1 Introduction

The blow-up solutions of parabolic p-Laplician equations have been studied by many authors (see, for instance, [1–10]). In this work, we research the blow-up solutions of the following parabolic p-Laplacian equations with a gradient source term:

$$ \textstyle\begin{cases} (b(u) )_{t} =\nabla \cdot ( \vert \nabla u \vert ^{p-2}\nabla u )+f(x,u, \vert \nabla u \vert ^{2},t) &\text{in } \varOmega \times (0,t^{*}), \\ \frac{\partial u}{\partial n}=0 &\text{on } \partial \varOmega \times (0,t^{*}),\\ u(x,0)=u_{0}(x)\geq 0 & \text{in } \overline{\varOmega }. \end{cases} $$
(1.1)

In (1.1), \(p>2\), the spatial domain \(\varOmega \subset \mathbb{R}^{N}\) (\(N\geq 2\)) is bounded, the boundary ∂Ω is smooth, and \(t^{*}\) is blow-up time, \(b(s)\) is a \(C^{3}(\overline{\mathbb{R}_{+}})\) function with \(b'(s)>0\), \(s\in \overline{\mathbb{R}_{+}}\), \(f(x,s,r,t)\) is a nonnegative \(C^{1}(\varOmega \times \overline{\mathbb{R}_{+}}\times \overline{\mathbb{R}_{+}}\times \mathbb{R}_{+})\) function, and \(u_{0}(x)\) is a nonnegative \(C^{2}(\overline{\varOmega })\) function satisfying \(\frac{\partial u_{0}(x)}{\partial n}=0\), \(x\in \partial \varOmega \).

As far as I know, there are many papers on the blow-up problems of parabolic equations with a gradient term (see, for instance, [11–23]). For the sake of research (1.1), we mainly focus on the papers [12, 22]. In [22], Zhang et al. researched the following problems:

$$ \textstyle\begin{cases} (b(u) )_{t} =\nabla \cdot (\rho ( \vert \nabla u \vert ^{p}) \vert \nabla u \vert ^{p-2} \nabla u)+h(x)k(t)f(u) &\text{in } \varOmega \times (0,t^{*}),\\ u(x,t)=0 &\text{on } \partial \varOmega \times (0,t^{*}),\\ u(x,0)=u_{0}(x)\geq 0 & \text{in } \overline{\varOmega }. \end{cases} $$
(1.2)

In (1.2), \(p\geq 2\), the spatial domain \(\varOmega \subset \mathbb{R}^{N}\) (\(N\geq 2\)) is bounded, and the boundary ∂Ω is smooth. They derived some conditions which ensure the solution of (1.2) blows up in a finite \(t^{*}\) or exists globally. Moreover, an upper bound and a lower bound of the blow-up time were also specified when the blow-up occurs. Their research relied on using differential inequality techniques. In [12], Ding et al. studied the following problems:

$$ \textstyle\begin{cases} (b(u) )_{t} =\Delta u+f(x,u, \vert \nabla u \vert ^{2},t) &\text{in } \varOmega \times (0,t^{*}),\\ \frac{\partial u}{\partial n}=r(u) & \text{on } \partial \varOmega \times (0,t^{*}),\\ u(x,0)=u_{0}(x)>0 & \text{in } \overline{\varOmega }. \end{cases} $$
(1.3)

In (1.3), the spatial domain \(\varOmega \subset \mathbb{R}^{N}\) (\(N\geq 2\)) is bounded, and the boundary ∂Ω is smooth. By combining the parabolic maximum principles and the differential inequality techniques, they obtained sufficient conditions for the existence of the blow-up solution and the global solution of (1.3). In addition, the upper bounds of the blow-up time \(t^{*}\) and the upper estimates of the blow-up rate were also given.

Inspired by the above two research works, in this paper we study the blow-up solutions of (1.1). Since the source term of the equation in (1.1) contains a gradient term, and the source term of the equation in (1.2) does not, the research methods in [22] cannot be used to research (1.1). In this paper, we use the method in [12] to study (1.1). In other words, we combine the parabolic maximum principles with the differential inequality techniques to study (1.1). The difficulty in using this research method is that some suitable auxiliary functions need to be constructed. We note that although the source terms in equations of (1.1) and (1.3) are the same, the boundary conditions are different. Therefore, the auxiliary functions in the paper [12] are not suitable for researching (1.1). In this paper, we need to construct some auxiliary functions that are completely different from those in [12] to study the blow-up solution of (1.1). We give sufficient conditions to guarantee that the solution of (1.1) blows up at a finite time \(t^{*}\). The upper bounds of the blow-up time \(t^{*}\) and the upper estimates of the blow-up rate are also obtained.

For convenience, in this paper we use a comma to represent partial derivative and adopt the summation convention, for example,

$$ u,_{i}u,_{j}u,_{ij}=\sum ^{N}_{i=1}\sum^{N}_{j=1} \frac{\partial u}{\partial x_{i}}\frac{\partial u}{\partial x_{j}} \frac{\partial ^{2}u}{\partial x_{i}\partial x_{j}}. $$

2 The main result and its proof

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:

  1. (i)
    $$ \beta \geq \alpha >0. $$
    (2.5)
  2. (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)
  3. (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)
  4. (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:

  1. (i)
    $$ \beta \geq \alpha >0. $$
  2. (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

$$ u(x,t)\leq \ln \frac{1}{\alpha (t^{*}-t)}. $$

3 Application

In the following, we give an example to illustrate the application of Theorem 2.1

Example 3.1

Let u be a nonnegative \(C^{3}(\varOmega \times (0,t^{*}))\cap C^{2}(\overline{\varOmega }\times [0,t^{*}))\) solution of the following problem:

$$ \textstyle\begin{cases} ({\mathrm{e}}^{\frac{u}{2}} )_{t} =\nabla \cdot ( \vert \nabla u \vert \nabla u )+ (1+\sum^{3}_{i=1}x^{2}_{i}+ \vert \nabla u \vert ^{2}t ){\mathrm{e}}^{u} & \text{in }\varOmega \times (0,t^{*}), \\ \frac{\partial u}{\partial n}=0 & \text{on }\partial \varOmega \times (0,t^{*}), \\ u(x,0)=1 & \text{in }\overline{\varOmega }, \end{cases} $$

where \(\varOmega = \{ x=(x_{1},x_{2},x_{3}) \vert \sum^{3}_{i=1}x^{2}_{i}<1 \} \). We now have

$$ p=3, \qquad b(u)={\mathrm{e}}^{\frac{u}{2}}, \qquad f\bigl(x,u, \vert \nabla u \vert ^{2},t\bigr)= \Biggl(1+\sum^{3}_{i=1}x^{2}_{i}+ \vert \nabla u \vert ^{2}t \Biggr){\mathrm{e}}^{u}, \qquad u_{0}(x)=1. $$

It follows from (2.1) and (2.2) that

$$ \alpha =\min_{x\in \overline{\varOmega }} \frac{\nabla \cdot ( \vert \nabla u_{0} \vert ^{p-2}\nabla u_{0} )+f(x,u_{0}, \vert \nabla u_{0} \vert ^{2},0)}{{\mathrm{e}}^{u_{0}}} =\min _{x\in \overline{\varOmega }} \Biggl(1+ \sum^{3}_{i=1}x^{2}_{i} \Biggr)=1 $$

and

$$ \beta =\inf_{(x,s,t)\in \varOmega \times \overline{\mathbb{R}_{+}}\times \mathbb{R}_{+}} \frac{f(x,s,0,t)}{{\mathrm{e}}^{s}} =\inf _{(x,s,t)\in \varOmega \times \overline{\mathbb{R}_{+}}\times \mathbb{R}_{+}} \Biggl(1+ \sum^{3}_{i=1}x^{2}_{i} \Biggr)=1, $$

from which we know that the assumption (2.5) holds. We also easily check that the assumptions (2.6)–(2.8) hold true. Hence, Theorem 2.1 implies that \(u(x,t)\) must blow up at a finite time \(t^{*}\) and

$$ t^{*}\leq \frac{1}{\alpha } \int ^{+\infty }_{M_{0}} \frac{b'(\tau )}{{\mathrm{e}}^{\tau }}\,\mathrm{d}\tau = \frac{1}{2} \int ^{+ \infty }_{1}{{\mathrm{e}}^{-\frac{\tau }{2}}} { \mathrm{d}}\tau ={\mathrm{e}}^{-\frac{1}{2}}, $$

as well as

$$ u(x,t)\leq H^{-1} \bigl(\alpha \bigl(t^{*}-t\bigr) \bigr) = \ln \frac{1}{(t^{*}-t)^{2}}, \quad (x,t)\in \overline{\varOmega }\times [0,t^{*}). $$

4 Conclusions

In this paper, we combine parabolic maximum principles with the differential inequality techniques to study the blow-up solution of problem (1.1). The key to our research is constructing two auxiliary functions (2.3) and (2.4). With their help, we obtain sufficient conditions for the existence of the blow-up solution of problem (1.1). In addition, we also give an upper bound on the blow-up time and an upper estimate of the blow-up rate.