Blow-up of solutions for a system of nonlocal singular viscoelastic equations with sources and distributed delay terms

Abstract

In this paper, we investigate a scenario concerning a coupled nonlocal singular viscoelastic equation with sources and distributed delay terms. By establishing suitable conditions, we have proved that a finite-time blow-up occurs in the solution.

1 Introduction

In this paper, we examine the subsequent system comprising two singular nonlinear and nonlocal equations, inspired by a one-dimensional viscoelastic system:

$$ \textstyle\begin{cases} u_{tt}-\frac{1}{x}(xu_{x})_{x}+\int _{0}^{t}g_{1}(t-s) \frac{1}{x}(xu_{x}(x,s))_{x}\,ds+\mu _{1} u_{t} \\ \quad {}+\int _{\tau _{1}}^{\tau _{2}}\mu _{2}(s)u_{t} ( x, t-s )\,ds=f_{1} ( u,v ) ,\quad \text{in }Q, \\ v_{tt}-\frac{1}{x}(xv_{x})_{x}+\int _{0}^{t}g_{2}(t-s) \frac{1}{x}(xv_{x}(x,s))_{x}\,ds+\mu _{3} v_{t} \\ \quad {}+\int _{\tau _{1}}^{\tau _{2}}\mu _{4}(s)v _{t} ( x, t-s )\,ds=f_{2} ( u,v ) ,\quad \text{in }Q, \\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x), \quad x\in (0,L), \\ v(x,0)=v_{0}(x),\qquad v_{t}(x,0)=v_{1}(x),\quad x\in (0,L), \\ u_{t}(x,-t)=h_{0}(x,t),\qquad v_{t}(x,-t)=k_{0}(x,t), \quad (x,t)\in (0,L)\times (0,\tau _{2}), \\ u(L,t)=v(L,t)=0,\qquad \int _{0}^{L}xu(x,t)\,dx=\int _{0}^{L}xv(x,t)\,dx=0,\end{cases} $$

(1)

where

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

(2)

and $Q=(0,L)\times (0,T)\times (\tau _{1},\tau _{2}))$, $L<\infty$, $T<\infty$, $\mu _{1},\mu _{3} >0$, $g_{1}(.)$, $g_{2}(.): \mathbb{R} ^{+}\rightarrow \mathbb{R} ^{+}$ and $f_{1},f_{2}:\mathbb{R}^{2}\longrightarrow \mathbb{R}$ are functions stated in (8).

These issues arise in one-dimensional or longitudinal elasticity when considering long-term memory viscosity. The second integral represents the distributed delay terms, where $\tau _{1}, \tau _{2} >0$ denote time delays, $\mu _{2}$, $\mu _{4}$ are $L^{\infty}$ functions.

Our work is motivated by the findings presented in the following papers:

In [13], the authors examined a model depicting the motion of a viscoelastic two-dimensional body on the unit disc, focusing on radial solutions. Furthermore, they established the uniqueness and existence of the generalized solution for the below stated nonlocal problem

$$ \textstyle\begin{cases} u_{tt}-\frac {1}{x}(xu_{x})_{x}+\int _{0}^{t}g(t-s) \frac{1}{x}(xu_{x}(x,s))_{x}\,ds=f(x,t,u,u_{x}), \quad \text{in }Q, \\ u_{x}(1,t)=0, \qquad \int _{0}^{1}xu(x,t)\,dx=0, \quad t \in ( 0,T ) , \\ u(x,0)=\varphi (x),\qquad u_{t}(x,0)=\psi (x),\quad x\in (0,1),\end{cases} $$

where $Q=(0,1)\times ( 0,T )$ with right hand side f is a Lipshitzian function.

The findings presented in [14] indicate the occurrence of blow-up for large initial data and demonstrate decay outcomes for sufficiently small initial data, which are applicable to the subsequent nonlocal singular problem

$$ \textstyle\begin{cases} u_{tt}-\frac{1}{x}(xu_{x})_{x}+\int _{0}^{t}g(t-s) \frac{1}{x}(xu_{x}(x,s))_{x}\,ds= \vert u \vert ^{p-2}u, \\ u(a,t)=0,\qquad \int _{0}^{a}xu(x,t)\,dx=0, \\ u(x,0)=\varphi (x),\qquad u_{t}(x,0)=\psi (x).\end{cases} $$

They acquired the blow-up properties of local solutions utilizing Georgiev–Todorova method, even with negative initial energy. Authors in [11], employed the direct method to prove blow-up of solutions under appropriate conditions on initial data [23]. The authors in [26], extended the previous result to systems with higher dimensions and get additional blow-up results. The authors in [15], substituted the source terms $f_{1}(u,v)$ and $f_{2}(u,v)$ in the studied system in [26], respectively by $\vert v \vert ^{q+1} \vert u \vert ^{p-1}u$, $\vert u \vert ^{p+1} \vert v \vert ^{q-1}v$ and the Bessel operator $\frac {1}{x}\frac {\partial }{\partial x} ( x \frac {\partial }{\partial x} )$ instead of Laplace operator Δ and considered the nonlocal boundary condition

$$ \int _{0}^{L}xu(x,t)\,dx= \int _{0}^{L}xv(x,t)\,dx=0,\quad L< \infty , p, q>1 $$

and with two different functions $g(.)$. Moreover, it is augmented by both nonlocal and classical condition. Furthermore, in [17], the authors investigated the identical problem presented in [15], where they derived a nonlinear source of polynomial nature. This source is capable of inducing solutions to blow up within a finite time frame, even in the existence of enhanced damping $u_{t}$. They considered three different cases regarding the sign of the initial energy. Pişkin and Ekinci [22] have addressed problem (1) by substituting Bessel operator with the Kirchhoff operator featuring degenerate damping terms. They employed a technique identical to that used to establish the global existence and provide a decay rate for solutions, as well as demonstrate a finite-time blow-up when the behavior of decreasing relaxation functions is stated as:

$$ \begin{gathered} g_{1}^{\prime }(t)\leq -\xi (t)g_{1}(t), \quad t\geq 0, \\ g_{2}^{\prime }(t)\leq -\xi (t)g_{2}(t),\quad t\geq 0,\end{gathered}$$

and $\xi (t)$ satisfies

$$ \int _{0}^{\infty }\xi (s)\,ds=+\infty ,\quad \forall t>0. $$

Boulaaras et al. [18] investigated the subsequent system, which consists of two singular one-dimensional nonlinear equations arising in generalized viscoelasticity, featuring long-term memory, nonlocal boundary conditions, and general source terms

$$ \textstyle\begin{cases} u_{tt}-\frac{1}{x}(xu_{x})_{x}+\int _{0}^{t}g_{1}(t-s) \frac{1}{x}(xu_{x}(x,s))_{x}\,ds=f_{1} ( u,v ) ,\quad \text{in }Q, \\ v_{tt}-\frac{1}{x}(xv_{x})_{x}+\int _{0}^{t}g_{2}(t-s) \frac{1}{x}(xv_{x}(x,s))_{x}\,ds=f_{2} ( u,v ) ,\quad \text{in }Q. \end{cases} $$

By employing potential-well theory the authors proved the existence of a global solution for the problem. And in the same vein, in [5] Boularaas and Mezouar proved the existence and decay of solutions of a singular nonlocal viscoelastic system featuring nonlocal boundary conditions, localized damping term and linear source term. In domain of blow-up phenomena, the authors in [27] investigated the finite-time blow-up of solutions for an initial boundary value problem with nonlocal boundary conditions, pertaining to a system of nonlinear singular viscoelastic equations. Other works in the same vein can be found here [1–4, 7, 10–12, 24, 25].

The influence of delay frequently emerges in numerous applications and practical issues, transforming various systems into distinct problems warranting investigation. Recently, numerous authors have examined asymptotic behavior, stability, and blow-up phenomena of solutions in evolution systems with time delay. Refer to works by [6, 8, 9, 19] for further details.

Motivated by the aforementioned works, in this study, we expand upon the earlier investigation outlined in [15] to encompass singular one-dimensional nonlinear viscoelastic equations with source and distributed delay terms. Specifically, we delve into the blow-up phenomenon of solutions with negative initial energy for problem (1).

In the following, let $c,c_{i},C>0$, are positive constants.

Our paper is structured as follows: In the subsequent section, we establish concepts, lemmas, and hypotheses essential for our analysis. In Sect. 3, we state and prove the blow-up phenomenon of solutions.

2 Preliminaries

In this section, we present the following definitions, symbols, spaces and lemmas that we utilize throughout the paper.

Let $L_{x}^{p}=L_{x}^{p}(0,L)$ represent the weighted Banach space having norm

$$ \Vert u \Vert _{L_{x}^{p}}= \biggl( \int _{0}^{L }x \vert u \vert ^{p}\,dx \biggr) ^{\frac {1}{p}}. $$

Hilbert space of square integral functions is denoted by $H=L_{x}^{2}(0,L)$ having the finite norm

$$ \Vert u \Vert _{H}= \biggl( \int _{0}^{L }xu^{2}\,dx \biggr) ^{ \frac {1}{2}}. $$

Hilbert space is represented by $V=V_{x}^{1}((0,L))$ equipped with the norm

$$ \Vert u \Vert _{V}= \bigl( \Vert u \Vert _{H}^{2}+ \Vert u_{x} \Vert _{H}^{2} \bigr) ^{\frac {1}{2}}, $$

and

$$ V_{0}= \bigl\{ u\in V\text{ such that }u(L )=0 \bigr\} . $$

Lemma 1

(Poincare-type inequality) For any v in $V_{0}$, we have

$$ \int _{0}^{L }xv^{2}(x)\,dx\leq C_{p} \int _{0}^{L }x\bigl(v_{x}(x) \bigr)^{2}\,dx $$

and

$$ V_{0}= \bigl\{ v\in V\textit{ such that }v(L )=0 \bigr\} . $$

Remark 1

Clearly $\Vert u \Vert _{V_{0}}= \Vert u_{x} \Vert _{H}$ defines an equivalent norm on $V_{0}$.

Theorem 1

(See [1]) For any v in $V_{0}$ and $2< p<4$, we have

$$ \int _{0}^{L }x \vert v \vert ^{p}\,dx \leq C_{\ast } \Vert v_{x} \Vert _{H=L_{x}^{p}(0,L )}^{p}, $$

where $C_{\ast }$ is a constant depending on L and p only.

We demonstrate the blow-up outcome given the following appropriate assumptions.

(A1) $g_{1},g_{2}: \mathbb{R}_{+}\rightarrow \mathbb{R}_{+}$ are decreasing and differentiable functions such that

$$\begin{aligned}& g_{1}(t)\geq 0 , 1- \int _{0}^{\infty }g_{1} ( s )\,ds=l_{1}>0, \\& g_{2}(t)\geq 0 , 1- \int _{0}^{\infty }g_{2} ( s )\,ds=l_{2}>0. \end{aligned}$$

(3)

(A2) There exists a constants $\xi _{1},\xi _{2}>0$ such that

$$\begin{aligned}& g_{1}^{\prime } ( t ) \leq -\xi _{1} g_{1} ( t ) , \quad t\geq 0, \\& g_{2}^{\prime } ( t ) \leq -\xi _{2} g_{2} ( t ) , \quad t\geq 0. \end{aligned}$$

(4)

(A3) $\mu _{2},\mu _{4}:[\tau _{1}, \tau _{2}]\rightarrow \mathbb{R}$ is a bounded function satisfying

$$\begin{aligned} \frac{2\delta +1}{2} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \,ds< \mu _{1}, \quad \text{and}\quad \frac{2\delta +1}{2} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \,ds< \mu _{3}, \quad \delta >\frac{1}{2}. \end{aligned}$$

(5)

By combining arguments of other studies [16, 20, 22] with used the Faedo–Galerkin method, we get the local existence theorem.

Theorem 2

Let (3), (4), and (5) hold. Assume that

$$ \textstyle\begin{cases} -1< r< \frac {4-n}{n-2},\quad n\geq 3; \\ r\geq -1,\quad n=1,2 \end{cases} $$

(6)

Then, for any $(u_{0},v_{0})\in V_{0}^{2}$, $(v_{1},v_{2})\in H^{2}$ and $(h_{0},k_{0})\in \mathcal{H,}$ problem (1) has a unique local solution

$$ u\in C\bigl(\bigl(0,T^{*}\bigr);V_{0}\bigr)\cap C^{1}\bigl(\bigl(0,T^{*}\bigr);H\bigr), $$

for $T^{*}>0$ small enough, where $\mathcal{H}=L^{2}_{x}((0,L)\times (0,1)\times (\tau _{1},\tau _{2}))$.

Lemma 2

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

$$\begin{aligned} F(u, v) =&\frac{1}{2(r+2)} \bigl[u f_{1}(u, v)+v f_{2}(u, v) \bigr] \\ =&\frac{1}{2(r+2)} \bigl[a_{1} \vert u+v \vert ^{2(r+2)}+2 b_{1} \vert u v \vert ^{r+2} \bigr] \geq 0, \end{aligned}$$

where

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

We take $a_{1}=b_{1} = 1$ for convenience.

Lemma 3

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

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

(7)

Taking new variable as in [19],

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

so we get

$$ \textstyle\begin{cases} sz_{t}(x, \rho , s, t)+z_{\rho}(x, \rho , s, t)=0, \\ sy_{t}(x, \rho , s, t)+y_{\rho}(x, \rho , s, t)=0, \\ z(x, 0, s, t)=u_{t}(x, t), \\ y(x, 0, s, t)=v_{t}(x, t). \end{cases} $$

Consequently, the problem (1) is equivalent to

$$ \textstyle\begin{cases} u_{tt}-\frac{1}{x}(xu_{x})_{x}+\int _{0}^{t}g_{1}(t-s) \frac{1}{x}(xu_{x}(x,s))_{x}\,ds+\mu _{1} u_{t} \\ \quad {} +\int _{\tau _{1}}^{\tau _{2}}\mu _{2}(s)z ( x, 1,s,t )\,ds=f_{1} ( u,v ) ,\quad \text{in }Q, \\ v_{tt}-\frac{1}{x}(xv_{x})_{x}+\int _{0}^{t}g_{2}(t-s) \frac{1}{x}(xv_{x}(x,s))_{x}\,ds+\mu _{3} v_{t} \\ \quad {}+\int _{\tau _{1}}^{\tau _{2}}\mu _{4}(s)v _{t} ( x, 1,s,t )\,ds=f_{2} ( u,v ) ,\quad \text{in }Q, \\ sz_{t}(x, \rho , s, t)+z_{\rho}(x, \rho , s, t)=0, \\ sy_{t}(x, \rho , s, t)+y_{\rho}(x, \rho , s, t)=0, \\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x),\quad x\in (0,L), \\ v(x,0)=v_{0}(x), \qquad v_{t}(x,0)=v_{1}(x),\quad x\in (0,L), \\ z(x,\rho ,s,0)=h_{0}(x,\rho s), \qquad (x,\rho ,s)\in (0,L)\times (0,1)\times (0,\tau _{2}) \\ y(x,\rho ,s,0)=k_{0}(x,\rho s) \qquad (x,\rho ,s)\in (0,L)\times (0,1)\times (0,\tau _{2}) \\ u(L,t)=v(L,t)=0,\qquad \int _{0}^{L}xu(x,t)\,dx=\int _{0}^{L}xv(x,t)\,dx=0,\end{cases} $$

(8)

We define the energy functional.

Lemma 4

Assume (3), (4), (5), and (6) hold, let $(u,v)$ be a solution of (1), then $E(t)$ is non-increasing, that is

$$\begin{aligned} E(t) =&\frac{1}{2} \Vert u_{t} \Vert _{H}^{2}+ \frac{1}{2} \Vert v_{t} \Vert _{H}^{2}+ \frac{1}{2}\biggl(1-\frac{1}{2} \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2} \\ &{}+\frac{1}{2}\biggl(1-\frac{1}{2} \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2}+ \frac{1}{2}(g_{1}o u_{x})+\frac{1}{2}(g_{2}o v_{x}) \\ &{}+\mathcal{G}(z,y)- \int _{0}^{L}xF(u,v)\,dx \end{aligned}$$

(9)

satisfies

$$\begin{aligned} E'(t) =&-C_{0} \biggl\{ \Vert u_{t} \Vert ^{2}_{H}+ \int _{\tau _{1}}^{ \tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \bigl\Vert z(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds \\ &{}+ \Vert v_{t} \Vert ^{2}_{H}+ \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \bigl\Vert y(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds \biggr\} \\ &{}+\frac{1}{2}g'_{1}\circ u_{x}+ \frac{1}{2}g'_{2}\circ v_{x}- \int _{0}^{t}g_{1}(s)\,ds \Vert u_{x} \Vert ^{2}_{H}- \int _{0}^{t}g_{2}(s)\,ds \Vert v_{x} \Vert ^{2}_{H} \\ \leq & 0, \end{aligned}$$

(10)

where

$$\begin{aligned}& \int _{0}^{L}xF(u,v)\,dx=\frac{1}{2(r+2)} \bigl( \Vert u+v \Vert ^{2(r+2)}_{L^{2(r+2)}_{x}}+2 \Vert uv \Vert ^{(r+2)}_{L^{(r+2)}_{x}} \bigr), \\& (g\circ u_{x}) (t)= \int _{0}^{L } \int _{0}^{t}xg(t-s) \bigl\vert u_{x}(x,t)-u_{x}(x,s) \bigr\vert ^{2}\,ds\,dx, \end{aligned}$$

(11)

and

$$ \mathcal{G}(z,y):=\frac{1}{2} \int _{0}^{1 } \int _{\tau _{1}}^{\tau _{2}}s \bigl\{ \vert \mu _{2} \vert \bigl\Vert z(x,\rho ,s,t) \bigr\Vert ^{2}_{H}+ \vert \mu _{4} \vert \bigl\Vert y(x,\rho ,s,t) \bigr\Vert ^{2}_{H} \bigr\} \,ds\,dx, $$

Proof

By multiplying (1)₁, (1)₂ by $xu_{t}$, $xv_{t}$, respectively, and integrating over $(0,L)$, we get

$$\begin{aligned}& \frac {d}{dt} \biggl\{ \frac{1}{2} \Vert u_{t} \Vert _{H}^{2}+ \frac{1}{2} \Vert v_{t} \Vert _{H}^{2}+\frac{1}{2}l_{1} \Vert u_{x} \Vert _{H}^{2}+\frac{1}{2}l_{2} \Vert v_{x} \Vert _{H}^{2} + \frac{1}{2}(g_{1} \circ u_{x}) \\& \qquad {}+\frac{1}{2}(g_{2}\circ u_{x})- \int _{0}^{L}xF(u,v)\,dx \biggr\} \\& \quad = -\mu _{1} \int _{0}^{L}x u_{t}^{2}\,dx-\mu _{3} \int _{0}^{L}x v_{t}^{2}\,dx+ \frac{1}{2}g'_{1}\circ u_{x}+ \frac{1}{2}g'_{2}\circ v_{x} \\& \qquad {} -\biggl( \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2}-\biggl( \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2} \\& \qquad {} - \int _{0}^{L}xu_{t} \int _{\tau _{1}}^{\tau _{2}} \mu _{2}(s) z ( x, 1, s, t )\,ds \,dx \\& \qquad {} - \int _{0}^{L}xv_{t} \int _{\tau _{1}}^{\tau _{2}} \mu _{2}(s) y ( x, 1, s, t )\,ds \,dx. \end{aligned}$$

(12)

Now, multiplying the Equation (8)₃ by $xz\vert \mu _{2}(s)\vert$ and integrating the result over $(0, L)\times (0, 1)\times (\tau _{1}, \tau _{2})$

$$\begin{aligned}& \frac{d}{dt }\frac{1}{2} \int _{0}^{L} \int _{0}^{1} \int _{\tau _{1}}^{ \tau _{2}}xs \bigl\vert \mu _{2}(s) \bigr\vert z^{2}(x, \rho , s, t)\,ds \,d\rho \,dx \\& \quad = - \int _{0}^{L} \int _{0}^{1} \int _{\tau _{1}}^{\tau _{2}}x \bigl\vert \mu _{2}(s) \bigr\vert zz_{\rho } ( x, \rho , s, t )\,ds \,d\rho \,dx \\& \quad = -\frac{1 }{2 } \int _{0}^{L} \int _{0}^{1} \int _{\tau _{1}}^{\tau _{2}}x \bigl\vert \mu _{2}(s) \bigr\vert \frac{d}{ \,d\rho}z^{2} ( x, \rho , s, t )\,ds \,d\rho \,dx \\& \quad = \frac{1 }{2 } \int _{0}^{L} \int _{\tau _{1}}^{\tau _{2}}x \bigl\vert \mu _{2}(s) \bigr\vert \bigl(z^{2} ( x, 0 , s, t ) -z^{2}(x, 1, s, t) \bigr)\,ds \,dx \\& \quad = \frac{1 }{2 } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \,ds \Vert u_{t} \Vert ^{2}_{H}- \frac{1 }{2} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \bigl\Vert z(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds, \end{aligned}$$

(13)

using Young’s inequality, we achieve

$$\begin{aligned}& - \int _{0}^{L}xu_{t} \int _{\tau _{1}}^{\tau _{2}} \mu _{2}(s) z( x, 1, s, t)\,ds\,dx \\& \quad \leq \delta \biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \,ds\biggr) \Vert u_{t} \Vert ^{2}_{H}+ \frac{1}{4\delta} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \bigl\Vert z(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds. \end{aligned}$$

(14)

Similarly, we get

$$\begin{aligned}& \frac{d}{dt }\frac{1}{2} \int _{0}^{L} \int _{0}^{1} \int _{\tau _{1}}^{ \tau _{2}}xs \bigl\vert \mu _{4}(s) \bigr\vert y^{2}(x, \rho , s, t)\,ds \,d\rho \,dx \\& \quad = \frac{1 }{2 } \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \,ds \Vert v_{t} \Vert ^{2}_{H}- \frac{1 }{2} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \bigl\Vert y(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds, \end{aligned}$$

(15)

and

$$\begin{aligned}& - \int _{0}^{L}xv_{t} \int _{\tau _{1}}^{\tau _{2}} \mu _{4}(s) y ( x, 1, s, t )\,ds \,dx \\& \quad \leq \delta \biggl( \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \,ds\biggr) \Vert v_{t} \Vert ^{2}_{H}+ \frac{1}{4\delta} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \bigl\Vert y(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds. \end{aligned}$$

(16)

By remplacing (13)–(16) into (12), we obtain (9) and

$$\begin{aligned} E'(t) =& - \biggl(\mu _{1}-\biggl(\delta + \frac{1}{2}\biggr) \int _{\tau _{1}}^{ \tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \,ds \biggr) \Vert u_{t} \Vert ^{2}_{H} \\ &{}- \biggl(\mu _{3}-\biggl(\delta +\frac{1}{2}\biggr) \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \,ds \biggr) \Vert v_{t} \Vert ^{2}_{H} \\ &{}- \biggl(\frac{2\delta -1}{4\delta} \biggr) \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \bigl\Vert z(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds \\ &{}- \biggl(\frac{2\delta -1}{4\delta} \biggr) \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \bigl\Vert y(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds \\ & - \int _{0}^{t}g_{1}(s)\,ds \Vert u_{x} \Vert ^{2}_{H}- \int _{0}^{t}g_{2}(s)\,ds \Vert v_{x} \Vert ^{2}_{H}+\frac{1}{2}g'_{1} \circ u_{x}+\frac{1}{2}g'_{2} \circ v_{x}. \end{aligned}$$

(17)

Hence, according to (4) and (5), we get (10), where $C_{0}=\min \{\mu _{1}-(\delta +\frac{1}{2})\int _{\tau _{1}}^{ \tau _{2}}\vert \mu _{2}(s)\vert \,ds,\mu _{3}-(\delta +\frac{1}{2}) \int _{\tau _{1}}^{\tau _{2}}\vert \mu _{4}(s)\vert \,ds , \frac{2\delta -1}{4\delta} \}>0$.

The proof is completed. □

3 Blow-up

In this segment, we establish the blow-up outcome for the solution of problem (1).

Following this, we introduce the functional

$$\begin{aligned} \mathbb{H}(t) =&-E(t) \\ =&-\frac{1}{2} \Vert u_{t} \Vert _{H}^{2}- \frac{1}{2} \Vert v_{t} \Vert _{H}^{2}-\frac{1}{2}\biggl(1-\frac{1}{2} \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2} \\ &{}-\frac{1}{2}\biggl(1-\frac{1}{2} \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2}- \frac{1}{2}(g_{1}o u_{x})-\frac{1}{2}(g_{2}o v_{x}) \\ &{}+\mathcal{G}(z,y)+\frac{1}{2(r+2)} \bigl[ \Vert u+v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+2 \Vert uv \Vert _{L_{x}^{r+2}}^{r+2} \bigr]. \end{aligned}$$

(18)

Theorem 3

Let (3)–(5), and (6) holds and $E(0)<0$, then solution of problem (1) blow-up in finite time.

Proof

From (10), we have

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

(19)

Therefore

$$\begin{aligned} \mathbb{H}'(t)=-E'(t) \geq & 0, \end{aligned}$$

(20)

this implies that

$$\begin{aligned}& \mathbb{H}'(t) \geq C_{0} \biggl\{ \Vert u_{t} \Vert ^{2}_{H}+ \int _{ \tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \bigl\Vert z(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds \biggr\} \\& \mathbb{H}'(t) \geq C_{0} \biggl\{ \Vert v_{t} \Vert ^{2}_{H}+ \int _{ \tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \bigl\Vert y(x,1,s,t) \bigr\Vert ^{2}_{H}\,ds \biggr\} . \end{aligned}$$

(21)

From (11) and (7), we get

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

(22)

We set

$$\begin{aligned} \mathcal{K}(t) =&\mathbb{H}^{1-\alpha}+ \varepsilon \int _{0}^{L}x(uu_{t}+vv_{t})\,dx+ \frac{\varepsilon}{2} \int _{0}^{L}x\bigl(\mu _{1}u^{2}+ \mu _{3}v^{2}\bigr)\,dx, \end{aligned}$$

(23)

where

$$ 0< \alpha < \min \biggl\{ \frac{r+1}{r+2}, \frac {r+1}{2(r+2)} \biggr\} . $$

(24)

By multiplying (1)₁, (1)₂ by xu, xv and derivative of (23), we achieve

$$\begin{aligned} \mathcal{K}'(t) = &(1-\alpha )\mathbb{H}^{-\alpha} \mathbb{H}'(t)+ \varepsilon \bigl( \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}\bigr)- \varepsilon \bigl( \Vert u_{x} \Vert _{H}^{2}+ \Vert v_{x} \Vert _{H}^{2}\bigr) \\ & {}\underbrace{+\varepsilon \int _{0}^{L} u_{x} \int ^{t}_{0}g_{1}(t-s) x u_{x}(s)\,ds\,dx}_{I_{1}} \\ &{} \underbrace{+\varepsilon \int _{0}^{L} v_{x} \int ^{t}_{0}g_{2}(t-s) x v_{x}(s)\,ds\,dx}_{I_{2}} \\ & {}\underbrace{- \int _{0}^{L}xu \int _{\tau _{1}}^{\tau _{2}} \mu _{2}(s) z ( x, 1, s, t )\,ds \,dx}_{I_{3}} \\ &{} \underbrace{- \int _{0}^{L}xv \int _{\tau _{1}}^{\tau _{2}} \mu _{2}(s) y ( x, 1, s, t )\,ds \,dx}_{I_{4}} \\ &{}+\varepsilon \bigl[ \Vert u+v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+2 \Vert uv \Vert _{L_{x}^{(r+2)}}^{r+2} \bigr]. \end{aligned}$$

(25)

We have

$$\begin{aligned}& I_{1} = \varepsilon \int _{0}^{t}g_{1}(t-s)\,ds \int _{0}^{L} u_{x}.\bigl(x u_{x}(s)-x u_{x}(t)\bigr)\,dx\,ds+\varepsilon \biggl( \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2} \\& \hphantom{I_{1}} \geq \varepsilon \biggl(\frac{1}{2} \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2}-\frac{\varepsilon}{2}(g_{1} \circ u_{x}), \end{aligned}$$

(26)

$$\begin{aligned}& I_{2} = \varepsilon \int _{0}^{t}g_{2}(t-s)\,ds \int _{0}^{L} v_{x}.\bigl(x v_{x}(s)-x v_{x}(t)\bigr)\,dx\,ds+\varepsilon \biggl( \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2} \\& \hphantom{I_{2}}\geq \varepsilon \biggl(\frac{1}{2} \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2}-\frac{\varepsilon}{2}(g_{2} \circ u_{x}), \end{aligned}$$

(27)

Next, by Young’s inequality, we find for $\delta _{1}>0$

$$\begin{aligned} I_{3}\geq -\varepsilon \delta _{1}\mu _{1} \Vert u \Vert _{H}^{2}- \frac{\varepsilon}{4\delta _{1}} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \bigl\Vert z ( x, 1, s, t ) \bigr\Vert _{H}^{2}\,ds, \end{aligned}$$

(28)

and

$$\begin{aligned} I_{4}\geq -\varepsilon \delta _{1}\mu _{3} \Vert v \Vert _{H}^{2}- \frac{\varepsilon}{4\delta _{1}} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \bigl\Vert y ( x, 1, s, t ) \bigr\Vert _{H}^{2}\,ds, \end{aligned}$$

(29)

from (25), we get

$$\begin{aligned} \mathcal{K}'(t) \geq &(1-\alpha )\mathbb{H}^{-\alpha} \mathbb{H}'(t)+ \varepsilon \bigl( \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}\bigr) \\ &{}-\varepsilon \biggl(\biggl(1-\frac{1}{2} \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2}+\biggl(1- \frac{1}{2} \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2} \biggr) \\ &{}-\frac{\varepsilon}{2}(g_{1}\circ u_{x})- \frac{\varepsilon}{2}(g_{2} \circ v_{x})+\varepsilon \bigl[ \Vert u+v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+2 \Vert uv \Vert _{L_{x}^{(r+2)}}^{r+2} \bigr] \\ &{}-\varepsilon \delta _{1}\mu _{1} \Vert u \Vert _{H}^{2}- \frac{\varepsilon}{4\delta _{1}} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{2}(s) \bigr\vert \bigl\Vert z ( x, 1, s, t ) \bigr\Vert _{H}^{2}\,ds \\ &{}-\varepsilon \delta _{1}\mu _{3} \Vert v \Vert _{H}^{2}- \frac{\varepsilon}{4\delta _{1}} \int _{\tau _{1}}^{\tau _{2}} \bigl\vert \mu _{4}(s) \bigr\vert \bigl\Vert y ( x, 1, s, t ) \bigr\Vert _{H}^{2}\,ds. \end{aligned}$$

(30)

At this stage, we choose $\delta _{1}$ so that, for large κ to be chosen later

$$ \frac{1}{4\delta _{1}C_{0}}=\kappa \mathbb{H}^{-\alpha}(t), $$

by (21) and putting in (30), we get

$$\begin{aligned} \mathcal{K}'(t) \geq &\bigl[(1-\alpha )-\varepsilon \kappa \bigr] \mathbb{H}^{- \alpha}\mathbb{H}'(t)+\varepsilon \bigl( \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}\bigr) \\ &{}-\varepsilon \biggl(\biggl(1-\frac{1}{2} \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2}+\biggl(1- \frac{1}{2} \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2} \biggr) \\ &{}-\frac{\varepsilon}{2}(g_{1}\circ u_{x})- \frac{\varepsilon}{2}(g_{2} \circ v_{x})+\varepsilon \bigl[ \Vert u+v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+2 \Vert uv \Vert _{L_{x}^{(r+2)}}^{r+2} \bigr] \\ &{}-\frac{\varepsilon \mu _{1}\mathbb{H}^{\alpha}}{4\kappa C_{0}} \Vert u \Vert _{H}^{2}- \frac{\varepsilon \mu _{3}\mathbb{H}^{\alpha}}{4\kappa C_{0}} \Vert v \Vert _{H}^{2}. \end{aligned}$$

(31)

For $0< a<1$ and from (18), we obtain

$$\begin{aligned} \varepsilon \bigl[ \Vert u+v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+2 \Vert uv \Vert _{L_{x}^{(r+2)}}^{r+2} \bigr] =& \varepsilon a \bigl[ \Vert u+v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+2 \Vert uv \Vert _{L_{x}^{(r+2)}}^{r+2} \bigr] \\ &{}+ 2\varepsilon (r+2) (1-a)\mathbb{H}(t) \\ &{}+ \varepsilon (r+2) (1-a) \bigl( \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}\bigr) \\ &{}+\varepsilon (r+2) (1-a) \biggl(1- \int _{0}^{t}g_{1}(s)\,ds\biggr) \Vert u_{x} \Vert _{H}^{2} \\ &{}+ \varepsilon (p+2) (1-a) \biggl(1- \int _{0}^{t}g_{2}(s)\,ds\biggr) \Vert v_{x} \Vert _{H}^{2} \\ &{}+\varepsilon (r+2) (1-a) (g_{1}\circ u_{x}) \\ &{}+\varepsilon (r+2) (1-a) (g_{2}\circ v_{x}) \\ &{}+2\varepsilon (r+2) (1-a)\mathcal{G}(z,y). \end{aligned}$$

(32)

Substituting in (31) and by (7), we achieve

$$\begin{aligned} \mathcal{K}'(t) \geq &\bigl[(1-\alpha )-\varepsilon \kappa \bigr] \mathbb{H}^{- \alpha}\mathbb{H}'(t)+\varepsilon \bigl[(r+2) (1-a)+1\bigr]\bigl( \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}\bigr) \\ &{}+\varepsilon \biggl[(r+2) (1-a) \biggl(1- \int _{0}^{t}g_{1}(s)\,ds\biggr)-\biggl(1- \frac{1}{2} \int _{0}^{t}g_{2}(s)\,ds \biggr) \biggr] \Vert u_{x} \Vert _{H}^{2} \\ &{}+\varepsilon \biggl[(r+2) (1-a) \biggl(1- \int _{0}^{t}g_{2}(s)\,ds\biggr)-\biggl(1- \frac{1}{2} \int _{0}^{t}g_{2}(s)\,ds \biggr) \biggr] \Vert v_{x} \Vert _{H}^{2} \\ &{}+\varepsilon \biggl[(r+2) (1-a)-\frac{1}{2} \biggr](g_{1}o u_{x}+g_{2}o v_{x})+2 \varepsilon (r+2) (1-a) \mathcal{G}(z,y) \\ &{}+\frac{\varepsilon ac_{0}}{2r+4} \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)} \bigr]+ 2\varepsilon (r+2) (1-a) \mathbb{H}(t) \\ &{}-\frac{\varepsilon \mu _{1}\mathbb{H}^{\alpha}}{4\kappa C_{0}} \Vert u \Vert _{H}^{2}- \frac{\varepsilon \mu _{3}\mathbb{H}^{\alpha}}{4\kappa C_{0}} \Vert v \Vert _{H}^{2} \end{aligned}$$

(33)

According to (22), we get

$$\begin{aligned} \mathbb{H}^{\alpha} \Vert u \Vert _{H}^{2} \leq &c \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2) \alpha}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)\alpha} \bigr] \Vert u \Vert _{H}^{2}, \\ \leq &c \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)\alpha +2}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)\alpha} \Vert u \Vert _{H}^{2} \bigr], \end{aligned}$$

by using the fact that

$$ \Phi ^{\varrho}\leq \Phi +1\leq \biggl(1+\frac{1}{a}\biggr) (\Phi +a), \quad \forall \Phi \geq 0, 0< \varrho \leq 1, a\geq 0, $$

(34)

we have

$$\begin{aligned} \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)\alpha +2} \Vert u \Vert _{H}^{2} \leq &d \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+\mathcal{H}(0) \bigr], \\ \leq &d \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+\mathcal{H}(t) \bigr], \end{aligned}$$

where $d=(1+\frac{1}{\mathcal{H}(0)}$.

On the other hand, using the following inequality

$$ (\Psi +\Theta )^{\gamma}\leq c\bigl(\Psi ^{\gamma}+\Theta ^{\gamma}\bigr), \quad \forall \Psi ,\Theta \geq 0, $$

we have

$$\begin{aligned} \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)\alpha} \Vert u \Vert _{H}^{2} \leq &c \bigl[ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)\alpha} \Vert u \Vert _{L_{x}^{2(r+2)}}^{2} \bigr], \\ \leq &c \bigl[ \Vert v \Vert _{L_{x}^{2(r+2)}}^{\alpha} \Vert u \Vert _{L_{x}^{2(r+2)}}^{ \frac{2}{2(r+2)}} \bigr]^{2(r+2)}, \\ \leq &c \bigl[ \Vert v \Vert _{L_{x}^{2(r+2)}}^{ \frac{\alpha (2r+4)+2}{2(r+2)}}+ \Vert u \Vert _{L_{x}^{2(r+2)}}^{ \frac{2\alpha (r+2)+2}{2(r+2)}} \bigr]^{2(r+2)}, \end{aligned}$$

from (24), we get

$$\begin{aligned} \mathbb{H}^{\alpha}(t) \Vert u \Vert _{H}^{2} \leq &c_{2} \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \mathcal{H}(t) \bigr]. \end{aligned}$$

(35)

Similarly, we find

$$\begin{aligned} \mathbb{H}^{\alpha}(t) \Vert v \Vert _{H}^{2} \leq &c_{3} \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \mathcal{H}(t) \bigr]. \end{aligned}$$

(36)

Substituting (35) and (36) into (33), we get

$$\begin{aligned} \mathcal{K}'(t) \geq &\bigl[(1-\alpha )-\varepsilon \kappa \bigr] \mathbb{H}^{- \alpha}\mathbb{H}'(t)+\varepsilon \bigl[(r+2) (1-a)+1\bigr]\bigl( \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}\bigr) \\ &{}+\varepsilon \underbrace{ \biggl[ \bigl((r+2) (1-a)-1 \bigr)- \int _{0}^{t}g_{1}(s)\,ds \biggl((r+2) (1-a)-\frac{1}{2} \biggr) \biggr]}_{J_{1}} \Vert u_{x} \Vert _{H}^{2} \\ &{}+\varepsilon \underbrace{ \biggl[ \bigl((r+2) (1-a)-1 \bigr)- \int _{0}^{t}g_{2}(s)\,ds \biggl((r+2) (1-a)-\frac{1}{2} \biggr) \biggr]}_{J_{2}} \Vert v_{x} \Vert _{H}^{2} \\ &{}+\varepsilon \biggl[(r+2) (1-a)-\frac{1}{2} \biggr](g_{1}o u_{x}+g_{2}o v_{x})+2 \varepsilon (r+2) (1-a) \mathcal{G}(z,y) \\ &{}+\varepsilon \biggl(\frac{ac_{0}}{2r+4}-\frac{c_{4}}{4\kappa C_{0}} \biggr) \bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)} \bigr] \\ &{}+ \varepsilon \biggl(2(r+2) (1-a)-\frac{c_{4}}{4\kappa C_{0}} \biggr) \mathbb{H}(t), \end{aligned}$$

(37)

where $c_{4}=\min \{c_{2}\mu _{1},c_{3}\mu _{3}\}$.

Now, taking $a>0$ small enough such that

$$ J_{0}=(r+2) (1-a)-1>0 $$

let

$$ \max \biggl\{ \int _{0}^{\infty}g_{1}(s)\,ds, \int _{0}^{\infty}g_{2}(s)\,ds \biggr\} < \frac {(r+2)(1-a)-1}{ ((r+2)(1-a)-\frac {1}{2} )}= \frac {2J_{0}}{2J_{0}+1}, $$

(38)

gives

$$\begin{aligned}& J_{1} = \biggl\{ (r+2) (1-a)-1)- \int _{0}^{t}g_{1}(s)\,ds\biggl((r+2) (1-a)- \frac{1}{2}\biggr) \biggr\} >0, \\& J_{2} = \biggl\{ (r+2) (1-a)-1)- \int _{0}^{t}g_{2}(s)\,ds\biggl((r+2) (1-a)- \frac{1}{2}\biggr) \biggr\} >0. \end{aligned}$$

Next, we pick κ in a way that

$$\begin{aligned} J_{3} =&2(r+2) (1-a)-\frac{c_{4}}{4\kappa C_{0}}>0, \\ J_{4} =&\frac{ac_{0}}{2r+4}-\frac{c_{4}}{4\kappa C_{0}}>0, \end{aligned}$$

Chose fixed values for κ and a and select ε small enough such that

$$ J_{5}=(1-\alpha )-\varepsilon \kappa >0, $$

and

$$ \mathcal{K}(0)=\mathbb{H}^{1-\alpha}(0)+ \varepsilon \int _{0}^{L}x(u_{0}u_{1}+v_{0}v_{1})\,dx+ \frac{\varepsilon}{2} \int _{0}^{L}x\bigl(\mu _{1}u_{0}^{2}+ \mu _{3}v_{0}^{2}\bigr)\,dx>0. $$

Thus, for some $\beta _{1}>0$, estimate (37) becomes

$$\begin{aligned} \mathcal{K}'(t) \geq &\beta _{1} \bigl\{ \mathbb{H}(t)+ \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2} + \Vert u_{x} \Vert _{H}^{2}+ \Vert v_{x} \Vert _{H}^{2}+\mathcal{G}(z,y) \\ &{}+(g_{1}o u_{x})+(g_{2}o v_{x})+ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(p+2)}+ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)} \bigr\} . \end{aligned}$$

(39)

and

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

(40)

Now, utilizing Young’s and Holder’s inequalities, we obtain

$$\begin{aligned} \biggl\vert \int _{0}^{L}x(uu_{t}+vv_{t})\,dx \biggr\vert ^{\frac{1}{1-\alpha}} \leq &c\bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{\frac{\theta}{1-\alpha}}+ \Vert u_{t} \Vert _{H}^{\frac{\mu}{1-\alpha}} \\ &{}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{\frac{\theta}{1-\alpha}}+ \Vert v_{t} \Vert _{H}^{\frac{\mu}{1-\alpha}}\bigr] , \end{aligned}$$

(41)

where $\frac{1}{\mu}+\frac{1}{\theta}=1$.

We chose $\theta =2(1-\alpha )$ to achieve

$$ \frac{\mu}{1-\alpha}=\frac{2}{1-2\alpha}\leq 2(r+2). $$

For $s=\frac{2}{(1-2\alpha )}$ and by using (24) and (34), we obtain

$$\begin{aligned}& \Vert u \Vert _{L_{x}^{2(r+2)}}^{\frac{2}{1-2\alpha}} \leq \,d\bigl( \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+\mathbb{H}(t)\bigr) \\& \Vert v \Vert _{L_{x}^{2(r+2)}}^{\frac{2}{1-2\alpha}} \leq \,d\bigl( \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+\mathbb{H}(t)\bigr), \quad \forall t\geq 0. \end{aligned}$$

Therefore,

$$ \biggl\vert \int _{0}^{L}x(uu_{t}+vv_{t})\,dx \biggr\vert ^{\frac{1}{1-\alpha}}\leq c\bigl[ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}+\mathbb{H}(t)\bigr]. $$

(42)

Next,

$$\begin{aligned} \mathcal{K}^{\frac{1}{1-\alpha}}(t) =& \biggl(\mathbb{H}^{1-\alpha}+ \varepsilon \int _{0}^{L}x(uu_{t}+vv_{t})\,dx+ \frac{\varepsilon}{2} \int _{0}x\bigl(\mu _{1}u^{2}+\mu _{3}v^{2}\bigr)\,dx \biggr)^{\frac{1}{1-\alpha}} \\ \leq &c \biggl(\mathbb{H}(t)+ \biggl\vert \int _{0}^{L}x(uu_{t}+vv_{t})\,dx \biggr\vert ^{\frac{1}{1-\alpha}}+ \Vert u \Vert _{H}^{\frac{2}{1-\alpha}}+ \Vert v \Vert _{H}^{\frac{2}{1-\alpha}} \biggr) \\ \leq &c \bigl(\mathbb{H}(t)+ \Vert u_{t} \Vert _{H}^{2}+ \Vert v_{t} \Vert _{H}^{2}+ \Vert u_{x} \Vert _{H}^{2}+ \Vert v_{x} \Vert _{H}^{2}+(g_{1}o u_{x}) \\ &{}+(g_{2}o v_{x})+\mathcal{G}(z,y)+ \Vert u \Vert _{L_{x}^{2(r+2)}}^{2(r+2)}+ \Vert v \Vert _{L_{x}^{2(r+2)}}^{2(r+2)} \bigr) . \end{aligned}$$

(43)

From (39) and (43), we get

$$ \mathcal{K}'(t)\geq \lambda \mathcal{K}^{\frac{1}{1-\alpha}}(t), $$

(44)

where $\lambda > 0$, which depends on c and $\beta _{1}$ only.

By integrating (44), we get

$$ \mathcal{K}^{\frac{\alpha}{1-\alpha}}(t)\geq \frac{1}{\mathcal{K}^{\frac{-\alpha}{1-\alpha}}(0)-\lambda \frac{\alpha}{(1-\alpha )} t}. $$

Thus, the solution blows up in a finite time $T^{*}$, with

$$ T^{*}\leq \frac{1-\alpha}{\lambda \alpha \mathcal{K}^{\alpha /(1-\alpha )}(0)}. $$

Then the proof is complete. □