Blow-up phenomena for a nonlinear parabolic problem with p-Laplacian operator under nonlinear boundary condition

Abstract

In this paper, we study the blow-up phenomena for a positive solution of a nonlinear parabolic problem with p-Laplacian operator under a nonlinear boundary condition. The sufficient conditions which ensure that the blow-up does occur at finite time are presented by constructing some appropriate auxiliary functions and using first-order differential inequality technique. Moreover, a lower bound and an upper bound for the blow-up time are derived when blow-up happens.

Introduction

The mathematical investigation of the blow-up phenomena of a solution to nonlinear parabolic equations and systems has received a great deal of attention during the last few decades [1–6]. The authors in [7, 8] considered an initial-boundary value problem for parabolic equations of the form

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{{\partial u}}{{\partial t}} = \Delta u + {u^{p}} - {\vert {\nabla u} \vert ^{q}} &\mbox{in } \mathcal{O} \times(0,\infty),\\ u = 0 &\mbox{on } \partial\mathcal{O} \times(0,\infty),\\ u(x,0) = h(x) \ge0 &\mbox{in } \mathcal{O}. \end{array}\displaystyle \right . $$

(1)

Here $\mathcal{O}$ is a bounded domain in $\mathbb{R}^{3}$, △ is the Laplace operator, ∇ is the gradient operator, $\partial\mathcal{O}$ is the boundary of $\mathcal{O}$. They proved that problem (1) blows up at finite time $T^{*}$ if $1 < p \le5$ and $1 < q < \frac{{2p}}{{p + 1}}$. Soon et al. in [1] gave a lower bound for the blow-up time $T^{*}$ under the above condition. Shortly afterwards, the relative result in [1] was extended to the case with nonlinear boundary condition by Liu [9]. Further, Enache in [10] considered a more complicated case, in which he investigated the following class of quasilinear initial-boundary value problems:

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} {u_{t}} = \operatorname{div} ( {b(u)\nabla u} ) + f(u) &\mbox{in }\mathcal{O} \times(0,\infty),\\ \frac{{\partial u}}{{\partial n}} + \kappa u = 0 &\mbox{on } \partial\mathcal{O} \times(0,\infty),\\ u(x,0) = h(x) \ge0 &\mbox{in } \mathcal{O}. \end{array}\displaystyle \right . $$

(2)

Here n is the unit outer normal vector of $\partial\mathcal{O}$, and $\frac{{\partial u}}{{\partial n}}$ is outward normal derivative of u on the boundary $\partial\mathcal{O}$ which is assumed to be sufficiently smooth. Under the suitable assumptions on the functions b, f, and h, the author established a sufficient condition to guarantee the occurrence of the blow-up. Moreover, a lower bound for the blow-up time was obtained.

However, there are few papers on blow-up phenomena of the problem with a p-Laplacian operator except [11], in which Zhou considered the following:

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} {u_{t}} = \operatorname{div} ( {u{{\vert {\nabla u} \vert }^{p - 2}}\nabla u} ) + (\gamma + 1){\vert {\nabla u} \vert ^{p}} &\mbox{in }\mathcal{O} \times(0,\infty),\\ \frac{{\partial u}}{{\partial n}} = 0 &\mbox{on } \partial\mathcal{O} \times (0,\infty),\\ u(x,0) = h(x) \ge0 & \mbox{in } \mathcal{O}. \end{array}\displaystyle \right . $$

(3)

He proved that problem (3) blows up at finite time $T^{*}$ when $0 < \gamma < 1$. But he did not give any bounds to the scale ${T^{*}}$.

In this text, we consider the more complicated case than the ones in (1)-(3),

$$ { \bigl( {a(u)} \bigr)_{t}} = \operatorname{div} \bigl( {b(u){{\vert {\nabla u} \vert }^{p - 2}}\nabla u} \bigr) + \gamma b'(u){\vert { \nabla u} \vert ^{p}} + f(u) $$

(4)

with the following nonlinear boundary condition:

$$ \frac{{\partial u}}{{\partial n}} + g(u) = 0 $$

(5)

and the initial condition

$$ u(x,0) = h(x) \ge0. $$

(6)

In the process of deriving the lower bound, we make the following assumptions:

• (A1) The parameters of problem (4) satisfy $0 \le\gamma \le2$, $p > 2$.

• (A2) The function $g(s)$ satisfies

  $$g(s) = \sum_{i = 1}^{n} {{ \kappa_{i}} {s^{{\sigma_{i}}}}}, $$

  where ${\kappa_{i}}$s and ${\sigma_{i}}$s are nonnegative constants.

Since the initial data $h(x)$ in (6) is nonnegative, it is easy to see that the solution u to problem (4)-(6) is nonnegative in $\mathcal{O} \times ( {0,\infty} )$ by the parabolic maximum principles [12, 13]. In Section 2, we plan to present the sufficient conditions which guarantee the occurrence of the blow-up. In Section 3, we will find a lower bound for the blow-up time when blow-up occurs.

The blow-up solution

In this section we mainly seek the sufficient conditions for the blow-up. To this end, we define some auxiliary functions of the form

$$\begin{aligned} &G(s) = 2 \int_{0}^{s} {yb{{(y)}^{ ( {p - 1} )p - 1}}a'(y)\,\mathrm{d}y}, \\ &A(t) = \int_{O} {G \bigl( {u(x,t)} \bigr)\,\mathrm{d}x}, \\ &{H_{i}}(s) = \int_{0}^{s} {{y^{p{\sigma_{i}} - {\sigma_{i}}}}b{{(y)}^{p(p - 1)}} \,\mathrm{d}y} , \quad i = 1,2, \ldots,n,\\ &\sigma = \max \{ {{\sigma_{i}}, i = 1,2, \ldots,n} \},\qquad F(s) = \int_{0}^{s} {f(s)b{{(s)}^{ ( {p - 1} )p - 1}} \,\mathrm{d}s}, \\ &B(t) = \int_{\mathcal{O}} {F(u)\,\mathrm{d}x} - \frac{1}{p} \int _{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2}}} \,\mathrm{d}x} - \sum _{i = 1}^{n} {\kappa_{i}^{p - 1} \int _{\partial\mathcal{O}} {{H_{i}} ( u )\,\mathrm{d}x,} } \end{aligned}$$

(7)

where $u(x,t)$ is the solution of problem (3).

The main result of this section is formulated in the following theorem.

Theorem 2.1

Let $u(x,t)$ be the solution of problem (4)-(6). Assume that

$$\begin{aligned}& sf(s)b{(s)^{ ( {p - 1} )p - 1}} \ge p(1 + \alpha)F(s),\quad s>0, \end{aligned}$$

(8)

$$\begin{aligned}& \lim_{y \to\infty} {y^{\sigma p - \sigma + 1}}b{(y)^{p(p - 1)}} = 0 \quad \textit{and}\quad B(0) \ge0, \end{aligned}$$

(9)

where α is a positive constant. Then $u(x,t)$ blows up as some finite time $T^{*}$ such that

$${T^{*}} \le{M^{ - 1}}A{(0)^{1 - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}, $$

where M is a positive constant to be determined later.

Proof

We first compute

$$\begin{aligned} A'(t) ={}& \int_{\mathcal{O}} {G' \bigl( {u(x,t)} \bigr){u_{t}} \,\mathrm{d}x} \\ = {}&2 \int_{\mathcal{O}} {ub{{(u)}^{ ( {p - 1} )p - 1}} \bigl[ {\operatorname{div} \bigl( {b(u){{\vert {\nabla u} \vert }^{p - 2}}\nabla u} \bigr) + \gamma b'(u){{\vert {\nabla u} \vert }^{p}} + f(u)} \bigr]\,\mathrm{d}x} \\ = {}&2 \int_{\mathcal{O}} {uf(u)b{{(u)}^{ ( {p - 1} )p - 1}} {\,\mathrm {d}}x} \\ &{} + \bigl[ {\gamma - 2 \bigl( { ( {p - 1} )p - 1} \bigr)} \bigr] \int_{\mathcal{O}} {ub{{(u)}^{ ( {p - 1} )p - 1}}b'(u){{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2}}} \,\mathrm{d}x} \\ &{} - 2 \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2}}} \,\mathrm{d}x} - 2\sum _{i = 1}^{n} {\kappa_{i}^{p - 1} \int_{\partial\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {u^{p{\sigma_{i}} - {\sigma_{i}} + 1}} \,\mathrm{d}x} } . \end{aligned}$$

Noting that $b' \le0$ and $\gamma \le2$, we drop the nonnegative terms to obtain

$$\begin{aligned} A'(t) \ge{}&2 \int_{\mathcal{O}} {uf(u)b{{(u)}^{ ( {p - 1} )p - 1}} \,\mathrm{d}x} - 2 \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2}}} \,\mathrm{d}x} \\ &{} - 2\sum_{i = 1}^{n} { \kappa_{i}^{p - 1} \int _{\partial\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {u^{p{\sigma _{i}} - {\sigma_{i}} + 1}} \,\mathrm{d}x} } . \end{aligned}$$

(10)

Next, we prove

$$ ( {p{\sigma_{i}} - {\sigma_{i}} + 1} )H ( u ) \ge {u^{p{\sigma_{i}} - {\sigma_{i}} + 1}}b{(u)^{p(p - 1)}}. $$

(11)

Use the method of integration by parts and consider condition (9). Then we obtain

$$\begin{aligned} {H_{i}} ( u ) ={}& \int_{0}^{u} {{y^{p{\sigma_{i}} - {\sigma _{i}}}}b{{(y)}^{p(p - 1)}} \,\mathrm{d}y} \\ ={}& {y^{p{\sigma_{i}} - {\sigma_{i}} + 1}}b{(y)^{p(p - 1)}} \int_{0}^{u} {} - (p{\sigma_{i}} - { \sigma_{i}}) \int_{0}^{u} {{y^{p{\sigma _{i}} - {\sigma_{i}}}}b{{(y)}^{p(p - 1)}} \,\mathrm{d}y} \\ &{}- p(p - 1) \int_{0}^{u} {{y^{p}}b{{(y)}^{p(p - 1) - 1}}b'(y)\,\mathrm{d}y} \\ \ge{}&{u^{p{\sigma_{i}} - {\sigma_{i}} + 1}}b{(u)^{p(p - 1)}} - (p{\sigma_{i}} - { \sigma_{i}}) \int_{0}^{u} {{y^{p{\sigma_{i}} - {\sigma _{i}}}}b{{(y)}^{p(p - 1)}} \,\mathrm{d}y} \\ ={}& {u^{p{\sigma_{i}} - {\sigma_{i}} + 1}}b{(u)^{p(p - 1)}} - (p{\sigma_{i}} - { \sigma_{i}}){H_{i}} ( u ). \end{aligned}$$

Thus, we prove (11). Further, inserting (8) and (11) into (10) gives

$$\begin{aligned} A'(t) \ge{}&2 ( {p\sigma - \sigma + 1} ) (1 + \alpha) \int _{\mathcal{O}} {F(u)\,\mathrm{d}x} \\ &{} - 2(1 + \alpha) \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2}}} {\,\mathrm {d}}x} \\ &{} - 2 ( {p\sigma - \sigma + 1} ) (1 + \alpha)\sum _{i = 1}^{n} {\kappa_{i}^{p - 1} \int_{\partial\mathcal {O}} {{H_{i}} ( u )\,\mathrm{d}x} } \\ \ge{}&2 ( {p\sigma - \sigma + 1} ) (1 + \alpha)B(t). \end{aligned}$$

(12)

On the other hand, computing $B(t)$ in (12) gives

$$\begin{aligned} B'(t) ={}& \int_{\mathcal{O}} {f(u)b{{(u)}^{ ( {p - 1} )p - 1}} {u_{t}} \,\mathrm{d}x} \\ &{} - ( {p - 1} ) \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p - 1}}b'(u){u_{t}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2}}} \,\mathrm{d}x} \\ &{} - \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2} - 1}}\nabla u\nabla{u_{t}} \,\mathrm{d}x} \\ &{}- \sum_{i = 1}^{n} {\kappa_{i}^{p - 1} \int _{\partial\mathcal{O}} {{{H'}_{i}} ( u ){u_{t}} \,\mathrm{d}x} } \\ = {}& \int_{\mathcal{O}} {f(u)b{{(u)}^{ ( {p - 1} )p - 1}} {u_{t}} \,\mathrm{d}x} \\ &{} - ( {p - 1} ) \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p - 1}}b'(u){u_{t}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2}}} \,\mathrm{d}x} \\ &{} - \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p}} {{ \bigl[ {{{(\nabla u)}^{2}}} \bigr]}^{\frac{p}{2} - 1}}\nabla u\nabla{u_{t}} \,\mathrm{d}x} \\ &{}- \sum_{i = 1}^{n} {\kappa_{i}^{p - 1} \int _{\partial\mathcal{O}} {{u^{p{\sigma_{i}} - {\sigma_{i}}}}b{{(u)}^{p(p - 1)}} {u_{t}} \,\mathrm{d}x} } \\ = {}& \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p - 1}} {u_{t}} \bigl\{ {f(u) + b'(u){{ \bigl( {{{(\nabla u)}^{2}}} \bigr)}^{\frac {p}{2}}}} } \\ &{} { + b(u) \cdot{\operatorname{div}} \bigl[ {{{ \bigl( {{{(\nabla u)}^{2}}} \bigr)}^{\frac{p}{2}}}} \bigr]} \bigr\} \,\mathrm{d}x \\ \ge{}& \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p - 1}} {u_{t}} \bigl\{ {f(u) + ( {\gamma + 1} )b'(u){{ \bigl( {{{(\nabla u)}^{2}}} \bigr)}^{\frac{p}{2}}}} } \\ &{} { + b(u) \cdot{\operatorname{div}} \bigl[ {{{ \bigl( {{{(\nabla u)}^{2}}} \bigr)}^{\frac{p}{2}}}} \bigr]} \bigr\} \,\mathrm{d}x \\ ={}& \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p - 1}} {u_{t}} {{ \bigl( {a(u)} \bigr)}_{t}} \,\mathrm{d}x} \\ ={}& \int_{\mathcal{O}} {b{{(u)}^{ ( {p - 1} )p - 1}}a'(u){{ ( {{u_{t}}} )}^{2}} \,\mathrm{d}x.} \end{aligned}$$

Since $a' > 0$ and $B(0) \ge0$, we see that $B(t)$ is a nondecreasing function satisfying

$$B(t) \ge0. $$

Multiplying (12) by $B(t)$ and using the Hölder inequality, we obtain

$$\begin{aligned} 0 &\le(1 + \alpha)A'(t)B(t) \\ &\le\frac{1}{{2 ( {p\sigma - \sigma + 1} )}}{ \bigl( {A'(t)} \bigr)^{2}} \\ & = \frac{2}{{ ( {p\sigma - \sigma + 1} )}}{ \biggl( { \int_{\mathcal{O}} {ub{{(u)}^{ ( {p - 1} )p - 1}}a'(u){u_{t}} {\,\mathrm {d}}x} } \biggr)^{2}} \\ & \le\frac{2}{{ ( {p\sigma - \sigma + 1} )}}B'(t) \biggl( { \int_{\mathcal{O}} {ub{{(u)}^{ ( {p - 1} )p - 1}}a'(u){u^{2}} \,\mathrm{d}x} } \biggr). \end{aligned}$$

(13)

We further prove that

$$ G(u) \ge{u^{2}}b{(u)^{ ( {p - 1} )p - 1}}a'(u). $$

(14)

Noting $b' \le0$, $a' > 0$, and $a'' \le0$, and using the method of integration by parts, we derive

$$\begin{aligned} G(u) ={}& {s^{2}}b{(s)^{ ( {p - 1} )p - 1}}a'(s) \int_{0}^{u} {} - \int _{0}^{u} {sb{{(s)}^{ ( {p - 1} )p - 1}}a'(s)\,\mathrm{d}s} \\ &{} - \bigl( { ( {p - 1} )p - 1} \bigr) \int_{0}^{u} {{s^{2}}b{{(s)}^{ ( {p - 1} )p - 2}}b'(s)a'(s)\,\mathrm{d}s} \\ &{} - \int_{0}^{u} {{s^{2}}b{{(s)}^{ ( {p - 1} )p - 1}}a''(s)\,\mathrm{d}s} \\ \ge{}&{u^{2}}b{(u)^{ ( {p - 1} )p - 1}}a'(u) - G(u). \end{aligned}$$

Thus, we prove (14) and substitute it into (13). Then we get

$$\begin{aligned} (1 + \alpha)A'(t)B(t) &\le\frac{2}{{p\sigma - \sigma + 1}}B'(t) \biggl( { \int_{\mathcal{O}} {G(u)\,\mathrm{d}x} } \biggr) \\ & = \frac{2}{{p\sigma - \sigma + 1}}B'(t)A(t), \end{aligned}$$

which leads to

$$ \frac{\mathrm{d}}{{\mathrm{d}t}} \bigl( {{A^{ - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}B} \bigr) \ge0. $$

(15)

Integrating (15) from 0 to t gives

$$\frac{{B(t)}}{{B(0)}} \ge{ \biggl( {\frac{{A(t)}}{{A(0)}}} \biggr)^{\frac {1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}. $$

This and (12) imply that

$$\begin{aligned} A'(t) \ge{}&2 ( {p\sigma - \sigma + 1} ) (1 + \alpha)B(0) \\ &{} \cdot A{(0)^{ - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}A{(t)^{\frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}} \end{aligned}$$

or

$$ \frac{{A'(t)}}{{A{{(t)}^{\frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}}} \ge2 ( {p\sigma - \sigma + 1} ) (1 + \alpha)B(0)A{(0)^{ - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}. $$

(16)

Use the fact that $p > 2$, $\sigma > 0$ and integrate (16) from 0 to t. Then we deduce that

$$\begin{aligned} &A{(t)^{1 - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}} \\ &\quad\le A{(0)^{1 - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}} - Mt, \end{aligned}$$

(17)

where

$$\begin{aligned} M ={}& 2 \biggl[ {\frac{1}{2} ( {p\sigma - \sigma + 1} ) (1 + \alpha) - 1} \biggr] ( {p\sigma - \sigma + 1} ) \\ &{}\cdot(1 + \alpha)B(0)A{(0)^{ - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}. \end{aligned}$$

Inequality (17) cannot hold for $A{(0)^{1 - \frac{p}{2}(1 + \alpha)}} - Mt \le0$, that is, for

$$t \ge{M^{ - 1}}A{(0)^{1 - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}. $$

Hence, we conclude that the solution u of problem (4)-(6) blows up at some finite time $T^{*}$ with upper bound ${M^{ - 1}}A{(0)^{1 - \frac{1}{2} ( {p\sigma - \sigma + 1} )(1 + \alpha)}}$. The proof is complete. □

Lower bound for blow-up time

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).