Blow-Up Results for Viscoelastic Wave Equations with Damping Terms on Stratified Groups

Abstract

In this paper, by means of Poincáre and Folland–Stein inequalities and Green’s identities for the sub-Laplacian on stratified Lie groups, we prove blow-up results in finite time for the viscoelastic wave equations both with strong and weak damping terms on stratified Lie groups.

1 Introduction

The following viscoelastic wave equation with weak damping was considered by Messaoudi in [14].

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle u_{tt}-\Delta u+\int _{0}^{t}k(t-\tau )\Delta u\text {d}\tau +a |u_{t}|^{q-2}u_{t}=|u|^{p-2}u,\,\,\,(x,t)\in \Omega \times [0,T],\\ u(x,t)=0,\,\,\,\,x\in \partial \Omega ,\\ u(x,0)=u_{0}(x),\,\,u_{t}(x,0)=u_{1}(x), \end{array}\right. } \end{aligned}$$

(1.1)

where $u_{0}\in W^{1,2}_{0}(\Omega ),\,\,\,u_{1}\in L^{2}(\Omega )$ and $k\in C^{1}[0,T]$ satisfying $1-\int _{0}^{\infty }k(\tau )\text {d}\tau =r>0.$ The author proved that any solution with negative initial energy $p>q$ blows up in finite-time and extended the result by considering positive initial energy in [15]. We refer [1, 3, 8, 11, 22] for the further discussions in this topic. The main motivation of this paper is to establish blow-up result on stratified Lie groups.

For the convenience, we give the definition of a stratified Lie group. A Lie group ${\mathbb {G}}=({\mathbb {R}}^{n},\circ )$ is called a stratified Lie group (or a homogeneous Carnot group) if it satisfies the following two conditions:

• (i) For natural numbers $N+\cdots +N_{r}=n$ where $N_{1}=N$, the decomposition ${\mathbb {R}}^{n}={\mathbb {R}}^{N}\times \cdots \times {\mathbb {R}}^{N_{r}}$ holds, and for each $\lambda >0$ the dilation $\delta _{\lambda }: {\mathbb {R}}^{n}\rightarrow {\mathbb {R}}^{n}$ given by

  $$\begin{aligned} \delta _{\lambda }(x)\equiv \delta _{\lambda }(x^{(1)},\ldots ,x^{(r)}):=(\lambda x^{(1)},\ldots ,\lambda ^{r}x^{(r)}) \end{aligned}$$

  is an automorphism of the group ${\mathbb {G}}.$ Here $x^{(k)}\in {\mathbb {R}}^{N_{k}}$ for $k=1,\ldots ,r.$

• (ii) Let $N_{1}$ be as in (i) and let $X_{1},\ldots ,X_{N_{1}}$ be the left invariant vector fields on ${\mathbb {G}}$ such that $X_{k}(0)=\frac{\partial }{\partial x_{k}}\big |_{0}$ for $k=1,\ldots ,N_{1}.$ Then, the Hörmander condition

  $$\begin{aligned} \mathrm{rank}(\mathrm{Lie}\{X_{1},\ldots ,X_{N_{1}}\})=n \end{aligned}$$

  holds for every $x\in {\mathbb {R}}^{n},$ i.e., the iterated commutators of $X_{1},\ldots ,X_{N_{1}}$ span the Lie algebra of ${\mathbb {G}}.$ That is, we say that the triple ${\mathbb {G}}=({\mathbb {R}}^{n},\circ , \delta _{\lambda })$ is a stratified Lie group (or a stratified group, in short). The number r above is called the step of ${\mathbb {G}}$ and the left invariant vector fields $X_{1},\ldots ,X_{N_{1}}$ are called the (Jacobian) generators of ${\mathbb {G}}$. The number

  $$\begin{aligned} Q=\sum _{k=1}^{r}kN_{k},\,\,\,N_{1}=N, \end{aligned}$$

  is called the homogeneous dimension of ${\mathbb {G}}$. We will also use the notation

  $$\begin{aligned} \nabla _{{\mathbb {G}}}:=(X_{1},\ldots , X_{N_{1}}) \end{aligned}$$

  for the (horizontal) gradient. Recall that the standard Lebesgue measure $\text {d}x$ on ${\mathbb {R}}^{N}$ is the Haar measure for ${\mathbb {G}}$.

The sum of squares of vector fields operators has been studied intensively since Lars Hörmander’s fundamental work [10] on this subject. Much of the progress has been connected to the development of analysis on the stratified (Lie) groups, following the pioneering work of Gerald B. Folland [5]. For the further discussions in this direction, we refer to the recent published monographs [4, 6], and [19] as well as references therein.

In the present paper, in addition to the “subelliptic" viscoelastic wave equation with weak damping terms, we also consider the viscoelastic wave equation with strong damping terms for the sub-Laplacian.

In [21], Song and Xue established blow-up result for the viscoelastic wave equation with strong damping term in the following form.

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle u_{tt}-\Delta u+\int _{0}^{t}k(t-\tau )\Delta u\text {d}\tau + \Delta u_{t}=|u|^{p-2}u,\,\,\,(x,t)\in \Omega \times [0,T],\\ u(x,t)=0,\,\,\,\,x\in \partial \Omega ,\\ u(x,0)=u_{0}(x),\,\,u_{t}(x,0)=u_{1}(x). \end{array}\right. } \end{aligned}$$

(1.2)

To the best of our knowledge, lower bounds for everywhere blow-up time of the problem (1.2) have not yet been considered on stratified groups. Thus, in the last section of this paper, we focus on studying this problem. We refer an interested reader to [2, 9, 12, 13, 16, 20] for related studies on the subject.

The present paper is structured as follows. A brief discussion about the preliminary facts on the stratified groups is given in Sect. 2. In Sect. 3, we investigate blow-up result for the viscoelastic wave equation with strong damping terms. Finally, in Sect. 4 we present blow-up result for the viscoelastic wave equation with weak damping terms on the stratified groups.

2 Preliminaries

Throughout the paper, we use the following notations $\Vert u\Vert _{L^{p}(\Omega )}=\Vert u\Vert _{p}$, $\Vert u\Vert _{2}=\Vert u\Vert$, $(g,f)=\int _{\Omega }g(x)f(x)\text {d}x$ and $(\nabla _{{\mathbb {G}}}g,\nabla _{{\mathbb {G}}}f)=\int _{\Omega }\nabla _{{\mathbb {G}}} g\cdot \nabla _{{\mathbb {G}}} f\text {d}x$.

Let $\Omega \subset {\mathbb {G}}$ be an open set. The notation $u\in C^{1}(\Omega )$ means $\nabla _{{\mathbb {G}}} u\in C(\Omega )$. That is sub-Laplacian is given by:

$$\begin{aligned} \Delta _{{\mathbb {G}}} f:={\mathcal {L}}_{2}f=\nabla _{{\mathbb {G}}}\cdot (\nabla _{{\mathbb {G}}}f). \end{aligned}$$

(2.1)

Let us consider Sobolev space

$$\begin{aligned} S^{1,2}(\Omega ):=\{u:u\in L^{2}(\Omega ),\,\,|\nabla _{{\mathbb {G}}}u|\in L^{2}(\Omega )\}, \end{aligned}$$

(2.2)

supplemented with the norm

$$\begin{aligned} \Vert u\Vert _{S^{1,2}(\Omega )}:=\left( \int _{\Omega }|u|^{2}+|\nabla _{{\mathbb {G}}}u|^{2}\text {d}x\right) ^\frac{1}{2}. \end{aligned}$$

Then, we define the functional class $S_{0}^{1,2}(\Omega )$ to be the completion of $C^{1}_{0}(\Omega )$ in the norm

$$\begin{aligned} \Vert u\Vert _{S_{0}^{1,2}(\Omega )}:=\left( \int _{\Omega }|\nabla _{{\mathbb {G}}}u|^{2}\text {d}x\right) ^\frac{1}{2}. \end{aligned}$$

Let $Q\ge 3$ be the homogeneous dimension of a stratified Lie group ${\mathbb {G}}$ and $\text {d}x$ be the volume element on ${\mathbb {G}}$. The following auxiliary results will be useful in what follows. We start with the analogue of the Green’s identity on stratified Lie groups (see [17]).

Proposition 2.1

(Green’s first identity) Let $1<p<\infty .$ Let $v\in C^{1}(\Omega )\bigcap C({\overline{\Omega }})$ and $u\in C^{2}(\Omega )\bigcap C^{1}({\overline{\Omega }})$. Then,

$$\begin{aligned} \int _{\Omega }\left( (|\nabla _{{\mathbb {G}}}u|^{p-2}\mathcal {{\widetilde{\nabla }}}v) u+v{\mathcal {L}}_{p}u\right) \text {d}x=\int _{\partial \Omega }|\nabla _{{\mathbb {G}}}u|^{p-2}v\langle \mathcal {{\widetilde{\nabla }} }u,\text {d}x\rangle , \end{aligned}$$

(2.3)

where

$$\begin{aligned} \mathcal {{\widetilde{\nabla }} }u=\sum _{k=1}^{N_{1}}\left( X_{k}u\right) X_{k}, \end{aligned}$$

and

$$\begin{aligned} {\mathcal {L}}_{p}f=\nabla _{{\mathbb {G}}}\cdot (|\nabla _{{\mathbb {G}}}|^{p-2}\nabla _{{\mathbb {G}}}f). \end{aligned}$$

Further, let us recall $L^{p}(\Omega )$-Poincaré inequality on stratified Lie groups (see [18]).

Theorem 2.2

Assume that $\Omega \subset {\mathbb {G}}$ and $f\in C^{\infty }_{0}(\Omega \setminus \{x'=0\})$ and $R'=\sup \limits _{x\in \Omega }|x'|$. Then, we have

$$\begin{aligned} R\Vert f\Vert _{p}\le \Vert \nabla _{{\mathbb {G}}} f\Vert _{p},\,\,\,\,1<p<\infty , \end{aligned}$$

(2.4)

where $R=\frac{|N-p|}{R'p}.$

Finally, we have the following Folland–Stein embedding theorem on the stratified Lie groups (see [5], also, e.g., [7]).

Theorem 2.3

Let ${\mathbb {G}}$ be a stratified Lie group and $\Omega \subset {\mathbb {G}}$ be an open set. Then, there exists a constant $C=C({\mathbb {G}})>0$ such that for all $f\in C_{0}^{\infty }(\Omega )$ we have

$$\begin{aligned} \Vert f\Vert _{L^{p^{*}}(\Omega )}\le C\left( \int _{\Omega }|\nabla _{{\mathbb {G}}} f|^{p}\text {d}x\right) ^{\frac{1}{p}},\; 1<p<Q, \end{aligned}$$

(2.5)

where $p^{*}=\frac{Qp}{Q-p}.$

3 Blow-Up with Strong Damping

In this section, we consider the following nonlinear viscoelastic wave equation on stratified Lie groups:

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle u_{tt}-\Delta _{{\mathbb {G}}} u+\int _{0}^{t}k(t-\tau )\Delta _{{\mathbb {G}}} u\text {d}\tau -a\Delta _{{\mathbb {G}}} u_{t}=|u|^{p-2}u,\,\,\,(x,t)\in \Omega \times [0,T],\\ u(x,t)=0,\,\,\,\,x\in \partial \Omega ,\\ u(x,0)=u_{0}(x),\,\,u_{t}(x,0)=u_{1}(x), \end{array}\right. } \end{aligned}$$

(3.1)

where $\Omega \subset {\mathbb {G}}$ is a Haar measurable set with a smooth boundary $\partial \Omega ,$ $N\ge 3$, where N is defined in (i), $u_{0}\in S^{1,2}_{0}(\Omega )$, $u_{1}\in L^{2}(\Omega )$, a is a positive constant and $p>2$ satisfies the following condition.

$$\begin{aligned} \frac{2Q}{Q-2}>p>2,\,\,\,Q\ge 3. \end{aligned}$$

(3.2)

We assume that the function $C^{1}(0,\infty )\ni k:{\mathbb {R}}_{+}\rightarrow {\mathbb {R}}_{+}$ has the following properties:

$$\begin{aligned} 1-\int _{0}^{+\infty }k(s)\text {d}s=r>\frac{1}{(p-1)^{2}} \end{aligned}$$

(3.3)

and

$$\begin{aligned} k(s)\ge 0,\,\,k'(s)\le 0. \end{aligned}$$

(3.4)

Let us define the following functional

$$\begin{aligned} I(t)=\frac{1}{2}\left( \Vert u_{t}(t)\Vert ^{2}_{2}+\left( 1-\int _{0}^{t}k(s)\text {d}s\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}+k\circ \nabla _{{\mathbb {G}}} u\right) -\frac{1}{p}\Vert u(t)\Vert ^{p}_{p}, \end{aligned}$$

(3.5)

where $\displaystyle k\circ \nabla _{{\mathbb {G}}} u=\int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(\cdot ,t)-\nabla _{{\mathbb {G}}} u(\cdot ,\tau )\Vert ^{2}_{2}\text {d}\tau .$

We give the main tools for obtaining blow-up result.

Lemma 3.1

Suppose that (3.3)–(3.4) hold true. Let u be a weak solution of (3.1), then we have

• (a) I(t) is a non-increasing function, i.e.,

  $$\begin{aligned} I'(t)\le 0,\,\,\,\forall t\in [0,T]; \end{aligned}$$

  (3.6)
• (b)
  $$\begin{aligned} I(t)+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}(\tau )\Vert ^{2}\text {d}\tau \le I(0),\,\,\,\,\,t\in [0,T],\,\,\,a>0. \end{aligned}$$

  (3.7)

Proof

Let us rewrite the equation in (3.1) as follows:

$$\begin{aligned} u_{tt}-\Delta _{{\mathbb {G}}} u+\int _{0}^{t}k(t-\tau )\Delta _{{\mathbb {G}}} u\text {d}\tau -a\Delta _{{\mathbb {G}}} u_{t}-|u|^{p-2}u=0. \end{aligned}$$

Multiplying both sides by $u_{t}$ and integrating over $\Omega$ , we compute

$$\begin{aligned} \begin{aligned} 0&=\int _{\Omega }u_{tt}u_{t}\text {d}x-\int _{\Omega }u_{t}\Delta _{{\mathbb {G}}} u\text {d}x+\int _{0}^{t}k(t-\tau )\int _{\Omega }u_{t}\Delta _{{\mathbb {G}}} u\text {d}x\text {d}\tau -a\int _{\Omega }u_{t}\Delta _{{\mathbb {G}}} u_{t}\text {d}x\\ {}&\quad -\int _{\Omega }u_{t}|u|^{p-2}u\text {d}x\\ {}&{\mathop {=}\limits ^{(2.3)}}\int _{\Omega }u_{tt}u_{t}\text {d}x-\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}\cdot \nabla _{{\mathbb {G}}} u\text {d}x-\int _{0}^{t}k(t-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}\cdot \nabla _{{\mathbb {G}}}u\text {d}x\text {d}\tau \\ {}&\quad +a\int _{\Omega }|\nabla _{{\mathbb {G}}} u_{t}|^{2}\text {d}x-\int _{\Omega }u_{t}|u|^{p-2}u\text {d}x\\ {}&=\frac{\text {d}}{\text {d}t}\left( \frac{1}{2}\int _{\Omega }|u_{t}|^{2}\text {d}x+\frac{1}{2}\int _{\Omega }|\nabla _{{\mathbb {G}}} u|^{2}\text {d}x-\frac{1}{p}\int _{\Omega }|u|^{p}\text {d}x\right) \\ {}&\quad -\int _{0}^{t}k(t-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}\cdot \nabla _{{\mathbb {G}}} u\text {d}x\text {d}\tau +a\int _{\Omega }|\nabla _{{\mathbb {G}}} u_{t}|^{2}\text {d}x\\ {}&=\frac{\text {d}}{\text {d}t}\left( \frac{1}{2}\Vert u_{t}\Vert ^{2}+\frac{1}{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-\frac{1}{p}\Vert u\Vert _{p}^{p}\text {d}x\right) -\int _{0}^{t}k(t-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}\cdot \nabla _{{\mathbb {G}}}u\text {d}x\text {d}\tau \\ {}&\quad +a\Vert \nabla _{{\mathbb {G}}}u_{t}\Vert ^{2}. \end{aligned} \end{aligned}$$

(3.8)

Next, we calculate the following integral.

$$\begin{aligned} \begin{aligned} \int _{0}^{t}k(t&-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}(t)\cdot \nabla _{{\mathbb {G}}} u(\tau )\text {d}x\text {d}\tau \\ {}&=\int _{0}^{t}k(t-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}(t)\cdot (\nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)+\nabla _{{\mathbb {G}}} u(t))\text {d}x\text {d}\tau \\ {}&=\int _{0}^{t}k(t-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}(t)\cdot (\nabla _{{\mathbb {G}}}u(\tau )-\nabla _{{\mathbb {G}}} u(t))\text {d}x\text {d}\tau \\ {}&\quad +\int _{0}^{t}k(t-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}(t)\cdot \nabla _{{\mathbb {G}}} u(t)\text {d}x\text {d}\tau \\ {}&=-\frac{1}{2}\int _{0}^{t}k(t-\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau \\ {}&\quad +\frac{1}{2}\int _{0}^{t}k(t-\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau \\ {}&=-\frac{1}{2}\int _{0}^{t}k(t-\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau \\ {}&\quad +\frac{1}{2}\int _{0}^{t}k(\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau . \end{aligned} \end{aligned}$$

(3.9)

Moreover, a direct calculation shows

$$\begin{aligned} \begin{aligned} -\frac{1}{2}\int _{0}^{t}k(t&-\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau \\ {}&=-\frac{1}{2}\frac{\text {d}}{\text {d}t}\int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}\text {d}\tau \\ {}&\quad +\frac{1}{2}\int _{0}^{t}k'(t-\tau )\int _{\Omega }|\nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau \\ {}&=-\frac{1}{2}\frac{\text {d} k\circ \nabla _{{\mathbb {G}}} u}{\text {d}t}+\frac{k'\circ \nabla _{{\mathbb {G}}} u}{2} \end{aligned} \end{aligned}$$

(3.10)

and

$$\begin{aligned} \frac{1}{2}\int _{0}^{t}k(\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau =\frac{1}{2}\frac{\text {d}}{\text {d}t}\left( \int _{0}^{t}k(\tau )\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}_{2}\text {d}\tau \right) -\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}. \end{aligned}$$

(3.11)

By substituting the last expressions in (3.9), we have

$$\begin{aligned} \begin{aligned} \int _{0}^{t}k(t&-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}(t)\cdot \nabla _{{\mathbb {G}}} u(\tau )\text {d}x\text {d}\tau \\ {}&=-\frac{1}{2}\int _{0}^{t}k(t-\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau \\ {}&\quad +\frac{1}{2}\int _{0}^{t}k(\tau )\frac{\text {d}}{\text {d}t}\int _{\Omega }|\nabla _{{\mathbb {G}}} u(t)|^{2}\text {d}x\text {d}\tau \\ {}&=-\frac{1}{2}\frac{\text {d} k\circ \nabla _{{\mathbb {G}}} u}{\text {d}t}+\frac{k'\circ \nabla _{{\mathbb {G}}} u}{2}+\frac{1}{2}\frac{\text {d}}{\text {d}t}\left( \int _{0}^{t}k(\tau )\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\text {d}\tau \right) \\ {}&\quad -\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}. \end{aligned} \end{aligned}$$

(3.12)

Next, using (3.12) in (3.8) yields

$$\begin{aligned} \begin{aligned} 0&=\frac{\text {d}}{\text {d}t}\left( \frac{1}{2}\Vert u_{t}\Vert ^{2}_{2}+\frac{1}{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}_{2}-\frac{1}{p}\Vert u\Vert _{p}^{p}\text {d}x\right) -\int _{0}^{t}k(t-\tau )\int _{\Omega }\nabla _{{\mathbb {G}}} u_{t}\cdot \nabla _{{\mathbb {G}}} u\text {d}x\text {d}\tau \\ {}&\quad +a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\\ {}&=\frac{\text {d}}{\text {d}t}\left( \frac{1}{2}\Vert u_{t}\Vert ^{2}+\frac{1}{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-\frac{1}{p}\Vert u\Vert _{p}^{p}\text {d}x\right) +\frac{1}{2}\frac{\text {d} k\circ \nabla _{{\mathbb {G}}} u}{\text {d}t}-\frac{k'\circ \nabla _{{\mathbb {G}}} u}{2}\\ {}&\quad -\frac{1}{2}\frac{\text {d}}{\text {d}t}\left( \int _{0}^{t}k(\tau )\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\text {d}\tau \right) +\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\\ {}&=\frac{\text {d}}{\text {d}t}\left( \frac{1}{2}\Vert u_{t}\Vert ^{2}+\frac{1}{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-\frac{1}{2}\int _{0}^{t}k(\tau )\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\text {d}\tau +\frac{1}{2}k\circ \nabla _{{\mathbb {G}}} u-\frac{1}{p}\Vert u\Vert _{p}^{p}\text {d}x\right) \\ {}&\quad +\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}-\frac{k'\circ \nabla _{{\mathbb {G}}} u}{2}\\ {}&=\frac{\text {d}I}{\text {d}t}+\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}_{2}+a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}-\frac{k'\circ \nabla _{{\mathbb {G}}} u}{2}, \end{aligned} \end{aligned}$$

(3.13)

that is,

$$\begin{aligned} \begin{aligned} \frac{\text {d}I}{\text {d}t}&=-\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}+\frac{k'\circ \nabla _{{\mathbb {G}}} u}{2}=-\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\\&+\frac{1}{2}\int _{0}^{t}k'(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(t)-\nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau -a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\\ {}&{\mathop {\le }\limits ^{(3.4)}}-a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}. \end{aligned} \end{aligned}$$

(3.14)

Therefore, we have

$$\begin{aligned} \frac{\text {d}I}{\text {d}t}\le -a\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\le 0, \end{aligned}$$

(3.15)

that is,

$$\begin{aligned} I'(t)\le 0. \end{aligned}$$

This proves the statement (a). The part (b) follows from integrating (3.15) over (0, t)

$$\begin{aligned} I(t)-I(0)\le -a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau , \end{aligned}$$

(3.16)

which is equivalent to

$$\begin{aligned} I(t)+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \le I(0). \end{aligned}$$

$\square$

Now, we present the main result of this section.

Theorem 3.2

Suppose that $p>2$ satisfies (3.2), $a>0$ and $k\in C^{1}[0,T]$ satisfies the conditions (3.3) and (3.4). Let u be a solution of (3.1), satisfying

$$\begin{aligned} \left( 2(u,u_{t})+a\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}_{2}\right) |_{t=0}>\frac{2p}{\theta }I(0), \end{aligned}$$

(3.17)

where $\theta =\max _{\mu _{1}\in (0,1)}\theta (\mu _{1})=\theta (\mu _{1}^{*})$ with

$$\begin{aligned} \theta (\mu _{1})=\min \left( \left( (p+2)a\alpha \mu _{1}R\right) ^{\frac{1}{2}},\frac{\alpha (1-\mu _{1})}{a}\right) . \end{aligned}$$

(3.18)

Then, u blows up at a finite time.

Proof

Let us define the following function:

$$\begin{aligned} Z(t)=2(u_{t},u)+a\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}-\mu I(0), \end{aligned}$$

(3.19)

where $\mu$ is a positive constant to be specified. By multiplying u(t) Eq. (3.1) and integrating over $\Omega$, we get

$$\begin{aligned} \begin{aligned} (u_{tt},u)+a(\nabla _{{\mathbb {G}}} u,\nabla _{{\mathbb {G}}} u_{t})=-\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-\int _{0}^{t}\int _{\Omega }k(t-\tau )\Delta _{{\mathbb {G}}} u(\tau )\text {d}\tau u(t)\text {d}x+\Vert u\Vert ^{p}_{p}. \end{aligned} \end{aligned}$$

(3.20)

Then, by using this fact, we have

$$\begin{aligned} \begin{aligned} Z'(t)&=2\Vert u_{t}\Vert ^{2}+2(u_{tt},u)+2a(\nabla _{{\mathbb {G}}} u,\nabla _{{\mathbb {G}}} u_{t})\\ {}&=2\Vert u_{t}\Vert ^{2}-2\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2\int _{\Omega }k(t-\tau )\Delta _{{\mathbb {G}}} u(\tau )\text {d}\tau u(t)\text {d}x+2\Vert u\Vert ^{p}_{p}. \end{aligned}\nonumber \\ \end{aligned}$$

(3.21)

Next, by means of Green’s identity, we compute

$$\begin{aligned} \begin{aligned} \int _{0}^{t}&\int _{\Omega }k(t-\tau )u(t)\Delta _{{\mathbb {G}}} u \text {d}\tau \text {d}x=-\int _{0}^{t}\int _{\Omega }k(t-\tau )(\nabla _{{\mathbb {G}}}u(t)\cdot \nabla _{{\mathbb {G}}} u(\tau ))\text {d}x\text {d}\tau \\ {}&=-\int _{0}^{t}\int _{\Omega }k(t-\tau )\nabla _{{\mathbb {G}}} u(t)\cdot (\nabla _{{\mathbb {G}}} (u(\tau )-u(t)))\text {d}x\text {d}\tau -\int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\text {d}\tau \\ {}&=-\int _{0}^{t}\int _{\Omega }k(t-\tau )\nabla _{{\mathbb {G}}} u(t)\cdot (\nabla _{{\mathbb {G}}} (u(\tau )-u(t)))\text {d}x\text {d}\tau -\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau \\ {}&=-\int _{0}^{t}k(t-\tau )(\nabla _{{\mathbb {G}}}u(t),\nabla _{{\mathbb {G}}} (u(\tau )-u(t)))\text {d}\tau -\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau . \end{aligned}\nonumber \\ \end{aligned}$$

(3.22)

This yields

$$\begin{aligned} \begin{aligned} Z'(t)&=2\Vert u_{t}\Vert ^{2}+2(u_{tt},u)+2a(\nabla _{{\mathbb {G}}} u,\nabla _{{\mathbb {G}}} u_{t})\\ {}&=2\Vert u_{t}\Vert ^{2}-2\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2\int _{\Omega }k(t-\tau )\Delta _{{\mathbb {G}}} u(\tau )\text {d}\tau u(t)\text {d}x+2\Vert u\Vert ^{p}_{p}\\ {}&=2\Vert u_{t}\Vert ^{2}-2\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+2\int _{0}^{t}k(t-\tau )(\nabla _{{\mathbb {G}}} u(t),\nabla _{{\mathbb {G}}} (u(\tau )-u(t)))\text {d}\tau \\ {}&\quad +2\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau +2\Vert u\Vert ^{p}_{p}. \end{aligned}\nonumber \\ \end{aligned}$$

(3.23)

On the other hand, by using Young’s inequality, we have

$$\begin{aligned} \begin{aligned} \int _{0}^{t}k(t-\tau )(\nabla _{{\mathbb {G}}} u(t),\nabla _{{\mathbb {G}}} (u(t)-u(\tau )))\text {d}\tau&\le \frac{p}{2}\int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\text {d}\tau \\ {}&+\frac{1}{2p}\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau , \end{aligned}\nonumber \\ \end{aligned}$$

(3.24)

that is,

$$\begin{aligned} \begin{aligned} \int _{0}^{t}&k(t-\tau )(\nabla _{{\mathbb {G}}} u(t),\nabla _{{\mathbb {G}}} (u(\tau )-u(t)))\text {d}\tau \\ {}&\quad \ge -\frac{p}{2}\int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\text {d}\tau -\frac{1}{2p}\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau . \end{aligned}\nonumber \\ \end{aligned}$$

(3.25)

Thus, in the view of (3.25), we get

$$\begin{aligned} \begin{aligned} Z'(t)&=2\Vert u_{t}\Vert ^{2}-2\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+2\int _{0}^{t}k(t-\tau )(\nabla _{{\mathbb {G}}} u(t),\nabla _{{\mathbb {G}}} (u(\tau )-u(t)))\text {d}\tau \\ {}&\quad +2\Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau +\Vert u\Vert ^{p}_{p}\\ {}&\ge 2\Vert u_{t}\Vert ^{2}-2\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-p\int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\text {d}\tau \\ {}&\quad -\frac{1}{p}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau +2\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau +2\Vert u\Vert ^{p}_{p}\\ {}&=(p+2)\Vert u_{t}\Vert ^{2}+(p-2)\left( 1-\int _{0}^{t}k(\tau )\text {d}\tau \right) \Vert \nabla _{{\mathbb {G}}}u(t)\Vert ^{2}\\ {}&\quad -p\Vert u_{t}\Vert ^{2}-p\left( 1-\int _{0}^{t}k(\tau )\text {d}\tau \right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}+2\Vert u\Vert _{p}^{p}\\ {}&\quad +2\left( -\frac{p}{2}\int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u(\tau )-\nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\text {d}\tau -\frac{1}{2p}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau \right) \\ {}&\ge (p+2)\Vert u_{t}\Vert ^{2}+(p-2)\left( 1-\int _{0}^{t}k(\tau )\text {d}\tau \right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\\ {}&\quad -\frac{1}{p}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau -2pI(t). \end{aligned} \end{aligned}$$

(3.26)

Now from the part (b) of Lemma 3.1, it follows that

$$\begin{aligned} Z'(t)\ge & {} (p+2)\Vert u_{t}\Vert ^{2}+(p-2)\left( 1-\int _{0}^{t}k(\tau )\text {d}\tau \right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\nonumber \\&\quad -\frac{1}{p}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau -2pI(t)\nonumber \\&{\mathop {\ge }\limits ^{(3.7)}}(p+2)\Vert u_{t}\Vert ^{2}+(p-2)\left( 1-\int _{0}^{t}k(\tau )\text {d}\tau \right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\nonumber \\&\quad -\frac{1}{p}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau +2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{\tau }(\tau )\Vert ^{2}\text {d}\tau -2pI(0)\nonumber \\&{\mathop {\ge }\limits ^{(3.3)}}(p+2)\Vert u_{t}\Vert ^{2}+\left( (p-2)r-\frac{1}{p}(1-r)\right) \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\nonumber \\&\quad -2pI(0)+2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}(\tau )\Vert ^{2}\text {d}\tau \nonumber \\&=(p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)+2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \nonumber \\&{\mathop {\ge }\limits ^{a>0}}(p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0), \end{aligned}$$

(3.27)

where $\alpha =\left( (p-2)r-\frac{1}{p}(1-r)\right) .$ Note that $\alpha >0$ since the condition (3.3). Further, by using Young’s inequality, we have

$$\begin{aligned} 2\left( (p+2)a\alpha \mu _{1}R\right) ^{\frac{1}{2}}|(u_{t},u)|\le (p+2)\Vert u_{t}\Vert ^{2}+a\alpha \mu _{1}R\Vert u\Vert ^{2}. \end{aligned}$$

(3.28)

Combining Theorem 2.2 with this fact, we get

$$\begin{aligned} \begin{aligned} Z'(t)&\ge (p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)+2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&=(p+2)\Vert u_{t}\Vert ^{2}+\alpha a\mu _{1}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+(1-a\mu _{1})\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)\\ {}&\quad +2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&{\mathop {\ge }\limits ^{(2.4)}}(p+2)\Vert u_{t}\Vert ^{2}+\alpha Ra\mu _{1}\Vert u\Vert ^{2}+(1-a\mu _{1})\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)\\ {}&\quad +2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&{\mathop {\ge }\limits ^{(3.28)}}2\left( (p+2)\alpha \mu _{1}R\right) ^{\frac{1}{2}}|(u_{t},u)|+(1-a\mu _{1})\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)\\ {}&\quad +2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&{\mathop {\ge }\limits ^{a>0}}2\left( (p+2)a\alpha \mu _{1}R\right) ^{\frac{1}{2}}|(u_{t},u)|+(1-a\mu _{1})\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)\\ {}&\ge \theta (\mu _{1})\left( 2(u_{t},u)+a\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-\frac{2p}{\theta (\mu _{1})}I(0)\right) , \end{aligned} \end{aligned}$$

(3.29)

where R is defined in Theorem 2.2, $\mu _{1}\in (0,1)$ is to be specified later and

$$\begin{aligned} \theta (\mu _{1})=\min \left( \left( (p+2)a\alpha \mu _{1}R\right) ^{\frac{1}{2}},\frac{\alpha (1-\mu _{1})}{a}\right) . \end{aligned}$$

(3.30)

It is straightforward to show that $K_{1}(\mu _{1})=\left( (p+2)a\alpha \mu _{1}R\right) ^{\frac{1}{2}}$ is strictly increasing function for $\mu _{1}\in [0,1]$ with $K_{1}(0)=0$ and $K_{1}(1)=\left( (p+2)a\alpha R\right) ^{\frac{1}{2}}$. Similarly, $K_{2}(\mu _{2})=\frac{\alpha (1-\mu _{1})}{a}$ is strictly decreasing function for $\mu _{1}\in [0,1]$ with $K_{2}(0)=\frac{\alpha }{a}$ and $K_{2}(1)=0$. Thus, $\theta (\mu _{1})$ attains its maximum at the point $\mu _{1}=\mu ^{*}_{1}$, where $\mu ^{*}_{1}$ is the root of the $\left( (p+2)a\alpha \mu _{1}R\right) ^{\frac{1}{2}}=\frac{\alpha (1-\mu _{1})}{a}$. Setting

$$\begin{aligned} \theta =\sup _{\mu _{1}\in (0,1)}\theta (\mu _{1})=\theta (\mu ^{*}_{1}) \ \ \text {and} \ \ \mu =\frac{2p}{\theta } \end{aligned}$$

in (3.19) implies that $Z(0)\ge 0.$ Therefore, we get

$$\begin{aligned} Z'(t)\ge \theta Z(t), \end{aligned}$$

which implies

$$\begin{aligned} Z(t)\ge Z(0)\exp (\theta t), \end{aligned}$$

that is,

$$\begin{aligned} Z(t)\rightarrow +\infty \,\,\,\text {as}\,\,\,t\rightarrow +\infty . \end{aligned}$$

By introducing a new function

$$\begin{aligned} \xi (t)=\Vert u\Vert ^{2}+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau +a(T-t)\Vert \nabla _{{\mathbb {G}}} u_{0}\Vert ^{2},\,\,\,\,t\in [0,T], \end{aligned}$$

(3.31)

we compute

$$\begin{aligned} \begin{aligned} \xi '(t)=2(u_{t},u)+a\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}&-a\Vert \nabla _{{\mathbb {G}}} u_{0}\Vert ^{2}=2(u_{t},u)+a\int _{0}^{t}\frac{\text {d}}{\text {d}\tau }\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\text {d}\tau \\ {}&=2(u_{t},u)+2a\int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{\tau }(\tau ),\nabla _{{\mathbb {G}}} u(\tau ))\text {d}\tau . \end{aligned} \end{aligned}$$

(3.32)

It easy to see that $\xi ''(t)=Z'(t)$, so we have

$$\begin{aligned} \begin{aligned} \xi ''(t)\ge (p+2)\Vert u_{t}\Vert ^{2}&+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)+2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&\ge (p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0). \end{aligned} \end{aligned}$$

(3.33)

Let $0< \gamma <1,\varepsilon>0,T_{B}>0$ be such that $\gamma (p+2)>4+\varepsilon =\nu ,$ and

$$\begin{aligned} (p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)\ge \gamma ((p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2})\,\,\,\,t>T_{B}.\nonumber \\ \end{aligned}$$

(3.34)

Thus, by using these results, we get

$$\begin{aligned} \begin{aligned} \xi ''(t)&\ge (p+2)\Vert u_{t}\Vert ^{2} +\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)+2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&\ge (p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-2pI(0)+2ap\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&\ge \gamma ((p+2)\Vert u_{t}\Vert ^{2}+\alpha \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2})+a\gamma (p+2)\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \\ {}&\ge (4+\varepsilon )(\Vert u_{t}\Vert ^{2}+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau )\\ {}&=\nu (\Vert u_{t}\Vert ^{2}+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau ),\,\,\,t>T_{B}. \end{aligned} \end{aligned}$$

(3.35)

Next, Cauchy—Schwarz–Bunyakovsky inequality yields the following estimates:

$$\begin{aligned}&|(u_{t},u)|\le \Vert u_{t}\Vert ^{2}\Vert u\Vert ^{2}, \end{aligned}$$

(3.36)

$$\begin{aligned}&\left( \int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t},\nabla _{{\mathbb {G}}} u)\text {d}\tau \right) ^{2}\le \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\text {d}\tau . \end{aligned}$$

(3.37)

Thus, we obtain

$$\begin{aligned} \begin{aligned} 2(u_{t},&u)\int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t},\nabla _{{\mathbb {G}}} u)\text {d}\tau \\ {}&{\mathop {\le }\limits ^{(3.36),(3.37)}}2\Vert u_{t}\Vert \Vert u\Vert \left( \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \right) ^{\frac{1}{2}}\left( \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\text {d}\tau \right) ^{\frac{1}{2}}\\ {}&\le \Vert u\Vert ^{2}\left( \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \right) +\Vert u_{t}\Vert ^{2}\left( \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\text {d}\tau \right) . \end{aligned} \end{aligned}$$

(3.38)

Therefore, for $t>T_{B}$ we have

$$\begin{aligned} \xi ''(t)\xi (t)- & {} \frac{\nu }{4}(\xi '(t))^{2}>\nu \left( \Vert u_{t}\Vert ^{2}+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \right) \left( \Vert u\Vert ^{2}+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau \right) \nonumber \\&\quad -\nu \left( 2(u_{t},u)+2a\int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t}(\tau ),\nabla _{{\mathbb {G}}} u(\tau ))\text {d}\tau \right) ^{2}\nonumber \\&=\nu \Big (\Vert u_{t}\Vert ^{2}\Vert u\Vert ^{2}+a\Vert u_{t}\Vert ^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau +a\Vert u\Vert ^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \nonumber \\&\quad +a^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau \Big )\nonumber \\&\quad -\nu \Big ((u_{t},u)^{2}+2a(u_{t},u)\int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t}(\tau ),\nabla _{{\mathbb {G}}} u(\tau ))\text {d}\tau \nonumber \\&\quad +a^{2}\left( \int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t}(\tau ),\nabla _{{\mathbb {G}}} u(\tau ))\text {d}\tau \right) ^{2}\Big )\nonumber \\&{\mathop {\ge }\limits ^{(3.36)}}\nu \Big (a\Vert u_{t}\Vert ^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau +a\Vert u\Vert ^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \nonumber \\&\quad +a^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau \Big )\nonumber \\&\quad -\gamma \Big (2a(u_{t},u)\int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t}(\tau ),\nabla _{{\mathbb {G}}} u(\tau ))\text {d}\tau \nonumber \\&\quad +a^{2}\left( \int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t}(\tau ),\nabla _{{\mathbb {G}}} u(\tau ))\text {d}\tau \right) ^{2}\Big )\nonumber \\&{\mathop {\ge }\limits ^{(3.37)}}a\nu \left( \Vert u_{t}\Vert ^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau +\Vert u\Vert ^{2}\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u_{t}\Vert ^{2}\text {d}\tau \right) \nonumber \\&\quad -2\nu a(u_{t},u)\int _{0}^{t}(\nabla _{{\mathbb {G}}} u_{t}(\tau ),\nabla _{{\mathbb {G}}} u(\tau ))\text {d}\tau \nonumber \\&{\mathop {\ge }\limits ^{(3.38)}}0. \end{aligned}$$

(3.39)

By setting $\phi (s)=\xi (t-T_{B})$, where $s=t-T_{B}$, it is easy to see that

$$\begin{aligned} \phi ''\phi -\frac{\gamma }{4}(\phi ')^{2}\ge 0. \end{aligned}$$

Thus, there exists $T_{B}<t<T$ such that

$$\begin{aligned} \lim _{t\rightarrow T_{B}}\phi (s)=+\infty , \end{aligned}$$

(3.40)

i.e.,

$$\begin{aligned} \lim _{t\rightarrow T_{B}}\left( \Vert u\Vert ^{2}+a\int _{0}^{t}\Vert \nabla _{{\mathbb {G}}} u(\tau )\Vert ^{2}\text {d}\tau +(T-t)\Vert \nabla _{{\mathbb {G}}} u_{0}\Vert ^{2}\right) =+\infty . \end{aligned}$$

(3.41)

Therefore, in the view of the last expression we have

$$\begin{aligned} \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\rightarrow +\infty ,\,\,\,\,t\rightarrow T_{B}. \end{aligned}$$

$\square$

4 Blow-Up with Weak Damping

In this section, we consider the viscoelastic wave equation with weak damping for the sub-Laplacian:

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle u_{tt}-\Delta _{{\mathbb {G}}} u+\int _{0}^{t}k(t-\tau )\Delta _{{\mathbb {G}}} u\text {d}\tau +a |u_{t}|^{q-2}u_{t}=|u|^{p-2}u,\,\,\,(x,t)\in \Omega \times [0,T],\\ u(x,t)=0,\,\,\,\,x\in \partial \Omega ,\\ u(x,0)=u_{0}(x),\,\,u_{t}(x,0)=u_{1}(x), \end{array}\right. }\nonumber \\ \end{aligned}$$

(4.1)

where $\Omega \subset {\mathbb {G}}$, is a Haar measurable set with a smooth boundary $\partial \Omega ,$ $a>0$, $p>2$ $q\ge 1$, $u_{0}\in S^{1,2}_{0}(\Omega ),$ and $u_{1}\in L^{2}(\Omega )$. The function I(t) is defined as in (3.5) and the function k satisfies (3.3)–(3.4). Further, let p and q be such that

$$\begin{aligned} \max \{p,q\}\le \frac{2(Q-1)}{Q-2}. \end{aligned}$$

(4.2)

We state the following lemmata which will be useful in proving blow-up result for (4.1).

Lemma 4.1

Suppose that p, q satisfy (4.2). Then, we have

$$\begin{aligned} \Vert u\Vert ^{\gamma }_{p}\le C\left( \Vert u\Vert ^{p}_{p}+\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\right) ,\,\,\,\,\,2\le \gamma \le p, \end{aligned}$$

(4.3)

where C is a positive constant which depends only on the Haar measure of $\Omega$.

Proof

Assume that $\Vert u\Vert _{p}>1$. Since $2\le \gamma \le p$, Sobolev Embedding Theorem 2.3 with $2^{*}=\frac{2Q}{Q-2}$ yields

$$\begin{aligned} \begin{aligned} \Vert u\Vert ^{\gamma }_{p}&\le \Vert u\Vert ^{p}_{p}\le \Vert u\Vert ^{p}_{p}+\Vert u\Vert ^{2}_{2^{*}}{\mathop {\le }\limits ^{(2.5)}}\Vert u\Vert ^{p}_{p}+C\Vert \nabla _{{\mathbb {G}}}u\Vert ^{2}_{2}\le C\left( \Vert u\Vert ^{p}_{p}+\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\right) . \end{aligned} \end{aligned}$$

(4.4)

Now assume $\Vert u\Vert _{p}\le 1$. Let $p=\frac{Qp'}{Q-p'}$ with $1<p'<Q$. Then we have the $1<p'<p$ yielding continuous embedding, i.e., $L^{p}(\Omega )\hookrightarrow L^{p'}(\Omega )$. Thus, we have

$$\begin{aligned} \Vert \nabla _{{\mathbb {G}}}u\Vert _{p'}\le C\Vert \nabla _{{\mathbb {G}}}u\Vert _{p}. \end{aligned}$$

(4.5)

Since $2\le \gamma$, we have

$$\begin{aligned} \begin{aligned} \Vert u\Vert ^{\gamma }_{p}\le \Vert u\Vert ^{2}_{p}{\mathop {\le }\limits ^{(2.5)}}C\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}_{p'}{\mathop {\le }\limits ^{(4.5)}}C\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}_{p}&\le C\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}_{p}+\Vert u\Vert ^{p}_{p} \\ {}&\le C\left( \Vert u\Vert ^{p}_{p}+\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\right) . \end{aligned} \end{aligned}$$

(4.6)

$\square$

Lemma 4.2

Let u be a weak solution of (4.1) with (4.2). Then, we have

$$\begin{aligned} \Vert u\Vert ^{\gamma }_{p}\le C\left( I(t)-\Vert u_{t}\Vert ^{2}-(k\circ \nabla _{{\mathbb {G}}} u)+\Vert u\Vert ^{p}_{p}\right) ,\,\,\,\,\forall t\in [0,T], \end{aligned}$$

(4.7)

where $2\le \gamma \le p$ and C is a positive constant.

Proof

The function I(t) is given by

$$\begin{aligned} I(t)=\frac{1}{2}\left( \Vert u_{t}(t)\Vert ^{2}_{2}+\left( 1-\int _{0}^{t}k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}+k\circ \nabla _{{\mathbb {G}}} u\right) -\frac{1}{p}\Vert u(t)\Vert ^{p}_{p}. \end{aligned}$$

(4.8)

Hence, by using (3.3) and (3.4), we compute

$$\begin{aligned} \begin{aligned} r\Vert \nabla _{{\mathbb {G}}}u(t)\Vert ^{2}_{2}&{\mathop {=}\limits ^{(3.3)}}\left( 1-\int _{0}^{\infty }k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}\\ {}&{\mathop {\le }\limits ^{(3.4)}}\left( 1-\int _{0}^{t}k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}\\ {}&=2I(t)-\Vert u_{t}(t)\Vert ^{2}_{2}-k\circ \nabla _{{\mathbb {G}}} u+\frac{2}{p}\Vert u\Vert ^{p}_{p}. \end{aligned} \end{aligned}$$

(4.9)

Now we apply Lemma 4.1 with $2\le \gamma \le p$, to obtain

$$\begin{aligned} \begin{aligned} \Vert u\Vert ^{\gamma }_{p}\le C\left( \Vert u\Vert ^{p}_{p}+\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\right) {\mathop {\le }\limits ^{(4.9)}}C\left( I(t)-\Vert u_{t}\Vert ^{2}-(k\circ \nabla _{{\mathbb {G}}} u)+\Vert u\Vert ^{p}_{p}\right) . \end{aligned} \end{aligned}$$

(4.10)

$\square$

Lemma 4.3

Suppose that (3.3)–(3.4) are satisfied. Let u be a weak solution of (4.1), then I(t) is a non-increasing function for $t\in [0,T]$, i.e.,

$$\begin{aligned} I'(t)\le 0,\,\,\,\forall t\in [0,T]. \end{aligned}$$

(4.11)

We omit the proof of Lemma 4.3 since it is similar to that of Lemma 3.1. The main result of this section is the following theorem.

Theorem 4.4

Assume that $q > 1$ and $p> \max \{2, q\}$ satisfy the condition (4.2). If (3.3) and (3.4) hold with $I(0)<0$, then solution u of (4.1) blows up at a finite time.

Proof

By Lemma 4.3, we have

$$\begin{aligned} I'(t)\le 0, \end{aligned}$$

(4.12)

hence,

$$\begin{aligned} I(t)\le I(0),\,\,\,\forall t\in [0,T]. \end{aligned}$$

Let us define by $Z(t)=-I(t)$. Then, we get

$$\begin{aligned} \begin{aligned} 0&<Z(0)\le Z(t)=-I(t)\\ {}&=-\frac{1}{2}\left( \Vert u_{t}(t)\Vert ^{2}_{2}+\left( 1-\int _{0}^{t}k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}+k\circ \nabla _{{\mathbb {G}}} u\right) +\frac{1}{p}\Vert u(t)\Vert ^{p}_{p}\\ {}&=-\frac{1}{2}\Vert u_{t}(t)\Vert ^{2}_{2}-\frac{1}{2}\left( 1-\int _{0}^{t}k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}-\frac{1}{2}k\circ \nabla _{{\mathbb {G}}} u+\frac{1}{p}\Vert u(t)\Vert ^{p}_{p}\\ {}&{\mathop {\le }\limits ^{(3.3),(3.4)}}\frac{1}{p}\Vert u(t)\Vert ^{p}_{p}. \end{aligned} \end{aligned}$$

(4.13)

Similarly by Lemma 3.1, we obtain

$$\begin{aligned} Z'(t)=-I'(t)=a\Vert u_{t}\Vert ^{q}_{q}-\frac{1}{2}(k'\cdot \nabla _ {{\mathbb {G}}}u)+\frac{1}{2}k(t)\Vert \nabla _{{\mathbb {G}}}u\Vert ^{2} {\mathop {\ge }\limits ^{(3.3),(3.4)}}0. \end{aligned}$$

(4.14)

Let us also define the following function

$$\begin{aligned} A(t)=Z^{1-\beta }(t)-\varepsilon (u_{t},u), \end{aligned}$$

(4.15)

where $0<\beta \le \min \{\frac{p-2}{2p},\frac{p-q}{p(q-1)}\}$. By means of direct calculations and the Cauchy–Bunyakovsky–Schwarz inequality, we get

$$\begin{aligned} \begin{aligned} A'(t)&=(1-\beta )Z^{-\beta }(t)Z'(t)-\varepsilon (u_{tt},u)-\varepsilon \Vert u_{t}\Vert ^{2}\\ {}&=(1-\beta )Z^{-\beta }(t)Z'(t)-\varepsilon \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \int _{0}^{t}k(t-\tau )(\nabla _{{\mathbb {G}}} u(\tau ),\nabla _{{\mathbb {G}}} u(t))\text {d}\tau \\ {}&\quad +\varepsilon \Vert u\Vert ^{p}_{p}-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x+\varepsilon \Vert u_{t}\Vert ^{2}\\ {}&{\mathop {\ge }\limits ^{(3.22),(4.14) }}a(1-\beta )Z^{-\beta }(t)\Vert u_{t}\Vert ^{q}_{q}-\varepsilon \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \Vert u\Vert ^{p}_{p}-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x\\ {}&\quad +\varepsilon \int _{0}^{t}\int _{\Omega }k(t-\tau )(\nabla _{{\mathbb {G}}} u(t),\nabla _{{\mathbb {G}}} (u(\tau )-u(t)))\text {d}x\text {d}\tau \\ {}&+\varepsilon \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau +\varepsilon \Vert u_{t}\Vert ^{2}\\ {}&{\mathop {\ge }\limits ^{C-B-S}}a(1-\beta )Z^{-\beta }(t)\Vert u_{t}\Vert ^{q}_{q}-\varepsilon \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \Vert u\Vert ^{p}_{p}-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x+\varepsilon \Vert u_{t}\Vert ^{2}\\ {}&\quad -\varepsilon \int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\Vert \nabla _{{\mathbb {G}}} (u(\tau )-u(t))\Vert ^{2}\text {d}\tau +\varepsilon \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau . \end{aligned} \end{aligned}$$

(4.16)

In the view of (3.5), we have

$$\begin{aligned} \begin{aligned} \frac{1}{p}\Vert u\Vert ^{p}_{p}&=Z(t)+\frac{1}{2}\left( \Vert u_{t}(t)\Vert ^{2}+\left( 1-\int _{0}^{t}k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}+k\circ \nabla _{{\mathbb {G}}} u\right) . \end{aligned} \end{aligned}$$

(4.17)

On the other hand, from (4.16) with (3.25), we obtain

$$\begin{aligned} \begin{aligned} A'(t)&\ge a(1-\beta )Z^{-\beta }(t)\Vert u_{t}\Vert ^{q}_{q}-\varepsilon \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \Vert u\Vert ^{p}_{p}-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x+\varepsilon \Vert u_{t}\Vert ^{2}\\ {}&\quad -\varepsilon \int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\Vert \nabla _{{\mathbb {G}}} (u(\tau )-u(t))\Vert ^{2}\text {d}\tau +\varepsilon \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau \\ {}&{\mathop {=}\limits ^{(4.17)}}a(1-\beta )Z^{-\beta }(t)\Vert u_{t}\Vert ^{q}_{q}-\varepsilon \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x+\varepsilon \Vert u_{t}\Vert ^{2}\\ {}&\quad +\frac{\varepsilon p}{2}\left( 2Z(t)+\Vert u_{t}\Vert ^{2}+\left( 1-\int _{0}^{t}k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+k\circ \nabla _{{\mathbb {G}}} u\right) \\ {}&\quad -\varepsilon \int _{0}^{t}k(t-\tau )\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\Vert \nabla _{{\mathbb {G}}} (u(\tau )-u(t))\Vert ^{2}\text {d}\tau +\varepsilon \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}\int _{0}^{t}k(\tau )\text {d}\tau \\ {}&{\mathop {\ge }\limits ^{(3.25)}}a(1-\beta )Z^{-\beta }(t)\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon p Z(t)-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x\\ {}&\quad +\left( \frac{\varepsilon p}{2}-\varepsilon \delta \right) (k\circ \nabla _{{\mathbb {G}}} u)+\left( \left( \frac{p}{2}-1\right) -\varepsilon \left( \frac{p}{2}-1+\frac{1}{4\delta }\right) \int _{0}^{t}k(\tau )\text {d}\tau \right) \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}, \end{aligned} \end{aligned}$$

(4.18)

where $\delta \in (0,\frac{p}{2}).$ We apply Young’s inequality to estimate the fourth term on the right hand side of the (4.18) to obtain

$$\begin{aligned} \begin{aligned} A'(t)&\ge (1-\beta )Z^{-\beta }(t)\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon p Z(t)-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x\\ {}&\quad +\left( \frac{\varepsilon p}{2}-\varepsilon \delta \right) (k\circ \nabla _{{\mathbb {G}}} u)+\left( \left( \frac{p}{2}-1\right) -\left( \frac{p}{2}-1+\frac{1}{4\delta }\right) \int _{0}^{t}k(\tau )\text {d}\tau \right) \Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}\\ {}&=(1-\beta )Z^{-\beta }(t)\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon p Z(t)-a\varepsilon \int _{\Omega }|u_{t}|^{q-2}u_{t}u\text {d}x\\ {}&\quad +\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)+\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon p Z(t)\\ {}&\ge a\left( (1-\beta )Z^{-\beta }-\frac{\varepsilon \lambda ^{-q'}}{q'}\right) \Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)\\ {}&\quad +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-\frac{\varepsilon a \lambda ^{q}}{q}\Vert u\Vert ^{q}_{q}+\varepsilon p Z(t),\,\,\,\,\forall \lambda >0, \end{aligned}\nonumber \\ \end{aligned}$$

(4.19)

where

$$\begin{aligned} C_{1}=\frac{\varepsilon p}{2}-\varepsilon \delta>0,\,\,\,\,C_{2}=\left( \frac{p}{2}-1\right) -\left( \frac{p}{2}-1+\frac{1}{4\delta }\right) \int _{0}^{t}k(\tau )\text {d}\tau >0. \end{aligned}$$

Then, by setting $\lambda ^{-q'}=\chi Z^{-\beta }(t)$ we get

$$\begin{aligned} A'(t)&\ge a\left( (1-\beta )Z^{-\beta }-\frac{\varepsilon \lambda ^{-q'}}{q'}\right) \Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}\nonumber \\&\quad +\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)+\varepsilon p Z(t) +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}-\frac{\varepsilon a \lambda ^{q}}{q}\Vert u\Vert ^{q}_{q}\nonumber \\&=a\left( (1-\beta )-\frac{\varepsilon \chi }{q'}\right) Z^{-\beta }\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)\nonumber \\&\quad +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \left( p Z(t)-\frac{\varepsilon a \chi ^{1-q}}{q}Z^{-\beta (1-q)}\Vert u\Vert ^{q}_{q}\right) . \end{aligned}$$

(4.20)

Next, by using (4.13) and the fact that $L^{p}(\Omega )\hookrightarrow L^{q}(\Omega )$ for $p>q$, we obtain

$$\begin{aligned} \Vert u\Vert ^{q}_{q}\le C\left( \frac{1}{p}\right) ^{\beta (q-1)}\Vert u\Vert ^{q+\beta p(q-1)}_{p}. \end{aligned}$$

(4.21)

The last inequality applied to (4.20) yields

$$\begin{aligned} \begin{aligned} A'(t)&\ge a\left( (1-\beta )-\frac{\varepsilon \chi }{q'}\right) Z^{-\beta }\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)\\ {}&\quad +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \left( p Z(t)-\frac{ a \chi ^{1-q}}{q}\Vert u\Vert ^{q}_{q}\right) \\ {}&\ge a\left( (1-\beta )-\frac{\varepsilon \chi }{q'}\right) Z^{-\beta }\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)\\ {}&\quad +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \left( p Z(t)-C\frac{ a \chi ^{1-q}}{q}\left( \frac{1}{p}\right) ^{\beta (q-1)}\Vert u\Vert ^{q+\beta p(q-1)}_{p}\right) . \end{aligned}\nonumber \\ \end{aligned}$$

(4.22)

Now, we apply Lemma 4.2 with $\gamma =q+\beta p(q-1)\le p$.

$$\begin{aligned} A'(t)\ge & {} a\left( (1-\beta )-\frac{\varepsilon \chi }{q'}\right) Z^{-\beta }\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)\nonumber \\&\quad +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \left( p Z(t)-C\frac{a \chi ^{1-q}}{q}\left( \frac{1}{p}\right) ^{\beta (q-1)}\Vert u\Vert ^{\gamma }_{p}\right) \nonumber \\&{\mathop {\ge }\limits ^{(4.7)}}a\left( (1-\beta )-\frac{\varepsilon \chi }{q'}\right) Z^{-\beta }\Vert u_{t}\Vert ^{q}_{q}+\left( \varepsilon +\frac{\varepsilon p}{2}\right) \Vert u_{t}\Vert ^{2}+\varepsilon C_{1}(k\circ \nabla _{{\mathbb {G}}} u)\nonumber \\&\quad +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \left( p Z(t)+C'_{1} \chi ^{1-q}\left( Z(t)+\Vert u_{t}\Vert ^{2}+(k\circ \nabla _{{\mathbb {G}}} u)-\Vert u\Vert ^{p}_{p}\right) \right) \nonumber \\&\ge a\left( (1-\beta )-\frac{\varepsilon \chi }{q'}\right) Z^{-\beta }\Vert u_{t}\Vert ^{q}_{q}+\varepsilon \left( \frac{p}{2}+1+C_{1}\chi ^{1-q}\right) \Vert u_{t}\Vert ^{2}-\varepsilon C'_{1}\chi ^{1-q}\Vert u\Vert ^{p}_{p}\nonumber \\&\quad +\varepsilon \left( C_{1}+C'_{1}\chi ^{1-q}\right) (k\circ \nabla _{{\mathbb {G}}} u) +\varepsilon C_{2}\Vert \nabla _{{\mathbb {G}}} u\Vert ^{2}+\varepsilon \left( p+C'_{1}\chi ^{1-q}\right) Z(t), \nonumber \\ \end{aligned}$$

(4.23)

where $C'_{1}=\frac{aC\left( \frac{1}{p}\right) ^{\beta (q-1)}}{q}$. By the assumption $I(t)<0$, that is,

$$\begin{aligned} Z(t)\ge -\frac{1}{2}\left( \Vert u_{t}(t)\Vert ^{2}_{2}+\left( 1-\int _{0}^{t}k(s)ds\right) \Vert \nabla _{{\mathbb {G}}} u(t)\Vert ^{2}_{2}+k\circ \nabla _{{\mathbb {G}}} u\right) +\frac{1}{p}\Vert u(t)\Vert ^{p}_{p}. \end{aligned}$$

(4.24)

By setting $p=2b+(p-2b)$ where $b=\min \{C_{1},C_{2}\}$ and letting $\chi$ to be large enough in (4.23) we have

$$\begin{aligned} \begin{aligned} A'(t)\ge a\left( (1-\beta )-\frac{\varepsilon \chi }{q'}\right) Z^{-\beta }\Vert u_{t}\Vert ^{q}_{q}+\varepsilon \sigma \left( Z(t)+\Vert u_{t}\Vert ^{2}+\Vert u\Vert ^{p}_{p}+k\circ \nabla _{{\mathbb {G}}} u\right) , \end{aligned} \end{aligned}$$

(4.25)

where $\sigma >0.$ Next, we choose sufficiently small $\varepsilon$ so that $(1-\beta )-\frac{\varepsilon \chi }{q'}>0$. Thus, we have

$$\begin{aligned} A'(t)>\varepsilon \sigma \left( Z(t)+\Vert u_{t}\Vert ^{2}+\Vert u\Vert ^{p}_{p}+k\circ \nabla _{{\mathbb {G}}} u\right) , \end{aligned}$$

(4.26)

and

$$\begin{aligned} A(0)=Z^{1-\beta }(0)+\varepsilon (u_{0},u_{1})>0. \end{aligned}$$

Hence,

$$\begin{aligned} 0<A(0)\le A(t),\,\,\,\forall t\in [0,T]. \end{aligned}$$

Now, by using the Cauchy–Bunyakovsky–Schwarz inequality, embedding of spaces and Young’s inequalities, we get

$$\begin{aligned} |(u_{t},u)|^{\frac{1}{1-\beta }}\le \Vert u_{t}\Vert ^{\frac{1}{1-\beta }}\Vert u\Vert ^{\frac{1}{1-\beta }}\le C\Vert u_{t}\Vert ^{\frac{1}{1-\beta }}\Vert u\Vert ^{\frac{1}{1-\beta }}_{p}\le C \left( \Vert u\Vert ^{\gamma }_{p}+\Vert u_{t}\Vert ^{2}\right) , \end{aligned}$$

(4.27)

with $\frac{1}{(1-\beta )\gamma }+\frac{1}{2(1-\beta )}=1$. By Lemma 4.2, we obtain

$$\begin{aligned} |(u_{t},u)|^{\frac{1}{1-\beta }}\le C\left( Z(t)+\Vert u\Vert ^{p}_{p}+\Vert u_{t}\Vert ^{2}+k\circ \nabla _{{\mathbb {G}}} u\right) . \end{aligned}$$

(4.28)

By using this fact, we calculate

$$\begin{aligned} \begin{aligned} A(t)&=\left( Z^{1-\beta }(t)+\varepsilon (u_{t},u)\right) ^{\frac{1}{1-\beta }}\le 2^\frac{1}{1-\beta }\left( Z(t)+|(u_{t},u)|^{\frac{1}{1-\beta }}\right) \\ {}&\le 2^\frac{1}{1-\beta }\left( Z(t)+C\left( Z(t)+\Vert u\Vert ^{p}_{p}+\Vert u_{t}\Vert ^{2}+k\circ \nabla _{{\mathbb {G}}} u\right) \right) \\ {}&\le C\left( Z(t)+\Vert u\Vert ^{p}_{p}+\Vert u_{t}\Vert ^{2}+k\circ \nabla _{{\mathbb {G}}} u\right) \\ {}&\le C A'(t),\,\,\,\,\,\forall t\in [0,T]. \end{aligned} \end{aligned}$$

(4.29)

Thus,

$$\begin{aligned} A^{\frac{\beta }{1-\beta }}(t)\ge \frac{C(1-\beta )}{C(1-\beta )A^{-\frac{\beta }{1-\beta }}(0)-t\beta }, \end{aligned}$$

(4.30)

hence, we arrive at

$$\begin{aligned} T_{B}\le \frac{C(1-\beta )}{\beta (A(0))^{\frac{\beta }{1-\beta }}}. \end{aligned}$$

(4.31)

Hence, A(t) blows up in finite time. That is,

$$\begin{aligned} \lim _{t\rightarrow T_{B}}\Vert \nabla _{{\mathbb {G}}} u\Vert =+\infty . \end{aligned}$$

(4.32)

$\square$