Optimal decay result for Kirchhoff plate equations with nonlinear damping and very general type of relaxation functions

Abstract

In this paper, we consider plate equations with viscoelastic damping localized on a part of the boundary and nonlinear damping in the domain. We establish general and optimal decay rate results for a wider class of relaxation functions. These results are obtained without imposing any restrictive growth assumption on the frictional damping term. Our results are more general than the earlier results.

1 Introduction

In this paper, we consider the following Kirchhoff plate equations:

$$\begin{aligned}& u_{tt}+\Delta ^{2}u+\eta (t)h(u_{t})=0, \quad \text{in } \varOmega \times (0,\infty ), \end{aligned}$$

(1)

$$\begin{aligned}& u=\frac{\partial u}{\partial \nu }=0,\quad \text{on } \varGamma _{0}\times (0,\infty ), \end{aligned}$$

(2)

$$\begin{aligned}& -u+ \int _{0}^{t}k_{1}(t-s)\varPhi _{2}u(s)\,ds=0,\quad \text{on } \varGamma _{1}\times (0,\infty ), \end{aligned}$$

(3)

$$\begin{aligned}& \frac{\partial u}{\partial \nu }+ \int _{0}^{t}k_{2}(t-s)\varPhi _{1}u(s)\,ds=0,\quad \text{on } \varGamma _{1}\times (0,\infty ), \end{aligned}$$

(4)

$$\begin{aligned}& u(x,0)=u_{0}(x,t),\qquad u_{t}(x,0)=u_{1}(x), \quad \text{in } \varOmega . \end{aligned}$$

(5)

In system (1)–(5), $u = u(x, t)$ is the transversal displacement of a thin vibrating plate subjected to boundary viscoelastic damping and an internal time-dependent fractional damping. The integral terms in (3) and (4) describe the memory effects. The causes of these memory effects are, for example, the interaction with another viscoelastic element. In the above system, $\eta \in C^{1} (0, \infty )$ is a positive nonincreasing function called the time-dependent coefficient of the frictional damping and $u_{0}$ and $u_{1}$ are the initial data. The functions $k_{1}, k _{2} \in C^{1} (0, \infty )$ are positive and nonincreasing, called relaxation functions, and h is a function that satisfies some conditions. Denoting by $\varPhi _{1}$, $\varPhi _{2}$ the differential operators

$$ \varPhi _{1}u=\Delta u+(1-\rho )D_{1}u,\qquad \varPhi _{2}u=\frac{\partial \Delta u}{\partial \nu }+(1- \rho )\frac{\partial D_{2}u}{\partial \delta }, $$

where

$$ D_{1}u=2\nu _{1}\nu _{2} u_{xy}-\nu _{1}^{2}u_{yy}-\nu _{2}^{2}u_{xx}, \qquad D_{2}u= \bigl(\nu _{1}^{2}-\nu _{2}^{2} \bigr)u_{xy}+\nu _{1}\nu _{2} (u_{yy}-u_{xx} ), $$

and $\rho \in (0,\frac{1}{2})$ represents the Poisson coefficient. The vector $\nu =(\nu _{1}, \nu _{2})$ denotes the unit outward normal and $\delta =(-\nu _{2}, \nu _{1})$ denotes the external unit normal to the boundary of the domain. The stability of the Kirchhoff plate equations in which the boundary (internal) feedback is linear or nonlinear has been studied by several authors, such as Lagnese [1], Komornik [2], Lasiecka [3], Cavalcanti et al. [4], Ammari and Tucsnak [5], Komornik [6], Guzman and Tucsnak [7], Vasconcellos and Teixeira [8] and Pazoto et al. [9]. For the existence, multiplicity and asymptotic behavior of nonnegative solutions for a fractional Schrödinger–Poisson–Kirchhoff type system, we refer to Xiang and Wang [10]. There exist a large number of papers which discuss the plate equations when the memory effects are in the domain or at the boundary. Here, we refer to Lagnese [11] and Rivera et al. [12] for the internal viscoelastic damping. They proved that the energy decays exponentially (polynomially) if the relaxation function k decays exponentially (polynomially). Alabau-Boussouira et al. [13] obtained the same results but for an abstract problem. Regarding the internal damping, if the viscoelastic term does not exist and $\eta \equiv 1$, the problem (1) was studied and analyzed in the literature such as by Enrike [14] who established an exponential decay for the wave equation with linear damping term. This result was extended by Komornik [15] and Nakao [16] who used different methods and treated the problem when the damping term is nonlinear. For the boundary damping, Santos and Junior [17] showed that the energy decays exponentially if the resolvent kernels r decays exponentially and polynomially if r decays polynomially. In the presence of $\eta (t)$, Benaissa et al. [18] established energy decay results which depend on h and $\eta (t)$. In all the above work, the rates of decay in the relaxation function were either of exponential or of polynomial type. In 2008, Messaoudi in [19] and [20] gave general decay rates for an extended class of relaxation functions for which the exponential (polynomial) decay rates are just special cases. However, the optimal decay rates in the polynomial decay case were not obtained. Specifically, he considered a relaxation function k that satisfies

$$ k^{\prime }(t)\le -\xi (t) k^{p}(t), \quad t\ge 0, $$

(6)

where $p=1$ and ξ is a positive nonincreasing differentiable function. Furthermore, he showed that the decay rates of the energy are the same rates of decay of the kernel k. However, the decay rate is not necessarily of exponential or polynomial decay type. After that, different papers appeared and used the condition (6) where $p=1$; see, for instance, [21,22,23,24,25,26,27,28,29,30]. Lasiecka and Tataru [31] took one step forward and considered the following condition:

$$ k^{\prime }(t)\le -G\bigl(k(t)\bigr), $$

(7)

where G is a positive, strictly increasing and strictly convex function on $(0,R_{0}]$, and G satisfies $G(0)=G^{\prime }(0)=0$. Using the above condition and imposing additional constraints conditions on G, several authors in different approaches obtained general decay results in terms of G; see for example [32,33,34,35,36], and [37]. Later, the condition (6) was extended by Messaoudi and Al-Khulaifi [38] to the case $1\le p <\frac{3}{2}$ only and they obtained general and optimal decay results. In [34], Lasiecka et al. established optimal decay rate for all $1 \le p < 2$, but with $\gamma (t)=1$. Very recently, Mustafa [39] obtained optimal exponential and polynomial decay rates for all $1 \le p < 2$ and γ is a function of t. The work most closely related to our study is by Kang [40], Mustafa and Abusharkh [41] and Mustafa [42]. Kang [40] investigated the system (1)–(5) whereas $\eta (t) \equiv 1$ and

$$ G_{i} \bigl(-r_{i}'(t) \bigr)=-r_{i}'(t),\quad \forall i=1,2, $$

(8)

and established general decay results. Mustafa and Abusharkh [41] considered the system (1)–(5). But with the condition

$$ r_{i}''(t) \geq G\bigl(- r_{i}^{\prime }(t)\bigr),\quad \forall i=1,2, $$

(9)

and $h(t) \equiv 0$. They established explicit and general decay rate results. Very recently, Mustafa [42] studied system (1)–(5). However, under the same condition (9) he obtained a general decay rate result. Our contribution in this paper is to investigate the system (1)–(5) under a very general assumption on the resolvent kernels $r_{i}$. This assumption is more general as it comprises the earlier results in [40, 41] and [42] in the presence of $\xi (t)$ and the very general assumption on the relaxation functions. Furthermore, we obtain our results without imposing any restrictive growth assumption on the damping and take into account the effect of the time-dependent coefficient $\eta (t)$. The rest of the paper is as follows: In Sect. 2, we give a literature review and in Sect. 3, we state our main results and provide some examples. In Sect. 4, some technical lemmas are presented and established. Finally, we prove and discuss our decay results.

2 Preliminaries

In this section, some important materials in the proofs of our results will be presented. In this paper, $L^{2}(\varOmega )$ stands for the standard Lebesgue space and $H_{0}^{1}(\varOmega )$ the Sobolev space. We use those spaces with their usual scalar products and norms. Moreover, we denote by W the following space: $W=\{w\in H^{2}(\varOmega ): w=\frac{\partial w}{\partial \nu }=0 \text{ on } \varGamma _{0}\}$, and $r_{i}$ is the resolvent kernel of $\frac{-k_{i}^{\prime }}{k_{i}(0)}$, which satisfies

$$ r_{i}(t)+\frac{1}{k_{i}(0)}\bigl(k_{i}^{\prime }*r_{i} \bigr) (t)=- \frac{1}{k_{i}(0)}k^{\prime }_{i}(t),\quad \forall i=1,2, $$

where ∗ represents the convolution product

$$ (f*g) (t)= \int _{0}^{t}f(t-s)g(s)\,ds. $$

From (3) and (4), we get the following Volterra equations:

$$\begin{aligned}& \varPhi _{2}u+\frac{1}{k_{1}(0)}k_{1}^{\prime }*\varPhi _{2}u= \frac{1}{k_{1}(0)}u_{t}, \\& \varPhi _{1}u+\frac{1}{k_{2}(0)}k^{\prime }_{2}*\varPhi _{1}u=- \frac{1}{k_{2}(0)}\frac{\partial u_{t}}{\partial \nu }. \end{aligned}$$

Taking $\tau _{i}=\frac{1}{k_{i}(0)}$, for $i=1,2$, and using the Volterra’s inverse operator, we get

$$\begin{aligned}& \varPhi _{2}u=\tau _{1}\{u_{t}+r_{1}*u_{t} \},\quad \text{on } \varGamma _{1}\times (0,\infty ), \\& \varPhi _{1}u=-\tau _{2}\biggl\{ \frac{\partial u_{t}}{\partial \nu }+r_{2}* \frac{ \partial u_{t}}{\partial \nu }\biggr\} ,\quad \text{on } \varGamma _{1}\times (0,\infty ), \end{aligned}$$

In our paper, we assume that $u_{0} \equiv 0$, so we have

$$\begin{aligned}& \varPhi _{2}u=\tau _{1}\bigl\{ u_{t}+r_{1}(0) u+r_{1}'*u\bigr\} , \quad \text{on } \varGamma _{1}\times (0,\infty ), \end{aligned}$$

(10)

$$\begin{aligned}& \varPhi _{1}u=-\tau _{2}\biggl\{ \frac{\partial u_{t}}{\partial \nu }+r_{2}(0)\frac{ \partial u}{\partial \nu }+r_{2}'* \frac{\partial u}{\partial \nu }\biggr\} ,\quad \text{on } \varGamma _{1}\times (0,\infty ). \end{aligned}$$

(11)

Throughout the paper, c is a generic positive constant and we use (10) and (11) instead of (3) and (4).

2.1 Assumptions

$(A{1})$ :

We assume that $\varOmega \subset \mathbb{R}^{2}$ is a bounded domain with a smooth boundary $\varGamma =\varGamma _{0}\cup \varGamma _{1}$. Here, the partitions $\varGamma _{0}$ and $\varGamma _{1}$ are closed and disjoint. We also assume that $\operatorname{meas}(\varGamma _{0})> 0$, and there exists a fixed point $x_{0}\in \mathbb{R}^{2}$ such that $m\cdot \nu \le 0$ on $\varGamma _{0}$ and $m \cdot \nu > 0$ on $\varGamma _{1}$ where $m(x):=x-x _{0}$ and ν is the unit outward normal vector. This assumption leads to positive constants $\delta _{0}$ and R such that

$$ m \cdot \nu \ge \delta _{0} > 0\quad \text{on } \varGamma _{1} \quad \text{and}\quad \bigl\vert m(x) \cdot \nu \bigr\vert \le R, \quad \forall x\in \varOmega . $$

$(A{2})$ :

We assume that $h:\mathbb{R} \rightarrow \mathbb{R}$ is a $C^{0}$ nondecreasing function and there exists a strictly increasing function $h_{0}\in C^{1}(\mathbb{R}^{+})$ with $h_{0}(0)=0$ such that

$$ \begin{aligned} &h_{0}\bigl( \vert s \vert \bigr)\le \bigl\vert h(s) \bigr\vert \le h_{0}^{-1} \bigl( \vert s \vert \bigr)\quad \text{for all } \vert s \vert \le \epsilon , \\ & c_{1} \vert s \vert \le \bigl\vert h(s) \bigr\vert \le c_{2} \vert s \vert \quad \text{for all } \vert s \vert \ge \epsilon , \end{aligned} $$

(12)

where $c_{1}$, $c_{2}$, ϵ are positive constants. In the case $h_{0}$ is nonlinear, we assume that the function H defined by $H(s)=\sqrt{s}h_{0} (\sqrt{s})$ is a strictly convex $C^{2}$ on $(0,r_{0}]$, where $r_{0} > 0$.

$(A{3})$ :

We assume that $r_{i} : \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$, for $i=1,2$, is a $C^{2}$ function satisfies

$$ \lim_{t \rightarrow \infty } r_{i}(t)=0, \qquad r_{i}(0) > 0 , \qquad r_{i}'(t) \leq 0, $$

(13)

and there exists a positive, differentiable and nonincreasing function $\xi _{i} : \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$. We also assume that there exists a positive function $G_{i} \in C^{1}(\mathbb{R}^{+})$, $G_{i}$ being a linear or strictly increasing and strictly convex $C^{2}$ function on $(0,R_{i}]$, $R_{i} > 0$, with $G_{i}(0) = G_{i}'(0) = 0$, such that

$$ r_{i}''(t) \geq \xi _{i}(t) G_{i} \bigl(-r_{i}'(t) \bigr), \quad (i = 1, 2) \ \forall t > 0. $$

(14)

Furthermore, we assume that the system ((1)–(5) has a unique solution

$$ u\in L^{\infty }\bigl(\mathbb{R}^{+};H^{4}(\varOmega ) \cap W\bigr)\cap W^{1,\infty }\bigl(\mathbb{R}^{+};W\bigr)\cap W^{2,\infty }\bigl(\mathbb{R}^{+};L^{2}(\varOmega )\bigr). $$

This result can be obtained by using the Galerkin method as in Park and Kang [43] and Santos et al. [17].

Remark 2.1

It is worth noting that condition (12) was considered first in [31].

Remark 2.2

Using Assumption $(A2)$, one may notice that $\operatorname{sh}(s)>0$, for all $s\ne 0$.

Remark 2.3

If G is a strictly increasing and strictly convex $C^{2}$ function on $(0, r_{1}]$, with $G(0) = G'(0) = 0$, then it has an extension G̅, which is a strictly increasing and strictly convex $C^{2}$ function on $(0,\infty )$. For instance, if $G(r_{1}) = a$, $G'(r _{1}) = b$, $G''(r_{1}) = c$, we can define G̅, for $t > r_{1}$, by

$$ \overline{G}(t)=\frac{c}{2} t^{2}+ (b-cr_{1})t+ \biggl(a+\frac{c}{2} {r_{1}}^{2}-b r_{1} \biggr). $$

(15)

The same remark can be established for H̅.

Now, we define the bilinear form $a(\cdot , \cdot )$ as follows:

$$ a(u,v)= \int _{\varOmega }\bigl\{ u_{xx}v_{xx}+u_{yy}v_{yy}+ \rho (u_{xx}v_{yy}+u _{yy}v_{xx})+2(1-\rho )u_{xy}v_{xy}\bigr\} \,dx\,dy. $$

(16)

It is well known that $\sqrt{a(u , u)}$ is an equivalent norm on W, that is,

$$ \beta _{1} \Vert u \Vert ^{2}_{H^{2}(\varOmega )} \leq a(u,u) \leq \beta _{2} \Vert u \Vert ^{2}_{H^{2}(\varOmega )}, $$

(17)

for some positive constants $\beta _{1}$ and $\beta _{2}$. From (17) and the Sobolev embedding theorem, we have, for some positive constants $c_{p}$ and $c_{s}$,

$$ \Vert u \Vert ^{2} \leq c_{p} a(u,u), \quad \text{and} \quad \Vert \nabla u \Vert ^{2} \leq c_{s} a(u,u), \quad \forall u \in H^{2}(\varOmega ). $$

(18)

The energy functional associated with (1)–(5) is

$$ \begin{aligned}[b]E(t)&:=\frac{1}{2} \biggl[ \int _{\varOmega }{ \vert u_{t} \vert }^{2}+a(u,u)+ \tau _{1} \int _{\varGamma _{1}}\bigl(r_{1}(t){ \vert u \vert }^{2}-\bigl(r_{1}^{\prime }\circ u\bigr)\bigr)\,d\varGamma \biggr] \\ &\quad {}+\frac{1}{2} \biggl[\tau _{2} \int _{\varGamma _{1}} \biggl(r _{2}(t) \biggl\vert \frac{\partial u}{\partial \nu } \biggr\vert - \biggl(r_{2} ^{\prime } \circ \frac{\partial u}{\partial \nu } \biggr) \biggr) \,d \varGamma \biggr], \end{aligned} $$

(19)

where $(f \circ g)(t)=\int _{0}^{t} f(t-s) |g(t)-g(s)|^{2} \,ds$.

Our main stability results are in the following two theorems.

3 The main results

Theorem 3.1

Assume that $(A1)$–$(A3)$ are satisfied and $h_{0}$ is linear. Then the solution of (1)–(5) satisfies, for all $t \geq t_{1}$,

$$\begin{aligned}& E (t)\le c_{1} e^{-c_{2}\int _{t_{1}}^{t}\sigma (s)\,ds},\quad \textit{if }G\textit{ is linear}, \end{aligned}$$

(20)

$$\begin{aligned}& E (t)\le m_{2} G_{4}^{-1} \biggl(m_{1} \int _{t_{1}}^{t}\sigma (s)\,ds \biggr),\quad \textit{if }G\textit{ is nonlinear}, \end{aligned}$$

(21)

where $c_{1}$, $c_{2}$, $m_{1}$ and $m_{2}$ are strictly positive constants. $G_{4}(t)=\int _{t}^{r}\frac{1}{sG^{\prime }(s)}\,ds$, $G= \min \{G_{1}, G_{2} \}$, and $\sigma (t)=\min \{\eta (t),\xi (t)\}$ where $\xi (t)=\min \{\xi _{1}(t), \xi _{2}(t) \}$. $G_{1}$, $G_{2}$ and $\xi _{1}(t)$, $\xi _{2}(t)$ are defined in $(A3)$.

Theorem 3.2

Assume that $(A1)$–$(A3)$ are satisfied and $h_{0}$ is nonlinear. Then there exist strictly positive constants $c_{3}$, $c_{4}$, $m_{3}$, $m_{4}$, $\varepsilon _{1}$ and $\varepsilon _{2}$ such that the solution of (1)–(5) satisfies, for all $t \geq t_{1}$,

$$ E(t)\le H_{1}^{-1} \biggl(c_{3} \int _{t_{1}}^{t}\sigma (s)\,ds+c_{4} \biggr),\quad \textit{if }G\textit{ is linear}, $$

(22)

where $H_{1}(t)=\int _{t}^{1}\frac{1}{H_{2}(s)}\,ds$ and $H_{2}(t)=t H ^{\prime }(\varepsilon _{1}t)$.

$$ E(t) \leq m_{4} (t-t_{1}) {W_{2}}^{-1} \biggl( \frac{m_{3}}{(t-t_{1}) \int _{t_{1}}^{t}\sigma (s) \,ds } \biggr),\quad \textit{if }G \textit{ is nonlinear}, $$

(23)

where $W_{2}(t)=tW'(\varepsilon _{2} t)$, $W= (\overline{G}^{-1}+ \overline{H}^{-1} )^{-1}$ and G̅, H̅ are introduced in Remark 2.3 .

Remark 3.1

([44])

In (21), one can see that the decay rate of $E(t)$ is consistent with the decay rate of $(-r_{i}'(t))$ given by (14). So, the decay rate of $E(t)$ is optimal.

In fact, using the general assumption (14), and taking into account the fact that $G=\min \{G_{1}, G_{2}\}$ and $\sigma (t)= \min \{\eta (t), \xi (t)\}$, we have

$$ -r_{i}'(t)\le G_{5}^{-1} \biggl( \int _{-r_{i}^{\prime -1}(r)}^{t}\sigma (s)\,ds \biggr),\quad \forall t \ge -r_{i}^{\prime -1}(r), $$

where $G_{5}(t)=\int _{t}^{r}\frac{1}{G(s)} \,ds$. Using the properties of G, we get

$$ G_{4}(t)= \int _{t}^{r}\frac{1}{s G^{\prime }(s)}\,ds\le \int _{t}^{r} \frac{1}{G(s)}\,ds=G_{5}(t). $$

Also, using the properties of $G_{4}$ and $G_{5}$, we have

$$ G_{4}^{-1}(t)\le G_{5}^{-1}(t). $$

This shows that (21) provides the best decay rates expected under the very general assumption (14).

Example 3.3

• (1.A) $h_{0}$ and G are linear and $\eta (t) \equiv 1$.

  Let $r_{i}'(t)= - a_{i} e^{-b_{i}(1+t)}$, where $b_{i} > 0$ and $a_{i} > 0$, $\forall i=1,2$, so that Assumption $(A3)$ is satisfied, then $r_{i}^{\prime \prime }(t)=\xi _{i}(t) G_{i}(-r_{i}'(t))$. We take $a=\min \{ a_{1}, a_{2}\}$, $b=\min \{ b_{1}, b_{2}\}$, $G=\min \{ G _{1}, G_{2}\}$, $\xi (t)=\min \{ \xi _{1}(t), \xi _{2}(t)\}$ and $\sigma (t) = \min \{ \eta (t), \xi (t)\}$. Hence, $G(t)=t$, $\xi (t)=b$ and we let $\sigma (t):=b_{0}=\min \{1,b\}$. For the nonlinear case, assume that $h_{0}(t)=ct$ and $H(t)=\sqrt{t} h_{0}( \sqrt{t})=ct$. Therefore, we can use (20) to deduce

  $$ E(t) \leq c_{1} e^{-c_{2}t} , $$

  (24)

  which is the exponential decay.

• (1.B) $h_{0}$ and G are linear and $\eta (t)= \frac{b}{t+1}$.

  Let $r_{i}'(t)= - a_{i} e^{-b_{i}(1+t)}$, where $b_{i} > 0$ and $a_{i} > 0$, $\forall i=1,2$, so that Assumption $(A3)$ is satisfied, then $r_{i}^{\prime \prime }(t)=\xi _{i}(t) G_{i}(-r_{i}'(t))$. We take $a=\min \{ a_{1}, a_{2}\}$, $b=\min \{ b_{1}, b_{2}\}$, $G=\min \{ G _{1}, G_{2}\}$, $\xi (t)=\min \{ \xi _{1}(t), \xi _{2}(t)\}$ and $\sigma (t) = \min \{ \eta (t), \xi (t)\}$. Hence, $G(t)=t$, $\xi (t)=b$ and $\sigma (t)=\frac{b}{t+1}$. For the nonlinear case, assume that $h_{0}(t)=ct$ and $H(t)=\sqrt{t} h_{0}(\sqrt{t})=ct$. Therefore, we can use (20) to deduce

  $$ E(t) \leq \frac{c }{1+\ln (t+1)} . $$

  (25)

• (2) $h_{0}$ is linear, G is nonlinear and $\eta (t) \equiv 1$.

  Let $r_{i}'(t)= - a_{i} e^{-t^{q}}$, where $0< q<1$ and $a_{i}>0$, $\forall i=1,2$, so that Assumption $(A3)$ is satisfied, then $r_{i}''(t)= \xi _{i}(t)G_{i} ( -r_{i}'(t) )$. We take $a=\min \{ a_{1}, a_{2}\}$, $G=\min \{ G_{1}, G_{2}\}$, $\xi (t)= \min \{ \xi _{1}(t), \xi _{2}(t)\}$ and $\sigma (t) = \min \{ \eta (t), \xi (t)\}$. Hence, $\xi (t)=1$ and $G(t)=\frac{q^{t}}{ ( \ln (a/t) ) ^{\frac{1}{q}-1}}$. In this case, $\sigma (t) \equiv 1$. For, the boundary feedback, let $h_{0}(t)=ct$, and $H(t)=\sqrt{t} h_{0}( \sqrt{t})=ct$. Since

  $$ G'(t)=\frac{(1-q)+q \ln (a/t)}{ ( \ln (a/t) )^{1/q}} $$

  and

  $$ G''(t)=\frac{(1-q) ( \ln (a/t)+1/q )}{ ( \ln (a/t) )^{\frac{1}{q}+1}}, $$

  the function G satisfies the condition $(A3)$ on $(0,r]$ for any $r>0$. We have

  $$ \begin{aligned}[b]G_{4}(t)&= \int _{t}^{r}\frac{1}{sG^{\prime }(s)}\,ds= \int _{t}^{r}\frac{ [\ln {\frac{a}{s}} ]^{\frac{1}{q}}}{s [1-q+q\ln {\frac{a}{s}} ]}\,ds \\ & = \int _{\ln {\frac{a}{r}}}^{\ln {\frac{a}{t}}}\frac{u ^{\frac{1}{q}}}{1-q+qu}\,du \\ & =\frac{1}{q} \int _{\ln {\frac{a}{r}}}^{\ln {\frac{a}{t}}} u^{\frac{1}{q}-1} \biggl[ \frac{u}{\frac{1-q}{q}+u} \biggr]\,du \\ & \le \frac{1}{q} \int _{\ln {\frac{a}{r}}}^{\ln {\frac{a}{t}}} u^{\frac{1}{q}-1}\,du \le \biggl(\ln { \frac{a}{t}} \biggr)^{\frac{1}{q}}. \end{aligned} $$

  Then (21) gives

  $$ E(t)\leq k e^{-k t^{q}}, $$

  (26)

  which is the optimal decay.

• (3) $h_{0}$ is nonlinear, G is linear and $\eta (t)=\frac{b}{(t+e) \ln (t+e)}$.

  Let $r_{i}'(t)= - a_{i} e^{-b_{i}(1+t)}$, where $b_{i} > 0$ and $a_{i} > 0$, $\forall i=1,2$, so that Assumption $(A3)$ is satisfied, then $r_{i}^{\prime \prime }(t)=\xi _{i}(t) G_{i}(-r_{i}'(t))$. We take $a=\min \{ a_{1}, a_{2}\}$, $b=\min \{ b_{1}, b_{2}\}$, $G=\min \{ G _{1}, G_{2}\}$, $\xi (t)=\min \{ \xi _{1}(t), \xi _{2}(t)\}$ and $\sigma (t) = \min \{ \eta (t), \xi (t)\}$. Hence, $G(t)=t$ and $\xi (t)= b$. In this case, $\sigma (t)=\frac{b}{(t+e)\ln (t+e)}$. Also, assume that $h_{0}(t)=ct^{q}$, where $q>1$ and $H(t)=\sqrt{t} h_{0}( \sqrt{t})=ct^{\frac{q+1}{2}}$. Then

  $$ {H_{1}}^{-1}(t)= (ct+1 )^{\frac{-2}{q-1}}. $$

  Therefore, applying (22), we obtain

  $$ E(t)\leq \frac{c}{ [1+\ln ( \ln (t+e)) ]^{\frac{2}{q-1}}}. $$

  (27)

• (4) $h_{0}$ is nonlinear, G is non-linear and $\eta (t) \equiv 1$.

  Let $r_{i}'(t)=\frac{- a_{i}}{(1+t)^{2}}$, where $a_{i} > 0$, $\forall i=1,2$, is chosen so that Assumption $(A3)$ holds. We choose $a=\min \{ a_{1}, a_{2}\}$, then $r_{i}^{\prime \prime }(t)= b_{i} G _{i}(-r_{i}'(t))$. We select $b=\min \{ b_{1}, b_{2}\}$, $G=\min \{ G _{1}, G_{2}\}$, $\xi (t)=\min \{ \xi _{1}(t), \xi _{2}(t)\}$ and $\sigma (t) = \min \{ \eta (t), \xi (t)\}$. In this example, $G(s)=s^{\frac{3}{2}}$, $\xi (t)=b$. For the boundary feedback, let $h_{0}(t)=c t^{5}$ and $H(t)=c t^{3}$. Then

  $$ W(s)=\bigl(G^{-1}+H^{-1}\bigr)^{-1}= \biggl( \frac{-1+\sqrt{1+4s}}{2} \biggr) ^{3} $$

  and

  $$ \begin{aligned}[b]W_{2}(s)&=\frac{3s}{\sqrt{1+4s}} \biggl( \frac{-1+\sqrt{1+4s}}{2} \biggr) ^{2} \\ & =\frac{3s}{2\sqrt{1+4s}}+\frac{3s^{2}}{\sqrt{1+4s}}- \frac{3s}{2} \\ & \le \frac{3s}{2}+\frac{3 s^{2}}{2\sqrt{s}}- \frac{3s}{2}=c s^{\frac{3}{2}}. \end{aligned} $$

  Therefore, applying (23), we obtain

  $$ E(t)\le \frac{c}{(t-t_{1})^{\frac{1}{3}}}. $$

For the proofs of our main results, we state and establish several lemmas in the following section.

4 Technical lemmas

In this section, we introduce some lemmas which are important in our proofs of our main results.

Lemma 4.1

([1]) Let u and v be functions in $H^{4}( \varOmega )$ and $\rho \in \mathbb{R}$. Then we have

$$ \int _{\varOmega }\bigl(\Delta ^{2}u\bigr)v\,dx=a(u,v)+ \int _{\varGamma _{1}}\biggl\{ (\varPhi _{2}u)v-( \varPhi _{1}u)\frac{\partial u}{\partial \nu }\biggr\} \,d\varGamma $$

(28)

and

$$ \begin{aligned}[b] \int _{\varOmega }(m.\nabla v)\Delta ^{2}v \,dx&=a(v,v)+ \frac{1}{2} \int _{ \varGamma }m.v \bigl[v_{xx}^{2}+v_{yy}^{2}+2 \rho v_{xx}v_{yy}+2(1-\rho )v _{xy}^{2} \bigr]\,d\varGamma \\ &\quad {}+ \int _{\varGamma } \biggl[(\varPhi _{2}v)m.\nabla v-(\varPhi _{1}v)\frac{ \partial }{\partial \nu }(m.\nabla \nu ) \biggr]\,d\varGamma . \end{aligned} $$

(29)

Lemma 4.2

Under Assumptions $(A1)$–$(A3)$ and considering Remark 2.2, the energy functional E satisfies, along the solution of (1)–(5), the estimate

$$ \begin{aligned}[b]E^{\prime }(t)&=- \frac{\tau _{1}}{2} \int _{\varGamma _{1}}\bigl(2{ \vert u_{t} \vert }^{2}-r_{1}^{\prime }(t){ \vert u \vert }^{2}+r_{1}^{\prime \prime } \circ u\bigr)\,d\varGamma \\ &\quad {}-\frac{\tau _{2}}{2} \int _{\varGamma _{1}} \biggl(2{ \biggl\vert \frac{\partial u_{t}}{\partial \nu } \biggr\vert }^{2}-r_{2}^{ \prime }(t){ \biggl\vert \frac{\partial u}{\partial \nu } \biggr\vert }^{2}+r_{2}^{\prime \prime } \circ \frac{\partial u}{\partial \nu } \biggr)\,d \varGamma -\eta (t) \int _{\varOmega } u_{t} h( u_{t} )\,dx \\ & \le 0. \end{aligned} $$

(30)

Proof

The proof can be established by multiplying Eq. (1) by $u_{t}$, integrating by parts over Ω, and using (28) and the boundary conditions (10) and (11). With the help of the ideas in [44], one can establish the following two helpful lemmas.

Lemma 4.3

For $i=1,2$, $0 < \alpha _{i} < 1$, and for

$$ C_{\alpha _{i}}:= \int _{0}^{\infty }\frac{r_{i}^{\prime 2}(s)}{r_{i}''-\alpha _{i} r_{i}'(s)}\,ds\quad \textit{and} \quad \theta _{i}(t):=r_{i}''(t)- \alpha _{i} r_{i}'(t), $$

(31)

we have

$$\begin{aligned}& \begin{aligned}[b] & \biggl( \int _{0}^{t} r_{1}'(t-s) \bigl\vert u(s)- u(t) \bigr\vert \,ds \biggr)^{2} \le C_{\alpha _{1}}( \theta _{1} \circ u) (t), \end{aligned} \end{aligned}$$

(32)

$$\begin{aligned}& \begin{aligned}[b] & \biggl( \int _{0}^{t} r_{2}'(t-s) \biggl\vert \frac{\partial u(s)}{ \partial \nu }- \frac{\partial u(t)}{\partial \nu } \biggr\vert \,ds \biggr) ^{2}\le C_{\alpha _{2}}\biggl( \theta _{2} \circ \frac{\partial u}{\partial \nu }\biggr) (t). \end{aligned} \end{aligned}$$

(33)

Lemma 4.4

There exist positive constants $d_{1}$, $d_{2}$ and $t_{1}$ such that

$$ r_{i}''(t) \geq -d_{i} r_{i}'(t), \quad (i = 1, 2)\ \forall t \in [0, t_{1}]. $$

(34)

Lemma 4.5

Under Assumptions $(A1)$–$(A3)$, the functional

$$ \psi _{1}(t):= \int _{\varOmega }(m.\nabla u)u_{t}\,dx $$

(35)

satisfies, along the solution of (1)–(5), the estimate

$$\begin{aligned} \psi _{1}^{\prime }(t) &\le \frac{1}{2} \int _{\varGamma _{1}} m.\nu { \vert u _{t} \vert }^{2}\,d\varGamma -\frac{1}{2} \int _{\varOmega }{ \vert u_{t} \vert } ^{2}\,dx- \biggl(1-\frac{c_{0}}{2}-\frac{\varepsilon c}{2} \biggr)a(u,u) \\ &\quad {}+\frac{ {\tau _{1}}^{2}}{2\varepsilon } \int _{\varGamma _{1}} \bigl[ \vert u_{t} \vert ^{2}+r_{1}^{2}(t) \vert u \vert ^{2} \bigr]\,d \varGamma +\frac{{\tau _{1}}^{2}C_{\alpha _{1}}}{2\varepsilon } \int _{\varGamma _{1}}(\theta _{1}\circ u)\,d\varGamma \\ &\quad {}+\frac{\tau _{2}^{2}}{2\varepsilon } \int _{\varGamma _{1}} \biggl[ \biggl\vert \frac{\partial u_{t}}{\partial \nu } \biggr\vert +r_{2}^{2}(t) \biggl\vert \frac{\partial u}{\partial \nu } \biggr\vert \biggr]\,d\varGamma +\frac{ {\tau _{2}}^{2}C_{\alpha _{2}}}{2\varepsilon } \int _{\varGamma _{1}} \biggl(\theta _{2}\circ \frac{\partial u}{\partial \nu } \biggr)\,d\varGamma + \frac{c}{2} \int _{\varOmega } h^{2}(u_{t})\,dx \\ &\quad {}- \biggl[\frac{1}{2}-\frac{\varepsilon c}{2} \biggr] \int _{\varGamma _{1}}m.\nu \bigl[u_{xx}^{2}+u_{yy}^{2}+2 \rho u_{xx}u_{yy}+2(1- \rho )u_{xy}^{2} \bigr]\,d\varGamma . \end{aligned}$$

(36)

Proof

By direct integrations, using (1), and using (29) with $v=u$, we obtain

$$ \begin{aligned}[b]\psi _{1}^{\prime }(t)&= \int _{\varOmega }(m \cdot \nabla u_{t})u_{t}\,dx+ \int _{\varOmega }(m \cdot \nabla u)u_{tt}\,dx \\ & =\frac{1}{2} \int _{\varGamma _{1}}m \cdot \nu { \vert u_{t} \vert }^{2}\,d\varGamma -\frac{1}{2} \int _{\varOmega }{ \vert u_{t} \vert }^{2}\,dx-a(u,u)- \eta (t) \int _{\varOmega } h(u_{t}) (m \cdot \nabla u) \,dx \\ &\quad {}- \int _{\varGamma } \biggl[(\varPhi _{2}u) (m \cdot \nabla u)-( \varPhi _{1}u)\frac{\partial }{\partial \nu }(m \cdot \nabla u) \biggr]\,d \varGamma \\ &\quad {}-\frac{1}{2} \int _{\varGamma }m.\nu \bigl[u^{2}_{xx}+u^{2} _{yy}+2\rho u_{xx}u_{yy}+2(1-\rho )u^{2}_{xy} \bigr]\,d\varGamma . \end{aligned} $$

(37)

Since $u_{xx}u_{yy}-(u_{xy})^{2}=0$ on $\varGamma _{0}$, we have

$$ \begin{aligned}[b] u_{xx}u_{yy}+2(1- \rho )u_{xy}^{2}=(\Delta u)^{2}\quad \text{on } \varGamma _{0} \end{aligned} . $$

(38)

Now, as $u=\frac{\partial u}{\partial \nu }=0$ on $\varGamma _{0}$, we have $D_{1}u=D_{2}u=0$ on $\varGamma _{0}$ and

$$ \begin{aligned}[b] \frac{\partial }{\partial \nu }(m.\nabla u)=(m. \nu )\Delta u. \end{aligned} $$

(39)

Combining (37), (38) and (39), (37) becomes

$$\begin{aligned} \psi _{1}^{\prime }(t)&= \frac{1}{2} \int _{\varGamma _{1}}m.\nu { \vert u_{t} \vert }^{2}\,d\varGamma -\frac{1}{2} \int _{\varOmega } \vert u_{t} \vert ^{2}\,dx-a(u,u)- \eta (t) \int _{\varOmega } (m \cdot \nabla u) h(u_{t}) \,dx \\ &\quad {}+\frac{1}{2} \int _{\varGamma _{0}}m.\nu (\Delta u)^{2}\,d \varGamma - \frac{1}{2} \int _{\varGamma _{1}}m.\nu \bigl[u_{xx}^{2}+u_{yy} ^{2}+2\rho u_{xx}u_{yy}+2(1-\rho )u_{xy}^{2} \bigr]\,d\varGamma \\ &\quad {}- \int _{\varGamma _{1}}(\varPhi _{2}u) (m.\nabla u)\,d\varGamma + \int _{\varGamma _{1}}(\varPhi _{1}u)\frac{\partial }{\partial \nu }(m.\nabla u)\,d \varGamma . \end{aligned}$$

(40)

Now, Young’s inequality leads to

$$\begin{aligned}& \begin{aligned}[b] \biggl\vert \int _{\varGamma _{1}}(\varPhi _{2} u) (m. \nabla u)\,d\varGamma \biggr\vert \le \frac{1}{2\varepsilon } \int _{\varGamma _{1}} \vert \varPhi _{2} u \vert ^{2}\,d\varGamma +\frac{\varepsilon }{2} \int _{\varGamma _{1}} \vert m. \nabla u \vert ^{2}\,d\varGamma , \end{aligned} \end{aligned}$$

(41)

$$\begin{aligned}& \begin{aligned}[b] \biggl\vert \int _{\varGamma _{1}}(\varPhi _{1} u)\frac{\partial }{\partial \nu }(m. \nabla u)\,d\varGamma \biggr\vert \le \frac{1}{2\varepsilon } \int _{\varGamma _{1}} \vert \varPhi _{1} u \vert ^{2}\,d\varGamma +\frac{\varepsilon }{2} \int _{\varGamma _{1}} \biggl\vert \frac{\partial }{\partial \nu }(m. \nabla u) \biggr\vert ^{2}\,d\varGamma , \end{aligned} \end{aligned}$$

(42)

where ε is a positive constant. Using (17) and (18), the fact $\mid m(x)\mid \leq R$, and the trace theory, we obtain

$$ \begin{aligned}[b] & \int _{\varGamma _{1}} \vert m.\nabla u \vert ^{2}\,d\varGamma + \int _{\varGamma _{1}} \biggl\vert \frac{\partial }{\partial \nu }(m.\nabla u) \biggr\vert ^{2}\,d\varGamma \\ &\quad\leq R^{2} c_{s} a(u,u)+R \int _{\varGamma _{1}}m.\nu \bigl[u_{xx}^{2}+u_{yy} ^{2}+2\rho u_{xx}u_{yy}+2(1-\rho )u_{xy}^{2} \bigr]\,d\varGamma . \end{aligned} $$

(43)

Furthermore, using (17) and (18) and the property of the function $\eta (t)$, we have

$$ \begin{aligned}[b] \biggl\vert \eta (t) \int _{\varOmega } h(u_{t}) m.\nabla u \,dx \biggr\vert \leq \frac{c}{2} \int _{\varOmega } h^{2}(u_{t}) \,dx + \frac{R^{2} c _{s}}{2} a(u,u). \end{aligned} $$

(44)

Combining (40)–(44), we have

$$ \begin{aligned}[b]\psi _{1}^{\prime }(t) &\le \frac{1}{2} \int _{\varGamma _{1}}m.\nu { \vert u _{t} \vert }^{2}\,d\varGamma -\frac{1}{2} \int _{\varOmega }{ \vert u_{t} \vert } ^{2}\,dx- \biggl(1- \frac{\lambda _{0}}{2}- \frac{\varepsilon \lambda _{0}}{2} \biggr)a(u,u) \\ &\quad {}+\frac{1}{2\varepsilon } \int _{\varGamma _{1}} \vert \varPhi _{1} u \vert ^{2}\,d\varGamma +\frac{1}{2\varepsilon } \int _{\varGamma _{1}} \vert \varPhi _{2} u \vert ^{2} \,d\varGamma + \frac{c}{2} \int _{\varOmega } h^{2}(u _{t})\,dx \\ &\quad {}- \biggl[\frac{1}{2}-\frac{\varepsilon R}{2} \biggr] \int _{\varGamma _{1}}m.\nu \bigl[u_{xx}^{2}+u_{yy}^{2}+2 \rho u_{xx}u_{yy}+2(1- \rho )u_{xy}^{2} \bigr]\,d\varGamma , \end{aligned} $$

(45)

where $\lambda _{0}= R^{2} c_{s}$. By direct computation and using (4.3), we arrive at

$$ \begin{aligned}[b]\bigl(r_{1}' \ast u\bigr) (t)&= \int _{0}^{t} r_{1}'(t-s)u(s)\,ds= \int _{0}^{t} r_{1}'(t-s) \bigl[u(s)-u(t)+u(t)\bigr]\,ds \\ & = \int _{0}^{t} r_{1}'(t-s) \bigl[u(s)-u(t)\bigr]\,ds+ \int _{0}^{t} r _{1}'(t-s)u(t)\,ds \\ & = - \int _{0}^{t} r_{1}'(t-s) \bigl[u(t)-u(s)\bigr]\,ds+ \int _{0}^{t} r _{1}'(t-s)u(t)\,ds \\ & =- \int _{0}^{t} r_{1}'(t-s) \bigl[u(t)-u(s)\bigr]\,ds+ r_{1}(t)u(t)-r _{1}(0)u(t) \\ & \leq \bigl[ C_{\alpha _{1}} (\theta _{1} \circ u) (t) \bigr] ^{\frac{1}{2}}+r_{1}(t)u(t)-r_{1}(0)u(t), \end{aligned} $$

(46)

similarly, we can show that

$$ \begin{aligned}[b] & \biggl(r_{2}' \ast \frac{\partial u}{\partial \nu } \biggr) (t)\leq \biggl[ C_{\alpha _{2}} \biggl(\theta _{2} \circ \frac{\partial u}{ \partial \nu } \biggr) (t) \biggr]^{\frac{1}{2}}+r_{2}(t) \frac{\partial u(t)}{\partial \nu }-r_{2}(0)\frac{\partial u(t)}{\partial \nu }, \end{aligned} $$

(47)

then from the boundary conditions (10), (11) and using (46) and (47), we have

$$ \begin{aligned}[b] &\varPhi _{2}u \leq \tau _{1} \bigl\{ u_{t} + r_{1}(t) u+ \bigl[C_{\alpha _{1}} (\theta _{1} \circ u) (t) \bigr]^{\frac{1}{2}} \bigr\} , \\ &\varPhi _{1}u \leq -\tau _{2} \biggl\{ \frac{\partial u_{t}}{\partial \nu }+r _{2}(t)\frac{\partial u}{\partial \nu }+ \biggl[ C_{\alpha _{2}} \biggl(\theta _{2} \circ \frac{\partial u_{t}}{\partial \nu }\biggr) (t) \biggr]^{ \frac{1}{2}} \biggr\} . \end{aligned} $$

(48)

Substituting the inequalities (48) in (45) and using the fact $m.\nu \le 0$ on $\varGamma _{0}$, (36) is achieved. □

Lemma 4.6

Under Assumptions $(A1)$–$(A3)$, the functionals

$$ \begin{aligned} &\psi _{2}(t)= \int _{\varGamma _{1}} \int _{0}^{t}\mu _{1}(t-s) \bigl\vert u(s) \bigr\vert ^{2}\,ds \,dx, \\ & \psi _{3}(t)= \int _{\varGamma _{1}} \int _{0}^{t}\mu _{2}(t-s) \biggl\vert \frac{\partial u(s)}{\partial \nu } \biggr\vert ^{2} \,ds \,dx, \end{aligned} $$

(49)

satisfy, along the solution of (1)–(5), the estimates

$$ \begin{aligned} &\psi _{2}^{\prime }(t) \le \frac{1}{2} \bigl(r_{1}' \circ u\bigr) (t)+ 3 r_{1}(0) \int _{\varGamma _{1}} \bigl\vert u(t) \bigr\vert ^{2}\,dx, \\ &\psi _{3}^{\prime }(t) \le \frac{1}{2} \biggl(r_{2}' \circ \frac{\partial u}{\partial \nu } \biggr) (t)+ 3r_{2}(0) \int _{\varGamma _{1}} \biggl\vert \frac{\partial u(t)}{ \partial \nu } \biggr\vert ^{2}\,dx, \end{aligned} $$

(50)

where $\mu _{i}(t)= \int _{t}^{+\infty }(-r_{i}'(s)) \,ds$, $i=1,2$.

Proof

Taking the derivative of the first equation in (49) and using the fact $\mu _{1}^{\prime }(t)=r_{1}'(t)$, we have

$$ \begin{aligned}[b]\psi _{2}^{\prime }(t)&=r_{1}(0) \int _{\varGamma _{1}} \bigl\vert u(t) \bigr\vert ^{2}\,dx+ \int _{\varGamma _{1}} \int _{0}^{t}r_{1}'(t-s) \bigl\vert u(s) \bigr\vert ^{2} \,dx \\ & = \int _{\varGamma _{1}} \int _{0}^{t}r_{1}'(t-s) \bigl\vert u(s)- u(t) \bigr\vert ^{2} \,ds \,dx \\ &\quad {}+2 \int _{\varGamma _{1}}u(t) \int _{0}^{t}r_{1}'(t-s) \bigl(u(s)-u(t)\bigr)\,ds\,dx+r _{1}(t) \int _{\varGamma _{1}} \bigl\vert u(t) \bigr\vert ^{2}\,dx. \end{aligned} $$

(51)

Using the fact $\lim_{t \rightarrow \infty } r_{1}(t) = 0$, and Young’s inequality we have the following:

$$ \begin{aligned}[b] &2 \int _{\varGamma _{1}} u(t) \int _{0}^{t}r_{1}'(t-s) \bigl(u(s)-u(t)\bigr)\,ds\,dx \\ &\quad \le 2\gamma \int _{\varGamma _{1}} \bigl\vert u(s) \bigr\vert ^{2} \,dx+ \frac{ \int _{0}^{t}r_{1}'(s)}{2\gamma } \int _{\varGamma _{1}} \int _{0}^{t}r_{1}'(t-s) \bigl\vert u(s)- u(t) \bigr\vert ^{2} \,ds \,dx \\ &\quad \le 2\gamma \int _{\varGamma _{1}} \bigl\vert u(s) \bigr\vert ^{2} \,dx+ \frac{ \int _{0}^{\infty }r_{1}'(s)}{2\gamma } \int _{\varGamma _{1}} \int _{0}^{t}r _{1}'(t-s) \bigl\vert u(s)- u(t) \bigr\vert ^{2} \,ds \,dx \\ &\quad \le 2 \gamma \int _{\varGamma _{1}} \bigl\vert u(t) \bigr\vert ^{2} \,dx- \frac{r _{1}(0)}{2\gamma } \int _{\varGamma _{1}} \int _{0}^{t}r_{1}'(t-s) \bigl\vert u(s)- u(t) \bigr\vert ^{2} \,ds\,dx \\ &\quad \le 2 r_{1}(0) \int _{\varGamma _{1}} \bigl\vert u(t) \bigr\vert ^{2} \,dx- \frac{1}{2} \int _{\varGamma _{1}} \int _{0}^{t}r_{1}'(t-s) \bigl\vert u(s)- u(t) \bigr\vert ^{2} \,ds\,dx. \end{aligned} $$

(52)

Combining (51) and (52) and using the fact that $\mu _{1}(t) \leq \mu _{1}(0)=r_{1}(0)$, the first estimate in (50) is established. Similarly, we can establish the second estimate in (50). □

Lemma 4.7

Under Assumptions $(A1)$–$(A3)$, the functional $L(t):=NE(t)+N_{1}\psi _{1}(t)+n_{0} E(t)$, where $N, N_{1}, n_{0} > 0$, satisfies along the solution of (1)–(5) the following estimate:

$$ \begin{aligned}[b]L^{\prime }(t)&\le -m E(t)- \frac{1}{4} \int _{t_{1}}^{t} r_{1}'(t-s) \int _{\varGamma _{1}} \bigl\vert u(t)-u(s) \bigr\vert ^{2} \,dx \,d\varGamma \\ &\quad {}-\frac{1}{4} \int _{t_{1}}^{t} r_{2}'(t-s) \int _{\varGamma _{1}} \biggl\vert \frac{\partial u(t)}{\partial \nu }-\frac{\partial u(s)}{ \partial \nu } \biggr\vert ^{2} \,d\varGamma +c \int _{\varOmega } h^{2}(u_{t})\,dx,\quad \forall t \ge t_{1}. \end{aligned} $$

(53)

Proof

Using $L'(t)=NE'(t)+N_{1} \psi _{1}'(t)+n_{0} E'(t)$, combining (30) and (36), using the properties of $r_{i}$ and $r_{i}'$ given in Assumption $(A3)$ and using $\mid m \cdot \nu \mid \leq R$, we obtain

$$ \begin{aligned}[b]L'(t) &\leq - \biggl( \tau _{1} N-\frac{R N_{1}}{2}-\frac{N_{1} \tau _{1} ^{2}}{2\varepsilon } \biggr) \int _{\varGamma _{1}}{ \vert u_{t} \vert }^{2}\,d \varGamma - \biggl(\tau _{2} N-\frac{N_{1} \tau _{1}^{2}}{2\varepsilon } \biggr) \int _{\varGamma _{1}}{ \biggl\vert \frac{\partial u_{t}}{\partial \nu } \biggr\vert }^{2}\,d\varGamma \\ &\quad {}-N_{1} \biggl(1-\frac{\lambda _{0}}{2}-\frac{\varepsilon \lambda _{0}}{2} \biggr) a(u,u)+\frac{N_{1} \tau _{1}^{2}}{2\varepsilon } \int _{\varGamma _{1}}r_{1}^{2}(t){ \vert u \vert }^{2}\,d\varGamma\\ &\quad {} +\frac{N_{1} \tau _{2}^{2}}{2\varepsilon } \int _{\varGamma _{1}}r_{2}^{2}(t){ \biggl\vert \frac{\partial u_{t}}{\partial \nu } \biggr\vert }^{2}\,d\varGamma -\frac{N_{1}}{2} \int _{\varOmega }{ \vert u_{t} \vert }^{2}\,dx\\ &\quad {}+ \frac{N _{1} \tau _{1}^{2} C_{\alpha _{1}}}{2\epsilon } \int _{\varGamma _{1}}(\theta _{1} \circ u ) \,d\varGamma + \frac{N_{1} \tau _{2}^{2} C_{\alpha _{1}}}{2 \epsilon } \int _{\varGamma _{1}} \biggl(\theta _{2} \circ \frac{\partial u _{t}}{\partial \nu } \biggr)\,d\varGamma \\ &\quad {}-N_{1} \biggl(\frac{1}{2}-\frac{\varepsilon R}{2} \biggr) \int _{\varGamma _{1}}m.\nu \bigl[u_{xx}^{2}+u_{yy}^{2}+2 \mu u_{xx}u_{yy}+2(1- \mu )u_{xy}^{2} \bigr]\,d\varGamma \\ &\quad {}-\frac{N_{1}\tau _{1}}{2} \int _{\varGamma _{1}}\bigl(r_{1}^{\prime \prime } \circ u\bigr)\,d \varGamma -\frac{N_{1}\tau _{2}}{2} \int _{\varGamma _{1}}\biggl(r _{2}^{\prime \prime } \circ \frac{\partial u}{\partial \nu }\biggr)\,d\varGamma\\ &\quad {} +n_{0} E'(t)+ \frac{N_{1} c}{2} \int _{\varOmega } h^{2}(u_{t})\,dx. \end{aligned} $$

(54)

Then choosing $0 < \varepsilon \leq \min \{ \frac{1}{R}, \frac{2- \lambda _{0}}{\lambda _{0}}\}$ so that $\frac{1}{2}-\frac{\varepsilon R}{2} > 0$ and $c_{0}:=1-\frac{\lambda _{0}}{2}-\frac{\varepsilon \lambda _{0}}{2} > 0$ and using $\lim_{t \rightarrow \infty } r_{i}(t) = 0$, for $i = 1, 2$, we obtain

$$\begin{aligned} L'(t) &\leq - \biggl( \tau _{1} N-\frac{R N_{1}}{2}-\frac{N_{1} \tau _{1} ^{2}}{2\varepsilon } \biggr) \int _{\varGamma _{1}}{ \vert u_{t} \vert }^{2}\,d \varGamma - \biggl(\tau _{2} N-\frac{N_{1} \tau _{1}^{2}}{2\varepsilon } \biggr) \int _{\varGamma _{1}}{ \biggl\vert \frac{\partial u_{t}}{\partial \nu } \biggr\vert }^{2}\,d\varGamma \\ &\quad {}-\frac{N_{1}}{2} \int _{\varOmega }{ \vert u_{t} \vert }^{2}\,dx- N_{1}c_{0} a(u,u)+\frac{N_{1} \tau _{1}^{2} C_{\alpha _{1}}}{2\epsilon } \int _{\varGamma _{1}}(\theta _{1} \circ u ) \,d\varGamma \\ &\quad {} + \frac{N_{1} c}{2} \int _{\varOmega } h^{2}(u_{t})\,dx +\frac{N_{1} \tau _{2}^{2} C_{\alpha _{2}}}{2\epsilon } \int _{\varGamma _{1}} \biggl(\theta _{2} \circ \frac{\partial u_{t}}{\partial \nu } \biggr)\,d\varGamma -\frac{N_{1}\tau _{1}}{2} \int _{\varGamma _{1}}\bigl(r_{1} ^{\prime \prime } \circ u\bigr)\,d \varGamma \\ &\quad {} -\frac{N_{1}\tau _{2}}{2} \int _{\varGamma _{1}}\biggl(r_{2}^{\prime \prime } \circ \frac{\partial u}{ \partial \nu }\biggr)\,d\varGamma +n_{0} E'(t). \end{aligned}$$

(55)

In this case, we choice N large enough so that

$$ \begin{aligned}[b] &\tau _{2} N- \frac{N_{1} \tau _{1}^{2}}{2\varepsilon }> 0, \\ &\tau _{1} N-\frac{R N_{1}}{2}-\frac{N_{1} \tau _{1}^{2}}{2\varepsilon }> 0. \end{aligned} $$

(56)

Then (55) reduces to

$$ \begin{aligned}[b] L'(t) &\leq - \frac{N_{1}}{2} \int _{\varOmega }{ \vert u_{t} \vert }^{2}\,dx-N _{1} c_{0} a(u,u)+\frac{N_{1} \tau _{1}^{2} C_{\alpha _{1}}}{2\epsilon } \int _{\varGamma _{1}}(\theta _{1} \circ u )\,d\varGamma \\ &\quad {}+ \frac{N_{1} c}{2} \int _{\varOmega } h^{2}(u_{t})\,dx+\frac{N_{1} \tau _{2}^{2} C_{\alpha _{2}}}{2\epsilon } \int _{\varGamma _{1}} \biggl(\theta _{2} \circ \frac{\partial u_{t}}{\partial \nu } \biggr)\,d\varGamma \\ &\quad {} -\frac{N_{1}\tau _{1}}{2} \int _{\varGamma _{1}}\bigl(r_{1} ^{\prime \prime } \circ u\bigr)\,d \varGamma-\frac{N_{1}\tau _{2}}{2} \int _{\varGamma _{1}}\biggl(r_{2}^{\prime \prime } \circ \frac{\partial u}{ \partial \nu }\biggr)\,d\varGamma+n_{0} E'(t). \end{aligned} $$

(57)

Recall that $r_{i}''=\alpha r_{i}' + \theta _{i}$, $i=1,2$, and use (19), to obtain

$$ \begin{aligned}[b]L^{\prime }(t)&\le - \frac{N_{1}}{2} \int _{\varOmega }{ \vert u_{t} \vert } ^{2}\,dx-N_{1} c_{0} a(u,u)- \biggl( \frac{N_{1} \tau _{1}}{2}-\frac{N_{1} \tau _{1}^{2} C_{\alpha _{1}}}{2\epsilon } \biggr) \int _{\varGamma _{1}}(\theta _{1} \circ u)\,d\varGamma \\ &\quad {}- \biggl(\frac{N_{1}\tau _{2}}{2}-\frac{N_{1} \tau _{2}^{2} C_{\alpha _{2}}}{2 \epsilon } \biggr) \int _{\varGamma _{1}} \biggl(\theta _{2} \circ \frac{\partial u}{\partial \nu } \biggr)\,d\varGamma +\frac{N_{1} c}{2} \int _{\varOmega } h^{2}(u _{t})\,dx, \\ &\quad {}-\frac{\tau _{1} N_{1}\alpha }{2} \int _{\varGamma _{1}}\bigl(r_{1} ^{\prime } \circ u\bigr)\,d \varGamma -\frac{\tau _{2} N_{1} \alpha }{2} \int _{\varGamma _{1}}\biggl(r_{2}^{\prime } \circ \frac{\partial u}{\partial \nu }\biggr)\,d\varGamma +n_{0} E'(t)\quad \forall t\ge t_{1}. \end{aligned} $$

(58)

Now, our purpose is to have, for $i=1,2$,

$$ \frac{N_{i} \tau _{i}}{2}-C_{\alpha _{i}} \biggl(\frac{N_{i} \tau _{i} ^{2} }{2\epsilon } \biggr) > \frac{N_{i} \tau _{i}}{4}. $$

(59)

As in [44], we can deduce that $\alpha _{i} C _{\alpha _{i}} \rightarrow 0$ when $\alpha _{i} \rightarrow 0$. Then there exists $0 < {\alpha _{0}}_{i} < 1$ such that if $\alpha _{i} < {\alpha _{0}}_{i}$, then

$$ C_{\alpha _{i}} < \frac{\epsilon }{4\alpha _{i} \tau _{i}^{2} N_{i}}. $$

Now, we choose $0 < \alpha _{i} = \frac{1}{2 N_{i} \tau _{i}} < 1$, to obtain

$$ C_{\alpha _{i}} \biggl(\frac{N_{i} \tau _{i}^{2} }{2\epsilon } \biggr) < \frac{1}{8 \alpha _{i}}= \frac{N_{i} \tau _{i}}{4}, $$

(60)

and hence, we have

$$ \begin{aligned}[b] &N_{1} \biggl( \frac{\tau _{i}}{2}-\frac{\tau _{i}^{2} C_{\alpha _{i}}}{2 \epsilon } \biggr)> 0, \quad i=1,2, \end{aligned} $$

(61)

and then (58) becomes

$$ \begin{aligned}[b] L^{\prime }(t)&\le - \frac{N_{1}}{2} \int _{\varOmega }{ \vert u_{t} \vert } ^{2}\,dx-N_{1} c_{0} a(u,u) + \frac{N_{1} c}{2} \int _{\varOmega } h^{2}(u _{t})\,dx, \\ &\quad {}-\frac{1}{4} \int _{\varGamma _{1}}\bigl(r_{1}^{\prime } \circ u\bigr)\,d \varGamma -\frac{1}{4} \int _{\varGamma _{1}}\biggl(r_{2}^{\prime } \circ \frac{ \partial u}{\partial \nu }\biggr)\,d\varGamma + n_{0} E'(t). \end{aligned} $$

(62)

From (34) and (30), we notice that, for all $t \geq t_{1}$,

$$ \begin{aligned} &{-} \int _{0}^{t_{1}}r_{1}'(s) \int _{\varGamma _{1}} \bigl\vert u(t)-u(t-s) \bigr\vert ^{2} \,d\varGamma \,ds \\ &\quad {}\leq \frac{1}{d_{1}} \int _{0}^{t_{1}}r_{1}''(s) \int _{\varGamma _{1}} \bigl\vert u(t)-u(t-s) \bigr\vert ^{2} \,d\varGamma \,ds \le -cE'(t), \\ &{-} \int _{0}^{t_{1}}r_{2}'(s) \int _{\varGamma _{1}} \biggl\vert \frac{ \partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{\partial \nu } \biggr\vert ^{2} \,d\varGamma \,ds \\ &\quad \leq \frac{1}{d_{2}} \int _{0}^{t_{1}}r_{2}''(s) \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{ \partial \nu } \biggr\vert ^{2} \,d\varGamma \,ds \le -cE'(t). \end{aligned} $$

(63)

Then, using (62) and (63), we have for all $t \geq t_{1}$

$$ \begin{aligned}[b]L^{\prime }(t)&\le - \frac{N_{1}}{2} \int _{\varOmega }{ \vert u_{t} \vert } ^{2}\,dx-N_{1} c_{0} a(u,u)-\frac{1}{4} \int _{t_{1}}^{t}r_{1}'(s) \int _{\varGamma _{1}} \bigl\vert u(t)-u(t-s) \bigr\vert ^{2} \,d\varGamma \,ds \\ &\quad {}-\frac{1}{4} \int _{t_{1}}^{t}r_{2}'(s) \int _{\varGamma _{1}} \biggl\vert \frac{\partial u(t)}{\partial \nu }-\frac{\partial u(t-s)}{ \partial \nu } \biggr\vert \,d\varGamma \,ds\\&\quad{} +\frac{N_{1} c}{2} \int _{\varOmega } h^{2}(u_{t})\,dx +(n_{0}-c) E'(t). \end{aligned} $$

(64)

Now, we choose $n_{0}$ so that $n_{0}-c >0$, then (53) is established. Moreover, we can choose N even larger (if needed) so that

$$ L(t) \sim E(t). $$

(65)

 □

Lemma 4.8

[45] Under Assumptions $(A1)$–$(A3)$, the solution satisfies the estimates

$$\begin{aligned}& \begin{aligned}[b] \int _{\varOmega _{1}}h^{2}(u_{t})\,dx \le c \int _{\varOmega _{1}}u_{t} h(u_{t})\,dx,\quad \textit{if }h_{0}\textit{ is linear}, \end{aligned} \end{aligned}$$

(66)

$$\begin{aligned}& \begin{aligned}[b] \int _{\varOmega _{1}}h^{2}(u_{t})\,dx \le cH^{-1}\bigl(J(t)\bigr)-cE^{\prime }(t),\quad \textit{if }h_{0}\textit{ is nonlinear}, \end{aligned} \end{aligned}$$

(67)

where

$$ J(t):= \int _{\varOmega _{1}}u_{t}(t)h\bigl(u_{t}(t)\bigr)\,dx \le -cE^{\prime }(t) $$

(68)

and

$$ \varOmega _{1}=\bigl\{ x\in \varOmega : \bigl\vert u_{t}(t) \bigr\vert \le \varepsilon _{1} \bigr\} . $$

Lemma 4.9

Assume that $(A1)$–$(A3)$ hold and $h_{0}$ is linear. Then the energy functional satisfies the following estimate:

$$ \int _{0}^{+\infty }E(s)\,ds < \infty . $$

(69)

Proof

Let $F(t)=L(t)+\psi _{2}(t)+ \psi _{3}(t)$, using (50) and (64), and using the trace theory, we obtain for all $t \geq t_{1}$

$$ \begin{aligned}[b]F^{\prime }(t)&\le - \frac{N_{1}}{2} \int _{\varOmega } \vert u_{t} \vert \,dx-N _{1} c_{0} a (u,u)+\frac{1}{4}\bigl(r_{1}' \circ u\bigr) (t) +\frac{1}{4} \biggl(r_{2}' \circ \frac{\partial u}{\partial \nu }\biggr) (t) \\ &\quad {}+ \frac{N_{1} c}{2} \int _{\varGamma _{1}} h^{2}(u_{t})\,d \varGamma +3 r_{1}(0) \int _{\varOmega } \bigl\vert u(t) \bigr\vert ^{2}\,dx+3r_{2}(0) \int _{\varOmega } \biggl\vert \frac{\partial u(t)}{\partial \nu } \biggr\vert ^{2}\,dx. \end{aligned} $$

(70)

Using (17) and (18), we arrive at

$$ \begin{aligned}[b]F^{\prime }(t)&\le - \frac{N_{1}}{2} \int _{\varOmega } \vert u_{t} \vert \,dx-(N _{1} c_{0} - c_{r} ) a (u,u)+\frac{1}{4} \bigl(r_{1}' \circ u\bigr) (t) + \frac{1}{4} \biggl(r_{2}' \circ \frac{\partial u}{\partial \nu }\biggr) (t) \\ &\quad {}+ c \int _{\varGamma _{1}} h^{2}(u_{t})\,d \varGamma , \end{aligned} $$

(71)

where $c_{r}= (3 c_{p} r_{1}(0) + 3 c_{s} r_{2}(0))$ and $c_{p}$, $c _{s}$ are given in (18). Here, we choose $N_{1}$ large enough so that $N_{1}c_{0} -c_{r} > 0$. After that, we can choose N even larger (if needed) so that (56) holds. Now, we have

$$ \begin{aligned}[b]F^{\prime }(t)&\le -b E(t)+c \int _{\varOmega }u_{t} h(u_{t})\,dx \\ & \le -bE(t)-cE^{\prime }(t), \end{aligned} $$

where b is a positive constant. Therefore,

$$ b \int _{t_{1}}^{t}E(s)\,ds\le F_{1}(t_{1})-F_{1}(t) \le F_{1}(t_{1})< \infty , $$

(72)

where $F_{1}(t)=F(t)+cE(t)\sim E$. □

Now, we define

$$ \begin{aligned} &I_{1}(t):= \int _{t_{1}}^{t}r_{1}^{\prime \prime }(s) \int _{\varGamma _{1}} { \bigl\vert u(t)-u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds\le -cE^{\prime }(t), \\ &I _{2}(t):= \int _{t_{1}}^{t}r_{2}^{\prime \prime }(s) \int _{\varGamma _{1}} { \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{ \partial \nu } \biggr\vert }^{2}\,d \varGamma \,ds\le -cE^{\prime }(t). \end{aligned} $$

(73)

Lemma 4.10

Under Assumptions $(A1)$–$(A3)$, and if $h_{0}$ is linear, we have the following estimates:

$$ \int _{t_{1}}^{t}-r_{1}' (s) \int _{\varOmega }{ \bigl\vert u(t)-u(t-s) \bigr\vert } ^{2}\,dx\,ds\le \frac{1}{q} \overline{G_{1}}^{-1} \biggl(\frac{q I_{1}(t)}{ \xi _{1}(t)} \biggr) $$

(74)

and

$$ \int _{t_{1}}^{t}-r_{2}' (s) \int _{\varOmega } { \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{\partial \nu } \biggr\vert }^{2} \,d \varGamma \,ds\le \frac{1}{q} \overline{G_{2}}^{-1} \biggl( \frac{q I_{2}(t)}{\xi _{2}(t)} \biggr), $$

(75)

and if $h_{0}$ is nonlinear, we have the following estimates:

$$\begin{aligned}& \int _{t_{1}}^{t}-r_{1}' (s) \int _{\varOmega }{ \bigl\vert u(t)-u(t-s) \bigr\vert } ^{2}\,dx\,ds\le \frac{(t-t_{1})}{q} \overline{G_{1}}^{-1} \biggl(\frac{q I _{1}(t)}{(t-t_{1})\xi _{1}(t)} \biggr), \end{aligned}$$

(76)

$$\begin{aligned}& \int _{t_{1}}^{t}-r_{2}' (s) \int _{\varOmega }{ \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{\partial \nu } \biggr\vert }^{2} \,d \varGamma \,ds\le \frac{(t-t_{1})}{q} \overline{G_{2}}^{-1} \biggl( \frac{q I_{2}(t)}{(t-t_{1})\xi _{2}(t)} \biggr), \end{aligned}$$

(77)

where $q \in (0,1)$, $\overline{G_{1}}$ and $\overline{G_{2}}$ are the extensions of $G_{1}$ and $G_{2}$, respectively, such that $\overline{G _{1}}$ and $\overline{G_{2}}$ are strictly increasing and strictly convex $C^{2}$ functions on $(0,\infty )$

Proof

Case I: if $h_{0}$ is linear: we define the following quantities:

$$ \begin{aligned}[b] &\lambda _{1} (t):=q \int _{t_{1}}^{t} \int _{\varGamma _{1}} { \bigl\vert u(t)-u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds, \\ &\lambda _{2} (t):=q \int _{t_{1}}^{t} \int _{\varGamma _{1}} { \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{ \partial u(t-s)}{\partial \nu } \biggr\vert }^{2}\,d \varGamma \,ds, \end{aligned} $$

(78)

where by (69), (19) and (16) we can choose q so small such that $\forall t \ge t_{1}$,

$$ \lambda _{i} (t)< 1,\quad i=1,2. $$

(79)

Since $G_{i}$ is strictly convex on $(0,R_{i}]$ and $G_{i}(0)=0$, we have

$$ G_{i}(\theta z)\le \theta G_{i}(z),\quad 0\le \theta \le 1\text{ and }z\in (0,r], $$

(80)

where $r=\min \{R_{1}, R_{2}\}$. Without loss of generality, for all $t \ge t_{1}$, we assume that $I_{i}(t)> 0$, $i=1,2$, otherwise we get an exponential decay from (53). Using (14), (79), (80) and Jensen’s inequality, we have

$$ \begin{aligned}[b]I_{1}(t)&=\frac{1}{q \lambda _{1}(t)} \int _{t_{1}}^{t}\lambda _{1} (t) r _{1}''(s) \int _{\varGamma _{1}}{q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ & \ge \frac{1}{q \lambda _{1}(t)} \int _{t_{1}}^{t} \lambda _{1} (t) \xi _{1} (s) G_{1}\bigl(-r_{1}'(s)\bigr) \int _{\varGamma _{1}}{ q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ & \ge \frac{1}{q \lambda _{1}(t)} \int _{t_{1}}^{t} \xi _{1} (s) G_{1} \bigl(-\lambda _{1} (t) r_{1}'(s)\bigr) \int _{\varGamma _{1}}{ q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ & \ge \frac{ \xi _{1} (t)}{q \lambda _{1}(t)} \int _{t_{1}} ^{t} G_{1}\bigl(-\lambda _{1} (t) r_{1}'(s)\bigr) \int _{\varGamma _{1}}{ q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ &\ge \frac{ \xi _{1} (t)}{q \lambda _{1}(t)} \lambda _{1}(t) G_{1}\biggl(q \int _{t_{1}}^{t} - r_{1}'(s) \int _{\varGamma _{1}}{ \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds\biggr) \\ & = \frac{ \xi _{1} (t)}{q } \overline{G_{1}} \biggl( q \int _{t _{1}}^{t} -r_{1}'(s) \int _{\varGamma _{1}}{ \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \biggr). \end{aligned} $$

This gives

$$ \int _{t_{1}}^{t} -r_{1}' (s) \int _{\varGamma _{1}}{ \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \leq \frac{1}{q } \overline{G_{1}}^{-1} \biggl(\frac{qI _{1}(t)}{\xi _{1} (t)} \biggr). $$

Similarly, we can show that

$$ \int _{t_{1}}^{t}-r_{2}' (s) \int _{\varGamma _{1}} { \biggl\vert \frac{ \partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{\partial \nu } \biggr\vert }^{2} \,d \varGamma \,ds\le \frac{1}{q} \overline{G_{2}}^{-1} \biggl( \frac{q I_{2}(t)}{\xi _{2}(t)} \biggr). $$

Case II: if $h_{0}$ is nonlinear: we introduce the following functionals:

$$ \begin{aligned}[b] &\lambda _{3} (t):= \frac{q }{t-t_{1}} \int _{t_{1}}^{t} \int _{\varGamma _{1}} { \bigl\vert u(t)-u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds, \\ & \lambda _{4} (t):= \frac{q }{t-t_{1}} \int _{t_{1}}^{t} \int _{\varGamma _{1}} { \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{\partial \nu } \biggr\vert }^{2}\,d \varGamma \,ds, \end{aligned} $$

(81)

then using (16), (19) and (69), we can choose q so small enough so that $\forall t \ge t_{1}$,

$$ \lambda _{i} (t)< 1, \quad i=3,4. $$

(82)

Using (14), (80), (82) and Jensen’s inequality, we get

$$ \begin{aligned}[b]I_{1}(t) &=\frac{1}{q \lambda _{3}(t)} \int _{t_{1}}^{t}\lambda _{3} (t) r _{1}''(s) \int _{\varGamma }{q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ & \ge \frac{1}{q \lambda _{3}(t)} \int _{t_{1}}^{t} \lambda _{3} (t) \xi _{1} (s) G_{1}\bigl(-r_{1}'\bigr) \int _{\varGamma }{ q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ & \ge \frac{1}{q \lambda _{3}(t)} \int _{t_{1}}^{t} \xi _{1} (s) G_{1} \bigl(-\lambda _{3} (t) r_{1}'(s)\bigr) \int _{\varGamma }{ q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ & \ge \frac{ \xi _{1} (t)}{q \lambda _{3}(t)} \int _{t_{1}} ^{t} G_{1}\bigl(-\lambda _{3} (t) r_{1}'(s)\bigr) \int _{\varGamma }{ q \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ & \ge \frac{(t-t_{1}) \xi _{1} (t)}{q \lambda _{3}(t)} \lambda _{3}(t) G_{1} \biggl(\frac{q}{(t-t_{1})} \int _{t_{1}}^{t} - r _{1}'(s) \int _{\varGamma }{ \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \biggr) \\ & = \frac{ (t-t_{1}) \xi _{1} (t)}{q } \overline{G_{1}} \biggl( \frac{q}{(t-t_{1})} \int _{t_{1}}^{t} -r_{1}'(s) \int _{\varGamma } { \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \biggr). \end{aligned} $$

This gives

$$ \int _{t_{1}}^{t} -r_{1}' \int _{\varGamma }{ \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \leq \frac{(t-t_{1})}{q} \overline{G_{1}}^{-1} \biggl(\frac{q I_{1}(t)}{(t-t_{1})\xi _{1}(t)} \biggr). $$

Similarly, we can have

$$ \int _{t_{1}}^{t} -r_{2}' (s) \int _{\varGamma _{1}} { \biggl\vert \frac{ \partial u(t)}{\partial \nu }- \frac{\partial u(t-s)}{\partial \nu } \biggr\vert }^{2} \,d \varGamma \,ds\le \frac{(t-t_{1})}{q} \overline{G_{2}} ^{-1} \biggl( \frac{q I_{2}(t)}{(t-t_{1})\xi _{1}(t)} \biggr). $$

 □

5 Proofs of our main results

Here, we prove the main results of our work given in Theorem 3.1 and 3.2.

Proof of Theorem 3.1, case 1, G is linear

We multiply (53) by the nonincreasing function $\sigma (t)$. We use (14), (30) and (66), and invoke (14) to have

$$ \begin{aligned}[b] \sigma (t) L^{\prime }(t)&\le -m \sigma (t) E(t)-c\sigma (t) \int _{t _{1}}^{t}r_{1}^{\prime }(s) \int _{\varGamma _{1}}{ \bigl\vert u(t)- u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ &\quad {}-c\sigma (t) \int _{t_{1}}^{t}r_{2}^{ \prime }(s) \int _{\varGamma _{1}}{ \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{u(t-s)}{ \partial \nu } \biggr\vert }^{2}\,d \varGamma \,ds+c\sigma (t) \int _{\varOmega } h^{2}(u_{t})\,dx\quad \forall t\ge t_{1} \\ & \le -m \sigma (t) E(t)+c \int _{\varGamma _{1}} \biggl[\bigl(r_{1}^{\prime \prime } \circ u \bigr) (t) + \biggl(r_{2}^{\prime \prime } \circ \frac{\partial u(t)}{ \partial \nu } \biggr) \biggr] \,d \varGamma +c\sigma (t) \int _{\varOmega } h^{2}(u _{t})\,dx \\ & \le -m \sigma (t) E(t)-2c E'(t). \end{aligned} $$

This gives

$$ (\sigma L +2c E)' \leq -m \sigma (t) E(t), \quad \forall t \geq t_{1}. $$

(83)

Using the fact $\sigma '(t) \leq 0$, we have $\sigma L + 2cE \sim E$, and we can obtain

$$ E (t)\le c_{1} e^{- c_{2} \int _{t_{1}}^{t} \sigma (s) \,ds}. $$

(84)

 □

Proof of Theorem 3.1, case 2, G is nonlinear

Using (53), (66), (75) and (74), we get

$$ \begin{aligned}[b]L^{\prime }(t)&\le -mE(t)-c \int _{t_{1}}^{t}r_{1}^{\prime }(s) \int _{\varGamma _{1}}{ \bigl\vert u(t)-u(t-s) \bigr\vert }^{2}\,d \varGamma \,ds \\ &\quad {}-c \int _{t_{1}}^{t}r_{2}^{\prime }(s) \int _{\varGamma _{1}} { \biggl\vert \frac{\partial u(t)}{\partial \nu }- \frac{u(t-s)}{ \partial \nu } \biggr\vert }^{2}\,d \varGamma \,ds+c \int _{\varOmega } h^{2}(u _{t})\,dx\quad \forall t \ge t_{1} \\ & \le -m E(t)+\frac{1}{q } \overline{G}^{-1} \biggl( \frac{qI_{1}(t)}{ \sigma (t)} \biggr)+\frac{1}{q } \overline{G}^{-1} \biggl( \frac{qI_{2}(t)}{ \sigma (t)} \biggr)-cE'(t) \\ &\le -m E(t)+\frac{1}{q } \overline{G}^{-1} \biggl( \frac{qI(t)}{\sigma (t)} \biggr)-cE'(t), \end{aligned} $$

(85)

where $I(t)=\max \{I_{1}(t),I_{2}(t)\}$ $\forall t \geq t_{1}$. Let $\mathcal{F}_{1}(t)=L(t)+cE(t)\sim E$, then (85) becomes

$$ \mathcal{F}_{1}^{\prime }(t)\le -m E(t)+c ( \overline{G} ) ^{-1} \biggl(\frac{qI(t)}{\sigma (t)} \biggr), $$

(86)

we notice that the functional $\mathcal{F}_{2}$, defined by

$$ \mathcal{F}_{2}(t):=\overline{G}^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)\mathcal{F}_{1}(t) $$

satisfies

$$ \alpha _{1}\mathcal{F}_{2}(t)\le E(t)\le \alpha _{2}\mathcal{F}_{2}(t) $$

(87)

where $\alpha _{1},\alpha _{2}>0$, and

$$ \begin{aligned}[b]\mathcal{F}_{2}^{\prime }(t)&= \varepsilon _{0} \frac{E^{\prime }(t)}{E(0)}\overline{G}^{\prime \prime } \biggl( \varepsilon _{0} \frac{E(t)}{E(0)} \biggr)\mathcal{F}_{1}(t)+ \overline{G}^{\prime } \biggl(\varepsilon _{0}\frac{E(t)}{E(0)} \biggr){\mathcal{F}_{1}}^{ \prime }(t) \\ & \le -m E(t)\overline{G}^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)+c \overline{G}^{\prime } \biggl(\varepsilon _{0}\frac{E(t)}{E(0)} \biggr)\overline{G}^{-1} \biggl( \frac{qI(t)}{ \sigma (t)} \biggr). \end{aligned} $$

(88)

As in the sense of Young (see [46]), let $\overline{G}^{*}$ be the convex conjugate of G̅, then

$$ \overline{G}^{*}(a)=a\bigl(\overline{G}^{\prime } \bigr)^{-1}(a)-\overline{G} \bigl[\bigl(\overline{G}^{\prime } \bigr)^{-1}(a) \bigr],\quad \text{if } a\in \bigl(0,\overline{G}^{\prime }(r)\bigr] $$

(89)

and $\overline{G}^{*}$ satisfies the generalized Young inequality

$$ A B\le \overline{G}^{*}(A)+\overline{G}(B),\quad \text{if } A\in \bigl(0,\overline{G}^{\prime }(r)\bigr], B\in (0,r]. $$

(90)

So, with $A=\overline{G}^{\prime } (\varepsilon _{0}\frac{E^{ \prime }(t)}{E(0)} )$ and $B=\overline{G}^{-1} (\frac{qI(t)}{ \sigma (t)} )$ and using (19) and (88)–(90), we arrive at

$$ \begin{aligned}[b]\mathcal{F}_{2}^{\prime }(t) &\le -m E(t)\overline{G}^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)+c \overline{G}^{*} \biggl(\overline{G} ^{\prime } \biggl(\varepsilon _{0}\frac{E(t)}{E(0)} \biggr) \biggr)+c \biggl(\frac{qI(t)}{\sigma (t)} \biggr) \\ & \le -m E(t)\overline{G}^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)+c\varepsilon _{0}\frac{E(t)}{E(0)} \overline{G}^{\prime } \biggl(\varepsilon _{0}\frac{E(t)}{E(0)} \biggr)+c \biggl(\frac{qI(t)}{\sigma (t)} \biggr). \end{aligned} $$

(91)

So, multiplying (91) by $\sigma (t)$ and using the fact that $\varepsilon _{0}\frac{E(t)}{E(0)}< r$, $\overline{G}^{\prime } (\varepsilon _{0}\frac{E(t)}{E(0)} )=G^{\prime } (\varepsilon _{0}\frac{E(t)}{E(0)} )$, gives

$$ \begin{aligned}[b]\sigma (t) \mathcal{F}_{2}^{\prime }(t) &\le -m \sigma (t) E(t)G^{ \prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)+c \sigma (t) \varepsilon _{0} \frac{E(t)}{E(0)}G^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)+c q I(t) \\ &\le -m \sigma (t) E(t)G^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)+c \sigma (t)\varepsilon _{0} \frac{E(t)}{E(0)}G^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)-c E^{\prime }(t). \end{aligned} $$

Now, for all $t\ge t_{1}$ and with a good choice of $\varepsilon _{0}$, we obtain

$$ \mathcal{F}_{3}^{\prime }(t)\le -m_{0} \sigma (t) \biggl(\frac{E(t)}{E(0)} \biggr)G ^{\prime } \biggl(\varepsilon _{0}\frac{E(t)}{E(0)} \biggr)=-m_{0} \sigma (t) G_{3} \biggl(\frac{E(t)}{E(0)} \biggr), $$

(92)

where $\mathcal{F}_{3}=\sigma \mathcal{F}_{2}+c E \sim E$ satisfies, for any $\beta _{3},\beta _{4}> 0$,

$$ \beta _{3}\mathcal{F}_{3}(t)\le E(t)\le \beta _{4}\mathcal{F}_{3}(t), $$

(93)

and $G_{3}(t)=t G^{\prime }(\varepsilon _{0}t)$. Since $G^{\prime } _{3}(t)=G^{\prime }(\varepsilon _{0}t)+\varepsilon _{0}t G^{\prime \prime }(\varepsilon _{0}t)$. Since G is strictly convex over $(0,r]$, we find that $G_{3}^{\prime }(t), G_{3}(t)>0$ on $(0,1]$. Then, with

$$ R(t)= \frac{\beta _{3} \mathcal{F}_{3}(t)}{E(0)}, $$

using (93) and (92), we obtain

$$ R(t)\sim E(t) $$

(94)

and then

$$ R^{\prime }(t)\le -m_{1}\sigma (t) G_{3}\bigl(R(t) \bigr),\quad \forall t\ge t_{1}, $$

where $m_{1} > 0$. We, after integration over $(t_{1},t)$, get

$$ \begin{aligned}[b] \int _{t_{1}}^{t}\frac{-R^{\prime }(s)}{G_{3}(R(s))}\,ds \ge m_{1} \int _{t_{1}}^{t}\sigma (s)\,ds. \end{aligned} $$

Hence, by an appropriate change of variable, we get

$$ \begin{aligned}[b] \int _{\varepsilon _{0} R(t)}^{\varepsilon _{0} R(t_{1})}\frac{1}{\tau G ^{\prime }(\tau )}\,d\tau \ge m_{1} \int _{t_{1}}^{t} \sigma (s)\,ds. \end{aligned} $$

Thus, we have

$$ R(t)\le \frac{1}{\varepsilon _{0}}G_{4}^{-1} \biggl(m_{1} \int _{t_{1}} ^{t}\sigma (s)\,ds \biggr), $$

(95)

where $G_{4}(t)=\int _{t}^{r}\frac{1}{sG^{\prime }(s)}\,ds$. Here, we used the strictly decreasing property of $G_{4}$ over $(0,r]$. Therefore (21) is established by virtue of (94) and hence we finished the proof of Theorem 3.1. □

Proof of Theorem 3.2, case 1, G is linear

Multiplying (53) by $\sigma (t)$, using (67), gives, as $\sigma (t)$ is nonincreasing, the following:

$$\begin{aligned}& \begin{aligned}[b]\sigma (t) L^{\prime }(t) &\le -m \sigma (t) E(t)+c \int _{\varGamma _{1}} \biggl[\bigl(r_{1}^{\prime \prime } \circ u \bigr) (t) + \biggl(r_{2}^{\prime \prime } \circ \frac{\partial u(t)}{\partial \nu } \biggr) \biggr] \,d \varGamma +c\sigma (t) \int _{\varOmega } h^{2}(u_{t})\,dx \\ & \le -m \sigma (t) E(t)-2c E'(t)+c\sigma (t) \int _{\varOmega } h^{2}(u_{t})\,dx \\ & \le -m \sigma (t) E(t)-2c E'(t)+c \sigma (t) \bigl(H^{-1}\bigl(J(t)\bigr)-cE'(t) \bigr) \\ & \le -m \sigma (t) E(t)-3c E'(t)+c \sigma (t) H^{-1} \bigl(J(t)\bigr), \end{aligned} \\& (\sigma L +3 c E)' \leq -m \sigma (t) E(t)+ c \sigma (t) H^{-1}\bigl(J(t)\bigr),\quad \forall t \geq t_{1}. \end{aligned}$$

(96)

Therefore, (96) becomes

$$ {\mathcal{L}}'(t) \leq -m \sigma (t) E(t)+ c \sigma (t) H^{-1}\bigl(J(t)\bigr),\quad \forall t \geq t_{1}, $$

(97)

where $\mathcal{L}:=\sigma L +3c E \sim E$. Now, for $\varepsilon _{1}< r _{0}$ and $c_{0}>0$, using (97) and the fact that $E^{\prime }\le 0$, $H^{\prime }>0$, $H^{\prime \prime }>0$ on $(0,r_{0}]$, we notice that the functional $\mathcal{L}_{1}$, defined by

$$ \mathcal{L}_{1}(t):=H' \biggl(\varepsilon _{1} \frac{E(t)}{E(0)} \biggr) \mathcal{L}(t)+c_{0}E(t) $$

satisfies, for some $\alpha _{3},\alpha _{4}>0$.

$$ \alpha _{3} \mathcal{L}_{1}(t)\le E(t)\le \alpha _{4}\mathcal{L}_{1}(t) $$

(98)

and

$$ \begin{aligned}[b]\mathcal{L}_{1}^{\prime }(t)&= \varepsilon _{0} \frac{E^{\prime }(t)}{E(0)}H^{\prime \prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr) \mathcal{L}(t)+ H^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr){ \mathcal{L}}^{\prime }(t)+c_{0}E^{\prime }(t) \\ & \le -m E(t)H^{\prime } \biggl(\varepsilon _{0} \frac{E(t)}{E(0)} \biggr)+c \sigma (t) H^{\prime } \biggl(\varepsilon _{0}\frac{E(t)}{E(0)} \biggr)H^{-1}\bigl(J(t) \bigr)+c_{0}E^{\prime }(t). \end{aligned} $$

(99)

Now, let $H^{*}$ be the convex conjugate of H (see [46]), then, as in (89) and (90), with $A=H^{\prime } (\varepsilon _{1} \frac{E(t)}{E(0)} )$ and $B=H^{-1}(J(t))$, (99) gives

$$ \begin{aligned}[b]\mathcal{L}_{1}^{\prime }(t)&\le -m E(t)H^{\prime } \biggl(\varepsilon _{1}\frac{E(t)}{E(0)} \biggr)+c \sigma (t) H^{*} \biggl(H^{\prime } \biggl(\varepsilon _{1}\frac{E(t)}{E(0)} \biggr) \biggr)+c \sigma (t) J(t)+c _{0}E^{\prime }(t) \\ & \le -m E(t)H^{\prime } \biggl(\varepsilon _{1} \frac{E(t)}{E(0)} \biggr)+c\varepsilon _{1} \sigma (t) \frac{E(t)}{E(0)}H^{\prime } \biggl(\varepsilon _{1} \frac{E(t)}{E(0)} \biggr)-c E^{\prime }(t)+c_{0} E^{\prime }(t). \end{aligned} $$

Choosing suitable $\varepsilon _{1}$ and $c_{0}$, we find, for all $t\ge t_{1}$,

$$ \mathcal{L}_{1}^{\prime }(t)\le -c \sigma (t) \frac{E^{\prime }(t)}{E(0)}H^{\prime } \biggl(\varepsilon _{1} \frac{E(t)}{E(0)} \biggr)=-c \sigma (t) H_{2} \biggl(\varepsilon _{1} \frac{E(t)}{E(0)} \biggr), $$

(100)

where $H_{2}(t)=t H^{\prime }(\varepsilon _{1}t)$. We have $H^{\prime }_{2}(t)=H^{\prime }(\varepsilon _{1}t)+\varepsilon _{1}t H^{\prime \prime }(\varepsilon _{1}t)$. Since H is strictly convex over $(0,r_{0}]$, we find that $H_{2}^{\prime }(t), H_{2}(t)>0$ on $(0,1]$. Then, with

$$ R_{1}(t)= \frac{\alpha _{3}{\mathcal{L}_{1}}(t)}{E(0)}, $$

using (98) and (100), we have

$$\begin{aligned}& R_{1}(t)\sim E(t), \\& R_{1}^{\prime }(t)\le -c_{3} \sigma (t) H_{2} \bigl(R_{1}(t)\bigr),\quad \forall t\ge t_{1}, \end{aligned}$$

(101)

where $c_{3}>0$. Thus, we integrate over $(t_{1},t)$ to get

$$ R_{1}(t)\le H_{1}^{-1} \biggl(c_{3} \int _{t_{1}}^{t} \sigma (s) \,ds +c_{4} \biggr),\quad \forall t\ge t_{1}, $$

(102)

where $c_{4}>0$, and $H_{1}(t)=\int _{t}^{1}\frac{1}{H_{2}(s)}\,ds$. □

Proof of Theorem 3.2, case 2, G is nonlinear

Using (53), (67) and (77), we obtain

$$ L^{\prime }(t)\le -m E(t)+c(t-t_{1}) (\overline{G} )^{-1} \biggl(\frac{qI(t)}{(t-t_{1})\sigma (t)} \biggr)+c H^{-1}\bigl(J(t) \bigr)-cE ^{\prime }(t). $$

(103)

Since $\lim_{t\to +\infty } \frac{1}{t-t_{1}}=0$, there exists $t_{2} > t_{1}$ such that $\frac{1}{t-t_{1}} < 1$ whenever $t > t_{2}$. By setting $\theta =\frac{1}{t-t_{1}} < 1$ and using (80), we obtain

$$ \overline{H}^{-1}\bigl(J(t)\bigr) \leq (t-t_{1}) \overline{H}^{-1} \biggl(\frac{J(t)}{(t-t _{1})} \biggr),\quad \forall t \ge t_{2}, $$

and then (103) becomes

$$ \begin{aligned}[b]L^{\prime }(t)&\le -m E(t)+c(t-t_{1}) (\overline{G} )^{-1} \biggl( \frac{qI(t)}{(t-t_{1})\sigma (t)} \biggr)+c(t-t_{1}) \overline{H}^{-1} \biggl(\frac{J(t)}{(t-t_{1})} \biggr) \\ &\quad {}-cE^{\prime }(t),\quad \forall t\ge t_{2}. \end{aligned} $$

(104)

Let $L_{1}(t)=L(t)+cE(t)\sim E$, then (104) takes the form

$$\begin{aligned}& L_{1}^{\prime }(t)\le -mE(t)+c(t-t_{1}) (\overline{G} ) ^{-1} \biggl(\frac{qI(t)}{(t-t_{1})\sigma (t)} \biggr)+c(t-t_{1}) \overline{H}^{-1} \biggl(\frac{J(t)}{(t-t_{1})}\biggr) ,\quad \end{aligned}$$

(105)

Let $r_{3}=\min {\{r,r_{0}\}}$, $\chi (t)=\max { \{\frac{qI(t)}{(t-t _{1})\sigma (t)},\frac{J(t)}{(t-t_{1})}} \}$ and $W= ( ( \overline{G} ) ^{-1}+\overline{H}^{-1} )^{-1}$.

So, (105) reduces to

$$ L_{1}^{\prime }(t)\le -m E(t)+c(t-t_{1}) W^{-1}\bigl(\chi (t)\bigr),\quad\forall t \ge t_{2} . $$

(106)

Now, for $\varepsilon _{2}< r_{3}$ and using (103) and the fact that $E^{\prime }\le 0$, $W^{\prime }>0$, $W^{\prime \prime }>0$ on $(0,r_{3}]$, we find that the functional $L_{2}$, defined by

$$ L_{2}(t):=W^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr)L_{1}(t),\quad \forall t\ge t_{2}, $$

satisfies, for some $\alpha _{5},\alpha _{6}>0$.

$$ \alpha _{5} L_{2}(t)\le E(t)\le \alpha _{6}L_{2}(t) $$

(107)

and, for all $t\ge t_{2}$,

$$ \begin{aligned}[b]L_{2}^{\prime }(t)&= \biggl(\frac{-\varepsilon _{2}}{(t-t_{1})^{2}} \frac{E(t)}{E(0)}+\frac{\varepsilon _{2}}{(t-t_{1})} \frac{E^{\prime }(t)}{E(0)} \biggr)W^{\prime \prime } \biggl(\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr){L_{1}}(t) \\ &\quad {}+W^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr){L}^{\prime }_{1}(t) \\ & \le -m E(t)W^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t _{1}} \cdot \frac{E(t)}{E(0)} \biggr)+c(t-t_{1})W^{\prime } \biggl( \frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr)W^{-1}\bigl( \chi (t)\bigr). \end{aligned} $$

(108)

Let $W^{*}$ be the convex conjugate of W (see [46]), then as in (89) and (90),

$$ W^{*}(a)=a\bigl(W^{\prime }\bigr)^{-1}(a)-W \bigl[\bigl(W^{\prime }\bigr)^{-1}(a) \bigr],\quad \text{if } a \in \bigl(0,W^{\prime }(r_{3})\bigr] $$

(109)

and $W^{*}$ satisfies the Young inequality,

$$ A B\le W^{*}(A)+W(B),\quad \text{if } A\in \bigl(0,W^{\prime }(r_{3})\bigr], B\in (0,r_{3}]. $$

(110)

Therefore, taking $A=W^{\prime } (\frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} )$ and $B=W^{-1}(\chi (t))$, (108) gives

$$ \begin{aligned}[b]L_{2}^{\prime }(t) &\le -m E(t)W^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t _{1}} \cdot \frac{E(t)}{E(0)} \biggr)+c (t-t_{1})W^{*} \biggl(W^{ \prime } \biggl( \frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr) \biggr) \\ &\quad {}+c (t-t_{1})\chi (t) \\ &\le -m E(t)W^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t _{1}} \cdot \frac{E(t)}{E(0)} \biggr)+c(t-t_{1})\frac{\varepsilon _{2}}{t-t _{1}} \cdot \frac{E(t)}{E(0)}W^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t _{1}} \cdot \frac{E(t)}{E(0)} \biggr) \\ &\quad {}+c(t-t_{1})\chi (t). \end{aligned} $$

(111)

Using (68) and (73), we observe that

$$ \begin{aligned}[b] &(t-t_{1})\sigma (t) \chi (t)\le -cE^{\prime }(t). \end{aligned} $$

So, multiplying (111) by $\sigma (t)$, using the fact that $\varepsilon _{2}\frac{E(t)}{E(0)}< r_{3}$, gives

$$ \begin{aligned}[b]\sigma (t)L_{2}^{\prime }(t)&\le -m \sigma (t) E(t)W^{\prime } \biggl(\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr)+c \varepsilon _{2} \sigma (t) \cdot \frac{E(t)}{E(0)}W^{\prime } \biggl(\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr) \\ &\quad {}-cE^{\prime }(t),\quad \forall t\ge t_{2}. \end{aligned} $$

Using the property of $\sigma (t)$, we obtain, for all $t \ge t_{2}$,

$$ \begin{aligned}[b]\bigl(\sigma (t)L_{2}+cE \bigr)^{\prime }(t)&\le -m \sigma (t) E(t)W^{\prime } \biggl( \frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr) \\ &\quad {}+c \varepsilon _{2}\sigma (t) \frac{E(t)}{E(0)}W^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr). \end{aligned} $$

Therefore, by setting $L_{3}(t):=\sigma (t)L_{2}(t)+cE(t) \sim E(t)$, we get

$$ \begin{aligned}[b] &L_{3}^{\prime }(t)\le -m \sigma (t) E(t)W^{\prime } \biggl(\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr)+c \varepsilon _{2} \sigma (t) \cdot \frac{E(t)}{E(0)}W^{\prime } \biggl(\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr). \end{aligned} $$

This gives, for a suitable choice of $\varepsilon _{2}$,

$$ \begin{aligned}[b] &L_{3}^{\prime }(t)\le -m_{0} \sigma (t) \biggl(\frac{E(t)}{E(0)} \biggr)W ^{\prime } \biggl(\frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr),\quad \forall t\ge t_{2}, \end{aligned} $$

or

$$ m_{0} \biggl(\frac{E(t)}{E(0)} \biggr)W^{\prime } \biggl(\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr)\sigma (t) \leq - L_{3}^{\prime }(t),\quad \forall t\ge t_{2}. $$

(112)

An integration of (112) yields

$$ \int _{t_{2}}^{t} m_{0} \biggl( \frac{E(s)}{E(0)} \biggr)W^{\prime } \biggl(\frac{\varepsilon _{2}}{s-t_{1}} \cdot \frac{E(s)}{E(0)} \biggr) \sigma (s) \,ds \leq - \int _{t_{2}}^{t} L_{3}^{\prime }(s)\,ds\le L_{3}(t _{2}). $$

(113)

Using the facts that $W',W'' > 0$ and the nonincreasing property of E, we deduce that the map $t \mapsto E(t)W^{\prime } (\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} )$ is nonincreasing and, consequently, we have

$$ \begin{aligned}[b] & m_{0} \biggl( \frac{E(t)}{E(0)} \biggr)W^{\prime } \biggl(\frac{ \varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr) \int _{t_{2}} ^{t} \sigma (s) \,ds \\ &\quad \leq \int _{t_{2}}^{t} m_{0} \biggl( \frac{E(s)}{E(0)} \biggr)W ^{\prime } \biggl(\frac{\varepsilon _{2}}{s-t_{1}} \cdot \frac{E(s)}{E(0)} \biggr)\sigma (s) \,ds\le L_{3}(t_{2}). \end{aligned} $$

(114)

Multiplying each side of (114) by $\frac{1}{t-t_{1}}$, we have

$$ m_{0} \biggl(\frac{1}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr) W^{ \prime } \biggl(\frac{\varepsilon _{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr) \int _{t_{2}}^{t} \sigma (s) \,ds \leq \frac{m _{3}}{t-t_{1}}. $$

(115)

Next, we set $W_{2}(t)=t W^{\prime }(\varepsilon _{2}t)$ which is strictly increasing, then we obtain

$$ m_{0} W_{2} \biggl(\frac{1}{t-t_{1}} \cdot \frac{E(t)}{E(0)} \biggr) \int _{t_{2}}^{t} \sigma (s) \,ds \leq \frac{m_{3}}{t-t_{1}}. $$

(116)

Finally, for two positive constants $m_{3}$ and $m_{4}$, we obtain

$$ E(t) \leq m_{4} (t-t_{1}) {W_{2}}^{-1} \biggl( \frac{m_{3}}{(t-t_{1}) \int _{t_{2}}^{t}\sigma (s) \,ds } \biggr). $$

(117)

This finishes the proof of Theorem 3.2. □