Due to the need to study the blow-up solution of (1.1), we define the following two constants:
$$\begin{aligned}& \alpha =\min_{x\in \overline{\varOmega }} \frac{\nabla \cdot ( \vert \nabla u_{0} \vert ^{p-2}\nabla u_{0} )+f(x,u_{0}, \vert \nabla u_{0} \vert ^{2},0)}{{\mathrm{e}}^{u_{0}}}, \end{aligned}$$
(2.1)
$$\begin{aligned}& \beta =\inf_{(x,s,t)\in \varOmega \times \overline{\mathbb{R}_{+}}\times \mathbb{R}_{+}}{\frac{f(x,s,0,t)}{{\mathrm{e}}^{s}},} \end{aligned}$$
(2.2)
where \(u_{0}\) is the initial value of (1.1). For the same purpose, we also construct two auxiliary functions as follows:
$$\begin{aligned}& G(x,t)=b'(u)u_{t}-\alpha { \mathrm{e}}^{u}, \quad (x,t)\in \overline{\varOmega } \times [0,t^{*}), \end{aligned}$$
(2.3)
$$\begin{aligned}& H(s)= \int ^{+\infty }_{s}\frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau , \quad s\in \overline{\mathbb{R}_{+}}, \end{aligned}$$
(2.4)
where \(u(x,t)\) is a nonnegative \(C^{3}(\varOmega \times (0,t^{*}))\cap C^{2}(\overline{\varOmega }\times [0,t^{*}))\) solution of (1.1). Now we have
$$ H'(s)=-\frac{b'(s)}{{\mathrm{e}}^{s}}< 0, \quad s\in \mathbb{R}_{+}, $$
which implies that the function H has an inverse function \(H^{-1}\). The following Theorem 2.1 is the main result of the blow-up solution to (1.1).
Theorem 2.1
Let u be a nonnegative \(C^{3}(\varOmega \times (0,t^{*}))\cap C^{2}(\overline{\varOmega }\times [0,t^{*}))\)solution of (1.1). Assume the following four assumptions are true:
-
(i)
$$ \beta \geq \alpha >0. $$
(2.5)
-
(ii)
$$ \int ^{+\infty }_{M_{0}}\frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau < + \infty , \quad M_{0}=\max_{x\in \overline{\varOmega }}u_{0}(x). $$
(2.6)
-
(iii)
For \(s\in \overline{\mathbb{R}_{+}}\),
$$ (p-1) \biggl(\frac{1}{b'(s)} \biggr)'+(p-2) \frac{1}{b'(s)}\geq 0, \qquad \biggl(\frac{1}{b'(s)} \biggr)''+2 \biggl(\frac{1}{b'(s)} \biggr)'+ \frac{1}{b'(s)}\geq 0. $$
(2.7)
-
(iv)
For \((x,s,r,t)\in \varOmega \times \overline{\mathbb{R}_{+}}\times \overline{\mathbb{R}_{+}}\times \mathbb{R}_{+}\),
$$ \begin{gathered}[b] f_{r}(x,s,r,t)\geq 0, \qquad f_{t}(x,s,r,t)\geq 0, \\ \frac{f_{s}(x,s,r,t)}{b'(s)}-(p-1)f(x,s,r,t) \biggl[ \biggl( \frac{1}{b'(s)} \biggr)'+\frac{1}{b'(s)} \biggr]\geq 0. \end{gathered} $$
(2.8)
Then, \(u(x,t)\)must blow up at a finite time \(t^{*}\)and
$$ t^{*}\leq \frac{1}{\alpha } \int ^{+\infty }_{M_{0}} \frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau , $$
as well as
$$ u(x,t)\leq H^{-1} \bigl(\alpha \bigl(t^{*}-t\bigr) \bigr), \quad (x,t)\in \overline{\varOmega }\times [0,t^{*}). $$
Proof
By directly calculating the auxiliary function \(G(x,t)\) defined in (2.3), we have
$$ G,_{i}=b''u_{t}u,_{i}+b'u_{t},_{i}- \alpha {\mathrm{e}}^{u}u,_{i} $$
(2.9)
and
$$ G,_{ij}=b'''u_{t}u,_{i}u,_{j}+b''u_{t},_{j}u,_{i}+b''u_{t},_{i}u,_{j}+b''u_{t}u,_{ij}+b'u_{t},_{ij} -\alpha {\mathrm{e}}^{u}u,_{i}u,_{j}-\alpha { \mathrm{e}}^{u}u,_{ij}. $$
(2.10)
By (2.10), we get
$$ \Delta G=G_{ii}=b''' \vert \nabla u \vert ^{2}u_{t}+2b'' (\nabla u\cdot \nabla u_{t} )+b''u_{t} \Delta u+b'\Delta u_{t} -\alpha {\mathrm{e}}^{u} \vert \nabla u \vert ^{2}-\alpha {\mathrm{e}}^{u}\Delta u. $$
(2.11)
Making use of the first equation of (1.1), we derive
$$ \begin{aligned}[b] G_{t}&= \bigl(b'(u)u_{t} \bigr)_{t}-\alpha {\mathrm{e}}^{u}u_{t}= \bigl[ \bigl(b(u) \bigr)_{t} \bigr]_{t}-\alpha { \mathrm{e}}^{u}u_{t} \\ &= \bigl[ \nabla \cdot \bigl( \vert \nabla u \vert ^{p-2}\nabla u \bigr)+f(x,u,q,t) \bigr]_{t}- \alpha {\mathrm{e}}^{u}u_{t} \\ &= \bigl[ \vert \nabla u \vert ^{p-2} \Delta u+(p-2) \vert \nabla u \vert ^{p-4}u,_{i}u,_{j}u,_{ij}+f(x,u,q,t) \bigr]_{t}- \alpha {\mathrm{e}}^{u}u_{t} \\ &=(p-2) \vert \nabla u \vert ^{p-4} ( \nabla u\cdot \nabla u_{t} )\Delta u+ \vert \nabla u \vert ^{p-2}\Delta u_{t} \\ &\quad {}+(p-2) (p-4) \vert \nabla u \vert ^{p-6} (\nabla u\cdot \nabla u_{t} )u,_{i}u,_{j}u,_{ij} +2(p-2) \vert \nabla u \vert ^{p-4}u_{t},_{i}u,_{j}u,_{ij} \\ &\quad {}+(p-2) \vert \nabla u \vert ^{p-4}u,_{i}u,_{j}u_{t},_{ij} +f_{u}u_{t}+2f_{q} (\nabla u\cdot \nabla u_{t} ) +f_{t}-\alpha { \mathrm{e}}^{u}u_{t}, \end{aligned} $$
(2.12)
where \(q=|\nabla u|^{2}\). It follows from (2.10)–(2.12) that
$$ \begin{aligned}[b] &\frac{ \vert \nabla u \vert ^{p-2}}{b'}\Delta G+(p-2) \frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{j}G,_{ij}-G_{t} \\ &\quad =(p-1) \frac{b'''}{b'} \vert \nabla u \vert ^{p}u_{t} +2(p-1)\frac{b''}{b'} \vert \nabla u \vert ^{p-2} (\nabla u\cdot \nabla u_{t} ) +\frac{b''}{b'} \vert \nabla u \vert ^{p-2}u_{t} \Delta u \\ &\qquad {}-\alpha (p-1)\frac{{\mathrm{e}}^{u}}{b'} \vert \nabla u \vert ^{p}- \alpha \frac{{\mathrm{e}}^{u}}{b'} \vert \nabla u \vert ^{p-2} \Delta u +(p-2) \frac{b''}{b'} \vert \nabla u \vert ^{p-4}u_{t}u,_{i}u,_{j}u,_{ij} \\ &\qquad {}- \alpha (p-2)\frac{{\mathrm{e}}^{u}}{b'} \vert \nabla u \vert ^{p-4}u,_{i}u,_{j}u,_{ij} -(p-2) \vert \nabla u \vert ^{p-4} (\nabla u\cdot \nabla u_{t} )\Delta u \\ &\qquad {}-(p-2) (p-4) \vert \nabla u \vert ^{p-6} (\nabla u\cdot \nabla u_{t} )u,_{i}u,_{j}u,_{ij}-2(p-2) \vert \nabla u \vert ^{p-4}u_{t},_{i}u,_{j}u,_{ij} \\ &\qquad {}+ \bigl(\alpha {\mathrm{e}}^{u}-f_{u} \bigr)u_{t}-2f_{q} (\nabla u\cdot \nabla u_{t} )-f_{t}. \end{aligned} $$
(2.13)
With (2.9), we obtain
$$ u_{t},_{i}=\frac{1}{b'}G,_{i}- \frac{b''}{b'}u_{t}u,_{i}+\alpha \frac{{\mathrm{e}}^{u}}{b'}u,_{i} $$
(2.14)
and
$$ \nabla u_{t}=\frac{1}{b'}\nabla G- \frac{b''}{b'}u_{t}\nabla u+\alpha \frac{{\mathrm{e}}^{u}}{b'}\nabla u. $$
(2.15)
We insert (2.14) and (2.15) into (2.13) to derive
$$\begin{aligned} \begin{aligned}[b] &\frac{ \vert \nabla u \vert ^{p-2}}{b'}\Delta G+(p-2) \frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{j}G,_{ij} \\ &\qquad {}+ \frac{1}{b'} \biggl((p-2) (p-4) \vert \nabla u \vert ^{p-6}u,_{i}u,_{j}u,_{ij} +(p-2) \vert \nabla u \vert ^{p-4}\Delta u -2(p-1)\frac{b''}{b'} \vert \nabla u \vert ^{p-2} \\ &\qquad {}+2f_{q} \biggr) (\nabla u\cdot \nabla G )+2(p-2)\frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{ij}G,_{i}-G_{t} \\ &\quad = \biggl((p-1)\frac{b'''}{b'}-2(p-1) \frac{(b'')^{2}}{(b')^{2}} \biggr) \vert \nabla u \vert ^{p}u_{t} + \biggl(2 \alpha (p-1) \frac{b''{\mathrm{e}}^{u}}{(b')^{2}}-\alpha (p-1) \frac{{\mathrm{e}}^{u}}{b'} \biggr) \vert \nabla u \vert ^{p} \\ &\qquad {}+(p-1) \frac{b''}{b'} \vert \nabla u \vert ^{p-2}u_{t}\Delta u-\alpha (p-1) \frac{{\mathrm{e}}^{u}}{b'} \vert \nabla u \vert ^{p-2}\Delta u \\ &\qquad {}+(p-1) (p-2) \frac{b''}{b'} \vert \nabla u \vert ^{p-4}u_{t}u,_{i}u,_{j}u,_{ij} -\alpha (p-1) (p-2) \frac{{\mathrm{e}}^{u}}{b'} \vert \nabla u \vert ^{p-4}u,_{i}u,_{j}u,_{ij} \\ &\qquad {}+ \bigl(\alpha {\mathrm{e}}^{u}-f_{u} \bigr)u_{t}+2\frac{f_{q}b''}{b'} \vert \nabla u \vert ^{2}u_{t} -2\alpha \frac{f_{q}{\mathrm{e}}^{u}}{b'} \vert \nabla u \vert ^{2}-f_{t}. \end{aligned} \end{aligned}$$
(2.16)
By the first equation of (1.1), we have
$$ \vert \nabla u \vert ^{p-2}\Delta u=b'u_{t}-(p-2) \vert \nabla u \vert ^{p-4}u,_{i}u,_{j}u,_{ij}-f. $$
(2.17)
We insert (2.17) into (2.16) to get
$$\begin{aligned} \begin{aligned}[b] &\frac{ \vert \nabla u \vert ^{p-2}}{b'}\Delta G+(p-2) \frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{j}G,_{ij} \\ &\qquad {}+ \frac{1}{b'} \biggl((p-2) (p-4) \vert \nabla u \vert ^{p-6}u,_{i}u,_{j}u,_{ij} +(p-2) \vert \nabla u \vert ^{p-4}\Delta u -2(p-1)\frac{b''}{b'} \vert \nabla u \vert ^{p-2} \\ &\qquad {}+2f_{q} \biggr) (\nabla u\cdot \nabla G )+2(p-2)\frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{ij}G,_{i}-G_{t} \\ &\quad = \biggl((p-1)\frac{b'''}{b'}-2(p-1) \frac{(b'')^{2}}{(b')^{2}} \biggr) \vert \nabla u \vert ^{p}u_{t} + \biggl(2 \alpha (p-1) \frac{b''{\mathrm{e}}^{u}}{(b')^{2}}-\alpha (p-1) \frac{{\mathrm{e}}^{u}}{b'} \biggr) \vert \nabla u \vert ^{p} \\ &\qquad {}+(p-1)b''(u_{t})^{2}- \biggl(\alpha (p-2){\mathrm{e}}^{u} +f_{u}+(p-1) \frac{fb''}{b'} \biggr)u_{t} \\ &\qquad {}+\alpha (p-1)\frac{f{\mathrm{e}}^{u}}{b'}+2 \frac{f_{d}b''}{b'} \vert \nabla u \vert ^{2}u_{t} -2\alpha \frac{f_{d}{\mathrm{e}}^{u}}{b'} \vert \nabla u \vert ^{2}-f_{t}. \end{aligned} \end{aligned}$$
(2.18)
It follows from (2.3) that
$$ u_{t}=\frac{1}{b'}G+\alpha \frac{{\mathrm{e}}^{u}}{b'}. $$
(2.19)
We insert (2.19) into (2.18) to obtain
$$\begin{aligned} \begin{aligned}[b] &\frac{ \vert \nabla u \vert ^{p-2}}{b'}\Delta G+(p-2) \frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{j}G,_{ij} \\ &\qquad {}+ \frac{1}{b'} \biggl((p-2) (p-4) \vert \nabla u \vert ^{p-6}u,_{i}u,_{j}u,_{ij} +(p-2) \vert \nabla u \vert ^{p-4}\Delta u -2(p-1)\frac{b''}{b'} \vert \nabla u \vert ^{p-2} \\ &\qquad {}+2f_{q} \biggr) (\nabla u\cdot \nabla G )+2(p-2)\frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{ij}G,_{i} \\ &\qquad {}+ \biggl\{ \frac{1}{(b')^{2}} \biggl[ \biggl(2(p-1) \frac{(b'')^{2}}{b'}-(p-1)b''' \biggr) \vert \nabla u \vert ^{p}-2f_{q}b'' \vert \nabla u \vert ^{2} \biggr] \\ &\qquad {}+\frac{1}{b'} \biggl[ \alpha {\mathrm{e}}^{u} \biggl(p-2-2(p-1)\frac{b''}{b'} \biggr)+f_{u} +(p-1) \frac{fb''}{b'}-(p-1)\frac{b''}{b'}G \biggr] \biggr\} G -G_{t} \\ &\quad =-\alpha (p-1){\mathrm{e}}^{u} \biggl[ \biggl( \frac{1}{b'} \biggr)''+2 \biggl(\frac{1}{b'} \biggr)'+\frac{1}{b'} \biggr] \vert \nabla u \vert ^{p} - \alpha ^{2}{\mathrm{e}}^{2u} \biggl[(p-1) \biggl(\frac{1}{b'} \biggr)'+(p-2) \frac{1}{b'} \biggr] \\ &\qquad {}-\alpha {\mathrm{e}}^{u} \biggl\{ \frac{f_{u}}{b'}-(p-1)f \biggl[ \biggl(\frac{1}{b'} \biggr)'+ \frac{1}{b'} \biggr] \biggr\} -2\alpha {\mathrm{e}}^{u}f_{q} \biggl[ \biggl( \frac{1}{b'} \biggr)'+\frac{1}{b'} \biggr] \vert \nabla u \vert ^{2}-f_{t}. \end{aligned} \end{aligned}$$
(2.20)
From assumptions (2.7) and (2.8) we know that the right-hand side of equality (2.20) is nonpositive. So now we have
$$\begin{aligned} \begin{aligned}[b] &\frac{ \vert \nabla u \vert ^{p-2}}{b'}\Delta G+(p-2) \frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{j}G,_{ij} \\ &\qquad {}+ \frac{1}{b'} \biggl((p-2) (p-4) \vert \nabla u \vert ^{p-6}u,_{i}u,_{j}u,_{ij} +(p-2) \vert \nabla u \vert ^{p-4}\Delta u -2(p-1)\frac{b''}{b'} \vert \nabla u \vert ^{p-2} \\ &\qquad {}+2f_{q} \biggr) (\nabla u\cdot \nabla G )+2(p-2)\frac{ \vert \nabla u \vert ^{p-4}}{b'}u,_{i}u,_{ij}G,_{i} \\ &\qquad {}+ \biggl\{ \frac{1}{(b')^{2}} \biggl[ \biggl(2(p-1) \frac{(b'')^{2}}{b'}-(p-1)b''' \biggr) \vert \nabla u \vert ^{p}-2f_{q}b'' \vert \nabla u \vert ^{2} \biggr] \\ &\qquad {}+\frac{1}{b'} \biggl[ \alpha {\mathrm{e}}^{u} \biggl(p-2-2(p-1)\frac{b''}{b'} \biggr)+f_{u} +(p-1) \frac{fb''}{b'}-(p-1)\frac{b''}{b'}G \biggr] \biggr\} G \\ &\qquad {}-G_{t} \leq 0 \quad \mbox{in } \varOmega \times \bigl(0,t^{*}\bigr). \end{aligned} \end{aligned}$$
(2.21)
Combining (2.21) and parabolic maximum principles ([24], Theorems 2.7–2.9, pp. 20–21), it follows that G may take its nonpositive minimum value under the following three possible cases:
(1) for \(t=0\); (2) at a point \((\hat{x},\hat{t})\in \varOmega \times (0,t^{*})\) where \(|\nabla u(\hat{x},\hat{t})|=0\); (3) on the boundary \(\partial \varOmega \times (0,t^{*})\).
We first study the first case. By (2.1), we derive
$$\begin{aligned} \begin{aligned}[b] &\min_{x\in \overline{\varOmega }}G(x,0) \\ &\quad =\min_{x\in \overline{\varOmega }} \bigl\{ \nabla \cdot \bigl( \vert \nabla u_{0} \vert ^{p-2} \nabla u_{0} \bigr) +f \bigl(x,u_{0}, \vert \nabla u_{0} \vert ^{2},0 \bigr)-\alpha {\mathrm{e}}^{u_{0}} \bigr\} \\ &\quad =\min_{x\in \overline{\varOmega }} \biggl\{ {\mathrm{e}}^{u_{0}} \biggl( \frac{\nabla \cdot ( \vert \nabla u_{0} \vert ^{p-2}\nabla u_{0} ) +f(x,u_{0}, \vert \nabla u_{0} \vert ^{2},0)}{{\mathrm{e}}^{u_{0}}}-\alpha \biggr) \biggr\} =0 . \end{aligned} \end{aligned}$$
(2.22)
Then, we study the second case. With (2.5), we have
$$\begin{aligned} \begin{aligned}[b] G(\hat{x},\hat{t})&= \bigl[\nabla \cdot \bigl( \vert \nabla u \vert ^{p-2} \nabla u \bigr) +f\bigl(x,u, \vert \nabla u \vert ^{2},t\bigr)-\alpha {\mathrm{e}}^{u} \bigr] \bigg|_{(\hat{x},\hat{t})} \\ &= \bigl[ \vert \nabla u \vert ^{p-2}\Delta u+(p-2) \vert \nabla u \vert ^{p-4}u,_{i}u,_{j}u,_{ij} +f\bigl(x,u, \vert \nabla u \vert ^{2},t\bigr)-\alpha { \mathrm{e}}^{u} \bigr]\bigg|_{(\hat{x}, \hat{t})} \\ &\geq \bigl[- \vert \nabla u \vert ^{p-2} \vert \Delta u \vert -(p-2) \vert \nabla u \vert ^{p-4} \vert \nabla u \vert \vert \nabla u \vert \vert u,_{ij} \vert +f\bigl(x,u, \vert \nabla u \vert ^{2},t\bigr)- \alpha {\mathrm{e}}^{u} \bigr] \bigg|_{(\hat{x},\hat{t})} \\ &= \biggl[- \vert \nabla u \vert ^{p-2} \vert \Delta u \vert -(p-2) \vert \nabla u \vert ^{p-2} \vert u,_{ij} \vert +{\mathrm{e}}^{u} \biggl(\frac{f(x,u, \vert \nabla u \vert ^{2},t)}{{\mathrm{e}}^{u}}- \alpha \biggr) \biggr] \bigg|_{(\hat{x},\hat{t})} \\ &={\mathrm{e}}^{u(\hat{x},\hat{t})} \biggl( \frac{{f(\hat{x},u(\hat{x},\hat{t}),0,\hat{t})}}{{\mathrm{e}}^{u(\hat{x},\hat{t})}} -\alpha \biggr)\geq { \mathrm{e}}^{u(\hat{x},\hat{t})}(\beta -\alpha )\geq 0. \end{aligned} \end{aligned}$$
(2.23)
Finally, we study the third case. Using of the boundary condition of (1.1), we obtain
$$ \frac{\partial G}{\partial n}=b'' \frac{\partial u}{\partial n}u_{t}+b' \frac{\partial u_{t}}{\partial n} -\alpha { \mathrm{e}}^{u} \frac{\partial u}{\partial n}=b' \biggl( \frac{\partial u}{\partial n} \biggr)_{t}=0 \quad \text{on } \partial \varOmega \times \bigl(0,t^{*}\bigr). $$
(2.24)
It follows from parabolic maximum principles and (2.22)–(2.24) that the minimum value of G in \(\overline{\varOmega }\times [0,t^{*})\) is zero. In fact, if the minimum value of G in \(\overline{\varOmega }\times [0,t^{*})\) is negative, then this minimum value must be taken on \(\partial \varOmega \times (0,t^{*})\). So there is \((\tilde{x},\tilde{t})\in \partial \varOmega \times (0,t^{*})\) such that \(G(\tilde{x},\tilde{t})=\min_{(x,t)\in \overline{\varOmega } \times [0,t^{*})}G(x,t)<0\). The parabolic maximum principle means
$$ \frac{\partial G}{\partial n}\bigg|_{(\tilde{x},\tilde{t})}< 0, $$
which contradicts (2.24). Hence, the minimum value of G in \(\overline{\varOmega }\times [0,t^{*})\) is zero. In other words, we have
$$ G(x,t)\geq 0 \quad \text{in } \overline{\varOmega }\times [0,t^{*}), $$
from which we get the following differential inequality:
$$ \frac{b'(u)}{\alpha {\mathrm{e}}^{u}}u_{t}\geq 1 \quad \text{in } \overline{\varOmega }\times [0,t^{*}). $$
(2.25)
At the point \(\tilde{x}\in \overline{\varOmega }\) where \(u_{0}(\tilde{x})=M_{0}\), we integrate (2.25) from 0 to t to get
$$ \frac{1}{\alpha } \int ^{t}_{0}\frac{b'(u)}{{\mathrm{e}}^{u}}u_{t}\,\mathrm{d}t = \frac{1}{\alpha } \int ^{u(\tilde{x},t)}_{M_{0}} \frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau \geq t. $$
(2.26)
It follows from (2.26) that u must blow up at some finite time \(t^{*}\). In fact, assuming that the solution u does not blow up, we have, for any \(t>0\),
$$ \frac{1}{\alpha } \int ^{+\infty }_{M_{0}} \frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau > \frac{1}{\alpha } \int ^{u( \tilde{x},t)}_{M_{0}}\frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau \geq t. $$
(2.27)
Letting \(t\rightarrow +\infty \) in (2.27), we derive
$$ \frac{1}{\alpha } \int ^{+\infty }_{M_{0}} \frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau =+ \infty , $$
which contradicts (2.6). Hence, u must blow up at some finite time \(t^{*}\). Furthermore, letting \(t\rightarrow t^{*}\) in (2.26), we obtain
$$ t^{*}\leq \frac{1}{\alpha } \int ^{+\infty }_{M_{0}} \frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau . $$
At each fixed point \(x\in \overline{\varOmega }\), we integrate (2.25) from t to ť (\(0< t<\check{t}<t^{*}\)) and use (2.4) to deduce
$$ H\bigl(u(x,t)\bigr)\geq H\bigl(u(x,t)\bigr)-H\bigl(u(x,\check{t}) \bigr) = \int ^{u(x,\check{t})}_{u(x,t)} \frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau \geq \alpha (\check{t}-t). $$
(2.28)
Letting \(\check{t}\rightarrow t^{*}\) in (2.28), we have
$$ H\bigl(u(x,t)\bigr)\geq \alpha \bigl(t^{*}-t \bigr) $$
and
$$ u(x,t)\leq H^{-1} \bigl(\alpha \bigl(t^{*}-t\bigr) \bigr). $$
The proof is complete. □
In Theorem 2.1, we have the following conclusions when \(b(u)\equiv u\):
Corollary 2.1
Let u be a nonnegative \(C^{3}(\varOmega \times (0,t^{*}))\cap C^{2}(\overline{\varOmega }\times [0,t^{*}))\)solution of the following problem:
$$ \textstyle\begin{cases} u_{t} =\nabla \cdot ( \vert \nabla u \vert ^{p-2}\nabla u )+f(x,u, \vert \nabla u \vert ^{2},t) &\textit{in } \varOmega \times (0,t^{*}),\\ \frac{\partial u}{\partial n}=0 &\textit{on } \partial \varOmega \times (0,t^{*}), \\ u(x,0)=u_{0}(x)\geq 0 & \textit{in } \overline{\varOmega }, \end{cases} $$
where \(p>2\), the spatial domain \(\varOmega \subset \mathbb{R}^{N}\) (\(N\geq 2\)) is bounded, and the boundary ∂Ω is smooth. Assume the following two assumptions are true:
-
(i)
$$ \beta \geq \alpha >0. $$
-
(ii)
For \((x,s,r,t)\in \varOmega \times \overline{\mathbb{R}_{+}}\times \overline{\mathbb{R}_{+}}\times \mathbb{R}_{+}\),
$$ f_{r}(x,s,r,t)\geq 0, \qquad f_{t}(x,s,r,t)\geq 0, \qquad f_{s}(x,s,r,t)-(p-1)f(x,s,r,t) \geq 0. $$
Then \(u(x,t)\)must blow up at a finite time \(t^{*}\)and
$$ t^{*}\leq \frac{{\mathrm{e}}^{-M_{0}}}{\alpha }, \quad M_{0}=\max _{x \in \overline{\varOmega }}u_{0}(x), $$
as well as
$$ u(x,t)\leq \ln \frac{1}{\alpha (t^{*}-t)}. $$