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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In this paper, we study the following initial boundary problem
{utt−Δu+∫t0g(t−s)Δu(s)ds−Δut=|u|p−2u,(x,t)∈Ω×[0,T),u(x,t)=0, x∈∂Ω,u(x,0)=u0(x),ut(x,0)=u1(x), |
where
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
utt−Δu+∫t0g(t−s)Δu(s)ds−|ut|m−2ut=|u|p−2u. |
Tian [20] considered a semilinear parabolic equation with viscoelastic term
ut−Δu+∫t0g(t−s)Δu(s)ds=|u|p−2u. |
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
Throughout the paper, we use
Theorem 2.1 ([6]). Let
1−∫∞0g(s)ds=l>0. | (2.1) |
Let p be such that
{2<p<∞ n=1,2,2<p⩽2nn−2n⩾3. | (2.2) |
Then problem (1.1) has a unique local solution
u∈C([0,Tm);H10(Ω), ut∈C([0,Tm);L2(Ω))∩L2([0,Tm);H10(Ω)), |
for some
Define the energy functional
E(t)=12‖ut‖22+12(1−∫t0g(s)ds)‖∇u‖22+12(g∘∇u)(t)−1p‖u‖pp, |
where
(g∘v)(t)=∫t0g(t−s)‖v(t)−v(s)‖22ds. |
Theorem 2.2 ([6]). Assume that
g(s)⩾0,g′(s)⩽0,∫∞0g(s)ds<1−1(p−1)2. | (2.3) |
Let
(2∫Ωuutdx+‖∇u(t)‖22)|t=0>2pκE(0), | (2.4) |
then
κ=max |
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
\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
(1) If
\begin{equation*} t^*\geqslant \frac{2n-np+4p}{2K_1(p-2)}[\varphi(0)]^{\frac{4-2p}{2n-np+4p}}, \end{equation*} |
where
(2) If
\begin{equation*} t^*\geqslant \frac{2(p-1)}{K_2(p-2)}[\varphi(0)]^{\frac{2-p}{2(p-1)}}, \end{equation*} |
where
(3) If
\begin{equation*} t^*\geqslant \frac{p-2}{K_3(p+2)}[\varphi(0)]^{\frac{2-p}{p+2}}. \end{equation*} |
where
Proof. According to Theorem 2.2, the solution
\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
\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
\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
\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
\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
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
\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
\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
\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
\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
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
\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
\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
\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
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
\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
\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
Theorem 2.4. Let
\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
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
\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
\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
We now focus our attention on the second term on the right in (2.36). Since
\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
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
Integrating (2.39) from
\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
\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.
The authors declare that there are no conflicts of interest in this paper.
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