Bridge the gap between fixed-length and variable-length evolutionary neural architecture search algorithms

Yunhong Gong; Yanan Sun; Dezhong Peng; Xiangru Chen; Yunhong Gong; Yanan Sun; Dezhong Peng; Xiangru Chen

doi:10.3934/era.2024013

Electronic Research Archive

2024, Volume 32, Issue 1: 263-292. doi: 10.3934/era.2024013

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Research article Special Issues

Bridge the gap between fixed-length and variable-length evolutionary neural architecture search algorithms

1.
College of Computer Science, Sichuan University, Chengdu 610065, China
2.
National Innovation Center for UHD Video Technology, Chengdu 610095, China

Received: 18 October 2023 Revised: 26 November 2023 Accepted: 12 December 2023 Published: 22 December 2023

Evolutionary neural architecture search (ENAS) aims to automate the architecture design of deep neural networks (DNNs). In recent years, various ENAS algorithms have been proposed, and their effectiveness has been demonstrated. In practice, most ENAS methods based on genetic algorithms (GAs) use fixed-length encoding strategies because the generated chromosomes can be directly processed by the standard genetic operators (especially the crossover operator). However, the performance of existing ENAS methods with fixed-length encoding strategies can also be improved because the optimal depth is regarded as a known priori. Although variable-length encoding strategies may alleviate this issue, the standard genetic operators are replaced by the developed operators. In this paper, we proposed a framework to bridge this gap and to improve the performance of existing ENAS methods based on GAs. First, the fixed-length chromosomes were transformed into variable-length chromosomes with the encoding rules of the original ENAS methods. Second, an encoder was proposed to encode variable-length chromosomes into fixed-length representations that can be efficiently processed by standard genetic operators. Third, a decoder cotrained with the encoder was adopted to decode those processed high-dimensional representations which cannot directly describe architectures into original chromosomal forms. Overall, the performances of existing ENAS methods with fixed-length encoding strategies and variable-length encoding strategies have both improved by the proposed framework, and the effectiveness of the framework was justified through experimental results. Moreover, ablation experiments were performed and the results showed that the proposed framework does not negatively affect the original ENAS methods.

Keywords:

Citation: Yunhong Gong, Yanan Sun, Dezhong Peng, Xiangru Chen. Bridge the gap between fixed-length and variable-length evolutionary neural architecture search algorithms[J]. Electronic Research Archive, 2024, 32(1): 263-292. doi: 10.3934/era.2024013

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Abstract

1. Introduction

In this research, we mainly focused on wave equation to study and examine the coupled system. In this system, we assumed a bounded domain $\Omega\in{\bf R}^{N}$ where $\partial\Omega$ indicates sufficiently smooth boundary of $\Omega\in{\bf R}^{N}$ and take the positive constants $\xi_{0}, \xi_{1}, \sigma, \beta_{1}, \beta_{3}$ where $m\geq 1$ for $N = 1, 2$ , and $1 < m\leq\frac{N+2}{N-2}$ for $N\geq 3$ . The coupled system with these terms is given by

$\begin{eqnarray} \left\{ \begin{array}{l} v_{tt}-\bigg(\xi_{0}+\xi_{1}\Vert\nabla v\Vert^{2}_{2}+\delta(\nabla v, \nabla v_{t})_{L^{2}(\Omega)}\bigg)\Delta v(t)+ \int_{0}^{\infty}g_{1}(s)\Delta v(t-s)ds\\ \quad +\beta_{1}\vert v_{t}(t)\vert^{m-2} v_{t}(t)+ \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\vert\vert v_{t}(t-r) \vert^{m-2} v_{t}(t-r)dr+f_{1}(v, w) = 0.\\ w_{tt}-\bigg(\xi_{0}+\xi_{1}\Vert\nabla w\Vert^{2}_{2}+\delta(\nabla w, \nabla w_{t})_{L^{2}(\Omega)}\bigg)\Delta w(t)+ \int_{0}^{\infty}g_{2}(s)\Delta w(t-s)ds\\ \quad +\beta_{3}\vert w_{t}(t)\vert^{m-2} w_{t}(t)+ \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\vert\vert w_{t}(t-r) \vert^{m-2} w_{t}(t-r)dr+f_{2}(v, w) = 0.\\ v( z, -t) = v_{0}( z), \quad v_{t}( z, 0) = v_{1}( z), w( z, -t) = w_{0}( z), \quad w_{t}( z, 0) = w_{1}( z), \quad in \quad \Omega \\ v_{t}( z, -t) = j_{0}( z, t), \quad w_{t}( z, -t) = \varrho_{0}( z, t), \quad in \quad \Omega\times(0, \tau_{2})\\ v( z, t) = w(z, t) = 0, \quad in \quad \partial\Omega\times(0, \infty) \end{array} \right. \end{eqnarray}$

(1.1)

in which $G = \Omega\times(\tau_{1}, \tau_{2})\times(0, \infty)$ and $\tau_{1} < \tau_{2}$ are taken to be non-negative constants in a manner that $\beta _{2}$ , $\beta_{4} : [\tau_{1}, \tau_{2}] \rightarrow \mathbb{R}$ indicates distributive time delay while $g_{i}$ , $i = 1, 2$ are positive.

The viscoelastic damping term, whose kernel is the function $g$ , is a physical term used to describe the link between the strain and stress histories in a beam that was inspired by the Boltzmann theory. There are several publications that discuss this subject and produce a lot of fresh and original findings ^[1,2,3,4,5], particularly the hypotheses regarding the initial condition ^{[6,7,8,9,10,11,12]} and the kernel. See ^{[13,14,15,16,17]}. As it concerns to the plate equation and the span problem, Balakrishnan and Taylor introduced a novel damping model in ^[18] that they dubbed the Balakrishnan-Taylor damping. Here are a few studies that specifically addressed the research of this dampening for further information ^{[18,19,20,21,22,23]}.

Several applications and real-world issues are frequently affected by the delay, which transforms numerous systems into interesting research topics. Numerous writers have recently studied the stability of the evolution systems with time delays, particularly the effect of distributed delay. See ^[24,25,26].

In ^[1], the authors presented the stability result of the system over a considerably broader class of kernels in the absence of delay and Balakrishnan-Taylor damping $\xi_{0} = 1, \xi_{1} = \delta = \beta_{i} = 0, i = 1, \dots, 4$ .

Based on everything said above, one specific problem may be solved by combining these damping terms (distributed delay terms, Balakrishnan-Taylor damping and infinite memory), especially when the past history and the distributed delay

$\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{i}(r)\vert\vert u_{t}(t-r) \vert^{m-2} u_{t}(t-r)dr, \ \ \ \ i = 2, 4$

are added. We shall attempt to throw light on it since we think it represents a fresh topic that merits investigation and analysis in contrast to the ones mentioned before. Our study is structured into multiple sections: in the second section, we establish the assumptions, notions, and lemmas we require; in the final section, we substantiate our major finding.

2. Fundamental knowledge

In this section of the paper, we will introduce some basic results related to the theory for the analysis of our problem. Let us take the below:

(G1) $h_{i}:{\bf R}_{+}\rightarrow {\bf R}_{+}$ are a non-increasing $C^{1}$ functions fulfills the following

$\begin{equation} g_{i}(0) > 0, \quad , \quad \xi_{0}-\int_{0}^{\infty}h_{i}(s)ds = l_{i} > 0, \quad i = 1, 2, \end{equation}$

(2.1)

and

$g_{0} = \int_{0}^{\infty}h_{1}(s)ds, \quad \widehat{g}_{0} = \int_{0}^{\infty}g_{2}(s)ds,$

(G2) One can find a function $C^{1}$ functions $G_{i}:{\bf R}_{+}\rightarrow {\bf R}_{+}$ holds true $G_{i}(0) = G_{i}'(0) = 0$ .

The functions $G_{i}(t)$ are strictly increasing and convex of class $C^{2}({\bf R}_{+})$ on $(0, \varrho], \quad r\leq g_{i}(0)$ or linear in a manner that

$\begin{equation} g_{i}'(t)\leq -\zeta_{i}(t)G_{i}(g_{i}(t)), \quad \forall t\geq 0, \quad {\rm{for}} \quad i = 1, 2, \end{equation}$

(2.2)

in which $\zeta_{i}(t)$ are a $C^{1}$ functions fulfilling the below

$\begin{equation} \zeta_{i}(t) > 0, \quad \zeta_{i}'(t)\leq 0, \quad \forall t\geq 0. \end{equation}$

(2.3)

(G3) $\beta _{2}$ , $\beta _{4}:[\tau_{1}, \tau_{2}]\rightarrow {\bf R}$ are a bounded function fulfilling the below

$\begin{eqnarray} &&\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\vert dr < \beta_{1}, \\ &&\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\vert dr < \beta_{3}. \end{eqnarray}$

(2.4)

(G4) $f_{i}:{\bf R}^{2}\rightarrow {\bf R}$ are $C^{1}$ functions with $f_{i}(0, 0) = 0$ , and one can find a function $F$ in a way that

$\begin{eqnarray} &&f_{1}(c, e) = \frac{dF}{dc}(c, e), \quad f_{2}(c, e) = \frac{dF}{de}(c, e), \\ &&F\geq 0, \quad af_{1}(c, e)+ef_{2}(c, e) = F(c, e)\geq0, \end{eqnarray}$

(2.5)

and

$\begin{equation} \frac{df_{i}}{dc}(c, e)+\frac{df_{i}}{de}(c, e)\leq d(1+c^{p_{i}-1}+e^{p_{i}-1}). \quad \forall(c, e)\in{\bf R}^{2}. \end{equation}$

(2.6)

Take the below

$\begin{equation*} (g\circ\phi)(t): = \int_{\Omega}\int_{0}^{\infty}h(r)\vert \phi(t)-\phi(t-r)\vert^{2}dr dz, \label{A2.0} \end{equation*}$

and

$\begin{eqnarray*} M_{1}(t)&: = &\bigg(\xi_{0}+\xi_{1}\Vert\nabla v\Vert^{2}_{2}+\delta(\nabla v(t), \nabla v_{t}(t))_{L^{2}(\Omega)}\bigg), \notag\\ M_{2}(t)&: = &\bigg(\xi_{0}+\xi_{1}\Vert\nabla w\Vert^{2}_{2}+\delta(\nabla w(t), \nabla w_{t}(t))_{L^{2}(\Omega)}\bigg). \end{eqnarray*}$

Lemma 2.1. (Sobolev-Poincare inequality ^[27]). Assume that $2\leq q < \infty$ for $n = 1, 2$ and $2\leq q < \frac{2n}{n-2}$ for $n\geq3$ . Then, one can find $c_{*} = c(\Omega, q) > 0$ in a manner that

$\begin{equation*} \Vert v\Vert_{q}\leq c_{*}\Vert\nabla v\Vert_{2}, \quad \forall v\in G^{1}_{0}(\Omega). \end{equation*}$

Moreover, choose the below as in ^[26]:

$\begin{eqnarray*} &&x(z, \rho, r, t) = v_{t}(z, t-r\rho), \\ &&y(z, \rho, r, t) = w_{t}(z, t-r\rho) \end{eqnarray*}$

with

$\begin{equation} \left\{ \begin{array}{l} r x_{t}(z, \rho, r, t)+x_{\rho}(z, \rho, r, t) = 0, \quad s y_{t}(z, \rho, r, t)+y_{\rho}(z, \rho, r, t) = 0 \\ x(z, 0, r, t) = v_{t}(z, t), \quad y(z, 0, r, t) = w_{t}(z, t). \end{array} \right. \end{equation}$

(2.7)

Take the auxiliary variable (see ^[28])

$\begin{eqnarray*} &&\eta^{t}\left( z, s \right) = v \left( z, t\right) -v \left( z, t-s \right), s \geq 0, \\ &&\vartheta ^{t}\left( z, s \right) = w \left( z, t \right) -w \left( z, t- s \right), s \geq 0. \end{eqnarray*}$

Then

$\begin{eqnarray} && \eta _{t}^{t}\left( z, s \right) +\eta_{s}^{t}\left( z, s \right) = v _{t}\left( z, t\right), \\ &&\vartheta _{t}^{t}\left( z, s \right) +\vartheta_{s}^{t}\left( z, s \right) = w _{t}\left( z, t\right). \end{eqnarray}$

(2.8)

Rewrite the problem (1.1) as follows

$\begin{equation} \left\{ \begin{array}{l} v_{tt}-\bigg(l_{1}+\xi_{1}\Vert\nabla v\Vert^{2}_{2}+\delta (\nabla v, \nabla v_{t})_{L^{2}(\Omega)}\bigg)\Delta v(t)+ \int_{0}^{\infty}g_{1}(s)\Delta \eta^{t}(s)ds \\ \quad +\beta_{1}\vert v_{t}(t)\vert^{m-2} v_{t}(t)+ \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(s)\vert\vert x(z, 1, r, t) \vert^{m-2} x(z, 1, r, t)dr+f_{1}(v, w) = 0, \\ w_{tt}-\bigg(l_{2}+\xi_{1}\Vert\nabla w\Vert^{2}_{2}+\delta(\nabla w, \nabla w_{t})_{L^{2}(\Omega)}\bigg)\Delta w(t)+ \int_{0}^{\infty}g_{2}(s)\Delta \vartheta^{t}(s)ds\\ \quad +\beta_{3}\vert w_{t}(t)\vert^{m-2} w_{t}(t)+ \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\vert\vert y(z, 1, r, t) \vert^{m-2} y(z, 1, r, t)dr+f_{2}(v, w) = 0, \\ r x_{t}(z, \rho, r, t)+x_{\rho}(z, \rho, r, t) = 0, \\ r y_{t}(z, \rho, r, t)+y_{\rho}(z, \rho, r, t) = 0, \\ \eta _{t}^{t}\left( z, s \right) +\eta _{s}^{t}\left( z, s \right) = v_{t}\left( z, t\right)\\ \vartheta _{t}^{t}\left( z, s \right) +\vartheta _{s}^{t}\left( z, s \right) = w_{t}\left( z, t\right), \end{array} \right. \end{equation}$

(2.9)

where

$\begin{equation*} (z, \rho, r, t)\in \Omega\times(0, 1)\times(\tau_{1}, \tau_{2})\times(0, \infty). \end{equation*}$

with

$\begin{equation} \left\{ \begin{array}{l} v( z, -t) = v_{0}( z), \quad v_{t}( z, 0) = v_{1}( z), \quad w( z, -t) = w_{0}( z), \quad w_{t}( z, 0) = w_{1}( z), \quad in \quad \Omega \\ x(z, \rho, r, 0) = j_{0}(z, \rho r), \quad y(z, \rho, r, 0) = \varrho_{0}(z, \rho r), \quad in \quad \Omega\times(0, 1)\times(0, \tau_{2})\\ v( z, t) = \eta ^{t}\left( z, s \right) = 0, \quad z \in \quad \partial\Omega, \quad t, s \in(0, \infty), \\ \eta ^{t}\left( z, 0\right) = 0, \quad \forall t\geq 0 , \quad \eta ^{0}\left( z, s \right) = \eta _{0}\left(s \right) = 0, \quad \forall s \geq 0, \\ w( z, t) = \vartheta ^{t}\left( z, s \right) = 0, \quad z \in \quad \partial\Omega, \quad t, s \in(0, \infty), \\ \vartheta ^{t}\left( z, 0\right) = 0, \quad \forall t\geq 0 , \quad \vartheta ^{0}\left( z, s \right) = \vartheta _{0}\left(s \right) = 0, \quad \forall s \geq 0. \end{array} \right. \end{equation}$

(2.10)

In the upcoming Lemma, the energy functional will be introduced.

Lemma 2.2. Let the energy functional is symbolized by $E$ , then it is given by

$\begin{eqnarray} E( t) & = &\frac{1}{2}(\Vert v_{t}\Vert_{2}^{2}+\Vert w_{t}\Vert_{2}^{2})+\frac{\xi_{1}}{4}(\Vert \nabla v(t)\Vert_{2}^{4}+\Vert \nabla w(t)\Vert_{2}^{4})+\int_{\Omega}F(v, w)dz \\ &&+\frac{1}{2}\bigg(l_{1}\Vert\nabla v(t)\Vert_{2}^{2}+l_{2}\Vert\nabla w(t)\Vert_{2}^{2}\bigg)+\frac{1}{2}((g_{1}\circ \nabla v)(t)+(g_{2}\circ \nabla w)(t))\notag\\ &&+\frac{m-1}{m}\int_{0}^{1}\int_{ \tau_{1}}^{\tau_{2}}s\bigg(\vert \beta_{2}(r)\vert\Vert x(z, \rho, r, t)\Vert^{m}_{m}+\vert \beta_{4}(r)\vert\Vert y(z, \rho, r, t)\Vert^{m}_{m}\bigg) dr d\rho.\\ \end{eqnarray}$

(2.11)

The above fulfills the below

$\begin{eqnarray} E^{\prime }\left( t\right) &\leq&-\gamma_{0}\bigg(\Vert v_{t}(t)\Vert^{m}_{m}+\Vert w_{t}(t)\Vert^{m}_{m}\bigg)+\frac{1}{2}\bigg((g_{1}^{\prime }\circ\nabla v)(t)+(g_{2}^{\prime }\circ\nabla w)(t)\bigg)\\ &&-\frac{\delta}{4} \bigg\{\bigg(\frac{d}{dt}\bigg\{\Vert \nabla v(t)\Vert^{2}_{2}\bigg\}\bigg)^{2}+\bigg(\frac{d}{dt}\bigg\{\Vert \nabla w(t)\Vert^{2}_{2}\bigg\}\bigg)^{2}\bigg\}\leq 0, \end{eqnarray}$

(2.12)

in which $\gamma_{0} = \min\{\beta _{1}-\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert dr, \quad \beta _{3}-\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{4}(r)\vert dr \}$ .

Proof. To prove the result, we take the inner product of $(2.9)$ with $v_{t}, w_{t}$ and after that integrating over $\Omega$ , the following is obtained

$\begin{eqnarray} &&( v_{tt}(t), v_{t}(t))_{L^{2}(\Omega)}-(M_{3}(t)\Delta v(t), v_{t}(t))_{L^{2}(\Omega)} \\ &&+(\int_{0}^{\infty}h_{1}(s)\Delta \eta^{t}(s)ds, v_{t}(t))_{L^{2}(\Omega)}+\beta_{1}( \vert v_{t}\vert^{m-2}v_{t}, v_{t})_{L^{2}(\Omega)}\\ &&+\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(s)\vert(\vert x(z, 1, r, t)\vert^{m-2} x(z, 1, r, t), v_{t}(t))_{L^{2}(\Omega)}dr\\ &&+( w_{tt}(t), w_{t}(t))_{L^{2}(\Omega)}-(M_{4}(t)\Delta w(t), w_{t}(t))_{L^{2}(\Omega)} \\ &&+(\int_{0}^{\infty}h_{2}(s)\Delta \vartheta^{t}(s)ds, w_{t}(t))_{L^{2}(\Omega)}+\beta_{3}( \vert w_{t}\vert^{m-2}w_{t}, w_{t})_{L^{2}(\Omega)}\\ &&+\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{4}(s)\vert(\vert y(z, 1, r, t)\vert^{m-2} y(z, 1, r, t), w_{t}(t))_{L^{2}(\Omega)}dr\\ &&+( f_{1}(v, w), v_{t}(t))_{L^{2}(\Omega)}+( f_{2}(v, w), w_{t}(t))_{L^{2}(\Omega)} = 0. \end{eqnarray}$

(2.13)

in which

$\begin{eqnarray*} M_{3}(t)&: = &\bigg(l_{1}+\xi_{1}\Vert\nabla v\Vert^{2}_{2}+\delta(\nabla v(t), \nabla v_{t}(t))_{L^{2}(\Omega)}\bigg), \notag\\ M_{4}(t)&: = &\bigg(l_{2}+\xi_{1}\Vert\nabla w\Vert^{2}_{2}+\delta(\nabla w(t), \nabla w_{t}(t))_{L^{2}(\Omega)}\bigg). \end{eqnarray*}$

Using mathematical skills, the following is obtained

$\begin{eqnarray} (v_{tt}(t), v_{t}(t))_{L^{2}(\Omega)} = \dfrac{1}{2}\dfrac{d}{dt}\bigg(\Vert v_{t}(t)\Vert^{2}_{2}\bigg), \end{eqnarray}$

(2.14)

further simplification leads us to the following

$\begin{eqnarray} &&-(M_{3}(t)\Delta v(t), v_{t}(t))_{L^{2}(\Omega)} \\ & = &-(\bigg(l_{1}+\xi_{1}\Vert\nabla v\Vert^{2}_{2}+\delta(\nabla v(t), \nabla v_{t}(t))_{L^{2}(\Omega)}\bigg)\Delta v(t), v_{t}(t))_{L^{2}(\Omega)}\\ & = &\bigg(l_{1}+\xi_{1}\Vert\nabla v\Vert^{2}_{2}+\delta(\nabla v(t), \nabla v_{t}(t))_{L^{2}(\Omega)}\bigg)\int_{\Omega}\nabla v(t).\nabla v_{t}(t)dz\\ & = &\bigg(l_{1}+\xi_{1}\Vert\nabla v\Vert^{2}_{2}+\delta(\nabla v(t), \nabla v_{t}(t))_{L^{2}(\Omega)}\bigg)\frac{d}{dt}\bigg\{\int_{\Omega}\vert\nabla v(t)\vert^{2}dz\bigg\}\\ & = &\dfrac{d}{dt}\bigg\{\frac{1}{2}\bigg(l_{1}+\frac{\xi_{1}}{2}\Vert\nabla v\Vert^{2}_{2}\bigg)\Vert\nabla v(t)\Vert^{2}_{2}\bigg\}+\frac{\delta}{4}\dfrac{d}{dt}\bigg\{\Vert\nabla v(t)\Vert^{2}_{2}\bigg\}^{2}.\\ \end{eqnarray}$

(2.15)

The following is obtained after calculation

$\begin{eqnarray} (\int_{0}^{\infty}g_{1}(s)\Delta \eta^{t}(s)ds, v_{t}(t))_{L^{2}(\Omega)}& = &\int_{\Omega}\nabla v_{t}\int_{0}^{\infty}g_{1}(s)\nabla \eta^{t}(s)ds dz\\ & = &\int_{0}^{\infty}g_{1}(s)\int_{\Omega}\nabla v_{t}\nabla\eta^{t}(s) dz ds\\ & = &\int_{0}^{\infty}g_{1}(s)\int_{\Omega}(\nabla\eta^{t}_{t}+\nabla\eta^{t}_{s})\nabla\eta^{t}(s)dz ds \\ & = &\int_{0}^{\infty} g_{1}(s)\int_{\Omega}\nabla\eta^{t}_{t}\nabla\eta^{t}(s) dz ds \\ &&+\int_{\Omega}\int_{0}^{\infty}g_{1}(s)\nabla\eta^{t}_{s}\nabla\eta^{t}(\nabla)d\nabla dz \\ & = &\frac{1}{2}\frac{d}{dt}(g_{1}\circ\nabla v)(t)-\frac{1}{2}(g_{1}'\circ\nabla v)(t). \end{eqnarray}$

(2.16)

In the same way, we have

$\begin{eqnarray} (w_{tt}(t), w_{t}(t))_{L^{2}(\Omega)}& = &\dfrac{1}{2}\dfrac{d}{dt}\bigg(\Vert w_{t}(t)\Vert^{2}_{2}\bigg), \\ -(M_{4}(t)\Delta w(t), w_{t}(t))_{L^{2}(\Omega)}& = &\dfrac{d}{dt}\bigg\{\frac{1}{2}\bigg(l_{2}+\frac{\xi_{1}}{2}\Vert\nabla w \Vert^{2}_{2}\bigg)\Vert\nabla w(t)\Vert^{2}_{2}\bigg\}\\ &&+\frac{\delta}{4}\dfrac{d}{dt}\bigg\{\Vert\nabla w(t)\Vert^{2}_{2}\bigg\}^{2}, \\ (\int_{0}^{\infty}g_{2}(s)\Delta \vartheta^{t}(s)ds, w_{t}(t))_{L^{2}(\Omega)}& = &\frac{1}{2}\frac{d}{dt}(g_{2}\circ\nabla w)(t)-\frac{1}{2}(g_{2}'\circ\nabla w)(t). \end{eqnarray}$

(2.17)

Now, multiplying the equation $(2.9)$ by $- x \vert \beta_{2}(r)\vert, \quad - y\vert \beta_{4}(r)\vert$ , and integrating over $\Omega\times(0, 1)\times (\tau_{1}, \tau_{2})$ and utilizing $(2.7)$ , the below is obtained

$\begin{eqnarray} &\dfrac{d}{dt }&\frac{m-1}{m}\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}}r \vert \beta_{2}(r)\vert. \vert x(z, \rho, r, t)\vert^{m}dr d\rho dz \\ & = & -(m-1)\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert .\vert y\vert^{m-1} x_{\rho }dr d\rho dz \\ & = &-\frac{m-1}{m}\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert\frac{d}{ d\rho}\vert x(z, \rho, r, t)\vert^{m}dr d\rho dz \\ & = &\frac{m-1}{m}\int_{\Omega}\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert \bigg(\vert x( z, 0 , r, t)\vert^{m} -\vert x(z, 1, r, t)\vert^{m}\bigg)dr dz \\ & = &\frac{m-1}{m}\bigg(\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert dr \bigg)\int_{\Omega}\vert v_{t}(t)\vert^{m}dz\\ &&-\frac{m-1}{m} \int_{\Omega}\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert.\vert x\left( z, 1, r, t\right)\vert^{m} dr dz\\ & = &\frac{m-1}{m}\bigg(\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert dr\bigg) \Vert v_{t}(t)\Vert_{m}^{m}\\ &&-\frac{m-1}{m} \int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert \Vert x\left( z, 1, r, t\right)\Vert_{m}^{m} dr. \end{eqnarray}$

(2.18)

Similarly, we have

$\begin{eqnarray} &&\dfrac{d}{dt }\frac{m-1}{m}\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}}r \vert \beta_{4}(r)\vert. \vert y(z, \rho, r, t)\vert^{m}dr d\rho dz \\ & = &\frac{m-1}{m}\bigg(\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{4}(r)\vert dr\bigg) \Vert w_{t}(t)\Vert_{m}^{m}-\frac{m-1}{m} \int_{\tau_{1}}^{\tau_{2}}\vert \beta_{4}(r)\vert \Vert y\left( z, 1, r, t\right)\Vert_{m}^{m} dr . \end{eqnarray}$

(2.19)

Here, we utilize the inequalities of Young as

$\begin{eqnarray} &&\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert\bigg(\vert x(z, 1, r, t)\vert^{m-2}x(z, 1, r, t), v_{t}(t)\bigg)_{L^{2}(\Omega)}ds \\ &\leq&\frac{1}{m}\bigg(\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert dr\bigg)\Vert v_{t}(t)\Vert_{m}^{m}+\frac{m-1}{m}\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{2}(r)\vert \Vert x\left( z, 1, r, t\right)\Vert_{m}^{m} dr, \end{eqnarray}$

(2.20)

and

$\begin{eqnarray} &&\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{4}(r)\vert\bigg(\vert y(z, 1, r, t)\vert^{m-2}y(z, 1, r, t), w_{t}(t)\bigg)_{L^{2}(\Omega)}dr \\ &\leq&\frac{1}{m}\bigg(\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{4}(r)\vert dr\bigg)\Vert w_{t}(t)\Vert_{m}^{m}+\frac{m-1}{m}\int_{\tau_{1}}^{\tau_{2}}\vert \beta_{4}(r)\vert \Vert y\left( z, 1, r, t\right)\Vert_{m}^{m} dr. \end{eqnarray}$

(2.21)

Finally, we have

$\begin{equation} ( f_{1}(v, w), v_{t}(t))_{L^{2}(\Omega)}+( f_{2}(v, w), w_{t}(t))_{L^{2}(\Omega)} = \frac{d}{dt}\int_{\Omega}F(v, w)dz. \end{equation}$

(2.22)

Thus, after replacement of (2.14)–(2.22) into (2.13), we determined (2.11) and (2.12). As a result, we obtained that $E$ is a non-increasing function by (2.2)–(2.5), which is required.

Theorem 2.3. Take the function ${\bf U} = (v, v_{t}, w, w_{t}, x, y, \eta^{t}, \vartheta^{t})^{T}$ and assume that (2.1)–(2.5) holds true. Then, for any ${\bf U}_{0}\in \mathcal{H}$ , then one can find a unique solution ${\bf U}$ of problems (2.9) and (2.10) in a manner that

$\begin{eqnarray*} &&U\in C({\bf R}_{+}, \mathcal{G}). \end{eqnarray*}$

If ${\bf U}_{0}\in \mathcal{G}_{1}$ , then $U$ fulfills the following

$\begin{eqnarray*} &&U\in C^{1}({\bf R}_{+}, \mathcal{G})\cap C({\bf R}_{+}, \mathcal{G}_{1}), \end{eqnarray*}$

in which

$\begin{eqnarray*} \mathcal{G}& = &(G^{1}_{0}(\Omega)\times L^{2}(\Omega))^{2}\times (L^{2}(\Omega, (0, 1), (\tau_{1}, \tau_{2})))^{2}\times (L_{g_{1}}\times L_{g_{2}}).\\ \mathcal{G}_{1}& = &\bigg\{U\in\mathcal{G}/v, w\in G^{2}\cap G^{1}_{0}, v_{t}, w_{t}\in G^{1}_{0}(\Omega), x, y, x_{\rho}, y_{\rho}\in L^{2}(\Omega, (0, 1), (\tau_{1}, \tau_{2})), \notag\\ && \quad (\eta^{t}, \vartheta^{t})\in L_{g_{1}}\times L_{g_{2}}, \eta^{t}(z, 0) = \vartheta^{t}(z, 0) = 0, x(z, 0, r, t) = v_{t}, \notag\\ && \quad y(z, 0, r, t) = w_{t} \bigg\}. \end{eqnarray*}$

3. Analysis of stability

Here, the stability of the systems (2.9) and (2.10) will be established and investigated. For which the following lemma is needed

Lemma 3.1. Let us suppose that (2.1) and (2.2) fulfills.

$\begin{eqnarray} \int_{\Omega}\bigg(\int_{0}^{\infty}g_{i}(s)(v(t)-v(t-s))ds \bigg)^{2}dz\leq C_{\kappa, i}(h_{i}\circ v)(t), \quad i = 1, 2. \end{eqnarray}$

(3.1)

where

$\begin{eqnarray*} C_{\kappa_{i}} &: = & \int_{0}^{\infty}\frac{g_{i}^{2}(s)}{\kappa g_{i}(s)-g_{i}'(s)}ds \notag\\ h_{i}(t)&: = &\kappa g_{i}(t)-g_{i}'(t), \quad i = 1, 2. \end{eqnarray*}$

Proof.

$\begin{eqnarray} &&\int_{\Omega}\bigg(\int_{0}^{\infty}g_{i}(s)(v(t)-v(t-s))ds \bigg)^{2}dz\\ & = &\int_{\Omega}\bigg(\int_{-\infty}^{t}g_{i}(t-s)(v(t)-v(t-s))ds \bigg)^{2}dz\\ & = &\int_{\Omega}\bigg(\int_{-\infty}^{t}\frac{g_{i}(t-s)}{\sqrt{\kappa g_{i}(t-s)-g_{i}'(t-s)}}\sqrt{\kappa g_{i}(t-s)-g_{i}'(t-s)}\\ && \quad (v(t)-v(s))ds \bigg)^{2}dz\\ \end{eqnarray}$

(3.2)

which is obtained through Young's inequality (Eq 3.1).

Lemma 3.2. (Jensens inequality). Let $f : \Omega \rightarrow [c, e]$ and $h : \Omega \rightarrow {\bf R}$ are integrable functions in a manner that for any $z \in \Omega$ , $h(z) > 0$ and $\int_{\Omega}h(z)dz = k > 0$ . Furthermore, assume a convex function G such that $G : [c, e]\rightarrow {\bf R}$ . Then

$\begin{equation} G\bigg(\frac{1}{k}\int_{\Omega}f(z)h(z)dz\bigg) < \frac{1}{k}\int_{\Omega}G( f(z))h(z)dz. \end{equation}$

(3.3)

Lemma 3.3. It is mentioned in ^[12] that one can find a positive constant $\beta$ , $\widehat{\beta}$ in a manner that

$\begin{eqnarray} I_{1}(t)& = &\int_{\Omega}\int_{t}^{\infty}g_{1}(s)\vert\nabla\eta^{t}(\delta)\vert^{2}ds dz\leq\beta\mu(t), \\ I_{2}(t)& = &\int_{\Omega}\int_{t}^{\infty}g_{2}(s)\vert\nabla\vartheta^{t}(\delta)\vert^{2}ds dz\leq\widehat{\beta}\widehat{\mu}(t), \end{eqnarray}$

(3.4)

in which

$\begin{eqnarray*} \mu(t)& = &\int_{0}^{\infty}g_{1}(t+s)\bigg(1+\int_{\Omega}\nabla v_{0}^{2}(z, s)dz\bigg)ds, \notag\\ \widehat{\mu}(t)& = &\int_{0}^{\infty}g_{2}(t+s)\bigg(1+\int_{\Omega}\nabla w_{0}^{2}(z, s)dz\bigg)ds. \end{eqnarray*}$

Proof. As the function $E(t)$ is decreasing and utilizing (2.11), we have the following

$\begin{eqnarray} \int_{\Omega}\vert\nabla\eta^{t}(s)\vert^{2} dz & = & \int_{\Omega}(\nabla v(z, t)-v(z, t-s)^{2} dz\\ &\leq&2\int_{\Omega}\nabla v^{2}(z, t) dz+2\int_{\Omega}\nabla v^{2}(z, t-s) dz\\ &\leq&2\sup\limits_{s > 0}\int_{\Omega}\nabla v^{2}(z, s) dz+2\int_{\Omega}\nabla v^{2}(z, t-x) dz\\ &\leq&\frac{4E(0)}{l_{1}}+2\int_{\Omega}\nabla v^{2}(z, t-s) dz, \end{eqnarray}$

(3.5)

for any $t, s \geq 0$ . Further, we have

$\begin{eqnarray} I_{1}(t) &\leq&\frac{4E(0)}{l_{1}}\int_{t}^{\infty}g_{1}(s)ds+2\int_{t}^{\infty}g_{1}(s)\int_{\Omega}\nabla v^{2}(z, t-s) dz ds \\ &\leq&\frac{4E(0)}{l_{1}}\int_{0}^{\infty}g_{1}(t+s)ds +2\int_{0}^{\infty}g_{1}(t+s)\int_{\Omega}\nabla v_{0}^{2}(z, s) dz ds \\ &\leq&\beta\mu(t), \end{eqnarray}$

(3.6)

in which $\beta = \max\{\frac{4E(0)}{l_{1}}, 2\}$ and $\mu(t) = \int_{0}^{\infty}g_{1}(t+s)(1+\int_{\Omega}\nabla u_{0}^{2}(z, s) dz)ds$ .

In the same way, we can deduce that

$\begin{eqnarray} I_{2}(t) &\leq&\frac{4E(0)}{l_{2}}\int_{0}^{\infty}g_{2}(t+s)ds+2\int_{0}^{\infty}g_{2}(t+s)\int_{\Omega}\nabla w_{0}^{2}(z, s) dz ds \\ &\leq&\widehat{\beta}\widehat{\mu}(t), \end{eqnarray}$

(3.7)

in which $\widehat{\beta} = \max\{\frac{4E(0)}{l_{2}}, 2\}$ and $\widehat{\mu}(t) = \int_{0}^{\infty}g_{2}(t+s)(1+\int_{\Omega}\nabla w_{0}^{2}(z, s) dz)ds$ . In the upcoming part, we set the following

$\begin{eqnarray} \Psi(t)&: = &\int_{\Omega}\bigg( v(t)v _{t}(t)+w(t)w _{t}(t)\bigg) dz+\frac{\delta}{4}\bigg(\Vert\nabla v(t)\Vert_{2}^{4}+\Vert\nabla w(t)\Vert_{2}^{4}\bigg), \end{eqnarray}$

(3.8)

and

$\begin{eqnarray} \Phi(t)&: = &-\int_{\Omega}v_{t}\int_{0}^{\infty}g_{1}(s)(v(t)-v(t-s))ds dz\\ &&-\int_{\Omega}w_{t}\int_{0}^{\infty}g_{2}(s)(w(t)-w(t-s))ds dz, \end{eqnarray}$

(3.9)

and

$\begin{eqnarray} \Theta(t)&: = &\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}}r e^{-\rho r}\bigg(\vert \beta_{2}(r)\vert. \Vert x(z, \rho, r, t)\Vert_{m}^{m}+\vert \beta_{4}(r)\vert. \Vert y(z, \rho, r, t)\Vert_{m}^{m}\bigg) dr d\rho .\\ \end{eqnarray}$

(3.10)

Lemma 3.4. In (3.8), the functional $\Psi(t)$ fulfills the following

$\begin{eqnarray} \Psi'(t)&\leq &\Vert v_{t}\Vert_{2}^{2}+\Vert w_{t}\Vert_{2}^{2} -(l-\varepsilon(c_{1}+c_{2})-\sigma_{1})\bigg(\Vert \nabla v \Vert_{2}^{2}+\Vert \nabla w\Vert_{2}^{2}\bigg) \\ &&-\xi_{1}\bigg(\Vert\nabla v\Vert_{2}^{4}+\Vert\nabla w\Vert_{2}^{4}\bigg)+c(\varepsilon)\bigg(\Vert v_{t}\Vert_{m}^{m}+\Vert w_{t}\Vert_{m}^{m}\bigg)\\ &&+c(\sigma_{1})\bigg(C_{\kappa, 1}(g_{1}\circ\nabla v)(t)+C_{\kappa, 2}(h_{2}\circ\nabla w)(t)\bigg)-\int_{\Omega}F(v, w)dz\\ &&+c(\varepsilon)\int_{\tau_{1}}^{\tau_{2}}\bigg(\vert\beta_{2}(r)\Vert x(z, 1, r, t)\Vert_{m}^{m}+\vert\beta_{4}(r)\Vert y(z, 1, r, t)\Vert_{m}^{m}\bigg)dr. \end{eqnarray}$

(3.11)

for any $\varepsilon, \sigma_{1} > 0$ with $l = \min\{l_{1}, l_{2}\}$ .

Proof. To prove the result, differentiate (3.8) first and then apply $(2.9)$ , we have the following

$\begin{eqnarray} \Psi'(t)& = &\Vert v_{t}\Vert_{2}^{2}+\int_{\Omega}v_{tt}vdz+\delta\Vert\nabla v\Vert_{2}^{2}\int_{\Omega}\nabla v_{t}\nabla vdz\\ &&+\Vert w_{t}\Vert_{2}^{2}+\int_{\Omega}w_{tt}wdz+\delta\Vert\nabla w\Vert_{2}^{2}\int_{\Omega}\nabla w_{t}\nabla wdz\\ & = &\Vert v_{t}\Vert_{2}^{2}+\Vert w_{t}\Vert_{2}^{2}-\xi_{0}(\Vert\nabla v\Vert_{2}^{2}+\Vert\nabla w\Vert_{2}^{2})-\xi_{1}(\Vert\nabla v\Vert_{2}^{4}+\Vert\nabla w\Vert_{2}^{4})\\ &&-\underbrace{\beta_{1}\int_{\Omega}\vert v_{t}\vert^{m-2} v_{t}vdz}_{I_{11}}-\underbrace{\beta_{3}\int_{\Omega}\vert w_{t}\vert^{m-2} w_{t}wdz}_{I_{12}}\\ &&+\underbrace{\int_{\Omega}\nabla v(t)\int_{0}^{\infty}g_{1}(s)\nabla v(t-s)ds dz}_{I_{21}}\\ &&+\underbrace{\int_{\Omega}\nabla w(t)\int_{0}^{\infty}g_{2}(s)\nabla w(t-s)ds dz}_{I_{22}}\\ &&-\underbrace{ \int_{\Omega}\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\vert\vert x(z, 1, r, t) \vert^{m-2} x(z, 1, r, t)v dr dz}_{I_{31}}\\ &&-\underbrace{ \int_{\Omega}\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\vert\vert y(z, 1, r, t) \vert^{m-2} y(z, 1, r, t)w dr dz}_{I_{32}}\\ &&-\underbrace{\int_{\Omega}(vf_{1}(v, w)+wf_{2}(v, w))dz}_{I_{4}}. \end{eqnarray}$

(3.12)

We estimate the last 6 terms of the RHS of (3.12). The following is obtained by applying Young's, Sobolev-Poincare and Hölder's inequalities on (2.1) and (2.11), we have

$\begin{eqnarray} I_{11}&\leq&\varepsilon \beta_{1}^{m}\Vert v\Vert_{m}^{m}+c(\varepsilon)\Vert v_{t}\Vert_{m}^{m}\\ &\leq&\varepsilon \beta_{1}^{m}c_{p}^{m}\Vert\nabla v\Vert_{2}^{m}+c(\varepsilon)\Vert v_{t}\Vert_{m}^{m}\\ &\leq&\varepsilon \beta_{1}^{m}c_{p}^{m}\bigg(\frac{E(0)}{l_{1}}\bigg)^{(m-2)/2}\Vert\nabla v\Vert_{2}^{2}+c(\varepsilon)\Vert v_{t}\Vert_{m}^{m}\\ &\leq&\varepsilon c_{11}\Vert\nabla v\Vert_{2}^{2}+c(\varepsilon)\Vert v_{t}\Vert_{m}^{m}. \end{eqnarray}$

(3.13)

In addition to this, for any $\sigma_{1} > 0$ , by Lemma 3.1, we have the below

$\begin{eqnarray} I_{21}&\leq&(\int_{0}^{\infty}g_{1}(s)ds)\Vert\nabla v\Vert_{2}^{2}-\int_{\Omega}\nabla v(t)\int_{0}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))ds dz\\ &\leq&(\xi_{0}-l_{1}+\sigma_{1})\Vert\nabla v\Vert_{2}^{2}+\frac{c}{\sigma_{1}}C_{\kappa, 1}(h_{1}\circ\nabla v)(t). \end{eqnarray}$

(3.14)

Taking same steps to $I_{12}$ , the below is obtained

$\begin{eqnarray} I_{31}&\leq&\varepsilon c_{21}\Vert\nabla v\Vert_{2}^{2}+c(\varepsilon)\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\vert.\Vert x(z, 1, r, t)\Vert_{m}^{m}dr. \end{eqnarray}$

(3.15)

Same steps for $I_{11}, I_{21}$ and $I_{31}$ , we have

$\begin{eqnarray} I_{12}&\leq&\varepsilon c_{12}\Vert\nabla w\Vert_{2}^{2}+c(\varepsilon)\Vert w_{t}\Vert_{m}^{m}\\ I_{22}&\leq&(\xi_{0}-l_{2}+\sigma_{1})\Vert\nabla w\Vert_{2}^{2}+\frac{c}{\sigma_{1}}C_{\kappa, 2}(h_{2}\circ\nabla w)(t), \\ I_{32}&\leq&\varepsilon c_{22}\Vert\nabla w\Vert_{2}^{2}+c(\varepsilon)\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\vert.\Vert y(z, 1, r, t)\Vert_{m}^{m}dr. \end{eqnarray}$

(3.16)

Combining (3.13)–(3.21), (3.12) and (2.5), the required (3.11) is obtained.

Lemma 3.5. For any $\sigma, \sigma_{2}, \sigma_{3} > 0$ , the functional $\Phi(t)$ introduced in (3.9) holds true

$\begin{eqnarray} \Phi'(t)&\leq&-\bigg(l_{0}-\sigma_{3}\bigg)\bigg(\Vert v_{t}\Vert_{2}^{2}+\Vert w_{t}\Vert_{2}^{2}\bigg)+\xi_{1}\sigma\bigg(\Vert\nabla v\Vert_{2}^{4}+\Vert\nabla w\Vert_{2}^{4}\bigg)\\ &&+\sigma\bigg(\xi_{0}+\widehat{l_{0}}^{2}+c\widehat{l}\bigg)\Vert\nabla v\Vert_{2}^{2}+\sigma\bigg(\xi_{0}+\widehat{h}_{0}^{2}+cl_{2}\bigg)\Vert\nabla w\Vert_{2}^{2}\\ &&+\sigma_{2}2\delta E(0)\bigg(\frac{1}{l_{1}}\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla v\Vert_{2}^{2}\bigg)^{2}+\frac{1}{l_{2}}\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla w\Vert_{2}^{2}\bigg)^{2}\bigg)\\ &&+c(\sigma, \sigma_{2}, \sigma_{3})\bigg(C_{\kappa, 1} (h_{1}\circ\nabla v)(t)+C_{\kappa, 2} (h_{2}\circ\nabla w)(t)\bigg)\\ &&+c(\sigma)\bigg(\Vert v_{t}\Vert_{m}^{m}+\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\Vert x(z, 1, r, t)\Vert_{m}^{m}dr\bigg)\\ &&+c(\sigma)\bigg(\Vert w_{t}\Vert_{m}^{m}+\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\Vert y(z, 1, r, t)\Vert_{m}^{m}dr\bigg). \end{eqnarray}$

(3.17)

where $\widehat{l} = \max\{l_{1}, l_{2}\}$ , $l_{0} = \min\{g_{0}, \widehat{g}_{0}\}$ and $\widehat{l_{0}} = \max\{g_{0}, \widehat{g}_{0}\}$ .

Proof. To prove the result, simplification of (3.9) and $(2.9)$ through mathematical skills leads us to the following

$\begin{eqnarray} \Phi'(t)& = &-\int_{\Omega}v_{tt}\int_{0}^{\infty}g_{1}(s)(v(t)-v(t-s))ds dz \\ &&-\int_{\Omega}v_{t}\frac{\partial}{\partial t}\bigg(\int_{0}^{\infty}g_{1}(s)(v(t)-v(t-s))ds \bigg) dz\\ &&-\int_{\Omega}w_{tt}\int_{0}^{\infty}g_{2}(s)(w(t)-w(t-s))ds dz\\ &&-\int_{\Omega}w_{t}\frac{\partial}{\partial t}\bigg(\int_{0}^{\infty}g_{2}(s)(w(t)-w(t-s))ds \bigg) dz\\ & = &\underbrace{(\xi_{0}+\xi_{1}\Vert\nabla v\Vert_{2}^{2})\int_{\Omega}\nabla v\int_{0}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))ds dz}_{J_{11}}\\ &&+\underbrace{(\xi_{0}+\xi_{1}\Vert\nabla w\Vert_{2}^{2})\int_{\Omega}\nabla w\int_{0}^{\infty}g_{2}(s)(\nabla w(t)-\nabla w(t-s))ds dz}_{J_{12}}\\ &&\underbrace{+\delta\int_{\Omega}\nabla v\nabla v_{t}dz.\int_{\Omega}\nabla v\int_{0}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))ds dz}_{J_{21}}\\ &&\underbrace{+\delta\int_{\Omega}\nabla w\nabla w_{t}dz.\int_{\Omega}\nabla w\int_{0}^{\infty}g_{2}(s)(\nabla w(t)-\nabla w(t-s))ds dz}_{J_{22}}\\ &&\underbrace{-\int_{\Omega}\bigg(\int_{0}^{\infty}g_{1}(s)\nabla v(t-s)ds \bigg).\bigg(\int_{0}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))ds \bigg)dz}_{J_{31}}\\ &&\underbrace{-\int_{\Omega}\bigg(\int_{0}^{\infty}g_{2}(s)\nabla w(t-s)ds \bigg).\bigg(\int_{0}^{\infty}g_{2}(s)(\nabla w(t)-\nabla w(t-s))ds \bigg)dz}_{J_{32}}\\ &&\underbrace{-\beta_{1}\int_{\Omega}\vert v_{t}\vert^{m-2} v_{t}\bigg(\int_{0}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))ds \bigg) dz}_{J_{41}}\\ &&\underbrace{-\beta_{3}\int_{\Omega}\vert w_{t}\vert^{m-2} w_{t}\bigg(\int_{0}^{\infty}g_{2}(s)(\nabla w(t)-\nabla w(t-s))ds \bigg) dx}_{J_{42}}\\ &&- \int_{\Omega}\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\vert\vert x(z, 1, r, t) \vert^{m-2} x(z, 1, r, t)\\ &&\underbrace{\times\int_{0}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))ds \bigg) ds dz}_{J_{51}}\\ &&- \int_{\Omega}\int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\vert\vert y(z, 1, r, t) \vert^{m-2} y(z, 1, r, t)\\ &&\underbrace{\times\int_{0}^{\infty}g_{2}(s)(\nabla w(t)-\nabla w(t-s))ds \bigg) ds dz}_{J_{51}}\\ &&\underbrace{-\int_{\Omega}v_{t}\frac{\partial}{\partial t}\bigg(\int_{0}^{\infty}g(s)(v(t)-v(t-s))ds \bigg) dz}_{J_{61}}\\ &&\underbrace{-\int_{\Omega}w_{t}\frac{\partial}{\partial t}\bigg(\int_{0}^{\infty}g_{2}(s)(w(t)-w(t-s))ds \bigg) dz}_{J_{62}}\\ &&\underbrace{-\int_{\Omega}f_{1}(v, w).\bigg(\int_{0}^{\infty}g_{1}(s)( v(t)- v(t-s))ds \bigg)dz}_{J_{71}}\\ &&\underbrace{-\int_{\Omega}f_{2}(v, w).\bigg(\int_{0}^{\infty}g_{2}(s)( w(t)- w(t-s))ds \bigg) dz}_{J_{72}}. \end{eqnarray}$

(3.18)

Here, we will find our the approximation of the terms of the RHS of (3.18). Using the well-known Young's, Sobolev-Poincare and Hölder's inequalities on (2.1), (2.11) and Lemma 3.1, we proceed as follows

$\begin{eqnarray} \vert J_{11}\vert&\leq&(\xi_{0}+\xi_{1}\Vert\nabla v\Vert_{2}^{2})\bigg(\sigma\Vert\nabla v\Vert_{2}^{2}+\frac{1}{4\sigma}C_{\kappa, 1}(h_{1}\circ\nabla v)(t)\bigg)\\ &\leq&\sigma\xi_{0}\Vert\nabla v\Vert_{2}^{2}+\sigma\xi_{1}\Vert\nabla v\Vert_{2}^{4}+\bigg(\frac{\xi_{0}}{4\sigma}+\frac{\xi_{1}E(0)}{4l_{1}\xi}\bigg)C_{\kappa, 1}(h_{1}\circ\nabla v)(t), \\ \end{eqnarray}$

(3.19)

and

$\begin{eqnarray} J_{21}&\leq&\sigma_{2}\delta\bigg(\int_{\Omega}\nabla v\nabla v_{t}dz\bigg)^{2}\Vert\nabla v\Vert_{2}^{2}+\frac{\delta}{4\sigma_{2}}C_{\kappa, 1}(h_{1}\circ\nabla v)(t)\\ &\leq&\sigma_{2}\frac{2\delta E(0)}{l_{1}}\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla v\Vert_{2}^{2}\bigg)^{2}+\frac{\delta}{4\sigma_{2}}C_{\kappa, 1}(h_{1}\circ\nabla v)(t), \end{eqnarray}$

(3.20)

$\begin{eqnarray} \vert J_{31}\vert&\leq&\int_{\Omega}\bigg(\int_{0}^{\infty}g_{1}(s)\nabla v(t)ds \bigg)\bigg(\int_{0}^{\infty}g_{1}(s)(\nabla v(t-s)-\nabla v(t))ds \bigg)dz \\ &&-\int_{\Omega}\bigg(\int_{0}^{\infty}g_{1}(s)( \nabla v(t)-\nabla v(t-s))ds \bigg)^{2}dz \\ &\leq&\delta g_{0}^{2}\Vert\nabla v\Vert_{2}^{2}+\bigg(1+\frac{1}{4\delta}\bigg)C_{\kappa, 1}(h_{1}\circ\nabla v)(t), \end{eqnarray}$

(3.21)

$\begin{eqnarray} \vert J_{41}\vert&\leq&c(\sigma)\Vert\nabla v_{t}\Vert_{m}^{m}+\sigma\beta_{1}^{m}\int_{\Omega}\bigg(\int_{0}^{\infty}g_{1}(s)( v(t)- v(t-s))ds \bigg)^{m}dz\\ &\leq&c(\sigma)\Vert\nabla v_{t}\Vert_{m}^{m}+\sigma\bigg(\beta_{1}^{m}c_{p}^{m}[\frac{4g_{0} E(0)}{l_{1}}]^{(m-2)}\bigg)C_{\kappa, 1}(h_{1}\circ\nabla v)(t)\\ &\leq&c(\sigma)\Vert\nabla v_{t}\Vert_{m}^{m}+\sigma c_{3}C_{\kappa, 1}(h_{1}\circ\nabla v)(t). \end{eqnarray}$

(3.22)

In the same, we obtained the following

$\begin{eqnarray} J_{51}&\leq&c(\sigma)\Vert x(z, 1, r, t)\Vert_{m}^{m}+\sigma c_{4}C_{\kappa, 1}(h_{1}\circ\nabla v)(t), \end{eqnarray}$

(3.23)

and to find the approximation of $J_{61}$ , we have

$\begin{eqnarray*} \frac{\partial}{\partial t}\bigg(\int_{0}^{\infty}g_{1}(s)(v(t)-v(t-s))ds\bigg)& = &\frac{\partial}{\partial t}\bigg(\int_{-\infty}^{t}g_{1}(t-s)(v(t)-v(s))ds \bigg)\notag\\ & = &\int_{-\infty}^{t}g_{1}'(t-s)(v(t)-v(s))ds \notag\\ &&+\bigg(\int_{-\infty}^{t}g_{1}(t-s)ds \bigg)v_{t}(t)\notag\\ & = &\int_{0}^{\infty}g_{1}'(s)(v(t)-v(t-s))ds+ g_{0}v_{t}(t), \end{eqnarray*}$

the (2.2) implies that

$\begin{eqnarray} J_{61}&\leq&-(g_{0}-\sigma_{3})\Vert v_{t}\Vert_{2}^{2}+\frac{c}{\sigma_{3}}C_{\kappa, 1}(h_{1}\circ\nabla v)(t). \end{eqnarray}$

(3.24)

In the same steps, the estimation of $J_{i2}$ , $i = 1, .., 6$ are obtained and

$\begin{eqnarray} J_{71}&\leq&c\sigma l_{1}\Vert\nabla v\Vert_{2}^{2}+c(\sigma)C_{\kappa, 1}(h_{1}\circ\nabla v)(t)\\ J_{72}&\leq&c\sigma l_{2}\Vert\nabla w \Vert_{2}^{2}+c(\sigma)C_{\kappa, 2}(h_{2}\circ\nabla v)(t). \end{eqnarray}$

(3.25)

Here, put (3.19)–(3.25) into (3.18), the required result is obtained.

Lemma 3.6. The functional $\Theta(t)$ introduced in (3.10) fulfills the below

$\begin{eqnarray} \Theta'(t)&\leq&-\gamma_{1}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} r\bigg(\vert \beta_{2}(r)\vert.\Vert x(z, \rho, r, t)\Vert_{m}^{m}+\vert \beta_{4}(r)\vert.\Vert y(z, \rho, r, t)\Vert_{m}^{m}\bigg) dr d\rho \\ &&-\gamma_{1}\int_{\tau_{1}}^{\tau_{2}}\bigg( \vert \beta_{2}(s)\vert. \Vert x\left( z, 1, r, t\right)\Vert_{m}^{m}+\vert \beta_{4}(r)\vert. \Vert y\left( z, 1, r, t\right)\Vert_{m}^{m}\bigg) dr \\ && +\beta_{5}\bigg(\Vert v_{t}(t)\Vert_{m}^{m} +\Vert w_{t}(t)\Vert_{m}^{m}\bigg). \end{eqnarray}$

(3.26)

in which $\beta_{5} = \max\{\beta_{1}, \beta_{3}\}$ .

Proof. To prove the result, using $\Theta(t)$ , and $(2.9)$ , we obtained the following

$\begin{eqnarray*} \Theta'( t) & = &-m\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} e^{-r\rho} \vert \beta_{2}(r)\vert.\vert x\vert^{m-1} x_{\rho}\left( z, \rho, r, t\right) dr d\rho dz \\ &&-m\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} e^{-r\rho} \vert \beta_{4}(r)\vert.\vert y\vert^{m-1} y_{\rho}\left( z, \rho, r, t\right) dr d\rho dz \\ & = &-\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} r e^{-r\rho} \vert \beta_{2}(r)\vert.\vert x(z, \rho, r, t)\vert^{m} dr d\rho dz \\ &&-\int_{\Omega}\int_{\tau_{1}}^{\tau_{2}} \vert \beta_{2}(r)\vert\bigg[e^{-r} \vert x\left( z, 1, r, t\right)\vert^{m}-\vert x\left( z, 0, r, t\right)\vert^{m}\bigg] dr dz \\ &&-\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} r e^{-r\rho} \vert \beta_{4}(r)\vert.\vert y(z, \rho, r, t)\vert^{m} dr d\rho dz \\ &&-\int_{\Omega}\int_{\tau_{1}}^{\tau_{2}} \vert \beta_{4}(r)\vert\bigg[e^{-r} \vert y\left( z, 1, r, t\right)\vert^{m}-\vert y\left( z, 0, r, t\right)\vert^{m}\bigg] dr dz \end{eqnarray*}$

Utilizing $x(z, 0, r, t) = v_{t}(z, t), y(z, 0, r, t) = w_{t}(z, t)$ , and $e^{-r}\leq e^{-r\rho}\leq 1$ , for any $0 < \rho < 1$ , moreover, select $\gamma_{1} = e^{-\tau_{2}}$ , we have

$\begin{eqnarray*} \Theta'( t) &\leq &-\gamma_{1}\int_{\Omega}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}}r\bigg(\vert \beta_{2}(r)\vert.\vert z(z, \rho, r, t)\vert^{m}+\vert \beta_{4}(r)\vert.\vert y(z, \rho, r, t)\vert^{m}\bigg) dr d\rho dz \\ &&-\gamma_{1}\int_{\Omega}\int_{\tau_{1}}^{\tau_{2}} \bigg(\vert \beta_{2}(r)\vert \vert x(z, 1, r, t)\vert^{m}+\vert \beta_{4}(r)\vert \vert y(z, 1, r, t)\vert^{m}\bigg) dr dz\\ && +\int_{\tau_{1}}^{\tau_{2}} \vert \beta_{2}(r)\vert dr\int_{\Omega}\vert v_{t}\vert^{m}(t)dz+\int_{\tau_{1}}^{\tau_{2}} \vert \beta_{4}(r)\vert dr\int_{\Omega}\vert w_{t}\vert^{m}(t)dz, \end{eqnarray*}$

applying (2.4), the required proof is obtained. In the next step, we below functional are introduced

$\begin{eqnarray} \mathcal{A}_{1}(t)&: = &\int_{\Omega}\int_{0}^{t}\varphi_{1}(t-s)\nabla v(s)^{2}ds dz, \\ \mathcal{A}_{2}(t)&: = &\int_{\Omega}\int_{0}^{t}\varphi_{2}(t-s)\nabla w(s)^{2}ds dz, \end{eqnarray}$

(3.27)

in which $\varphi_{1}(t) = \int_{t}^{\infty}g_{1}(s)ds, \varphi_{2}(t) = \int_{t}^{\infty}g_{2}(s)ds$ .

Lemma 3.7. Let us suppose that (2.1) and (2.2) satisfied. Then, the functional $F_{1} = \mathcal{A}_{1}+\mathcal{A}_{2}$ and fulfills the following

$\begin{eqnarray} F_{1}'(t)&\leq&-\frac{1}{2}\bigg((g_{1}\circ\nabla v)(t)+(g_{2}\circ\nabla w)(t)\bigg)\\ &&+3g_{0}\int_{\Omega}\nabla v^{2}dz+3\widehat{g}_{0}\int_{\Omega}\nabla w^{2}dz\\ &&+\frac{1}{2}\int_{\Omega}\int_{t}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))^{2}ds dz\\ &&+\frac{1}{2}\int_{\Omega}\int_{t}^{\infty}g_{2}(s)(\nabla w(t)-\nabla w(t-s))^{2}ds dz. \end{eqnarray}$

(3.28)

Proof. We can easily prove this lemma with the help of Lemma 3.7 in ^[13] and Lemma 3.4 in ^[15].

Now, we have sufficient mathematical tools to prove the below mentioned Theorem.

Theorem 3.8. Take (2.1)–(2.5), then one can find positive constants $\varsigma_{i}, i = 1, 2, 3$ and positive function $\varsigma_{4}(t)$ in a way that the energy functionalmentioned in (2.11) fulfills

$\begin{equation} E\left( t\right) \leq \varsigma_{1}D_{2}^{-1} \bigg(\frac{\varsigma_{2}+\varsigma_{3}\int_{0}^{t}\widehat{\zeta}(\nu)D_{4}(\varsigma_{4}(\nu)\mu_{0}(\nu))d\nu}{\int_{0}^{t}\zeta_{0}(\nu)d\nu}\bigg), \end{equation}$

(3.29)

in which

$\begin{equation} D_{2}(t) = tD'(\varepsilon_{0}t), \quad D_{3}(t) = tD'^{-1}(t), \quad D_{4}(t) = \overline{D}^{*}_{3}(t), \quad \end{equation}$

(3.30)

and

$\mu_{0} = \max\{\mu, \widehat{\mu}\}, \quad \widehat{\zeta} = \max\{\zeta_{1}, \zeta_{2}\}, \quad \zeta_{0} = \min\{\zeta_{1}, \zeta_{2}\},$

which are increasing and convex in $(0$ , $\varrho]$ .

Proof. For the proof, we define the below functional

$\begin{eqnarray} \mathcal{G}(t)&: = & NE(t)+N_{1}\Psi(t)+N_{2}\Phi(t)+N_{3}\Theta(t), \end{eqnarray}$

(3.31)

we determined the positive constants $N, N_{i}, i = 1, 2, 3$ . Simplifying (3.36) and utilizing 2.12, the Lemmas 3.4–3.6, we have

$\begin{eqnarray} \mathcal{G}'(t)&: = &NE'(t)+N_{1}\Psi'(t)+N_{2}\Phi'(t) +N_{3}\Theta'(t)\\ &\leq&-\bigg\{N_{2}(l_{0}-\sigma_{3})-N_{1}\bigg\}\bigg(\Vert v_{t}\Vert_{2}^{2}+\Vert w_{t}\Vert_{2}^{2}\bigg)\\ &&-\bigg\{N_{3}\xi_{1}-N_{2}\xi_{1}\sigma\bigg\}\bigg(\Vert\nabla v\Vert_{2}^{4}+\Vert\nabla w\Vert_{2}^{4}\bigg)\\ &&-\bigg\{N_{1}(l-\varepsilon(c_{1}+c_{2})-\sigma_{1})-N_{2}\sigma(\xi{0}+\widehat{l_{0}}^{2}+c\widehat{l})\bigg\}\bigg(\Vert\nabla v\Vert_{2}^{2}+\Vert\nabla w\Vert_{2}^{2}\bigg)\\ &&-\bigg\{\frac{N\delta}{4}-N_{2}\sigma_{2}\frac{2\delta E(0)}{l}\bigg\}\bigg[\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla v\Vert_{2}^{2}\bigg)^{2}+\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla w\Vert_{2}^{2}\bigg)^{2}\bigg]\\ &&+\bigg\{N_{1}c(\sigma_{1})+N_{2}c(\sigma, \sigma_{2}, \sigma_{3})\bigg\}\bigg(C_{\kappa, 1}(h_{1}\circ\nabla v)(t)+C_{\kappa, 2}(h_{2}\circ\nabla w)(t)\bigg)\\ &&+\frac{N}{2}\bigg( (g_{1}'\circ\nabla v)(t)+(g_{2}'\circ\nabla w)(t)\bigg)\\ &&-\bigg\{\gamma_{0}N-N_{1}c(\varepsilon)-N_{2}c(\sigma)-N_{3}\beta_{5}\bigg\}\bigg(\Vert v_{t}\Vert_{m}^{m}+\Vert w_{t}\Vert_{m}^{m}\bigg)\\ &&-\bigg(\gamma_{1}N_{3}- N_{1}c(\varepsilon)-N_{2}c(\sigma)\bigg) \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\Vert x(z, 1, r, t)\Vert_{m}^{m}ds\bigg)\\ &&-N_{3}\gamma_{1}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} r\vert \beta_{2}(r)\vert.\Vert x(z, \rho, r, t)\Vert_{m}^{m} dr d\rho\\ &&-\bigg(\gamma_{1}N_{3}- N_{1}c(\varepsilon)-N_{2}c(\sigma)\bigg) \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\Vert y(z, 1, r, t)\Vert_{m}^{m}dr \bigg)\\ &&-N_{3}\gamma_{1}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} r \vert \beta_{4}(r)\vert.\Vert y(z, \rho, r, t)\Vert_{m}^{m} dr d\rho-N_{1}\int_{\Omega}F(v, w)dz . \end{eqnarray}$

(3.32)

We select the various constants at this point such that the values included in parenthesis are positive in this stage. Here, putting

$\sigma_{3} = \frac{l_{0}}{2}, \quad \varepsilon = \frac{l}{4(c_{1}+c_{2})}, \quad \sigma_{1} = \frac{l}{4}, \quad \sigma_{2} = \frac{lN}{16 E(0)N_{2}}, \quad N_{1} = \frac{l_{0}}{4}N_{2}.$

Thus, we arrive at

$\begin{eqnarray} \mathcal{H}'(t)&\leq&-\frac{l_{0}}{4} N_{2}\bigg(\Vert w_{t}\Vert_{2}^{2}+\Vert w_{t}\Vert_{2}^{2}\bigg)-\zeta_{1}N_{2}\bigg(\frac{l_{0}}{4}-\delta\bigg)\bigg(\Vert\nabla w\Vert_{2}^{4}+\Vert\nabla u\Vert_{2}^{4}\bigg)\\ &&-N_{2}\bigg(\frac{ll_{0}}{8}-\delta(\zeta_{0}+\widehat{h_{0}}^{2}+c\widehat{l})\bigg)\bigg(\Vert\nabla w\Vert_{2}^{2}+\Vert\nabla u\Vert_{2}^{2}\bigg)\\ &&-\frac{N\delta}{8}\bigg[\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla v\Vert_{2}^{2}\bigg)^{2}+\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla w\Vert_{2}^{2}\bigg)^{2}\bigg]\\ &&+N_{2}c(\sigma, \sigma_{1}, \sigma_{2}, \sigma_{3})\bigg(C_{\kappa, 1}(h_{1}\circ\nabla v)(t)+C_{\kappa, 2}(h_{2}\circ\nabla w)(t)\bigg)\\ &&+\frac{N}{2}\bigg( (g_{1}'\circ\nabla v)(t)+(g_{2}'\circ\nabla v)(t)\bigg)-N_{1}\int_{\Omega}F(v, w)dz\\ &&-\bigg(\gamma_{0}N-N_{2}c(\sigma, \varepsilon)-N_{3}\beta_{5}\bigg)\bigg(\Vert v_{t}\Vert_{m}^{m}+\Vert w_{t}\Vert_{m}^{m}\bigg)\\ &&-\bigg(\gamma_{1}N_{3}-N_{2}c(\sigma, \varepsilon)\bigg) \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{2}(r)\Vert x(z, 1, r, t)\Vert_{m}^{m}ds\bigg)\\ &&-N_{3}\gamma_{1}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} r\vert \beta_{2}(r)\vert.\Vert x(z, \rho, r, t)\Vert_{m}^{m} dr d\rho\\ &&-\bigg(\gamma_{1}N_{3}-N_{2}c(\sigma, \varepsilon)\bigg) \int_{\tau_{1}}^{\tau_{2}}\vert\beta_{4}(r)\Vert y(z, 1, r, t)\Vert_{m}^{m}dr\bigg)\\ &&-N_{3}\gamma_{1}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} r\vert \beta_{4}(r)\vert.\Vert y(z, \rho, r, t)\Vert_{m}^{m} dr d\rho . \end{eqnarray}$

(3.33)

In the upcoming, we select $\sigma$ in a manner that

$\sigma < \min\bigg\{\frac{l_{0}}{4}, \frac{ll_{0}}{8(\xi_{0}+\widehat{g_{0}}^{2}+c\widehat{l})}\bigg\}.$

After that, we take $N_{2}$ in a way that

$N_{2}\bigg(\frac{ll_{0}}{8}-\sigma(\xi_{0}+\widehat{g_{0}}^{2}+c\widehat{l})\bigg) > 4l_{0},$

and take $N_{3}$ large enough in a way that

$\gamma_{1}N_{3}-N_{2}c(\sigma, \varepsilon) > 0.$

As a result, for positive constants $d_{i}, i = 1, 2, 3, 4, 5$ , (3.33) can be written as

$\begin{eqnarray} \mathcal{H}'(t)&\leq&-d_{1}(\Vert v_{t}\Vert_{2}^{2}+\Vert w_{t}\Vert_{2}^{2})-d_{2}(\Vert\nabla v\Vert_{2}^{4}+\Vert\nabla w\Vert_{2}^{4})-4l_{0}(\Vert\nabla v\Vert_{2}^{2}+\Vert\nabla w\Vert_{2}^{2})\\ &&-\frac{N\delta}{8}\bigg[\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla v\Vert_{2}^{2}\bigg)^{2}+\bigg(\frac{1}{2}\frac{d}{dt}\Vert\nabla w\Vert_{2}^{2}\bigg)^{2}\bigg]\\ &&-\bigg(\frac{N}{2}-d_{3}C_{\kappa}\bigg)\bigg((h_{1}\circ\nabla v)(t)+(h_{2}\circ\nabla w)(t)\bigg)\\ &&+\frac{N\kappa}{2} \bigg((g_{1}\circ\nabla v)(t)+(g_{2}\circ\nabla w)(t)\bigg) \\ &&-(\gamma_{0}N-c)\bigg(\Vert v_{t}\Vert_{m}^{m}+\Vert w_{t}\Vert_{m}^{m}\bigg)-d_{5}\int_{\Omega}F(v, w)dz\\ &&-d_{4}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}} s \bigg(\vert\beta_{2}(r)\vert.\Vert x(z, \rho, r, t)\Vert_{m}^{m}+\vert\beta_{4}(r)\vert.\Vert y(z, \rho, r, t)\Vert_{m}^{m}\bigg) dr d\rho, \end{eqnarray}$

(3.34)

in which $C_{\kappa} = \max\{C_{\kappa, 1}, C_{\kappa, 2}\}$ .

We know that $\frac{\kappa g_{i}^{2}(s)}{\kappa g_{i}(s)-g_{i}(s)}\leq g_{i}(s)$ , then from from Lebesgue Dominated Convergence, we have the below

$\begin{equation} \lim\limits_{\kappa\rightarrow 0^{+}}\kappa C_{\kappa, i} = \lim\limits_{\kappa\rightarrow 0^{+}}\int_{0}^{\infty}\frac{\kappa g_{i}^{2}(s)}{\kappa g_{i}(s)-g_{i}(s)}ds = 0, \quad i = 1, 2 \end{equation}$

(3.35)

which leads to

$\lim\limits_{\kappa\rightarrow 0^{+}}\kappa C_{\kappa} = 0.$

As a result of this, one can find $0 < \kappa_{0} < 1$ in a manner that if $\kappa < \kappa_{0}$ , then

$\begin{equation} \kappa C_{\kappa}\leq \frac{1}{d_{3}}. \end{equation}$

(3.36)

From (3.8)–(3.10) through mathematical skills, we have the following

$\begin{eqnarray} \vert\mathcal{H}(t)-NE(t)\vert&\leq&\frac{N_{1}}{2}\bigg(\Vert v_{t}(t)\Vert_{2}^{2}+\Vert w_{t}(t)\Vert_{2}^{2}+c_{p}\Vert\nabla w(t)\Vert_{2}^{2}+c_{p}\Vert\nabla w(t)\Vert_{2}^{2}\bigg)\\ &&+\delta\frac{N_{1}}{4}\bigg(\Vert\nabla v(t)\Vert_{2}^{4}+\Vert\nabla w(t)\Vert_{2}^{4}\bigg)+\frac{N_{2}}{2}\bigg(\Vert v_{t}(t)\Vert_{2}^{2}+\Vert w_{t}(t)\Vert_{2}^{2}\bigg)\\ &&+\frac{N_{2}}{2}c_{p}\bigg(C_{\kappa, 1}(g_{1}\circ\nabla v)(t)+C_{\kappa, 2}(g_{2}\circ\nabla w)(t)\bigg) \\ &&+N_{3}\int_{0}^{1}\int_{\tau_{1}}^{\tau_{2}}r e^{-\rho r}\bigg(\vert \beta_{2}(r)\vert. \Vert x(z, \rho, r, t)\Vert_{m}^{m}+\vert \beta_{4}(r)\vert. \Vert y(z, \rho, r, t)\Vert_{m}^{m}\bigg) dr d\rho. \end{eqnarray}$

(3.37)

By the fact $e^{-\rho r} < 1$ and (2.2), we have the below

$\begin{eqnarray} \vert\mathcal{H}(t)-NE(t)\vert&\leq&C(N_{1}, N_{2}, N_{3})E(t) = C_{1}E(t). \end{eqnarray}$

(3.38)

that is

$\begin{equation} \left( N-C_{1}\right) E\left( t\right) \leq \mathcal{H}\left( t\right) \leq \left( N+C_{1}\right) E\left( t\right). \end{equation}$

(3.39)

Here, set $\kappa = \frac{1}{2N}$ and take $N$ large enough in a manner that

$\begin{eqnarray*} &&N-C_{1} > 0, \quad , \quad \gamma_{0}N-c > 0, \quad \frac{1}{2}N-\frac{1}{2\kappa_{0}} > 0, \quad \kappa = \frac{1}{2N} < \kappa_{0}, \end{eqnarray*}$

we find

$\begin{equation} \mathcal{H}^{\prime }\left( t\right) \leq -k_{2}E(t)+\frac{1}{4}(( g_{1}\circ\nabla v)(t)+(g_{2}\circ\nabla w)(t) ) \end{equation}$

(3.40)

for some $k_{2} > 0$ , and

$\begin{equation} c_{5}E\left( t\right) \leq \mathcal{H}\left( t\right) \leq c_{6}E\left( t\right) , \forall t\geq 0 \end{equation}$

(3.41)

for some $c_{5}, c_{6} > 0$ , we have

$\mathcal{H}(t)\sim E(t).$

After that, the below cases are considered:

Case 3.9. $G_{i}, i = 1, 2$ are linear. Multiplying (3.40) by $\zeta_{0}(t) = \min\{\zeta_{1}(t), \zeta_{2}(t)\}$ , we find

$\begin{eqnarray} \zeta_{0}(t)\mathcal{H}^{\prime }\left( t\right) &\leq& -k_{2}\zeta_{0}(t)E(t)+\frac{1}{4}\zeta_{0}(t)(( g_{1}\circ\nabla v)(t)+(g_{2}\circ\nabla w)(t) )\\ &\leq&-k_{2}\zeta_{0}(t)E(t)+\frac{1}{4}\zeta_{1}(t)( g_{1}\circ\nabla v)(t)+\frac{1}{4}\zeta_{2}(t)(g_{2}\circ\nabla w)(t) . \end{eqnarray}$

(3.42)

The last two terms in (3.42), we have

$\begin{eqnarray} \frac{\zeta_{1}(t)}{4}(g_{1}\circ\nabla v)(t) & = & \frac{\zeta_{1}(t)}{4}\int_{\Omega}\int_{0}^{\infty}g_{1}(\delta)\vert\nabla\eta^{t}(s)\vert^{2}ds dz \\ & = &\underbrace{ \frac{\zeta_{1}(t)}{4}\int_{\Omega}\int_{0}^{t}g_{1}(s)\vert\nabla\eta^{t}(s)\vert^{2}ds dz}_{I_{1}}\\ &&+\underbrace{ \frac{\zeta_{1}(t)}{4}\int_{\Omega}\int_{t}^{\infty}g_{1}(s)\vert\nabla\eta^{t}(s)\vert^{2}ds dz}_{I_{2}} \end{eqnarray}$

(3.43)

To estimate $I_{1}$ , using (2.11),

$\begin{eqnarray} I_{1} &\leq& \frac{1}{4}\int_{\Omega}\int_{0}^{t}\zeta_{1}(s)g_{1}(s)\vert\nabla\eta^{t}(s)\vert^{2}ds dz\\ & = &-\frac{1}{4}\int_{\Omega}\int_{0}^{t}g_{1}'(s)\vert\nabla\eta^{t}(s)\vert^{2}ds dz\\ &\leq&-\frac{1}{2l_{1}}E'(t), \end{eqnarray}$

(3.44)

and by (3.4), we get

$\begin{eqnarray} I_{2} &\leq& \frac{\beta}{4}\zeta_{1}(t)\mu(t). \end{eqnarray}$

(3.45)

In the same way, we obtained

$\begin{eqnarray} \frac{\zeta_{2}(t)}{4}(g_{2}\circ\nabla w)(t) &\leq&-\frac{1}{2l_{2}}E'(t)+\frac{\widehat{\beta}}{4}\zeta_{2}(t)\widehat{\mu}(t). \end{eqnarray}$

(3.46)

As a result of this, we get

$\begin{equation} \zeta_{0}(t)\mathcal{H}^{\prime }\left( t\right) \leq -k_{2}\zeta_{0}(t)E(t)-\frac{1}{\widehat{l}}E'(t)+2\beta_{0} w(t), \end{equation}$

(3.47)

where $\beta_{0} = \max\{\frac{\beta}{4}, \frac{\widehat{\beta}}{4}\}$ and $w(t) = \widehat{\zeta}(t)\mu_{0}(t)$ .

Applying $\zeta_{i}'(t)\leq0$ , we get

$\begin{eqnarray} \mathcal{H}_{1}^{\prime }\left( t\right)\leq-k_{2}\zeta_{0}(t)E(t)+2\beta_{0} w(t), \end{eqnarray}$

(3.48)

with

$\mathcal{H}_{1}(t) = \zeta_{0}(t)\mathcal{H}\left( t\right)+\frac{1}{\widehat{l}} E(t)\sim E(t),$

we have

$\begin{equation} k_{4}E(t)\leq \mathcal{H}_{1}(t)\leq k_{5}E(t), \end{equation}$

(3.49)

then, the following is obtained from (3.48)

$\begin{eqnarray*} k_{2}E(T)\int_{0}^{T}\zeta_{0}(t)dt&\leq&k_{2}\int_{0}^{T}\zeta_{0}(t)E(t)dt\notag\\ &\leq&\mathcal{H}_{1}(0)-\mathcal{H}_{1}(T)+2\beta_{0}\int_{0}^{T}w(t)dt\notag\\ &\leq&\mathcal{H}_{1}(0)+2\beta_{0}\int_{0}^{T}\widehat{\zeta}(t)\mu_{0}(t)dt. \end{eqnarray*}$

Further analysis implies that

$\begin{equation*} E(T)\leq\frac{1}{k_{2}}\bigg(\frac{\mathcal{G}_{1}(0)+2\beta_{0} \int_{0}^{T}\widehat{\xi}(t)\mu_{0}(t)dt}{\int_{0}^{T}\xi_{0}(t)dt}\bigg), \end{equation*}$

From the linearity of $D$ , the linearity of the functions $D_{2}, D'_{2}$ and $D_{4}$ can easily be determined. This implies that

$\begin{equation} E(T)\leq\lambda_{1}D_{2}^{-1}\bigg(\frac{\frac{\mathcal{H}_{1}(0)}{k_{2}}+\frac{2\beta_{0}}{k_{2}} \int_{0}^{T}\widehat{\zeta}(t)\mu_{0}(t)dt}{\int_{0}^{T}\zeta_{0}(t)dt}\bigg), \end{equation}$

(3.50)

which gives (3.29) with $\varsigma_{1} = \lambda_{1}$ , $\varsigma_{2} = \frac{\mathcal{H}_{1}(0)}{k_{2}}$ , $\varsigma_{3} = \frac{2\beta_{0}}{\lambda_{2}k_{2}}$ , and $\varsigma_{4}(t) = Id(t) = t$ . Hence, the required proof is completed.

Case 3.10. Let $H_{i}, i = 1, 2$ are nonlinear. Then, with the help of (3.28) and (3.40). Assume the positive functional

$\begin{equation*} \mathcal{H}_{2}(t) = \mathcal{H}(t)+F_{1}(t) \end{equation*}$

then for all $t\geq 0$ and for some $k_{3} > 0$ , the following holds true

$\begin{eqnarray} \mathcal{H}'_{2}(t)&\leq& -k_{3}E(t)+\frac{1}{2}\int_{\Omega}\int_{t}^{\infty}g_{1}(s)(\nabla v(t)-\nabla v(t-s))^{2}ds dz\\ &&+\frac{1}{2}\int_{\Omega}\int_{t}^{\infty}g_{2}(s)(\nabla w(t)-\nabla w(t-s))^{2}ds dz, \end{eqnarray}$

(3.51)

with the help of (3.4), we have

$\begin{eqnarray} k_{3}\int_{0}^{t}E(x)dx&\leq& \mathcal{H}_{2}(0)-\mathcal{H}_{2}(t)+\beta_{0}\int_{0}^{t}\mu_{0}(\varsigma)d\varsigma\\ &\leq&\mathcal{H}_{2}(0)+\beta_{0}\int_{0}^{t}\mu_{0}(\varsigma)d\varsigma. \end{eqnarray}$

(3.52)

Therefore

$\begin{eqnarray} \int_{0}^{t}E(x)dx&\leq&k_{6}\mu_{1}(t), \end{eqnarray}$

(3.53)

where $k_{6} = \max\{\frac{\mathcal{H}_{2}(0)}{k_{3}}, \frac{\beta_{0}}{k_{3}}\}$ and $\mu_{1}(t) = 1+\int_{0}^{t}\mu_{0}(\varsigma)d\varsigma$ .

Corollary 3.11. The following is obtained from (2.11) and (3.53):

$\begin{eqnarray} &&\int_{0}^{t}\int_{\Omega}\vert\nabla v(t)-\nabla v(t-s)\vert^{2}dz ds \\ &&+\int_{0}^{t}\int_{\Omega}\vert\nabla w(t)-\nabla w(t-s)\vert^{2}dz ds \\ &\leq&2\int_{0}^{t}\int_{\Omega}\nabla v^{2}(t)-\nabla v^{2}(t-s)dzds\\ &&+2\int_{0}^{t}\int_{\Omega}\nabla w^{2}(t)-\nabla w^{2}(t-s)dz ds \\ &\leq&\frac{4}{l_{0}}\int_{0}^{t}E(t)-E(t-s)ds \\ &\leq&\frac{8}{l_{0}}\int_{0}^{t}E(x)dx\leq\frac{8k_{6}}{l_{0}}\mu_{1}(t). \end{eqnarray}$

(3.54)

Now, we define $\phi_{i}(t), i = 1, 2$ by

$\begin{eqnarray} \phi_{1}(t)&: = &\mathcal{B}(t)\int_{0}^{t}\int_{\Omega}\vert\nabla v(t)-\nabla v(t-s)\vert^{2}dzds, \\ \phi_{2}(t)&: = &\mathcal{B}(t)\int_{0}^{t}\int_{\Omega}\vert\nabla w(t)-\nabla w(t-s)\vert^{2}dz ds \end{eqnarray}$

(3.55)

where $\mathcal{B}(t) = \frac{\mathcal{B}_{0}}{\mu_{1}(t)}$ and $0 < \mathcal{B}_{0} < \min\{1, \frac{l}{8k_{6}}\}$ .

Then, by (3.53), we have

$\begin{equation} \phi_{i}(t) < 1, \quad \forall t > 0, \quad i = 1, 2 \end{equation}$

(3.56)

Further, we suppose that $\phi_{i}(t) > 0, \quad \forall t > 0, \quad i = 1, 2$ . In addition to this, we define another functional $\Gamma_{1}, \Gamma_{2}$ by

$\begin{eqnarray} \Gamma_{1}(t)&: = &-\int_{0}^{t}g_{1}'(s)\int_{\Omega}\vert\nabla v(t)-\nabla v(t-s)\vert^{2}dz ds, \\ \Gamma_{2}(t)&: = &-\int_{0}^{t}g_{2}'(s)\int_{\Omega}\vert\nabla w(t)-\nabla w(t-s)\vert^{2}dz ds \end{eqnarray}$

(3.57)

Here, obviously $\Gamma_{i}(t)\leq -cE'(t), \quad i = 1, 2$ . As $G_{i}(0) = 0, \quad i = 1, 2$ and $G_{i}(t)$ are convex strictly on $(0$ , $\varrho]$ , then

$\begin{equation} G_{i}(\lambda z)\leq\lambda G_{i}(z), \quad 0 < \lambda < 1, \quad z\in(0, \varrho], \quad i = 1, 2. \end{equation}$

(3.58)

Applying (2.3) and (3.56), we get

$\begin{eqnarray} \Gamma_{1}(t)& = &\frac{-1}{\mathcal{B}(t) \phi_{1}(t)}\int_{0}^{t} \phi_{1}(t)(g_{1}'(s))\int_{\Omega}\mathcal{B}(t)\vert \nabla v(t)-\nabla v(t-s)\vert^{2}dzds\\ &\geq&\frac{1}{\mathcal{B}(t) \phi_{1}(t)}\int_{0}^{t} \phi_{1}(t)\zeta_{1}(s)G_{1}(g_{1}(s))\int_{\Omega}\mathcal{B}(t)\vert \nabla v(t)-\nabla v(t-s)\vert^{2}dzds\\ &\geq &\frac{\zeta_{1}(t)}{\mathcal{B}(t) \phi_{1}(t)}\int_{0}^{t} G_{1}(\phi_{1}(t)g_{1}(s))\int_{\Omega}\mathcal{B}(t)\vert \nabla v(t)-\nabla v(t-s)\vert^{2}dzds\\ &\geq &\frac{\zeta_{1}(t)}{\mathcal{B}(t) }G_{1}\bigg(\frac{1}{\phi_{1}(t)}\int_{0}^{t} \phi_{1}(t)g_{1}(s)\int_{\Omega}\mathcal{B}(t)\vert \nabla v(t)-\nabla v(t-s)\vert^{2}dzds\bigg)\\ & = &\frac{\zeta_{1}(t)}{\mathcal{B}(t) }G_{1}\bigg(\mathcal{B}(t)\int_{0}^{t} g_{1}(s)\int_{\Omega}\vert\nabla v(t)-\nabla v(t-s)\vert^{2}dzds\bigg)\\ & = &\frac{\zeta_{1}(t)}{\mathcal{B}(t) }\overline{G_{1}}\bigg(\mathcal{B}(t)\int_{0}^{t} g_{1}(s)\int_{\Omega}\vert \nabla v(t)-\nabla v(t-s)\vert^{2}dzds\bigg). \end{eqnarray}$

(3.59)

$\begin{eqnarray} \Gamma_{2}(t)&\geq &\frac{\zeta_{2}(t)}{\mathcal{B}(t) }\overline{G_{2}}\bigg(\mathcal{B}(t)\int_{0}^{t} g_{2}(s)\int_{\Omega}\vert \nabla w(t)-\nabla w(t-s)\vert^{2}dzds\bigg). \end{eqnarray}$

(3.60)

Taking the same steps, $\overline{G_{i}}, i = 1, 2$ are $C^{2}$ -extension of $G_{i}$ that are convex strictly and increasing strictlyon ${\bf R}_{+}$ . From (3.59), we have the following

$\begin{eqnarray} \int_{0}^{t} g_{1}(s)\int_{\Omega}\vert\nabla v(t)-\nabla v(t-s)\vert^{2}dzds&\leq&\frac{1}{\mathcal{B}(t)}\overline{G_{1}}^{-1}\bigg(\frac{\mathcal{B}(t) \Gamma_{1}(t)}{\zeta_{1}(t)}\bigg)\\ \int_{0}^{t} g_{2}(s)\int_{\Omega}\vert\nabla w(t)-\nabla w(t-s)\vert^{2}dzds&\leq&\frac{1}{\mathcal{B}(t)}\overline{G_{2}}^{-1}\bigg(\frac{\mathcal{B}(t) \Gamma_{2}(t)}{\zeta_{2}(t)}\bigg). \end{eqnarray}$

(3.61)

Putting (3.61) and (3.4) into (3.40), we have

$\begin{eqnarray} \mathcal{H}^{\prime }\left( t\right) &\leq& -k_{2}E(t)+\frac{c}{\mathcal{B}(t)}\overline{G_{1}}^{-1}\bigg(\frac{\mathcal{B}(t) \Gamma_{1}(t)}{\zeta_{1}(t)}\bigg)\\ &&+\frac{c}{\mathcal{B}(t)}\overline{G_{2}}^{-1}\bigg(\frac{\mathcal{B}(t) \Gamma_{2}(t)}{\zeta_{2}(t)}\bigg)+k_{6}\mu_{0}(t) \end{eqnarray}$

(3.62)

Here, introduce $\mathcal{K}_{1}(t)$ for $\varepsilon_{0} < r$ by

$\begin{eqnarray} \mathcal{K}_{1}(t) = D^{\prime}\left(\varepsilon_{0} \frac{\mathcal{B}(t)E(t)}{E(0)}\right) \mathcal{H}(t)+E(t), \end{eqnarray}$

(3.63)

in which $D' = \min\{G_{1}, G_{2}\}$ and is equivalent to $E(t)$ . Because of this $E^{\prime}(t) \leq 0, \overline{G_{i}}^{\prime} > 0,$ and $\overline{G_{i}}^{\prime \prime} > 0, i = 1, 2$ . Also applying (3.62), we obtained that

$\begin{eqnarray} \mathcal{K}_{1}^{\prime}(t)& = & \varepsilon_{0}\bigg( \frac{E^{\prime}(t) \mathcal{B}(t)}{E(0)}+\frac{E(t) \mathcal{B}'(t)}{E(0)}\bigg) D^{\prime \prime}\left(\varepsilon_{0} \frac{E(t) \mathcal{B}(t)}{E(0)}\right) \mathcal{H}(t) \\ &&+D^{\prime}\left(\varepsilon_{0} \frac{E(t) \mathcal{B}(t)}{E(0)}\right) \mathcal{H}^{\prime}(t)+E^{\prime}(t) \\ &\leq &-k_{2} E(t) D^{\prime}\left(\varepsilon_{0} \frac{\mathcal{B}(t) E(t)}{E(0)}\right)+k_{6}\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{\mathcal{B}(t) E(t)}{E(0)}\right) \\ &&\left.+\frac{c}{\mathcal{B}(t)} \overline{G_{1}}^{-1}\left(\frac{\mathcal{B}(t) \Gamma_{1}(t)}{\zeta_{1}(t)}\right)\right) D^{\prime}\left(\varepsilon_{0} \frac{\mathcal{B}(t) E(t)}{E(0)}\right)\\ &&\left.+\frac{c}{\mathcal{B}(t)} \overline{G_{2}}^{-1}\left(\frac{\mathcal{B}(t) \Gamma_{2}(t)}{\zeta_{2}(t)}\right)\right) D^{\prime}\left(\varepsilon_{0} \frac{\mathcal{B}(t) E(t)}{E(0)}\right)+E^{\prime}(t) \end{eqnarray}$

(3.64)

According to ^[29], we introduce the conjugate function of $\overline{G_{i}}$ by $\overline{G_{i}}^{*},$ which fulfills

$\begin{eqnarray} A B \leq \overline{G_{i}}^{*}\left(A\right)+\overline{G_{i}}\left(B\right), \quad i = 1, 2 \end{eqnarray}$

(3.65)

For $A = D^{\prime}\left(\varepsilon_{0}(E(t)\mathcal{B}(t)) /(E(0)))\right) \text { and } B_{i} = \overline{G_{i}}^{-1}((\mathcal{B}(t) \Gamma_{i}(t))/(\zeta_{i}(t))), \quad i = 1, 2$ and applying (3.64), we have

$\begin{eqnarray} \mathcal{K}_{1}^{\prime}(t) &\leq &-k_{2} E(t) D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)+k_{6}\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right) \\ &&+\frac{c}{\mathcal{B}(t)} \overline{G_{1}}^{*}\left(D^{\prime}\left(\varepsilon_{0} \frac{ E(t)\mathcal{B}(t)}{E(0)}\right)\right)+\frac{c}{\mathcal{B}(t)} \frac{\mathcal{B}(t) \Gamma_{1}(t)}{\zeta_{1}(t)}\\ &&+\frac{c}{\mathcal{B}(t)} \overline{G_{2}}^{*}\left(D^{\prime}\left(\varepsilon_{0} \frac{ E(t)\mathcal{B}(t)}{E(0)}\right)\right)+\frac{c}{\mathcal{B}(t)} \frac{\mathcal{B}(t) \Gamma_{2}(t)}{\zeta_{2}(t)}+E^{\prime}(t) \\ &\leq &-k_{2} E(t) D^{\prime}\left(\varepsilon_{0} \frac{ E(t)\mathcal{B}(t)}{E(0)}\right)+k_{6}\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{E(t) \mathcal{B}(t) }{E(0)}\right)\\ &&+\frac{c}{\mathcal{B}(t)}D'\bigg( \varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\bigg) (\overline{G_{1}}^{\prime})^{-1}\bigg[D'\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)\bigg] \\ &&+\frac{c}{\mathcal{B}(t)}D'\bigg( \varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\bigg) (\overline{G_{2}}^{\prime})^{-1}\bigg[D'\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)\bigg]\\ &&+\frac{c \Gamma_{1}(t)}{\zeta{1}(t)} +\frac{c \Gamma_{2}(t)}{\zeta_{2}(t)}. \end{eqnarray}$

(3.66)

Here, we multiply (3.66) by $\zeta_{0}(t)$ and get

$\begin{eqnarray} \zeta_{0}(t) \mathcal{K}_{1}^{\prime}(t) &\leq &-k_{2}\zeta_{0}(t) E(t) D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)+k_{6}\zeta_{0}(t)\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)\\ &&+\frac{2c\zeta_{0}(t)}{\mathcal{B}(t)} \varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)+c \Gamma_{1}(t) +c \Gamma_{2}(t) \\ &\leq &-k_{2}\zeta_{0}(t) E(t) D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)+k_{6}\zeta_{0}(t)\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{E(t) \mathcal{B}(t) }{E(0)}\right)\\ &&+\frac{2c\zeta_{0}(t)}{\mathcal{B}(t)} \varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)} D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)-c E^{\prime}(t) \end{eqnarray}$

(3.67)

where we utilized the following $\varepsilon_{0}(\mathcal{B}(t) E(t) / E(0)) < r$ , $D^{\prime} = \min \{G_{1}, G_{2}\}$ and $\Gamma_{i} < -cE'(t), i = 1, 2$ , and define the functional $\mathcal{K}_{2}(t)$ as

$\begin{eqnarray} \mathcal{K}_{2}(t) = \zeta_{0}(t) \mathcal{K}_{1}(t)+c E(t) \end{eqnarray}$

(3.68)

Effortlessly, one can prove that $\mathcal{K}_{2}(t) \sim E(t)$ , i.e., one can find two positive constants $m_{1}$ and $m_{2}$ in a manner that

$\begin{eqnarray} m_{1} \mathcal{K}_{2}(t) \leq E(t) \leq m_{2} \mathcal{K}_{2}(t), \end{eqnarray}$

(3.69)

then, we have

$\begin{eqnarray} \mathcal{K}_{2}^{\prime}(t) &\leq&-\beta_{6} \zeta_{0}(t) \frac{ E(t)}{E(0)} D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)+k_{6}\zeta_{0}(t)\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)\\ & = &-\beta_{6} \frac{\zeta_{0}(t)}{\mathcal{B}(t)} D_{2}\left(\frac{ E(t) \mathcal{B}(t)}{E(0)}\right)+k_{6}\zeta_{0}(t)\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right), \end{eqnarray}$

(3.70)

where $\beta_{6} = (k_{2}E(0)-2c\varepsilon_{0})$ and $D_{2}(t) = t D^{\prime}\left(\varepsilon_{0} t\right)$ .

Choosing $\varepsilon_{0}$ so small such that $\beta_{6} > 0$ , since $D_{2}^{\prime}(t) = D^{\prime}\left(\varepsilon_{0} t\right)+\varepsilon_{0} t D^{\prime \prime}\left(\varepsilon_{0} t\right)$ . As $D_{2}^{\prime}(t), D_{2}(t) > 0$ on $(0$ , $1]$ and $G_{i}$ on $(0$ , $\varrho]$ are strictly increasing. Applying Young's inequality (3.65) on the last term in (3.70)

with $A = D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)$ and $B = \frac{k_{6}}{\delta}\mu(t)$ , we find

$\begin{eqnarray} k_{6}\mu_{0}(t)D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)& = &\frac{\sigma}{\mathcal{B}(t)}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg)\bigg(D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)\bigg)\\ & < &\frac{\sigma}{\mathcal{B}(t)}D_{3}^{*}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg)+\frac{\sigma}{\mathcal{B}(t)}D_{3}\bigg(D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)\bigg)\\ & < &\frac{\sigma}{\mathcal{B}(t)}D_{4}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg)\\ &&+\frac{\sigma}{\mathcal{B}(t)}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)D^{\prime}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right)\\ & < &\frac{\sigma}{\mathcal{B}(t)}D_{4}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg)+\frac{\sigma\varepsilon_{0} }{\mathcal{B}(t)}D_{2}\left(\varepsilon_{0} \frac{ E(t) \mathcal{B}(t)}{E(0)}\right). \end{eqnarray}$

(3.71)

Here, choose $\sigma$ small enough in a manner that $\beta_{6}-\sigma\varepsilon_{0} > 0$ andcombining (3.70) and (3.71), we have

$\begin{eqnarray} \mathcal{K}_{2}^{\prime}(t) &\leq&-\beta_{7} \frac{\zeta_{0}(t)}{\mathcal{B}(t)} D_{2}\left(\frac{ E(t) \mathcal{B}(t)}{E(0)}\right)+\frac{\sigma\zeta_{0}(t)}{\mathcal{B}(t)}D_{4}\bigg(\frac{k_{6}}{\delta}\mathcal{B}(t)\mu_{0}(t)\bigg). \end{eqnarray}$

(3.72)

where $\beta_{7} = \beta_{6}-\sigma \varepsilon_{0} > 0$ , $D_{3}(t) = t D'^{-1}\left(t\right)$ and $D_{4}(t) = \overline{D}_{3}^{*}\left(t\right)$ .

In light of fact $E' < 0$ and $\mathcal{B}' < 0$ , then $D_{2}(\frac{E(t) \mathcal{B}(t)}{E(0)})$ is decreasing. As a consequences of this, for $0\leq t\leq T$ , we have

$\begin{equation} D_{2}\bigg(\frac{E(T) \mathcal{B}(T)}{E(0)}\bigg) < D_{2}\bigg(\frac{E(t) \mathcal{B}(t)}{E(0)}\bigg). \end{equation}$

(3.73)

In the next step, combine (3.72) with (3.73) and multiply by $\mathcal{B}(t)$ , the following is obtained

$\begin{equation} \mathcal{B}(t)\mathcal{K}_{2}^{\prime}(t)+\beta_{7}\zeta_{0}(t) D_{2}\left(\frac{ E(T) \mathcal{B}(T)}{E(0)}\right) < \sigma\zeta_{0}(t)D_{4}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg). \end{equation}$

(3.74)

Since $\mathcal{B}' < 0$ , then for any $0 < t < T$

$\begin{eqnarray} (\mathcal{B}\mathcal{K}_{2})^{\prime}(t)+\beta_{7}\zeta_{0}(t) D_{2}\left(\frac{ E(T) \mathcal{B}(T)}{E(0)}\right)& < &\sigma \zeta_{0}(t)D_{4}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg)\\ & < &\sigma\widehat{\zeta}(t)D_{4}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg). \end{eqnarray}$

(3.75)

Simplify (3.75) over $[0, T]$ and apply $\mathcal{B}(0) = 1$ , the following is obtained

$\begin{equation} D_{2}\left(\frac{ E(T) \mathcal{B}(T)}{E(0)}\right)\int_{0}^{T}\zeta_{0}(t)dt < \frac{\mathcal{K}_{2}(0)}{\beta_{7}}+\frac{\sigma}{\beta_{7}}\int_{0}^{T}\widehat{\zeta}(t)D_{4}\bigg(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t)\bigg)dt. \end{equation}$

(3.76)

Consequently, we have

$\begin{equation} D_{2}\left(\frac{ E(T) \mathcal{B}(T)}{E(0)}\right) < \frac{\frac{\mathcal{K}_{2}(0)}{\beta_{7}}+\frac{\sigma}{\beta_{7}}\int_{0}^{T}\widehat{\zeta}(t)D_{4}(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t))dt}{\int_{0}^{T}\zeta_{0}(t)dt}. \end{equation}$

(3.77)

As a results of this, we obtain

$\begin{equation} \left(\frac{ E(T) \mathcal{B}(T)}{E(0)}\right) < D_{2}^{-1}\bigg(\frac{\frac{\mathcal{K}_{2}(0)}{\beta_{7}}+\frac{\sigma}{\beta_{7}}\int_{0}^{T}\widehat{\zeta}(t)D_{4}(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t))dt}{\int_{0}^{T}\zeta_{0}(t)dt}\bigg). \end{equation}$

(3.78)

As a result of this, we get

$\begin{equation} E(T) < \frac{ E(0)}{\mathcal{B}(T)}D_{2}^{-1}\bigg(\frac{\frac{\mathcal{K}_{2}(0)}{\beta_{7}}+\frac{\sigma}{\beta_{7}}\int_{0}^{T}\widehat{\zeta}(t)D_{4}(\frac{k_{6}}{\sigma}\mathcal{B}(t)\mu_{0}(t))dt}{\int_{0}^{T}\zeta_{0}(t)dt}\bigg). \end{equation}$

(3.79)

where, we have (3.29) with $\varsigma_{1} = \frac{ E(0)}{\mathcal{B}(T)}$ , $\varsigma_{2} = \frac{\mathcal{K}_{2}(0)}{\beta_{7}}$ , $\varsigma_{3} = \frac{\sigma}{\beta_{7}}$ , and $\varsigma_{4}(t) = \frac{k_{6}}{\sigma}\mathcal{B}(t)$ .

Hence, the required result is obtained 3.8.

4. Conclusions

The purpose of this work was to study when the coupled system of nonlinear viscoelastic wave equations with distributed delay components, infinite memory and Balakrishnan-Taylor damping. Assume the kernels $g_{i} :{\bf R}_{+}\rightarrow {\bf R}_{+}$ holds true the below

$g_{i}'(t)\leq-\zeta_{i}(t)G_{i}(g_{i}(t)), \quad \forall t\in {\bf R}_{+}, \quad \text{for} \quad i = 1, 2,$

in which $\zeta_{i}$ and $G_{i}$ are functions. We prove the stability of the system under this highly generic assumptions on the behaviour of $g_i$ at infinity and by dropping the boundedness assumptions in the historical data. This type of problem is frequently found in some mathematical models in applied sciences. Especially in the theory of viscoelasticity. What interests us in this current work is the combination of these terms of damping, which dictates the emergence of these terms in the problem. In the next work, we will try to using the same method with same problem. But in added of other dampings.

Acknowledgments

The researchers would like to thank the Deanship of Scientific Research, Qassim University for funding the publication of this project.

Conflict of interest

The authors declare there is no conflicts of interest.

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