Blow-Up Phenomena for a Class of Parabolic or Pseudo-parabolic Equation with Nonlocal Source

Abstract

In this paper, we consider a class of parabolic or pseudo-parabolic equation with nonlocal source term:

$$\begin{aligned} u_t-\nu \triangle u_t-\hbox {div}(\rho (|\nabla u|)^2\nabla u)=u^p(x,t)\int _{\Omega }k(x,y)u^{p+1}(y,t)dy, \end{aligned}$$

where $\nu \ge 0$ and $p>0$. Using some differential inequality techniques, we prove that blow-up cannot occur provided that $q>p$, also, we obtain some finite-time blow-up results and the lifespan of the blow-up solution under some different suitable assumptions on the initial energy. In particular, we prove finite-time blow-up of the solution for the initial data at arbitrary energy level. Furthermore, the lower bound for the blow-up time is determined if blow-up does occur.

1 Introduction

In this paper, we deal with the following the initial boundary value problem of a class of parabolic or pseudo-parabolic equation with nonlocal source term:

$$\begin{aligned} {\left\{ \begin{array}{ll} u_t-\nu \triangle u_t-\hbox {div}(\rho (|\nabla u|)^2\nabla u)\\ \qquad =u^p(x,t)\int _{\Omega }k(x,y)u^{p+1}(y,t)dy, &{} (x, t)\in \Omega \times (0,T)\\ u(x,t)=0, &{}(x, t)\in \partial \Omega \times (0,T),\\ u(x,t)=u_0(x)\ge 0, &{}x\in \Omega , \end{array}\right. } \end{aligned}$$

(1.1)

where $\Omega \subset {\mathbb {R}}^n(n\ge 3)$ is a bounded domain with smooth boundary $\partial \Omega$, $\nu \ge 0$, $p>0$, $q>0$ and $T\in (0, \infty ]$ is the maximal existence time of the solution, k(x, y) is an integrable, real-valued function satisfying:

$$\begin{aligned} k(x,y)= & {} k(y,x),\quad \int _{\Omega }\int _{\Omega }k^2(x,y)dxdy<+\infty , \\&\quad \int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy>0. \end{aligned}$$

This type of equations describes a variety of important physical and biological phenomena, such as the analysis of heat conduction in materials with memory, the aggregation of population [12], and so on (see [1] and the references therein). In population dynamics theory, the nonlocal term indicates that the individuals are competing not only with others at their own point in space but also with individual at other points in the domain [11, 12]. If $\nu =0$, $\rho =1,$ Eq. (1.1) reduces to the following semilinear parabolic equation:

$$\begin{aligned} u_{t}-\triangle u=f(u). \end{aligned}$$

(1.2)

The global existence, asymptotic behavior, and finite-time blow-up for the solutions to (1.2)(especially $f(u)=|u|^{p-1}u$) have been studied by many researchers, see [2, 3, 8]and the references therein. Recently, Eq. (1.2) with the nonlocal source $f(u)=\big (\frac{1}{|x|^{n-2}}*|u|^p\big )|u|^{p-2}u$ was considered in [5, 7].

If $\nu >0$ (for the sake of simplicity, $\nu =1$ in this paper), $\rho =1,$ Eq. (1.1) reduces to the following semilinear pseudo-parabolic equation:

$$\begin{aligned} u_{t}-\triangle u_t -\triangle u=f(u). \end{aligned}$$

(1.3)

There are many works about Eq. (1.3) with f(u) being polynomial, such as the existence and uniqueness in [16], blow-up in [10, 17,18,19], asymptotic behavior in [18], and so on.

Recently, Yang and Liang [20] considered a special case of (1.1), that is:

$$\begin{aligned} u_t-\triangle u_t-\triangle u=u^p(x,t)\int _{\Omega }k(x,y)u^{p+1}(y,t)dy. \end{aligned}$$

(1.4)

They proved the finite-time blow-up result provided the initial energy is negative, as well as a nonblow-up criterion. We also mention the paper [1], where Di and Shang considered a four order pseudo-parabolic equation (i.e., an extra term $\triangle ^2u$ in RHS of (1.4)) and obtained a blow-up result of the solutions under suitable initial energy.

For the gentle case $\rho$:

$$\begin{aligned} u_t-\nu \triangle u_t-\hbox {div}(\rho (|\nabla u|)^2\nabla u)=f(u), \end{aligned}$$

(1.5)

where $f(u)\approx u^p,$ the initial boundary problem of (1.5) was investigated in [14, 15]. The blow-up results and the lifespan provided that the initial energy is negative as well as the nonblow-up criterion were established by Payne et al. [14] (parabolic case $\nu =0$), Liu et al [6], and Peng et al. [15](pseudo-parabolic case $\nu =1$).

Recently, Long and Chen [9] considered the following pseudo-parabolic equation with nonlocal source:

$$\begin{aligned} u_t-\triangle u_t-\hbox {div}(|\nabla u|^{2q}\nabla u)=u^p(x,t)\int _{\Omega }k(x,y)u^{p+1}(y,t)dy. \end{aligned}$$

(1.6)

Under $q\rightarrow 0$ and $|\nabla u|\ne 0$, the limit equation of (1.6) is (1.4). When (i) $q<p$ and $J(u_0)\le 0$ or (ii) $q=p$ and $J(u_0)<0$, they proved that the solutions blow up in finite time and the upper and the lower bound.

To our knowledge, no results have been obtained about finite-time blow-up and the lifespan for the solution of the gentle problem (1.1), especially, when the solution with high energy level. Moreover, there is little information about (1.1) under parabolic case ($\nu =0$) with nonlocal source term. The aim of this paper is to present a comprehensive study for the finite blow-up and nonblow-up criterion of problem (1.1).

2 Preliminaries

Throughout this paper, the Banach spaces $L^p=L^p(\Omega )$ and $W_0^{1, p}=W_0^{1, p}(\Omega )$ are endowed with the norms by $\Vert \cdot \Vert _p=\big (\int _{\Omega }|\cdot |^pdx\big )^{\frac{1}{p}}$, and $\Vert \cdot \Vert _{W_0^{1,p}}=\big (\int _{\Omega }(|\cdot |^p+|\nabla \cdot |^p)dx\big )^{\frac{1}{p}}$ as usual. We assume that $\rho$ is a positive $C^1$ function satisfying:

$$\begin{aligned} \rho (s)+2s\rho '(s)\ge 0, \quad s>0, \end{aligned}$$

(2.1)

so that $\hbox {div}(\rho (|\nabla \cdot |)^2\nabla \cdot )$ is elliptic. We also claim that $\rho$ satisfies the condition:

$$\begin{aligned} \rho (s)\ge b_1+b_2s^q, \quad s>0, \end{aligned}$$

(2.2)

where $q>0$ and $b_1, b_2$ are positive constants. Furthermore, we assume that $u_0$ satisfies the compatibility condition $u_0(x)=0$ on $\partial \Omega$.

We first state the local existence theorem for the weak solution to problem (1.1) as follow. See similar result in [1, 9, 15].

Theorem 2.1

Assume $0<p\le \frac{2}{n-2}$, $u_0\in W_0^{1,2q+2}(\Omega )$, and (2.1) hold, there exists a $T>0$, such that the problem (1.1) has a unique local solution $u\in L^{\infty } (0,T;W_0^{1, 2q+2}(\Omega ))$ with $u_t\in L^2(0,T; H_0^1(\Omega ))$ (resp. $u_t\in L^2(0,T; L^2(\Omega ))$ ) for the case of $\nu =1$ (resp. $\nu =0$),

• (i) for a.e. $t\in [0, T]$, the following identity:

  $$\begin{aligned} \langle u_t, v\rangle +\nu \langle \nabla u_t, \nabla v\rangle +\langle \rho (|\nabla u|^2\nabla u,\nabla v\rangle =\langle u^p(x,t)\int _{\Omega }k(x,y)u^{p+1}(y,t)dy, v\rangle \nonumber \\ \end{aligned}$$

  (2.3)

  holds for all $v\in W_0^{1, 2q+2}(\Omega )$.

• (ii) $u(0)=u_0$.

Before processing our main results, we will make some calculations on the nonlocal term:

$$\begin{aligned} F(u)= & {} \int _0^1(f(su), u)ds\nonumber \\= & {} \int _0^1\int _{\Omega } s^pu^p(x,t)\big (\int _{\Omega }k(x,y)s^{p+1}u^{p+1}(y,t)dy\big )u(x,t)dxds\nonumber \\= & {} \frac{1}{2p+2}\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy. \end{aligned}$$

(2.4)

Differentiating (2.4) with respect to t, using the symmetry of k(x, y), we have:

$$\begin{aligned} \frac{d}{dt}F(u)&=\displaystyle \frac{1}{2p+2}\frac{d}{dt}\int \limits _{\Omega }\int \limits _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy\nonumber \\&\quad =\displaystyle \int \limits _{\Omega }\int \limits _{\Omega }k(x,y)u^p(x,t)u^{p+1}(y,t)u_t(x,t)dxdy. \end{aligned}$$

(2.5)

Now, we give some useful inequalities which will be used throughout the paper. Let $\lambda _1$ be the principal eigenvalue of the problem:

$$\begin{aligned}&\triangle w+\lambda w=0\quad \text {in} \quad \Omega ,\nonumber \\&w=0\quad \text {on} \quad \partial \Omega ,\nonumber \\&w>0 \quad \text {in} \quad \Omega ; \end{aligned}$$

(2.6)

then we have:

$$\begin{aligned} \lambda _1\Vert u\Vert _2^2\le \Vert \nabla u\Vert _2^2,\quad \Vert \nabla u\Vert _2^2\ge \frac{\lambda _1}{1+\lambda _1}\Vert u\Vert _{H_0^1}^2, \quad u\in H_0^1(\Omega ). \end{aligned}$$

(2.7)

Using of Hölder’s inequality and (2.7) , we get:

$$\begin{aligned} \Vert \nabla u(t)\Vert _{2q+2}^{2q+2}\ge |\Omega |^{-q}\big (\frac{\lambda _1}{1+\lambda _1}\big )^{q+1} \Vert u(t)\Vert _{H_0^1}^{2q+2}, \end{aligned}$$

(2.8)

where $|\Omega |$ denotes the volume of $\Omega$.

3 Nonblow-Up Case

In this section, we prove that the solution u(t) of problem (1.1) cannot blow-up at any finite time provided that $q>p>0.$

We define the auxiliary function:

$$\begin{aligned} \varphi (t)=\varphi _{\nu }(t)=\Vert u(t)\Vert _2^2+\nu \Vert \nabla u(t)\Vert _2^2,\quad \nu =0,1. \end{aligned}$$

(3.1)

Differentiating (3.1) with respect to t, and using (1.1), (2.2) and (2.3), we obtain:

$$\begin{aligned} \varphi '(t)= & {} 2\int _{\Omega }uu_tdx+2\nu \int _{\Omega }\nabla u\cdot \nabla u_tdx\nonumber \\= & {} -2\int _{\Omega }\rho (|\nabla u|^2)|\nabla u|^2dx+2\int _{\Omega }\nonumber \\&\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy\nonumber \\\le & {} -2b_1\int _{\Omega }|\nabla u|^2dx-2b_2\int _{\Omega }|\nabla u|^{2q+2}dx+2\int _{\Omega }\nonumber \\&\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy. \end{aligned}$$

(3.2)

Using the Hölder and Sobolev inequalities, we estimate the last term in the right-hand side of (3.2) as:

$$\begin{aligned}&\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy\nonumber \\&\quad \le \int _{\Omega }u^{p+1}(x,t)\left( \int _{\Omega }k^2(x,y)dy\right) ^{\frac{1}{2}}\left( \int _{\Omega }u^{2p+2}(y,t)dy\right) ^{\frac{1}{2}}dx\nonumber \\&\quad =\Vert u\Vert ^{p+1}_{2p+2}\int _{\Omega }u^{p+1}(x,t)\left( \int _{\Omega }k^2(x,y)dy\right) ^{\frac{1}{2}}dx\nonumber \\&\quad \le \Vert u\Vert ^{2p+2}_{2p+2}\left( \int _{\Omega }\int _{\Omega }k^2(x,y)dydx\right) ^{\frac{1}{2}}\nonumber \\&\quad \le \kappa \Vert u\Vert ^{2p+2}_{2p+2}\le \kappa C_*^{2p+2}\Vert u\Vert _{H_0^1}^{2p+2}, \end{aligned}$$

(3.3)

where $\kappa =\big (\int _{\Omega }\int _{\Omega }k^2(x,y)dydx\big )^{\frac{1}{2}}<\infty$, and $C_*$ is the best embedding constant: $\Vert u\Vert _{2p+2}\le C_*\Vert u\Vert _{H_0^1}$.

Now, we will process our calculations in two cases: $\nu =1$ and $\nu =0$ respectively.

Pseudo-parabolic case: $\nu =1.$ It follows from (2.7), (2.8),(3.2), and (3.3) that:

$$\begin{aligned} \varphi _1'(t)\le -A_1\varphi _1(t)-B_1[\varphi _1{(t)}]^{q+1}+C_1[\varphi _1(t)]^{p+1},\end{aligned}$$

(3.4)

where:

$$\begin{aligned} A_1=\frac{2b_1\lambda _1}{1+\lambda _1},\quad B_1=2b_2|\Omega |^{-q} \left( \frac{\lambda _1}{1+\lambda _1}\right) ^{q+1},\quad C_1=2\kappa C_*^{2p+2}. \end{aligned}$$

We conclude from (3.4) and $q>p>0$ that the solution cannot blow-up in finite time. In fact, Let $h_1(s)=-A_1s-B_1s^{q+1}+C_1s^{p+1}$, and then, $h_1(0)=0$ and $\displaystyle {\lim }_{s\rightarrow +\infty }h_(s)=-\infty$, since $q>p$. By the continuity of $\varphi _1(t)$ and the properties of polynomials, we can deduce that (1) if $h_1(s)\le 0$ for all $s\ge 0$, then $\varphi _1(t)\le \varphi _1(0)$; (2) if there exists some finite time $t_1$, such that $h_1(t_1)>0$, we denote $S_1$ to be the largest positive root of $h_1(s)=0$, then $\varphi _1(t)$ could not be larger than the value $S_1$; otherwise, $\varphi _1'(t)$ would be negative which is impossible. Moreover:

$$\begin{aligned} \varphi _1(t)\le \max \big \{\varphi _1(0), S_1 \big \}. \end{aligned}$$

Parabolic case: $\nu =0.$ Since

$$\begin{aligned} |\nabla u^{q+1}|^2=(q+1)^2u^{2q}|\nabla u|^2, \end{aligned}$$

it follows from Hölder’s inequality that:

$$\begin{aligned} \int _{\Omega }|\nabla u^{q+1}|^2dx\le (q+1)^2\left( \int _{\Omega }|\nabla u|^{2q+2}dx\right) ^{\frac{1}{q+1}}\left( \int _{\Omega }{u^{2q+2}}dx\right) ^{\frac{q}{q+1}}. \end{aligned}$$

Letting $w=u^{q+1}$ in the Poincaré’s inequality $\lambda _1\Vert w\Vert _2^2\le \Vert \nabla w\Vert _2^2$, we obtain that:

$$\begin{aligned} \int _{\Omega }u^{2q+2}dx\le \left[ \frac{(q+1)^2}{\lambda _1}\right] ^{q+1}\int _{\Omega }|\nabla u|^{2q+2}dx. \end{aligned}$$

Using $q>p$ and Hölder’s inequality again, we have:

$$\begin{aligned} \int _{\Omega }u^{2p+2}dx\le \left( \int _{\Omega }u^{2q+2}dx\right) ^{\frac{p+1}{q+1}}|\Omega |^{\frac{q-p}{q+1}}. \end{aligned}$$

and

$$\begin{aligned} \int _{\Omega }u^2dx\le \left( \int _{\Omega }u^{2p+2}dx\right) ^{\frac{1}{p+1}}|\Omega |^{\frac{p}{p+1}}. \end{aligned}$$

It follows from (2.7),(3.2),(3.3) and the above inequalities that:

$$\begin{aligned} \varphi _0'(t)\le & {} -2b_1\int _{\Omega }|\nabla u|^2dx-2b_2\left[ \frac{\lambda _1}{(q+1)^2}\right] ^{q+1}\int _{\Omega }u^{2q+2}dx+2\kappa \int _{\Omega }u^{2p+2}dx \\\le & {} ^-2b_1\lambda _1\varphi _0(t)-2b_2\left[ \frac{\lambda _1}{(q+1)^2}\right] ^{q+1}\left( \int _{\Omega }u^{2p+2}dx\right) \\&\left( \int _{\Omega }u^{2p+2}dx\right) ^{\frac{q-p}{p+1}}|\Omega |^{-\frac{q-p}{p+1}}+2\kappa \int _{\Omega }u^{2p+2}dx\\\le & {} -2b_1\lambda _1\varphi _0(t)-2b_2\left[ \frac{\lambda _1}{(q+1)^2}\right] ^{q+1}\left( \int _{\Omega }u^{2p+2}dx\right) \\&\left( \int _{\Omega }u^2dx\right) ^{q-p}|\Omega |^{p-q}+2\kappa \int _{\Omega }u^{2p+2}dx\\= & {} -2b_1\lambda _1\varphi _0(t)\!+\!2\int _{\Omega }u^{2p+2}dx\left( \kappa \!-\!b_2\left[ \frac{\lambda _1}{(q+1)^2}\right] ^{q+1}|\Omega |^{p-q}[\varphi _0(t)]^{q-p}\right) . \end{aligned}$$

It follows from the above inequality that the solution u(t) cannot blow-up in finite time. In fact, if $\varphi _0(t)$ were to be sufficiently large at some time $t_0$, then $\varphi _0'(t)$ would be negative, so that $\varphi _0(t)$ could not be larger than that value. Moreover:

$$\begin{aligned} \varphi _0(t)\le \max \left\{ \varphi _0(0), \left[ \frac{\kappa (q+1)^{2q+2}|\Omega |^{q-p}}{b_2\lambda _1^{q+1}}\right] ^{\frac{1}{q-p}} \right\} . \end{aligned}$$

We summarize the above discussions in the following theorem.

Theorem 3.1

If $0<p<q$ and u is the nonnegative solution of problem (1.1), then u cannot blow-up at finite time in $H_0^1$-norm for $\nu =1$ (resp. $L^2$-norm for $\nu =0$).

4 Criterions of Blow-Up

In this section, we consider a specific class of problem (1.1), for which we can obtain the finite-time blow-up results provided that the initial energy satisfies different conditions. We also establish the lower and the upper bounds for the blow-up time. Let u be the nontrivial solution of:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} u_t-\nu \triangle u_t-b_1\triangle u -b_2\text{ div }(|\nabla u|^{2q}\nabla u)\\ \qquad =u^p(x,t)\int _{\Omega }k(x,y)u^{p+1}(y,t)dy, &{}{} (x, t)\in \Omega \times (0,T),\\ u(x,t)=0, &{}{}(x, t)\in \partial \Omega \times (0,T),\\ u(x,0)=u_0(x)\ge 0, &{}{}x\in \Omega , \end{array}\right. } \end{aligned}\end{aligned}$$

(4.1)

where $b_1$ and $b_2$ are positive constants and the parameters p and q satisfy the condition $0\le q \le p.$

To obtain the blow-up results, we introduce the functions:

$$\begin{aligned} J(u(t))= & {} \frac{b_1}{2}\Vert \nabla u\Vert _2^2+\frac{b_2}{2q+2}\Vert \nabla u\Vert _{2q+2}^{2q+2}\nonumber \\&\quad -\frac{1}{2p+2}\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy \end{aligned}$$

(4.2)

and

$$\begin{aligned} I(u(t))=b_1\Vert \nabla u\Vert _2^2+b_2\Vert \nabla u\Vert _{2q+2}^{2q+2}-\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy, \end{aligned}$$

where we have used (2.4).

Lemma 4.1

Let u be the solution of the problem (4.1), and then, J(u(t)) is non-increasing function, that is, $\frac{d}{dt}J(u(t))\le 0$. Moreover, it holds that:

$$\begin{aligned} J(u(t))+\int _0^t(\Vert u_t\Vert ^2+\nu \Vert \nabla u_t\Vert ^2)dt=J(u_0). \end{aligned}$$

Proof

Similar to the proof of Lemma 2.1 in [1] and using (2.4) and (2.5), we can obtain the proof. $\square$

4.1 Finite-Time Blow-Up for Nonpositive Initial Energy

In this subsection, we will establish the blow-up results for $J(u_0)\le 0$ and the upper bounds for the maximal existence time T.

Theorem 4.1

Let $0\le q\le p$, u be the nonnegative solution of (4.1) with $J(u_0)<0$, and then, u blows up in finite time T with:

$$\begin{aligned} T\le T_{11}=\frac{\Vert u_0\Vert ^2_{H_0^1}}{-4p(p+1)J(u_0)}, \quad \text {for}\,\, \nu =1, \\ T\le T_{10}=\frac{\Vert u_0\Vert _2^2}{-4p(p+1)J(u_0)}, \quad \text {for}\,\, \nu =0. \end{aligned}$$

Proof

Given $\varphi (t)$ be the function defined in (3.1), since $0\le q \le p$, we compute:

$$\begin{aligned} \varphi '(t)= & {} 2\int _{\Omega }uu_tdx+2\nu \int _{\Omega }\nabla u\cdot \nabla u_tdx\nonumber \\= & {} -2b_1\Vert \nabla u\Vert _2^2-2b_2\Vert \nabla u\Vert ^{2q+2}_{2q+2}+2\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy\nonumber \\&\ge \psi (t), \end{aligned}$$

(4.3)

where:

$$\begin{aligned} \psi (t)= & {} -4(p+1)J(u(t))\\= & {} -2(p+1)b_1\Vert \nabla u\Vert _2^2-\frac{2(p+1)b_2}{q+1}\Vert \nabla u\Vert _{2q+2}^{2q+2}\\&+2\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy. \end{aligned}$$

In view of Lemma 4.1, it holds that:

$$\begin{aligned} \psi '(t)=-4(p+1)\frac{d}{dt}J(u(t))=4(p+1)\int _{\Omega }(|u_t|^2+\nu |\nabla u_t|^2)dx\ge 0. \end{aligned}$$

It follows from $J(u_0)<0$ that $\psi (t)>0$ for any $t\ge 0$. In view of Hölder’s and Schwarz’s inequalities and $\psi (t)>0$, we conclude from the above inequalities that:

$$\begin{aligned} \varphi (t)\psi '(t)= & {} {} 4(p+1)\big (\int _{\Omega }(|u|^2+\nu |\nabla u|^2)dx\big )\left( \int _{\Omega }(|u_t|^2+\nu |\nabla u_t|^2)dx\right) \\&\ge 4(p+1)\left( \int _{\Omega }(uu_t+\nu \nabla u\cdot \nabla u_t)dx\right) ^2 \\= & {} {} (p+1)[\varphi '(t)]^2\ge (p+1)\varphi '(t)\psi (t),\end{aligned}$$

which can be rewritten as:

$$\begin{aligned} \frac{\psi '(t)}{\psi (t)}\ge (p+1)\frac{\varphi '(t)}{\varphi (t)}. \end{aligned}$$

(4.4)

Integrating (4.4) on [0, t], noticing $\varphi '(t)\ge \psi (t)$, we have:

$$\begin{aligned} \frac{\psi (t)}{[\varphi (t)]^{p+1}}\ge \frac{\psi (0)}{[\varphi (0)]^{p+1}}\Rightarrow \frac{\varphi '(t)}{[\varphi (t)]^{p+1}}\ge \frac{\psi (0)}{[\varphi (0)]^{p+1}}. \end{aligned}$$

(4.5)

Then, a further integration results in:

$$\begin{aligned} \frac{1}{\varphi ^p(t)}\le \frac{1}{\varphi ^p(0)}-p\frac{\psi (0)}{[\varphi (0)]^{p+1}}t. \end{aligned}$$

(4.6)

It is obvious that (4.6) cannot holds for all time t and u blows up at some finite T, i.e., $\displaystyle {\lim }_{t\rightarrow T^-}\varphi (t)=+\infty$, where:

$$\begin{aligned} T\le T_1=\frac{\varphi (0)}{p\psi (0)}, \end{aligned}$$

which implies the conclusions of this theorem for both $\nu =1$ and $\nu =0$ cases. $\square$

Moreover, integrating the second inequality of (4.5) from t to T, we can obtain the following blow-up rate:

$$\begin{aligned} \varphi (t)\le \left[ \frac{p\psi (0)}{[\varphi (0)]^{p+1}}\right] ^{-\frac{1}{p}}(T-t)^{-\frac{1}{p}}, \end{aligned}$$

that is:

$$\begin{aligned} \Vert u(t)\Vert _{H_0^1}\le \left[ \frac{-4p(p+1)J(u_0)}{[\Vert u_0\Vert _{H_0^1}^{2p+2}]}\right] ^{-\frac{1}{2p}}(T-t)^{-\frac{1}{2p}},\quad \,\, \nu =1; \end{aligned}$$

$$\begin{aligned} \Vert u(t)\Vert _{2}\le \left[ \frac{-4p(p+1)J(u_0)}{[\Vert u_0\Vert _{2}^{2p+2}]}\right] ^{-\frac{1}{2p}}(T-t)^{-\frac{1}{2p}}, \quad \,\, \nu =0. \end{aligned}$$

Now, we will give the blow-up result for the case $J(u_0)=0$. Moreover, it is also valid for $J(u_0)\le 0$ from the proof of the theorem.

Theorem 4.2

Let $0< q< p$ and u be the nonnegative solution of (4.1) with $J(u_0)\le 0$, and then, u blows up in finite time T with:

$$\begin{aligned} T\le T_{21}=\int _{\Vert u_0\Vert _{H_0^1}^2}^{+\infty }\frac{d\eta }{A_2\eta +B_2\eta ^{q+1}}, \quad \text {for}\,\, \nu =1, \\ T\le T_{20}=\int _{\Vert u_0\Vert _{2}^2}^{+\infty }\frac{d\eta }{A_3\eta +B_3\eta ^{q+1}}, \quad \text {for}\,\, \nu =0, \end{aligned}$$

where the positive constants $A_2, B_2$ and $A_3, B_3$ are defined in (4.9) and (4.10), respectively.

Proof

Given $\varphi (t)$ be the function defined in (3.1), in view of(4.2) and Lemma 4.1, we have:

$$\begin{aligned}&\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy\nonumber \\&\quad =2(p+1)\bigg (\int _0^t(\Vert u_t\Vert _2^2+\nu \Vert \nabla u_t\Vert _2^2)dt+\frac{b_1}{2}\Vert \nabla u\Vert _2^2\nonumber \\&\quad +\frac{b_2}{2q+2}\Vert \nabla u\Vert _{2q+2}^{2q+2}-J(u_0)\bigg ). \end{aligned}$$

(4.7)

Substituting (4.7) into $\varphi '(t)$ (see (4.3)), in view of $J(u_0)\le 0$, we deduce that:

$$\begin{aligned} \varphi '(t)\ge 2p b_1\Vert \nabla u\Vert _2^2+\frac{2(p-q)b_2}{q+1}\Vert \nabla u\Vert _{2q+2}^{2q+2}. \end{aligned}$$

(4.8)

Pseudo-parabolic case: $\nu =1.$ It follows from (2.7) and (2.8), (4.8) reduces to:

$$\begin{aligned} \varphi _1'(t)\ge A_2 \varphi _1(t)+B_2[\varphi _1(t)]^{q+1}, \end{aligned}$$

where:

$$\begin{aligned} A_2=\frac{2pb_1\lambda _1}{1+\lambda _1},\quad B_2=\frac{2(q-p)b_2}{(q+1)|\Omega |^q}\left( \frac{\lambda _1}{1+\lambda _1}\right) ^{q+1}. \end{aligned}$$

(4.9)

On integrating the above inequality on [0, t], we have:

$$\begin{aligned} t\le \int _{\varphi _1(0)}^{\varphi _1(t)}\frac{d\eta }{A_2\eta +B_2\eta ^{q+1}}\le \int _{\varphi _1(0)}^{+\infty }\frac{d\eta }{A_2\eta +B_2\eta ^{q+1}}<+\infty . \end{aligned}$$

It follows from that the solution u blows up at some finite time in $H_0^1$-norm, since the above inequality cannot hold for all time t.

Parabolic case: $\nu =0.$ It follows from (2.7), (2.8) and Hölder’s inequality , (4.8) reduces to:

$$\begin{aligned} \varphi _0'(t)\ge & {} 2p b_1\Vert \nabla u\Vert _2^2+\frac{2(p-q)b_2}{(q+1)|\Omega |^q}\Vert \nabla u\Vert _2^{2q+2} \\\ge & {} A_3\varphi _0(t)+B_3[\varphi _0(t)]^{q+1}, \end{aligned}$$

where:

$$\begin{aligned} A_3=2pb_1\lambda _1,\quad B_3=\frac{2(p-q)b_2}{(q+1)|\Omega |^q}\lambda _1^{q+1}. \end{aligned}$$

(4.10)

Similarly, we have

$$\begin{aligned} t\le \int _{\varphi _0(0)}^{\varphi _0(t)}\frac{d\eta }{A_3\eta +B_3\eta ^{q+1}}\le \int _{\varphi _0(0)}^{+\infty }\frac{d\eta }{A_3\eta +B_3\eta ^{q+1}}<+\infty . \end{aligned}$$

It follows from that the solution u blows up at some finite time in $L^2$-norm, since the above inequality cannot hold for all time t. $\square$

Remark 4.1

(1) The similar result of Theorem 4.2 was obtained in [9] for the case $b_1=0, \nu =1.$

(2) For the case $q=0$ (assume $b_2=0$ for convenience), from the proof of Theorem 4.2, we can obtain that the solution u(x, t) increases at least exponentially, that is:

$$\begin{aligned} \Vert u\Vert _{H_0^1}\ge \Vert u_0\Vert _{H_0^1}e^{\frac{pb_1\lambda _1}{1+\lambda _1}t},\quad \text {for} \,\,\nu =1; \\ \Vert u\Vert _{2}\ge \Vert u_0\Vert _{2}e^{pb_1\lambda _1t},\quad \text {for} \,\,\nu =0. \end{aligned}$$

4.2 Finite-Time Blow-Up for Arbitrary Initial Energy

In this subsection, we will establish the blow-up results for arbitrary initial energy and the upper bounds for the maximal existence time T for both pseudo-parabolic case $\nu =1$ and parabolic case $\nu =0$.

Theorem 4.3

Let $0\le q< p$ and u be the nonnegative solution of (4.1) with the initial energy satisfies:

• (1)
  $$\begin{aligned} J(u_0)<\frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)}\Vert u_0\Vert _{H_0^1}^2,\quad \text {for} \,\,\nu =1; \end{aligned}$$

  (4.11)

  then, u blows up at some finite in $H_0^1$-norm. Moreover, the upper bound can be estimated by:

  $$\begin{aligned} T\le T_{31}=\frac{2(p+1)(1+\lambda _1)\Vert u_0\Vert ^2_{H_0^1}}{p^3b_1\lambda _1\Vert u_0\Vert ^2_{H_0^1}-2p^2(p+1)(1+\lambda _1)J(u_0)}. \end{aligned}$$

  (4.12)
• (2)
  $$\begin{aligned} J(u_0)<\frac{pb_1\lambda _1}{2(p+1)}\Vert u_0\Vert _2^2,\quad \text {for}\,\,\nu =0; \end{aligned}$$

  (4.13)

  then, u blows up at some finite in $L^2$-norm. Moreover, the upper bound can be estimated by:

  $$\begin{aligned} T\le T_{30}=\frac{2(p+1)\Vert u_0\Vert ^2_2}{p^3b_1\lambda _1\Vert u_0\Vert ^2_2-2p^2(p+1)J(u_0)}. \end{aligned}$$

  (4.14)

Proof

Let u(t) be the solution of the problem (4.1) with the initial energy satisfying (4.11) when $\nu =1$ (resp. (4.13) when $\nu =0$). We may assume $J(u(t))\ge 0$; otherwise, there exists some $t_0\ge 0$, such that $J(u(t_0))<0$, then u(t) will blow up at some finite time by Theorem 4.1, the proof is complete. Therefore, in the following, we give our proof by contradiction, and assume that u(t) exists globally and $J(u(t))\ge 0$ for all $t\ge 0$.

In view of (4.1) and Lemma 4.1, we have the following equalities:

$$\begin{aligned} \frac{d}{dt}J(u(t))=-\Vert u_t\Vert _2^2-\nu \Vert \nabla u_t\Vert _2^2, \\ \varphi '(t)=\frac{d}{dt}(\Vert u\Vert _2^2+\nu \Vert \nabla u\Vert _2^2)=-2I(u(t). \end{aligned}$$

Pseudo-parabolic case: $\nu =1.$ Since:

$$\begin{aligned}&\int _0^t\Vert u_s(s)\Vert _{H_0^1}ds\\&\qquad \ge \Vert \int _0^tu_s(s)ds\Vert _{H_0^1}=\Vert u(t)-u_0\Vert _{H_0^1}\ge \Vert u(t)\Vert _{H_0^1}-\Vert u_0\Vert _{H_0^1}, \quad t\ge 0, \end{aligned}$$

by Hölder’s inequality and $J(u_0)\ge J(u(t))\ge 0$, we obtain that:

$$\begin{aligned} \begin{aligned} \Vert u(t)\Vert _{H_0^1}&\le \Vert u_0\Vert _{H_0^1}+t^{\frac{1}{2}}[\int _0^t \Vert u_s(s)\Vert _{H_0^1}^2ds]^{\frac{1}{2}}\\&= \Vert u_0\Vert _{H_0^1}+t^{\frac{1}{2}}[J(u_0)-J(u(t))]^{\frac{1}{2}}\\&\le \Vert u_0\Vert _{H_0^1}+t^{\frac{1}{2}}(J(u_0))^{\frac{1}{2}}, \quad t\ge 0. \end{aligned} \end{aligned}$$

(4.15)

On other hand, in view of (2.7), (4.3), and $0\le q<p$, we have:

$$\begin{aligned} \frac{d}{dt}\big (\Vert u(t)\Vert _{H^1_0}^2\big )= & {} 2pb_1\Vert \nabla u\Vert ^2_2+\frac{2(p-q)b_2}{q+1}\Vert \nabla u\Vert _{2q+2}^{2q+2}-4(p+1)J(u(t))\\\ge & {} \frac{2pb_1\lambda _1}{1+\lambda _1}\Vert u\Vert _{H_0^1}^2-4(p+1)J(u(t))\\= & {} \frac{2pb_1\lambda _1}{1+\lambda _1}\big [\Vert u\Vert _{H_0^1}^2-\frac{2(p+1)(1+\lambda _1)}{pb_1\lambda _1}J(u(t))\big ]. \end{aligned}$$

Since $\frac{d}{dt}(J(u(t)))\le 0$, then we can deduce that

$$\begin{aligned} \frac{d}{dt}H_1(t)\ge \frac{2pb_1\lambda _1}{1+\lambda _1} H_1(t) \end{aligned}$$

for all $t\ge 0,$ where:

$$\begin{aligned} H_1(t)=\Vert u\Vert _{H_0^1}^2-\frac{2(p+1)(1+\lambda _1)}{pb_1\lambda _1}J(u(t)). \end{aligned}$$

Using Gronwall’s inequality, we obtain that:

$$\begin{aligned} \Vert u\Vert _{H_0^1}^2\ge \frac{2(p+1)(1+\lambda _1)}{pb_1\lambda _1}J(u(t))+e^{\frac{2pb_1\lambda _1}{1+\lambda _1}t}H_1(0), \end{aligned}$$

where $H_1(0)=\Vert u_0\Vert _{H_0^1}^2-\frac{2(p+1)(1+\lambda _1)}{pb_1\lambda _1}J(u_0)>0$ due to (4.11). By the assumption $J(u(t))\ge 0$ for all $t\ge 0$, we get:

$$\begin{aligned} \Vert u\Vert _{H_0^1}\ge \sqrt{H_1(0)}e^{\frac{pb_1\lambda _1}{1+\lambda _1}t}, \end{aligned}$$

which contradicts (4.15) for t sufficiently large. Hence, u(t) blows up at some finite time, i.e., $T<\infty .$

Next, we establish an upper bound estimate of T. To this end, we first claim that:

$$\begin{aligned} I(u(t))= & {} b_1\Vert \nabla u\Vert _2^2+b_2\Vert \nabla u\Vert _{2q+2}^{2q+2}\\&-\int _{\Omega }\int _{\Omega }k(x,y)u^{p+1}(x,t)u^{p+1}(y,t)dxdy<0,\quad t\in [0, T). \end{aligned}$$

Indeed, in view of the definitions of J(u(t)) and I(u(t)), after a simple calculation, we obtain:

$$\begin{aligned} J(u(t))= & {} \frac{pb_1}{2(p+1)}\Vert \nabla u(t)\Vert _2^2+\frac{(p-q)b_2}{2(q+1)(p+1)}\Vert \nabla u(t)\Vert _{2q+2}^{2q+2}\nonumber \\&\quad +\frac{1}{2(p+1)}I(u(t)),\quad t\in [0,T). \end{aligned}$$

(4.16)

It follows from (2.7), (4.11), and (4.16) that:

$$\begin{aligned} \frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)}\Vert u_0\Vert _{H_0^1}^2>J(u_0)\ge \frac{pb_1}{2(p+1)}\frac{\lambda _1}{1+\lambda _1}\Vert u_0\Vert _{H_0^1}^2+\frac{1}{2(p+1)}I(u_0), \end{aligned}$$

where we use $0\le q<p$, which implies $I(u_0)<0.$ We assume there exists a $t_0\in (0, T)$, such that $I(u(t_0))=0$, $I(u(t))<0,$ for $t\in [0,t_0).$ Hence, $\Vert u(t)\Vert _{H_0^1}^2$ is strictly increasing on $[0, t_0)$. Then, it follows from (4.11) that:

$$\begin{aligned} J(u_0)<\frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)}\Vert u_0\Vert _{H_0^1}^2 <\frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)}\Vert u(t_0)\Vert _{H_0^1}^2. \end{aligned}$$

(4.17)

On the other hand, since J(u(t)) is non-increasing with respect to t, and combining $0\le q<p$ and (4.16), we get:

$$\begin{aligned} J(u_0)\ge J(u(t_0))= & {} \frac{pb_1}{2(p+1)}\Vert \nabla u(t_0)\Vert _2^2\\&+\frac{(p-q)b_2}{2(q+1)(p+1)}\Vert \nabla u(t_0)\Vert _{2q+2}^{2q+2}+\frac{1}{2(p+1)}I(u(t_0))\\&\ge \frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)} \Vert u(t_0)\Vert ^2_{H_0^1}, \end{aligned}$$

which contradicts (4.17). Hence, $I(u(t)<0$ and $\Vert u(t)\Vert _{H_0^1}^2$ is strictly increasing on [0, T).

For any ${\tilde{T}}\in (0, T)$, we define the functional:

$$\begin{aligned} F(t)=\int _0^t\Vert u(s)\Vert _{H_0^1}^2ds+(T-t)\Vert u_0\Vert _{H_0^1}^2+\beta (t+\gamma )^2,\quad t\in [0,{\tilde{T}}], \end{aligned}$$

with two positive constants $\beta , \gamma$ to be chosen later. Since $\Vert u(t)\Vert _{H_0^1}^2$ is strictly increasing, we have:

$$\begin{aligned} \begin{aligned} F'(t)&=\Vert u(t)\Vert _{H_0^1}^2-\Vert u_0\Vert _{H_0^1}^2+2\beta (t+\gamma )\\&=\int _0^t\frac{d}{ds}\Vert u(s)\Vert _{H_0^1}^2ds+2\beta (t+\gamma )>0. \end{aligned}\end{aligned}$$

(4.18)

In view of (2.7), Lemma 4.1, and $0\le q<p$, we have:

$$\begin{aligned} \begin{aligned} F''(t)&=\frac{d}{dt}\Vert u(t)\Vert _{H_0^1}^2+2\beta \\&=2pb_1\Vert \nabla u\Vert ^2_2+\frac{2(p-q)b_2}{q+1}\Vert \nabla u\Vert _{2q+2}^{2q+2}-4(p+1)J(u(t))+2\beta \\&\ge \frac{2pb_1\lambda _1}{1+\lambda _1}\Vert u(t)\Vert _{H_0^1}^2+4(p+1)\int _0^t\Vert u_s\Vert _{H_0^1}^2ds-4(p+1)J(u_0). \end{aligned} \end{aligned}$$

(4.19)

By the definition of F(t), we have $F(0)=T\Vert u_0\Vert ^2_{H_0^1}+\beta \gamma ^2>0.$ Since $\Vert u(t)\Vert _{H_0^1}^2$ is strictly increasing on [0, T), it follows from (4.18) that $F'(t)>0$ for all $t\in [0, {\tilde{T}}]$, which implies that $F(t)>0$ and F(t) is strictly increasing for any $t\in [0, {\tilde{T}}]$.

Now, for any $t\in [0, {\tilde{T}}],$ we define:

$$\begin{aligned} \xi (t):= & {} \bigg (\int _0^t\Vert u(s)\Vert _{H_0^1}^2ds\\&+\beta (t+\gamma )^2\bigg )\left( \int _0^t\Vert u_s\Vert _{H_0^1}^2ds+\beta \right) \\&-\left( \int _0^t\frac{1}{2}\frac{d}{ds}\Vert u(s)\Vert _{H_0^1}^2ds+\beta (t+\gamma )\right) ^2. \end{aligned}$$

Using Hölder’s inequality and the element algebraic inequality:

$$\begin{aligned} ab+cd\le \sqrt{a^2+c^2}\sqrt{b^2+d^2}, \end{aligned}$$

we can deduce:

$$\begin{aligned} \begin{aligned} \frac{1}{2}\frac{d}{ds}\Vert u(s)\Vert _{H_0^1}^2=&{} \int _{\Omega }(uu_s+\nabla u\cdot \nabla u_s)dx\\\le&{} \Vert u\Vert _2\Vert u_s\Vert _2+\Vert \nabla u\Vert _2\Vert \nabla u_s\Vert _2\\\le&{} \left( \Vert u\Vert _2^2+\Vert \nabla u\Vert _2^2\right) ^{\frac{1}{2}}\big (\Vert u_s\Vert _2^2+\Vert \nabla u_s\Vert _2^2\big )^{\frac{1}{2}}\\=&{} \Vert u\Vert _{H_0^1}\Vert u_s\Vert _{H_0^1}. \end{aligned} \end{aligned}$$

It follows from Hölder’s inequality that:

$$\begin{aligned} \int _0^t\frac{1}{2}\frac{d}{ds}\Vert u(s)\Vert _{H_0^1}^2ds\le \left( \int _0^t\Vert u\Vert _{H_0^1}^2ds\right) ^{\frac{1}{2}} \left( \int _0^t\Vert u_s\Vert _{H_0^1}^2ds\right) ^{\frac{1}{2}}. \end{aligned}$$

In view of the above inequalities and the definition of $\xi (t)$ that:

$$\begin{aligned} \xi (t)\ge & {} \left( \int _0^t\Vert u\Vert _{H_0^1}^2ds+\beta (t+\gamma )^2\right) \left( \int _0^t\Vert u_s\Vert _{H_0^1}^2ds+\beta \right) \\&-\left( [\int _0^t\Vert u\Vert _{H_0^1}^2ds]^{\frac{1}{2}}[\int _0^t\Vert u_s\Vert _{H_0^1}^2ds]^{\frac{1}{2}}+\beta (t+\gamma )\right) \\= & {} \beta \left( [\int _0^t\Vert u\Vert _{H_0^1}^2ds]^{\frac{1}{2}}-[\int _0^t\Vert u_s\Vert _{H_0^1}^2ds]^{\frac{1}{2}}(t+\gamma )\right) ^2\ge 0. \end{aligned}$$

Hence, using the above inequality and (4.18), we have:

$$\begin{aligned} -(F'(t))^2= & {} -4\left[ \frac{1}{2}\int _0^t\frac{d}{ds}\Vert u(s)\Vert _{H_0^1}^2ds+\beta (t+\gamma )\right] ^2\\= & {} 4\bigg (\xi (t)-\left( F(t)-(T-t)\Vert u_0\Vert _{H_0^1}^2\right) \\&\left( \int _0^t\Vert u_s(s)\Vert _{H_0^1}^2ds+\beta \right) \bigg )\\\ge & {} -4F(t)\left( \int _0^t\Vert u_s(s)\Vert _{H_0^1}^2ds+\beta \right) . \end{aligned}$$

It follows from the above inequality and (4.19) that:

$$\begin{aligned} F(t)F''(t) -(p+1)(F'(t))^2\ge & {} F(t)\left( F''(t)-4(p+1)\int _0^t\Vert u_s(s)\Vert _{H_0^1}^2ds-4(p+1)\beta \right) \\\ge & {} F(t)\left( \frac{2pb_1\lambda _1}{1+\lambda _1}\Vert u(t)\Vert _{H_0^1}^2-4(p+1)J(u_0)-4(p+1)\beta \right) \\= & {} F(t)\bigg [4(p+1)\left( \frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)}\Vert u(t)\Vert _{H_0^1}^2-J(u_0)\right) \\&-4(p+1)\beta \bigg ]. \end{aligned}$$

In view of (4.11), letting $\beta$ sufficiently small, such that:

$$\begin{aligned} 0<\beta \le \beta _0:=\frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)}\Vert u_0\Vert _{H_0^1}^2-J(u_0), \end{aligned}$$

(4.20)

we get:

$$\begin{aligned} F(t)F''(t) -(p+1)(F'(t))^2\ge 0, \quad t\in [0, {\tilde{T}}]. \end{aligned}$$

Define $G(t)=F^{-p}(t)$ for $t\in [0, {\tilde{T}}]$. After a simple calculation, by $F(t)>0,$ $F'(t)>0$, and $p>0$, we obtain:

$$\begin{aligned} G'(t)= & {} -pF^{-p-1}(t)F'(t)<0,\\ G''(t)= & {} -pF^{-p-2}[F(t)F''(t)-(p+1)(F'(t))^2]\le 0, \end{aligned}$$

holds for all $t\in [0, {\tilde{T}}]$, which means that G(t) is concave on $[0, {\tilde{T}}]$. Hence, it holds that:

$$\begin{aligned} G({\tilde{T}})\le G(0)+G'(0){\tilde{T}}. \end{aligned}$$

(4.21)

By the definition of G(t), we obtain:

$$\begin{aligned} G'(0)=-pF^{-p-1}(0)F'(0)=-pG(0)\frac{F'(0)}{F(0)}<0. \end{aligned}$$

Hence, it follows from (4.21) that:

$$\begin{aligned} {\tilde{T}}\le -\frac{G(0)}{G'(0)}=\frac{T\Vert u\Vert _{H_0^1}^2+\beta \gamma ^2}{2p\beta \gamma }=\frac{\Vert u\Vert _{H_0^1}^2}{2p\beta \gamma }T+\frac{\gamma }{2p} \end{aligned}$$

for any ${\tilde{T}}\in [0, T)$. Letting ${\tilde{T}}\rightarrow T^-$, we get:

$$\begin{aligned} T\le \frac{\Vert u\Vert _{H_0^1}^2}{2p\beta \gamma }T+\frac{\gamma }{2p}. \end{aligned}$$

(4.22)

Fixing an arbitrary $\beta$ satisfying (4.20), then let $\gamma$ be sufficiently large, such that:

$$\begin{aligned} \frac{\Vert u_0\Vert _{H_0^1}^2}{2p\beta }<\gamma <+\infty . \end{aligned}$$

Then, in view of (4.22), we have:

$$\begin{aligned} T\le \frac{\beta \gamma ^2}{2p\beta \gamma -\Vert u_0\Vert _{H_0^1}^2}. \end{aligned}$$

(4.23)

Define a function $T_{\beta }(\gamma )$ by:

$$\begin{aligned} T_{\beta }(\gamma )=\frac{\beta \gamma ^2}{2p\beta \gamma -\Vert u_0\Vert _{H_0^1}^2},\quad \gamma \in \left( \frac{\Vert u_0\Vert _{H_0^1}^2}{2p\beta }, +\infty \right) . \end{aligned}$$

It is easy to verify that the function $T_{\beta }(\gamma )$ has a unique minimum at:

$$\begin{aligned} \gamma _{\beta }:=\frac{\Vert u_0\Vert _{H_0^1}^2}{p\beta }\in (\frac{\Vert u_0\Vert _{H_0^1}^2}{2p\beta }, +\infty ). \end{aligned}$$

Then, it follows from (4.23) that:

$$\begin{aligned} T\le \inf _{\gamma \in \left( \frac{\Vert u_0\Vert _{H_0^1}^2}{2p\beta }, +\infty \right) }T_{\beta }(\gamma )=T_{\beta }(\gamma _{\beta })=\frac{\Vert u_0\Vert _{H_0^1}^2}{p^2\beta }, \end{aligned}$$

for any $\beta$ satisfying (4.20). Hence, it holds that:

$$\begin{aligned} T\le \inf _{\beta \in (0,\beta _0]}\frac{\Vert u_0\Vert _{H_0^1}^2}{p^2\beta }=\frac{\Vert u_0\Vert _{H_0^1}^2}{p^2\beta _0} =\frac{2(p+1)(1+\lambda _1)\Vert u_0\Vert ^2_{H_0^1}}{p^3b_1\lambda _1\Vert u_0\Vert ^2_{H_0^1}-2p^2(p+1)(1+\lambda _1)J(u_0)}. \end{aligned}$$

This completes the proof of Theorem 4.3 for the pseudo-parabolic ($\nu =1$) case.

Parabolic case: $\nu =0.$ In this case, we need to replace the norm $H_0^1$ by the norm $L^2$, and the inequality $\Vert \nabla u\Vert _2^2\ge \frac{\lambda _1}{1+\lambda _1}\Vert u\Vert _{H_0^1}^2$ by the inequality $\Vert \nabla u\Vert _2^2\ge \lambda _1\Vert u\Vert _2^2.$ By modifying the previous proof, we can easily deduce the proof. $\square$

Corollary 4.1

For $0\le q<p$ and any $M\in {\mathbb {R}}$, then there exist initial data $u_{0M}\in W_0^{1, 2q+2}(\Omega )$ satisfying $J(u_{0M})=M$, such that the weak solution for the corresponding problem (4.1) will blows up at finite time in $H_0^1$-norm for $\nu =1$ (resp. $L^2$-norm for $\nu =0$).

Proof

According to Theorem 4.1, it is easy to see that the result is valid for $M<0$. We only need to verify the result for the case of $M\ge 0.$ In view of Theorem 4.3, it is sufficiently to check that there exists $u_0\in W_0^{1, 2q+2}(\Omega )$ satisfying (4.11) for $\nu =1$ (resp. (4.13) for $\nu =0$).

Let $\Omega _1$ and $\Omega _2$ be two arbitrary disjoint open subdomains of $\Omega$. We assume $v\in W_0^{1, 2q+2}(\Omega _1)\subset W_0^{1, 2q+2}(\Omega )\subset H_0^1(\Omega )$ be an arbitrary nonzero function, and then, we can take $\alpha _1>0$ sufficiently large, such that:

$$\begin{aligned} \alpha _1^2\left( \int _{\Omega }v^2dx+\int _{\Omega }\nu |\nabla v|^2dx\right) =\alpha _1^2\left( \int _{\Omega _1}v^2dx+\int _{\Omega _1}\nu |\nabla v|^2dx\right) >\frac{M}{C}, \end{aligned}$$

where $C=\frac{pb_1\lambda _1}{2(p+1)(1+\lambda _1)}$ when $\nu =1$, and $C=\frac{pb_1\lambda _1}{2(p+1)}$ when $\nu =0$.

We claim that there exists $w\in W_0^{1, 2q+2}(\Omega _2)\subset W_0^{1, 2q+2}(\Omega )$ and $\alpha _0>\alpha _1$ such that $J(w)=M-J(\alpha _0 v)$. Indeed, we choose a function $w_k\in C_0^1(\Omega _2)$, such that $\Vert \nabla w_k\Vert _2\ge k$ and $\Vert w_k\Vert _{\infty }\le c_0.$ In view of (3.3), we have:

$$\begin{aligned}&\frac{b_1}{2}\int _{\Omega _2}|\nabla w_k|^2dx+\frac{b_2}{2q+2}\int _{\Omega _2}|\nabla w_k|^{2q+2}dx-\frac{1}{2p+2}\\&\int _{\Omega _2}\int _{\Omega _2}k(x,y)w_k^{p+1}(x,t)w_k^{p+1}(y,t)dxdy\\&\ge \frac{b_1}{2}\int _{\Omega _2}|\nabla w_k|^2dx+\frac{b_2}{2q+2}|\Omega _2|^{-q}\big (\int _{\Omega _2}|\nabla w_k|^2dx\big )^{q+1}-\frac{\kappa }{2p+2}c_0^{2p+2}|\Omega _2|. \end{aligned}$$

On the other hand, since $0\le q<p$, it holds that:

$$\begin{aligned}&M-J(\alpha v)\\&\quad =M- \frac{b_1\alpha ^2}{2}\int _{\Omega _1}|\nabla v|^2dx-\frac{b_2\alpha ^{2q+2}}{2q+2}\int _{\Omega _1}|\nabla v|^{2q+2}dx\\&+\frac{\alpha ^{2p+2}}{2p+2}\int _{\Omega _2}\int _{\Omega _2}k(x,y)w_k^{p+1}(x,t)w_k^{p+1}(y,t)dxdy\rightarrow +\infty ,\quad as \quad \alpha \rightarrow +\infty . \end{aligned}$$

Therefore, there exist $k_0>0$ and $\alpha _0>\alpha _1$, such that both are sufficiently large, such that:

$$\begin{aligned} M-J(\alpha _0v)= & {} \frac{b_1}{2}\int _{\Omega _2}|\nabla w_{k_0}|^2dx+\frac{b_2}{2q+2}\int _{\Omega _2}|\nabla w_{k_0}|^{2q+2}dx\\&-\frac{1}{2p+2}\int _{\Omega _2}\int _{\Omega _2}k(x,y)w_{k_0}^{p+1}(x,t)w_{k_0}^{p+1}(y,t)dxdy. \end{aligned}$$

Then, choosing $w=w_{k_0}$, and denoting $u_{0M}:=\alpha _0 v+w$, we obtain:

$$\begin{aligned} \int _{\Omega }|u_{0M}|^2dx+\nu \int _{\Omega }|\nabla u_{0M}|^2dx=\alpha _0^2\int _{\Omega _1}v^2dx+\nu \alpha _0^2\int _{\Omega _1}|\nabla v|^2dx>\frac{M}{C}, \end{aligned}$$

and

$$\begin{aligned} M=J(\alpha _0v)+J(w)=J(u_{0M}), \end{aligned}$$

which implies that:

$$\begin{aligned} J(u_{0M})< C\left( \int _{\Omega }|u_{0M}|^2dx+\nu \int _{\Omega }|\nabla u_{0M}|^2dx\right) . \end{aligned}$$

In view of Theorem 4.3, the proof is complete. $\square$

Remark 4.2

For the case $J(u_0)<0$, the initial condition given in (4.11) (resp. (4.13)) is obviously satisfied. However, we obtain the upper bounds $T_{11}$ and $T_{31}$ (resp. $T_{10}$ and $T_{30}$) for the blow-up time T in Theorem 4.1 and Theorem 4.3 using different methods. In fact, $T_{11}$ is more accurate provided that $J(u_0)<\frac{p^2b_1\lambda _1\Vert u_0\Vert _{H_0^1}}{2p(p+1)(3p+4)(1+\lambda _1)}$, and $T_{31}$ is more accurate provided that $\frac{p^2b_1\lambda _1\Vert u_0\Vert _{H_0^1}}{2p(p+1)(3p+4)(1+\lambda _1)}\le J(u_0)<0$; Resp. $T_{10}$ is more accurate provided that $J(u_0)<\frac{p^2b_1\lambda _1\Vert u_0\Vert _{L^2}}{2p(p+1)(3p+4)}$, and $T_{30}$ is more accurate provided that $\frac{p^2b_1\lambda _1\Vert u_0\Vert _{L^2}}{2p(p+1)(3p+4)}\le J(u_0)<0$;

4.3 Lower Bound for the Blow-Up Time

First, we consider problem (1.1) with $\nu =1$, and determine a lower bound for T if the solution u blows up at finite time $t=T$ in $H_0^1$-norm.

Theorem 4.4

Let $0<q\le p\le \frac{2}{n-2}$, and u be the nonnegative solution of the problem (1.1)($\nu =1$) which blows up at finite time T in $H_0^1$-norm, and then, T is bounded from below as:

$$\begin{aligned} T\ge \frac{C_*^{-2(p+1)}}{2\kappa p\Vert u_0\Vert _{H_0^1}^{2p}}. \end{aligned}$$

Proof

The proof is similar as that of Theorem 3.1 in [9]. It follows from (3.1) to (3.3) that:

$$\begin{aligned} \varphi '_1(t)\le 2\kappa C_*^{-2(p+1)} [\varphi _1(t)]^{p+1}, \end{aligned}$$

which is equivalent to:

$$\begin{aligned}{}[\varphi _1^{-p}(t)]'\ge -2\kappa p C_*^{-2(p+1)}. \end{aligned}$$

Integrating the above inequality from 0 to t leads to:

$$\begin{aligned} \varphi _1^{-p}(t)\ge \varphi _1^{-p}(0)-2\kappa p C_*^{-2(p+1)}t. \end{aligned}$$

Passing to the limit as $t\rightarrow T^-$, we obtain that the conclusion of Theorem 4.4 holds. $\square$

Then, using the technique of differential inequality (see [13, 14]), we obtain a lower bound for the blow-up time if the blow-up does occurs for the problem (1.1) with $\nu =0.$

We define the auxiliary function $\chi (t)$:

$$\begin{aligned} \chi (t)=\int _{\Omega }u^{mp}dx, \end{aligned}$$

(4.24)

for some positive constant m to be chosen later.

Theorem 4.5

Let $0<q<p,$ $m>\frac{2}{p}+2$, and u(x, t) be the nonnegative solution of the problem (1.1)($\nu =0$) which blows up at finite time T in the measure $\chi$ given in (4.24), and then T is bounded from below by:

$$\begin{aligned} T\ge \int _{\chi (0)}^{\infty }\frac{d\chi }{K_1\chi ^{\frac{2p+2}{mp}}+ K_2\chi ^{\frac{(n-2)\theta _1}{n\theta _1-2}}+K_3\chi ^{\frac{(n-2)\theta _2}{n\theta _2-2}}}, \end{aligned}$$

(4.25)

where $K_1, K_2$ and $K_3$ are positive constants that will be determined in (4.36).

Proof

In view of (1.1) and (2.2), we compute:

$$\begin{aligned} \chi '(t)= & {} \int _{\Omega }u^{mp-1}[\hbox {div} (\rho (|\nabla u|^2)\nabla u)+u^p(x,t)\int _{\Omega }k(x,y)u^{p+1}(y,t)dy]dx\\= & {} -mp(mp-1)\int _{\Omega }u^{mp-2}\rho (|\nabla u^2)|\nabla u|^2dx+mp\\&\int _{\Omega }\int _{\Omega }k(x,y)u^{mp-1+p}(x,t)u^{p+1}(y,t)dxdy\\\le & {} -mp(mp-1)\int _{\Omega }u^{mp-2}[b_1+b_2|\nabla u|^{2q}]|\nabla u|^2dx\\&+mp \int _{\Omega }\int _{\Omega }k(x,y)u^{mp-1+p}(x,t)u^{p+1}(y,t)dxdy. \end{aligned}$$

By the similar argument as (3.3), we have:

$$\begin{aligned}&\int _{\Omega }\int _{\Omega }k(x,y)u^{mp-1+p}(x,t)u^{p+1}(y,t)dxdy\\&\quad \le \kappa \left( \int _{\Omega }u^{2p+2}(y,t)dy\right) ^{\frac{1}{2}} \left( \int _{\Omega }u^{2mp-2+2p}(x,t)dx\right) ^{\frac{1}{2}}\\&\quad \le \frac{\kappa }{2} \left( \int _{\Omega }u^{2p+2}dx+\int _{\Omega }u^{2mp-2+2p}dx\right) . \end{aligned}$$

It follows from Hölder’s inequality and $m>\frac{2}{p}+2$ that:

$$\begin{aligned} \int _{\Omega }u^{2p+2}dx\le |\Omega |^{\frac{mp-2-2p}{mp}}\left( \int _{\Omega }u^{mp}dx\right) ^{\frac{2p+2}{mp}}. \end{aligned}$$

(4.26)

Noticing that:

$$\begin{aligned} |\nabla u^{\alpha }|^{2(q+1)}=|\alpha u^{\alpha -1}\nabla u|^{2(q+1)}=\alpha ^{2(q+1)}u^{mp-2}|\nabla u|^{2(q+1)}, \end{aligned}$$

where $\alpha =\frac{mp+2q}{2(q+1)}$. Denoting $\delta =\frac{mp-2+2(p-q)}{\alpha }>0$ and $v=u^{\alpha }$, we obtain:

$$\begin{aligned} \int _{\Omega }u^{2mp-2+2p}dx=\int _{\Omega }v^{2(q+1)+\delta }dx, \end{aligned}$$

Using Hölder’s and Schwarz’s inequalities, we have:

$$\begin{aligned} \int _{\Omega }|\nabla v^{q+1}|^2dx\le & {} (q+1)^2\left( \int _{\Omega }|\nabla v|^{2(q+1)}dx\right) ^{\frac{1}{q+1}}\left( \int _{\Omega }v^{2(q+1)}dx\right) ^{\frac{q}{q+1}}\\\le & {} (q+1)\int _{\Omega }|\nabla v|^{2(q+1)}dx+q(q+1)\int _{\Omega }v^{2(q+1)}dx. \end{aligned}$$

Hence, in view of (4.26), by dropping the $b_1$ term, then we obtain:

$$\begin{aligned} \begin{aligned}\chi '(t)\le&-\frac{mp(mp-1)b_2}{(q+1)\alpha ^{2(q+1)}}\int _{\Omega }|\nabla v^{q+1}|^2dx+\frac{mp\kappa }{2} |\Omega |^{\frac{mp-2-2p}{mp}}[\chi (t)]^{\frac{2p+2}{mp}}\\&+\frac{qmp(mp-1)b_2}{\alpha ^{2(q+1)}}\int _{\Omega }v^{2(q+1)}dx+\frac{mp\kappa }{2}\int _{\Omega }v^{2(q+1)+\delta }dx. \end{aligned} \end{aligned}$$

(4.27)

To estimate the last two terms in (4.27), we will use the following Hölder’s inequality:

$$\begin{aligned} \int _{\Omega }v^{r+s}dx\le \left( \int _{\Omega }v^{\frac{r}{\theta }}dx\right) ^{\theta } \left( \int _{\Omega }v^{\frac{s}{1-\theta }}dx\right) ^{1-\theta }, \end{aligned}$$

(4.28)

where $0<\theta <1$ and r, s are positive constants.

We choose $r_1, s_1$ and $\theta _1$, such that

$$\begin{aligned} r_1+s_1=2(q+1), \quad \frac{r_1}{\theta _1}=\frac{mp}{\alpha },\quad \frac{s_1}{1-\theta _1}=(q+1)\frac{2n}{n-2}, \end{aligned}$$

that is:

$$\begin{aligned}&r_1=\frac{mp}{\alpha }\frac{2(q+1)\frac{2}{n-2}}{2(q+1)\frac{n}{n-2}-\frac{mp}{\alpha }},\\&\quad s_1=2(q+1)-\frac{mp}{\alpha }\frac{2(q+1)\frac{2}{n-2}}{2(q+1)\frac{n}{n-2}-\frac{mp}{\alpha }},\\&\quad \theta _1=\frac{2(q+1)\frac{2}{n-2}}{2(q+1)\frac{n}{n-2}-\frac{mp}{\alpha }}. \end{aligned}$$

Then, it follows from (4.28) that:

$$\begin{aligned} \int _{\Omega }v^{2(q+1)}dx\le \left( \int _{\Omega }v^{\frac{mp}{\alpha }}dx\right) ^{\theta _1} \left( \int _{\Omega }v^{(q+1)\frac{2n}{n-2}}dx\right) ^{1-\theta _1}. \end{aligned}$$

(4.29)

By the same argument, we choose $r_2, s_2$ and $\theta _2$, such that:

$$\begin{aligned} r_2+s_2=2(q+1)+\delta , \quad \frac{r_2}{\theta _2}=\frac{mp}{\alpha },\quad \frac{s_2}{1-\theta _2}=(q+1)\frac{2n}{n-2}, \end{aligned}$$

that is:

$$\begin{aligned}&r_2=\frac{mp}{\alpha }\frac{2(q+1)\frac{2}{n-2}-\delta }{2(q+1)\frac{n}{n-2}-\frac{mp}{\alpha }},\\&\quad s_2=2(q+1)+\delta -\frac{mp}{\alpha }\frac{2(q+1)\frac{2}{n-2}-\delta }{2(q+1)\frac{n}{n-2}-\frac{mp}{\alpha }},\\&\quad \theta _2=\frac{2(q+1)\frac{2}{n-2}-\delta }{2(q+1)\frac{n}{n-2}-\frac{mp}{\alpha }}, \end{aligned}$$

and obtain:

$$\begin{aligned} \int _{\Omega }v^{2(q+1)+\delta }dx\le \left( \int _{\Omega }v^{\frac{mp}{\alpha }}dx\right) ^{\theta _2} \left( \int _{\Omega }v^{(q+1)\frac{2n}{n-2}}dx\right) ^{1-\theta _2}. \end{aligned}$$

(4.30)

Letting $c_*$ be the best embedding constant: $\Vert w\Vert _{\frac{2n}{n-2}}\le c_*\Vert \nabla w\Vert _2$, we obtain:

$$\begin{aligned} \Vert v^{q+1}\Vert _{\frac{2n}{n-2}}^{\frac{2n}{n-2}(1-\theta _1)}\le c_*^{\frac{2n}{n-2}(1-\theta _1)}\Vert \nabla v^{q+1}\Vert _2^{\frac{2n}{n-2}(1-\theta _1)}, \end{aligned}$$

(4.31)

and

$$\begin{aligned} \Vert v^{q+1}\Vert _{\frac{2n}{n-2}}^{\frac{2n}{n-2}(1-\theta _2)}\le c_*^{\frac{2n}{n-2}(1-\theta _2)}\Vert \nabla v^{q+1}\Vert _2^{\frac{2n}{n-2}(1-\theta _2)}. \end{aligned}$$

(4.32)

Combining (4.29) and (4.31), using Schwarz’s inequality, we get:

$$\begin{aligned} \int _{\Omega }v^{2(q+1)}dx\le & {} c_*^{\frac{2n}{n-2}(1-\theta _1)}\big (\int _{\Omega }v^{\frac{mp}{\alpha }}dx\big )^{\theta _1} \left( \int _{\Omega }|\nabla v^{q+1}|^2dx\right) ^{\frac{n}{n-2}(1-\theta _1)}\nonumber \\\le & {} \frac{n\theta _1-2}{n-2}(c_*^2\varepsilon _1^{-1})^{\frac{n(1-\theta _1)}{n\theta _1-2}}\left( \int _{\Omega }v^{\frac{mp}{\alpha }}\right) ^{\frac{(n-2)\theta _1}{n\theta _1-2}} \nonumber \\&+\frac{n(1-\theta _1)}{n-2}\varepsilon _1\int _{\Omega }|\nabla v^{q+1}|^2dx, \end{aligned}$$

(4.33)

where $\varepsilon _1$ is a positive constant to be chosen later. Similarly, it follows from (4.30) and (4.32) that:

$$\begin{aligned}&\int _{\Omega }v^{2(q+1)+\delta }dx \le \frac{n\theta _2-2}{n-2}(c_*^2\varepsilon _2^{-1})^{\frac{n(1-\theta _2)}{n\theta _2-2}}\nonumber \\&\quad \left( \int _{\Omega }v^{\frac{mp}{\alpha }}\right) ^{\frac{(n-2)\theta _2}{n\theta _2-2}}\nonumber \\&\quad +\frac{n(1-\theta _2)}{n-2}\varepsilon _2\int _{\Omega }|\nabla v^{q+1}|^2dx, \end{aligned}$$

(4.34)

where $\varepsilon _1$ is a positive constant to be chosen later.

Combining (4.33) and (4.34) with (4.27) gives:

$$\begin{aligned} \chi '(t)\le & {} -\bigg [\frac{mp(mp-1)b_2}{(q+1)\alpha ^{2(q+1)}}-\frac{qmp(mp-1)b_2}{\alpha ^{2(q+1)}}\\&\frac{n(1-\theta _1)}{n-2}\varepsilon _1 -\frac{mp\kappa }{2}\frac{n(1-\theta _2)}{n-2}\varepsilon _2\bigg ]\int _{\Omega }|\nabla v^{q+1}|^2dx\\&+\frac{mp\kappa }{2} |\Omega |^{\frac{mp-2-2p}{mp}}[\chi (t)]^{\frac{2p+2}{mp}}\\&\quad + \frac{qmp(mp-1)b_2}{\alpha ^{2(q+1)}}\frac{n\theta _1-2}{n-2}\\&\quad (c_*^2\varepsilon _1^{-1})^{\frac{n(1-\theta _1)}{n\theta _1-2}}[\chi (t)]^{\frac{(n-2)\theta _1}{n\theta _1-2}}\\&+\frac{mp\kappa }{2}\frac{n\theta _2-2}{n-2}\\&\quad (c_*^2\varepsilon _2^{-1})^{\frac{n(1-\theta _2)}{n\theta _2-2}}[\chi (t)]^{\frac{(n-2)\theta _2}{n\theta _2-2}}. \end{aligned}$$

By choosing $\varepsilon _1$ and $\varepsilon _2$ sufficiently small, such that:

$$\begin{aligned} \frac{mp(mp-1)b_2}{(q+1)\alpha ^{2(q+1)}}-\frac{mp(mp-1)b_2}{\alpha ^{2(q+1)}}\frac{n(1-\theta _1)}{n-2}\varepsilon _1 -\frac{mp\kappa }{2}\frac{n(1-\theta _2)}{n-2}\varepsilon _2\ge 0, \end{aligned}$$

then, we can obtain the differential inequality:

$$\begin{aligned} \chi '(t)\le K_1[\chi (t)]^{\frac{2p+2}{mp}}+ K_2[\chi (t)]^{\frac{(n-2)\theta _1}{n\theta _1-2}}+K_3[\chi (t)]^{\frac{(n-2)\theta _2}{n\theta _2-2}}, \end{aligned}$$

or equivalently:

$$\begin{aligned} \frac{d\chi }{K_1\chi ^{\frac{2p+2}{mp}}+ K_2\chi ^{\frac{(n-2)\theta _1}{n\theta _1-2}}+K_3\chi ^{\frac{(n-2)\theta _2}{n\theta _2-2}}}\le dt, \end{aligned}$$

(4.35)

where:

$$\begin{aligned} \begin{aligned} K_1=\frac{mp\kappa }{2}&|\Omega |^{\frac{mp-2-2p}{mp}}; \quad K_2=\frac{mp(mp-1)b_2}{\alpha ^{2(q+1)}}\frac{n\theta _1-2}{n-2}(c_*^2\varepsilon _1^{-1})^{\frac{n(1-\theta _1)}{n\theta _1-2}};\\&K_3=\frac{mp\kappa }{2}\frac{n\theta _2-2}{n-2}(c_*^2\varepsilon _2^{-1})^{\frac{n(1-\theta _2)}{n\theta _2-2}}. \end{aligned}\end{aligned}$$

(4.36)

Integrating of the differential inequality (4.35) from 0 to t leads to:

$$\begin{aligned} \int _{\chi (0)}^{\chi (t)}\frac{d\chi }{K_1\chi ^{\frac{2p+2}{mp}}+ K_2\chi ^{\frac{(n-2)\theta _1}{n\theta _1-2}}+K_3\chi ^{\frac{(n-2)\theta _2}{n\theta _2-2}}}\le t. \end{aligned}$$

Passing to the limit as $t\rightarrow T^-$, we obtain:

$$\begin{aligned} \int _{\chi (0)}^{\infty }\frac{d\chi }{K_1\chi ^{\frac{2p+2}{mp}}+ K_2\chi ^{\frac{(n-2)\theta _1}{n\theta _1-2}}+K_3\chi ^{\frac{(n-2)\theta _2}{n\theta _2-2}}}\le T. \end{aligned}$$

Thus, the proof is complete. $\square$