General decay of the solution to a nonlinear viscoelastic beam with delay

Abstract

In this paper we consider a nonlinear viscoelastic beam with a linear delay term and infinite memory term. The well posedness of solutions is proved using the semigroup method. We establish a general decay results by using minimal and general conditions on the relaxation function, from which the usual exponential and polynomial decay rates are only special cases.

1 Introduction

In this paper, we consider the following one-dimensional version of a system which describes the vibrations of shallow shells with time delay and infinite memory term:

$$\begin{aligned} \left\{ \begin{array}{lll} u_{tt}-\left[ g(u,w)\right] _{x}+\alpha _{1}u_{t}+\alpha _{2}u_{t}(t-\tau ) \qquad &{} =0 &{} \text {in }\mathscr {Q}, \\ \mathscr {D}w_{tt}+w_{xxxx}-\left[ f(u,w)\right] _{x}+k\left( x\right) g(u,w)+h *w_{xxxx}\text { } &{} =0 &{} \text {in }\mathscr {Q}, \end{array} \right. \end{aligned}$$

(1)

where $\mathscr {Q}=I \times \mathbb {R}_{+}$, $I=]0,L[$ is an interval, $\partial I$ its boundary ($\partial I =\{0\}\cup \{L\}$), $\Sigma =\left\{ 0\right\} \times \mathbb {R}_{+}\cup \left\{ L\right\} \times \mathbb {R}_{+}$, $\mathscr {D}$ is the operator $\mathscr {I}-{\partial }_{x}^{2}$. The functions f and g are defined by:

$$\begin{aligned} f(u,w)= & {} \frac{2}{1-\mu }\left[ w_{x}\left( u_{x}+\frac{1}{2} w^{2}+k(x)w\right) \right] , \end{aligned}$$

(2)

$$\begin{aligned} g(u,w)= & {} \frac{2}{1-\mu }\left[ u_{x}+\frac{1}{2}w^{2}+k(x)w\right] . \end{aligned}$$

(3)

In (1), subscripts mean partial derivatives, the space variable x runs in the interval $0< x < L$ and t denotes the positive time variable. The functions $u = u(x, t)$ and $w = w(x, t)$ are, respectively, the longitudinal and transversal displacements of the beam at the point x at time t. Additionally, $\mu$ is a constant, $0< \mu < 1$ and $k= k(x)$ represents the curvature of the beam at the point x.

In the system (1), $\alpha _{1}u _{t}$ represents a frictional damping. The time delay is given by $\alpha _{2}u_{t}(t-\tau )$, where $\alpha _{1}$, $\alpha _{2}$, $\tau$ are positive constants.

In (1), $(h*u)(t)$ is defined by

$$\begin{aligned} (h*u)(t)=\int _{0}^{ \infty }h(t-s)u(x,s)ds. \end{aligned}$$

The viscoelastic damping term that appears in the equations describes the relationship between the stress and the history of the strain in the beam, according to Boltzmann theory. The function h represents the kernel of the memory term or the relaxation function.

The system (1) is subjected with the boundary conditions

$$\begin{aligned} w(0,t)=w(L,t)= & {} 0,\nonumber \\ w_{x}(0,t)=w_{x}(L,t)= & {} 0,\nonumber \\ u(0,t)=u(L,t)= & {} 0,\quad t\in \mathbb {R}_{+}^{*} \end{aligned}$$

(4)

and the initial conditions

$$\begin{aligned} \left( u(x,0),u_{t}(x,0),w(x,-s),w_{t}(x,0)\right) =\left( u_{0}(x),u_{1}(x),w_{0}(x),w_{1}(x)\right) \text {,}\left( x,s\right) \in \mathscr {Q}. \end{aligned}$$

(5)

The main purpose about problem (1)–(5) is to deal with the well posedness and asymptotic behavior of solutions. Before stating and proving our results, let us recall some other results related to our work.

Several authors have studied the Mindlin–Timoshenko system of equations (see, e.g., [16]). This Model is a widely used and fairly complete mathematical model for describing the transverse vibrations of beams. It is a more accurate model than the Euler-Bernoulli one, since it also takes into account transverse shear effects.

For a beam of length $L > 0$, this one-dimensional nonlinear system reads as

$$\begin{aligned} \frac{h_{0}^{3}}{12}\phi _{tt}-\phi _{xx}+k\left[ \phi + \psi _{x})\right]= & {} 0,\ \text { in }\quad Q, \nonumber \\ \rho h\psi _{tt}-k\left[ \phi + \psi _{x})\right] _{x}+\left[ \psi _{x}\left( \eta _{x}+ \frac{1}{2}\psi _{x}^{2}\right) \right] _{x}= & {} 0,\ \text { in }\quad Q, \nonumber \\ \rho h\eta _{tt}-\left[ \eta _{x}+\frac{1}{2}(\psi _{x})^{2}\right] _{x}= & {} 0,\ \text { in }\quad Q, \end{aligned}$$

(6)

where $Q = (0,L) \times (0, T )$ and T is a given positive time. Here, the unknown $\phi = \phi (x,t)$ represent the angle of rotation. The parameter k is the so called modulus of elasticity in shear. It is given by the expression $k = \widehat{k}Eh_{0}/2 (1 + \epsilon )$, where $\widehat{k}$ is a shear correction coefficient, E is the Young’s modulus and $\epsilon$ is the Poisson’s ratio, $0< \epsilon < 1/2$.

For Mindlin–Timoshenko system, there is a large literature, addressing problems of existence, uniqueness and asymptotic behavior in time when some damping effects are considered, as well as some other important properties (see [13, 26, 28] and references therein).

When one assumes the linear filament of the beam to remain orthogonal to the deformed middle surface, the transverse shear effects are neglected, and one obtains, from the Mindlin–Timoshenko system of equations, the following von Kármán system (see [28]).

$$\begin{aligned} \left\{ \begin{array}{l} \rho h_{0}\mathscr {D}\psi _{tt}+\psi _{xxxx}-\left[ \psi _{x}(\eta _{x}+\frac{1}{2}\psi _{x}^{2})\right] _{x}=0,\text { in }]0,L[\times \mathbb {R}_{+}^{*}, \\ \rho h\eta _{tt}-[\eta _{x}+\frac{1}{2}(\psi _{x})^{2}]_{x}=0,\text { in }]0,L[\times \mathbb {R}_{+}^{*}. \end{array} \right. \end{aligned}$$

(7)

There is also an extensive literature about system (7) (see [13, 18, 26, 27, 38, 51, 53,54,55,56] and references therein).

Lagnese and Leugering [27] considered a one-dimensional version of the von Kármán system describing the planar motion of a uniform prismatic beam of length L. More precisely, in [27] the following system was considered:

$$\begin{aligned} \left\{ \begin{array}{l} \psi _{tt}+\psi _{xxxx}-h_{0}\psi _{xxtt}-\left[ \psi _{x}(\eta _{x}+\frac{1}{2}\psi _{x}^{2})\right] _{x}=0\text { in }]0,L[\times \mathbb {R}_{+}^{*} \\ \eta _{tt}-\left[ \eta _{x}+\frac{1}{2}(\psi _{x})^{2}\right] _{x}=0\text { in }]0,L[\times \mathbb {R}_{+}^{*} \end{array} \right. \end{aligned}$$

(8)

In [27], suitable dissipative boundary conditions at $x = 0$, $x = L$ and initial conditions at $t = 0$ were given and the stabilization problem was studied.

In [4], Araruna et al. have showed how the so called von Kármán model (8) can be obtained as a singular limit of a modified Mindlin–Timoshenko system (6) when the modulus of elasticity in shear k tends to infinity, provided a regularizing term through a fourth order dispersive operator is added. Introducing damping mechanisms, the authors also show that the energy of solutions for this modified Mindlin–Timoshenko system decays exponentially, uniformly with respect to the parameter k. As $k\longrightarrow \infty$, the authors obtain the damped von Kármán model with associated energy exponentially decaying to zero as well.

The subject of stability of von Kármán system has received a lot of attention in the last years, see [11, 12, 18, 18, 25, 26, 33, 39, 49, 50, 52] and references therein.

Delay effects are very important because most natural phenomena are in many cases very complicated and do not depend only on the current state but also on the past history of the system. The presence of delay can be a source of instability. In recent years, the stabilization of PDEs with delay effects has draw attention for many author and become an active area of research, see [11, 15, 24, 46,47,48, 57, 59,60,61].

For the stability of other kind of wave equation, let us mention the following problem:

$$\begin{aligned} u_{tt}(x,t)-\Delta u(x,t)+\int _{0}^{t}h(t-s)\Delta u(x,s)ds\nonumber \\ +\alpha _{1}h_{1}(u_{t}(x,t))+\alpha _{2}h_{2}(u_{t}(x,t-\tau ))= & {} 0, \quad \text { in}\quad \Omega \times \mathbb {R}_{+},\nonumber \\ u(x,t)= & {} 0, \quad \text { in}\quad \Gamma \times \mathbb {R}_{+},\nonumber \\ u(x,0)=u_{0}(x)\quad u_{t}(x,0)= & {} u_{1}(x), \quad \text { in}\quad \Omega ,\nonumber \\ u_{t}(x,t-\tau )= & {} f_{0}(x,t-\tau ),\quad \text {in}\quad \Omega \times ]0,\tau [, \end{aligned}$$

(9)

where $\Omega$ is a bounded domain in $\mathbb {R}^{n}$, $n \in \mathbb {N}$, with a smooth boundary $\partial \Omega =\Gamma$, h is a positive non-increasing function defined on $\mathbb {R}^{n}$, $h_{1}$ and $h_{2}$ are two functions, $\tau > 0$ is a time delay, $\alpha _{1}$ and $\alpha _{2}$ are positive real numbers and the initial data $( u_{0}, u_{1}, f_{0})$ belong to a suitable function space.

In the case $h \equiv 0$, problem (9) has been studied by many authors (see [6,7,8, 10, 46, 61]).

For a wider class of relaxation functions, Messaoudi [36, 37] considered

$$\begin{aligned} u_{tt}-\Delta u +\int _{0}^{t}h(t-s)\Delta u(s)ds=b\mid u\mid ^{\gamma }u, \end{aligned}$$

(10)

for $\gamma > 0$ and $b = 0$ or $b = 1$, and the relaxation function satisfies

$$\begin{aligned} h^{\prime }(t)\le -\zeta (t)h(t), \end{aligned}$$

(11)

where $\zeta$ is a differentiable nonincreasing positive function. He established a more general decay result, from which the usual exponential and polynomial decay results are only special cases. Such a condition was then employed in a series of papers, see for instance [3, 22, 23, 41, 42, 52].

Recently, Mustafa and Messaoudi [45] studied the problem (10) with $b = 0$ for the relaxation functions satisfying

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

(12)

where H is a nonnegative function, with $H(0) = H^{\prime }(0)= 0$ and H is strictly increasing and strictly convex on ]0, k[ for some $k_{0}> 0$. The authors showed a general relation between the decay rate for the energy and that of the relaxation function h without imposing restrictive assumptions on the behavior of h at infinity. On the other hand, a condition of the form (12) where H is a convex function satisfying some smoothness properties, was introduced by Alabau-Boussouira and Cannarsa [2] and used then by several authors with different approaches. We refer to [32] where not only general but also optimal result was established by Lasiecka and Wang.

The main objective of this work is to investigate the problem (1) with the following very general class of relaxation functions

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

(13)

where H is increasing and convex without any additional constraints on H and the coefficients. We will establish a general decay rate for the energy associated to the system for linear damping, time delay terms and finite memory. We would like to see the influence of frictional and viscoelastic dampings on the rate of decay of solutions in the presence of linear degenerate delay term.

To prove decay estimates, we shall pursue a strategy based on an adaptation of non linear differential inequalities technique developed in [40, 43, 44] and we use a perturbed energy method and some properties of convex functions which were introduced and developed by many authors [1, 9, 14, 17, 30, 31, 34].

Our work is organized as follows. In the next section, we prepare some material needed in the proof of our result, like some lemmas (Poincaré’s and Young’s inequalities) and some useful notations. We introduce the different functionals by which we modify the classical energy to get an equivalent useful one. In Sect. 4, we state and prove the well-posedness of the problem. Finally, in Sect. 5, we will prove our main results concerning the exponential decay of the energy associated to the solutions of the problem.

2 Statement of results

In order to deal with the delay feedback term, motivated by [46, 47], we define the following new dependent variables $\eta$ and z:

$$\begin{aligned} \begin{array}{cc} z(x,t,p)=u_{t}(x,t-p\tau ) \qquad &{} \text {in } \mathscr {Q}\times (0,1), \\ \eta (x,t,s)=w(x,t)-w(x,t-s) &{} \text { in}\mathscr {Q}\times \mathbb {R}_{+}, \end{array} \end{aligned}$$

(14)

consequently, we obtain

$$\begin{aligned} \begin{array}{rr} \eta (x,t,0)=0 &{} \text {in }\mathscr {Q}, \\ \eta (x,t,s)=0 &{} \text {in }\Sigma {\times }\mathbb {R}_{+}^{2}, \\ \eta _{0}(x,s):=w_{0}(x)-w(x,-s) &{} \text {in }\mathscr {Q}{\times }\mathbb {R}_{+},\\ z(x,t,0)=u_{t}(x,t) &{} \text {in }\mathscr {Q},\\ z(x,0,p)=z_{0}(x,-p\tau ) &{} \text {in }I\times \left( 0,1\right) , \end{array} \end{aligned}$$

(15)

clearly, (14) gives

$$\begin{aligned} \begin{array}{rr} \tau z_{t}(x,t,p)-z_{p}(x,t,p)=0 &{} \text {in }\ \mathscr {Q}\times (0,1), \\ \eta _{t}(x,t,s)+\eta _{s}(x,t,s)-w_{t}(x,t)=0 &{} \text {in}\mathscr {Q}\times \mathbb {R}_{+}, \ \end{array} \end{aligned}$$

(16)

where $z_{p}=\partial _{p}z$ and $\eta _{s}=\partial _{s}\eta$.

Therefore, problem (1)–(5) is equivalent to

$$\begin{aligned} \left\{ \begin{array}{lll} u_{tt}-\left[ g(u,w)\right] _{x}+\alpha _{1}u_{t}+\alpha _{2}z(1) =0 &{} \text {in }\mathscr {Q}, \quad \\ \mathscr {D}w_{tt}+lw_{xxxx}-\left[ f(u,w)\right] _{x}+g(u,w)+\int _{0}^{ \infty }h(s)\eta _{xxxx}(s)ds =0 &{} \text {in}\mathscr {Q},\\ \tau z_{t}(p)-z_{p}(p) =0 &{} \text {in }\mathscr {Q}\times (0,1), \\ \eta _{t}(s)+\eta _{s}(s)-w_{t} =0 &{} \text {in }\mathscr {Q}\times \mathbb {R} _{+}, \end{array} \right. \nonumber \\ \end{aligned}$$

(17)

where $l=1-\int _{0}^{\infty }h(s)ds$, with boundary conditions

$$\begin{aligned} \begin{array}{ll} u(x,t)=0 &{} \text {in }\Sigma ,\\ w(x,t)=w_{x}(x,t)=0 &{} \text {in }\Sigma ,\\ \eta (x,t,s)=\eta _{x}(x,t,s)=0 &{} \text {in }\Sigma {\times } \mathbb {R} _{+}^{2}, \end{array} \end{aligned}$$

(18)

and initial conditions

$$\begin{aligned} \begin{array}{ll} \left( u(x,0),u_{t}(x,0)\right) =\left( u_{0}(x),u_{1}(x)\right) &{} \text {on }I, \\ \left( w(x,-s),w_{t}(x,0)\right) =\left( w_{0}(x,s),w_{1}(x)\right) &{} \text { in }\mathscr {Q},\\ z_{0}(x,p)=z(x,-p\tau ) &{} \text {in }I\times \left( 0,1\right) , \\ \eta _{0}(x,s)=w_{0}(x)-w(x,-s)&{} \text {in }\mathscr {Q}. \end{array} \end{aligned}$$

(19)

In what follow, we assume that the function k belong to the sobolev space $H^{1}(0,L)$.

3 Preliminaries

In this section, we state our stability results for problem (17)–(19). For this purpose, we start with the following hypotheses:

($\mathscr {H}_{1}$) $h: \mathbb {R}_{+}\rightarrow \mathbb {R}_{+}$ is a non-increasing differentiable function such that $h (0)>0$ and

$$\begin{aligned} l =1 -\int _0^{+\infty }h (s)ds >0. \end{aligned}$$

(20)

Also assume that there exist a positive constant $\alpha$such that

$$\begin{aligned} h^{\prime }(s)\le \alpha h(s),\quad \forall s\ge 0, \end{aligned}$$

(21)

($\mathscr {H}_{1}$) There exists an increasing strictly convex function $G:\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}$ of class $C^1 (\mathbb {R}_{+})\cap C^2(]0,+\infty [)$ satisfying

$$\begin{aligned} G(0)=G'(0)=0\quad \text {and}\quad \lim _{t\rightarrow +\infty }G'(t)=+\infty \end{aligned}$$

such that

$$\begin{aligned} \int _0^{+\infty }{\frac{{h(s)}}{{G^{-1}(-h'(s))}}}ds +\sup _{s\in \mathbb {R}_{+}}{\frac{{h(s)}}{{G^{-1}(-h'(s))}}}<+\infty . \end{aligned}$$

(22)

Remark 1

The condition (22) introduced in [19] is satisfied by any positive function h of class $C^{1}(\mathbb {R}_{+})$ with $h^{\prime }< 0$ and h is integrable on $\mathbb {R}_{+}$ (see [19,20,21] for explicit examples).

C and c denote some general positive constants, which may be different in different estimates.

3.1 Functional setting and assumptions.

In order to prove the well-posedness of (17) by using the semigroups theory, we introduce some functional spaces.

Let us introduce the energy space $\mathscr {H}$ by

$$\begin{aligned} \mathscr {H}=H_{0}^{1}(I)\times L^{2}(I)\times H_{0}^{2}(I)\times H_{0}^{1}(I)\times L^{2}(I\times \left( 0,1\right) )\times L_{h}(I) \end{aligned}$$

where $L_{h}(I)$ is the weighted Sobolev space defined by

$$\begin{aligned} L_{h}(I)=\left\{ v\in L^{2}\left( 0,\infty ;H_{0}^{2}(I)\right) \mid \int _{0}^{\infty }h(s)\left\| v_{xx}(s)\right\| ^{2}ds<\infty \right\} . \end{aligned}$$

The space $L_{h}(I)$ is endowed with the inner product

$$\begin{aligned} \left\langle v,\overline{v}\right\rangle _{h}=\int _{0}^{\infty }h (s)\left\langle v_{xx}(s),\overline{v}_{xx}(s)\right\rangle ds. \end{aligned}$$

Also, we denote by $\left\langle \left\langle .,.\right\rangle \right\rangle$ the natural inner product on the space $L^{2}(I\times \left( 0,1\right) )$, we note that the norms

$$\begin{aligned} \int _{0}^{L}\int _{0}^{1}z^{2}(x,t,p)dpdx \quad \text {and }\quad \int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z^{2}(x,t,p)dpdx \end{aligned}$$

are equivalent in $L^{2}(I\times \left( 0,1\right) ).$

Let $\xi$ be any positive number which satisfy

$$\begin{aligned} \alpha _{2}\tau<\xi <(2\alpha _{1}-\alpha _{2})\tau . \end{aligned}$$

Also, we define

$$\begin{aligned} L_{h}^{*}(0,L)=\left\{ v\in L^{2}\left( 0,\infty ;\left( H^{4}\cap H_{0}^{2}\right) (I)\right) \mid \int _{0}^{\infty }h(s)\left\| v_{xx}(s)\right\| ^{2}ds<\infty \right\} . \end{aligned}$$

$\mathscr {H}$ is endowed with the norm

$$\begin{aligned} \left\| \mathscr {U}\right\| _{\mathscr {H}}=\frac{2}{1-\mu }\left\| u_{x}\right\| ^{2}+\left\| u_{t}\right\| ^{2}+l\left\| w_{xx}\right\| ^{2}+\left\| w\right\| ^{2}+\left\| w_{tx}\right\| ^{2}+\xi |||z|||^{2}+h \circ \eta _{xx}\nonumber \\ \end{aligned}$$

(23)

where

$$\begin{aligned} \mathscr {U}= & {} \left( u,u_{t},w,w_{t},z,\eta \right) \\ |||z|||^{2}= & {} \int _{0}^{1}e^{-2\tau p}\left\| z\left( p\right) \right\| ^{2}dp \end{aligned}$$

and

$$\begin{aligned} h \circ \eta _{xx}=\int _{0}^{\infty }h(s)\left\| \eta _{xx}(s)\right\| ^{2}ds. \end{aligned}$$

3.2 Energy identity

We start by the following lemma:

Lemma 1

For $\left( h,\phi \right) \in \left( \mathscr {C}^{1}\cap L^{1}\right) \left( \mathbb {R}_{+}\right) \times \mathscr {C}^{1}\left( \mathbb {R}\right)$, we have

$$\begin{aligned} 2\left( h *\phi \right) \phi _{t}= & {} \frac{d}{dt}\left\{ -\int _{0}^{\infty }h(s)\left( \phi (t)-\phi (t-s)\right) ^{2}ds+h_{\infty }\left| \phi (t)\right| ^{2}\right\} \nonumber \\{} & {} +\int _{0}^{\infty }h(s)\left( \phi (t)-\phi (t-s)\right) ^{2}ds \end{aligned}$$

(24)

Lemma 2

Assume that $(\psi ,\psi _{t},\eta ,\eta _{t},z)$ is a strong solution of the problem (17)-(19). Then we have

$$\begin{aligned} \frac{\xi }{\tau }\frac{d}{dt}\int _{0}^{L}\int _{0}^{1}z^{2}(x,p,t)dp dx= & {} -\frac{\xi }{2\tau }\int _{0}^{L}\int _{0}^{1}\frac{\partial }{\partial p}z^{2}(x,p,t)dp dx\nonumber \\= & {} \frac{\xi }{2\tau }\int _{0}^{L}\left\{ z^{2}(x,0,t)-z^{2}(x,1,t)\right\} dx. \end{aligned}$$

(25)

Proof

We multiply the third equation in (17) by $\frac{\xi }{\tau }z$ and integrate the result over $(0,L) \times (0,1)$ with respect to p and x, respectively, to get

$$\begin{aligned} \frac{\xi }{\tau }\frac{d}{dt}\int _{0}^{L}\int _{0}^{1}z^{2}(x,p,t)d\rho dx= & {} -\frac{\xi }{2\tau }\int _{0}^{L}\int _{0}^{1}\frac{\partial }{\partial p}z^{2}(x,p,t)dp dx\\= & {} \frac{\xi }{2\tau }\int _{0}^{L}\left\{ z^{2}(x,0,t) -z^{2}(x,1,t)\right\} dx \end{aligned}$$

which gives (25). $\square$

We define the energy associated with the solution of system (17)–(19) by

$$\begin{aligned} 2\mathscr {E}(t)= & {} \int _{0}^{L}\left\{ u_{t}^{2}+\frac{2}{1-\mu }\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) ^{2}+lw_{xx}^{2}+w_{xt}^{2}dx\right\} \nonumber \\{} & {} +h \circ \eta _{xx}+\xi \int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z^{2}\left( x,p\right) dpdx \end{aligned}$$

(26)

where $\xi$ is a positive constant such that

$$\begin{aligned} \alpha _{2}\tau<\xi <(2\alpha _{1}-\alpha _{2})\tau \end{aligned}$$

(27)

and $\alpha _{1}$ and $\alpha _{1}$ satisfying

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

(28)

Lemma 3

Assume that $(\psi ,\psi _{t},\eta ,\eta _{t},z)$ is a strong solution of the problem (17)–(19). Then the derivative of $\mathscr {E}(t)$ satisfies

$$\begin{aligned} 2\frac{d\mathscr {E}(t)}{dt}= & {} h^{\prime }\circ \eta _{xx}+(\frac{\xi }{2\tau }-\alpha _{1})\int _{0}^{L}u_{t}^{2}(s)dx \nonumber \\{} & {} -\frac{\xi }{2\tau }\int _{0}^{L}z^{2}(1)dx-\alpha _{2}\int _{0}^{L}z(1)u_{t}dx. \end{aligned}$$

(29)

Moreover, for all $t \ge 0$, we have

$$\begin{aligned} 2\frac{d}{dt}\mathscr {E}(t)= & {} h^{\prime }\circ \eta _{xx}+(\frac{\xi }{2\tau }-\alpha _{1})\int _{0}^{L}u_{t}^{2}(s)dx \nonumber \\{} & {} -\frac{\xi }{2\tau }\int _{0}^{L}z^{2}(1)dx-\alpha _{2}\int _{0}^{L}z(1)u_{t}dx \nonumber \\\le & {} h ^{\prime }\circ \eta _{xx}+(\frac{\alpha _{2}}{2}-\frac{\xi }{2\tau })\int _{0}^{L}z^{2}(1)dx \nonumber \\{} & {} +(\frac{\alpha _{2}}{2}+\frac{\xi }{2\tau }-\alpha _{1})\int _{0}^{L}u_{t}^{2}(s)dx \nonumber \\\le & {} 0. \end{aligned}$$

(30)

Proof

Multiplying the first equation in (17) by $u_{t},$ the second by $w_{t}$ and the third by $\xi z(p),$ integrating by part and using boundary condition in (18) and Lemma 2 yields (30). $\square$

Lemma 4

(Jensen inequality) Let F be a convex function on [a, b],  $r_{1}:\Omega \rightarrow [a,b]$ and $r_{2}$ are integrable functions on $\Omega ,$ $r_{2}(x)\ge 0$, and $\int _{\Omega }r_{2}(x)dx=k_{0}>0$, then Jensen’s inequality states that

$$\begin{aligned} F\left[ \frac{1}{k_{0}}\int _{\Omega }r_{1}(x)r_{2}(x)dx\right] \le \frac{1}{k_{0}}\int _{\Omega }F[r_{1}(x)]r_{2}(x)dx. \end{aligned}$$

(31)

4 Global well-posedness

In this section we show the existence and regularity of solutions of the one dimensional viscoelastic Marguerre–Vlasov system (17)–(19).

Then problem (17)–(19) is reduced to the following problem for an abstract first-order evolutionary equation:

$$\begin{aligned} \mathscr {U}_{t}= & {} \mathscr {A}\mathscr {U}+\mathscr {B}(\mathscr {U}) \nonumber \\ \mathscr {U}(0)= & {} \mathscr {U}_{0}=(u_{0},u_{1},w_{0},w_{1},z_{0},\eta _{0})^{T}, \end{aligned}$$

(32)

where $\mathscr {U} = (u,u_{t},w,w_{t}, z,\eta )^{T}$, and

$$\begin{aligned} \mathscr {A}\mathscr {U}= & {} \left[ \begin{array}{c} u_{t} \\ \frac{2}{1-\mu }u_{xx}-\alpha _{1}u_{t}-\alpha _{2}z(1) \\ w_{t} \\ \mathscr {D}^{-1}\left[ -lw_{xxxx}-\int _{0}^{\infty }h(s)\eta _{xxxx}(s)ds\right] \\ \frac{1}{\tau }z_{p}(p) \\ -\eta _{s}(s)+w \end{array} \right] ,\\ \mathscr {B}\mathscr {U}= & {} \left[ \begin{array}{c} 0 \\ \frac{2}{1-\mu }\left[ \frac{1}{2}w^{2}+kw\right] _{x} \\ 0 \\ \mathscr {D}^{-1}\left[ \left[ f(u,w)\right] _{x}-kg(u,w)\right] \\ 0 \\ 0 \end{array} \right] . \end{aligned}$$

with the domain

$$\begin{aligned} D(\mathscr {A})= & {} \left( H^{2}\cap H_{0}^{1}\right) (I)\times L^{2}(I)\times \left( H^{4}\cap H_{0}^{2}\right) (I)\times H_{0}^{2}(I) \\{} & {} \times L^{2}(I\times H^{1}\left( 0,1\right) )\times L_{h}^{*}(0,L) \end{aligned}$$

Lemma 5

The operator $\mathscr {A}$ defined in (32) is the infinitesimal generator of a $C_{0}$-semigroup in $\mathscr {H}$.

Proof

For all $\mathscr {U}(t) \in D(\mathscr {A})$, one has

$$\begin{aligned} \left\langle A\mathscr {U},\mathscr {U}\right\rangle _{\mathscr {H}}= & {} -\alpha _{1}\int _{0}^{L}u_{t}^{2}dx-\alpha _{2}\int _{0}^{L}u_{t}z(1)dx +l\int _{0}^{L}w_{xx}w_{txx}dx \nonumber \\{} & {} -l\int _{0}^{L}w_{t}.\mathscr {D}\mathscr {D}^{-1}w_{xxxx}dx-\int _{0}^{\infty }h(s)\int _{0}^{L}w_{t}.\mathscr {D}\mathscr {D}^{-1}\eta _{xxxx}(s)dsdx \nonumber \\{} & {} +\frac{\xi }{\tau }\left\langle \left\langle z_{p},z\right\rangle \right\rangle -\left\langle \eta _{s},\eta \right\rangle _{h}+\left\langle w,\eta \right\rangle _{h}. \end{aligned}$$

(33)

Integrating by parts, we have

$$\begin{aligned} -\left\langle \eta _{s},\eta \right\rangle _{h}=h ^{\prime }\circ \eta _{xx} \end{aligned}$$

(34)

and

$$\begin{aligned} 2\left\langle \left\langle z_{p},z\right\rangle \right\rangle =\int _{0}^{L}u_{t}^{2}dx-\int _{0}^{L}z^{2}(1)dx. \end{aligned}$$

(35)

Plugging (34) and (35) in (33), we get

$$\begin{aligned} \left\langle \mathscr {A}\mathscr {U},\mathscr {U}\right\rangle _{\mathscr {H}}=2\frac{d}{ dt}\mathscr {E}(t)\le 0. \end{aligned}$$

(36)

which implies that $\mathscr {A}$ is dissipative.

Next we will prove that the operator $\left( \mathscr {I}-\mathscr {A}\right)$: $D(\mathscr {A})\longrightarrow \mathscr {H}$ is onto, that is, given $\mathscr {F}=(f_{1},f_{2},f_{3},f_{4},f_{5},f_{6})\in \mathscr {H}$, we seek $\mathscr {U} = (u,u_{t},w,w_{t}, z,\eta )^{T} \in D(\mathscr {A})$ such that

$$\begin{aligned} \left( \mathscr {I}-\mathscr {A}\right) \mathscr {U} = \mathscr {F}. \end{aligned}$$

(37)

Equivalently, one must consider the system given by

$$\begin{aligned} u-u_{t}&=f_{1},&\left( 18.1\right) \end{aligned}$$

(38)

$$\begin{aligned} u_{t}-\frac{2}{1-\mu }u_{xx}+\alpha _{1}u_{t}+\alpha _{2}z(1)&=f_{2},&\left( 18.2\right) \end{aligned}$$

(39)

$$\begin{aligned} w-w_{t}&=f_{3},&\left( 18.3\right) \end{aligned}$$

(40)

$$\begin{aligned} \mathscr {D}w_{t}+lw_{xxxx}+\int _{0}^{\infty }h (s)\eta _{xxxx}(s)ds&= \mathscr {D}f_{4},&\left( 18.4\right) \end{aligned}$$

(41)

$$\begin{aligned} \tau z-z_{p}(p)&=\tau f_{5},&\left( 18.5\right) \end{aligned}$$

(42)

$$\begin{aligned} \eta (s)+\eta _{s}(s)-w&=f_{6}&\left( 18.6\right) . \end{aligned}$$

(43)

By integrating the Eqs. (42) and (43), we obtain

$$\begin{aligned} z(1)=u_{t}e^{-\tau }+\tau \int _{0}^{1}e^{\tau (1-r)}f_{4}\left( r\right) dr \end{aligned}$$

(44)

and

$$\begin{aligned} \mu (s)=(1-e^{-s})(w-f_{1})+\int _{0}^{s}e^{r-s}f_{5}(r)dr. \end{aligned}$$

(45)

Plugging (44) and (45) in (39) and (41) and keeping in mind that $u_{t}=u-f_{1}$ and $w_{t}=w-f_{3}$ we find that u and w fulfil the equations

$$\begin{aligned} \left\{ \begin{array}{lll} \left( 1+\alpha _{1}+\alpha _{2}e^{-\tau }\right) u-\frac{2}{1-\mu }u_{xx} &{} =\beta _{1}, &{} \\ \widetilde{l}w_{xxxx}+\mathscr {D}w \qquad &{} =\beta _{2},&{} \end{array} \right. \end{aligned}$$

(46)

where

$$\begin{aligned} \widetilde{l}= & {} l+\int _{0}^{\infty }h (s)(1-e^{-s})ds>0, \end{aligned}$$

(47)

$$\begin{aligned} \beta _{1}= & {} f_{2}+\left( 1+\alpha _{1}+\alpha _{2}e^{-\tau }\right) f_{1}-\alpha _{2}\tau \int _{0}^{1}e^{\tau (1-r)}f_{4}\left( r\right) dr, \end{aligned}$$

(48)

$$\begin{aligned} \beta _{2}= & {} \mathscr {D}\left( f_{3}+f_{4}\right) -\int _{0}^{\infty }h(s)\int _{0}^{s}e^{r-s}f_{5xxxx}(r)drds \nonumber \\{} & {} +\int _{0}^{\infty }h(s)(1-e^{-s})dsf_{1xxxx}. \end{aligned}$$

(49)

Since $x\mapsto \left( f_{1},f_{5}\right) \in H_{0}^{2}(I)\times H_{0}^{2}(I)$ then $\left( f_{1xxxx},f_{5xxxx}\right) \in H^{-2}(I)\times H^{-2}(I)$. To show that $\beta _{1}\in H^{-2}(I)$, we have to show that

$$\begin{aligned} x\longmapsto \int _{0}^{\infty }h (s)\int _{0}^{s}e^{r-s}f_{5xxxx}(r)drds\in H^{-2}(I). \end{aligned}$$

Applying Cauchy–Shwarz inequality and Fubini’s theorem, we get

$$\begin{aligned} \left\| \int _{0}^{\infty }h (s)e^{-s}\int _{0}^{s}e^{r}f_{5xx}(r)drds\right\| ^{2}= & {} \int _{0}^{1}\left( \int _{0}^{\infty }h (s)e^{-s}\int _{0}^{s}e^{r}f_{5xx}(r)drds\right) ^{2}dx \\\le & {} \int _{0}^{1}\left( \int _{0}^{\infty }h (s)e^{-s}\int _{0}^{s}e^{r}\left| f_{5xx}(r)\right| drds\right) ^{2}dx\\\le & {} \int _{0}^{1}\left( \int _{0}^{\infty }e^{r}\left| f_{5xx}(r)\right| \int _{r}^{\infty }h (s)e^{-s}drds\right) ^{2}dx \\\le & {} \int _{0}^{1}\left( \int _{0}^{\infty }h (r)e^{r}\left| f_{5xx}(r)\right| \int _{r}^{\infty }e^{-s}dsdr\right) ^{2}dx \\\le & {} \int _{0}^{1}\left( \int _{0}^{\infty }h (s)\left| f_{5xx}(r)\right| ds\right) ^{2}dx \\\le & {} \left( \int _{0}^{\infty }h (s)ds\right) \int _{0}^{1}\int _{0}^{\infty }h (s)f_{5xx}^{2}(r)(s)dsdx \\= & {} h _{\infty }\left\| f_{5}\right\| _{h }^{2}<\infty . \end{aligned}$$

Thus

$$\begin{aligned} x\longmapsto \int _{0}^{\infty }h (s)\int _{0}^{s}e^{r-s}f_{5xxxx}(r)drds\in H^{-2}(0,L) \end{aligned}$$

(50)

Therefore, by Lax–Milgram Theorem, system (38) admits a unique solution $\mathscr {U}\in D(\mathscr {A})$. This means that (37) holds and consequently $I-\mathscr {A}$ is onto. Thus, by using the Lumer–Phillips [35, Theorem 4.3], we deduce that the operator $\mathscr {A}$ generates a $C_{0}$ semigroup of contractions in $\mathscr {H}$. $\square$

Lemma 6

The operator $\mathscr {B}$ defined in (32) is locally Lipschitz in $\mathscr {H}$.

Proof

Let $\widetilde{\mathscr {U}}=(u,u_{t},w,w_{t},z,\eta )$ and $\widetilde{\mathscr {U}}=(\widetilde{u},\widetilde{u}_{t},\widetilde{w},\widetilde{w}_{t},\widetilde{z},\widetilde{\eta })$ two elements of $\mathscr {H}{\normalsize .}$ A direct calculation shows that

$$\begin{aligned} \mathscr {B}(\mathscr {U})-\mathscr {B}(\widetilde{\mathscr {U}})=\frac{2}{1-\mu }\left( 0,F,0,\mathscr {D}^{-1}G,0,0\right) \end{aligned}$$

where

$$\begin{aligned} \ F=\left[ \frac{1}{2}\left( w^{2}-\widetilde{w}^{2}\right) +k\left( w-\widetilde{w}\right) \right] _{x} \end{aligned}$$

and

$$\begin{aligned} \ G= & {} \left[ w_{x}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) -\widetilde{w} _{x}\left( \widetilde{u}_{x}+\frac{1}{2}\widetilde{w}^{2}+k(x)\widetilde{w} \right) \right] _{x} \\{} & {} -k\left[ \frac{1}{2}\left( w^{2}-\widetilde{w} ^{2}\right) +k\left( w-\widetilde{w}\right) \right] . \end{aligned}$$

So we have to estimate F and G in $L^{2}(I)$ and $H_{0}^{1}(I)$ norm respectively.

Since $k\in H^{1}(I)$, we can use the embedding $H^{1}(I)\hookrightarrow L^{\infty }(I)$ to prove

$$\begin{aligned} \left\| F\right\|= & {} \left\| \frac{1}{2}\left( w^{2}-\widetilde{w} ^{2}\right) +k(x)\left( w-\widetilde{w}\right) \right\| _{H_{0}^{1}} \nonumber \\= & {} \frac{1}{2}\left\| \left( w-\widetilde{w}\right) \left( w+\widetilde{w} \right) \right\| _{H_{0}^{1}}+\left\| k(x)\left( w-\widetilde{w} \right) \right\| _{H_{0}^{1}} \nonumber \\\le & {} C\left( \left\| w_{xx}\right\| +\left\| \widetilde{w} _{xx}\right\| +\left\| k\right\| _{H^{1}}\right) \left\| w_{x}- \widetilde{w}_{x}\right\| \nonumber \\\le & {} C\left( \left\| \mathscr {U}\right\| _{\mathscr {H}},\left\| \widetilde{\mathscr {U}}\right\| _{\mathscr {H}}\right) \left\| \mathscr {U}-\widetilde{\mathscr {U}}\right\| _{\mathscr {H}}. \end{aligned}$$

(51)

Now, let

$$\begin{aligned} G_{1}=\mathscr {D}^{-1}\partial _{x}\left[ w_{x}\left( u_{x}+\frac{1}{2} w^{2}+k(x)w\right) -\widetilde{w}_{x}\left( \widetilde{u}_{x}+\frac{1}{2} \widetilde{w}^{2}+k(x)\widetilde{w}\right) \right] \end{aligned}$$

(52)

Taking into account that the operator $\mathscr {D}^{-1}\partial _{x}$ is bounded from $L^{2}(I)$ into $H_{0}^{1}\left( I\right)$, we can write

$$\begin{aligned} \left\| G_{1}\right\| _{H_{0}^{1}}\le C\left\| w_{x}\left( u_{x}+ \frac{1}{2}w^{2}+k(x)w\right) -\widetilde{w}_{x}\left( \widetilde{u}_{x}+ \frac{1}{2}\widetilde{w}^{2}+k(x)\widetilde{w}\right) \right\| . \end{aligned}$$

(53)

Adding and subtracting the term $\widetilde{w}_{x}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right)$ inside the norm on the right hand side of (27) and proceed with the same manner, we find that

$$\begin{aligned} \left\| G_{1}\right\| _{H{0}^{1}}\le C\left( \left\| \mathscr {U} \right\| _{\mathscr {H}},\left\| \widetilde{\mathscr {U}}\right\| _{ \mathscr {H}}\right) \left\| \mathscr {U}-\widetilde{\mathscr {U}} \right\| _{\mathscr {H}}. \end{aligned}$$

Finally, let

$$\begin{aligned} G_{2}=\left[ \frac{1}{2}\left( w^{2}-\widetilde{w}^{2}\right) +k(x)\left( w- \widetilde{w}\right) \right] _{x}. \end{aligned}$$

Similarly, have

$$\begin{aligned} \left\| G_{2}\right\| \le C\left( \left\| \mathscr {U}\right\| _{ \mathscr {H}},\left\| \widetilde{\mathscr {U}}\right\| _{\mathscr {H} }\right) \left\| \mathscr {U}-\widetilde{\mathscr {U}}\right\| _{ \mathscr {H}}. \end{aligned}$$

(54)

Then the operator $\mathscr {B}$ is locally Lipschitz in $\mathscr {H}$. So problem (17)–(19) admits a local solution. The proof is hence complete. $\square$

The boundedness of the energy in (30) allows to extend the solution on [0, T] for an any arbitrary $T>0,$ so we have shown:

Theorem 1

Assume that $\left( \textbf{H}_{1}\right)$ holds. Let $\mathscr {U}_{0}\in D(\mathscr {A}),$ then (17)–(19) has a unique solution

$$\begin{aligned} \mathscr {U}\in \mathscr {C}\left( \mathbb {R}_{+},D(\mathscr {A})\right) . \end{aligned}$$

5 General decay

In this section we consider a wider class of kernel functions, and we establish a general decay result, where exponential and polynomial decay rates are special cases.

The main result of general decay is the following.

Theorem 2

Assume that (20) hold such that (21) hold or there exists a positive constant M such that

$$\begin{aligned} \underset{s>0}{\sup }\int _{0}^{L}\eta _{0xx}^{2}(s)dx\le M,\text { }\forall s\in \mathbb {R}_{+}, \end{aligned}$$

(55)

then there exists positive constants $c^{\prime },c^{\prime \prime }$ and $\epsilon _{0}$ for which $\mathscr {E}$ satisfies

$$\begin{aligned} \mathscr {E}(t)\le c^{\prime \prime }e^{-c^{\prime }t},\forall t\in \mathbb {R} _{+}, \end{aligned}$$

(56)

or

$$\begin{aligned} \mathscr {E}(t)\le c^{\prime \prime }G_{1}^{-1}(c^{\prime }t), \forall t\in \mathbb {R} _{+}, \end{aligned}$$

(57)

where

$$\begin{aligned} G_{1}(s)=\int _{s}^{1}\frac{dr}{rG^{\prime }(\epsilon _{0}r)},s\in (0,1], \end{aligned}$$

Remark 2

The previous theorem shows that exponential decay holds when (20) holds, otherwise, if (21) holds, we get a weak decay of energy. For precise examples illustrating (57) see [19,20,21].

To prove Theorem 2, we need some useful lemmas.

Lemma 7

The following inequalities holds

$$\begin{aligned} \left( \int _{0}^{\infty }h (s)\mu (s)ds\right) ^{2}\ {}\le & {} h _{\infty }\int _{0}^{\infty }h (s)\mu ^{2}(s)ds, \end{aligned}$$

(58)

$$\begin{aligned} \left( \int _{0}^{\infty }h (s)\mu (s)ds\right) ^{2}\le & {} -h (0)\int _{0}^{\infty }h ^{\prime }(s)\mu ^{2}(s)ds. \end{aligned}$$

(59)

Proof

For inequality (58), we have

$$\begin{aligned} \left( \int _{0}^{\infty }h (s)\mu (s)ds\right) ^{2}\ =\left( \int _{0}^{\infty }\sqrt{h (s)}(\sqrt{h (s)}\mu (s)ds\right) ^{2}. \end{aligned}$$

Cauchy–Schwarz inequality leads to

$$\begin{aligned} \left( \int _{0}^{\infty }h (s)\mu (s)ds\right) ^{2}\ \le \int _{0}^{\infty }h (s)ds\int _{0}^{\infty }h (s)\mu ^{2}(s)ds. \end{aligned}$$

Similarly, we prove (59) by replacing $\sqrt{h(t-s)}$ by $\sqrt{-h^{\prime }(t-s)}.$ $\square$

Let $\mathscr {F}$ be the functional defined by

$$\begin{aligned} \mathscr {F}(t)=\lambda \mathscr {E}(t)+\delta _{1}\mathscr {I}(t)+\delta _{2}\mathscr {I}_{2}(t)+\delta _{3}\mathscr {I}_{3}\left( t\right) \end{aligned}$$

(60)

where

$$\begin{aligned} \mathscr {I}_{1}(t)= & {} =\int _{0}^{L}u.u_{t}+w.\mathscr {D}w_{t}dx, \end{aligned}$$

(61)

$$\begin{aligned} \mathscr {I}_{2}(t)= & {} -\int _{0}^{L}w_{t}\int _{0}^{\infty }h (s)\eta (s)dsdx, \end{aligned}$$

(62)

$$\begin{aligned} \mathscr {I}_{3}(t)= & {} \int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z(p)dpdx, \end{aligned}$$

(63)

$\mathscr {E}(t)$ is defined in (26), $\lambda > 0$, $\delta _{1}$, $\delta _{2}$ and $\delta _{3}$ are positive constants that will be chosen later.

The following proposition gives the equivalence between $\mathscr {E}(t)$ and the functional $\mathscr {F}(t)$.

Proposition 1

Assume that $\left( \textbf{H}_{1}\right)$ holds, then there exists two positive constants $\beta _{1}$, $\beta _{2}$ such that

$$\begin{aligned} \beta _{1}\mathscr {E}(t)\le \mathscr {F}(t)\le \beta _{2}\mathscr {E}(t). \end{aligned}$$

(64)

Proof

To compare $\mathscr {F}(t)$ with $\mathscr {E}(t)$, we have to estimate the terms $\mathscr {I}\left( t\right)$, $\mathscr {J}\left( t\right)$ and $\mathscr {K}\left( t\right)$ of the right hand side of (60) and show that.

$$\begin{aligned} \mid \mathscr {F}(t)-\lambda \mathscr {E}(t)\mid \le c^{*}\mathscr {E}(t),\quad c^{*}>0. \end{aligned}$$

From (61), (62) and (63), we obtain

$\bullet$ Estimate for $\mathscr {I}_{1}(t)$

Using Poincaré’s and (72, we obtain

$$\begin{aligned} \delta _{1}\left| \mathscr {I}_{1}(t)\right|= & {} \delta _{1}\left| \int _{0}^{L}u_{t}u+\mathscr {D}w_{t}.wdx\right| \nonumber \\\le & {} \ \frac{\delta _{1}}{2}\int _{0}^{L}u_{t}^{2}dx+\frac{\delta _{1}}{2} \int _{0}^{L}u^{2}dx+\frac{\delta _{1}}{2}\int _{0}^{L}w_{t}^{2}dx\nonumber \\{} & {} +\frac{\delta _{1}}{2}\int _{0}^{L}w_{xt}^{2}dx+\frac{\delta _{1}}{2}L^{2}\int _{0}^{L}w_{xx}^{2}dx\nonumber \\\le & {} c_{1}\mathscr {E}(t). \end{aligned}$$

(65)

where $c_{1}$ is a positive constant.

$\bullet$ Estimate for $\mathscr {I}_{2}(t)$

Using Poincaré’s and (58, we obtain

$$\begin{aligned} \delta _{2}\left| \mathscr {I}_{2}(t)\right|= & {} \delta _{2}\left| \int _{0}^{L}\mathscr {D}w_{t}\int _{0}^{\infty }h (s)\mu (s)dsdx\right| \nonumber \\\le & {} \frac{\delta _{2}}{2}\int _{0}^{L}w_{t}^{2}dx+\frac{\delta _{2}}{2}\int _{0}^{L}w_{xt}^{2}dx\nonumber \\{} & {} +\frac{\delta _{2}}{2}L^{2}\int _{0}^{L}w_{xx}^{2}dx+Ch \circ \eta _{xx} \nonumber \\\le & {} c_{2}\mathscr {E}(t). \end{aligned}$$

(66)

where $c_{2}$ is a positive constant.

$\bullet$ Estimate for $\mathscr {I}_{3}(t):=\int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z^{2}dpdx$

Since $\mathscr {I}_{3}(t)$ defines a norm in $L^{2}(0,L;L^{2}(0,1))$ which is equivalent to the one induced by $L^{2}(0,L;L^{2}(0,1))$, then we have

$$\begin{aligned} \delta _{3}\left| \mathscr {I}_{3}(t)\right| \le \delta _{3}\int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z^{2}dpdx\le \delta _{3}\int _{0}^{L}\int _{0}^{1}z^{2}dpdx\le c_{3}\mathscr {E}(t), \end{aligned}$$

(67)

where $c_{3}$ is a positive constant.

According to (65), (66) and (67), we have

$$\begin{aligned} \mid \mathscr {F}(t)-\lambda \mathscr {E}(t)\mid \le c^{*}\mathscr {E}(t),\quad c^{*}>0, \end{aligned}$$

where

$$\begin{aligned} c^{*}=\max \left\{ c_{1},c_{2},c_{3}\right\} . \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \mid \mathscr {F}(t)-\lambda \mathscr {E}(t)\mid \le c^{*}\mathscr {E}(t), \end{aligned}$$

that is

$$\begin{aligned} (\lambda -c^{*})\mathscr {E}(t)\le \mathscr {F}(t)\le (\lambda +c^{*})\mathscr {E}(t). \end{aligned}$$

So, we choose $\lambda$ large enough such that $\beta _{1} = \lambda - c^{*}> 0$, $\beta _{2} = \lambda + c^{*}> 0$. Then (64) holds true.

This completes the proof. $\square$

In order to proof the main theorem, we need some additionals lemmas.

Lemma 8

Suppose that $(\psi ,\psi _{t},\eta ,\eta _{t},z)$ is the solution of (17)–(19). Then the derivative of the functional $\mathscr {I}_{1}(t)$ satisfies

$$\begin{aligned} \frac{d}{dt}\mathscr {I}_{1}(t)\le & {} C_{\varepsilon }\int _{0}^{L}u_{t}^{2}dx+\int _{0}^{L}w_{t}^{2}dx+\int _{0}^{L}w_{xt}^{2}dx-\left( l-\varepsilon C\right) \int _{0}^{L}w_{xx}^{2}dx\nonumber \\{} & {} -\left( \frac{8}{1-\mu }-\varepsilon C\right) \int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+kw\right) ^{2}dx\nonumber \\{} & {} +C_{\varepsilon }\int _{0}^{L}z^{2}(1)dx+C_{\varepsilon }h \circ \eta _{xx}. \end{aligned}$$

(68)

where $\varepsilon$ is an arbitrary positive constant.

Proof

Multiplying the first equation in (17) by u, the second by w and taking account that $\frac{d}{dt}\int _{0}^{L}h_{t}hdx=\int _{0}^{L}\left( h_{tt}h+h_{t}^{2}\right) dx,$ then integrating by part, we get

$$\begin{aligned} 0= & {} -\frac{d}{dt}\mathscr {I}_{1}(t)+\int _{0}^{L}u_{t}^{2}dx+ \int _{0}^{L}w_{t}^{2}dx+\int _{0}^{L}w_{xt}^{2}dx-l\int _{0}^{L}w_{xx}^{2}dx \nonumber \\{} & {} -\frac{8}{1-\mu }\int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+kw\right) ^{2}dx+\int _{0}^{L}\left( \alpha _{1}u_{t}+\alpha _{2}z(1)\right) udx \nonumber \\{} & {} +\int _{0}^{L}w_{xx}\int _{0}^{\infty }h (s)\eta _{xx}(s)dsdx. \end{aligned}$$

(69)

Using Young’s and Poincaré’s inequalities, the two last terms are estimated by

$$\begin{aligned} \int _{0}^{L}\left( \alpha _{1}u_{t}+\alpha _{2}z(1)\right) udx\le \varepsilon \int _{0}^{L}u^{2}dx+C_{\varepsilon }\int _{0}^{L}u_{t}^{2}dx+C_{\varepsilon }\int _{0}^{L}z^{2}(1)dx \end{aligned}$$

(70)

and using (58), we arrive at

$$\begin{aligned} \int _{0}^{L}w_{xx}\int _{0}^{\infty }h (s)\eta _{xx}(s)dsdx\le \varepsilon \int _{0}^{L}w_{xx}^{2}dx+C_{\varepsilon }h \circ \eta _{xx}. \end{aligned}$$

(71)

Now to estimate $\int _{0}^{L}u^{2}dx$, we have

$$\begin{aligned} \int _{0}^{L}u^{2}dx\le & {} L^{2}\int _{0}^{L}u_{x}^{2}dx\le L^{2}\int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) ^{2}dx \nonumber \\{} & {} +L^{6}\int _{0}^{L}w_{xx}^{4}dx+L^{6}\left\| k\right\| _{L^{\infty }}\int _{0}^{L}w_{xx}^{2}dx \nonumber \\\le & {} L^{2}\int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) ^{2}dx+\left( \frac{L^{6}C\mathscr {E}(0)}{4}+L^{6}\left\| k\right\| _{L^{\infty }}\right) \int _{0}^{L}w_{xx}^{2}dx \nonumber \\\le & {} C\left\{ \int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) ^{2}dx+\int _{0}^{L}w_{xx}^{2}dx\right\} . \end{aligned}$$

(72)

Plugging (70), (71) and (72) in (69) this proves (68). $\square$

Lemma 9

Assume that $(\psi ,\psi _{t},\eta ,\eta _{t},z)$ is the solution of (17)–(19). Then the derivative of the functional $\mathscr {I}_{2}(t)$ satisfies

$$\begin{aligned} \frac{d}{dt}\mathscr {I}_{2}(t)\le & {} -\left( h _{\infty }-3\varepsilon \right) \int _{0}^{L}w_{t}^{2}dx-\left( h _{\infty }-\varepsilon \right) \int _{0}^{L}w_{tx}^{2}dx+\left( \varepsilon +\varepsilon ^{2}\right) C\int _{0}^{L}w_{xx}^{2}dx \nonumber \\{} & {} +C\varepsilon \int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) ^{2}dx-C_{\varepsilon }h ^{\prime }\circ \eta _{xx}+C_{\varepsilon }h \circ \eta _{xx}. \end{aligned}$$

(73)

where $\varepsilon$ is an arbitrary positive constant.

Proof

Multiplying the second equation in (18) by $\int _{0}^{\infty }h(s)\mu (s)ds$ and integrating by parts, we get

$$\begin{aligned} 0= & {} \int _{0}^{L}\mathscr {D}w_{tt}\int _{0}^{\infty }h (s)\mu (s)dsdx+l\int _{0}^{L}w_{xx}\int _{0}^{\infty }h (s)\mu _{xx}(s)dsdx \nonumber \\{} & {} +\int _{0}^{L}f(u,w)\int _{0}^{\infty }h (s)\eta _{x}(s)dsdx+\int _{0}^{L}k\langle (u,w)wdx +\int _{0}^{L}\left( \int _{0}^{\infty }h(s)\eta _{xx}(s)ds\right) ^{2}dx.\nonumber \\ \end{aligned}$$

(74)

For the first term in the right hand side of (74), we have

$$\begin{aligned} \int _{0}^{L}\mathscr {D}w_{tt}\int _{0}^{\infty }h (s)\mu (s)ds=-\frac{d}{ dt}\mathscr {J}(t)-\int _{0}^{L}\mathscr {D}w_{t}\frac{d}{dt}\int _{0}^{\infty }h (s)\mu (s)dsdx. \end{aligned}$$

(75)

In the other hand, we also have

$$\begin{aligned} \frac{d}{dt}\int _{0}^{\infty }h (s)\eta (s)ds= & {} \frac{d}{dt} \int _{0}^{\infty }h (s)\left( w(t)-w(t-s)\right) ds \nonumber \\= & {} \frac{d}{dt}\int _{-\infty }^{t}h (t-s)\left( w(t)-w(s)\right) ds \nonumber \\= & {} \int _{-\infty }^{t}h ^{\prime }(t-s)\left( w(t)-w(t-s)\right) +h _{\infty }w_{t} \nonumber \\= & {} \int _{0}^{\infty }h ^{\prime }(s)\eta (s)ds+h _{\infty }w_{t}. \end{aligned}$$

(76)

Plugging (76) in (75), we infer

$$\begin{aligned} \int _{0}^{L}\mathscr {D}w_{t}\int _{0}^{\infty }h (s)\mu (s)dsdx= & {} - \frac{d}{dt}\mathscr {J}(t)-h _{\infty }\int _{0}^{L}w_{t}^{2}dx-h _{\infty }\int _{0}^{L}w_{tx}^{2}dx \nonumber \\{} & {} +\int _{0}^{L}\mathscr {D}w_{t}\int _{0}^{\infty }h ^{\prime }(s)\eta (s)dsdx. \end{aligned}$$

(77)

Integrating by parts and applying Young’s inequality and (59), then Poincaré’s inequality, we get

$$\begin{aligned} \int _{0}^{L}\mathscr {D}w_{t}\int _{0}^{\infty }h ^{\prime }(s)\mu (s)dsdx= & {} \int _{0}^{L}\left( w_{t}-w_{txx}\right) \int _{0}^{\infty }h ^{\prime }(s)\eta (s)dsdx \nonumber \\\le & {} 2\varepsilon \int _{0}^{L}w_{t}^{2}dx-C_{\varepsilon }h ^{\prime }\circ \eta _{xx}. \end{aligned}$$

(78)

Using the embedding $H^{1}(I)\hookrightarrow L^{\infty }\left( I\right)$, we estimate $-\int _{0}^{L}\left[ f(u,w)\right] _{x}\int _{0}^{\infty }h(s)\mu (s)dsdx$ as follows

$$\begin{aligned} -\int _{0}^{L}{} & {} \left[ f(u,w)\right] _{x}\int _{0}^{\infty }h (s)\mu (s)dsdx \nonumber \\= & {} -\frac{2}{1-\mu }\int _{0}^{L}w_{x}\left( u_{x}+\frac{1}{2} w^{2}+k(x)w\right) \int _{0}^{\infty }h (s)\eta _{x}(s)dsdx \nonumber \\\le & {} \varepsilon ^{2}\int _{0}^{L}\left[ w_{x}\left( u_{x}+\frac{1}{2} w^{2}+k(x)w\right) \right] ^{2}dx+C_{\varepsilon }h \circ \eta _{xx} \nonumber \\\le & {} C\varepsilon ^{2}\left\| w_{x}\right\| _{L\infty }\int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) ^{2}dx+C_{\varepsilon }h \circ \eta _{xx} \nonumber \\\le & {} C\varepsilon ^{3}\int _{0}^{L}w_{xx}^{2}dx+C\varepsilon \int _{0}^{L}\left( u_{x}+\frac{1}{2}w^{2}+k(x)w\right) ^{2}dx\nonumber \\{} & {} +C_{\varepsilon }h \circ \eta _{xx}. \end{aligned}$$

(79)

Similarly, we estimate $\int _{0}^{L}kg(u,w)wdx$ using the umbedding $H^{1}(I)\hookrightarrow L^{\infty }\left( I\right)$.

Gathering (77), (78) and (79), (73) is proven. $\square$

Lemma 10

Suppose that $(\psi ,\psi _{t},\eta ,\eta _{t},z)$ is the solution of (17)–(19). Then the time derivative of the functional $\mathscr {I}_{3}(t)$ satisfies

$$\begin{aligned} \frac{d}{dt}\mathscr {I}_{3}(t)=\frac{e^{-2\tau }}{\tau }\int _{0}^{L}z^{2}(1)-\frac{1}{\tau }\int _{0}^{L}\eta _{t}^{2}dx-2\mathscr {I}_{3}(t). \end{aligned}$$

(80)

Proof

Multiplying the third equation in (17) by $e^{-2\tau p}z$ and integrating over $I\times \left( 0,1\right)$, we arrive at

$$\begin{aligned} 0= & {} \tau \int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z(p)z_{t}(p)dpdx-\int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z(p)z_{p}(p)dpdx \\= & {} \frac{\tau }{2}\frac{d}{dt}\int _{0}^{L}\int _{0}^{1}e^{-2\tau p}z^{2}(p)dpdx-\frac{1}{2}\int _{0}^{L}\int _{0}^{1}e^{-2\tau p}\frac{d}{dp}\left[ z^{2}(p) \right] dpdx \\= & {} \frac{\tau }{2}\frac{d}{dt}\mathscr {I}_{3}(t)+\frac{1}{2}\int _{0}^{L}u_{t}^{2}dx-\frac{e^{-2\tau }}{2}\int _{0}^{L}z(1)^{2}dx+\tau \mathscr {I}_{3}(t), \end{aligned}$$

which gives (80) $\square$

Proposition 2

Assume that $\left( \textbf{H}_{1}\right)$ and $\left( \textbf{H}_{2}\right)$ hold, then there exists two positive constants $\beta _{1},\beta _{2}$ such that

$$\begin{aligned} \frac{d}{dt}\mathscr {F}(t)\le -\beta _{1}\mathscr {E}(t)+\beta _{2}h\circ \psi _{xx}. \end{aligned}$$

(81)

Proof

By using (60) and combining (30), (68), (73) and (80), we get

$$\begin{aligned} \frac{d}{dt}\mathscr {F}(t)\le & {} \left\{ -N_{1}\left( l-\varepsilon C\right) +N_{2}C\left( \varepsilon +\varepsilon ^{2} \right) \right\} \left\| \psi _{xx}\right\| ^{2} \nonumber \\{} & {} +\left\{ N_{1}- N_{2}\left( h_{\infty }-3\varepsilon \right) \right\} \left\| \psi _{t}\right\| ^{2}+\left\{ N_{1}C_{\varepsilon }+\lambda (\frac{\xi }{2\tau }+\frac{\alpha _{2}}{2}-\alpha _{1})\right\} \left\| \eta _{t}\right\| ^{2} \nonumber \\{} & {} +\left\{ N_{1}-N_{2}\left( h_{\infty }-\varepsilon \right) \right\} \nonumber \\{} & {} \times \left\| \psi _{tx}\right\| ^{2}+\left\{ N_{1}\left( \varepsilon C-\frac{8}{1-\mu }\right) +CN_{2}\varepsilon \right\} \left\| u_{x}+\frac{1}{2}w^{2}+k(x)w\right\| ^{2} \nonumber \\{} & {} -\left( N_{2}C_{\varepsilon }-\frac{\lambda }{2}\right) h^{\prime }\circ \psi _{xx}+C_{\varepsilon }\left( N_{1}+N_{2}\right) h\circ \psi _{xx}-2\mathscr {I}_{3}(t) \nonumber \\{} & {} +\left\{ N_{2}C_{\varepsilon }-\frac{e^{-2\tau }\alpha _{1}}{\tau } +\lambda (\frac{\alpha _{2}}{2}-\frac{\xi }{2\tau })\right\} \left\| z(1)\right\| ^{2}. \end{aligned}$$

(82)

We want to impose suitable conditions on the different parameters so that the coefficients on the right hand side of (82) are all strictly negative. That is to obtain the following inequalities

$$\begin{aligned} N_{2}C\left( \varepsilon +\varepsilon ^{2}\right) <N_{1}\left( l-\varepsilon C \right) , \end{aligned}$$

(83)

$$\begin{aligned} N_{1}<N_{2}\left( h_{\infty }-3\varepsilon \right) , \end{aligned}$$

(84)

$$\begin{aligned} N_{1}<N_{2}\left( h_{\infty }-\varepsilon \right) , \end{aligned}$$

(85)

$$\begin{aligned} CN_{2}\varepsilon <N_{1}\left( \frac{8}{1-\mu }-\varepsilon C\right) , \end{aligned}$$

(86)

$$\begin{aligned} N_{2}C_{\varepsilon }<\frac{\lambda }{2}, \end{aligned}$$

(87)

$$\begin{aligned} N_{2}C_{\varepsilon }<\frac{e^{-2\tau }\alpha _{1}}{\tau }-\lambda \left( \frac{\alpha _{2}}{2}-\frac{\xi }{2\tau }\right) , \end{aligned}$$

(88)

$$\begin{aligned} N_{1}C_{\varepsilon }<-\lambda \left( \frac{\xi }{2\tau }+\frac{\alpha _{2}}{2} -\alpha _{1}\right) . \end{aligned}$$

(89)

We observe that (83) and (84) will be satisfied if we choose $\varepsilon > 0$ small enough and such that

$$\begin{aligned} \varepsilon <\max \left\{ \frac{l}{C},\frac{h_{\infty }}{3},\frac{8}{C\left( 1-\mu \right) }\right\} . \end{aligned}$$

To make (84) and (85) hold we can choose

$$\begin{aligned} N_{1}<N_{2}\left( h_{\infty }-3\varepsilon \right) . \end{aligned}$$

Concerning (87), (88) and (89), we pick

$$\begin{aligned} \lambda =\max \left\{ 2N_{2}C_{\varepsilon },-\frac{N_{2}C_{\varepsilon }}{(\frac{\alpha _{2}}{2}-\frac{\xi }{2\tau })},-\frac{N_{1}C_{\varepsilon }}{(\frac{\xi }{2\tau }+\frac{\alpha _{2}}{2}-\alpha _{1})},c^{*}\right\} , \end{aligned}$$

This completes the proof. $\square$

We consider the following two cases.

Case I. H(t) is linear:

By multiplying (81) by $\xi (t)$ and using (30), we get

$$\begin{aligned} \frac{d}{dt}\mathscr {F}(t)\xi (t)\le & {} -\beta _{1}\xi (t)\mathscr {E}(t)+\beta _{2}\xi (t)h\circ \psi _{xx} \\\le & {} -\beta _{1}\xi (t)\mathscr {E}(t)+\beta _{2}h\xi \circ \psi _{xx} \\\le & {} -\beta _{1}\xi (t)\mathscr {E}(t)-\beta _{2}ah^{\prime }\circ \psi _{xx} \\\le & {} -\beta _{1}\xi (t)\mathscr {E}(t)-c\mathscr {E}^{\prime }(t) \end{aligned}$$

which gives, as $\xi$ is nonincreasing,

$$\begin{aligned} \frac{d}{dt}\left( \mathscr {F}(t)\xi (t)+c\mathscr {E}(t)\right) \le -\beta _{1}\xi (t)\mathscr {E}(t),\quad \forall t \ge t_{1}. \end{aligned}$$

Hence, using the fact that $\mathscr {F}(t)\xi (t)+c\mathscr {E}(t)$ is equivalent to $\mathscr {E}(t)$, it is easy to see that

$$\begin{aligned} \frac{d}{dt}\left( \mathscr {F}(t)\xi (t)+c\mathscr {E}(t)\right) \le -\beta _{1}\xi (t)(\mathscr {F}(t)\xi (t)+c\mathscr {E}(t)),\quad \forall t \ge t_{1}. \end{aligned}$$

for some $\beta _{1} > 0$. Then

$$\begin{aligned} \left( \mathscr {F}(t)\xi (t)+c\mathscr {E}(t)\right) \le \gamma _{2}e^{-\beta _{1}\int _{t_{1}}^{t}\xi (s)ds}, \quad \forall t \ge t_{1} \end{aligned}$$

from which we deduce

$$\begin{aligned} \mathscr {E}(t)\le \gamma _{2}e^{-\beta _{1}\int _{t_{1}}^{t}\xi (s)ds}, \forall t \ge t_{1} \end{aligned}$$

for some $\gamma _{2}>0.$

Furthermore, using the continuity and boundedness of $\mathscr {E} (t)$ in $[0, t_{1}]$, we get

$$\begin{aligned} \mathscr {E}(t)\le \gamma _{2}e^{-\beta _{1}\int _{0}^{t}\xi (s)ds},\quad \forall t \ge 0. \end{aligned}$$

Case II. H(t) is nonlinear:

Next, with $f(t) =\int _{t}^{\infty }h(s)ds$, we use the functional

$$\begin{aligned} \mathscr {K}(t)=\int _{0}^{t}f(t-s)\left\| \psi _{xx}(s)\right\| _{I}^{2}ds. \end{aligned}$$

(90)

Lemma 11

Suppose that $(\psi ,\psi _{t},\eta ,\eta _{t},z)$ is the solution of ((17))–(19). The functional $\mathscr {K}$ defined by (90) satisfies, for any $\varepsilon >0$, the estimate

$$\begin{aligned} \frac{d}{dt}\mathscr {K}(t)\le \left( 2\varepsilon -1\right) h\circ \psi _{xx}+\left( f(t)+C_{\varepsilon }\right) \left\| \psi _{xx}\right\| ^{2}. \end{aligned}$$

(91)

Proof

By Young’s inequality and the fact $f^{\prime }(t)=-h(t)$, we see that

$$\begin{aligned} \frac{d}{dt}\mathscr {K}(t)= & {} f(0)\left\| \psi _{xx}\right\| ^{2}-\int _{0}^{t}h(t-s)\left\| \psi _{xx}(s)\right\| ^{2}ds \nonumber \\= & {} -h\circ \psi _{xx}-2\left\langle \psi _{xx},h\diamond \psi _{xx}\right\rangle +f(t)\left\| \psi _{xx}\right\| ^{2}. \end{aligned}$$

(92)

But

$$\begin{aligned} -2\left\langle \psi _{xx},h\diamond \psi _{xx}\right\rangle \le C_{\varepsilon }\left\| \psi _{xx}\right\| ^{2}+2\varepsilon h\circ \psi _{xx}, \end{aligned}$$

(93)

Combining (92) and (93), we obtain (91). $\square$

Let us introduce the functional

$$\begin{aligned} \widetilde{\mathscr {F}}(t)=\mathscr {F}(t)+\sigma \mathscr {K}(t), \end{aligned}$$

where $\sigma$ is a positive constant. Then we have

$$\begin{aligned} \widetilde{\mathscr {F}}(t)\sim \mathscr {E}(t). \end{aligned}$$

Therefore, it is always possible to pick $N_{1}$ (in 82) and h large enough to get

$$\begin{aligned} \frac{d}{dt}\widetilde{\mathscr {F}}(t)\le -C\mathscr {E}(t). \end{aligned}$$

Integrating over $(t_{0},\infty )$, we get

$$\begin{aligned} C\int _{t_{0}}^{\infty }\mathscr {E}(s)ds\le \widetilde{\mathscr {F}}(t_{0})<\infty . \end{aligned}$$

(94)

Next, let us define the functional $\mathscr {L}(t)$

$$\begin{aligned} \mathscr {L}(t)=q\int _{t_{0}}^{t}\left\| \psi _{xx}(s)-\psi _{xx}(t-s)\right\| _{I}^{2}ds\text { },\forall t\ge t_{0}. \end{aligned}$$

where $q>0.$ Thanks to (94), we can always choose q such that

$$\begin{aligned} \mathscr {L}(t)<1,\forall t\ge t_{0}. \end{aligned}$$

(95)

Next we define

$$\begin{aligned} \mathscr {L}_{h}(t)=-\int _{t_{0}}^{t}h^{\prime }(s)\left\| {\psi }_{xx}(t)-{\psi }_{xx}(t-s)\right\| ^{2}ds,\text { }\forall t\ge t_{0}. \end{aligned}$$

Observe that

$$\begin{aligned} \mathscr {L}_{h}(t)\le -C\mathscr {E}^{\prime }(t), \end{aligned}$$

for some positive constant C.

Since H is strictly convexe on (0, r] and $H(0)=0$ we have

$$\begin{aligned} H(\theta x)\le \theta H(x),\text { }\left( \theta ,x\right) \in [0,1]\times (0,r]. \end{aligned}$$

(96)

Using $\left( \textbf{H}_{2}\right)$, we get:

$$\begin{aligned} \mathscr {L}_{h}(t)= & {} \frac{1}{q\mathscr {L}(t)}\int _{t_{0}}^{t}\mathscr {L} (t)\left( -h^{\prime }(s)\right) q\left\| {\psi }_{xx}(t)- {\psi }_{xx}(t-s)\right\| ^{2}ds \\\ge & {} \frac{1}{q\mathscr {L}(t)}\int _{t_{0}}^{t}\mathscr {L}(t)\xi (s)H\left( h(s)\right) q\left\| {\psi }_{xx}(t)-{\psi } _{xx}(t-s)\right\| ^{2}ds \\\ge & {} \frac{\xi (t)}{q\mathscr {L}(t)}\int _{t_{0}}^{t}H\left( \mathscr {L} (t)h(s)\right) q\left\| {\psi }_{xx}(t)-{\psi } _{xx}(t-s)\right\| ^{2}ds, \end{aligned}$$

Keeping in mind (95) and applying inequality (96) for $\theta :=\mathscr {L}(t)$ and $x=h(s)$, yields

$$\begin{aligned} \mathscr {L}_{h}(t)\ge \frac{\xi (t)}{q\mathscr {L}(t)}\int _{t_{0}}^{t}H \left( \mathscr {L}(t)h(s)\right) q\left\| {\psi }_{xx}(t)- {\psi }_{xx}(t-s)\right\| ^{2}ds. \end{aligned}$$

(97)

Applying Jensen’s inequality in (31) for $r_{1}(t)=\mathscr {L}(t)h(s)$ and $r_{2}(s)= q\left\| {\psi }_{xx}(t)-{\psi }_{xx}\right. \left. (t-s)\right\| ^{2},$ we obtain

$$\begin{aligned} \mathscr {L}_{h}(t)\ge & {} \frac{\xi (t)}{q\mathscr {L}(t)}\int _{t_{0}}^{t}H \left( \mathscr {L}(t)h(s)\right) q\left\| {\psi }_{xx}(t)- {\psi }_{xx}(t-s)\right\| ^{2}ds \\\ge & {} \frac{\xi (t)}{q}H\left( \frac{1}{\mathscr {L}(t)}\int _{t_{0}}^{t} \mathscr {L}(t)h(s)q\left\| {\psi }_{xx}(t)-{\psi } _{xx}(t-s)\right\| ^{2}ds\right) \\= & {} \frac{\xi (t)}{q}H\left( \int _{t_{0}}^{t}h(s)q\left\| {\psi }_{xx}(t)-{\psi }_{xx}(t-s)\right\| ^{2}ds\right) \\= & {} \frac{\xi (t)}{q}\overline{H}\left( \int _{t_{0}}^{t}h(s)q\left\| {\psi }_{xx}(t)-{\psi }_{xx}(t-s)\right\| ^{2}ds\right) , \end{aligned}$$

where $\overline{H}$ is an extention of H such that $\overline{H}$ is strictly increasing and strictly convexe $\mathscr {C}^{2}$ function on $(0,\infty )$ and this leads to

$$\begin{aligned} \int _{t_{0}}^{t}g(s)q\left\| {\psi }_{xx}(t)-{\psi } _{xx}(t-s)\right\| ^{2}ds\le \frac{1}{q}\overline{H}^{-1}\left( \frac{q \mathscr {L}_{g}(t)}{\xi (t)}\right) . \end{aligned}$$

So (81) becomes

$$\begin{aligned} \widetilde{\mathscr {F}}(t)\le -\beta _{1}\mathscr {E} (t)+\beta _{2}\frac{1}{q}\overline{H}^{-1}\left( \frac{q\mathscr {L}_{g}(t)}{ \xi (t)}\right) ,\text { }\forall t>t_{0}. \end{aligned}$$

(98)

Let $\epsilon _{0}<r$, using the fact that $\mathscr {E}^{\prime }\le 0,$ $\overline{H}^{\prime }>0,\overline{H}^{\prime \prime }>0$, we observe that the functional $\mathscr {N}$ defined by

$$\begin{aligned} \mathscr {N}(t):=\overline{H}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E} (t)}{\mathscr {E}(0)}\right) \mathscr {F}(t)+\mathscr {E}(t), \end{aligned}$$

is equivalent to $\mathscr {E}$.

Using (98), we find that $\mathscr {N}$ satisfies

$$\begin{aligned} \frac{d}{dt}\mathscr {N}(t)= & {} \epsilon _{0}\frac{\mathscr {E}^{\prime }(t)}{ \mathscr {E}(0)}{\overline{H}}^{{\prime }{\prime }}\left( \epsilon _{0}\frac{ \mathscr {E}(t)}{\mathscr {E}(0)}\right) \mathscr {F}(t)+{\overline{H}}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) \mathscr {F} ^{\prime }(t)+\mathscr {E}^{\prime }(t) \nonumber \\\le & {} \epsilon _{0}\frac{\mathscr {E}^{\prime }(t)}{\mathscr {E}(0)}{\overline{H}}^{{\prime }{\prime }}\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)} \right) \mathscr {F}(t)\nonumber \\{} & {} +{\overline{H}}^{\prime }\left( \epsilon _{0}\frac{ \mathscr {E}(t)}{\mathscr {E}(0)}\right) \left[ -\beta _{1}\mathscr {E} (t)+\beta _{2}\frac{1}{q}{\overline{H}}^{-1}\left( \frac{q\mathscr {L}_{g}(t)}{ \xi (t)}\right) \right] +\mathscr {E}^{\prime }(t) \nonumber \\\le & {} -\beta _{1}\mathscr {E}(t){\overline{H}}^{\prime }\left( \epsilon _{0} \frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) \nonumber \\{} & {} +\frac{\beta _{2}}{q}{\overline{H}}^{-1}\left( \frac{q\mathscr {L}_{g}(t)}{\xi (t)}\right) {\overline{H}}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) +\mathscr {E}^{\prime }(t). \end{aligned}$$

(99)

Let us denote by $G^{*}$ the conjugate function of the convex function G defined by $G^{*}(s)=Sup_{t\in \mathbb {R}^{+}}(st-G(t))$, then

$$\begin{aligned} st\le G^{*}(s)+G(t), \end{aligned}$$

(100)

and, thanks to the arguments given in [5, 9, 14, 29, 30]

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

This and the definition of H give

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

(101)

Taking $s:=\frac{C_{2}}{q}\overline{H}^{\prime }\left( \epsilon _{0} \frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right)$ and $t:=\overline{H} ^{-1}\left( \frac{q\mathscr {L}_{g}(t)}{\xi (t)}\right)$ in (100), then making use of (99), (100) and (101), we arrive at

$$\begin{aligned}{} & {} \frac{d}{dt}\mathscr {N}(t) \nonumber \\{} & {} \quad \le -\beta _{1}\mathscr {E}(t){\overline{H}}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) +\overline{ H}\left[ {\overline{H}}^{-1}\left( \frac{q \mathscr {L}_{g}(t)}{\xi (t)}\right) \right] +{\overline{H}}^{*}\left[ \frac{\beta _{2}}{q}{\overline{H}}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) \right] + \mathscr {E}^{\prime }(t) \nonumber \\{} & {} \quad \le -\beta _{1}\mathscr {E}(t){\overline{H}}^{\prime }\left( \epsilon _{0}\frac{ \mathscr {E}(t)}{\mathscr {E}(0)}\right) +\frac{q\mathscr {L}_{g}(t)}{\xi (t)}+ {\overline{H}}^{*}\left[ \frac{\beta _{2}}{q}{\overline{H}}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) \right] +\mathscr {E}^{\prime }(t) \nonumber \\{} & {} \quad \le -\beta _{1}\mathscr {E}(t){\overline{H}}^{\prime }\left( \epsilon _{0} \frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) +\frac{q\mathscr {L}_{g}(t)}{\xi (t)}+\frac{\beta _{2}\epsilon _{0}}{q}\frac{\mathscr {E}(t)}{\mathscr {E}(0)} {\overline{H}}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E} (0)}\right) +\mathscr {E}^{\prime }(t). \end{aligned}$$

(102)

Next, multiplying (102) by $\xi (t)$ and using the fact that $\epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}<r,$ $\overline{H} ^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) =H^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right)$, we get

$$\begin{aligned} \xi (t)\frac{d}{dt}\mathscr {N}(t)\le & {} -\beta _{1}\mathscr {E}(t)\xi (t) \overline{H}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E} (0)}\right) +q\mathscr {L}_{g}(t)\\{} & {} +\frac{\beta _{2}\epsilon _{0}}{q}\xi (t)\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\overline{H}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) +\xi (t)\mathscr {E}^{\prime }(t) \\\le & {} -\beta _{1}\mathscr {E}(t)\xi (t)\overline{H}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) +\frac{\beta _{2}\epsilon _{0}}{q}\xi (t)\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\overline{H}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) -c\mathscr {E}^{\prime }(t), \end{aligned}$$

where c is a positive constant [58].

Now, let us define the functional $\widetilde{\mathscr {N}}$

$$\begin{aligned} \widetilde{\mathscr {N}}(t)=\mathscr {N}(t)\xi (t)+\mathscr {E}(t). \end{aligned}$$

It is not difficult to see that there exist positive constants $\rho _{1}$ and $\rho _{2}$ for which we have

$$\begin{aligned} \rho _{1}\widetilde{\mathscr {N}}(t)\le \mathscr {E}(t)\le \rho _{2} \widetilde{\mathscr {N}}(t). \end{aligned}$$

(103)

Consequently, with an appropriate choice of $\epsilon _{0}$, then there exists a positive constant k such that

$$\begin{aligned} \frac{d}{dt}\widetilde{\mathscr {N}}(t)\le -k\xi (t)\overline{H}^{\prime }\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) =-k\xi (t)H_{2}\left( \epsilon _{0}\frac{\mathscr {E}(t)}{\mathscr {E}(0)}\right) \text { },\text { }\forall t\ge t_{0} \end{aligned}$$

(104)

where $H_{2}(s)=sH^{\prime }(\epsilon _{0}s).$

Since $H_{2}^{\prime }(s)=H^{\prime }(\epsilon _{0}s)+\epsilon _{0}sH^{\prime \prime }(\epsilon _{0}s)$, we use the strict convexity of H on [0, r), we observe that $H_{2}>0,$ $H_{2}^{\prime }>0$ on (0, r].

Defining now

$$\begin{aligned} \mathscr {R}(t)=\frac{\delta _{1}\widetilde{\mathscr {N}}(t)}{\mathscr {E}(0)}, \end{aligned}$$

thanks to (103) and (104) we have $\mathscr {E}\sim \mathscr {R}$ and for a positive constant $\widetilde{k}$

$$\begin{aligned} \frac{d}{dt}\mathscr {R}(t)\le -\widetilde{k}\xi (t)H_{2}(\mathscr {R}(t)) \text { },\text { }\forall t\ge t_{0}. \end{aligned}$$

Then, integrating over $(t_{0},t)$ yields

$$\begin{aligned} \int _{t_{0}}^{t}\frac{\mathscr {R}^{\prime }(s)}{H_{2}(\mathscr {R}(s))}\le -\int _{t_{0}}^{t}\widetilde{k}\xi (s)ds, \end{aligned}$$

and this leads to

$$\begin{aligned} \int _{\epsilon _{0}\mathscr {R}(t)}^{\epsilon _{0}\mathscr {R}(t_{0})}\frac{\mathscr {R}^{\prime }(s)}{H^{\prime }(\mathscr {R}(s))}\ge & {} \widetilde{k}\int _{t_{0}}^{t}\xi (s)ds, \end{aligned}$$

which gives us

$$\begin{aligned}&\mathscr {R}(t)\le \frac{1}{\epsilon _{0}}H_{1}^{-1}\left( \widetilde{k}\int _{t_{0}}^{t}\xi (s)ds\right) , \end{aligned}$$

where $H_{1}(t)=\int _{t}^{r}\frac{ds}{sH^{\prime }(s)}$.

This completes the proof.