In this section, we set up some conditions for blow-up to occur and derive an upper bound for blow-up time. In order to achieve this goal, we need to construct three auxiliary functions as follows:
$$\begin{aligned}& \phi(x,t)=g'(u)u_{t}-k(t)f(u), \quad (x,t) \in\overline{\Omega}\times[0,T), \end{aligned}$$
(2.1)
$$\begin{aligned}& \psi(s)= \int^{s}_{0}k(\tau)\,{\mathrm {d}}\tau, \quad s\in \overline{\mathbb{R}_{+}}, \end{aligned}$$
(2.2)
$$\begin{aligned}& \eta(s)= \int^{+\infty}_{s}\frac{g'(\tau)}{f(\tau)}\,{\mathrm {d}}\tau, \quad s\in\overline{\mathbb{R}_{+}}. \end{aligned}$$
(2.3)
Now we have
$$\psi'(s)=k(s)>0,\quad s\in\mathbb{R}_{+}, $$
which ensures the existence of the inverse function \(\psi^{-1}\) of ψ. Similarly,
$$\eta'(s)=-\frac{g'(s)}{f(s)}< 0,\quad s\in\mathbb{R}_{+} $$
implies that there exists the inverse function \(\eta^{-1}\). By using the above three auxiliary functions, we can get the following Theorem 2.1.
Theorem 2.1
Let
u
be a nonnegative classical solution of problem (1.1). Assume that the initial value
\(u_{0}\)
satisfies
$$ \nabla\cdot \bigl( \vert \nabla u_{0} \vert ^{p-2}\nabla u_{0} \bigr)\geq0,\quad x\in \overline{\Omega}. $$
(2.4)
In addition, we assume that the functions
f, g
and
k
satisfy
$$ \int^{\infty}_{M_{0}}\frac{g'(\tau)}{f(\tau)}\,{\mathrm {d}}\tau< \int^{+\infty }_{0}k(\tau)\,{\mathrm {d}}\tau, \quad M_{0}=\max_{\overline{\Omega}} u_{0}(x) $$
(2.5)
and
$$ g'(s) \biggl(\frac{f'(s)}{g'(s)} \biggr)'-f(s) \biggl(\frac{g''(s)}{ (g'(s) )^{2}} \biggr)'\geq0, \quad s\in\overline{\mathbb{R}_{+}}. $$
(2.6)
Then the solution
u
blows up at some finite time
T
and
$$\begin{aligned}& T\leq\psi \biggl( \int^{+\infty}_{M_{0}}\frac{g'(\tau)}{f(\tau)}\,{\mathrm {d}}\tau \biggr), \\& u(x,t)\leq\eta^{-1} \biggl( \int^{T}_{t}k(\tau) \, {\mathrm {d}}\tau \biggr). \end{aligned}$$
Proof
Consider the auxiliary function \(\phi(x,t)\) defined in (2.1). By direct calculation, we have
$$ \phi,{}_{i}=g''u_{t}u,{}_{i} + g'u_{t},{}_{i} - kf'u,{}_{i} $$
(2.7)
and
$$ \phi,{}_{ij}=g'''u_{t}u,{}_{i}u,{}_{j} + g''u_{t},{}_{j}u,{}_{i} + g''u_{t},{}_{i}u,{}_{j} + g''u_{t}u,{}_{ij} + g'u_{t},{}_{ij} - kf''u,{}_{i}u,{}_{j} -kf'u,{}_{ij}. $$
(2.8)
It follows from (2.8) that
$$ \Delta\phi=\phi,{}_{ii}=g''' \vert \nabla u \vert ^{2}u_{t}+2g'' (\nabla u\cdot \nabla u_{t} ) +g''u_{t} \Delta u+g'\Delta u_{t}-kf'' \vert \nabla u \vert ^{2}-kf'\Delta u. $$
(2.9)
Now we use the first equation of (1.1) to obtain
$$\begin{aligned} \phi_{t} =& \bigl[g'(u)u_{t}-k(t)f(u) \bigr]_{t}= \bigl[\bigl(g(u)\bigr)_{t}-k(t)f(u) \bigr]_{t}= \bigl[\nabla\cdot \bigl( \vert \nabla u \vert ^{p-2}\nabla u \bigr) \bigr]_{t} \\ =& \bigl[ \vert \nabla u \vert ^{p-2}\Delta u+(p-2) \vert \nabla u \vert ^{p-4}u,{}_{i}u,{}_{j}u,{}_{ij} \bigr]_{t} \\ =&(p-2) \vert \nabla u \vert ^{p-4} (\nabla u\cdot\nabla u_{t} )\Delta u+ \vert \nabla u \vert ^{p-2}\Delta u_{t} \\ &{}+(p-2) (p-4) \vert \nabla u \vert ^{p-6} (\nabla u \cdot\nabla u_{t} )u,{}_{i}u,{}_{j}u,{}_{ij} \\ &{}+2(p-2) \vert \nabla u \vert ^{p-4}u_{t},{}_{i}u,{}_{j}u,{}_{ij} +(p-2)|\nabla u|^{p-4}u,{}_{i}u,{}_{j}u_{t},{}_{ij}. \end{aligned}$$
(2.10)
Making use of (2.8), (2.9) and (2.10), we obtain
$$\begin{aligned} &\frac{|\nabla u|^{p-2}}{g'}\Delta\phi+(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{j} \phi,{}_{ij}-\phi_{t} \\ &\quad=(p-1)\frac{g'''}{g'} \vert \nabla u \vert ^{p}u_{t}+2(p-1) \frac{g''}{g'} \vert \nabla u \vert ^{p-2} (\nabla u\cdot\nabla u_{t} ) \\ &\qquad{}+\frac{g''}{g'} \vert \nabla u \vert ^{p-2}u_{t}\Delta u-(p-1)\frac{kf''}{g'} \vert \nabla u \vert ^{p} \\ &\qquad{}-\frac{kf'}{g'} \vert \nabla u \vert ^{p-2}\Delta u+(p-2)\frac{g''}{g'} \vert \nabla u \vert ^{p-4}u_{t}u,{}_{i}u,{}_{j}u,{}_{ij}-(p-2)\frac{kf'}{g'} \vert \nabla u \vert ^{p-4}u,{}_{i}u,{}_{j}u,{}_{ij} \\ &\qquad{}-(p-2) \vert \nabla u \vert ^{p-4} (\nabla u\cdot\nabla u_{t} )\Delta u -(p-2) (p-4) \vert \nabla u \vert ^{p-6} ( \nabla u\cdot\nabla u_{t} )u,{}_{i}u,{}_{j}u,{}_{ij} \\ &\qquad{}-2(p-2) \vert \nabla u \vert ^{p-4}u_{t},{}_{i}u,{}_{j}u,{}_{ij}. \end{aligned}$$
(2.11)
It follows from (2.7) that
$$ u_{t},{}_{i}=\frac{1}{g'} \phi,{}_{i}-\frac{g''}{g'}u_{t}u,{}_{i}+ \frac{kf'}{g'}u,{}_{i} $$
(2.12)
and
$$ \nabla u_{t}=\frac{1}{g'}\nabla\phi- \frac{g''}{g'}u_{t}\nabla u+\frac {kf'}{g'}\nabla u. $$
(2.13)
Substituting (2.12) and (2.13) into (2.11), we arrive at
$$\begin{aligned} &\frac{|\nabla u|^{p-2}}{g'}\Delta\phi+(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{j} \phi,{}_{ij} \\ &\qquad{}+\frac{|\nabla u|^{p-6}}{g'} \biggl((p-2) \vert \nabla u \vert ^{2}\Delta u+(p-2) (p-4)u,{}_{i}u,{}_{j}u,{}_{ij} \\ &\qquad{}-2(p-1) \frac{g''}{g'} \vert \nabla u \vert ^{4} \biggr) (\nabla u\cdot \nabla\phi ) \\ &\qquad{}+2(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{ij} \phi,{}_{i}-\phi_{t} \\ &\quad= \biggl((p-1)\frac{g'''}{g'}-2(p-1)\frac{ (g'' )^{2}}{ (g' )^{2}} \biggr) \vert \nabla u \vert ^{p}u_{t} + \biggl(2(p-1) \frac{kf'g''}{ (g' )^{2}}-(p-1)\frac{kf''}{g'} \biggr) \vert \nabla u \vert ^{p} \\ &\qquad{}+(p-1)\frac{g''}{g'} \vert \nabla u \vert ^{p-2}u_{t} \Delta u \\ &\qquad{}-(p-1)\frac {kf'}{g'} \vert \nabla u \vert ^{p-2}\Delta u +(p-1) (p-2)\frac{g''}{g'} \vert \nabla u \vert ^{p-4}u,{}_{i}u,{}_{j}u,{}_{ij}u_{t} \\ &\qquad{}-(p-1) (p-2)\frac{kf'}{g'} \vert \nabla u \vert ^{p-4}u,{}_{i}u,{}_{j}u,{}_{ij}. \end{aligned}$$
(2.14)
The first equation of (1.1) implies
$$ \vert \nabla u \vert ^{p-2}\Delta u=g'u_{t}-(p-2) \vert \nabla u \vert ^{p-4}u,{}_{i}u,{}_{j}u,{}_{ij}-kf. $$
(2.15)
Inserting (2.15) into (2.14), we derive
$$\begin{aligned} &\frac{|\nabla u|^{p-2}}{g'}\Delta\phi+(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{j} \phi,{}_{ij} \\ &\qquad{}+\frac{|\nabla u|^{p-6}}{g'} \biggl((p-2) \vert \nabla u \vert ^{2}\Delta u+(p-2) (p-4)u,{}_{i}u,{}_{j}u,{}_{ij} \\ &\qquad{}-2(p-1) \frac{g''}{g'} \vert \nabla u \vert ^{4} \biggr) (\nabla u\cdot \nabla\phi ) \\ &\qquad{}+2(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{ij} \phi,{}_{i}-\phi_{t} \\ &\quad= \biggl((p-1)\frac{g'''}{g'}-2(p-1)\frac{ (g'' )^{2}}{ (g' )^{2}} \biggr) \vert \nabla u \vert ^{p}u_{t} + \biggl(2(p-1) \frac{kf'g''}{ (g' )^{2}}-(p-1)\frac{kf''}{g'} \biggr) \vert \nabla u \vert ^{p} \\ &\qquad{}+(p-1)g'' (u_{t} )^{2}- \biggl((p-1)\frac{kfg''}{g'}+(p-1)kf' \biggr)u_{t}+(p-1)\frac{k^{2}ff'}{g'}. \end{aligned}$$
(2.16)
By (2.1), we have
$$ u_{t}=\frac{1}{g'}\phi+\frac{kf}{g'}. $$
(2.17)
Substituting (2.17) into (2.16), we deduce
$$\begin{aligned} &\frac{|\nabla u|^{p-2}}{g'}\Delta\phi+(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{j} \phi,{}_{ij} \\ &\qquad{}+\frac{|\nabla u|^{p-6}}{g'} \biggl((p-2) \vert \nabla u \vert ^{2}\Delta u+(p-2) (p-4)u,{}_{i}u,{}_{j}u,{}_{ij} \\ &\qquad{}-2(p-1) \frac{g''}{g'} \vert \nabla u \vert ^{4} \biggr) (\nabla u\cdot \nabla\phi ) \\ &\qquad{}+2(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{ij} \phi,{}_{i} \\ &\qquad{}+(p-1)\frac{1}{g'} \biggl\{ kf'-\frac{1}{g'} \biggl[g''(\phi+kf)+ \biggl(g'''-2 \frac{ (g'' )^{2}}{g'} \biggr) \vert \nabla u \vert ^{p} \biggr] \biggr\} \phi-\phi_{t} \\ &\quad=(1-p)k \biggl[g' \biggl(\frac{f'}{g'} \biggr)'-f \biggl(\frac{g''}{ (g' )^{2}} \biggr)' \biggr] \vert \nabla u \vert ^{p}. \end{aligned}$$
(2.18)
Assumption (2.6) implies that the right-hand side in equality (2.18) is nonpositive. In other words, we have
$$\begin{aligned} &\frac{|\nabla u|^{p-2}}{g'}\Delta\phi+(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{j} \phi,{}_{ij} \\ &\quad{}+\frac{|\nabla u|^{p-6}}{g'} \biggl((p-2) \vert \nabla u \vert ^{2}\Delta u+(p-2) (p-4)u,{}_{i}u,{}_{j}u,{}_{ij} \\ &\quad{}-2(p-1) \frac{g''}{g'} \vert \nabla u \vert ^{4} \biggr) (\nabla u\cdot \nabla\phi ) \\ &\quad{}+2(p-2)\frac{|\nabla u|^{p-4}}{g'}u,{}_{i}u,{}_{ij} \phi,{}_{i} \\ &\quad{}+(p-1)\frac{1}{g'} \biggl\{ kf'- \frac{1}{g'} \biggl[g''(\phi+kf) \\ &\quad{}+ \biggl(g'''-2\frac{ (g'' )^{2}}{g'} \biggr) \vert \nabla u \vert ^{p} \biggr] \biggr\} \phi-\phi_{t} \leq0 \quad \text{in } \Omega \times(0,T). \end{aligned}$$
(2.19)
Making use of the maximum principle [30], we have the following three possible cases where ϕ may take its minimum value:
-
(i)
for \(t=0\),
-
(ii)
at a point where \(|\nabla u|=0\),
-
(iii)
on the boundary \(\partial\Omega\times(0,T)\).
We first consider the first case. It follows from (2.4) that
$$\begin{aligned} \phi(x,0) =& g'(u_{0})u_{0t}-k(0)f(u_{0}) = \bigl(g (u_{0} ) \bigr)_{t}-k(0)f(u_{0}) \\ \\ =& \nabla\cdot \bigl( \vert \nabla u_{0} \vert ^{p-2}\nabla u_{0} \bigr)\geq0,\quad x\in \overline{\Omega}. \end{aligned}$$
(2.20)
Then we consider the second case. Assume that \((\bar{x},\bar{t})\in \Omega\times(0,T)\) is a point where \(|\nabla u(\bar{x},\bar{t})|=0\). Since
$$\begin{aligned} \bigl\vert \phi(x,t) \bigr\vert =& \bigl\vert \nabla\cdot \bigl( \vert \nabla u \vert ^{p-2}\nabla u \bigr) \bigr\vert = \bigl\vert \vert \nabla u \vert ^{p-2}\Delta u + \vert \nabla u \vert ^{p-4}u,{}_{i}u,{}_{j}u,{}_{ij} \bigr\vert \\ \leq& \vert \nabla u \vert ^{p-2} \vert \Delta u \vert + \vert \nabla u \vert ^{p-4} \vert \nabla u \vert \vert \nabla u \vert \vert u,{}_{ij} \vert = \vert \nabla u \vert ^{p-2} \bigl( \vert \Delta u \vert + \vert u,{}_{ij} \vert \bigr), \end{aligned}$$
the fact that \(p>2\) and \(|\nabla u(\bar{x},\bar{t})|=0\) imply
$$ \bigl\vert \phi(\bar{x},\bar{t}) \bigr\vert \leq \bigl\vert \nabla u(\bar{x},\bar{t}) \bigr\vert ^{p-2} \bigl( \bigl\vert \Delta u(\bar{x},\bar{t}) \bigr\vert + \bigl\vert u,{}_{ij}(\bar{x},\bar{t}) \bigr\vert \bigr)=0. $$
(2.21)
Hence, we have \(\phi(\bar{x},\bar{t})=0\). Finally, we consider the third case. Applying the boundary condition of (1.1), we get
$$ \frac{\partial\phi}{\partial n}= g''(u)u_{t} \frac{\partial u}{\partial n}+g'\frac{\partial u_{t}}{\partial n} -f'(u) \frac{\partial u}{\partial n} =g'(u) \biggl(\frac{\partial u}{\partial n} \biggr)_{t}=0\quad \text{on } \partial\Omega\times(0,T). $$
(2.22)
Combining (2.20)-(2.22) and the maximum principle, we now get that the minimum of ϕ in \(\overline{\Omega}\times[0,T)\) is nonnegative. Thus
$$\phi\geq0 \quad \text{in } \overline{\Omega}\times[0,T); $$
that is,
$$ \frac{g'(u)}{f(u)}u_{t}\geq k(t). $$
(2.23)
At the point \(x^{*}\in\overline{\Omega}\), where \(u_{0}(x^{*})=M_{0}\), we integrate (2.23) over \([0,t]\) to obtain
$$ \int^{t}_{0}\frac{g'(u)}{f(u)}u_{t} \, { \mathrm {d}}t = \int^{u(x^{*},t)}_{M_{0}}\frac{g'(\tau)}{f(\tau)} \, {\mathrm {d}}\tau\geq \int ^{t}_{0}k(\tau) \, {\mathrm {d}}\tau, $$
(2.24)
which ensures that u blows up in some finite time T. In fact, suppose that u remains global, then for any \(t>0\), we have
$$ \int^{+\infty}_{M_{0}}\frac{g'(\tau)}{f(\tau)} \, {\mathrm {d}}\tau\geq \int ^{u(x^{*},t)}_{M_{0}}\frac{g'(\tau)}{f(\tau)} \, {\mathrm {d}}\tau \geq \int^{t}_{0}k(\tau) \, {\mathrm {d}}\tau. $$
(2.25)
Letting \(t\rightarrow+\infty\) in (2.25), we deduce
$$\int^{+\infty}_{M_{0}}\frac{g'(\tau)}{f(\tau)}\,{\mathrm {d}}\tau\geq \int^{+\infty }_{0}k(\tau)\,{\mathrm {d}}\tau, $$
which contradicts assumption (2.5). This shows that u blows up in some finite time T. Furthermore, taking the limit as \(t\rightarrow T^{-}\) in (2.24), we arrive at
$$\lim_{t\rightarrow T^{-}} \int^{u(x^{*},t)}_{M_{0}}\frac{g'(\tau )}{f(\tau)}\,{\mathrm {d}}\tau\geq \lim_{t\rightarrow T^{-}} \int^{t}_{0}k(\tau)\,{\mathrm {d}}\tau $$
and
$$ \int^{+\infty}_{M_{0}}\frac{g'(\tau)}{f(\tau)}\,{\mathrm {d}}\tau\geq \int ^{T}_{0}k(\tau)\,{\mathrm {d}}\tau=\psi(T). $$
(2.26)
It follows from (2.26) that
$$T\leq\psi^{-1} \biggl( \int^{+\infty}_{M_{0}}\frac{g'(\tau)}{f(\tau)} \, {\mathrm {d}}\tau \biggr). $$
For each fixed x, we integrate inequality (2.23) over \([t,t^{*}]\)
\((0< t< t^{*}< T)\) to obtain
$$ \eta\bigl(u(x,t)\bigr)\geq\eta\bigl(u(x,t)\bigr)-\eta\bigl(u \bigl(x,t^{*}\bigr)\bigr) = \int^{u(x,t^{*})}_{u(x,t)}\frac{g'(\tau)}{f(\tau)} \, {\mathrm {d}}\tau \geq \int^{t^{*}}_{t}k(\tau) \, {\mathrm {d}}\tau. $$
(2.27)
Letting \(t^{*}\rightarrow T^{-}\) in (2.27), we deduce
$$\eta\bigl(u(x,t)\bigr)\geq \int^{T}_{t}k(\tau)\, {\mathrm {d}}\tau, $$
from which we have
$$u(x,t)\leq\eta^{-1} \biggl( \int^{T}_{t}k(\tau) \, {\mathrm {d}}\tau \biggr). $$
The proof of Theorem 2.1 is complete. □
Remark 2.1
When
$$\int^{+\infty}_{0}k(\tau) \, {\mathrm {d}}\tau=+\infty, $$
assumption (2.5) means
$$\int^{+\infty}_{M_{0}}\frac{g'(\tau)}{f(\tau)} \, {\mathrm {d}}\tau< + \infty, \quad M_{0}=\max_{\overline{\Omega}}u_{0}(x). $$