# Growth of solutions for a coupled nonlinear Klein–Gordon system with strong damping, source, and distributed delay terms

## Abstract

In this work, the exponential growth of solutions for a coupled nonlinear Klein–Gordon system with distributed delay, strong damping, and source terms is proved. Take into consideration some suitable assumptions.

## Introduction

In modeling in the biological, physical, and social sciences, it is sometimes necessary to take account of optimal control or time delays inherent in the phenomena (see for example [4, 16]). The inclusion of delays explicitly in the equations is often a simplification or idealization that is introduced because a detailed description of the underlying processes is too complicated to be modeled mathematically, or because some of the details are unknown. More generally, how does the qualitative behavior depend on the form and magnitude of the delays? In this paper we examine how we can apply the distributed delay term for knowing the behavior of growth of solutions for a coupled nonlinear Klein–Gordon system with strong damping, source terms.

We consider the following system:

$$\textstyle\begin{cases} u_{tt}+m_{1}u^{2}-\Delta u-\omega _{1}\Delta u_{t}+ \int _{0}^{t}g(t-s)\Delta u(s)\,ds \\ \quad{} +\mu _{1}u_{t}+ \int _{\tau _{1}}^{\tau _{2}} \vert \mu _{2}( \varrho ) \vert u_{t}(x,t-\varrho )\,d\varrho =f_{1}(u,v),\quad (x,t)\in \varOmega \times \mathbb{R}_{+}, \\ v_{tt}+m_{2}v^{2}-\Delta v-\omega _{2}\Delta v_{t}+ \int _{0}^{t}h(t-s)\Delta v(s)\,ds \\ \quad{} +\mu _{3}v_{t}+ \int _{\tau _{1}}^{\tau _{2}} \vert \mu _{4}( \varrho ) \vert v_{t}(x,t-\varrho )\,d\varrho =f_{2}(u,v), \quad (x,t)\in \varOmega \times \mathbb{R}_{+}, \\ u ( x,t ) =0, \qquad v ( x,t ) =0, \quad x\in \partial \varOmega, \\ u_{t} ( x,-t ) =f_{0} ( x,t ), \qquad v_{t} ( x,-t ) =k_{0} ( x,t ) \quad (x,t)\in \varOmega \times ( 0,\tau _{2} ), \\ u ( x,0 ) =u_{0} ( x ),\qquad u_{t} ( x,0 ) =u_{1} ( x ), \quad x\in \varOmega, \\ v ( x,0 ) =v_{0} ( x ),\qquad v_{t} ( x,0 ) =v_{1} ( x ),\quad x\in \varOmega,\end{cases}$$
(1.1)

where Ω is a bounded domain in $$\mathbb{R}^{n}$$ with smooth boundary ∂Ω and the source terms are defined as follows:

$$\textstyle\begin{cases} f_{1}(u,v)=a_{1} \vert u+v \vert ^{2(p+1)}(u+v)+b_{1} \vert u \vert ^{p}.u. \vert v \vert ^{p+2}, \\ f_{2}(u,v)=a_{1} \vert u+v \vert ^{2(p+1)}(u+v)+b_{1} \vert v \vert ^{p}.v. \vert u \vert ^{p+2}\end{cases}$$
(1.2)

and $$m_{1},m_{2},\omega _{1},\omega _{2},\mu _{1},\mu _{3},a_{1},b_{1}>0$$, and $$\tau _{1},\tau _{2}$$ are the time delay with $$0\leq \tau _{1}<\tau _{2}$$, and $$\mu _{2},\mu _{4}$$ are $$L^{\infty }$$ functions, and $$g,h$$ are differentiable functions.

Viscous materials are the opposite of elastic materials that possess the ability to store and dissipate mechanical energy. As the mechanical properties of these viscous substances are of great importance when they appear in many applications of natural sciences, many authors have given attention to this problem since the beginning of the new millennium.

In the case of only one equation and if $$\omega _{1}=0$$ (i.e., $$\Delta u_{t}=0$$), and $$\mu _{1}=\mu _{2}=0$$. Our problem (1.1) has been studied in . By using the Galerkin method they established the local existence result. Also, they showed the local solution is global in time under suitable conditions and with the same rate of decaying (polynomial or exponential) of the kernel g. They proved that the dissipation given by the viscoelastic integral term is strong enough to stabilize the oscillations of the solution. Moreover, their result has been obtained under weaker conditions than those used in . In , the authors proved the exponential decay of the following problem:

$$u_{tt}-\Delta u+ \int _{0}^{t}g(t-s)\Delta u(s)\,ds+a(x)u_{t}+ \vert u \vert ^{ \gamma }.u=0.$$
(1.3)

This later result has been improved in , in which they showed that the viscoelastic dissipation alone is strong enough to stabilize the problem even with an exponential rate.

In many works on this field under assumptions of the kernel g. For problem (1.1) and with $$\mu _{1}\neq 0$$, for example, in , the authors proved a blow-up result for the following problem:

$$\textstyle\begin{cases} u_{tt}-\Delta u+ \int _{0}^{\infty }g(t-s)\Delta u(s)\,ds+u_{t}= \vert u \vert ^{p-2}.u, \quad (x,t)\in \mathbb{R}^{n}\times (0,\infty ), \\ u ( x,0 ) =u_{0} ( x ),\qquad u_{t} ( x,0 ) =u_{1} ( x ),\end{cases}$$
(1.4)

where g satisfies $$\int _{0}^{\infty }g(s)\,ds<(2p-4)/(2p-3)$$, initial data were supported with negative energy like that $$\int u_{0}u_{1}\,dx>0$$.

If $$(w>0)$$. In , the authors considered the following problem:

$$\textstyle\begin{cases} u_{tt}-\Delta u+ \int _{0}^{\infty }g(t-s)\Delta u(s)\,ds- \Delta u_{t}= \vert u \vert ^{p-2}.u,\quad (x,t)\in \varOmega \times (0,\infty ), \\ u ( x,0 ) =u_{0} ( x ),\qquad u_{t} ( x,0 ) =u_{1} ( x ). \end{cases}$$
(1.5)

Under suitable assumptions on g that there were solutions of (1.5) with initial energy, they showed the blow-up in a finite time. For the same problem (1.5), in , Song et al. proved that there were solutions of (1.5) with positive initial energy that blows up in finite time. In addition, in  the authors showed a blow-up result if $$p>m$$ and established the global existence of the following problem:

$$\textstyle\begin{cases} u_{tt}-\Delta u+\int _{0}^{\infty }g(s)\Delta u(t-s)\,ds-\varepsilon _{1} \Delta u_{t}+\varepsilon _{2}u_{t} \vert u_{t} \vert ^{m-2}=\varepsilon _{3}u \vert u \vert ^{p-2}, \\ u(x,t)=0,\quad x\in \partial \varOmega,t>0, \\ u ( x,0 ) =u_{0} ( x ),\qquad u_{t} ( x,0 ) =u_{1} ( x ),\quad x\in \varOmega.\end{cases}$$
(1.6)

In the case of coupled of equations, in , the authors studied the following system of equations:

$$\textstyle\begin{cases} u_{tt}-\Delta u+u_{t} \vert u_{t} \vert ^{m-2}=f_{1}(u,v), \\ v_{tt}-\Delta v+v_{t} \vert v_{t} \vert ^{r-2}=f_{2}(u,v),\end{cases}$$
(1.7)

with nonlinear functions $$f_{1}$$ and $$f_{2}$$ satisfying appropriate conditions. Under certain restrictions imposed on the parameters and the initial data, they obtained numerous results on the existence of weak solutions. They also showed that any weak solution with negative initial energy blows up for a finite period of time by using the same techniques as in .

In , the authors considered the system:

$$\textstyle\begin{cases} u_{tt}-\Delta u+(a \vert u \vert ^{k}+b \vert v \vert ^{l})u_{t} \vert u_{t} \vert ^{m-2}=f_{1}(u,v), \\ v_{tt}-\Delta v+(a \vert u \vert ^{\theta }+b \vert v \vert ^{\vartheta })v_{t} \vert v_{t} \vert ^{r-2}=f_{2}(u,v),\end{cases}$$
(1.8)

where they stated and proved the blow-up in finite time of solution under some restrictions on the initial data and (with positive initial energy) for some conditions on the functions $$f_{1}$$ and $$f_{2}$$.

Later, in , the authors extended the result of , where they considered the following nonlinear viscoelastic system:

$$\textstyle\begin{cases} u_{tt}-\Delta u+\int _{0}^{\infty }g(s)\Delta u(t-s)\,ds+(a \vert u \vert ^{k}+b \vert v \vert ^{l})u_{t} \vert u_{t} \vert ^{m-2}=f_{1}(u,v), \\ v_{tt}-\Delta v+\int _{0}^{\infty }h(s)\Delta v(t-s)\,ds+(a \vert u \vert ^{\theta }+b \vert v \vert ^{ \varrho })v_{t} \vert v_{t} \vert ^{r-2}=f_{2}(u,v)\end{cases}$$
(1.9)

and proved that the solutions of the system of wave equations with viscoelastic term, degenerate damping, and strong nonlinear sources acting in both equations at the same time are globally nonexisting provided that the initial data are sufficiently large in a bounded domain of Ω.

To complement the above works, we are working to prove under appropriate assumptions that the solution of problem (1.1) grows exponentially:

$$\lim_{t\rightarrow \infty } \Vert u_{t} \Vert _{2(p+2)}^{2(p+2)}+ \Vert \nabla u \Vert _{2(p+2)}^{2(p+2)} \quad\text{goes to }\infty.$$
(1.10)

The paper is organized as follows. In Sect. 2, some necessary assumptions related to the problem are given. Then, in Sect. 3, the main result is proved.

## Assumptions

We consider the following assumptions:

1. (A1)

$$g,h:\mathbb{R}_{+}\rightarrow \mathbb{R}_{+}$$ are differentiable and decreasing functions such that

$$\textstyle\begin{cases} g(t)\geq 0, \qquad 1-\int _{0}^{\infty }g ( s ) \,ds=l_{1}>0, \\ h(t)\geq 0,\qquad 1-\int _{0}^{\infty }h ( s ) \,ds=l_{2}>0.\end{cases}$$
(2.1)
2. (A2)

There exist constants $$\xi _{1}, \xi _{2}>0$$ such that

$$\textstyle\begin{cases} g^{\prime } ( t ) \leq -\xi _{1}g ( t ),\quad t \geq 0, \\ h^{\prime } ( t ) \leq -\xi _{2}h ( t ),\quad t \geq 0.\end{cases}$$
(2.2)
3. (A3)

$$\mu _{2},\mu _{4}:[\tau _{1},\tau _{2}]\rightarrow \mathbb{R}$$ are $$L^{\infty }$$ functions so that, for all $$\delta >\frac{1}{2}$$,

\begin{aligned} \begin{aligned} &\biggl(\frac{2\delta -1}{2}\biggr) \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert \,d\varrho < \mu _{1}, \\ &\biggl(\frac{2\delta -1}{2}\biggr) \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert \,d\varrho < \mu _{3}. \end{aligned} \end{aligned}
(2.3)

## Main results

In this section, the blow-up result of solution of problem (1.1) is proved.

First, as in , we introduce the new variables:

\begin{aligned} &y(x,\rho,\varrho,t)=u_{t}(x,t-\varrho \rho ), \\ &z(x,\rho,\varrho,t)=v_{t}(x,t-\varrho \rho ), \end{aligned}

then

$$\textstyle\begin{cases} \varrho y_{t}(x,\rho,\varrho,t)+y_{\rho }(x,\rho,\varrho,t)=0, \\ y(x,0,\varrho,t)=u_{t}(x,t),\end{cases}$$
(3.1)

and

$$\textstyle\begin{cases} \varrho z_{t}(x,\rho,\varrho,t)+z_{\rho }(x,\rho,\varrho,t)=0, \\ z(x,0,\varrho,t)=v_{t}(x,t).\end{cases}$$
(3.2)

Let us denote

$$gou= \int _{\varOmega } \int _{0}^{t}g(t-s) \bigl\vert u(t)-u(s) \bigr\vert ^{2}\,ds\,dx.$$
(3.3)

Therefore, problem (1.1) takes the form

$$\textstyle\begin{cases} u_{tt}+m_{1}u^{2}-\Delta u-\omega _{1}\Delta u_{t}+ \int _{0}^{t}g(t-s)\Delta u(s)\,ds \\ \quad{} +\mu _{1}u_{t}+ \int _{\tau _{1}}^{\tau _{2}} \vert \mu _{2}( \varrho ) \vert y(x,1,\varrho,t)\,d\varrho =f_{1}(u,v),\quad x\in \varOmega,t\geq 0, \\ v_{tt}+m_{2}v^{2}-\Delta v-\omega _{2}\Delta v_{t}+ \int _{0}^{t}h(t-s)\Delta v(s)\,ds \\ \quad{} +\mu _{3}v_{t}+ \int _{\tau _{1}}^{\tau _{2}} \vert \mu _{4}( \varrho ) \vert z(x,1,\varrho,t)\,d\varrho =f_{2}(u,v),\quad x\in \varOmega,t\geq 0, \\ \varrho y_{t}(x,\rho,\varrho,t)+y_{\rho }(x,\rho,\varrho,t)=0, \\ \varrho z_{t}(x,\rho,\varrho,t)+z_{\rho }(x,\rho,\varrho,t)=0\end{cases}$$
(3.4)

with the initial and boundary condition

$$\textstyle\begin{cases} u(x,t)=0, \qquad v(x,t)=0,\quad x\in \partial \varOmega, \\ y(x,\rho,\varrho,0)=f_{0} ( x,\varrho \rho ),\qquad z(x,\rho,\varrho,0)=k_{0} ( x,\varrho \rho ), \\ u ( x,0 ) =u_{0} ( x ),\qquad u_{t} ( x,0 ) =u_{1} ( x ), \\ v ( x,0 ) =v_{0} ( x ),\qquad v_{t} ( x,0 ) =v_{1} ( x ),\end{cases}$$
(3.5)

where

$$(x,\rho,\varrho,t)\in \varOmega \times (0,1)\times (\tau _{1},\tau _{2}) \times (0,\infty ).$$

### Theorem 3.1

Assume (2.1), (2.2), and (2.3) hold. Let

$$\textstyle\begin{cases} -1< p< \frac{4-n}{n-2},& n\geq 3, \\ p\geq -1,& n=1,2.\end{cases}$$
(3.6)

Then, for any initial data,

$$(u_{0},u_{1},v_{0},v_{1},f_{0},k_{0}) \in \mathcal{H},$$

where

\begin{aligned} \mathcal{H} ={}&H_{0}^{1}(\varOmega )\times L^{2}(\varOmega )\times H_{0}^{1}( \varOmega ) \times L^{2}(\varOmega )\times L^{2}\bigl(\varOmega \times (0,1)\times ( \tau _{1},\tau _{2})\bigr) \\ &{}\times L^{2}\bigl(\varOmega \times (0,1)\times (\tau _{1},\tau _{2})\bigr), \end{aligned}

problem (3.4) has a unique solution

$$u\in C\bigl([0,T];\mathcal{H}\bigr)$$

for some$$T>0$$.

In the next theorem we give the global existence result, its proof is based on the potential well depth method in which the concept of so-called stable set appears, where we show that if we restrict our initial data in the stable set, then our local solution obtained is global in time. We will make use of arguments in .

### Theorem 3.2

Suppose that (2.1), (2.2), (2.3), and (3.6) hold. If$$u_{0},v_{0}\in W$$, $$u_{1},v_{1}\in H_{0}^{1}(\varOmega )$$, $$y,z\in L^{2}(\varOmega \times (0,1)\times (\tau _{1},\tau _{2}))$$, and

$$\frac{bC_{\ast }^{p}}{l}\biggl(\frac{2p}{(p-2)l}E(0)\biggr)^{\frac{p-2}{2}}< 1,$$
(3.7)

where$$C_{\ast }$$is the best Poincare constant, then the local solution$$(u,v,y,z)$$is global in time.

In order to achieve the main result, the following lemmas are needed.

### Lemma 3.1

There exists a function $$F(u,v)$$ such that

\begin{aligned} F(u,v) &=\frac{1}{2(\rho +2)} \bigl[ uf_{1}(u,v)+vf_{2}(u,v) \bigr] \\ &=\frac{1}{2(\rho +2)} \bigl[ a_{1} \vert u+v \vert ^{2(p+2)}+2b_{1} \vert uv \vert ^{p+2} \bigr] \geq 0, \end{aligned}

where

$$\frac{\partial F}{\partial u}=f_{1}(u,v), \qquad \frac{\partial F}{\partial v}=f_{2}(u,v),$$

taking $$a_{1}=b_{1}=1$$ for convenience.

### Lemma 3.2

()

There exist two positive constants $$c_{0}$$ and $$c_{1}$$ such that

$$\frac{c_{0}}{2(\rho +2)} \bigl( \vert u \vert ^{2(p+2)}+ \vert v \vert ^{2(p+2)} \bigr) \leq F(u,v)\leq \frac{c_{1}}{2(\rho +2)} \bigl( \vert u \vert ^{2(\rho +2)}+ \vert v \vert ^{2(p+2)} \bigr).$$
(3.8)

Define the energy functional as follows.

### Lemma 3.3

Assume that (2.1), (2.2), (2.3), and (3.6) hold, let$$(u,v,y,z)$$be a solution of (3.4), then$$E(t)$$is nonincreasing, that is,

\begin{aligned} E(t) ={}&\frac{1}{2} \Vert u_{t} \Vert _{2}^{2}+ \frac{1}{2} \Vert v_{t} \Vert _{2}^{2}+ \frac{m_{1}}{2} \Vert u \Vert _{2}^{2}+ \frac{m_{2}}{2} \Vert v \Vert _{2}^{2}+ \frac{1}{2}l_{1} \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{2}l_{2} \Vert \nabla v \Vert _{2}^{2} \\ &{}+\frac{1}{2}(go\nabla u)+\frac{1}{2}(ho\nabla v)+ \frac{1}{2}K(y,z)- \int _{\varOmega }F(u,v)\,dx \end{aligned}
(3.9)

satisfies

\begin{aligned} E^{\prime }(t) \leq{} &{-}c_{3}\biggl\{ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2}+ \int _{ \varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1, \varrho,t)\,d\varrho\, dx \\ &{}+ \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,1, \varrho,t)\,d\varrho\, dx\biggr\} \leq 0, \end{aligned}
(3.10)

where

$$K(y,z)= \int _{\varOmega } \int _{0}^{1} \int _{\tau _{1}}^{\tau _{2}} \varrho \bigl\{ \bigl\vert \mu _{2}(\varrho ) \bigr\vert y^{2}(x,\rho,\varrho,t)+ \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,\rho, \varrho,t)\bigr\} \,d\varrho\, d\rho\, dx.$$
(3.11)

### Proof

By multiplying the first and the second equation in (3.4) respectively by $$u_{t},v_{t}$$ and integrating over Ω, we get

\begin{aligned} &\frac{d}{dt}\biggl\{ \frac{1}{2} \Vert u_{t} \Vert _{2}^{2}+\frac{1}{2} \Vert v_{t} \Vert _{2}^{2}+\frac{m_{1}}{2} \Vert u \Vert _{2}^{2}+ \frac{m_{2}}{2} \Vert v \Vert _{2}^{2}+\frac{1}{2}l_{1} \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{2}l_{2} \Vert \nabla v \Vert _{2}^{2}+\frac{1}{2}(go\nabla u) \\ &\qquad{}+\frac{1}{2}(ho\nabla v)- \int _{\varOmega }F(u,v)\,dx\biggr\} \\ &\quad=-\mu _{1} \Vert u_{t} \Vert _{2}^{2}-m_{1} \Vert u \Vert _{2}^{2}- \int _{\varOmega }u_{t} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y(x,1, \varrho,t)\,d\varrho\, dx \\ &\qquad{}-\mu _{3} \Vert v_{t} \Vert _{2}^{2}-m_{2} \Vert v \Vert _{2}^{2}- \int _{ \varOmega }v_{t} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z(x,1, \varrho,t)\,d\varrho\, dx \\ &\qquad{}+\frac{1}{2}\bigl(g^{\prime }o\nabla u\bigr)-\frac{1}{2}g(t) \Vert \nabla u \Vert _{2}^{2}-\omega _{1} \Vert \nabla u_{t} \Vert _{2}^{2} \\ &\qquad{}+\frac{1}{2}\bigl(h^{\prime }o\nabla v\bigr)-\frac{1}{2}h(t) \Vert \nabla v \Vert _{2}^{2}-\omega _{2} \Vert \nabla v_{t} \Vert _{2}^{2}, \end{aligned}
(3.12)

and, from the initial and boundary condition in (3.4)

\begin{aligned} &\frac{d}{dt}\frac{1}{2} \int _{\varOmega } \int _{0}^{1} \int _{\tau _{1}}^{ \tau _{2}}\varrho \bigl\vert \mu _{2}(\varrho ) \bigr\vert y^{2}(x,\rho,\varrho,t)\,d \varrho\, d\rho\, dx \\ &\quad =-\frac{1}{2} \int _{\varOmega } \int _{0}^{1} \int _{ \tau _{1}}^{\tau _{2}}2 \bigl\vert \mu _{2}( \varrho ) \bigr\vert yy_{\rho }\,d\varrho\, d\rho\, dx \\ &\quad=\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,0,\varrho,t)\,d\varrho\, dx \\ &\qquad{}-\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1,\varrho,t)\,d\varrho\, dx \\ &\quad=\frac{1}{2}\biggl( \int _{\tau _{1}}^{\tau _{2}}|\mu _{2}(\varrho )\,d \varrho \biggr) \Vert u_{t} \Vert _{2}^{2} \\ &\qquad{}-\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1,\varrho,t)\,d\varrho\, dx \end{aligned}
(3.13)

and

\begin{aligned} &\frac{d}{dt}\frac{1}{2} \int _{\varOmega } \int _{0}^{1} \int _{\tau _{1}}^{ \tau _{2}}\varrho \bigl\vert \mu _{4}(\varrho ) \bigr\vert z^{2}(x,\rho,\varrho,t)\,d \varrho\, d\rho\, dx \\ &\quad=-\frac{1}{2} \int _{\varOmega } \int _{0}^{1} \int _{ \tau _{1}}^{\tau _{2}}2 \bigl\vert \mu _{4}( \varrho ) \bigr\vert zz_{\rho }\,d\varrho\, d\rho\, dx \\ &\quad=\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,0,\varrho,t)\,d\varrho\, dx \\ &\qquad{}-\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,1,\varrho,t)\,d\varrho\, dx \\ &\quad=\frac{1}{2}\biggl( \int _{\tau _{1}}^{\tau _{2}}|\mu _{4}(\varrho )\,d \varrho \biggr) \Vert v_{t} \Vert _{2}^{2} \\ &\qquad{}-\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,1,\varrho,t)\,d\varrho\, dx, \end{aligned}
(3.14)

then

\begin{aligned} \frac{d}{dt}E(t) ={}&{-}\mu _{1} \Vert u_{t} \Vert _{2}^{2}-m_{1} \Vert u \Vert _{2}^{2}- \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert u_{t}y(x,1,\varrho,t)\,d\varrho\, dx+ \frac{1}{2}\bigl(g^{ \prime }o\nabla u\bigr) \\ &{}-\frac{1}{2}g(t) \Vert \nabla u \Vert _{2}^{2}- \omega _{1} \Vert \nabla u_{t} \Vert _{2}^{2}+ \frac{1}{2}\biggl( \int _{\tau _{1}}^{\tau _{2}}| \mu _{2}(\varrho )\,d \varrho \biggr) \Vert u_{t} \Vert _{2}^{2} \\ &{}-\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1,\varrho,t)\,d\varrho\, dx \\ &{}-\mu _{3} \Vert v_{t} \Vert _{2}^{2}-m_{2} \Vert v \Vert _{2}^{2}- \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert v_{t}z(x,1, \varrho,t)\,d\varrho\, dx+ \frac{1}{2}\bigl(h^{\prime }o\nabla v\bigr) \\ &{}-\frac{1}{2}h(t) \Vert \nabla v \Vert _{2}^{2}- \omega _{2} \Vert \nabla v_{t} \Vert _{2}^{2}+ \frac{1}{2}\biggl( \int _{\tau _{1}}^{\tau _{2}}| \mu _{4}(\varrho )\,d \varrho \biggr) \Vert v_{t} \Vert _{2}^{2} \\ &{}-\frac{1}{2} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,1,\varrho,t)\,d\varrho\, dx. \end{aligned}
(3.15)

By (3.12)–(3.14), we get (3.9). Also, by using Young’s inequality, (2.1), (2.2), and (2.3) in (3.15), we obtain (3.10). □

Now, we define the functional

\begin{aligned} \mathbb{H}(t)=-E(t) ={}&{-}\frac{1}{2} \Vert u_{t} \Vert _{2}^{2}- \frac{1}{2} \Vert v_{t} \Vert _{2}^{2}-\frac{m_{1}}{2} \Vert u \Vert _{2}^{2}- \frac{m_{2}}{2} \Vert v \Vert _{2}^{2}-\frac{1}{2}l_{1} \Vert \nabla u \Vert _{2}^{2} \\ &{}-\frac{1}{2}l_{2} \Vert \nabla v \Vert _{2}^{2}-\frac{1}{2}(go\nabla u)-\frac{1}{2}(ho\nabla v)-\frac{1}{2}K(y,z) \\ &{}+\frac{1}{2(p+2)}\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr]. \end{aligned}
(3.16)

### Theorem 3.3

Assume that (2.1)(2.3) and (3.6) hold. Assume further that$$E(0)<0$$, then the solution of problem (3.4) grows exponentially.

### Proof

From (3.9 we have

$$E(t)\leq E(0)\leq 0.$$
(3.17)

Therefore,

\begin{aligned} \mathbb{H}^{\prime }(t)={}&{-}E^{\prime }(t) \\ \geq {}&c_{3}\biggl( \Vert u_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \int _{\varOmega } \int _{\tau _{1}}^{ \tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1,\varrho,t)\,d\varrho\, dx \\ &{}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2}+ \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,1,\varrho,t)\,d \varrho\, dx\biggr). \end{aligned}
(3.18)

Hence

\begin{aligned} \begin{aligned} &\mathbb{H}^{\prime }(t) \geq c_{3} \int _{\varOmega } \int _{\tau _{1}}^{ \tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1,\varrho,t)\,d\varrho\, dx\geq 0, \\ &\mathbb{H}^{\prime }(t) \geq c_{3} \int _{\varOmega } \int _{\tau _{1}}^{ \tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert z^{2}(x,1,\varrho,t)\,d\varrho\, dx\geq 0, \end{aligned} \end{aligned}
(3.19)

and

\begin{aligned} 0\leq \mathbb{H}(0)\leq \mathbb{H}(t) \leq {}&\frac{1}{2(p+2)}\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr] \\ \leq {}&\frac{c_{1}}{2(p+2)}\bigl[ \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)}^{2(p+2)}\bigr]. \end{aligned}
(3.20)

Setting

\begin{aligned} \mathcal{K}(t) ={}&\mathbb{H}+\varepsilon \int _{\varOmega }(uu_{t}+vv_{t})\,dx+\frac{\varepsilon }{2} \int _{\varOmega }\bigl(\mu _{1}u^{2}+\mu _{3}v^{2}\bigr)\,dx \\ &{}+\frac{\varepsilon }{2} \int _{\varOmega }\bigl(\omega _{1}(\nabla u)^{2}+ \omega _{2}(\nabla v)^{2}\bigr)\,dx, \end{aligned}
(3.21)

where $$\varepsilon >0$$ to be assigned later.

By multiplying the first and second equation on (3.4) respectively by $$u,v$$ and with a derivative of (3.21), we get

\begin{aligned} \mathcal{K}^{\prime }(t) ={}&\mathbb{H}^{\prime }(t)+\varepsilon \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2} \bigr)-\varepsilon \bigl( \Vert \nabla u \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{2}\bigr) \\ &{}+\varepsilon \int _{\varOmega }\nabla u \int _{0}^{t}g(t-s)\nabla u(s)\,ds\,dx+ \varepsilon \int _{\varOmega }\nabla v \int _{0}^{t}h(t-s)\nabla v(s)\,ds\,dx \\ &{}-\varepsilon \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert uy(x,1,\varrho,t)\,d\varrho\, dx-\varepsilon \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert vz(x,1,\varrho,t)\,d \varrho\, dx \\ &{}+\varepsilon {}\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr]. \end{aligned}
(3.22)

Using Young’s inequality, we get

\begin{aligned} &\varepsilon \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert uy(x,1,\varrho,t)\,d\varrho\, dx \\ &\quad\leq \varepsilon \biggl\{ \delta _{1}\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert \,d\varrho \biggr) \Vert u \Vert _{2}^{2} \\ &\qquad{}+\frac{1}{4\delta _{1}} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1,\varrho,t)\,d\varrho\, dx\biggr\} . \end{aligned}
(3.23)

Thus

\begin{aligned} &\varepsilon \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert vz(x,1,\varrho,t)\,d\varrho \,dx \\ &\quad \leq \varepsilon \biggl\{ \delta _{2}\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert \,d\varrho \biggr) \Vert v \Vert _{2}^{2} \\ &\qquad{}+\frac{1}{4\delta _{2}} \int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,1,\varrho,t)\,d\varrho\, dx\biggr\} . \end{aligned}
(3.24)

Also

\begin{aligned} &\varepsilon \int _{0}^{t}g(t-s)\,ds \int _{\varOmega }\nabla u.\nabla u(s)\,dx\,ds \\ &\quad=\varepsilon \int _{0}^{t}g(t-s)\,ds \int _{\varOmega }\nabla u.\bigl(\nabla u(s)- \nabla u(t)\bigr)\,dx\,ds \\ &\qquad{}+\varepsilon \int _{0}^{t}g(s)\,ds \Vert \nabla u \Vert _{2}^{2} \\ &\quad\geq \frac{\varepsilon }{2} \int _{0}^{t}g(s)\,ds \Vert \nabla u \Vert _{2}^{2}-\frac{\varepsilon }{2}(go\nabla u), \end{aligned}
(3.25)

so

\begin{aligned} &\varepsilon \int _{0}^{t}h(t-s)\,ds \int _{\varOmega }\nabla v.\nabla v(s)\,dx\,ds \\ &\quad=\varepsilon \int _{0}^{t}h(t-s)\,ds \int _{\varOmega }\nabla v.\bigl(\nabla v(s)- \nabla v(t)\bigr)\,dx\,ds \\ &\qquad{}+\varepsilon \int _{0}^{t}h(s)\,ds \Vert \nabla v \Vert _{2}^{2} \\ &\quad\geq \frac{\varepsilon }{2} \int _{0}^{t}h(s)\,ds \Vert \nabla v \Vert _{2}^{2}-\frac{\varepsilon }{2}(ho\nabla v). \end{aligned}
(3.26)

From (3.22)

\begin{aligned} \mathcal{K}^{\prime }(t) \geq {}&\mathbb{H}^{\prime }(t)+\varepsilon \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert u_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2} \bigr) \\ &{}-\varepsilon \biggl(\biggl(1-\frac{1}{2} \int _{0}^{t}g(s)\,ds\biggr) \Vert \nabla u \Vert _{2}^{2}+\biggl(1-\frac{1}{2} \int _{0}^{t}h(s)\,ds\biggr) \Vert \nabla v \Vert _{2}^{2}\biggr) \\ &{}-\varepsilon \delta _{1}\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert \,d\varrho \biggr) \Vert u \Vert _{2}^{2}- \varepsilon \delta _{2}\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert \,d\varrho \biggr) \Vert v \Vert _{2}^{2} \\ &{}-\frac{\varepsilon }{2}(go\nabla u)- \frac{\varepsilon }{4\delta _{1}}\int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert y^{2}(x,1, \varrho,t)\,d\varrho\, dx \\ &{}-\frac{\varepsilon }{2}(ho\nabla v)- \frac{\varepsilon }{4\delta _{2}}\int _{\varOmega } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert z^{2}(x,1, \varrho,t)\,d\varrho\, dx \\ &{}+\varepsilon {}\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr]. \end{aligned}
(3.27)

Therefore, using (3.19) and by setting $$\delta _{1},\delta _{1}$$ so that $$\frac{1}{4\delta _{1}c_{3}}=\frac{\kappa }{2}$$ and $$\frac{1}{4\delta _{2}c_{3}}=\frac{\kappa }{2}$$, substituting in (3.27), we get

\begin{aligned} \mathcal{K}^{\prime }(t) \geq{} &[1-\varepsilon \kappa ] \mathbb{H}^{ \prime }(t)+\varepsilon \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2}\bigr) \\ &{}-\varepsilon {}\biggl[ \biggl(1-\frac{1}{2} \int _{0}^{t}g(s)\,ds\biggr)\biggr] \Vert \nabla u \Vert _{2}^{2}-\varepsilon {}\biggl[ \biggl(1- \frac{1}{2} \int _{0}^{t}h(s)\,ds\biggr)\biggr] \Vert \nabla v \Vert _{2}^{2} \\ &{}-\varepsilon \frac{1}{2c_{3}\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert \,d\varrho \biggr) \Vert u \Vert _{2}^{2}- \frac{\varepsilon }{2}(go\nabla u) \\ &{}-\varepsilon \frac{1}{2c_{3}\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert \,d\varrho \biggr) \Vert v \Vert _{2}^{2}- \frac{\varepsilon }{2}(ho\nabla v) \\ &{}+\varepsilon {}\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr]. \end{aligned}
(3.28)

For $$0< a<1$$, from (3.16)

\begin{aligned} \varepsilon {}\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr] ={}&\varepsilon a\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr] \\ &{}+\varepsilon 2(p+2) (1-a)\mathbb{H}(t) \\ &{}+\varepsilon (p+2) (1-a) \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}\bigr) \\ &{}+\varepsilon (p+2) (1-a) \biggl(1- \int _{0}^{t}g(s)\,ds\biggr) \Vert \nabla u \Vert _{2}^{2} \\ &{}+\varepsilon (p+2) (1-a) \biggl(1- \int _{0}^{t}h(s)\,ds\biggr) \Vert \nabla v \Vert _{2}^{2} \\ &{}-\varepsilon (p+2) (1-a) (go\nabla u) \\ &{}-\varepsilon (p+2) (1-a) (ho\nabla v) \\ &{}+\varepsilon (p+2) (1-a)K(y,z). \end{aligned}
(3.29)

Substituting in (3.28), we get

\begin{aligned} \mathcal{K}^{\prime }(t) \geq {}&[1-\varepsilon \kappa ] \mathbb{H}^{ \prime }(t)+\varepsilon {}\bigl[ (p+2) (1-a)+1\bigr]\bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2} \bigr) \\ &{}+\varepsilon {}\biggl[ (p+2) (1-a) \biggl(1- \int _{0}^{t}g(s)\,ds\biggr)-\biggl(1- \frac{1}{2}\int _{0}^{t}g(s)\,ds\biggr)\biggr] \Vert \nabla u \Vert _{2}^{2} \\ &{}+\varepsilon {}\biggl[ (p+2) (1-a) \biggl(1- \int _{0}^{t}h(s)\,ds\biggr)-\biggl(1- \frac{1}{2}\int _{0}^{t}h(s)\,ds\biggr)\biggr] \Vert \nabla v \Vert _{2}^{2} \\ &{}-\varepsilon \frac{1}{2c_{3}\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}( \varrho ) \bigr\vert \,d\varrho \biggr) \Vert u \Vert _{2}^{2}- \varepsilon \frac{1}{2c_{3}\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}( \varrho ) \bigr\vert \,d \varrho \biggr) \Vert v \Vert _{2}^{2} \\ &{}+\varepsilon (p+2) (1-a)K(y,z)+\varepsilon {}\biggl[ (p+2) (1-a)- \frac{1}{2}\biggr](go\nabla u+ho\nabla v) \\ &{}+\varepsilon a\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr]+ \varepsilon 2(p+2) (1-a) \mathbb{H}(t). \end{aligned}
(3.30)

Using Poincare’s inequality, we obtain

\begin{aligned} \mathcal{K}^{\prime }(t) \geq {}&[1-\varepsilon \kappa ] \mathbb{H}^{ \prime }(t)+\varepsilon {}\bigl[ (p+2) (1-a)+1\bigr]\bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2} \bigr) \\ &{}+\varepsilon \biggl\{ \bigl[(p+2) (1-a)-1\bigr]-\biggl( \int _{0}^{t}g(s)\,ds\biggr)\biggl[(p+2) (1-a)- \frac{1}{2}\biggr] \\ &{}-\frac{c}{2\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \,ds\biggr)\biggr\} \Vert \nabla u \Vert _{2}^{2} \\ &{}+\varepsilon \biggl\{ \bigl[(p+2) (1-a)-1\bigr]-\biggl( \int _{0}^{t}h(s)\,ds\biggr)\biggl[(p+2) (1-a)- \frac{1}{2}\biggr] \\ &{}-\frac{c}{2\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \,ds\biggr)\biggr\} \Vert \nabla v \Vert _{2}^{2} \\ &{}+\varepsilon (p+2) (1-a)K(y,z)+\varepsilon {}\biggl[ (p+2) (1-a)- \frac{1}{2}\biggr](go\nabla u+ho\nabla v) \\ &{}+\varepsilon c_{0}a\bigl[ \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)}^{2(p+2)}\bigr] \\ &{}+\varepsilon 2(p+2) (1-a)\mathbb{H}(t). \end{aligned}
(3.31)

In this stage, we take $$a>0$$ small enough so that

$$\alpha _{1}=(p+2) (1-a)-1>0,$$

and we assume

$$\max \biggl\{ \int _{0}^{\infty }g(s)\,ds, \int _{0}^{\infty }h(s)\,ds\biggr\} < \frac{(p+2)(1-a)-1}{((p+2)(1-a)-\frac{1}{2})}= \frac{2\alpha _{1}}{2\alpha _{1}+1}.$$
(3.32)

Then we choose κ so large that

\begin{aligned} \alpha _{2} ={}&\biggl\{ (p+2) (1-a)-1)- \int _{0}^{t}g(s)\,ds\biggl((p+2) (1-a)- \frac{1}{2}\biggr) \\ &{} -\frac{c}{2\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \,ds\biggr)\biggr\} >0, \\ \alpha _{3} ={}&\biggl\{ (p+2) (1-a)-1)- \int _{0}^{t}h(s)\,ds\biggl((p+2) (1-a)- \frac{1}{2}\biggr) \\ &{} -\frac{c}{2\kappa }\biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \,ds\biggr)\biggr\} >0. \end{aligned}

We fixed κ and a, we appoint ε small enough so that

$$\alpha _{4}=1-\varepsilon \kappa >0,$$

and from (3.21) we get

\begin{aligned} \mathcal{K}(t) &\leq \frac{1}{2(p+2)}\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2} \bigr] \\ &\leq \frac{c_{1}}{2(p+2)}\bigl[ \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)}^{2(p+2)}\bigr]. \end{aligned}
(3.33)

Thus, for some $$\beta >0$$, estimate (3.31) becomes

\begin{aligned} \mathcal{K}^{\prime }(t) \geq {}&\beta \bigl\{ \mathbb{H}(t)+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2}+ \Vert \nabla u \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{2} \\ &{}+(go\nabla u)+(ho\nabla v)+K(y,z) \\ &{}+\bigl[ \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert u \Vert _{2(p+2)}^{2(p+2)}\bigr] \bigr\} . \end{aligned}
(3.34)

By (3.8), for some $$\beta _{1}>0$$,

\begin{aligned} \mathcal{K}^{\prime }(t) \geq{} &\beta _{1}\bigl\{ \mathbb{H}(t)+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2}+ \Vert \nabla u \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{2} \\ &{}+(go\nabla u)+(ho\nabla v)+K(y,z) \\ &{}+\bigl[ \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\bigr]\bigr\} \end{aligned}
(3.35)

and

$$\mathcal{K}(t)\geq \mathcal{K}(0)>0,\quad t>0.$$
(3.36)

Next, using Young’s and Poincare’s inequalities (), thus from (3.21), we have

\begin{aligned} \mathcal{K}(t) ={}&\biggl(\mathbb{H}^{1-\alpha }+\varepsilon \int _{\varOmega }(uu_{t}+vv_{t})\,dx+ \frac{\varepsilon }{2} \int _{\varOmega }\bigl(\mu _{1}u^{2}+\mu _{3}v^{2}\bigr)\,dx \\ &{}+\frac{\varepsilon }{2} \int _{\varOmega }\bigl(\omega _{1}\nabla u^{2}+ \omega _{2}\nabla v^{2}\bigr)\,dx\biggr) \\ \leq {}&c\biggl\{ \mathbb{H}(t)+ \biggl\vert \int _{\varOmega }(uu_{t}+vv_{t})\,dx \biggr\vert + \Vert u \Vert _{2}+ \Vert \nabla u \Vert _{2} \\ &{} + \Vert v \Vert _{2}+ \Vert \nabla v \Vert _{2}\biggr\} \\ \leq{} &c[\mathbb{H}(t)+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2}+ \Vert \nabla u \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{2} \\ \leq {}&c\bigl[\mathbb{H}(t)+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+ \Vert u \Vert _{2}^{2}+ \Vert v \Vert _{2}^{2}+ \Vert \nabla u \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{2}+(go \nabla u) \\ &{}+(ho\nabla v)+ \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)}^{2(p+2)}\bigr] \end{aligned}
(3.37)

for some $$c>0$$. From inequalities (3.34) and (3.37) we obtain the differential inequality

$$\mathcal{K}^{\prime }(t)\geq \lambda \mathcal{K}(t),$$
(3.38)

where $$\lambda >0$$, depending only on β and c.

A simple integration of (3.38) gives

$$\mathcal{K}(t)\geq \mathcal{K}(0)e^{(\lambda t)}\quad \text{for any }t>0.$$
(3.39)

From (3.21) and (3.33), then

$$\mathcal{K}(t)\leq \frac{c_{1}}{2(p+2)}\bigl[ \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)}^{2(p+2)} \bigr].$$
(3.40)

By (3.39) and (3.40) we have

$$\Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)}^{2(p+2)}\geq Ce^{( \lambda t)},\quad \forall t>0.$$

Therefore, we conclude that the solution grows exponentially. This completes the proof. □

## Conclusion

In this work, the growth of solutions for a coupled nonlinear Klein–Gordon system with strong damping, source, and distributed delay terms was studied. Next, motivated by last works in [1, 3, 5, 810, 1315, 20, 21, 2427, 31], and , we obtained the growth and blow-up for the studied problem (1.1) by constructing a type of cross-constrained variational problem and establishing so-called cross-invariant manifolds of the evolution flow. Then, the result of how small the initial data for which the solution exists globally was proved by using the scaling argument.

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### Acknowledgements

The authors would like to thank the anonymous referees and the handling editor for their careful reading and for relevant remarks/suggestions. The third author would like to thank his Professors/Scientists Pr. Mohamed Haiour and Pr. Azzedine Benchettah for the important content of masters and PhD courses in pure and applied mathematics which he received during his studies. Moreover, he thanks them for the additional help they provided to him during office hours in their office about the few concepts/difficulties he had encountered, and he appreciates their talent and dedication for their postgraduate students currently and previously!

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