Existence and general decay of solution for nonlinear viscoelastic two-dimensional beam with a nonlinear delay

Abstract

We investigate the longitudinal and transversal vibrations of the viscoelastic beam with nonlinear tension and nonlinear delay term under the general decay rate for relaxation function. The existence theorem is proved by the Faedo–Galerkin method and using suitable Lyapunov functional to establish the general decay result.

1 Introduction

Consider the following viscoelastic nonlinear beam in two-dimensional space:

$$\begin{aligned} \left\{ \begin{array}{c} \rho w_{tt}+Dw_{xxxxt}+E_{I}\left\{ w-h*w\right\} _{xxxx} -\left\{ \left[ T_{0}+E_{A}\left( v_{x}+\frac{1}{2}w_{x}^{2}\right) \right] w_{x}\right\} _{x}=0, \\ \rho v_{tt}-E_{A}\left( v_{x}+\frac{1}{2}w_{x}^{2}\right) _{x}+\mu _{1}g_{1}\left( v_{t}\right) +\mu _{2}g_{2}\left( v_{t}(t-\tau )\right) =0,\\ \forall \left( x,t\right) \in \left( 0,L\right) \times \left( 0,+\infty \left[ ,\right. \right. \end{array} \right. \end{aligned}$$

(1)

with homogeneous boundary conditions

$$\begin{aligned} \left\{ \begin{array}{c} w(0,t)=w(L,t)=w_{x}(0,t)=w_{x}(L,t)=0, \\ v(0,t)=v(L,t)=0,\qquad \forall t>0, \end{array} \right. \end{aligned}$$

(2)

and initial conditions

$$\begin{aligned} \left\{ \begin{array}{l} w(x,0)=w_{0}(x),w_{t}(x,0)=w_{1}(x), \\ v(x,0)=v_{0}(x),\quad v_{t}(x,0)=v_{1}(x),\qquad \forall x\in (0,L),\\ v_{t}(x,-t)=f(x,t),\forall \left( x,t\right) \in (0,L)\times (0,\tau ), \end{array} \right. \end{aligned}$$

(3)

where x and t denotes the space variable along the beam of length L and the time variable, respectively, w(x, t) and v(x, t) are the displacements in the transverse and longitudinal directions of the beam at the position x for time t, subscripts mean partial derivatives. $\rho$, D,  $E_{I},$ $T_{0}$ and $E_{A}$ are the uniform mass per unit length of the beam, the Kelvin–Voigt damping coefficient, bending stiffness, the tension and the stiffness of the beam, respectively. $\mu _{1}$ and $\mu _{2}$ are two positive real numbers, $g_{1}$ and $g_{2}$ are two functions, $\tau$ is a time delay, and f is history function, $h*w_{xx}$ is the viscoelastic damping term defined by

$$\begin{aligned} \left( h * w_{xx}\right) (t)=\int _{0}^{t} h(t-s) w_{xx}(x, s) \mathrm {d} s, \quad \forall t\ge 0, \end{aligned}$$

which describes the relationship between the stress and the strain from the Boltzmann Principle [6, 8]. The relaxation function h represents the kernel of the memory term.

The term $g_{2}\left( v_{t}(t-\tau )\right)$ represents distributed delay term. Time delay is the property of a physical system by which the response to an applied force is delayed in its effect [31]. The presence of the delay can become a source of instability. For example, the authors in [27, 28, 30] proved that the system is unstable under the condition $\mu _{1}<\mu _{2},$ but otherwise the system is stable.

In recent years, energy decay in viscoelastic systems has become an important research title, while the behaviour of the relaxation function influences the energy decay rate. These behaviour of the relaxation function are generalized by the following extended class of kernels, namely,

$$\begin{aligned} h^{\prime }(t) \le -H(h(t)), \qquad t \ge 0, \end{aligned}$$

where H satisfying some additional conditions imposed. We refer to previous studies [9, 12,13,14, 16,17,18, 32] that proved a general energy decay rate.

For a system with delay term and viscoelastic damping, It is important to mention that the authors in [1, 15, 19, 22, 24] established the existence of solutions and general decay rates under the assumption $\mu _{1}>\mu _{2}$.

In the absence of a delay, damping term ($\mu _{1}=\mu _{2}=0$) and the relaxation function satisfies

$$h^{\prime }(t) \leqslant -\gamma h(t),$$

where $\gamma$ is a positive constant. Lekdim et al. [20] investigated the problem (1)–(3),with the following boundary conditions

$$\begin{aligned} \left\{ \begin{array}{c} w(0,t)=w(L,t)=w_{x}(0,t)=w_{x}(L,t)=0, \\ v(0,t)=0, \qquad EAv_x(L,t)=U(t),\quad \forall t>0. \end{array} \right. \end{aligned}$$

(4)

They established an exponential stability under a suitable boundary control U(t).

Motivated by the previous works, in this work we consider (1)–(3) in which we generalize the results obtained in [20], without boundary control. By expanding the class of relaxation functions into which the existence and unconditional stability are established.

The rest of our paper is organized as follows. In Sect. 2, we present some notations, assumptions and technical lemmas which will be needed later. In Sect. 3, we establish the existence and uniqueness results. The general decay rate is provided in Sect. 4.

2 Hypothesis and preliminary results

In this section, we give some notations, hypotheses and lemmas necessary to prove our results.

Notation. Let $L^{2}(0,L)$ be the Hilbert space with the inner product $\left( \cdot ,\cdot \right)$ and norms $\left\| \cdot \right\|$.

We introduce the Hilbert spaces

$$\begin{aligned} V{=}H_{0}^{1}(0,L)\cap H^{2}(0,L),\ W{=}H_{0}^{2}(0,L)\cap H^{4}(0,L) \ \mathrm {and}{ \ } Z{=}L^{2}\left( 0,L;H^{1}\left( 0,1\right) \right) . \end{aligned}$$

As in [29], we introduce the new dependent variable

$$\begin{aligned} z(x,p,t)=v_{t}(x,t-\tau p),\qquad \left( x,p\right) \in \left( 0,L\right) \times \left( 0,1\right) ,\quad t\ge 0, \end{aligned}$$

(5)

which satisfies

$$\begin{aligned} \tau z_{t}(x,p,t)+z_{p}(x,p,t)=0,\quad \left( x,p\right) \in \left( 0,L\right) \times \left( 0,1\right) . \end{aligned}$$

(6)

The problem (1)–(3) is equivalent to

$$\begin{aligned} \left\{ \begin{array} [c]{l} \rho w_{tt}+Dw_{xxxxt}+E_{I}\left\{ w-h*w\right\} _{xxxx}-\left\{ \left[ T_{0}+E_{A}\left( v_{x}+\frac{1}{2}w_{x}^{2}\right) \right] w_{x}\right\} _{x}=0,\\ \rho v_{tt}-E_{A}\left( v_{x}+\frac{1}{2}w_{x}^{2}\right) _{x}+\mu _{1} g_{1}\left( v_{t}\right) +\mu _{2}g_{2}\left( z(x,p,t)\right) =0,\\ \begin{array} [c]{lll} \tau z_{t}(x,p,t)+z_{p}(x,p,t)=0, &{} &{} p\in \left( 0,1\right) ,\\ z(x,0,t)=v_{t}(x,t), &{} &{} \forall \left( x,t\right) \in \left( 0,L\right) \times \left( 0,+\infty \left[ ,\right. \right. \\ w(x,0)=w_{0}(x),w_{t}(x,0)=w_{1}(x), &{} &{} \\ v(x,0)=v_{0}(x),\quad v_{t}(x,0)=v_{1}(x),\qquad &{} &{} \forall x\in (0,L),\\ z(x,p,0)=f(x,p\tau ),\qquad &{} &{} \forall \left( x,p\right) \in (0,L)\times (0,1), \end{array} \end{array} \right. \end{aligned}$$

(7)

with boundary conditions (2).

Hypothesis on memory kernels, the damping and the delay functions

As in [1, 25], we make the following hypotheses on the kernel functions :

(H1):

$h \in \mathcal {C}^{2}(\mathbb {R}_{+},\mathbb {R}_{+})$ is differentiable function satisfying

$$\begin{aligned} h(0)>0,\qquad 1-\int _{0}^{\infty }h(s)ds=1-\bar{h}>0. \end{aligned}$$

(8)

(H2):

There exists a $\mathcal {C}^1$ function $H:\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}$ witch is linear or strictly increasing and strictly convex $\mathcal {C}^2$ function on $(0, \epsilon ], \ \epsilon <1$, with $H(0)=H^{\prime }(0)=0$, such that

$$\begin{aligned} h^{\prime }(t) \le -H(h(t)), \qquad \forall t \ge 0. \end{aligned}$$

(9)

For the weight of the delay, following [4, 10], we assume that

(H3):

$g_{1}\in \mathcal {C} (\mathbb {R},\mathbb {R})$ is nondecreasing function, such that there exist $\epsilon ,$ $c_{1},$ $c_{2}>0$, such that

$$\begin{aligned} \left\{ \begin{array}{lll} c_{1}\left| s\right| \le \left| g_{1}\left( s\right) \right| \le c_{2}\left| s\right| &{} \mathrm {if} &{} \left| s\right| \ge \epsilon , \\ s^{2}+g_{1}^{2}\left( s\right) \le H^{-1}\left( sg_{1}\left( s\right) \right) &{} \mathrm {if} &{} \left| s\right| \le \epsilon . \end{array} \right. \end{aligned}$$

(10)

(H4):

Let $g_{2}\in \mathcal {C}^{1}(\mathbb {R},\mathbb {R})$ odd nondecreasing function, such that there exist $c_{3},$ $a_{1},$ $a_{2}>0$ with

$$\begin{aligned} \left| g_{2}^{\prime }\left( s\right) \right|\le & {} c_{3} \end{aligned}$$

(11)

$$\begin{aligned} a_{1}sg_{2}\left( s\right)\le & {} G(s)\le a_{2}sg_{1}\left( s\right) , \end{aligned}$$

(12)

where $G(s)=\int _{0}^{s}g_{2}(r)dr$ and

$$\begin{aligned} a_{2}\mu _{2}<a_{1}\mu _{1}. \end{aligned}$$

(13)

Remark 1

(see [26]) By (H1), we have $\lim _{s \rightarrow +\infty } h(s)=0$ and assume that $\lim _{s \rightarrow +\infty } h'(s)=0$. This implies that there exists $t_0>0$ large enough such that

$$\begin{aligned} \max \left\{ h(t),-h^{\prime }(t)\right\} <\min \left\{ \epsilon , H(\epsilon ), H_{0}(\epsilon )\right\} , \quad \forall t \ge t_{0}, \end{aligned}$$

(14)

where $H_0(t)=H(D(t))$ provided that D is a positive $\mathcal {C}^1$ function, $D (0) = 0$, for which $H_0$ is strictly increasing and strictly convex $\mathcal {C}^2$ function on $(0,\epsilon ]$ and

$$\begin{aligned} \int _{0}^{+\infty } \frac{h(s)}{H_{0}^{-1}\left( -h^{\prime }(s)\right) } d s<+\infty . \end{aligned}$$

(15)

By the nonincreasing of h, we get

$$\begin{aligned} 0<h\left( t_{0}\right) \le h(t) \le h(0), \quad \forall t \in \left[ 0, t_{0}\right] , \end{aligned}$$

(16)

the continuity and positivity of H imply that

$$\begin{aligned} a \le H(h(t)) \le b, \quad \forall t \in \left[ 0, t_{0}\right] , \end{aligned}$$

(17)

for some positive constants a and b. Then there exists $\gamma >0$ such that

$$\begin{aligned} h^{\prime }(t) \le -\gamma h(t), \quad \forall t \in \left[ 0, t_{0}\right] . \end{aligned}$$

(18)

We define the energy functional of problem (1)–(3) by

$$\begin{aligned} E(t)= & {} \frac{\rho }{2}\left[ \left\| w_{t}\right\| ^{2}+\left\| v_{t}\right\| ^{2}\right] +\left( 1-\int \nolimits _{0}^{t}h(s)ds\right) \frac{E_{I}}{2}\left\| w_{xx}\right\| ^{2} \nonumber \\&+\frac{T_{0}}{2}\left\| w_{x}\right\| ^{2} +\frac{E_{A}}{2}\left\| v_{x}+\frac{1}{2}w_{x}^{2}\right\| ^{2}\nonumber \\&+\frac{E_{I}}{2}\left( h\circ w_{xx}\right) +\xi \int _{0}^{L}\int _{0}^{1}G\left( z(x,p,t)\right) dpdx. \end{aligned}$$

(19)

where $\xi$ is a positive constant such that

$$\begin{aligned} \tau \frac{\mu _{1}\left( 1-a_{1}\right) }{a_{1}}<\xi <\tau \frac{\mu _{1}-a_{2}\mu _{1}}{a_{2}} \end{aligned}$$

and

$$\begin{aligned} \left( h\circ u\right) (t) = \int \nolimits _{0}^{t}h(t-s)\left\| u(t)-u(s)\right\| ^{2}ds. \end{aligned}$$

Lemma 1

Let (w, v, z) be a solution of the problem (7), then, for $t\ge 0$, the time derivative of E(t) can be upper bounded by

$$\begin{aligned} E^{\prime }(t)\le & {} -D\left\| w_{xxt}\right\| ^{2}+\frac{E_{I}}{2} h^{\prime }\circ w_{xx}-\frac{E_{I}}{2}h(t)\left\| w_{xx}\right\| ^{2} \nonumber \\&-\eta _{1}\left( v_{t},g_{1}\left( v_{t}\right) \right) -\eta _{2}\left( z(x,1,t),g_{2}\left( z(x,1,t)\right) \right) \le 0. \end{aligned}$$

(20)

where $\eta _{1}=\mu _{1}-a_{2}\frac{\xi }{\tau }-a_{2}\mu _{2}$ and $\eta _{2}=a_{1}\frac{\xi }{\tau }-\mu _{2}\left( 1-a_{1}\right) .$

Proof

Taking the inner product in $L^{2}(0,L)$ of the first equation of (7) with $w_{t}$, the second equation with $v_{t},$ , then integrating by parts, we obtain

$$\begin{aligned}&\frac{d}{dt}\left\{ \frac{\rho }{2}\left[ \left\| w_{t}\right\| ^{2}+\left\| v_{t}\right\| ^{2}\right] +\frac{E_{I}}{2}\left\| w_{xx}\right\| ^{2}+\frac{T_{0}}{2}\left\| w_{x}\right\| ^{2}+\frac{ E_{A}}{2}\left\| v_{x}+\frac{1}{2}w_{x}^{2}\right\| ^{2}\right\} \nonumber \\&\qquad =-D\left\| w_{xxt}\right\| ^{2}+\left( h*w_{xx},w_{xxt}\right) -\mu _{1}\left( v_{t},g_{1}\left( v_{t}\right) \right) -\mu _{2}\left( v_{t},g_{2}\left( z(x,1,t)\right) \right) .\nonumber \\ \end{aligned}$$

(21)

The second term of the right hand side of the above equality gives

$$\begin{aligned} 2\left( h*w_{xx},w_{xxt}\right)= & {} h^{\prime }\circ w_{xx}-h(t)\left\| w_{xx}\right\| ^{2} \nonumber \\&-\frac{d}{dt}\left\{ \left( h\circ w_{xx}\right) -\left( \int \nolimits _{0}^{t}h(s)ds\right) \left\| w_{xx}\right\| ^{2}\right\} . \end{aligned}$$

(22)

By multiplying the third equation of (7) by $\xi g_{2}\left( z(x,p,t)\right)$ then integrating over $\left( 0,L\right) \times \left( 0,1\right) ,$ we find

$$\begin{aligned} \xi \int _{0}^{1}\int _{0}^{L}z_{t}g_{2}\left( z(x,p,t)\right) dpdx= & {} -\frac{ \xi }{\tau }\int _{0}^{1}\int _{0}^{L}\frac{\partial }{\partial p}G\left( z(x,p,t)\right) dpdx \nonumber \\= & {} -\frac{\xi }{\tau }\int _{0}^{L}\left[ G\left( z(x,1,t)\right) -G\left( z(x,0,t)\right) \right] dx.\nonumber \\ \end{aligned}$$

(23)

Combining (21)–(23), using the fact that $z(0,t)=v_{t}(t)$ and assumption (H4),  we get

$$\begin{aligned} \frac{d}{dt}E(t)= & {} -D\left\| w_{xxt}\right\| ^{2}+\frac{E_{I}}{2} h^{\prime }\circ w_{xx}-\frac{E_{I}}{2}h(t)\left\| w_{xx}\right\| ^{2}- \frac{\xi }{\tau }\int _{0}^{L}G\left( z(x,1,t)\right) dx \nonumber \\&-\left( \mu _{1}-a_{2}\frac{\xi }{\tau }\right) \left( v_{t},g_{1}\left( v_{t}\right) \right) -\mu _{2}\left( v_{t},g_{2}\left( z(x,1,t)\right) \right) . \end{aligned}$$

(24)

The conjugate function $G^{*}$ of the differentiable convex function G (see [7], pp. 9), i.e.

$$ G^{*}(s)=\sup _{t\ge 0}\left( st-G(t)\right) .$$

On the other hand $G^{*}$ is the Legendre transform of G. We refer to ( [2], pp. 61-62), that is given by

$$\begin{aligned} G^{*}(s)=s\left( G^{\prime }\right) ^{-1}(s)-G\left[ \left( G^{\prime }\right) ^{-1}(s)\right] ,\qquad \forall s\ge 0, \end{aligned}$$

(25)

which satisfies the generalized Young inequality :

$$\begin{aligned} st\le G^{*}(s)+G(t),\qquad \forall t,s\ge 0. \end{aligned}$$

(26)

By the definition of G, we obtain

$$\begin{aligned} G^{*}(s)=sg_{2}^{-1}(s)-G\left[ g_{2}^{-1}(s)\right] . \end{aligned}$$

(27)

Applying (26) with $s=g_{2}\left( z(1,t)\right)$ and $t=v_{t}(t)$, from the last term of (24), we obtain

$$\begin{aligned} \frac{d}{dt}E(t)\le & {} -D\left\| w_{xxt}\right\| ^{2}{+}\frac{E_{I}}{2} h^{\prime }\circ w_{xx}{-}\frac{E_{I}}{2}h(t)\left\| w_{xx}\right\| ^{2}{-}\left( \mu _{1}{-}a_{2}\frac{\xi }{\tau }\right) \left( v_{t},g_{1}\left( v_{t}\right) \right) \nonumber \\&+\mu _{2}\int _{0}^{L}\left( G(v_{t})+G^{*}(g_{2}\left( z\left( x,1,t\right) \right) )\right) dx-\frac{\xi }{\tau }\int _{0}^{L}G\left( z(x,1,t)\right) dx. \nonumber \\ \end{aligned}$$

(28)

The equality (27) and assumption (H4) imply that

$$\begin{aligned} G^{*}(g_{2}\left( z(x,1,t)\right) )= & {} z(x,1,t)g_{2}\left( z(x,1,t)\right) -G\left( z(x,1,t)\right) \nonumber \\\le & {} \left( 1-a_{1}\right) z(x,1,t)g_{2}\left( z(x,1,t)\right) . \end{aligned}$$

(29)

Combining (28) and (29), we have (20). $\square$

Remark 2

The Lemma 1 imply that E(t) nonincreasing, moreover

$$\begin{aligned} E(t)\le E(0),\qquad \forall t\ge 0. \end{aligned}$$

(30)

The following lemmas will be used frequently in the sequel

Lemma 2

([11]) Let $u\in C^{1}(\left[ 0,L\right] )$ satisfying $u(0,t)=0$. Then the following inequality hold:

$$\begin{aligned} \left\| u^{2}(t)\right\| _{\infty } \le 2\left\| u(t)\right\| \left\| u_{x}(t)\right\| ,\qquad \forall t\ge 0, \end{aligned}$$

where $\left\| .\right\| _{\infty }$ is the norm of $L^{\infty }(\left[ 0,L\right] ).$

Lemma 3

([23]) If w is a solution of problem (1)–(3), assuming that h satisfies (H1) and (H2). then we have

$$\begin{aligned} \int _{0}^{L}\left( h\diamond w\right) _{xx}^{2}dx\le & {} \bar{h}\left( h\circ w_{xx}\right) , \end{aligned}$$

(31)

$$\begin{aligned} \int _{0}^{L}\left( h^{\prime }\diamond w\right) _{xx}^{2}dx\le & {} -h(0)\left( h^{\prime }\circ w_{xx}\right) , \end{aligned}$$

(32)

where $\left( h\diamond u\right) (t) =\int \nolimits _{0}^{t}h(t-s)\left( u(s)-u(t)\right) ds.$

3 Well possedness

The main aim of this section is to prove the following existence and uniqueness theorem:

Theorem 1

Let $\left( w_{0},v_{0},z_{0}\right) \in W\times V\times Z$ and $\left( w_{1},v_{1}\right) \in H_{0}^{2}\left( 0,L\right) \times H_{0}^{1}\left( 0,L\right)$. Assumee that (H1)–(H4) hold and satisfy

$$\begin{aligned} z(x,p,t)=v_{t}(x,t-\tau p),\quad \left( x,p\right) \in \left( 0,L\right) \times \left( 0,1\right) ,t\ge 0. \end{aligned}$$

(33)

Then the system (1)–(3) has a unique solution $\left( w,v,z\right)$ in the sense that

$$\begin{aligned} w\in & {} L^{\infty }\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) ,\quad v\in L^{\infty }\left( 0,T;H_{0}^{1}\left( 0,L\right) \right) ,\quad z\in L^{\infty }\left( 0,T;Z\right) . \\ w_{t}\in & {} L^{\infty }\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) , \quad v_{t}{\in } L^{\infty }\left( 0,T;H_{0}^{1}\left( 0,L\right) \right) ,\quad z_{t}{\in } L^{\infty }\left( 0,T;L^{2}\left( \left( 0,L\right) \times \left( 0,1\right) \right) \right) . \\ w_{tt}\in & {} L^{\infty }\left( 0,T;L^{2}\left( 0,L\right) \right) \cap L^{2}\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) ,\quad v_{tt}\in L^{\infty }\left( 0,T;L^{2}\left( 0,L\right) \right) . \end{aligned}$$

Proof

We employ the Faedo–Galerkin technique to construct a solution.

Approximate solutions: Let $\left( w_{i},v_{i},z_{i}\right) _{i\le m}$ be a complete orthogonal system of $W\times V\times Z.$ For each $m\in \mathbb {N},$ let $W_{1}^{m}=span\left\{ w_{1},w_{2},\ldots ,w_{m}\right\} ,$ $W_{2}^{m}=span\left\{ v_{1},v_{2},\ldots ,v_{m}\right\}$ and $W_{3}^{m}=span\left\{ z_{1},z_{2},\ldots ,z_{m}\right\} ,$ such as the sequence $z_{i}(x,p)$ defined by $z_{i}(x,0)=v_{i}(x),$ we prolong $z_{i}(x,0)$ in Z by $z_{i}(x,p)$. For $w(0),w_{t}(0)\in W_{1}^{m},$ $v(0),v_{t}(0)\in W_{2}^{m}$ and $z(p,0,s)\in W_{3}^{m}$, searching for functions $w^{m}(x,t)=\sum _{i=1}^{m}k_{i}^{1}(t)w_{i}(x),$ $v^{m}(x,t)=\sum _{i=1}^{m}k_{i}^{2}(t)v_{i}(x)$ and $z^{m}(x,t,p)= \sum _{i=1}^{m}k_{i}^{3}(t)z_{i}(x,p),$ that satisfy the following equations

$$\begin{aligned}&\rho \left( w_{tt}^{m},\varphi \right) +D\left( w_{xxt}^{m},\varphi _{xx}\right) +E_{I}\left( w_{xx}^{m},\varphi _{xx}\right) +\left( T_{0}w_{x}^{m},\varphi _{x}\right) \nonumber \\&\qquad +E_{A}\left( \left( v_{x}^{m}+\frac{1}{2}\left( w_{x}^{m}\right) ^{2}\right) w_{x}^{m},\varphi _{x}\right) -E_{I}(h*w_{xx}^{m},\varphi _{xx})=0, \end{aligned}$$

(34)

$$\begin{aligned}&\rho \left( v_{tt}^{m},\phi \right) +E_{A}\left( v_{x}^{m},\phi _{x}\right) + \frac{E_{A}}{2}\left( \left( w_{x}^{m}\right) ^{2},\phi _{x}\right) \nonumber \\&\qquad +\left( \mu _{1}g_{1}\left( v_{t}^{m}\right) +\mu _{2}g_{2}\left( z^{m}(x,1,t)\right) ,\phi \right) =0 \end{aligned}$$

(35)

and

$$\begin{aligned} \int _{0}^{1}\left( \tau \left( z_{t}^{m},\psi \right) +\left( z_{p}^{m},\psi \right) \right) dp=0, \end{aligned}$$

(36)

for all $\left( \varphi ,\phi ,\psi \right) \in H_{0}^{2}\left( 0,L\right) \times H_{0}^{1}\left( 0,L\right) \times L^{2}\left( \left( 0,L\right) \times \left( 0,1\right) \right) ,$ with the initial conditions

$$\begin{aligned}&\left( w^{m}(0),v^{m}(0),z^{m}(p,0,s)\right) =\left( w_{0}^{m},v_{0}^{m},z_{0}^{m}\right) \rightarrow \left( w_{0},v_{0},z_{0}\right) \quad \mathrm {in }\quad W\times V\times Z, \\&\quad \left( w_{t}^{m}(0),v_{t}^{m}(0)\right) =\left( w_{1}^{m},v_{1}^{m}\right) \rightarrow \left( w_{1},v_{1}\right) \quad \mathrm {in }\quad H_{0}^{2}\left( 0,L\right) \times H_{0}^{1}\left( 0,L\right) . \end{aligned}$$

A Priori Estimates

Throughout this part, $B_{i},i=1,2,\ldots ,$ denote positive constants independent of m and $t\in \left[ 0,T\right] .$

Estimate 1: Let $E_{m}$ the energy defined by (19), for the solutions $w^{m},$ $v^{m}$ and $z^{m}.$ By utilizing the same steps used in the proof of Lemma 1, we find

$$\begin{aligned}&E_{m}(t)+\int _{0}^{t}\left[ D\left\| w_{xxt}^{m}\right\| ^{2}+\eta _{1}\left( v_{t}^{m},g_{1}\left( v_{t}^{m}\right) \right) \right] ds \nonumber \\&\quad +\int _{0}^{t}\eta _{2}\left( z^{m}(1,s),g_{2}\left( z^{m}(1,s)\right) \right) ds \le E_{m}(0)\le B_{1}. \end{aligned}$$

(37)

where $\eta _{1}$, $\eta _{2}$ positive constants.

Estimate 2: Firstly, we estimate $\left\| w_{tt}^{m}(0)\right\| ^{2},$ $\left\| v_{tt}^{m}(0)\right\| ^{2}$ and $\left\| z_{t}^{m}(0)\right\| ^{2}$.

Fixed $t=0$ and taking $\varphi =w_{tt}^{m}(0),$ $\phi =v_{tt}^{m}(0)$ and $\psi =z_{t}^{m}(0)$ in (34), (35) and (36), respectively, then integrate them by parts and apply Young’s inequality. From the assumptions (H3), (H4) and the initial data are sufficiently smooth, we can be infer that

$$\begin{aligned} \left\| w_{tt}^{m}(0)\right\| ^{2},\left\| v_{tt}^{m}(0)\right\| ^{2},\int _{0}^{1}\left\| z_{t}^{m}(0)\right\| ^{2}dp\le B_{2}. \end{aligned}$$

(38)

Now, we estimate $\left\| w_{tt}^{m}\right\| ^{2},$ $\left\| v_{tt}^{m}\right\| ^{2}$ and $\left\| z_{t}^{m}\right\| ^{2}$.

Let us fix t, $\zeta >0$ such that $\zeta <T-t$. Taking the difference of (34), (35) and (36) with $t=t+\zeta$ and $t=t$, and simultaneously replacing $\varphi ,$ $\phi$ and $\psi$ with $w_{t}^{m}(t+\zeta )-w_{t}^{m}(t),$ $v_{t}^{m}(t+\zeta )-v_{t}^{m}(t)$ and $\left( z^{m}(p,t+\zeta )-z^{m}(p,t)\right)$, respectively. Then gather the two first relations and integrating the last relation over (0, 1) , we get

$$\begin{aligned}&\frac{\rho }{2}\frac{d}{dt}\left\| w_{t}^{m}(t+\zeta )-w_{t}^{m}(t)\right\| ^{2}+D\left\| w_{xxt}^{m}(t+\zeta )-w_{xxt}^{m}(t)\right\| ^{2}\nonumber \\&\quad +\, \frac{E_{I}}{2}\frac{d}{dt}\left\| w_{xx}^{m}(t+\zeta )-w_{xx}^{m}(t)\right\| ^{2}+\frac{T_{0}}{2}\frac{d}{dt}\left\| w_{x}^{m}(t+\zeta )-w_{x}^{m}(t)\right\| ^{2}\nonumber \\&\quad +\, \frac{\rho }{2}\frac{d}{dt}\left\| v_{t}^{m}(t+\zeta )-v_{t}^{m}(t)\right\| ^{2}+\frac{E_{A}}{2}\frac{d}{dt}\left\| v_{x}^{m}(t+\zeta )-v_{x}^{m}(t)\right\| ^{2} \nonumber \\&\quad +\,\mu _{2}\left( v_{t}^{m}(t+\zeta )-v_{t}^{m}(t),g_{2}\left( z^{m}(1,t+\zeta )\right) -g_{2}\left( z^{m}(1,t)\right) \right) \nonumber \\&\quad +\, \mu _{1}\left( v_{t}^{m}(t+\zeta )-v_{t}^{m}(t),g_{1}\left( v_{t}^{m}(t+\zeta )\right) -g_{1}\left( v_{t}^{m}(t)\right) \right) =F_{1}+F_{2}, \end{aligned}$$

(39)

and

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\tau \int _{0}^{1}\left\| z^{m}(p,t{+}\zeta )-z^{m}(p,t)\right\| ^{2}dp= & {} {-}\frac{1}{2}\int _{0}^{1}\frac{d}{dp} \left\| z^{m}(p,t{+}\zeta ){-}z^{m}(p,t)\right\| ^{2}dp \nonumber \\= & {} \frac{1}{2}\left\| v_{t}^{m}(t+\zeta ) -v_{t}^{m}(t)\right\| ^{2}\nonumber \\&-\frac{1}{2}\left\| z^{m}(1,t+\zeta ) -z^{m}(1,t)\right\| ^{2}, \end{aligned}$$

(40)

where

$$\begin{aligned} F_{1} =&-\frac{E_{A}}{2}\int _{0}^{L}\left[ \left( w_{x}^{m}(t+\zeta )\right) ^{3}-\left( w_{x}^{m}(t)\right) ^{3}\right] \left[ w_{xt}^{m}(t+\zeta )-w_{xt}^{m}(t)\right] dx\\&-E_{A}\int _{0}^{L}\left[ w_{x}^{m}v_{x}^{m}(t+\zeta )-w_{x}^{m}v_{x}^{m}(t)\right] \left[ w_{xt}^{m}(t+\zeta )-w_{xt}^{m}(t)\right] dx \\&-\frac{E_{A}}{2}\int _{0}^{L}\left[ \left( w_{x}^{m}(t+\zeta )\right) ^{2}-\left( w_{x}^{m}(t)\right) ^{2}\right] \left[ v_{xt}^{m}(t+\zeta )-v_{xt}^{m}(t)\right] dx, \end{aligned}$$

and

$$\begin{aligned} F_{2}=E_{I}\left( \int _{0}^{t{+}\zeta }h(t{+}\zeta {-}s)w_{xx}(s)ds{-}\int _{0}^{t}h(t{-}s)w_{xx}(s)ds,w_{xxt}(t{+}\zeta ){-}w_{xxt}(t)\right) . \end{aligned}$$

Taking the first estimate, Young, Poincaré’s inequalities and Lemma 2 into account, we can estimate $F_1$ and $F_2$ as follows :

$$\begin{aligned} \left| F_{1}\right|\le & {} \frac{D}{4}\left\| w_{xxt}^{m}(t+\zeta )-w_{xxt}^{m}(t)\right\| ^{2}+B_{3}\left\| w_{x}^{m}(t+\zeta )-w_{x}^{m}(t)\right\| ^{2} \nonumber \\&+B_{3}\left( \left\| w_{xx}^{m}(t+\zeta )-w_{xx}^{m}(t)\right\| ^{2}+\left\| v_{t}^{m}(t+\zeta )-v_{t}^{m}(t)\right\| ^{2}\right) , \end{aligned}$$

(41)

$$\begin{aligned} \left| F_{2}\right|\le & {} B_{4}\int _{0}^{L}\left( \int \nolimits _{0}^{t+\zeta }h(t+\zeta -s)w_{xx}(s)ds-\int \nolimits _{0}^{t}h(t-s)w_{xx}(s)ds\right) ^{2}dx \nonumber \\&+\frac{D}{4}\left\| w_{xxt}(t+\zeta )-w_{xxt}(t)\right\| ^{2}. \end{aligned}$$

(42)

Combining (39)–(42), then dividing both sides by $\zeta ^{2}$ and taking the limit as $\zeta \rightarrow 0$, we get

$$\begin{aligned}&\frac{\rho }{2}\frac{d}{dt}\left\| w_{tt}^{m}(t)\right\| ^{2}+\frac{D }{2}\left\| w_{xxtt}^{m}(t)\right\| ^{2}+\frac{E_{I}}{2}\frac{d}{dt} \left\| w_{xxt}^{m}(t)\right\| ^{2} +\frac{T_{0}}{2}\frac{d}{dt}\left\| w_{xt}^{m}(t)\right\| ^{2} \nonumber \\&\qquad +\frac{ \rho }{2}\frac{d}{dt}\left\| v_{tt}^{m}(t)\right\| ^{2}+\frac{E_{A}}{2} \frac{d}{dt}\left\| v_{xt}^{m}(t)\right\| ^{2}+\mu _{2}\left( v_{tt}^{m}(t)z_{t}^{m}(1,t),g_{2}^{\prime }\left( z^{m}(1,t)\right) \right) \nonumber \\&\qquad +\mu _{1}\left( \left( v_{tt}^{m}(t)\right) ^{2},g_{1}^{\prime }\left( v_{t}^{m}(t)\right) \right) +\frac{1}{2}\frac{d}{dt}\tau \int _{0}^{1}\left\| z_{t}^{m}(p,t)\right\| ^{2}dp+\frac{1}{2}\left\| z_{t}^{m}(1,t)\right\| ^{2} \nonumber \\&\quad \le B_{5}\left( \left\| w_{xt}^{m}(t)\right\| ^{2}+\left\| w_{xxt}^{m}(t)\right\| ^{2}+\left\| v_{tt}^{m}(t)\right\| ^{2}\right) \nonumber \\&\qquad +B_{4}\int _{0}^{L}\left( h(0)w_{xx}^{m}+\int \nolimits _{0}^{t}h^{\prime }(t-s)w_{xx}^{m}(s)ds\right) ^{2}dx. \end{aligned}$$

(43)

On the other hand, we have

$$\begin{aligned}&\mu _{1}\left( \left( v_{tt}^{m}(t)\right) ^{2},g_{1}^{\prime }\left( v_{t}^{m}(t)\right) \right) \ge 0, \end{aligned}$$

(44)

$$\begin{aligned}&\mu _{2}\left( v_{tt}^{m}(t)z_{t}^{m}(1,t),g_{2}^{\prime }\left( z^{m}(1,t)\right) \right) \le \frac{\mu _{2}c_{3}}{2}\left( \left\| z_{t}^{m}(1,t)\right\| ^{2}+\left\| v_{tt}^{m}(t)\right\| ^{2}\right) , \end{aligned}$$

(45)

and

$$\begin{aligned} \int _{0}^{L}\int \nolimits _{0}^{t}h^{\prime }(t-s)w_{xx}^{m}(s)dsdx\le \sup _{ \left[ 0,T\right] }\left\| w_{xx}^{m}\right\| \left( \int \nolimits _{0}^{T}\left| h^{\prime }(s)\right| ds\right) \le B_{6}. \end{aligned}$$

(46)

Integrating (43) over $\left( 0,t\right) ,$ taking (37), (44)–(46) under consideration and noting that the initial data are sufficiently smooth. By Gronwall’s lemma, we conclude

$$\begin{aligned}&\left\| w_{tt}^{m}\right\| ^{2}+\int _{0}^{t}\left\| w_{xxtt}^{m}\right\| ^{2}ds+\left\| w_{xxt}^{m}\right\| ^{2}+\left\| w_{xt}^{m}\right\| ^{2}+\left\| v_{tt}^{m}\right\| ^{2} \nonumber \\&\quad +\left\| v_{xt}^{m}\right\| ^{2} +\tau \int _{0}^{1}\left\| z_{t}^{m}(p,t)\right\| ^{2}dp+\int _{0}^{t}\left\| z_{t}^{m}(1,s)\right\| ^{2}dpds \le B_{7}e^{B_{8}T}. \end{aligned}$$

(47)

Passage to the limit

The estimate (37) and (38) permits to deduce

$$\begin{aligned} \left\{ \begin{array}{l} \left( w^{m},v^{m}\right) ,\ \left( w_{t}^{m},v_{t}^{m}\right) \ \ \mathrm {are}\ \mathrm {bounded}\ \mathrm {in}\ L^{\infty }\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) \times L^{\infty }\left( 0,T;H_{0}^{1}\left( 0,L\right) \right) , \\ \left( w_{tt}^{m},v_{tt}^{m}\right) \ \mathrm {is}\ \mathrm {bounded}\ \mathrm {in}\ L^{\infty }\left( 0,T;L^{2}(0,L)\right) \cap L^{2}\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) \times L^{\infty }\left( 0,T;L^{2}(0,L)\right) , \\ \left( z^{m},z_{t}^{m}\right) \ \mathrm {is}\ \mathrm {bounded}\ \mathrm {in}\ L^{\infty }\left( 0,T;Z\right) \times L^{\infty }\left( 0,T;L^{2}(\left( 0,L\right) \times \left( 0,1\right) )\right) , \\ v_{t}^{m}g_{1}\left( v_{t}^{m}\right) , \ z^{m}\left( 1,t\right) g_{2}\left( z^{m}\left( 1,t\right) \right) \ \mathrm {are}\ \mathrm {bounded}\ \mathrm {in}\ L^{1}(\left( 0,L\right) \times \left( 0,T\right) ), \\ G\left( z^{m}\right) \ \mathrm {is}\ \mathrm {bounded}\ \mathrm {in}\ L^{\infty }\left( 0,T;L^{2}(\left( 0,L\right) \times \left( 0,1\right) )\right) \\ \left( v_{x}+\frac{1}{2}\left( w_{x}^{m}\right) ^{2}\right) \ \mathrm {is}\ \mathrm {bounded}\ \mathrm {in}\ L^{\infty }\left( 0,T;L^{2}(0,L)\right) . \end{array} \right. \end{aligned}$$

(48)

Therefore, there exists subsequences of $\left( w^{m}\right) ,$ $\left( v^{m}\right)$ and $\left( z^{m}\right)$, still denoted by $\left( w^{m}\right) ,$ $\left( v^{m}\right)$ and $\left( z^{m}\right) ,$ respectively, such that

$$\begin{aligned} \left\{ \begin{array}{l} \left( w^{m},w_{t}^{m}\right) {\rightarrow } \left( w,w_{t}\right) \ \mathrm {weak} \ \mathrm {star} \ \mathrm {in} \ L^{\infty }\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) {\times } L^{\infty }\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) , \\ \left( v^{m},v_{t}^{m}\right) \rightarrow \left( v,v_{t}\right) \ \mathrm {weak} \ \mathrm {star} \ \mathrm {in}{ \ } L^{\infty }\left( 0,T;H_{0}^{1}\left( 0,L\right) \right) \times L^{\infty }\left( 0,T;H_{0}^{1}\left( 0,L\right) \right) , \\ w_{tt}^{m}\rightarrow w_{tt} \ \mathrm {weak} \ \mathrm {star} \ \mathrm {in} \ L^{\infty }\left( 0,T;L^{2}(0,L)\right) \cap L^{2}\left( 0,T;H_{0}^{2}\left( 0,L\right) \right) , \\ v_{tt}^{m}\rightarrow v_{tt} \ \mathrm {weak} \ \mathrm {star} \ \mathrm {in} \ L^{\infty }\left( 0,T;L^{2}(0,L)\right) , \\ \left( z^{m},z_{t}^{m}\right) \rightarrow \left( z,z_{t}\right) \ \mathrm {weak} \ \mathrm {star} \ \mathrm {in} \ L^{\infty }\left( 0,T;Z\right) \times L^{\infty }\left( 0,T;L^{2}(\left( 0,L\right) \times \left( 0,1\right) )\right) , \\ \left( g_{1}\left( v^{m}\right) ,g_{2}\left( z_{t}^{m}\right) \right) \rightarrow \left( G_{1},G_{2}\right) \ \mathrm {weak}\, \mathrm {star} \ \mathrm {in} \ \left[ L^{2}(\left( 0,T\right) \times \left( 0,L\right) )\right] ^{2},\\ \left( v_{x}^{m}+\frac{1}{2}\left( w_{x}^{m}\right) ^{2}\right) \rightarrow \varGamma \ \mathrm {weak} \ \mathrm {star} \ \mathrm {in} \ L^{\infty }\left( 0,T;L^{2}(0,L)\right) . \end{array} \right. \end{aligned}$$

(49)

Analysis of the nonlinear terms

By Aubin-Lions compactness (see [3]), we conclude from (49), that

$$\begin{aligned} \begin{array}{l} w^{m}\rightarrow w \ \mathrm {strongly} \ \mathrm {in} {\ } W^{1,\infty }\left( 0,T;H_{0}^{1}\left( 0,L\right) \right) , \\ v^{m}\rightarrow v \ \mathrm {strongly} \ \mathrm {in}\,W^{1,\infty }\left( 0,T;L^{2}\left( 0,L\right) \right) , \end{array} \end{aligned}$$

(50)

therefore

$$\begin{aligned} z^{m}\rightarrow z \ \mathrm {strongly}\, \ \mathrm {and} \ \mathrm {a.e} \ \mathrm {in} \ \left[ 0,T\right] \times \left[ 0,L\right] . \end{aligned}$$

(51)

Lemma 4

([5]) We have the convergence $g_{1}\left( v_{t}^{m}\right) \rightarrow g_{1}\left( v_{t}\right)$ and $g_{2}\left( z^{m}\right) \rightarrow g_{2}\left( z\right)$ in $L^{1}\left( \left[ 0,T \right] \times \left[ 0,L\right] \right) .$ Hence,

$$\begin{aligned} \begin{array}{l} g_{1}\left( v_{t}^{m}\right) \rightarrow g_{1}\left( v_{t}\right) \ \mathrm {weak} \ \mathrm {in} \ L^{2}\left( \left[ 0,T\right] \times \left[ 0,L\right] \right) ,\\ g_{2}\left( z^{m}\right) \rightarrow g_{2}\left( z\right) \ \mathrm {weak} \ \mathrm {in} \ L^{2}\left( \left[ 0,T\right] \times \left[ 0,L\right] \right) . \end{array} \end{aligned}$$

(52)

From (50) Lions Lemma ([21], pp. 12), we concluded that $\varGamma =\left( v_{x}+\frac{1}{2}\left( w_{x}\right) ^{2}\right)$ and

$$\begin{aligned} \left( v_{x}^{m}+\frac{1}{2}\left( w_{x}^{m}\right) ^{2}\right) w_{x}^{m}\rightarrow \left( v_{x}+\frac{1}{2}\left( w\right) ^{2}\right) w_{x} \mathrm {\ weakly \ in \ } L^{2}\left( \left[ 0,T\right] \times \left[ 0,L \right] \right) . \end{aligned}$$

(53)

Now, we can pass to the limit in the approximate problem (34)–(35) to get a weak solution of the problem (1)–(3), (see [21, 33]).

Uniqueness The uniqueness can be proved by following the same procedures as in estimation 2. $\square$

4 Asymptotic behavior

The prove of energy decay relies heavily on the construction of Lyapunov functional and exploitation of convex analysis. For this intent, we start by constructing a Lyapunov functional :

$$\begin{aligned} \mathcal {L}(t)=E(t)+\beta _{1}\varPhi (t)+\beta _{2}\varPsi (t)+\beta _{3}\chi (t) \end{aligned}$$

(54)

where $\beta _{1}, \ \beta _{2}$ and $\beta _{3}$ are positive constants, E(t) is given by (19) and

$$\begin{aligned} \varPhi (t)= & {} \frac{\rho }{2}\left( w,w_{t}\right) +\rho \left( v,v_{t}\right) +\frac{D}{4}\left\| w_{xx}\right\| ^{2}, \end{aligned}$$

(55)

$$\begin{aligned} \varPsi (t)= & {} \rho \left( \left( h\diamond w\right) ,w_{t}\right) . \end{aligned}$$

(56)

$$\begin{aligned} \chi (t)= & {} \int _{0}^{L}\int _{0}^{1}e^{-2\tau p}G\left( z\left( x,p,t\right) \right) dpdx \end{aligned}$$

(57)

It is our aim to prove that functional $\mathcal {L}(t)$ satisfies an estimates. To pass this estimate to E(t), we will need the following proposition.

Proposition 1

Let $\mathcal {L}(t)$ and E(t) be the functional defined by (54) and (19), respectively. Then, for $\beta _{1}$ $\beta _{2}$ and $\beta _{3}$ small enough, we have

$$\begin{aligned} \alpha _{1}E(t)\le \mathcal {L}(t)\le \alpha _{2}E(t),\qquad \forall t>0. \end{aligned}$$

(58)

where $\alpha _{1}$ and $\alpha _{2}$ are positive constants.

Proof

By Young’s inequality, we have

$$\begin{aligned} \left| \varPhi (t)\right|\le & {} \frac{D}{4}\left\| w_{xx}\right\| ^{2}+\frac{\rho }{4}\left\| w_{t}\right\| ^{2}+\frac{ \rho L^{2}}{4}\left\| w_{x}\right\| ^{2} +\frac{\rho }{2}\left\| v_{t}\right\| ^{2}+\frac{\rho L^{2}}{2} \left\| v_{x}\right\| ^{2}. \end{aligned}$$

(59)

Obviously,

$$\begin{aligned} \left\| v_{x}\right\| ^{2}\le 2\left\| v_{x}+\frac{1}{2} w_{x}^{2}\right\| ^{2}+\frac{1}{2}\left\| w_{x}^{2}\right\| ^{2}, \end{aligned}$$

then using Holder’s inequality, Lemma 2 and inequality (30), we have

$$\begin{aligned} \left\| v_{x}\right\| ^{2}\le 2\left\| v_{x}+\frac{1}{2} w_{x}^{2}\right\| ^{2}+\frac{A_{1}}{2}\left\| w_{x}\right\| ^{2}, \end{aligned}$$

(60)

where $A_{1}=4E(0)/\sqrt{T_{0}\left( E_{I}\left( 1-\bar{h}\right) \right) }.$

Substituting (60) into (59), we obtain

$$\begin{aligned} \left| \varPhi (t)\right|\le & {} \frac{\rho }{4}\left\| w_{t}\right\| ^{2}+\frac{\rho }{2}\left\| v_{t}\right\| ^{2}+\frac{D}{4}\left\| w_{xx}\right\| ^{2}\nonumber \\&+\frac{\left( 1+A_{1}\right) \rho L^{2}}{4}\left\| w_{x}\right\| ^{2}+\rho L^{2}\left\| v_{x}+\frac{1}{2}w_{x}^{2}\right\| ^{2}. \end{aligned}$$

(61)

For $\varPsi$, by Young’s inequality and Lemma 3, we get

$$\begin{aligned} \left| \varPsi (t)\right| \le \frac{\rho }{2}\left\| w_{t}\right\| _{2}^{2}+\frac{L^{4}\rho \bar{h}}{2}\left( h\circ w_{xx}\right) . \end{aligned}$$

(62)

For $\chi$, from the decay of the function $e^{-2\tau p},$ we have

$$\begin{aligned} \left| \chi (t)\right| \le \frac{1}{\xi }E(t). \end{aligned}$$

(63)

By combining (61), (62) and (63), we deduce that

$$\begin{aligned} -\left( \lambda +\frac{\beta _{3}}{\xi }\right) E(t)\le \beta _{1}\varPhi (t)+\beta _{2}\varPsi (t)+ \beta _{3} \chi (t) \le \left( \lambda +\frac{\beta _{3}}{\xi }\right) E(t), \end{aligned}$$

where $\lambda =\max \left( \frac{\beta _{1}}{2}+\beta _{2},\beta _{1},\frac{ D}{2E_{I}\left( 1-\bar{h}\right) }\beta _{1},\frac{\left( 1+A_{1}\right) \rho L^{2}}{2T_{0}}\beta _{1},\frac{2\rho L^{2}}{E_{A}}\beta _{1},\frac{L^{4}\rho \bar{h}}{E_{I}}\beta _{2}\right) .$

We take $\beta _{1},$ $\beta _{2}$ and $\beta _{3}$ small, so that $\lambda + \frac{\beta _{3}}{\xi }<1,$ this complete proof. $\square$

Lemma 5

Let $\varPhi (t)$ be the functional given by (55), then, for $t\ge 0$,

$$\begin{aligned} \varPhi ^{\prime }(t)\le & {} \frac{\rho }{2}\left\| w_{t}\right\| ^{2}+\rho \left\| v_{t}\right\| ^{2}-\frac{E_{I}\left( 1-\bar{h}\right) }{ 4}\left\| w_{xx}\right\| ^{2} +\frac{E_{I}\bar{h}}{4\left( 1-\bar{h}\right) }\left( h\circ w_{xx}\right) \nonumber \\&-\,\left[ \frac{T_{0}}{2}-\theta _{1}A_{\mu }\frac{A_{1}}{2}\right] \left\| w_{x}\right\| ^{2} -\left[ E_{A}-2\theta _{1}A_{\mu }\right] \left\| v_{x}+\frac{1}{2} w_{x}^{2}\right\| ^{2} \nonumber \\&+\,\frac{\mu _{1}}{4\theta _{1}}\left\| g_{1}\left( v_{t}\right) \right\| ^{2} +\frac{\mu _{2}c_{3}}{4\theta _{1}}\left( z(x,1,t),g_{2}\left( z(x,1,t)\right) \right) . \end{aligned}$$

(64)

Proof

Differentiating $\varPhi (t)$ and using the equations (1), we obtain

$$\begin{aligned} \varPhi ^{\prime }(t)=\frac{\rho }{2}\left\| w_{t}\right\| ^{2}+\rho \left\| v_{t}\right\| ^{2}+J_{1}+J_{2}+J_{3}+J_{4}, \end{aligned}$$

(65)

where

$$\begin{aligned} \left\{ \begin{array}{l} J_{1}=-\frac{E_{I}}{2}\left( \left\{ w-h*w\right\} _{xxxx},w\right) , \\ J_{2}=\left( -\frac{D}{2}w_{xxxxt}+\frac{T_{0}}{2}w_{xx},w\right) +\frac{D}{2}\left( w_{xxt},w_{xx}\right) , \\ J_{3}=\frac{E_{A}}{2}\left( \left\{ \left( v_{x}+\frac{1}{2}w_{x}^{2}\right) w_{x}\right\} _{x},w\right) +{ E_{A}}\left( \left( v_{x}+\frac{1}{2} w_{x}^{2}\right) _{x},v\right) , \\ J_{4}=-\left( \mu _{1}g_{1}\left( v_{t}\right) +\mu _{2}g_{2}\left( z(x,1,t)\right) ,v\right) . \end{array} \right. \end{aligned}$$

Using integrating $J_{1}$ by parts twice and the boundary conditions (2), we get

$$\begin{aligned} J_{1}=-\frac{E_{I}}{2}\left\| w_{xx}\right\| ^{2}+\frac{E_{I}}{2}\left( h*w_{xx},w_{xx}\right) , \end{aligned}$$

add and subtract the term $\frac{E_{I}}{2}\left( \left( \int _{0}^{t}h(s)ds\right) w_{xx},w_{xx}\right)$ in the above equality then using Young’s inequality and Lemma 3, we have

$$\begin{aligned} J_{1}\le -\frac{E_{I}\left( 1-\bar{h}-\theta \right) }{2}\left\| w_{xx}\right\| ^{2}+\frac{E_{I}\bar{h}}{8\theta }\left( h\circ w_{xx}\right) . \end{aligned}$$

(66)

Integrating by parts the terms $J_{2}$ and $J_{3}$, we find

$$\begin{aligned} J_{2}= & {} -\frac{T_{0}}{2}\left\| w_{x}\right\| ^{2}, \end{aligned}$$

(67)

$$\begin{aligned} J_{3}= & {} -E_{A}\left\| v_{x}+\frac{1}{2}w_{x}^{2}\right\| ^{2}. \end{aligned}$$

(68)

By Young’s inequality, Poincaré’s inequality, inequality (60) and assumption (H4), we infer for $\theta _{1}>0$

$$\begin{aligned} J_{4}\le & {} \theta _{1}A_{\mu }\left\| v_{x}\right\| ^{2}+\frac{\mu _{1}}{4\theta _{1}}\left\| g_{1}\left( v_{t}\right) \right\| ^{2}+ \frac{\mu _{2}}{4\theta _{1}}\left\| g_{2}\left( z(x,1,t)\right) \right\| ^{2} \nonumber \\\le & {} \theta _{1}A_{\mu }\left( 2\left\| v_{x}+\frac{1}{2} w_{x}^{2}\right\| ^{2}+\frac{A_{1}}{2}\left\| w_{x}\right\| ^{2}\right) +\frac{\mu _{1}}{4\theta _{1}}\left\| g_{1}\left( v_{t}\right) \right\| ^{2}\nonumber \\&+\frac{\mu _{2}c_{3}}{4\theta _{1}}\left( z(x,1,t),g_{2}\left( z(x,1,t)\right) \right) , \end{aligned}$$

(69)

where $A_{\mu }=\left( \mu _{1}+\mu _{2}\right) L^{2}.$

Now, Substituting (66)–(69) into (65) and taking $\theta = \frac{\left( 1-\bar{h}\right) }{2},$ we have (64). $\square$

Lemma 6

Let $\varPsi (t)$ be the functional given by (56), then, for $t\ge 0$,

$$\begin{aligned} \varPsi ^{\prime }(t)\le & {} -\left[ \left( \int _{0}^{t}h(s)ds\right) -\frac{ \theta _{2}}{2}\right] \rho \left\| w_{t}\right\| ^{2}-\frac{\rho L^{4}h(0)}{2\theta _{2}}\left( h^{\prime }\circ w_{xx}\right) +A_{2}\bar{h}\left( h\circ w_{xx}\right) \nonumber \\&+\,\theta _{4}T_{0}\left\| w_{x}\right\| ^{2}+\theta _{3}D\left\| w_{xxt}\right\| ^{2} +\theta _{3}\left( 1+\bar{h}\right) E_{I}\left\| w_{xx}\right\| ^{2}\nonumber \\&+\,\theta _{4}E_{A}\left\| v_{x}+\frac{1}{2}w_{x}^{2}\right\| ^{2}, \end{aligned}$$

(70)

where $\theta _{i},$ $i=2,3,4.$ are positive constants and

$$\begin{aligned} A_{2}=\left( \theta _{3}+\frac{3}{4\theta _{3}}\right) E_{I}+\frac{D}{4\theta _{3}}+\frac{ L^{2}T_{0}+L^{2}E_{A}A_{1}}{4\theta _{4}}. \end{aligned}$$

Proof

Substituting the first equation in (1) into the derivative of $\varPsi (t)$, we obtain

$$\begin{aligned} \varPsi ^{\prime }(t)=-\rho \int _{0}^{t}h(s)ds\left\| w_{t}\right\| _{2}^{2}+K_{1}+K_{2}+K_{3}+K_{4}, \end{aligned}$$

(71)

where

$$\begin{aligned} \left\{ \begin{array}{l} K_{1}=\rho \left( h^{\prime }\diamond w,w_{t}\right) , \\ K_{2}=-\left( h\diamond w,Dw_{xxxxt}+E_{I}w_{xxxx}\right) , \\ K_{3}=E_{I}\left( h\diamond w,h*w_{xxxx}\right) , \\ K_{4}=E_{A}\left( h\diamond w,\left\{ \left[ T_{0}+\left( v_{x}+\frac{1}{2} w_{x}^{2}\right) \right] w_{x}\right\} _{x}\right) . \end{array} \right. \end{aligned}$$

Let’s estimate these terms, for $K_{1}$ similarly to (62), we get

$$\begin{aligned} K_{1}\le \frac{\rho \theta _{2}}{2}\left\| w_{t}\right\| _{2}^{2}- \frac{\rho L^{4}h(0)}{2\theta _{2}}\left( h^{\prime }\circ w_{xx}\right) . \end{aligned}$$

(72)

For $K_{2}$ and $K_{3}$, using integrating by parts twice, Young’s inequality and Lemma 3, we obtain

$$\begin{aligned} K_{2}\le & {} \theta _{3}D\left\| w_{xxt}\right\| ^{2}+\theta _{3}E_{I}\left\| w_{xx}\right\| ^{2}+\frac{\bar{h}\left( E_{I}+D\right) }{ 4\theta _{3}}\left( h\circ w_{xx}\right) , \end{aligned}$$

(73)

$$\begin{aligned} K_{3}\le & {} \theta _{3}E_{I}\bar{h}\left\| w_{xx}\right\| ^{2}+\left( \theta _{3}+\frac{1}{2\theta _{3}}\right) E_{I}\bar{h}\left( h\circ w_{xx}\right) . \end{aligned}$$

(74)

Integrating $K_{4}$ by parts, applying Young’s inequality, Holder’s inequality and Lemma 3,we get

$$\begin{aligned} K_{4}\le \theta _{4}T_{0}\left\| w_{x}\right\| ^{2}+\theta _{4}E_{A}\left\| v_{x}+\frac{1}{2}w_{x}^{2}\right\| ^{2}+\frac{ L^{2}T_{0}+L^{2}E_{A}A_{1}}{4\theta _{4}}\bar{h}\left( h\circ w_{xx}\right) . \end{aligned}$$

(75)

Combining (71)–(75), we obtain (70). $\square$

Lemma 7

Let $\chi (t)$ be the functional given by (55), then, for $t\ge 0$

$$\begin{aligned} \chi ^{\prime }(t)\le -2\chi (t)+\frac{a_{2}}{\tau }\left( v_{t},g_{1}\left( v_{t}\right) \right) +\frac{a_{1}e^{-2\tau }}{\tau }\left( z(x,1,t),g_{2}\left( z(x,1,t)\right) \right) . \end{aligned}$$

(76)

Proof

Take derivative of $\chi (t)$ with respect to t,  using the identity (62) and integrating by parts, we obtain

$$\begin{aligned} \chi ^{\prime }(t)= & {} -\frac{1}{\tau }\int _{0}^{L}\left[ e^{-2\tau }G\left( z\left( x,1,t\right) \right) -G\left( z\left( x,0,t\right) \right) \right] dx\\&-2\int _{0}^{L}\int _{0}^{1}e^{-2\tau p}G\left( z\left( x,p,t\right) \right) dpdx. \end{aligned}$$

In view of the hypotheses (H3) and the above equality, we have (70). $\square$

Theorem 2

For the system dynamics described by (1)–(3), under assumptions (H1)–(H4), given that $\left( w_{0},v_{0},z_{0}\right) \in W\times V\times Z$ and $\left( w_{1},v_{1}\right) \in H_{0}^{2}\left( 0,L\right) \times H_{0}^{1}\left( 0,L\right) ,$ where

$$\begin{aligned} z(x,p,t,s)=v_{t}(x,t-p\tau ),\quad p\in \left( 0,1\right) . \end{aligned}$$

(77)

Then, there exist strictly positive constants $\omega _{1},$ $\omega _{2},$ $\omega _{3}$ and $\varepsilon$ such that

$$\begin{aligned} E(t)\le \omega _{1}H_{1}^{-1}\left( \omega _{2}t+\omega _{3}\right) , \qquad \forall t>0, \end{aligned}$$

(78)

where

$$\begin{aligned} H_{1}(t)=\int _{t}^{1}\frac{ds}{H_{2}(s)}, \end{aligned}$$

and

$$\begin{aligned} H_{2}(t)=\left\{ \begin{array}{lll} t &{} \mathrm {if} &{} H\ \mathrm {is} \ \mathrm {linear } , \\ tH^{\prime }(\varepsilon t) &{} \mathrm {if} &{} H\ \mathrm {is} \ \mathrm {nonlinear } . \end{array} \right. \end{aligned}$$

(79)

Proof

Using the results (20), (64), (70) and (76), for all $t\ge t_{0}>0$, we obtain

$$\begin{aligned} \mathcal {L}^{\prime }(t)\le & {} \left[ \frac{E_{I}}{2}-\frac{\rho L^{4}h(0)}{ 2\theta _{2}}\beta _{2}\right] \left( h^{\prime }\circ w_{xx}\right) +\left[ \frac{E_{I}\bar{h}}{4\left( 1-\bar{h}\right) }\beta _{1}+A_{2}\bar{h}\beta _{2}\right] \left( h\circ w_{xx}\right) \nonumber \\&-\left[ \frac{\left( 1-\bar{h}\right) }{4}\beta _{1}-\theta _{3}\left( 1+ \bar{h}\right) \beta _{2}\right] E_{I}\left\| w_{xx}\right\| ^{2}-\left[ 1-\theta _{3}\beta _{2}\right] D\left\| w_{xxt}\right\| ^{2} \nonumber \\&-\left[ \left( h_{0}-\frac{\theta _{2}}{2}\right) \beta _{2}-\frac{\beta _{1}}{2}\right] \rho \left\| w_{t}\right\| ^{2} \nonumber \\&-\left[ \left( \frac{T_{0}}{2}-\theta _{1}A_{\mu }\frac{A_{1}}{2}\right) \beta _{1}-\theta _{4}T_{0}\beta _{2}\right] \left\| w_{x}\right\| ^{2}\nonumber \\&-\left[ \left( E_{A}-2\theta _{1}A_{\mu }\right) \beta _{1}-\theta _{4}E_{A}\beta _{2}\right] \left\| v_{x}+\frac{1}{2}w_{x}^{2}\right\| ^{2}\nonumber \\&-\left[ \eta _{1}-\frac{a_{2}}{\tau }\beta _{3}\right] \left( v_{t},g_{1}\left( v_{t}\right) \right) +\rho \beta _{1}\left\| v_{t}\right\| ^{2}+\frac{\mu _{1}}{4\theta _{1}}\beta _{1}\left\| g_{1}\left( v_{t}\right) \right\| ^{2} \nonumber \\&-\left[ \eta _{2}-\frac{a_{1}e^{-2\tau }}{\tau }\beta _{3}-\frac{\mu _{2}c_{2}}{4\theta _{1}}\beta _{1}\right] \left( z(x,1,t),g_{2}\left( z(x,1,t)\right) \right) -2\beta _{3}\chi (t)\nonumber \\ \end{aligned}$$

(80)

where $h_{0}=\int _{0}^{t_{0}}h(t)dt.$

Right now, we do select our parameters very carefully. First pick $\theta _{2}\le h_{0}$, after that take $\theta _{i},i=1,3,4.,$ $\beta _{1},$ $\beta _{2}$ and $\beta _{3}$ sufficiently small so that

$$\begin{aligned} \left\{ \begin{array}{l} \theta _{1}\le \frac{1}{2}\min \left\{ \frac{T_{0}}{A_{\mu }A_{1}},\frac{E_{A}}{2A_{\mu }}\right\} , \\ \theta _{3}<\min \left\{ \frac{\beta _{1}}{\beta _{2}}\frac{\left( 1-\bar{h}\right) }{\left( 1+\bar{h}\right) },\frac{1}{\beta _{2}}\right\} , \\ \theta _{4}<\frac{\beta _{1}}{4\beta _{2}}<\frac{h_{0}}{4}, \\ \beta _{2} \le \frac{E_{I}\theta _{2}}{\rho L^{4}h(0)},\\ \beta _{3}<\frac{\tau }{a_{2}}\eta _{1}, \\ \frac{a_{1}e^{-2\tau }}{\tau }\beta _{3}+\frac{\mu _{2}c_{2}}{4\theta _{1}}\beta _{1}<\eta _{2}, \\ A_{3}>0. \end{array} \right. \end{aligned}$$

(81)

Thus, (80) becomes

$$\begin{aligned} \mathcal {L}^{\prime }(t)\le -A_{3}E(t)+A_{4}\left( \left\| v_{t}\right\| ^{2}+\left\| g_{1}\left( v_{t}\right) \right\| ^{2}+\left( h\circ w_{xx}\right) (t)\right) , \quad \forall t\ge t_0, \end{aligned}$$

(82)

where $A_{3}$ and $A_{4}$ are two positive constants.

To complete the proof, we partition the interval $\left[ 0,L\right]$ into

$$\begin{aligned} L^{>}=\left\{ 0 \le x \le L:\left| v_{t}\right| >\epsilon \right\} , \qquad L^{<}=\left\{ 0\le x\le L:\left| v_{t}\right| \le \epsilon \right\} . \end{aligned}$$

From (H3), we estimate that

$$\begin{aligned} \int _{L^{>}}\left( v_{t}^{2}+g_{1}^{2}\left( v_{t}\right) \right) d x \le \left( 1 / c_{1}+c_{2}\right) \int _{L^{>}} v_{t} g_{1}\left( v_{t}\right) d x \le -A_5 E^{\prime }(t) \end{aligned}$$

(83)

where $A_5=\left( 1 / c_{1}+c_{2}\right) / \eta _{1}$.

For the estimate of the last term in the right hand side of (82) on $L^{<},$ we distinguish two cases

Case 1 ::

H is linear. From the assumption(H3), we deduce that there exists $c_{4},$ such that $s^{2}+g_{1}^{2}\left( s\right) \le c_{4}sg_{1}\left( s\right)$ in $\left[ -\epsilon ,\epsilon \right] ,$ and therefore

$$\begin{aligned} \int _{L^{<}}\left( v_{t}^{2}+g_{1}^{2}\left( v_{t}\right) \right) dx\le c_{4}\int _{L^{<}}v_{t}g_{1}\left( v_{t}\right) dx\le -A_6E^{\prime }(t) \end{aligned}$$

(84)

and the assumption (H2) gives

$$\begin{aligned} \int _{0}^{t} h(s) \int _{\varOmega }|w_{xx}(t)-w_{xx} u(t-s)|^{2} d x d s \le -A_6 E^{\prime }(t). \end{aligned}$$

(85)

Combining (82)–(84), we obtain

$$\begin{aligned} \left( \mathcal {L}(t)+\sigma E(t)\right) ^{\prime }\le -A_{3}H_{2}\left( E(t)\right) . \end{aligned}$$

(86)

where $\sigma =A_{4}\left( A_5+2A_6\right) .$

Case 2::

H is nonlinear. First, we deduce from (18) and (20) that

$$\begin{aligned} \int _{0}^{t_{0}}h(s)\left\| w_{xx}(t)-w_{xx}(t-s)\right\| ^{2}ds \le&\frac{-1}{\gamma }\int _{0}^{t_{0}}h^{\prime }(s)\left\| w_{xx}(t){-}w_{xx}(t{-}s)\right\| ^{2}ds \nonumber \\ \le&-\frac{2}{\gamma E_I}E^{\prime }(t). \end{aligned}$$

(87)

Next, we define the functions $\kappa _{p}$ and $\kappa$ by

$$\begin{aligned} \kappa _{p}(t)= & {} p\int _{t_0}^{t}\frac{h(s)}{H_{0}^{-1}\left( -h^{\prime }(s)\right) }\Vert w_{xx}(t)-w_{xx}(t-s)\Vert ^{2}ds \end{aligned}$$

(88)

$$\begin{aligned} \kappa (t)= & {} -\int _{t_0}^{t}h^{\prime }(s)\frac{h(s)}{H_{0}^{-1}\left( -h^{\prime }(s)\right) }\Vert w_{xx}(t)-w_{xx}(t-s)\Vert ^{2}ds, \end{aligned}$$

(89)

where $1/p>\frac{8 E(0)}{E_I(1-\bar{h})} \int _{0}^{+\infty } \frac{h(s)}{H_{0}^{-1}\left( -h^{\prime }(s)\right) } d s$ and $H_0$ is defined in (14), we find that $\kappa _{p}(t)<1,$ for all $t\ge 0.$

The properties of the functions $H_0$, D and h gives

$$\begin{aligned} \frac{h(s)}{H_{0}^{-1}\left( -h^{\prime }(s)\right) } \le \frac{h(s)}{H_{0}^{-1}(H(h(s)))}=\frac{h(s)}{D^{-1}(h(s))} \le \kappa _0, \end{aligned}$$

(90)

for some positive constant $\kappa _0.$

We can easily verify that

$$\begin{aligned} \kappa (t)\le & {} -\kappa _{0}\int _{t_{0}}^{t}h^{\prime }(s)\Vert w_{xx}(t)-w_{xx}(t-s)\Vert ^{2}ds \\\le & {} -\frac{8 \kappa _{0} E(0)}{E_I(1-\bar{h})}\int _{t_{0}}^{t}h^{\prime }(s)ds \\\le & {} \frac{8 \kappa _{0} E(0)}{E_I(1-\bar{h})}h(t_0) \le \frac{1}{2} \min \left\{ \epsilon , H(\epsilon ), H_{0}(\epsilon )\right\} ,\qquad \forall t\ge t_0. \end{aligned}$$

We have, by the convexity property of $H_0$ and $H_0(0)=0$ that

$$\begin{aligned} H_0(\nu x)\le \nu H_0(x) , \qquad x \in [0,\epsilon ], \ \nu \in [0,1]. \end{aligned}$$

(91)

By assumption (H2), identity (91) and Jensen’s inequality, we get

$$\begin{aligned} \kappa (t)\ge & {} \frac{1}{\kappa _p (t)}\int _{t_{0}}^{t}\frac{H_{0}\left( \kappa _p (t)H_{0}^{-1}\left( -h^{\prime }(s)\right) \right) }{H_{0}^{-1}\left( -h^{\prime }(s)\right) }h(s)\Vert w_{xx}(t)-w_{xx}(t-s)\Vert ^{2}ds \nonumber \\\ge & {} H_{0}\left( \frac{1}{\kappa _p (t)}\int _{t_{0}}^{t}\frac{\kappa _p (t)H_{0}^{-1}\left( -h^{\prime }(s)\right) }{H_{0}^{-1}\left( -h^{\prime }(s)\right) }h(s)\Vert w_{xx}(t)-w_{xx}(t-s)\Vert ^{2}ds\right) \nonumber \\= & {} H_{0}\left( \int _{t_{0}}^{t}h(s)\Vert w_{xx}(t)-w_{xx}(t-s)\Vert ^{2}ds\right) , \qquad \forall t\ge t_0, \end{aligned}$$

this is equivalent to

$$\begin{aligned} \int _{t_{0}}^{t}h(s)\Vert w_{xx}(t)-w_{xx}(t-s)\Vert ^{2}ds \le H_{0}^{-1}(\kappa (t)), \qquad \forall t\ge t_0. \end{aligned}$$

(92)

We can assume that $\epsilon$ is small enough such that $s g_{1}(s) \le \frac{1}{2} \min \left\{ \epsilon , H(\epsilon ), H_{0}(\epsilon )\right\}$ for all $|s| \le \epsilon$. With (H3) and reversed Jensen’s inequality for concave function and the concavity of $H^{-1},$ we obtain

$$\begin{aligned} \int _{L^{<}}\left( v_{t}^{2}+g_{1}^{2}\left( v_{t}\right) \right) dx\le \int _{L^{<}}H^{-1}\left( v_{t}g_{1}\left( v_{t}\right) \right) dx\le cH^{-1}\left( \vartheta (t)\right) , \end{aligned}$$

(93)

where $\vartheta (t)=\frac{1}{L}\int _{L^{<}}v_{t}g_{1}\left( v_{t}\right) dx.$

The inequalities (82), (83), (87), (92) and (93), gives

$$\begin{aligned} \left( \mathcal {L}(t)+A_{7}E(t)\right) ^{\prime }\le -A_{3}E(t)+A_{4}cH^{-1}\left( \vartheta (t)\right) +A_4 H_{0}^{-1}(\kappa (t)), \ \forall t\ge t_0. \end{aligned}$$

Since $H_0^{-1}(t){=}D^{-1}(H^{-1}(t))$, $D^{-1}(H^{-1}(0)=D^{-1}(0)){=}0$ and $H^{-1}(\kappa (t)){\le } \epsilon$. Moreover, the function $D^{-1}(H^{-1}(t))$ is concave, so its graph is below its tangent, that $H_0^{-1}(\kappa (t))\le cH^{-1}(\kappa (t))$. Therefore, for all $t{\ge } t_{0}$,

$$\begin{aligned} \left( \mathcal {L}(t)+A_{7}E(t)\right) ^{\prime }\le & {} -A_{3}E(t)+A_{4}cH^{-1}\left( \vartheta (t)\right) +A_{4}cH^{-1}(\kappa (t)), \nonumber \\\le & {} -A_{3}E(t)+A_{4}cH^{-1}\left( \vartheta (t)+\kappa (t)\right) . \end{aligned}$$

(94)

Take into account $E^{\prime }(t)\le 0,$ $H^{\prime }(t)>0,$ $H^{\prime \prime }(t)>0,$ and using the inequality (94), for $\varepsilon <\epsilon E(0)$, we infer

$$\begin{aligned}&\left[ H^{\prime }(\varepsilon E(t))\left\{ \mathcal {L}(t)+A_{7}E(t)\right\} +A_{8}E(t)\right] ^{\prime } \nonumber \\&\quad =\varepsilon E^{\prime }(t)H^{\prime \prime }(\varepsilon E(t))\left\{ \mathcal {L}(t){+}A_{7}E(t)\right\} {+}H^{\prime }(\varepsilon E(t))\left\{ \mathcal {L}^{\prime }(t){+}A_{7}E^{\prime }(t)\right\} {+}A_{8}E^{\prime }(t) \nonumber \\&\quad \le {-}A_{3}E(t)H^{\prime }(\varepsilon E(t)){+}A_{4}cH^{\prime }(\varepsilon E(t))H^{-1}\left( \vartheta (t){+}\kappa (t)\right) {+}A_{8}E^{\prime }(t),\quad \forall t{\ge } t_0 .\nonumber \\ \end{aligned}$$

(95)

Let $H^{*}$ by the convex conjugate of H, given by (25), then the increasing of the functions $(H^{\prime })^{-1}$, H and the fact that $H(0)=0,$ yield

$$\begin{aligned} H^{*}(s)\le s\left( H^{\prime }\right) ^{-1}(s). \end{aligned}$$

(96)

Applying inequalities (26) and (96) to the second term of the right hand side of (95), we obtain

$$\begin{aligned} H^{\prime }(\varepsilon E(t))H^{-1}\left( \vartheta (t)+\kappa (t)\right)\le & {} H^{*}(H^{\prime }(\varepsilon E(t)))-\left( \vartheta (t)+\kappa (t)\right) \nonumber \\\le & {} \varepsilon E(t)H^{\prime }(\varepsilon E(t))-A_{8}E^{\prime }(t), \end{aligned}$$

(97)

combining (95) and (97), we have

$$\begin{aligned} \left[ H^{\prime }(\varepsilon E(t))\left\{ \mathcal {L}(t)+A_{7}E(t) \right\} +A_{8}E(t)\right] ^{\prime }\le & {} -\left( A_{3}-A_{4}c\varepsilon \right) E(t)H^{\prime }(\varepsilon E(t)) \nonumber \\= & {} -A_{9}H_{2}(E(t)). \end{aligned}$$

(98)

Let us define

$$\begin{aligned} \tilde{\mathcal {L}}(t)=\left\{ \begin{array}{lll} \mathcal {L}(t)+\sigma E(t) &{} \mathrm {\ if\ } &{} H \ \mathrm {is} \ \mathrm {linear},\\ H^{\prime }(\varepsilon E(t))\left\{ \mathcal {L}(t)+A_{7}E(t)\right\} +A_{8}E(t) &{} \ \mathrm {if}\ &{} H\ \mathrm {is} \ \mathrm {nonlinear}. \end{array} \right. \end{aligned}$$

(99)

From (86) and (98), we conclude

$$\begin{aligned} \tilde{\mathcal {L}}(t)\le -\tilde{C}H_{2}(E(t)),\qquad t\ge t_{0}. \end{aligned}$$

Since $\mathcal {L}(t)$ and E(t) are equivalent and $0\le H^{\prime }(\varepsilon E(t))\le H^{\prime }(\varepsilon E(0)).$ So, there exist $\tilde{\alpha }_{1}$ and $\tilde{\alpha }_{2}$ two positive constants such that

$$\begin{aligned} \tilde{\alpha }_{1}E(t)\le \tilde{\mathcal {L}}(t)\le \tilde{\alpha }_{2}E(t). \end{aligned}$$

Now we put $\mathcal {L}_{\alpha }(t)=\alpha \tilde{\mathcal {L}}(t)$ for $\alpha \le 1/\tilde{\alpha }_{2},$ using the fact that $H_{2}$ is increasing, we obtain

$$\begin{aligned} \mathcal {L}_{\alpha }(t)= & {} \alpha \tilde{\mathcal {L}}(t)\le -\alpha \tilde{ C}H_{2}(\frac{1}{\tilde{\alpha }_{2}}\tilde{\mathcal {L}}(t)) \\\le & {} -\alpha \tilde{C}H_{2}(\mathcal {L}_{\alpha }(t)),\qquad t\ge t_{0}. \end{aligned}$$

Taking into consideration that $H_{1}^{\prime }=-1/H_{2},$ the above inequalities become

$$\begin{aligned} \mathcal {L}_{\alpha }(t)H_{1}^{\prime }(\mathcal {L}_{\alpha }(t))\ge \alpha \tilde{C},\qquad t\ge t_{0}, \end{aligned}$$

integrate this differential inequality over $(t_{0},t)$, we obtain

$$\begin{aligned} H_{1}\left( \mathcal {L}_{\alpha }(t)\right) \ge H_{1}\left( \mathcal {L}_{\alpha }\left( t_{0}\right) \right) +\alpha \tilde{C}\left( t-t_{0}\right) . \end{aligned}$$

Choosing $\alpha$ small enough such that $H_{1}(\mathcal {L}_{\alpha }(t_{0}))-\alpha \tilde{C}t_{0}>0.$ The decay of $H_{1}^{-1},$ yields

$$\begin{aligned} \mathcal {L}_{\alpha }(t)\le H_{1}^{-1}\left( \alpha \tilde{C}t+\left( H_{1}( \mathcal {L}_{\alpha }(t_{0}))-\alpha \tilde{C}t_{0}\right) \right) . \end{aligned}$$

finally, the equivalence of $\mathcal {L}(t),$ $\tilde{\mathcal {L}}(t),$ $\mathcal {L}_{\alpha }(t)$ and $E\left( t\right) ,$ result

$$\begin{aligned} E(t)\le \omega _{1}H_{1}^{-1}\left( \omega _{2}t+\omega _{3}\right) ,\qquad t\ge t_{0}. \end{aligned}$$

One can easily find a similar estimate over the interval $[0,t_0]$, by using decreasing of E and $H_{1}^{-1}$. This completes the proof. $\square$