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Research article

Lower bounds for the blow-up time to a nonlinear viscoelastic wave equation with strong damping

  • Received: 19 August 2018 Accepted: 11 October 2018 Published: 01 November 2018
  • This paper deals with a nonlinear viscoelastic wave equation with strong damping. By the means of the interpolation inequalities and di erential inequality technique, we obtain a lower bound for blow-up time of the solution. This result extends our earlier work Peng et al. [Appl. Math. Lett., 76, 2018].

    Citation: Xiaoming Peng, Xiaoxiao Zheng, Yadong Shang. Lower bounds for the blow-up time to a nonlinear viscoelastic wave equation with strong damping[J]. AIMS Mathematics, 2018, 3(4): 514-523. doi: 10.3934/Math.2018.4.514

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  • This paper deals with a nonlinear viscoelastic wave equation with strong damping. By the means of the interpolation inequalities and di erential inequality technique, we obtain a lower bound for blow-up time of the solution. This result extends our earlier work Peng et al. [Appl. Math. Lett., 76, 2018].


    1. Introduction

    In this paper, we study the following initial boundary problem

    {uttΔu+t0g(ts)Δu(s)dsΔut=|u|p2u,(x,t)Ω×[0,T),u(x,t)=0, xΩ,u(x,0)=u0(x),ut(x,0)=u1(x),

    where ΩRn is a bounded domain with a smooth boundary Ω.

    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

    uttΔu+t0g(ts)Δu(s)ds|ut|m2ut=|u|p2u.

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

    utΔu+t0g(ts)Δu(s)ds=|u|p2u.

    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<p2(n22)n(n2). Compared with the condition of blow up result for p in [6], there exists a gap for p between 2(n22)n(n2) and 2nn2. 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).


    2. Main results

    Throughout the paper, we use to denote the Lp norm for 1p. 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 (u0,u1)H10(Ω)×L2(Ω) be given. Let g be a C1 function satisfying

    10g(s)ds=l>0. (2.1)

    Let p be such that

    {2<p< n=1,2,2<p2nn2n3. (2.2)

    Then problem (1.1) has a unique local solution

    uC([0,Tm);H10(Ω), utC([0,Tm);L2(Ω))L2([0,Tm);H10(Ω)),

    for some Tm>0.

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

    E(t)=12ut22+12(1t0g(s)ds)u22+12(gu)(t)1pupp,

    where

    (gv)(t)=t0g(ts)v(t)v(s)22ds.

    Theorem 2.2 ([6]). Assume that p>2 satisfies (2.2) and let g be a C1 function satisfying

    g(s)0,g(s)0,0g(s)ds<11(p1)2. (2.3)

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

    (2Ωuutdx+u(t)22)|t=0>2pκE(0), (2.4)

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

    κ=max

    \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.


    Conflict of interest

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


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