1. Introduction

In this paper, we study the following initial boundary problem

$\begin{equation} \begin{cases} u_{tt}-\Delta u+\int_0^tg(t-s)\Delta u(s)ds-\Delta u_t = |u|^{p-2}u, (x, t)\in\Omega\times[0, T), \nonumber\\ u(x, t) = 0, \ x\in\partial\Omega, \label{eq1}\\ u(x, 0) = u_0(x), u_t(x, 0) = u_1(x), \nonumber \end{cases} \end{equation}$

where $\Omega\subset\mathbb{R}^n$ is a bounded domain with a smooth boundary $\partial\Omega$.

It is well known that viscoelastic materials present a natural damping, which is due to some properties of these materials to keep memory of their past trace. This type of equations with viscoelastic term describe a variety of important physical processes ^[1] and the reference therein. There is a vast literature on the existence or nonexistence of global solutions, blow up results in finite time, and the asymptotic behavior of the solutions for the viscoelastic equations, we refer the interested readers to ^{[2,3,4,5,6,7,8,9,10,11]} and the references therein. In particular, Song and Zhong ^[5] studied problem (1.1). They established a blow-up result for solutions with positive initial energy. Later, Song and Xue ^[6] extended this blow up result to solutions whose initial data have arbitrarily high initial energy.

Since Payne et al. ^[12,13] applied a differential inequality technique to obtain a lower bound on blow-up time for solutions of the semilinear heat equation. Many authors have given attention to this problem and obtained many profound results ^{[14,15,16,17,18]} and the references therein. However, there seems to have been little work devoted to obtaining lower bounds on blow-up time to solutions of viscoelastic problems. To our best knowledge, only few articles dealt with this questions, see ^[19,20,21]. Yang et al. ^[19] established a lower bound for the blow-up time of the following equation

$\begin{equation*} u_{tt}-\Delta u+\int_0^tg(t-s)\Delta u(s)ds-|u_t|^{m-2} u_t = |u|^{p-2}u. \end{equation*}$

Tian ^[20] considered a semilinear parabolic equation with viscoelastic term

$\begin{equation*} u_{t}-\Delta u+\int_0^tg(t-s)\Delta u(s)ds = |u|^{p-2}u. \end{equation*}$

By the means of differential inequality technique, they obtained a lower bound for blow-up time of the solution. Recently, Peng et al. ^[21] obtained a lower bound for the blow-up time to problem (1.1) by establishing a differential inequality. But they can only derived a lower bounds for blow up time $t^*$ when $2 < p\leqslant\frac{2(n^2-2)}{n(n-2)}$. Compared with the condition of blow up result for $p$ in ^[6], there exists a gap for $p$ between $\frac{2(n^2-2)}{n(n-2)}$ and $\frac{2n}{n-2}$. It is still open whether a lower bound estimate can be obtained if $p$ lies in this gap. Inspired by ^[18,20], the goal of this paper is to gives an answer to the problem unsolved in our earlier work Peng et. al ^[21]. By introducing a new auxiliary functional and using interpolation inequalities, we obtain lower bounds for the blow-up time for the problem (1.1).

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2. Main results

Throughout the paper, we use $\|\cdot\|$ to denote the $L^p-$ norm for $1\leqslant p\leqslant \infty$. Before stating our main results, let us recall the results on the local existence, uniqueness and blow-up in finite time of solutions to (1.1).

Theorem 2.1 (^[6]). Let $(u_0, u_1)\in H_0^1(\Omega)\times L^2(\Omega)$ be given. Let $g$ be a $C^1$ function satisfying

$\begin{equation}\label{eq2} 1-\int_0^{\infty}g(s)ds = l>0. \end{equation}$

(2.1)

Let p be such that

$\begin{equation}\label{eq3} \begin{cases} 2 < p < \infty \quad \qquad\ n = 1, 2, \\ 2 < p\leqslant \displaystyle\frac{2n}{n-2} \quad \quad n\geqslant 3. \end{cases} \end{equation}$

(2.2)

Then problem (1.1) has a unique local solution

$\begin{equation*} u\in C([0, T_m);H_0^1(\Omega), \ u_t\in C([0, T_m);L^2(\Omega))\cap L^2([0, T_m); H_0^1(\Omega)), \end{equation*}$

for some $T_m>0$.

Define the energy functional $E(t)$ associated to the problem (1.1)

$\begin{equation*} E(t) = \frac{1}{2}\|u_t\|_2^2+\frac{1}{2}\bigg(1-\int_0^{t}g(s)ds\bigg)\|\nabla u\|^2_2 +\frac{1}{2}(g\circ\nabla u)(t)-\frac{1}{p}\|u\|_p^p, \end{equation*}$

where

$\begin{equation*} (g\circ v)(t) = \int_0^tg(t-s)\|v(t)-v(s)\|_2^2ds. \end{equation*}$

Theorem 2.2 (^[6]). Assume that $p>2$ satisfies (2.2) and let $g$ be a $C^1$ function satisfying

$\begin{equation}\label{a1} g(s)\geqslant 0, \quad g'(s)\leqslant 0, \quad \int_0^{\infty}g(s)ds < 1-\frac{1}{(p-1)^2}. \end{equation}$

(2.3)

Let $u(t)$ be a solution of problem (1.1) satisfying

$\begin{equation}\label{a2} \bigg(2\int_{\Omega}uu_tdx+\|\nabla u(t)\|_2^2 \bigg)|_{t = 0}>\frac{2p}{\kappa}E(0), \end{equation}$

(2.4)

then $u(t)$ blow up in finite time, where

$\begin{align*} & \kappa = \max\limits_{\eta_1\in(0, 1)}\kappa(\eta_1) = \kappa(\eta^*), \\ & \kappa(\eta_1) = \min(\sqrt{(p+2)\delta\eta_1\lambda_1}, \delta(1-\eta_1)), \end{align*}$

$\lambda_1$ being the first eigenvalue of $-\Delta, \ \delta = (p-2)l-\frac{1}{p}(1-l)$, $\eta^*$ is the root of the equation $\sqrt{(p+2)\delta\eta_1\lambda_1} = \delta(1-\eta_1)$.

Let us introduce an auxiliary function

$\begin{equation}\label{eq6} \varphi(t) = \frac{1}{2}\|u_t\|_2^2+\frac{1}{2}\bigg(1-\int_0^{t}g(s)ds\bigg)\|\nabla u\|^2_2 +\frac{1}{2}(g\circ\nabla u)(t)+\frac{1}{p}\|u\|_p^p, \end{equation}$

(2.5)

with

$\begin{equation}\label{eq4} \varphi(0) = \frac{1}{2}\|u_1(x)\|_2^2+\frac{1}{2}\|\nabla u_0(x)\|_2^2 +\frac{1}{p}\|u_0(x)\|_p^p. \end{equation}$

(2.6)

Theorem 2.3. Under the conditions (2.3) and (2.4), assume $p$ satisfy

$\begin{equation*} \begin{cases} 2 < p < \infty, \quad \qquad\ n = 1, 2, \\ 2 < p < \displaystyle\frac{2n}{n-2}, \quad \quad n\geqslant 3, \end{cases} \end{equation*}$

then the solution $u(x, t)$ of problem (1.1) blows up in finite time $t^*$.

(1) If $n\geqslant3$, then $t^*$ is bounded below by

$\begin{equation*} t^*\geqslant \frac{2n-np+4p}{2K_1(p-2)}[\varphi(0)]^{\frac{4-2p}{2n-np+4p}}, \end{equation*}$

where $K_1$ is given in (2.22).

(2) If $n = 1$, then $t^*$ is bounded below by

$\begin{equation*} t^*\geqslant \frac{2(p-1)}{K_2(p-2)}[\varphi(0)]^{\frac{2-p}{2(p-1)}}, \end{equation*}$

where $K_2$ is given in (2.29).

(3) If $n = 2$, then $t^*$ is bounded below by

$\begin{equation*} t^*\geqslant \frac{p-2}{K_3(p+2)}[\varphi(0)]^{\frac{2-p}{p+2}}. \end{equation*}$

where $K_3$ is given in (2.34).

Proof. According to Theorem 2.2, the solution $u(x, t)$ of (1.1) blows up in a finite time $t^*$. Besides, Song and Xue ^[6] proved that

$\begin{equation*} \lim\limits_{t\rightarrow t^*}\bigg[ \|u_t\|_2^2+\bigg(1+\frac{1}{\lambda_1}\bigg) \|\nabla u\|_2^2 \bigg] = +\infty, \end{equation*}$

which implies that

$\begin{equation}\label{eq5} \lim\limits_{t\rightarrow t^*}\varphi(t) = +\infty. \end{equation}$

(2.7)

Multiplying Eq. (1.1) by $u_t$ and integrating over $\Omega$ yields

$\begin{equation}\label{eq7} \frac{1}{2}\frac{d}{dt}\bigg\{ \int_{\Omega}|u_t|^2dx+\int_{\Omega}|\nabla u|^2dx \bigg\} = \int_0^tg(t-s)\int_{\Omega} \nabla u_t \cdot \nabla udxds-\int_{\Omega}|\nabla u_t|^2dx +\int_{\Omega}|u|^{p-2}uu_tdx. \end{equation}$

(2.8)

For the first term on the right-hand side of (2.8), we have

$\begin{align}\label{eq8} \int_0^tg(t-s)\int_{\Omega} \nabla u_t \cdot \nabla udxds = &\frac{1}{2}\frac{d}{dt}\left\{\int_0^tg(s)ds\int_{\Omega}|\nabla u(t)|^2dx -\int_0^tg(t-s)ds\int_{\Omega}|\nabla u(s)-\nabla u(t)|^2dx \right\} \notag\\ &-\frac{1}{2}g(t)\int_{\Omega}|\nabla u(t)|^2dx +\frac{1}{2}\int_0^tg'(t-s)ds\int_{\Omega}|\nabla u(s)-\nabla u(t)|^2dx. \end{align}$

(2.9)

Inserting (2.9) into (2.8) gives

$\begin{align*} \frac{d}{dt}&\left\{ \frac{1}{2}\int_{\Omega}|u_t|^2dx+\frac{1}{2}\left( 1-\int_0^tg(s)ds \right) \int_{\Omega}|\nabla u|^2dx+\frac{1}{2}(g\circ\nabla u)(t) +\frac{1}{p}\int_{\Omega}|u|^pdx \right\} = -\int_{\Omega}|\nabla u_t(t)|^2dx \\ &+\frac{1}{2}\int_0^tg'(t-s)\int_{\Omega}|\nabla u(s) -\nabla u(t)|^2dxds-\frac{1}{2}g(t)\int_{\Omega}|\nabla u(t)|^2dx +2\int_{\Omega}|u|^{p-2}uu_tdx. \end{align*}$

From (2.5), the above identity can be rewritten as

$\begin{equation}\label{eq9} \varphi'(t) = -\|\nabla u_t(t)\|^2-\frac{1}{2}g(t)\|\nabla u(t)\|^2 +\frac{1}{2}(g'\circ\nabla u)(t)+2\int_{\Omega}|u|^{p-2}uu_tdx. \end{equation}$

(2.10)

Since $g'(s)\leqslant0$ and $g(s)\geqslant0$, it follows from (2.10) that

$\begin{equation*} \varphi'(t)\leqslant-\|\nabla u_t(t)\|^2+2\int_{\Omega}|u|^{p-2}uu_tdx. \end{equation*}$

Using Hölder inequality, we have

$\begin{equation}\label{eq10} \varphi'(t)\leqslant-\|\nabla u_t(t)\|^2+2\|u\|_p^{p-1}\|u_t\|_p. \end{equation}$

(2.11)

Next, we are going to estimate the second term on the right-hand side of (2.11).

Firstly, we consider the case $n\geqslant3$. Using interpolation inequality yields

$\begin{equation}\label{eq11} \|u_t\|_p\leqslant \|u_t\|_2^{\frac{2n-p(n-2)}{2p}}\|u_t\|_{\frac{2n}{n-2}}^{\frac{n(p-2)}{2p}}. \end{equation}$

(2.12)

For any $\varepsilon>0, r, s, \theta>1$, we have the following Young inequality

$\begin{equation}\label{eq12} abc\leqslant \frac{\varepsilon}{r}a^r+\frac{\varepsilon^{-\frac{s}{2r}}}{s}b^s +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}c^{\theta}, \ \frac{1}{r}+\frac{1}{s}+\frac{1}{\theta} = 1. \end{equation}$

(2.13)

Combing (2.12) with (2.13) gives

$\begin{align}\label{eq13} 2\|u\|_p^{p-1}\|u_t\|_p&\leqslant 2\|u\|_p^{p-1}\|u_t\|_2^{\frac{2n-p(n-2)}{2p}} \|u_t\|_{\frac{2n}{n-2}}^{\frac{n(p-2)}{2p}} \notag\\ &\leqslant \frac{\varepsilon}{r}\|u_t\|_{\frac{2n}{n-2}}^{2} +\frac{\varepsilon^{-\frac{s}{2r}}}{s}\|u_t\|_2^{\frac{2n-p(n-2)}{2p}s} +\frac{\varepsilon^{-\frac{\theta}{2r}}2^{\theta}}{\theta}\|u\|_p^{\theta(p-1)}, \end{align}$

(2.14)

with

$\begin{align*} & r = \frac{4p}{n(p-2)}>1, \\ & s = \frac{4p(2n-np+6p-4)}{(2n-np+2p)(2n-np+4p)} >\frac{4p}{2n-np+2p}>\frac{4p}{2p} = 2, \\ & \theta = \frac{2n-np+2p}{4(p-1)}s = \frac{p(2n-np+6p-4)}{(p-1)(2n-np+4p)} >\frac{2n-np+6p-4}{2n-np+4p}>1. \end{align*}$

Applying Sobolev inequality to the first term on the right-hand side of (2.13), we have

$\begin{equation}\label{eq14} \|u_t\|_{\frac{2n}{n-2}}^{2}\leqslant C_1^2\|\nabla u_t\|_{2}^{2}, \end{equation}$

(2.15)

where $C_1$ is the best constant of the Sobolev embedding $H_0^1(\Omega)\hookrightarrow L^{\frac{2n}{n-2}}(\Omega)$.

Recalling (2.5), we have

$\frac{1}{p}\|u\|_p^p \leqslant \frac{1}{2}\|u_t\|_2^2+\frac{1}{2}\bigg(1-\int_0^{t}g(s)ds\bigg) \|\nabla u\|^2_2+\frac{1}{2}(g\circ\nabla u)(t)+\frac{1}{p}\|u\|_p^p = \varphi(t),$

(2.16)

$\frac{1}{2}\|u_t\|_2^2 \leqslant\frac{1}{2}\|u_t\|_2^2+\frac{1}{2}\bigg(1-\int_0^{t}g(s)ds \bigg)\|\nabla u\|^2_2+\frac{1}{2}(g\circ\nabla u)(t) +\frac{1}{p}\|u\|_p^p = \varphi(t).$

(2.17)

Plugging (2.15)--(2.17) into (2.14), it follows that

$\begin{align}\label{eq17} 2\|u\|_p^{p-1}\|u_t\|_p&\leqslant \frac{\varepsilon C_1^2}{r}\|\nabla u_t\|_{2}^{2} +\frac{\varepsilon^{-\frac{s}{2r}}}{s}\|u_t\|_2^{\frac{2n-p(n-2)}{2p}s} +\frac{\varepsilon^{-\frac{\theta}{2r}}2^{\theta}}{\theta}\|u\|_p^{\theta(p-1)}\notag \\ &\leqslant \frac{\varepsilon C_1^2}{r}\|\nabla u_t\|_{2}^{2} +\frac{\varepsilon^{-\frac{s}{2r}}}{s}2^{\frac{(2n-np+2p)s}{4p}} [\varphi(t)]^{\frac{(2n-np+2p)s}{4p}} +\frac{\varepsilon^{-\frac{\theta}{2r}}2^{\theta}}{\theta} p^{\frac{\theta(p-1)}{p}}[\varphi(t)]^{\frac{\theta(p-1)}{p}}. \end{align}$

(2.18)

Noting that

$\begin{equation*} \frac{(2n-np+2p)s}{4p} = \frac{\theta(p-1)}{p} = \frac{2n-np+6p-4}{2n-np+4p}>1, \end{equation*}$

(2.18) can be rewritten as

$\begin{equation}\label{eq18} 2\|u\|_p^{p-1}\|u_t\|_p \leqslant \frac{\varepsilon C_1^2}{r}\|\nabla u_t\|_{2}^{2} +\left[\frac{2^{\gamma}}{s}\varepsilon^{-\frac{s}{2r}} +\frac{p^{\gamma}2^{\theta}}{\theta}\varepsilon^{-\frac{\theta}{2r}} \right][\varphi(t)]^{\gamma}. \end{equation}$

(2.19)

where

$\begin{equation}\label{eq19} \gamma = \frac{2n-np+6p-4}{2n-np+4p}>1. \end{equation}$

(2.20)

Inserting (2.19) into (2.11), we obtain

$\begin{equation}\label{eq20} \varphi'(t)\leqslant\left(\frac{\varepsilon C_1^2}{r}-1\right)\|\nabla u_t\|_{2}^{2} +\left[\frac{2^{\gamma}}{s}\varepsilon^{-\frac{s}{2r}} +\frac{p^{\gamma}2^{\theta}}{\theta}\varepsilon^{-\frac{\theta}{2r}} \right][\varphi(t)]^{\gamma}. \end{equation}$

(2.21)

Taking $\varepsilon = \frac{4p}{n(p-2)C_1^2}$ in (2.21) leads to

$\begin{equation}\label{eq21} \varphi'(t)\leqslant K_1[\varphi(t)]^{\gamma}. \end{equation}$

(2.22)

where

$\begin{equation}\label{eq22} K_1 = \frac{2n-np+2p}{p\gamma2^{2-\gamma}} \left[ \frac{n(p-2)C_1^2}{4p} \right]^{\frac{n\gamma(p-2)}{2n-np+2p}} +2^{\frac{p\gamma}{p-1}}\frac{p-1}{\gamma p^{1-\gamma}} \left[\frac{n(p-2)C_1^2}{4p}\right]^{\frac{n\gamma(p-2)}{8(p-1)}}. \end{equation}$

(2.23)

Integrating (2.22) from $0$ to $t$ results in

$\begin{equation} \frac{1}{1-\gamma}\bigg\{[\varphi(t)]^{1-\gamma}-[\varphi(0)]^{1-\gamma}\bigg\} \leqslant K_1t. \end{equation}$

(2.24)

Thus, letting $t\rightarrow t^*$ and taking into account (2.7), we have the lower bound for $t^*$

$\begin{equation*} t^*\geqslant \frac{1}{K_1(\gamma-1)}[\varphi(0)]^{1-\gamma} = \frac{2n-np+4p}{2K_1(p-2)}[\varphi(0)]^{\frac{2p-4}{2n-np+4p}}. \end{equation*}$

Next, we continue to estimate (2.11) for the case $n = 1$. Using Hölder inequality and Sobolev inequality, we have

$\begin{equation}\label{eq23} \|u_t\|_p^p\leqslant \|u_t\|_2^{p-2} \|u_t\|_{\infty}^2 \leqslant \|u_t\|_2^{p-2}(C_2\|\nabla u_t\|_2)^2, \end{equation}$

(2.25)

where $C_2$ is the best constant of the Sobolev embedding $H_0^1(\Omega)\hookrightarrow L^{\infty}(\Omega)$.

Using again (2.13), we arrive at

$\begin{align}\label{eq24} 2\|u\|_p^{p-1}\|u_t\|_p&\leqslant 2C_2^{\frac{2}{p}}\|u\|_p^{p-1}\|\nabla u_t\|_2^{\frac{p-2}{p}} \|u_t\|_2^{\frac{2p-np+2n}{2p}} \notag\\ &\leqslant \frac{\varepsilon}{p}\|\nabla u_t\|_2^{2} +\frac{\varepsilon^{-\frac{s}{2r}}}{s}\|u_t\|_2^{\frac{(2p-np+2n)s}{2p}} +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}2^{\theta}C_2^{\theta} \|u\|_p^{\theta(p-1)}, \end{align}$

(2.26)

with

$\begin{align*} & r = p>2, \\ & s = \frac{p(3p-4)}{(p-1)(p-2)} = \frac{2p(p-2)+p^2}{(p-1)(p-2)}>\frac{2p}{p-1} = 2, \\ & \theta = \frac{p(3p-4)}{2(p-1)^2}s = \frac{3(p-1)^2+2p-3}{2(p-1)^2} >\frac{3}{2}. \end{align*}$

Combining (2.11), (2.16), (2.17) with (2.26) yields

$\begin{align}\label{eq25} \varphi'(t)&\leqslant (\frac{\varepsilon}{p}-1)\|\nabla u_t\|_2^2 +\frac{\varepsilon^{-\frac{s}{2r}}}{s}\|u_t\|_2^{\frac{(p-2)s}{p}} +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}2^{\theta}C_2^{\frac{2\theta}{p}} \|u\|_p^{\theta(p-1)} \notag \\ & \leqslant (\frac{\varepsilon}{p}-1)\|\nabla u_t\|_2^2 +\left[\frac{\varepsilon^{-\frac{s}{2r}}}{s}2^{\frac{(p-2)s}{2p}} +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}2^{\theta}C_2^{\frac{2\theta}{p}} p^{\frac{\theta(p-1)}{p}}\right][\varphi(t)]^{\frac{3p-4}{2(p-1)}}. \end{align}$

(2.27)

Taking $\varepsilon = p$ in (2.27), we have

$\begin{equation}\label{eq26} \varphi'(t)\leqslant K_2[\varphi(t)]^{\frac{3p-4}{2(p-1)}}, \end{equation}$

(2.28)

where

$\begin{equation}\label{eq27} K_2 = \frac{(p-1)(p-2)}{p(3p-4)}(p2^{p-2})^{\frac{3p-4}{2(p-1)(p-2)}} \left[ 1+\frac{2(p-1)}{p-2}p^{\frac{2p^2-9p+8}{2(p-1)(p-2)}} (2C_2)^{\frac{3p-4}{2(p-1)^2}} \right]. \end{equation}$

(2.29)

Noting that $\frac{3p-4}{2(p-1)}>1$ and integrating (2.28) from $0$ to $t^*$ results in

$\begin{equation*} \frac{2(p-1)}{p-2}[\varphi(0)]^{\frac{2-p}{2(p-1)}} \leqslant K_2t^*, \end{equation*}$

which implies that

$\begin{equation*} t^*\geqslant \frac{2(p-1)}{K_2(p-2)}[\varphi(0)]^{\frac{2-p}{2(p-1)}}. \end{equation*}$

Finally, we estimate (2.11) for the case $n = 2$. Using interpolation theorem ^[22], we have

$\begin{equation}\label{eq28} \|u_t\|_p\leqslant N\|\nabla u_t\|_2^{\frac{p-2}{p}} \|u_t\|_2^{\frac{2}{p}}, \end{equation}$

(2.30)

where $K$ is an embedding constant.

Using again (2.13), we arrive at

$\begin{align}\label{eq29} 2\|u\|_p^{p-1}\|u_t\|_p&\leqslant 2N\|u\|_p^{p-1}\|\nabla u_t\|_2^{\frac{p-2}{p}} \|u_t\|_2^{\frac{2}{p}} \notag\\ &\leqslant \frac{\varepsilon}{r}\|\nabla u_t\|_2^{\frac{(p-2)r}{p}} +\frac{\varepsilon^{-\frac{s}{2r}}}{s}\|u_t\|_2^{\frac{2s}{p}} +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}2^{\theta}N^{\theta} \|u\|_p^{\theta(p-1)}, \notag\\ & = \frac{(p-2)\varepsilon}{2p}\|\nabla u_t\|_2^{2} +\frac{\varepsilon^{-\frac{s}{2r}}}{s}\|u_t\|_2^{\frac{4p}{p+2}} +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}2^{\theta}N^{\theta} \|u\|_p^{\frac{2p^2}{p+2}}, \end{align}$

(2.31)

with

$\begin{align*} & r = \frac{2p}{p-2}>2, \ s = \frac{2p^2}{p+2}>2, \ \theta = \frac{2p^2}{(p-1)(p+2)} >\frac{2p}{p+2}>1. \end{align*}$

Combining (2.11), (2.16), (2.17) with (2.31) yields

$\begin{align}\label{eq30} \varphi'(t)&\leqslant \bigg[\frac{(p-2)\varepsilon}{2p}-1\bigg]\|\nabla u_t\|_2^2 +\frac{\varepsilon^{-\frac{s}{2r}}}{s}\|u_t\|_2^{\frac{4p}{p+2}} +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}2^{\theta}K^{\theta} \|u\|_p^{\frac{2p^2}{p+2}} \notag \\ & \leqslant \bigg[\frac{(p-2)\varepsilon}{2p}-1\bigg] \|\nabla u_t\|_2^2 +\left[\frac{\varepsilon^{-\frac{s}{2r}}}{s}2^{\frac{2p}{p+2}} +\frac{\varepsilon^{-\frac{\theta}{2r}}}{\theta}2^{\theta}N^{\theta} p^{\frac{2p}{p+2}}\right][\varphi(t)]^{\frac{2p}{p+2}}. \end{align}$

(2.32)

Taking $\varepsilon = \frac{2p}{p-2}$ in (2.32), we have

$\begin{equation}\label{eq31} \varphi'(t)\leqslant K_3[\varphi(t)]^{\frac{2p}{p+2}}, \end{equation}$

(2.33)

where

$\begin{equation}\label{eq32} K_3 = \frac{p+2}{p^2}\left(\frac{2p}{p-2}\right)^{\frac{-p(p-2)}{2(p+2)}}2^{\frac{p-2}{p+2}} \left[ 1+(p-1) \left(\frac{2p}{p-2}\right)^{\frac{p(p-2)^2}{2(p-1)(p+2)}} 2^{\frac{2p}{(p-1)(p+2)}} N^{\frac{2p^2}{(p-1)(p+2)}}p^{\frac{2p}{p+2}} \right]. \end{equation}$

(2.34)

Noting that $\frac{2p}{p+2}>1$ and integrating (2.33) from $0$ to $t^*$ results in

$\begin{equation*} \frac{p-2}{p+2}[\varphi(0)]^{\frac{2-p}{p+2}} \leqslant K_3t^*, \end{equation*}$

which implies that

$\begin{equation*} t^*\geqslant \frac{p-2}{K_3(p+2)}[\varphi(0)]^{\frac{2-p}{p+2}}. \end{equation*}$

The proof is complete.

Remark 1. From the proof of (2.14), we observe that it is clear that $2n-np+2p = 0$ when $p = \frac{2n}{n-2}$ for $n\geqslant 3$. In this case, the inequality (2.14) doesn't hold. Thus we need to develop new ideas to restructure this inequality.

Theorem 2.4. Let $\varphi(t)$ and $\varphi(0)$ be defined in (2.5) and (2.6). Suppose that the conditions of Theorem 2.2 hold. Then the solution of (1.1) blows up in finite time $t^*$, which is bounded below by

$\begin{equation*} t^*\geqslant \int_{\varphi(0)}^{\infty}\frac{d\eta} {M_1\eta^{\frac{\alpha(p-1)}{(\alpha-1)p}}+M_2}, \end{equation*}$

where

$\begin{equation*} 1 < \alpha < 2, \quad M_1 = \frac{\alpha}{\alpha-1}2^{\frac{\alpha}{\alpha-1}} \alpha^{-\frac{1}{\alpha-1}}p^{\frac{\alpha(p-1)}{p(\alpha-1)}} , \quad M_2 = \frac{2}{2-\alpha}(\frac{\alpha}{2})^{\frac{\alpha}{2-\alpha}} B_s^{\frac{2\alpha}{2-\alpha}} \end{equation*}$

and $B_s$ is the best constant of the Sobolev embedding $H_0^1(\Omega)\hookrightarrow L^p(\Omega)$.

Proof. As already mentioned, going back to (2.11), we need to estimate the second term on the right hand side of (2.11). In what follows, we are going to estimate it in a different way.

For any $\varepsilon>0, r>1, s>1$, we have the following known Young inequality

$\begin{equation}\label{eq33} ab\leqslant \frac{\varepsilon}{r}a^r+\frac{\varepsilon^{-\frac{s}{r}}}{s}b^s, \quad \frac{1}{r}+\frac{1}{s} = 1. \end{equation}$

(2.35)

By means of the inequality (2.35) with $r = \alpha; s = \frac{\alpha}{\alpha-1}, \varepsilon = \alpha$, it follows that

$\begin{equation}\label{eq34} 2\|u\|_p^{p-1}\|u_t\|_p \leqslant C_3(\|u\|_p^p)^{\frac{\alpha(p-1)}{p(\alpha-1)}} +\|u_t\|_p^{\alpha}, \end{equation}$

(2.36)

where $C_3 = \frac{\alpha}{\alpha-1}2^{\frac{\alpha}{\alpha-1}}\alpha^{-\frac{1}{\alpha-1}}$.

We now focus our attention on the second term on the right in (2.36). Since $1 < \alpha < 2$, using Sobolev inequality and (2.35) with $r = \frac{2}{\alpha}, s = \frac{2}{2-\alpha}, \varepsilon = \frac{2}{\alpha}$, we arrive at

$\begin{equation}\label{eq35} \|u_t\|_p^{\alpha}\leqslant B_s^{\alpha}\|\nabla u_t\|_2^{\alpha} \leqslant \|\nabla u_t\|_2^2+M_2, \end{equation}$

(2.37)

where $M_2 = \frac{2}{2-\alpha}(\frac{\alpha}{2})^{\frac{\alpha}{2-\alpha}} B_s^{\frac{2\alpha}{2-\alpha}}, B_s$ is the best constant of the Sobolev embedding $H_0^1(\Omega)\hookrightarrow L^p(\Omega)$.

Inserting (2.37) into (2.36) yields

$\begin{equation}\label{eq36} 2\|u\|_p^{p-1}\|u_t\|_p\leqslant C_3(\|u\|_p^p)^{\frac{\alpha(p-1)}{p(\alpha-1)}} + \|\nabla u_t\|_2^2+M_2. \end{equation}$

(2.38)

Combining (2.11), (2.16) with (2.38), we get

$\begin{equation}\label{eq37} \varphi'(t) \leqslant M_1[\varphi(t)]^{\frac{\alpha(p-1)}{p(\alpha-1)}}+M_2, \end{equation}$

(2.39)

where $M_1 = C_3p^{\frac{\alpha(p-1)}{p(\alpha-1)}}$.

Integrating (2.39) from $0$ to $t$ yields

$\begin{equation*} \int_{\varphi(0)}^{\varphi(t)}\frac{d\eta}{M_1\eta^{\frac{\alpha(p-1)}{p(\alpha-1)}}+M_2} \leqslant t, \end{equation*}$

from which we deduce a lower bound for $t^*$, namely,

$\begin{equation*} \int_{\varphi(0)}^{\infty}\frac{d\eta}{M_1\eta^{\frac{\alpha(p-1)}{p(\alpha-1)}}+M_2} \leqslant t^*. \end{equation*}$

The proof is complete.

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Conflict of interest

The authors declare that there are no conflicts of interest in this paper.

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