In this section we seek the lower bound for the blow-up time \(T^{*}\). To this end, we define an auxiliary function of the form
$$\begin{aligned}& v(s) = \int_{0}^{s} {\frac{{a'(y)}}{{b(y)}}\,\mathrm{d}y,} \end{aligned}$$
(18)
$$\begin{aligned}& E(t) = \int_{\mathcal{O}} {{{ \bigl[ {v\bigl(u(x,t)\bigr)} \bigr]}^{\mu p + 2}} \,\mathrm {d}y} \quad\mbox{with } \mu \ge1. \end{aligned}$$
(19)
Moreover, we have to point out that (18) indicates
$$ \Delta v = \frac{{a'(u)}}{{b(u)}}\Delta u, $$
(20)
which is very important to prove the following theorem.
Theorem 3.1
Suppose that
\(\mathcal{O} \subset\mathbb{R}^{3}\)
is a bounded convex domain. Further, assume that the nonlinear functions
a, b, and
f
satisfy
$$ 0 < f(s) \le\delta b(s){ \biggl( { \int_{0}^{s} {v(y)\,\mathrm{d}y} } \biggr)^{p - 1}}, \quad s>0, $$
(21)
where
δ
is a positive constant independent of
a, b, and
f. Then the blow-up time
\(T^{*}\)
is bounded below by
$${T^{*}} \ge \int_{E(0)}^{ + \infty} {\frac{{\,\mathrm{d}\xi}}{{{A_{0}} + {A_{1}}\xi + {A_{2}}{\xi^{\frac{3}{2}}} + {A_{3}}{\xi^{3}} + {A_{4}}{\xi ^{\frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}}}}}, $$
where
\(A_{0}\), \(A_{1}\), \(A_{2}\), \(A_{3}\), and
\(A_{4}\)
are positive constants to be determined later.
Proof
We first compute
$$\begin{aligned} E'(t) ={}& ( {\mu p + 2} ) \int_{\mathcal{O}} {{v^{\mu p + 1}}\frac {{a'(u)}}{{b(u)}}{u_{t}} \,\mathrm{d}} x \\ ={}& ( {\mu p + 2} ) \int_{\mathcal{O}} {{v^{\mu p + 1}}\frac{1}{{b(u)}} \bigl[ {\operatorname{div} \bigl( {b(u){{\vert {\nabla u} \vert }^{p - 2}}\nabla u} \bigr)} } \\ &{} { + \gamma b'(u){{\vert {\nabla u} \vert }^{p}} + f(u)} \bigr]\,\mathrm{d}x \\ ={}& {-} {\kappa^{p - 1}} ( {\mu p + 2} ) \int _{\partial\mathcal{O}} {{v^{\mu p + 1}} {{\vert u \vert }^{ ( {p - 1} )\sigma}} \,\mathrm{d}} x \\ &{} - ( {\mu p + 2} ) ( {\mu p + 1} ) \int_{\mathcal{O}} {{v^{\mu p}}\nabla v{{\vert {\nabla u} \vert }^{p - 2}}\nabla u\,\mathrm{d}x} \\ &{} + ( {\mu p + 2} ) (1 + \gamma) \int _{\mathcal{O}} {{v^{\mu p + 1}}\frac{{b'(u)}}{{b(u)}}{{\vert { \nabla u} \vert }^{p}} \,\mathrm{d}x} \\ &{} + ( {\mu p + 2} ) \int_{\mathcal{O}} {{v^{\mu p + 1}}\frac{{f(u)}}{{b(u)}}\,\mathrm{d}x} \\ \le{}& {-} {\kappa^{p - 1}} ( {\mu p + 2} ) \int _{\partial\mathcal{O}} {{v^{\mu p + 1}} {{\vert u \vert }^{ ( {p - 1} )\sigma}} \,\mathrm{d}} x \\ &{} - ( {\mu p + 2} ) ( {\mu p + 1} ) \int_{\mathcal{O}} {{v^{\mu p}}\nabla v{{\vert {\nabla u} \vert }^{p - 2}}\nabla u\,\mathrm{d}x} \\ &{} + ( {\mu p + 2} ) (1 + \gamma) \int _{\mathcal{O}} {{v^{\mu p + 1}}\frac{{b'(u)}}{{b(u)}}{{\vert { \nabla u} \vert }^{p}} \,\mathrm{d}x} \\ &{} + \delta ( {\mu p + 2} ) \int_{\mathcal{O}} {{v^{\mu p + p}} \,\mathrm{d}x} . \end{aligned}$$
(22)
The last inequality holds due to condition (21). Further, in view of (20), (21), and \(b' \le0\), we drop some non-positive terms in (22) to get
$$\begin{aligned} E'(t) \le{}& {-} ( {\mu p + 2} ) ( {\mu p + 1} ) \int _{\mathcal{O}} {{{ \biggl( {\frac{{b(u)}}{{a'(u)}}} \biggr)}^{p - 1}} {v^{\mu p}} {{\vert {\nabla v} \vert }^{p}} \,\mathrm{d}x} \\ &{} + \delta ( {\mu p + 2} ) \int_{\mathcal{O}} {{v^{\mu p + p}} \,\mathrm{d}x}. \end{aligned}$$
(23)
Using the fact that \(b(s) \ge{b_{m}} > 0\) and \(0 < a'(s) \le{a'_{M}}\), (23) becomes
$$\begin{aligned} E'(t) \le{}& {-} ( {\mu p + 2} ) ( {\mu p + 1} ){(\mu + 1)^{ - p}} { \biggl( {\frac{{{b_{m}}}}{{{{a'}_{M}}}}} \biggr)^{p - 1}} \int _{\mathcal{O}} {{{\bigl\vert {\nabla{v^{\mu + 1}}} \bigr\vert }^{p}} \,\mathrm{d}x} \\ &{} + \delta ( {\mu p + 2} ) \int_{\mathcal{O}} {{v^{\mu p + p}} \,\mathrm{d}x} . \end{aligned}$$
(24)
Next, we seek to bound \(\delta ( {\mu p + 2} )\int_{\mathcal{O}} {{v^{\mu p + p}}\,\mathrm{d}x} \) in terms of \(E(t)\) and \(\int_{\mathcal{O}} {{{\vert {\nabla{v^{\mu + 1}}} \vert }^{p}}\,\mathrm{d}x} \). By means of the Hölder and Young inequalities, we have
$$\begin{aligned} \int_{\mathcal{O}} {{v^{\mu p + p}} \,\mathrm{d}x} \le{}&{\vert \mathcal{O} \vert ^{\frac{2}{{\mu p + p + 1}}}} { \biggl( { \int_{\mathcal{O}} {{v^{\mu p + p + 1}} \,\mathrm{d}x} } \biggr)^{\frac{{\mu p + p}}{{\mu p + p + 2}}}} \\ \le{}&\frac{2}{{\mu p + p + 2}}{\vert \mathcal{O} \vert } + \frac{{\mu p + p}}{{\mu p + p + 2}} \int_{\mathcal{O}} {{v^{\mu p + p + 2}} \,\mathrm{d}x} \\ \le{}&\frac{2}{{\mu p + p + 1}}{\vert \mathcal{O} \vert } + \frac{{\mu p + p}}{{\mu p + p + 2}} \\ &{} \cdot{ \biggl( { \int_{\mathcal{O}} {{v^{\frac {3}{2} ( {\mu p + 2} )}} \,\mathrm{d}x} } \biggr)^{\frac {{2p}}{{\mu p + 2}}}} { \biggl( { \int_{\mathcal{O}} {{v^{\mu p + 2}} \,\mathrm{d}x} } \biggr)^{\frac{{\mu p + 2 - 2p}}{{\mu p + 2}}}} \\ \le{}&\frac{2}{{\mu p + p + 2}}{\vert \mathcal{O} \vert } + \frac{{\mu p + p}}{{\mu p + p + 2}}\frac{{2p}}{{\mu p + 2}} \int_{\mathcal{O}} {{v^{\frac{3}{2} ( {\mu p + 2} )}} \,\mathrm{d}x} \\ &{} + \frac{{\mu p + p}}{{\mu p + p + 2}}\frac{{\mu p + 2 - 2p}}{{\mu p + 2}} \int_{\mathcal{O}} {{v^{\mu p + 2}} \,\mathrm{d}x} . \end{aligned}$$
(25)
Using the integral inequality derived in [1] (see (2.16)), namely
$$\int_{\mathcal{O}} {{u^{\frac{3}{2} ( {\mu p + 2} )}} \,\mathrm{d}x} \le \frac{{{3^{\frac{3}{4}}}}}{{2{\rho_{0}}^{\frac{3}{2}}}}E{(t)^{\frac {3}{2}}} + \frac{{\sqrt{2} }}{{{3^{\frac{3}{4}}}}}{ \biggl( { \frac{{{\rho _{1}}}}{{{\rho_{0}}}} + 1} \biggr)^{\frac{3}{2}}} \biggl[ {\frac {{E{{(t)}^{3}}}}{{4\chi ^{3}}} + \frac{3}{4}\chi \int_{\mathcal{O}} {{{\bigl\vert {\nabla{u^{\frac{1}{2} ( {\mu p + 2} )}}} \bigr\vert }^{2}} \,\mathrm{d}x} } \biggr], $$
(25) becomes
$$\begin{aligned} \int_{\mathcal{O}} {{v^{\mu p + p}} \,\mathrm{d}x} \le{}& \frac{2}{{\mu p + p + 2}}{\vert \mathcal{O} \vert } + \frac{{\mu p + p}}{{\mu p + p + 2}}\frac {{2p}}{{\mu p + 2}}\frac{{{3^{\frac{3}{4}}}}}{{2{\rho_{0}}^{\frac {3}{2}}}}E{(t)^{\frac{3}{2}}} \\ &{} + \frac{{\mu p + p}}{{\mu p + p + 2}}\frac {{2p}}{{\mu p + 2}}\frac{{\sqrt{2} }}{{{3^{\frac{3}{4}}}}}{ \biggl( { \frac {{{\rho_{1}}}}{{{\rho_{0}}}} + 1} \biggr)^{\frac{3}{2}}} \\ &{} \cdot \biggl[ {\frac{{E{{(t)}^{3}}}}{{4\chi ^{3}}} + \frac{3}{4}\chi \int_{\mathcal{O}} {{{\bigl\vert {\nabla{u^{\frac {1}{2} ( {\mu p + 2} )}}} \bigr\vert }^{2}} \,\mathrm{d}x} } \biggr] \\ &{} + \frac{{\mu p + p}}{{\mu p + p + 2}}\frac{{\mu p + 2 - 2p}}{{\mu p + 2}} \int_{\mathcal{O}} {{v^{\mu p + 2}} \,\mathrm{d}x} . \end{aligned}$$
(26)
For simplicity, let \(w = {v^{1 + ns}}\). Again by using the Hölder and Young inequalities, we obtain
$$\begin{aligned} & \int_{\mathcal{O}} {{{\bigl\vert {\nabla{v^{\frac{1}{2} ( {\mu p + 2} )}}} \bigr\vert }^{2}} \,\mathrm{d}x} \\ &\quad\le\frac{{{{ ( {\mu p + 1} )}^{2}}}}{{4{{ ( {\mu + 1} )}^{2}}}}{ \biggl( { \int_{\mathcal{O}} {{{\vert {\nabla w} \vert }^{p}}\,\mathrm{d}x} } \biggr)^{\frac{2}{p}}} { \biggl( { \int_{\mathcal{O}} {{w^{\frac{{p ( {\mu p + 2} )}}{{(p - 2) ( {\mu + 1} )}} - \frac{{2p}}{{p - 2}}}} \,\mathrm{d}x} } \biggr)^{\frac{{p - 2}}{p}}} \\ &\quad\le\frac{{{{ ( {\mu p + 1} )}^{2}}}}{{2p{{ ( {\mu + 1} )}^{2}}}} \int_{\mathcal{O}} {{{\vert {\nabla w} \vert }^{p}} \,\mathrm{d}x} \\ &\qquad{} + \frac{{p - 2}}{p}\frac{{{{ ( {\mu p + 2} )}^{2}}}}{{4{{ ( {\mu + 1} )}^{2}}}} \int_{\mathcal{O}} {{w^{\frac {{p ( {\mu p + 2} )}}{{(p - 2) ( {\mu + 1} )}} - \frac{{2p}}{{p - 2}}}} \,\mathrm{d}x} \\ &\quad\le\frac{{{{ ( {\mu p + 1} )}^{2}}}}{{2p{{ ( {\mu + 1} )}^{2}}}} \int_{\mathcal{O}} {{{\bigl\vert {\nabla{v^{1 + \mu}}} \bigr\vert }^{p}} \,\mathrm{d}x} \\ &\qquad{}+ \frac{{p - 2}}{p}{\vert \mathcal{O} \vert ^{1 - \frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}} \frac{{{{ ( {\mu p + 1} )}^{2}}}}{{4{{ ( {\mu + 1} )}^{2}}}}E{(t)^{\frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}}, \end{aligned}$$
combining which with (26) yields
$$\begin{aligned} &\delta ( {\mu p + 2} ) \int_{\mathcal{O}} {{u^{\mu p + p}} \,\mathrm{d}x} \\ &\quad\le{A_{0}} + {A_{1}}E(t) + {A_{2}}E{(t)^{\frac{3}{2}}} + {A_{3}}E{(t)^{3}} \\ &\qquad{}+ {A_{4}}E{(t)^{\frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}} + \chi{A_{5}} \int_{\mathcal{O}} {{{\bigl\vert {\nabla{v^{1 + \mu}}} \bigr\vert }^{p}} \,\mathrm{d}x}, \end{aligned}$$
(27)
where χ is a positive constant to be determined later,
$$\begin{aligned}& {A_{0}} = \frac{{2\delta ( {\mu p + 2} )}}{{\mu p + p + 2}}{\vert \mathcal{O} \vert },\qquad {A_{1}} = \delta ( {\mu p + 2} )\frac{{\mu p + p}}{{\mu p + p + 2}} \frac{{\mu p + 2 - 2p}}{{\mu p + 2}}, \\& {A_{2}} = \frac{{{3^{\frac{3}{4}}}}}{{2{\rho_{0}}^{\frac{3}{2}}}}\delta ( {\mu p + 2} )\frac{{\mu p + p}}{{\mu p + p + 2}} \frac {{2p}}{{\mu p + 2}}, \\& {A_{3}} = \frac{{\delta ( {\mu p + 2} )}}{{4\chi_{2}^{3}}}\frac {{\mu p + p}}{{\mu p + p + 2}}\frac{{2p}}{{\mu p + 2}} \frac{{\sqrt{2} }}{{{3^{\frac{3}{4}}}}}{ \biggl( {\frac{{{\rho_{1}}}}{{{\rho_{0}}}} + 1} \biggr)^{\frac{3}{2}}}, \\& \begin{aligned}[b] {A_{4}} ={}& \frac{3}{4}\frac{{\sqrt{2} }}{{{3^{\frac{3}{4}}}}}{ \biggl( {\frac {{{\rho_{1}}}}{{{\rho_{0}}}} + 1} \biggr)^{\frac{3}{2}}}\delta ( {\mu p + 2} ) \frac{{\mu p + p}}{{\mu p + p + 2}} \\ &{}\cdot\frac{{2p}}{{\mu p + 2}}\frac{{p - 2}}{p}{\vert \mathcal {O} \vert ^{1 - \frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}}\frac{{{{ ( {\mu p + 1} )}^{2}}}}{{4{{ ( {\mu + 1} )}^{2}}}}\chi, \end{aligned} \\& {A_{5}} = \frac{3}{4}\frac{{\sqrt{2} }}{{{3^{\frac{3}{4}}}}}{ \biggl( { \frac {{{\rho_{1}}}}{{{\rho_{0}}}} + 1} \biggr)^{\frac{3}{2}}}\delta ( {\mu p + 2} ) \frac{{\mu p + p}}{{\mu p + p + 2}}\frac{{2p}}{{\mu p + 2}}\frac{{{{ ( {\mu p + 1} )}^{2}}}}{{2p{{ ( {\mu + 1} )}^{2}}}}. \end{aligned}$$
Finally, inserting (27) into (24), we obtain
$$\begin{aligned} E'(t) \le{}& {-} ( {\mu p + 2} ) ( {\mu p + 1} ){(\mu + 1)^{ - p}}\frac{{{b_{m}}}}{{{{a'}_{M}}}} \int_{\mathcal{O}} {{{\bigl\vert {\nabla{v^{\mu + 1}}} \bigr\vert }^{p}} \,\mathrm{d}y} \\ &{} + {A_{0}} + {A_{1}}E(t) + {A_{2}}E{(t)^{\frac{3}{2}}} + {A_{3}}E{(t)^{3}} \\ &{} + {A_{4}}E{(t)^{\frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}} + \chi {A_{5}} \int_{\mathcal{O}} {{{\bigl\vert {\nabla{v^{1 + \mu}}} \bigr\vert }^{p}} \,\mathrm{d}x} . \end{aligned}$$
(28)
To make use of (28), we choose
$$\chi = A_{5}^{ - 1} ( {\mu p + 2} ) ( {\mu p + 1} ){(\mu + 1)^{ - p}} { \biggl( {\frac{{{b_{m}}}}{{{{a'}_{M}}}}} \biggr)^{p - 1}} $$
to arrive at
$$ \frac{\mathrm{d}}{{{\mathrm{d}}t}}E(t) \le{A_{0}} + {A_{1}}E(t) + {A_{2}}E{(t)^{\frac{3}{2}}} + {A_{3}}E{(t)^{3}} + {A_{4}}E{(t)^{\frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}}. $$
(29)
An integration of the differential inequality (29) from 0 to t implies that
$$\int_{E(0)}^{E(t)} {\frac{{{\mathrm{d}}\xi}}{{{A_{0}} + {A_{1}}\xi + {A_{2}}{\xi^{\frac{3}{2}}} + {A_{3}}{\xi^{3}} + {A_{4}}{\xi^{\frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}}}}} \le t $$
from which we derive a lower bound for \(T^{*}\), that is,
$${T^{*}} \ge \int_{E(0)}^{ + \infty} {\frac{{{\mathrm{d}}\xi}}{{{A_{0}} + {A_{1}}\xi + {A_{2}}{\xi^{\frac{3}{2}}} + {A_{3}}{\xi^{3}} + {A_{4}}{\xi ^{\frac{{2 ( {\mu p + 2} ) - p}}{{2 ( {p - 2} ) ( {\mu p + 2} )}}}}}}}. $$
Thus, the proof is complete. □
Remark 3.2
Theorem 3.1 remains valid if we assume that g is a positive \({L^{p}} ( \mathbb{R}_{+} )\) function replacing the one in Assumption (A2).