^{2}dsdx \\= & {} -\frac{1}{2}\left( g^{\prime }\circ \phi _{x}\right) \left( t\right) \ge 0. \end{aligned}$$

Thus, ${\mathcal {A}}$ is monotone. Next, we prove that the operator $\left( I+ {\mathcal {A}}\right)$ is surjective. Given $K=\left( k_{1},k_{2},k_{3},k_{4},k_{5}\right) ^{T}\in {\mathcal {H}}$, we prove that there exists a unique $\Phi \in D\left( {\mathcal {A}}\right)$ such that

$$\begin{aligned} \left( I+{\mathcal {A}}\right) \Phi =K. \end{aligned}$$

(3.2)

That is,

$$\begin{aligned} \left\{ \begin{array}{l} u-v=k_{1}\in H_{*}^{1}\left( 0,1\right) , \\ \rho v-\mu u_{xx}-b\phi _{x}=\rho k_{2}\in L_{*}^{2}\left( 0,1\right) , \\ \phi -\psi =k_{3}\in H_{0}^{1}\left( 0,1\right) , \\ J\psi -l\phi _{xx}+bu_{x}+\xi \phi -\int \nolimits _{0}^{\infty }g\left( s\right) \eta _{xx}^{t}\left( x,s\right) ds=Jk_{4}\in L^{2}\left( 0,1\right) , \\ \eta ^{t}+\eta _{s}^{t}-\psi =k_{5}\in L_{g}. \end{array} \right. \end{aligned}$$

(3.3)

Using (3.3)$_{5}$, we obtain

$$\begin{aligned} \eta ^{t}=e^{-s}\int _{0}^{s}e^{\varsigma }\left( \psi +k_{5}\left( \varsigma \right) \right) d\varsigma . \end{aligned}$$

(3.4)

Inserting $u-v=k_{1},$$\phi -\psi =k_{3}$ and (3.4) in (3.3)$_{2}$ and (3.3)$_{4},$ we obtain

$$\begin{aligned} \left\{ \begin{array}{l} \rho u-\mu u_{xx}-b\phi _{x}=h_{1}\in L_{*}^{2}\left( 0,1\right) , \\ \left( J+\xi \right) \phi +bu_{x}-\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \phi _{xx}=h_{2}\in L^{2}\left( 0,1\right) , \end{array} \right. \end{aligned}$$

(3.5)

where

$$\begin{aligned} \left\{ \begin{array}{l} h_{1}=\rho k_{2}+\rho k_{1}, \\ h_{2}=Jk_{4}+Jk_{3}+\int _{0}^{\infty }g\left( s\right) e^{-s}\int _{0}^{s}e^{\varsigma }\left( k_{5}-k_{3}\right) _{xx}d\varsigma ds. \end{array} \right. \end{aligned}$$

To solve (3.5), we consider the following variational formulation

$$\begin{aligned} B\left( \left( u,\phi \right) ,\left( u_{1},\phi _{1}\right) \right) = {\mathcal {G}}\left( u_{1},\phi _{1}\right) , \end{aligned}$$

(3.6)

where $B:\left[ H_{*}^{1}\left( 0,1\right) \times H_{0}^{1}\left( 0,1\right) \right] ^{2}\longrightarrow {\mathbb {R}}$ is the bilinear form defined by

$$\begin{aligned} B\left( \left( u,\phi \right) ,\left( u_{1},\phi _{1}\right) \right)&=\rho \int _{0}^{1}uu_{1}dx+\mu \int _{0}^{1}u_{x}u_{1x}dx+\left( J+\xi \right) \int _{0}^{1}\phi \phi _{1}dx \nonumber \\&\quad +b\int _{0}^{1}\left( u_{x}\phi _{1}+\phi u_{1x}\right) dx \nonumber \\&\quad +\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \int _{0}^{1}\phi _{x}\phi _{1x}dx, \end{aligned}$$

(3.7)

and ${\mathcal {G}}:\left[ H_{*}^{1}\left( 0,1\right) \times H_{0}^{1}\left( 0,1\right) \right] \longrightarrow {\mathbb {R}}$ is the linear functional given by

$$\begin{aligned} {\mathcal {G}}\left( u_{1},\phi _{1}\right) =\int _{0}^{1}h_{1}u_{1}dx+\int _{0}^{1}h_{2}\phi _{1}dx. \end{aligned}$$

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

$$\begin{aligned} \left\| \left( u,\phi \right) \right\| _{V}^{2}=\left\| u\right\| _{2}^{2}+\left\| \phi \right\| _{2}^{2}+\left\| u_{x}\right\| _{2}^{2}+\left\| \phi _{x}\right\| _{2}^{2}, \end{aligned}$$

we have

$$\begin{aligned} B\left( \left( u,\phi \right) ,\left( u,\phi \right) \right)&=\rho \int _{0}^{1}u^{2}dx+\mu \int _{0}^{1}u_{x}^{2}dx+\left( J+\xi \right) \int _{0}^{1}\phi ^{2}dx+2b\int _{0}^{1}\phi u_{x}dx \\&\quad +\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \int _{0}^{1}\phi _{x}^{2}dx. \end{aligned}$$

On the other hand, we can write

$$\begin{aligned} \mu u_{x}^{2}+2bu_{x}\phi +\left( J+\xi \right) \phi ^{2}&=\frac{1}{2}\left[ \mu \left( u_{x}+\frac{b}{\mu }\phi \right) ^{2}+\left( J+\xi \right) \left( \phi +\frac{b}{J+\xi }u_{x}\right) ^{2}\right. \\&\quad \left. +\left( \mu -\frac{b^{2}}{J+\xi }\right) u_{x}^{2}+\left( J+\xi - \frac{b^{2}}{\mu }\right) \phi ^{2}\right] , \end{aligned}$$

by using (1.3), we deduce that

$$\begin{aligned} B\left( \left( u,\phi \right) ,\left( u,\phi \right) \right)&\ge \rho \int _{0}^{1}u^{2}dx+\kappa _{1}\int _{0}^{1}u_{x}^{2}dx+\kappa _{2}\int _{0}^{1}\phi ^{2}dx \\&\quad +\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \int _{0}^{1}\phi _{x}^{2}dx, \end{aligned}$$

where

$$\begin{aligned} \kappa _{1}=\frac{1}{2}\left( \mu -\frac{b^{2}}{J+\xi }\right)>0, \kappa _{2}=\frac{1}{2}\left( J+\xi -\frac{b^{2}}{\mu }\right) >0, \end{aligned}$$

then, for some $M_{0}>0$

$$\begin{aligned} \left| B\left( \left( u,\phi \right) ,\left( u,\phi \right) \right) \right|&\ge \rho \left\| u\right\| _{2}^{2}+\kappa _{1}\left\| u_{x}\right\| _{2}^{2}+\kappa _{2}\left\| \phi \right\| _{2}^{2} \\&\quad +\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \left\| \phi _{x}\right\| _{2}^{2} \\&\ge M_{0}\left( \left\| u\right\| _{2}^{2}+\left\| \phi \right\| _{2}^{2}+\left\| u_{x}\right\| _{2}^{2}+\left\| \phi _{x}\right\| _{2}^{2}\right) \\&=M_{0}\left\| \left( u,\phi \right) \right\| _{V}^{2}. \end{aligned}$$

Thus, B is coercive. On the other hand, by using Cauchy-Schwarz and Poincaré’s inequalities, we obtain

$$\begin{aligned}&\left| B\left( \left( u,\phi \right) ,\left( u_{1},\phi _{1}\right) \right) \right| \\&\quad \le \rho \left\| u\right\| _{2}\left\| u_{1}\right\| _{2}+\mu \left\| u_{x}\right\| _{2}\left\| u_{1x}\right\| _{2}+\left( J+\xi \right) \left\| \phi \right\| _{2}\left\| \phi _{1}\right\| _{2} \\&\qquad +b\left( \left\| u_{x}\right\| _{2}\left\| \phi _{1}\right\| _{2}+\left\| \phi \right\| _{2}\left\| u_{1x}\right\| _{2}\right) \\&\qquad +\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \left\| \phi _{x}\right\| _{2}\left\| \phi _{1x}\right\| _{2} \\&\quad \le \zeta _{1}\left( \left\| u\right\| _{2}+\left\| \phi \right\| _{2}+\left\| u_{x}\right\| _{2}+\left\| \phi _{x}\right\| _{2}\right) \times \left( \left\| u_{1}\right\| _{2}+\left\| \phi _{1}\right\| _{2}+\left\| u_{1x}\right\| _{2}+\left\| \phi _{1x}\right\| _{2}\right) \\&\quad \le 16\zeta _{1}\left\| \left( u,\phi \right) \right\| _{V}\left\| \left( u_{1},\phi _{1}\right) \right\| _{V}. \end{aligned}$$

Similarly, we can show that

$$\begin{aligned} \left| {\mathcal {G}}\left( u_{1},\phi _{1}\right) \right| \le \zeta _{2}\left\| \left( u_{1},\phi _{1}\right) \right\| _{V}. \end{aligned}$$

Consequently, by the Lax-Milgram Lemma, the system (3.5) has a unique solution

$$\begin{aligned} \left( u,\phi \right) \in H_{*}^{1}\left( 0,1\right) \times H_{0}^{1}\left( 0,1\right) , \end{aligned}$$

satisfying

$$\begin{aligned} B\left( \left( u,\phi \right) ,\left( u_{1},\phi _{1}\right) \right) = {\mathcal {G}}\left( u_{1},\phi _{1}\right) , \forall \left( u_{1},\phi _{1}\right) \in V. \end{aligned}$$

The substitution of u and $\phi$ into (3.3)$_{1}$ and (3.3)$_{3}$ yields

$$\begin{aligned} \left( v,\psi \right) \in H_{*}^{1}\left( 0,1\right) \times H_{0}^{1}\left( 0,1\right) . \end{aligned}$$

Similarly, inserting $\psi$ in (3.4) and bearing in mind (3.3)$_{5}$, we obtain $\eta ^{t}\in L_{g}.$ Moreover, if we take $u_{1}=0$$\in H_{*}^{1}\left( 0,1\right)$ in (3.6), we get

$$\begin{aligned}&\left( J+\xi \right) \int _{0}^{1}\phi \phi _{1}dx+b\int _{0}^{1}u_{x}\phi _{1}dx \\&\quad +\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \int _{0}^{1}\phi _{x}\phi _{1x}dx \begin{array}{c} = \end{array} \int _{0}^{1}h_{2}\phi _{1}dx. \end{aligned}$$

Hence, we obtain

$$\begin{aligned}&\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \int _{0}^{1}\phi _{x}\phi _{1x}dx \nonumber \\&\quad =\int _{0}^{1}\left( h_{2}-\left( J+\xi \right) \phi -bu_{x}\right) \phi _{1}dx, \forall \phi _{1}\in H_{0}^{1}\left( 0,1\right) . \end{aligned}$$

(3.8)

By noting that $h_{2}-\left( J+\xi \right) \phi -bu_{x}\in L^{2}\left( 0,1\right) ,$ we obtain $\phi$$\in H^{2}\left( 0,1\right) \cap H_{0}^{1}\left( 0,1\right)$ and, consequently, (3.8) takes the form

$$\begin{aligned}&\int _{0}^{1}\left( -\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \phi _{xx}-h_{2}+\left( J+\xi \right) \phi +bu_{x}\right) \phi _{1}dx \\&\quad =0, \forall \phi _{1}\in H_{0}^{1}\left( 0,1\right) . \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} -\left( l+\int _{0}^{\infty }g\left( s\right) \left( 1-e^{-s}\right) ds\right) \phi _{xx}+\left( J+\xi \right) \phi +bu_{x}=h_{2}. \end{aligned}$$

This gives (3.5)$_{2}$. Similarly, if we take $\phi _{1}=0\in H_{0}^{1}\left( 0,1\right)$ in (3.6), we get

$$\begin{aligned} \mu u_{xx}=\rho u-b\phi _{x}-h_{1}\text { in }L_{*}^{2}\left( 0,1\right) , \end{aligned}$$

using the fact that $u_{x}\left( 0\right) =u_{x}\left( 1\right) =0,$ then we conclude

$$\begin{aligned} u\in H_{*}^{2}\left( 0,1\right) \cap H_{*}^{1}\left( 0,1\right) . \end{aligned}$$

Hence, there exists a unique $\Phi \in$$D\left( {\mathcal {A}}\right)$ such that (3.2) is satisfied. Therefore, ${\mathcal {A}}$ is a maximal monotone operator. Now, we prove that the operator $\Gamma$ defined in (3.1) is locally Lipschitz in ${\mathcal {H}}.$ Let $\Phi =\left( u,v,\phi ,\psi ,\eta ^{t}\right) ^{T}\in {\mathcal {H}}$ and $\Phi _{1}=\left( u_{1},v_{1},\phi _{1},\psi _{1},\eta _{1}^{t}\right) ^{T}\in {\mathcal {H}},$ then we have

$$\begin{aligned} \left\| \Gamma \left( \Phi \right) -\Gamma \left( \Phi _{1}\right) \right\| _{{\mathcal {H}}}\le \gamma _{1}\left\| f\left( \psi \right) -f\left( \psi _{1}\right) \right\| _{L^{2}}. \end{aligned}$$

By using (2.3), Hölder and Poincaré inequalities, we can get

$$\begin{aligned} \left\| f\left( \psi \right) -f\left( \psi _{1}\right) \right\| _{L^{2}}\le \gamma _{1}{k_{0}(}\left\| {\psi }\right\| _{2\varrho }^{\varrho }{+}\left\| {\psi }_{1}\right\| {_{2\varrho }^{\varrho })} \left\| {{\psi }-{\psi }_{1}}\right\| {\le }\gamma _{1}\left\| \psi _{x}-\psi _{1x}\right\| _{L^{2}}, \end{aligned}$$

which gives us

$$\begin{aligned} \left\| \Gamma \left( \Phi \right) -\Gamma \left( \Phi _{1}\right) \right\| _{{\mathcal {H}}}\le \gamma _{2}\left\| \Phi -\Phi _{1}\right\| _{{\mathcal {H}}}. \end{aligned}$$

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

4 Technical Lemmas

In this section, we use the multipliers method to construct the Lyapunov functional that must be equivalent to the energy of system (2.9). To achieve our goal we state and prove the following lemmas.

Lemma 2

The energy functional E, defined by

$$\begin{aligned} E(t)&=\frac{1}{2}\int _{0}^{1}\left\{ \rho u_{t}^{2}+\mu u_{x}^{2}+J\phi _{t}^{2}+2bu_{x}\phi +\xi \phi ^{2}+\left( \delta -g_{0}\right) \phi _{x}^{2}\right\} dx \nonumber \\&\quad +\frac{1}{2}\left( g\circ \phi _{x}\right) (t), \end{aligned}$$

(4.1)

satisfies

$$\begin{aligned} E^{\prime }(t)=-\chi \left( t\right) \int _{0}^{1}\phi _{t}f\left( \phi _{t}\right) dx+\frac{1}{2}\left( g^{^{\prime }}\circ \phi _{x}\right) (t)\le 0. \end{aligned}$$

(4.2)

Proof

Multiplying the first equation of (2.9) by $u_{t}$, the second equation of (2.9) by $\phi _{t}$, using (2.8), integrating over (0, 1) and summing them up, we obtain

$$\begin{aligned}&\frac{d}{2dt}\int _{0}^{1}\left\{ \rho u_{t}^{2}+\mu u_{x}^{2}+J\phi _{t}^{2}+2bu_{x}\phi ++\xi \phi ^{2}+\left( \delta -g_{0}\right) \phi _{x}^{2}\right\} dx \nonumber \\&\quad +\chi \left( t\right) \int _{0}^{1}\phi _{t}f\left( \phi _{t}\right) dx-\int _{0}^{1}\phi _{t}\int _{0}^{\infty }g(s)\eta _{xx}^{t}(x,s)dsdx \begin{array}{c} = \end{array} 0. \end{aligned}$$

(4.3)

We estimate the last term of (4.3) as follows

$$\begin{aligned}&-\int _{0}^{1}\phi _{t}\int _{0}^{\infty }g(s)\eta _{xx}^{t}(x,s)dsdx \\&\quad =-\int _{0}^{1}\left( \eta _{t}^{t}+\eta _{s}^{t}\right) \int _{0}^{\infty }g(s)\eta _{xx}^{t}(x,s)dsdx \\&\quad =-\int _{0}^{\infty }g(s)\int _{0}^{1}\eta _{t}^{t}\eta _{xx}^{t}(x,s)dxds-\int _{0}^{\infty }g(s)\int _{0}^{1}\eta _{s}^{t}\eta _{xx}^{t}(x,s)dxds, \end{aligned}$$

integrating by parts, we have

$$\begin{aligned} -\int _{0}^{1}\phi _{t}\int _{0}^{\infty }g(s)\eta _{xx}^{t}(x,s)dsdx=\frac{d}{ 2dt}\left( g\circ \phi _{x}\right) (t)-\frac{1}{2}\left( g^{^{\prime }}\circ \phi _{x}\right) (t). \end{aligned}$$

(4.4)

By substituting (4.4) in (4.3), bearing in mind (4.1), yields (4.2). $\square$

Remark 2

The energy E(t) defined by (4.1) is non-negative. In fact, by the same technique as in Remark (1), we can write

$$\begin{aligned} \mu u_{x}^{2}+2bu_{x}\phi +\xi \phi ^{2}>\mu _{1}u_{x}^{2}+\xi _{1}\phi ^{2}. \end{aligned}$$

Consequently,

$$\begin{aligned} E(t)>\frac{1}{2}\int _{0}^{1}\left\{ \rho u_{t}^{2}+\mu _{1}u_{x}^{2}+J\phi _{t}^{2}+\xi _{1}\phi ^{2}+\left( \delta -g_{0}\right) \phi _{x}^{2}\right\} dx+\frac{1}{2}\left( g\circ \phi _{x}\right) (t). \end{aligned}$$

Lemma 3

Let $(u,\phi )$ be the solution of (2.9). Then for any positive constant $\varepsilon _{1}$ the functional

$$\begin{aligned} F_{1}\left( t\right) =J\int _{0}^{1}\phi _{t}\phi dx+\frac{b\rho }{\mu } \int _{0}^{1}\phi \int _{0}^{x}u_{t}\left( y\right) dydx, \end{aligned}$$

satisfies

$$\begin{aligned} F_{1}^{\prime }\left( t\right)&\le -\frac{l}{2}\int _{0}^{1}\phi _{x}^{2}dx-\xi _{1}\int _{0}^{1}\phi ^{2}dx+\left( J+\frac{b^{2}\rho ^{2}}{ 4\mu ^{2}\varepsilon _{1}}\right) \int _{0}^{1}\phi _{t}^{2}dx \nonumber \\&\quad +k_{1}\int _{0}^{1}f^{2}\left( \phi _{t}\right) dx+\varepsilon _{1}\int _{0}^{1}u_{t}^{2}dx+\frac{g_{0}}{2l}\left( g\circ \phi _{x}\right) (t), \end{aligned}$$

(4.5)

where $c, k_{1}$ are positive constants.

Proof

By differentiating $F_{1}(t)$ with respect to t, using (2.9) and integration by parts

$$\begin{aligned} F_{1}^{\prime }\left( t\right)&=-\delta \int _{0}^{1}\phi _{x}^{2}dx-b\int _{0}^{1}u_{x}\phi dx-\xi \int _{0}^{1}\phi ^{2}dx \nonumber \\&\quad +\int _{0}^{1}\phi _{x}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( t-s\right) ds\right) dx-\chi \left( t\right) \int _{0}^{1}\phi f\left( \phi _{t}\right) dx+J\int _{0}^{1}\phi _{t}^{2}dx \nonumber \\&\quad +\frac{b\rho }{\mu }\int _{0}^{1}\phi _{t}\left( \int _{0}^{x}u_{t}\left( y\right) dy\right) dx+\frac{b\rho }{\mu }\int _{0}^{1}\phi \frac{d}{dt}\left( \int _{0}^{x}u_{t}\left( y\right) dy\right) dx. \end{aligned}$$

(4.6)

Using Cauchy-Schwarz inequality, we have

$$\begin{aligned} \int _{0}^{1}\left( \int _{0}^{x}u_{t}\left( y\right) dy\right) ^{2}dx\le \int _{0}^{1}\left( \int _{0}^{1}u_{t}dx\right) ^{2}dx\le \int _{0}^{1}u_{t}^{2}dx. \end{aligned}$$

By a simple computation it’s easy to prove that

$$\begin{aligned} \frac{d}{dt}\left( \int _{0}^{x}u_{t}\left( y\right) dy\right) =\frac{\mu }{ \rho }u_{x}+\frac{b}{\rho }\phi . \end{aligned}$$

Using Young’s inequality and (2.6), for any $\varepsilon _{1},\delta _{1},\delta _{2}>0$, we obtain

$$\begin{aligned}&\int _{0}^{1}\phi _{x}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( t-s\right) ds\right) dx \nonumber \\&\quad =-\int _{0}^{1}\phi _{x}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \left( \phi _{x}\left( t\right) -\phi _{x}\left( t-s\right) \right) ds\right) dx+g_{0}\int _{0}^{1}\phi _{x}^{2}dx \nonumber \\&\quad \le \left( \delta _{1}+g_{0}\right) \int _{0}^{1}\phi _{x}^{2}dx+\frac{1}{ 4\delta _{1}}\int _{0}^{1}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \left( \phi _{x}\left( t\right) -\phi _{x}\left( t-s\right) \right) ds\right) ^{2}dx \nonumber \\&\quad \le \left( \delta _{1}+g_{0}\right) \int _{0}^{1}\phi _{x}^{2}dx+\frac{1}{ 4\delta _{1}}g_{0}\left( g\circ \phi _{x}\right) (t), \end{aligned}$$

(4.7)

$$\begin{aligned} \frac{b\rho }{\mu }\int _{0}^{1}\phi _{t}\left( \int _{0}^{x}u_{t}\left( y\right) dy\right) dx\le \frac{b^{2}\rho ^{2}}{4\mu ^{2}\varepsilon _{1}} \int _{0}^{1}\phi _{t}^{2}dx+\varepsilon _{1}\int _{0}^{1}u_{t}^{2}dx, \end{aligned}$$

(4.8)

and

$$\begin{aligned} -\chi \left( t\right) \int _{0}^{1}\phi f\left( \phi _{t}\right) dx\le & {} \chi \left( t\right) \delta _{2}\int _{0}^{1}\phi ^{2}dx+\frac{\chi \left( t\right) }{4\delta _{2}}\int _{0}^{1}f^{2}\left( \phi _{t}\right) dx \nonumber \\\le & {} \chi \left( 0\right) \delta _{2}\int _{0}^{1}\phi ^{2}dx+\frac{\chi \left( 0\right) }{4\delta _{2}}\int _{0}^{1}f^{2}\left( \phi _{t}\right) dx. \end{aligned}$$

(4.9)

By substituting (4.7)–(4.9) into (4.6) and letting $\delta _{1}=\dfrac{l}{2},$$\delta _{2}=\dfrac{1}{\chi \left( 0\right) }\xi _{1},$ we obtain (4.5). $\square$

Lemma 4

Let $(u,\phi )$ be the solution of (2.9). Then for any positive constants $\varepsilon _{2},\varepsilon _{3}$ the functional

$$\begin{aligned} {F_{2}(t)=-J\int _{0}^{1}\phi _{t}\int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))dsdx,} \end{aligned}$$

satisfies

$$\begin{aligned} F_{2}^{\prime }(t)&\le -J\frac{g_{0}}{2}{\int _{0}^{1}\phi _{t}^{2}dx} +3\varepsilon _{2}\int _{0}^{1}\phi _{x}^{2}dx+\varepsilon _{3}\int _{0}^{1}u_{x}^{2}dx \nonumber \\&\quad +k_{2}\int _{0}^{1}{f}^{2}{\left( \phi _{t}\right) }dx-\frac{Jd_{2}}{2g_{0}} (g^{\prime }\circ \phi _{x})(t)+c\left( 1+\frac{1}{\varepsilon _{2}}+\frac{1 }{\varepsilon _{3}}\right) (g\circ \phi _{x})(t), \end{aligned}$$

(4.10)

where $k_{2}, c, d_{2}$are positive constants.

Proof

First, we note that

$$\begin{aligned}&\frac{\partial }{\partial t}\left( \int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))ds\right) \\&\quad =\frac{\partial }{\partial t}\left( \int _{-\infty }^{t}g(t-s)(\phi (t)-\phi (s))ds\right) \\&\quad =\int _{-\infty }^{t}g^{\prime }(t-s)(\phi (t)-\phi (s))ds+\int _{-\infty }^{t}g(t-s)\phi _{t}(t)ds \\&\quad =g_{0}\phi _{t}(t)+\int _{0}^{\infty }g^{\prime }(s)(\phi (t)-\phi (t-s))ds. \end{aligned}$$

Then, by a simple differentiation of $F_{2}(t)$ and using (2.9), we have

$$\begin{aligned} F_{2}^{\prime }(t)&={\delta \int _{0}^{1}\phi _{x}\int _{0}^{\infty }g(s)(\phi _{x}(t)-\phi _{x}(t-s))dsdx-Jg_{0}\int _{0}^{1}\phi _{t}^{2}dx} \nonumber \\&\quad {-J\int _{0}^{1}\phi _{t}}\int _{0}^{\infty }g^{\prime }(s)(\phi (t)-\phi (t-s))dsdx \nonumber \\&\quad +{b\int _{0}^{1}u_{x}\int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))dsdx}\nonumber \\&\quad +{\xi \int _{0}^{1}\phi \int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))dsdx} \nonumber \\&\quad -{\int _{0}^{1}\int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( x,t-s\right) ds\int _{0}^{\infty }g(s)(\phi _{x}(t)-\phi _{x}(t-s))dsdx} \nonumber \\&\quad +{\chi \left( t\right) \int _{0}^{1}f\left( \phi _{t}\right) \int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))dsdx,} \end{aligned}$$

(4.11)

and together with (2.9), by using Young’s inequality, (2.4), (2.6) and (2.7) we have the following estimations:

$$\begin{aligned}&\delta \int _{0}^{1}\phi _{x}\int _{0}^{\infty }g(s)(\phi _{x}(t)-\phi _{x}(t-s))dsdx \nonumber \\&\quad \le \varepsilon _{2}\int _{0}^{1}\phi _{x}^{2}dx+\frac{\delta ^{2}}{ 4\varepsilon _{2}}\int _{0}^{1}\left( \int _{0}^{\infty }g(s)(\phi _{x}(t)-\phi _{x}(t-s))ds\right) ^{2}dx \nonumber \\&\quad \le \varepsilon _{2}\int _{0}^{1}\phi _{x}^{2}dx+\frac{\delta ^{2}}{ 4\varepsilon _{2}}g_{0}(g\circ \phi _{x})(t), \end{aligned}$$

(4.12)

$$\begin{aligned}&\qquad -J\int _{0}^{1}\phi _{t}\int _{0}^{\infty }g^{\prime }(s)(\phi (t)-\phi (t-s))dsdx \nonumber \\&\quad \le J\delta _{3}\int _{0}^{1}\phi _{t}^{2}dx-\frac{J}{4\delta _{3}} \int _{0}^{1}\left( \int _{0}^{\infty }g^{\prime }(s)(\phi (t)-\phi (t-s))ds\right) ^{2}dx \nonumber \\&\quad \le J\delta _{3}\int _{0}^{1}\phi _{t}^{2}dx-\frac{Jd_{2}}{4\delta _{3}} (g^{\prime }\circ \phi _{x})(t), \end{aligned}$$

(4.13)

$$\begin{aligned}&b\int _{0}^{1}u_{x}\int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))dsdx \nonumber \\&\quad \le \varepsilon _{3}\int _{0}^{1}u_{x}^{2}dx+\frac{b^{2}}{4\varepsilon _{3} }\int _{0}^{1}\left( \int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))ds\right) ^{2}dx \nonumber \\&\quad \le \varepsilon _{3}\int _{0}^{1}u_{x}^{2}dx+\frac{b^{2}d_{1}}{ 4\varepsilon _{3}}(g\circ \phi _{x})(t), \end{aligned}$$

(4.14)

$$\begin{aligned}&\qquad -\int _{0}^{1}\int _{0}^{\infty }g(s)\phi _{x}(t-s)ds\int _{0}^{\infty }g(s)(\phi _{x}(t)-\phi _{x}(t-s))dsdx \nonumber \\&\quad =\int _{0}^{1}\left( \int _{0}^{\infty }g(s)(\phi _{x}(t)-\phi _{x}(t-s))ds\right) ^{2}dx \nonumber \\&\qquad -g_{0}\int _{0}^{1}\phi _{x}(t)\int _{0}^{\infty }g(s)(\phi _{x}(t)-\phi _{x}(t-s))dsdx \nonumber \\&\quad \le c\left( 1+\frac{1}{\varepsilon _{2}}\right) (g\circ \phi _{x})\left( t\right) +\varepsilon _{2}\int _{0}^{1}\phi _{x}^{2}dx, \end{aligned}$$

(4.15)

$$\begin{aligned}&\chi (t)\int _{0}^{1}{f\left( \phi _{t}\right) }\int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))dsdx \nonumber \\&\quad \le \chi (t)\varepsilon _{2}\int _{0}^{1}{f}^{2}{\left( \phi _{t}\right) } dx+\frac{\chi (t)d_{1}}{4\varepsilon _{2}}(g\circ \phi _{x})(t) \nonumber \\&\quad \le \chi (0)\varepsilon _{2}\int _{0}^{1}{f}^{2}{\left( \phi _{t}\right) } dx+\frac{\chi (0)d_{1}}{4\varepsilon _{2}}(g\circ \phi _{x})(t). \end{aligned}$$

(4.16)

By using Young’s and Poincaré inequalities and (2.4),

$$\begin{aligned} \xi \int _{0}^{1}\phi \int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))dsdx\le \varepsilon _{2}\int _{0}^{1}\phi _{x}^{2}dx+\frac{\xi ^{2}d_{1}}{ 4\varepsilon _{2}}(g\circ \phi _{x})(t). \nonumber \\ \end{aligned}$$

(4.17)

By substituting (4.12)–(4.17) into (4.11), we have

$$\begin{aligned} F_{2}^{\prime }(t)&\le -J(g_{0}-\delta _{3}){\int _{0}^{1}\phi _{t}^{2}dx} +3\varepsilon _{2}\int _{0}^{1}\phi _{x}^{2}dx+\varepsilon _{3}\int _{0}^{1}u_{x}^{2}dx \\&\quad +k_{2}\int _{0}^{1}{f}^{2}{\left( \phi _{t}\right) }dx-\frac{Jd_{2}}{ 4\delta _{3}}(g^{\prime }\circ \phi _{x})(t)+c\left( 1+\frac{1}{\varepsilon _{2}}+\frac{1}{\varepsilon _{3}}\right) (g\circ \phi _{x})(t). \end{aligned}$$

Finally, letting $\delta _{3}=\dfrac{g_{0}}{2},$ we obtain (4.10). $\square$

Lemma 5

Let $(u,\phi )$ be the solution of (2.9). Then for any positive constant $\varepsilon _{4}$ the functional

$$\begin{aligned} F_{3}\left( t\right) =J\int _{0}^{1}\phi _{x}u_{t}dx+J\int _{0}^{1}\phi _{t}u_{x}dx-\frac{\rho }{\mu }\int _{0}^{1}u_{t}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( t-s\right) ds\right) dx, \end{aligned}$$

satisfies

$$\begin{aligned} F_{3}^{\prime }\left( t\right)&\le -\frac{b}{2}\int _{0}^{1}u_{x}^{2}dx+c \int _{0}^{1}\phi _{x}^{2}dx+k_{3}\int _{0}^{1}f^{2}\left( \phi _{t}\right) dx \nonumber \\&\quad +c\left( g\circ \phi _{x}\right) \left( t\right) +c\varepsilon _{4}\int _{0}^{1}u_{t}^{2}dx-\frac{c}{\varepsilon _{4}}\left( g^{\prime }\circ \phi _{x}\right) \left( t\right) . \end{aligned}$$

(4.18)

Proof

First, we note that

$$\begin{aligned}&\frac{d}{dt}\left( \int _{0}^{\infty }g(s)\phi _{x}(t-s)ds\right) \\&\quad =\frac{d}{dt}\left( \int _{-\infty }^{t}g(t-s)\phi _{x}(s)ds\right) \\&\quad =g(0)\phi _{x}(t)+\int _{-\infty }^{t}g^{\prime }(t-s)\phi _{x}(s)ds \\&\quad =\int _{0}^{\infty }g^{\prime }(s)\left( \phi _{x}(t-s)-\phi _{x}(t)\right) ds+g(0)\phi _{x}(t)+\phi _{x}(t)\int _{0}^{\infty }g^{\prime }(s)ds \\&\quad =\int _{0}^{\infty }g^{\prime }(s)\left( \phi _{x}(t-s)-\phi _{x}(t)\right) ds. \end{aligned}$$

By differentiating $F_{3}(t)$, using (2.9) and then integrating by parts, we obtain

$$\begin{aligned} F_{3}^{\prime }\left( t\right)&=\frac{bJ}{\rho }\int _{0}^{1}\phi _{x}^{2}dx-b\int _{0}^{1}u_{x}^{2}dx-\xi \int _{0}^{1}\phi u_{x}dx-\chi \left( t\right) \int _{0}^{1}u_{x}f\left( \phi _{t}\right) dx \nonumber \\&\quad -\frac{b}{\mu }\int _{0}^{1}\phi _{x}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( t-s\right) ds\right) dx \nonumber \\&\quad -\frac{\rho }{\mu }\int _{0}^{1}u_{t}\left( \int \nolimits _{0}^{\infty }g^{\prime }\left( s\right) \left( \phi _{x}\left( t-s\right) -\phi _{x}\left( t\right) \right) ds\right) dx \end{aligned}$$

(4.19)

Using Young’s inequality, (2.5) and (2.6)

$$\begin{aligned}&-\frac{\rho }{\mu }\int _{0}^{1}u_{t}\left( \int \nolimits _{0}^{\infty }g^{\prime }\left( s\right) \left( \phi _{x}\left( t-s\right) -\phi _{x}\left( t\right) \right) ds\right) dx \nonumber \\&\quad \le \frac{\rho }{\mu }\varepsilon _{4}\int _{0}^{1}u_{t}^{2}dx-\frac{\rho }{4\mu \varepsilon _{4}}g(0)\left( g^{\prime }\circ \phi _{x}\right) \left( t\right) , \nonumber \\&-\frac{b}{\mu }\int _{0}^{1}\phi _{x}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( t-s\right) ds\right) dx \nonumber \\&\quad =\frac{b}{\mu }\int _{0}^{1}\phi _{x}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \left( \phi _{x}\left( t\right) -\phi _{x}\left( t-s\right) \right) ds\right) dx-\frac{b}{\mu }g_{0}\int _{0}^{1}\phi _{x}^{2}dx \nonumber \\&\quad \le \left( \delta _{4}-\frac{b}{\mu }g_{0}\right) \int _{0}^{1}\phi _{x}^{2}dx+\frac{b^{2}}{4\delta _{4}\mu ^{2}}\left( g\circ \phi _{x}\right) \left( t\right) . \end{aligned}$$

(4.20)

By letting $\delta _{4}=\dfrac{b}{\mu }g_{0},$ we get

$$\begin{aligned}&-\frac{b}{\mu }\int _{0}^{1}\phi _{x}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( t-s\right) ds\right) dx\le c\left( g\circ \phi _{x}\right) \left( t\right) , \end{aligned}$$

(4.21)

$$\begin{aligned}&-\chi \left( t\right) \int _{0}^{1}u_{x}f\left( \phi _{t}\right) dx \le \chi \left( t\right) \delta _{5}\int _{0}^{1}u_{x}^{2}dx+\frac{1}{4\delta _{5} }\chi \left( t\right) \int _{0}^{1}f^{2}\left( \phi _{t}\right) dx \nonumber \\&\quad \le \chi \left( 0\right) \delta _{5}\int _{0}^{1}u_{x}^{2}dx+\frac{1}{ 4\delta _{5}}\chi \left( 0\right) \int _{0}^{1}f^{2}\left( \phi _{t}\right) dx. \end{aligned}$$

(4.22)

By using Young’s and Poincaré inequalities, we have

$$\begin{aligned} -\xi \int _{0}^{1}\phi u_{x}dx\le \frac{\xi ^{2}}{b}\int _{0}^{1}\phi _{x}^{2}dx+\frac{b}{4}\int _{0}^{1}u_{x}^{2}dx. \end{aligned}$$

(4.23)

Insert (4.20)–(4.23) in (4.19). So, the estimate (4.19) becomes

$$\begin{aligned} F_{3}^{\prime }\left( t\right)&\le -\left( b-\chi \left( 0\right) \delta _{5}-\frac{b}{4}\right) \int _{0}^{1}u_{x}^{2}dx+c\int _{0}^{1}\phi _{x}^{2}dx+c\varepsilon _{4}\int _{0}^{1}u_{t}^{2}dx \\&\quad +\frac{1}{4\delta _{5}}\chi \left( 0\right) \int _{0}^{1}f^{2}\left( \phi _{t}\right) dx+c\left( g\circ \phi _{x}\right) \left( t\right) -\frac{c}{ \varepsilon _{4}}\left( g^{\prime }\circ \phi _{x}\right) \left( t\right) , \end{aligned}$$

and letting $\delta _{5}=\dfrac{b}{4\chi \left( 0\right) },$ the proof is, hence, complete. $\square$

Lemma 6

Let $(u,\phi )$ be the solution of (2.9). Then the functional

$$\begin{aligned} F_{4}\left( t\right) =-\rho \int _{0}^{1}u_{t}udx, \end{aligned}$$

satisfies

$$\begin{aligned} F_{4}^{\prime }\left( t\right) \le \frac{3\mu }{2}\int _{0}^{1}u_{x}^{2}dx+c \int _{0}^{1}\phi _{x}^{2}dx-\rho \int _{0}^{1}u_{t}^{2}dx. \end{aligned}$$

(4.24)

Proof

$$\begin{aligned} F_{4}^{\prime }\left( t\right)= & {} -\rho \int _{0}^{1}u_{t}^{2}dx+\rho \int _{0}^{1}u_{tt}udx \\= & {} -\rho \int _{0}^{1}u_{t}^{2}dx-\mu \int _{0}^{1}u_{xx}udx-b\int _{0}^{1}\phi _{x}udx \\= & {} \mu \int _{0}^{1}u_{x}^{2}dx+b\int _{0}^{1}\phi u_{x}dx-\rho \int _{0}^{1}u_{t}^{2}dx, \end{aligned}$$

by using Young’s and Poincaré inequalities, we obtain easily the estimation (4.24). $\square$

Now, we define the Lyapunov functional L(t) by

$$\begin{aligned} L(t)=NE(t)+N_{1}F_{1}(t)+N_{2}F_{2}(t)+N_{3}F_{3}(t)+F_{4}(t), \end{aligned}$$

(4.25)

where N,  $N_{1},$$N_{2},$$N_{3}$ are positive constants.

Lemma 7

Let $\left( u,\phi \right)$ be the solution of (2.9). Then, there exist two positive constants $b_{1}$ and $b_{2}$ such that the Lyapunov functional (4.25) satisfies

$$\begin{aligned} b_{1}E(t)\le L(t)\le b_{2}E(t), \forall t\ge 0, \end{aligned}$$

(4.26)

and

$$\begin{aligned} L^{\prime }(t)\le -c_{1}E(t)+c_{2}\left( g\circ \phi _{x}\right) (t)+c_{3}\int _{0}^{1}\left( \phi _{t}^{2}+f^{2}(\phi _{t})\right) dx. \end{aligned}$$

(4.27)

Proof

From (4.25), we get

$$\begin{aligned}&\left| L(t)-NE(t)\right| \\&\quad \le JN_{1}\int _{0}^{1}\left| \phi _{t}\phi \right| dx+\frac{b\rho }{\mu }N_{1}\int _{0}^{1}\left| \phi \int _{0}^{x}u_{t}\left( y\right) dy\right| dx \\&\quad \quad +{JN_{2}\int _{0}^{1}}\left| {\phi _{t}\int _{0}^{\infty }g(s)(\phi (t)-\phi (t-s))ds}\right| {dx+}JN_{3}\int _{0}^{1}\left| \phi _{x}u_{t}\right| dx \\&\quad \quad +JN_{3}\int _{0}^{1}\left| \phi _{t}u_{x}\right| dx+\frac{\rho }{ \mu }N_{3}\int _{0}^{1}\left| u_{t}\left( \int \nolimits _{0}^{\infty }g\left( s\right) \phi _{x}\left( t-s\right) ds\right) \right| dx \\&\quad \quad +\rho \int _{0}^{1}\left| u_{t}u\right| dx. \end{aligned}$$

By using Young’s, Cauchy-Schwarz, and Poincaré inequalities, we have

$$\begin{aligned} \left| L(t)-NE(t)\right| \le c{\int _{0}^{1}}\left( {\phi _{t}^{2}+} \phi _{x}^{2}+u_{x}^{2}+u_{t}^{2}+\phi ^{2}\right) dx+c\left( g\circ \phi _{x}\right) \left( t\right) \le cE(t), \end{aligned}$$

that is

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

Now, by choosing N (depending on $N_{1},$$N_{2}$ and $N_{3}$) sufficiently large we obtain (4.26).

By differentiating L(t),  we obtain

$$\begin{aligned} L^{\prime }(t)&\le -\left( \frac{Jg_{0}}{2}N_{2}-N_{1}\left( J+\frac{ b^{2}\rho }{\mu ^{2}}N_{1}\right) \right) {\int _{0}^{1}\phi _{t}^{2}dx} \nonumber \\&\quad -\left( N_{1}\frac{l}{4}-N_{3}c-c\right) \int _{0}^{1}\phi _{x}^{2}dx \nonumber \\&\quad -\left( N_{3}\frac{b}{4}-\frac{3\mu }{2}\right) \int _{0}^{1}u_{x}^{2}dx-N_{1}\xi _{1}\int _{0}^{1}\phi ^{2}dx-\frac{\rho }{2} \int _{0}^{1}u_{t}^{2}dx \nonumber \\&\quad +\left( \frac{N}{2}-\frac{4c^{2}}{\rho }N_{3}^{2}-N_{2}\frac{Jd_{2}}{2g_{0} }\right) \left( g^{\prime }\circ \phi _{x}\right) (t){+}k\int _{0}^{1}f^{2} \left( \phi _{t}\right) dx \nonumber \\&\quad +\left( \frac{g_{0}}{2l}N_{1}+c\left( N_{2}+\frac{12N_{2}^{2}}{lN_{1}}+ \frac{4N_{2}^{2}}{bN_{3}}\right) +cN_{3}\right) (g\circ \phi _{x})(t). \end{aligned}$$

(4.28)

By setting

$$\begin{aligned} \varepsilon _{1}=\frac{\rho }{4N_{1}}, \varepsilon _{2}=\frac{lN_{1}}{ 12N_{2}}, \varepsilon _{3}=\frac{bN_{3}}{4N_{2}}, \varepsilon _{4}=\frac{\rho }{4cN_{3}}, \end{aligned}$$

and by using (4.28), we get

$$\begin{aligned} L^{\prime }(t)&\le -\left( \frac{Jg_{0}}{2}N_{2}-N_{1}\left( J+\frac{ b^{2}\rho }{\mu ^{2}}N_{1}\right) \right) {\int _{0}^{1}\phi _{t}^{2}dx} \\&\quad -\left( N_{1}\frac{l}{4}-N_{3}c-c\right) \int _{0}^{1}\phi _{x}^{2}dx \\&\quad -\left( N_{3}\frac{b}{4}-\frac{3\mu }{2}\right) \int _{0}^{1}u_{x}^{2}dx-N_{1}\xi _{1}\int _{0}^{1}\phi ^{2}dx-\frac{\rho }{2} \int _{0}^{1}u_{t}^{2}dx \\&\quad +\left( \frac{N}{2}-\frac{4c^{2}}{\rho }N_{3}^{2}-N_{2}\frac{Jd_{2}}{2g_{0} }\right) \left( g^{\prime }\circ \phi _{x}\right) (t){+}k\int _{0}^{1}f^{2} \left( \phi _{t}\right) dx \\&\quad +\left( \frac{g_{0}}{2l}N_{1}+c\left( N_{2}+\frac{12N_{2}^{2}}{lN_{1}}+ \frac{4N_{2}^{2}}{bN_{3}}\right) +cN_{3}\right) (g\circ \phi _{x})(t). \end{aligned}$$

First, we choose $N_{3}$ large enough such that

$$\begin{aligned} N_{3}\frac{b}{4}-\frac{3\mu }{2}>0. \end{aligned}$$

For fixed $N_{3}>0$, we take $N_{1}$ large enough such that

$$\begin{aligned} N_{1}\frac{l}{4}-N_{3}c-c>0. \end{aligned}$$

Then we select $N_{2}$$>0$ large so that

$$\begin{aligned} \frac{Jg_{0}}{2}N_{2}-N_{1}\left( J+\frac{b^{2}\rho }{\mu ^{2}}N_{1}\right) >0. \end{aligned}$$

Finally, we choose N large enough such that

$$\begin{aligned} \frac{N}{2}-\frac{4c^{2}}{\rho }N_{3}^{2}-N_{2}\frac{Jd_{2}}{2g_{0}}>0. \end{aligned}$$

Consequently, we obtain the estimation (4.27) of $L^{\prime }(t)$. $\square$

5 Stability result

In this section, we state and prove our stability result.

Theorem 2

Assume that (H1)–(H3) hold. Let $h(t)=\alpha (t)\chi (t)$ be a positive non-increasing function, then, for any $\Phi _{0}\in D\left( {\mathcal {A}}\right)$satisfying, for some $c_{0}\ge 0,$

$$\begin{aligned} \int _{0}^{1}\phi _{0x}^{2}(x,s)dx\le c_{0},\ \forall s>0, \end{aligned}$$

(5.1)

there exist the positive constants $a_{1},$$a_{2},$$a_{3},$such that

$$\begin{aligned} E(t)\le a_{1}G_{0}^{-1}\left( \frac{a_{2}+a_{3}\int _{0}^{t}h(s)\int _{s}^{ \infty }g\left( \tau \right) d\tau ds}{\int _{0}^{t}h(s)ds}\right) , \end{aligned}$$

(5.2)

where

$$\begin{aligned} G_{0}\left( t\right) =tG^{\prime }\left( \varepsilon _{0}t\right) , \forall \varepsilon _{0}\ge 0. \end{aligned}$$

Proof

Multiplying (4.27) by h(t), we get

$$\begin{aligned} h(t)L^{\prime }(t)\le -c_{1}h(t)E(t)+c_{2}h(t)\left( g\circ \phi _{x}\right) (t)+c_{3}h(t)\int _{0}^{1}\left( \phi _{t}^{2}+f^{2}(\phi _{t})\right) dx. \nonumber \\ \end{aligned}$$

(5.3)

We distinguish two cases

1) Gis linear on$\left[ 0,\epsilon \right]$. By using (5.3) and the hypothesis $\left( H_{3}\right) ,$ we have

$$\begin{aligned} h(t)L^{\prime }(t)\le & {} -c_{1}h(t)E(t)+c_{2}h(t)\left( g\circ \phi _{x}\right) (t)+c_{3}h(t)\int _{0}^{1}\phi _{t}f(\phi _{t})dx \nonumber \\\le & {} -c_{1}h(t)E(t)+c_{2}h(t)\left( g\circ \phi _{x}\right) (t)-c_{3}\alpha (t)E^{\prime }(t). \end{aligned}$$

(5.4)

To estimate $h(t)\left( g\circ \phi _{x}\right) (t)$ we use the following technique

$$\begin{aligned}&h(t)\int _{0}^{1}\int _{0}^{t}g(s)(\phi _{x}(x,t)-\phi _{x}(x,t-s))^{2}\,ds\,dx \\&\quad \le \chi \left( t\right) {\int _{0}^{1}\int _{0}^{t}\alpha (s)g(s)(\phi _{x}(x,t)-\phi _{x}(x,t-s))^{2}\,ds\,dx} \\&\quad \le {-\chi \left( t\right) \int _{0}^{1}\int _{0}^{t}g^{\prime }(s)(\phi _{x}(x,t)-\phi _{x}(x,t-s))^{2}\,ds\,dx} \\&\quad \le {-\chi \left( t\right) \int _{0}^{1}\int _{0}^{\infty }g^{\prime }(s)(\phi _{x}(x,t)-\phi _{x}(x,t-s))^{2}\,ds\,dx} \\&\quad \le {-2\chi \left( t\right) E^{\prime }(t).} \end{aligned}$$

On the other hand, by using (5.1) and the fact that E(t) is non-increasing, for $t,s\in R_{+}$,

$$\begin{aligned} \int _{0}^{1}(\phi _{x}(x,t)-\phi _{x}(x,t-s))^{2}dx&{\le }{ \,2\int _{0}^{1}\phi _{x}^{2}(x,t)dx+2\int _{0}^{1}\phi _{x}^{2}(x,t-s)dx} \\&{\le }{\,4\sup _{s>0}\int _{0}^{1}\phi _{x}^{2}(x,s)dx+2\sup _{\tau >0}\int _{0}^{1}\phi _{0x}^{2}(x,\tau )dx} \\&{\le }{{\frac{8E(0)}{(\delta -g_{0})}}+2c_{0},} \end{aligned}$$

then, we obtain

$$\begin{aligned}&h(t)\int _{0}^{1}\int _{t}^{\infty }g(s)(\phi _{x}(x,t)-\phi _{x}(x,t-s))^{2}\,ds\,dx \\&\quad \le \left( \frac{8E(0)}{(\delta -g_{0})}+2c_{0}\right) h(t)\int _{t}^{\infty }g(s)ds. \end{aligned}$$

Therefore, we deduce that, for all $t\in R_{+}$,

$$\begin{aligned} h(t)\left( g\circ \phi _{x}\right) (t){\le }{-2\chi \left( t\right) E^{\prime }(t)+}\left( {{\frac{8E(0)}{(\delta -g_{0})}}+2c_{0}}\right) h(t){ \int _{t}^{\infty }g(s)ds.} \end{aligned}$$

(5.5)

Inserting (5.5) in (5.4) and using the fact that $E^{\prime }(t)\le 0,$ we get

$$\begin{aligned}&h(t)L^{\prime }(t)+\left( c_{3}\alpha (t)+2c_{2}{\chi \left( t\right) +\tau }_{1}\right) E^{\prime }(t) \nonumber \\&\quad \le h(t)L^{\prime }(t)+\left( c_{3}\alpha (t)+2c_{2}{\chi \left( t\right) +\tau }_{1}\right) E^{\prime }(t)-{\tau }_{1}E^{\prime }(t) \nonumber \\&\quad \le -c_{1}h(t)E(t)+\beta h(t)v\left( t\right) , \end{aligned}$$

(5.6)

where $\beta =c_{2}\left( {{\dfrac{8E(0)}{(\delta -g_{0})}}+2c_{0}}\right) {, }$$v\left( t\right) ={\int _{t}^{\infty }g(s)ds,}$${\tau }_{1}>0{.}$

Since $\alpha ^{\prime }(t)\le 0,$${\chi }^{\prime }{\left( t\right) \le 0,}$$h^{\prime }(t)\le 0,$ then (5.6) is equivalent to

$$\begin{aligned} L_{1}^{\prime }(t)\le -c_{1}h(t)E(t)+\beta h(t)v\left( t\right) , \end{aligned}$$

(5.7)

where

$$\begin{aligned} L_{1}(t)=h(t)L(t)+\left( c_{3}\alpha (t)+2c_{2}{\chi \left( t\right) +\tau } _{1}\right) E(t){\sim E(t)}. \end{aligned}$$

(5.8)

It’s easy to verify that this last relation holds. Indeed, we have from (4.26)

$$\begin{aligned} b_{1}E(t)\le L(t)\le b_{2}E(t),\forall t\ge 0, \end{aligned}$$

and because h(t),  $\alpha (t)$ and ${\chi \left( t\right) }$ are positive non-increasing functions, then for every $t\ge 0,$ we deduce that exists $m_{1},m_{2}>0,$ satisfying

$$\begin{aligned} m_{1}E(t)\le L_{1}(t)\le m_{2}E(t), \end{aligned}$$

with $m_{1}={\tau }_{1},$$m_{2}=b_{2}h(0)+c_{3}\alpha (0)+2c_{2}{\chi \left( 0\right) +\tau }_{1}.$ This proves that (5.8) is checked.

Because E(t) is a non-increasing function, for all $T\in R_{+},$ by using (5.7), we have

$$\begin{aligned} E(T)\int _{0}^{T}h(t)dt\le \left( \frac{L_{1}(0)}{c_{1}}+\frac{\beta }{c_{1}} \int _{0}^{T}h(t)v\left( t\right) dt\right) . \end{aligned}$$

(5.9)

Using the fact that $G_{0}^{-1}\left( t\right)$ is linear, then (5.9) can be rewritten as follows

$$\begin{aligned} E(T)\le \lambda G_{0}^{-1}\left( \frac{\dfrac{L_{1}(0)}{c_{1}}+\dfrac{\beta }{c_{1}}\int _{0}^{T}h(t)v\left( t\right) dt}{\int _{0}^{T}h(t)dt}\right) , \lambda >0, \end{aligned}$$

which gives (5.2) with $a_{1}=\lambda ,$$a_{2}=\dfrac{L_{1}(0)}{ c_{1}}$ and $a_{3}=\dfrac{\beta }{c_{1}}$. The proof is complete.

2) Gis nonlinear on$\left[ 0,\epsilon \right]$. In this case, we use the same estimation in the above case of $h(t)\left( g\circ \phi _{x}\right) (t)$ for the second term of (5.3). It’s left to estimate the last term of (5.3). For that we first choose $0\le \epsilon _{1}\le \epsilon ,$ such that $sf\left( s\right) \le \min \left( \epsilon ,G\left( \epsilon \right) \right) ,$$\forall \left| s\right| \le \epsilon _{1}$ and by using $\left( H_{3}\right)$, for $s\ne 0,$ it follows that

$$\begin{aligned} \left\{ \begin{array}{l} s^{2}+f^{2}(s)\le G^{-1}\left( sf\left( s\right) \right) , \forall \left| s\right| \le \epsilon _{1}, \\ \nu _{1}\left| s\right| \le \left| f(s)\right| \le \nu _{2}\left| s\right| , \forall \left| s\right| \ge \epsilon _{1}, \end{array} \right. \end{aligned}$$

and we consider the following two sets

$$\begin{aligned} I_{1}=\left\{ x\in \left( 0,1\right) :\left| \phi _{t}\right| \le \epsilon _{1}\right\} , I_{2}=\left\{ x\in \left( 0,1\right) :\left| \phi _{t}\right| >\epsilon _{1}\right\} . \end{aligned}$$

Now, we define $I\left( t\right)$ by

$$\begin{aligned} I\left( t\right) =\int _{I_{1}}\phi _{t}f(\phi _{t})dx, \end{aligned}$$

using Jensen’s inequality and the hypothesis $\left( H_{3}\right) ,$ we have

$$\begin{aligned} c_{3}h(t)\int _{0}^{1}\left( \phi _{t}^{2}+f^{2}(\phi _{t})\right) dx\le c_{3}^{\prime }h(t)G^{-1}\left( I\left( t\right) \right) -c_{3}^{\prime }\alpha (t)E^{\prime }(t). \end{aligned}$$

(5.10)

Inserting (5.10) in (5.3), we obtain

$$\begin{aligned} L_{1}^{\prime }(t)\le -c_{1}h(t)E(t)+\beta h(t)v\left( t\right) +c_{3}^{\prime }h(t)G^{-1}\left( I\left( t\right) \right) , \end{aligned}$$

(5.11)

where

$$\begin{aligned} L_{1}(t)=h(t)L(t)+\left( c_{3}\alpha (t)+2c_{2}{\chi \left( t\right) +\tau } _{1}\right) E(t){\sim E(t),} {\tau }_{1}>0, \end{aligned}$$

we use the same technique as in the precedent case to show that $L_{1}(t)$ is equivalent to E(t).

Now, for $\varepsilon _{0}<\epsilon _{1}$ and using the fact that $E^{\prime }(t)\le 0,$$G^{\prime }>0,$$G^{\prime \prime }>0$ on $\left( 0,\epsilon \right]$, we find that the functional $L_{2}\left( t\right)$, defined by

$$\begin{aligned} L_{2}\left( t\right) =G^{\prime }\left( \varepsilon _{0}E(t)\right) L_{1}\left( t\right) +{\tau }_{2}E\left( t\right) \sim E\left( t\right) , {\tau }_{2}>0, \end{aligned}$$

satisfies

$$\begin{aligned} L_{2}^{\prime }\left( t\right)&=E^{\prime }(t)\left( \varepsilon _{0}G^{^{\prime \prime }}\left( \varepsilon _{0}E(t)\right) L_{1}\left( t\right) +{\tau }_{2}\right) +L_{1}^{\prime }\left( t\right) G^{\prime }\left( \varepsilon _{0}E(t)\right) \nonumber \\&\le -c_{1}h(t)G_{0}\left( E(t)\right) ++\beta G^{\prime }\left( \varepsilon _{0}E(t)\right) h(t)v\left( t\right) \nonumber \\&\quad +c_{3}^{\prime }h(t)G^{\prime }\left( \varepsilon _{0}E(t)\right) G^{-1}\left( I\left( t\right) \right) . \end{aligned}$$

(5.12)

Note that, the equivalence between $L_{2}\left( t\right)$ and $E\left( t\right)$ is due to the fact that $G^{\prime }\left( \varepsilon _{0}E(t)\right)$ is positive non-increasing function and $L_{1}\left( t\right) \sim E(t).$ Indeed, we have for all $t\ge 0,$

$$\begin{aligned} m_{1}E(t)\le L_{1}(t)\le m_{2}E(t), \end{aligned}$$

and

$$\begin{aligned} 0<G^{\prime }\left( \varepsilon _{0}E(t)\right) \le G^{\prime }\left( \varepsilon _{0}E(0)\right) , \end{aligned}$$

then

$$\begin{aligned} {\tau }_{2}E\left( t\right) \le L_{2}\left( t\right) \le \left( G^{\prime }\left( \varepsilon _{0}E(0)\right) m_{2}+{\tau }_{2}\right) E(t). \end{aligned}$$

Therefore, there exists $\sigma _{1},\sigma _{2}>0,$ satisfying

$$\begin{aligned} \sigma _{1}E\left( t\right) \le L_{2}\left( t\right) \le \sigma _{2}E(t), \end{aligned}$$

with $\sigma _{1}={\tau }_{2},$$\sigma _{2}=G^{\prime }\left( \varepsilon _{0}E(0)\right) m_{2}+{\tau }_{2}.$

To estimate the last term of (5.12), we apply the following general Young’s inequality

$$\begin{aligned} AB\le G^{*}\left( A\right) +G\left( B\right) ,\text { if }A\in \left( 0,G^{\prime }\left( \epsilon \right) \right] , B\in \left( 0,\epsilon \right] , \end{aligned}$$

where

$$\begin{aligned} G^{*}\left( s\right) =s\left( G^{\prime }\right) ^{-1}\left( s\right) -G\left( \left( G^{\prime }\right) ^{-1}\left( s\right) \right) ,\text { if } s\in \left( 0,G^{\prime }\left( \epsilon \right) \right] , \end{aligned}$$

we deduce that

$$\begin{aligned} c_{3}^{\prime }h(t)G^{\prime }\left( \varepsilon _{0}E(t)\right) G^{-1}\left( I\left( t\right) \right) \le c_{3}^{\prime }\varepsilon _{0}h(t)G_{0}\left( E(t)\right) -c_{3}^{\prime }\alpha (t)E^{\prime }(t). \end{aligned}$$

(5.13)

Substituting (5.13) in (5.12) and letting $\varepsilon _{0}= \dfrac{c_{1}}{2c_{3}^{\prime }},$ we have

$$\begin{aligned} L_{2}^{\prime }\left( t\right) +c_{3}^{\prime }\alpha (t)E^{\prime }(t)\le -kh(t)G_{0}\left( E(t)\right) +\beta G^{\prime }\left( \varepsilon _{0}E(t)\right) h(t)v\left( t\right) , \end{aligned}$$

(5.14)

which can be rewritten as

$$\begin{aligned}&\left( L_{2}\left( t\right) +c_{3}^{\prime }\alpha (t)E(t)\right) ^{\prime }-c_{3}^{\prime }\alpha ^{\prime }(t)E(t) \nonumber \\&\quad \le -kh(t)G_{0}\left( E(t)\right) +\beta G^{\prime }\left( \varepsilon _{0}E(t)\right) h(t)v\left( t\right) , \end{aligned}$$

(5.15)

since $\alpha ^{\prime }(t)\le 0$, then (5.15) is equivalent to

$$\begin{aligned} L_{3}^{\prime }\left( t\right) \le -kh(t)G_{0}\left( E(t)\right) +\beta G^{\prime }\left( \varepsilon _{0}E(t)\right) h(t)v\left( t\right) . \end{aligned}$$

(5.16)

where

$$\begin{aligned} L_{3}\left( t\right) =L_{2}\left( t\right) +c_{3}^{\prime }\alpha (t)E(t)\sim E(t), \end{aligned}$$

this last relation is checked from the fact that $\alpha (t)$ is a positive non-increasing function and $L_{2}\left( t\right) \sim E(t)$. Indeed, for every $t\ge 0$, we have already

$$\begin{aligned} \sigma _{1}E\left( t\right) \le L_{2}\left( t\right) \le \sigma _{2}E(t), \end{aligned}$$

then

$$\begin{aligned} \sigma _{1}E(t)\le L_{3}(t)\le \sigma _{3}E(t), \end{aligned}$$

with $\sigma _{3}=\sigma _{2}+c_{3}^{\prime }\alpha (0).$

By using (5.16), because $G_{0}\left( E(t)\right)$ and $G^{\prime }\left( \varepsilon _{0}E(t)\right)$ are non-increasing functions, then for all $T\in R_{+},$ we have

$$\begin{aligned} kG_{0}\left( E(T)\right) \int _{0}^{T}h(t)dt\le \left( L_{3}(0)+\beta G^{\prime }\left( \varepsilon _{0}E(0)\right) \int _{0}^{T}h(t)v\left( t\right) dt\right) , \end{aligned}$$

that can be rewritten as follows

$$\begin{aligned} E(T)\le G_{0}^{-1}\left( \frac{\dfrac{L_{3}(0)}{k}+\dfrac{\beta G^{\prime }\left( \varepsilon _{0}E(0)\right) }{k}\int _{0}^{T}h(t)v\left( t\right) dt}{ \int _{0}^{T}h(t)dt}\right) , \end{aligned}$$

which gives (5.2) with $a_{1}=1,$$a_{2}=\dfrac{L_{3}(0)}{k}$ and $a_{3}=\dfrac{\beta G^{\prime }\left( \varepsilon _{0}E(0)\right) }{k}$. The proof is complete. $\square$