1 Introduction

In this paper, we study the blow-up and global solutions for the following nonlinear reaction-diffusion equations under Neumann boundary conditions:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} (g(u) )_{t} =\nabla\cdot(a(u)b(x)\nabla u)+f(x,u) & \mbox{in } D\times(0,T), \\ \frac{\partial u}{\partial n}=0 &\mbox{on } \partial D\times(0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.1)

where \(D\subset\mathbb{R}^{N}\) (\(N\geq2\)) is a bounded domain with smooth boundary ∂D, \(\partial/\partial n\) represents the outward normal derivative on ∂D, \(u_{0}\) is the initial value, T is the maximal existence time of u, and D̅ is the closure of D. In order to study the blow-up problem of (1.1) by using maximum principles, we make the following assumptions about the functions a, b, f, g, and \(u_{0}\). Set \(\mathbb{R}^{+}:=(0,+\infty)\). Throughout the paper, we assume that \(a(s)\) is a positive \(C^{2}(\mathbb{R}^{+})\) function, \(b(x)\) is a positive \(C^{1}(\overline{D})\) function, \(f(x,s)\) is a nonnegative \(C^{1}(D\times\mathbb{R}^{+})\) function, \(g(s)\) is a \(C^{3}(\mathbb{R}^{+})\) function, \(g'(s)>0\) for any \(s\in\mathbb{R}^{+}\), and \(u_{0}(x)\) is a positive \(C^{2}(\overline{D})\) function. Under these assumptions, the classical parabolic equation theory ensures that there exists a unique classical solution \(u(x,t)\) for problem (1.1) with some \(T>0\) and the solution is positive over \(\overline{D}\times[0,T)\). Moreover, by regularity theorem [1], \(u\in C^{3}(D\times(0,T))\cap C^{2}(\overline{D}\times[0,T))\).

During the past decades, the problems of the blow-up and global solutions for nonlinear reaction-diffusion equations have received considerable attention. The contributions in the filed can be found in [2–8] and the references therein. Many authors discussed the blow-up and global solutions for nonlinear reaction-diffusion equations under Neumann boundary conditions and obtained a lot of interesting results; we refer the reader to [9–19]. Some particular cases of (1.1) have been investigated already. Lair and Oxley [20] studied the following problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} u_{t}=\nabla\cdot(a(u)\nabla u)+f(u) & \mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 & \mbox{on } \partial D\times (0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.2)

where D is a bounded domain of \(\mathbb{R}^{N}\) (\(N\geq2\)) with smooth boundary ∂D. Necessary and sufficient conditions characterized by functions a and f were given for the existence of blow-up and global solutions. Zhang [21] discussed the following problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} (g(u))_{t}=\Delta u+f(u) & \mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 & \mbox{on } \partial D\times (0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.3)

where D is a bounded domain of \(\mathbb{R}^{N}\) (\(N\geq2\)) with smooth boundary ∂D. Sufficient conditions were developed there for the existence of blow-up and global solutions. Ding and Guo [22] considered the following problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} (g(u))_{t}=\nabla\cdot(a(u)\nabla u)\Delta u+f(u) & \mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 & \mbox{on } \partial D\times (0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.4)

where D is a bounded domain of \(\mathbb{R}^{N}\) (\(N\geq2\)) with smooth boundary ∂D. Sufficient conditions were given there for the existence of blow-up and global solutions. Meanwhile, an upper bound of the ‘blow-up time’, an upper estimate of ‘blow-up rate’, and an upper estimate of the global solution were also obtained.

The object of this paper is the blow-up and global solutions for problem (1.1). Since the reaction function \(f(x,u)\) contains not only the concentration variable u but also the space variable x, it seems that the methods of [20–22] are not applicable to problem (1.1). In this paper, we investigate problem (1.1) by constructing auxiliary functions completely different from those in [20–22] and technically using maximum principles and a first-order differential inequality technique. We obtain some existence theorems for a blow-up solution, an upper bound of ‘blow-up time’, an upper estimate of ‘blow-up rate’, existence theorems for a global solution, and an upper estimate of the global solution. Our results can be considered as extensions and supplements of those obtained in [20–22].

We proceed as follows. In Section 2 we study the blow-up solution of problem (1.1). Section 3 is devoted to the global solution of (1.1). A few examples are presented in Section 4 to illustrate the applications of the abstract results.

2 Blow-up solution

Our main result for the blow-up solution is stated in the following theorem.

Theorem 2.1

Let u be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are satisfied:

  1. (i)

    for any \(s\in\mathbb{R}^{+}\),

    $$ \biggl(\frac{a(s)}{g'(s)} \biggr)'\geq0,\quad \biggl[ \frac{1}{a(s)} \biggl(\frac{a(s)}{g'(s)} \biggr)' + \frac{1}{g'(s)} \biggr]'+ \biggl[\frac{1}{a(s)} \biggl( \frac {a(s)}{g'(s)} \biggr)' +\frac{1}{g'(s)} \biggr]\geq0; $$
    (2.1)
  2. (ii)

    for any \((x,s)\in D\times\mathbb{R}^{+}\),

    $$ \biggl(\frac{f(x,s)g'(s)}{a(s)} \biggr)_{s} -\frac{f(x,s)g'(s)}{a(s)} \geq0; $$
    (2.2)
  3. (iii)
    $$ \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\, {\mathrm{d}}s< + \infty,\qquad M_{0}:=\max_{\overline{D}}u_{0}(x); $$
    (2.3)
  4. (iv)
    $$ \beta:=\min_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{u_{0}}}>0. $$
    (2.4)

Then the solution u to problem (1.1) must blow up in a finite T, and

$$\begin{aligned}& T\leq\frac{1}{\beta} \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\, {\mathrm{d}}s, \end{aligned}$$
(2.5)
$$\begin{aligned}& u(x,t)\leq H^{-1} \bigl(\beta(T-t) \bigr), \quad \forall (x,t)\in {\overline{D}}\times[0,T), \end{aligned}$$
(2.6)

where

$$ H(z):= \int^{+\infty}_{z}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s, \quad z>0, $$
(2.7)

and \(H^{-1}\) is the inverse function of H.

Proof

Consider the auxiliary function

$$ \Psi(x,t):=g'(u)u_{t}-\beta{ \mathrm{e}}^{u}. $$
(2.8)

For brevity of notation, we write g in place of \(g(u)\), suppressing the symbol u. We find that

$$\begin{aligned}& \nabla\Psi=g''u_{t}\nabla u+g'\nabla u_{t}-\beta{\mathrm{e}}^{u}\nabla u, \end{aligned}$$
(2.9)
$$\begin{aligned}& \Delta\Psi=g'''u_{t}| \nabla u|^{2}+2g''\nabla u\cdot\nabla u_{t}+g''u_{t}\Delta u +g'\Delta u_{t}-\beta{\mathrm{e}}^{u}|\nabla u|^{2}-\beta{\mathrm{e}}^{u}\Delta u, \end{aligned}$$
(2.10)

and

$$\begin{aligned} \Psi_{t} =&g''(u_{t})^{2}+g'(u_{t})_{t}- \beta{\mathrm{e}}^{u}u_{t} \\ =&g''(u_{t})^{2}+g' \biggl(\frac{ab}{g'}\Delta u+\frac{a'b}{g'}|\nabla u|^{2} + \frac{a}{g'}\nabla b\cdot\nabla u+\frac{f}{g'} \biggr)_{t}- \beta{\mathrm{e}}^{u}u_{t} \\ =&g''(u_{t})^{2}+ \biggl(a'b-\frac{abg''}{g'} \biggr)u_{t}\Delta u +ab \Delta u_{t}+ \biggl(a''b-\frac{a'bg''}{g'} \biggr)u_{t}|\nabla u|^{2} \\ &{}+2a'b (\nabla u\cdot\nabla u_{t} ) + \biggl(a'- \frac{ag''}{g'} \biggr)u_{t} (\nabla b\cdot\nabla u )+a (\nabla b \cdot\nabla u_{t} ) \\ &{} + \biggl(f_{u}-\frac{fg''}{g'}-\beta{\mathrm{e}}^{u} \biggr)u_{t}. \end{aligned}$$
(2.11)

It follows from (2.10) and (2.11) that

$$\begin{aligned} \frac{ab}{g'}\Delta\Psi-\Psi_{t} =& \biggl( \frac{abg'''}{g'}+\frac{a'bg''}{g'}-a''b \biggr)u_{t}|\nabla u|^{2} + \biggl(2\frac{abg''}{g'}-2a'b \biggr) (\nabla u\cdot\nabla u_{t} ) \\ &{}+ \biggl(2\frac{abg''}{g'}-a'b \biggr)u_{t}\Delta u -\beta\frac{ab{\mathrm{e}}^{u}}{g'}|\nabla u|^{2} -\beta\frac{ab{\mathrm{e}}^{u}}{g'}\Delta u-g''(u_{t})^{2} \\ &{}+ \biggl(\frac{ag''}{g'}-a' \biggr)u_{t} (\nabla b \cdot\nabla u ) -a (\nabla b\cdot\nabla u_{t} ) + \biggl( \frac{fg''}{g'}-f_{u}+\beta{\mathrm{e}}^{u} \biggr)u_{t}. \end{aligned}$$
(2.12)

By (1.1) we have

$$ \Delta u=\frac{g'}{ab}u_{t}-\frac{a'}{a}| \nabla u|^{2}-\frac{1}{b} (\nabla b\cdot\nabla u )- \frac{f}{ab}. $$
(2.13)

Substituting (2.13) into (2.12), we get

$$\begin{aligned} \frac{ab}{g'}\Delta\Psi-\Psi_{t} =& \biggl( \frac{abg'''}{g'}-\frac{a'bg''}{g'}-a''b+ \frac {(a')^{2}b}{a} \biggr)u_{t}|\nabla u|^{2} + \biggl(2 \frac{abg''}{g'}-2a'b \biggr) (\nabla u\cdot\nabla u_{t} ) \\ &{}-\frac{(g')^{2}}{a} \biggl(\frac{a}{g'} \biggr)'(u_{t})^{2} -\frac{ag''}{g'}u_{t} (\nabla b\cdot\nabla u ) + \biggl( \frac{a'f}{a}-\frac{fg''}{g'}-f_{u} \biggr)u_{t} \\ &{}+ \biggl(\beta\frac{a'b{\mathrm{e}}^{u}}{g'}-\beta\frac{ab{\mathrm{e}}^{u}}{g'} \biggr)|\nabla u|^{2} +\beta\frac{a{\mathrm{e}}^{u}}{g'} (\nabla b\cdot\nabla u ) \\ &{}+\beta \frac{f{\mathrm{e}}^{u}}{g'}-a (\nabla b\cdot\nabla u_{t} ). \end{aligned}$$
(2.14)

In view of (2.9), we have

$$ \nabla u_{t}=\frac{1}{g'}\nabla\Psi- \frac{g''}{g'}u_{t}\nabla u+\beta \frac{{\mathrm{e}}^{u}}{g'}\nabla u. $$
(2.15)

Substitution of (2.15) into (2.14) results in

$$\begin{aligned}& \frac{ab}{g'}\Delta\Psi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Psi- \Psi_{t} \\& \quad = \biggl(\frac{abg'''}{g'}+\frac{a'bg''}{g'}-a''b+ \frac{(a')^{2}b}{a} -2\frac{ab(g'')^{2}}{(g')^{2}} \biggr)u_{t}|\nabla u|^{2} \\& \qquad {} + \biggl(2\beta\frac{abg''{\mathrm{e}}^{u}}{(g')^{2}}-\beta\frac{a'b{\mathrm{e}}^{u}}{g'} -\beta \frac{ab{\mathrm{e}}^{u}}{g'} \biggr)|\nabla u|^{2} -\frac{(g')^{2}}{a} \biggl( \frac{a}{g'} \biggr)'(u_{t})^{2} \\& \qquad {} + \biggl(\frac{a'f}{a}-\frac{fg''}{g'}-f_{u} \biggr)u_{t} +\beta\frac{f{\mathrm{e}}^{u}}{g'}. \end{aligned}$$
(2.16)

With (2.8), we have

$$ u_{t}=\frac{1}{g'}\Psi+\beta\frac{{\mathrm{e}}^{u}}{g'}. $$
(2.17)

Substituting (2.17) into (2.16), we obtain

$$\begin{aligned}& \frac{ab}{g'}\Delta\Psi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Psi \\& \qquad {} + \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} + \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Psi- \Psi_{t} \\& \quad = -\beta ab{\mathrm{e}}^{u} \biggl\{ \biggl[\frac{1}{a} \biggl(\frac {a}{g'} \biggr)'+\frac{1}{g'} \biggr]' + \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)'+\frac{1}{g'} \biggr] \biggr\} |\nabla u|^{2} -\frac{(g')^{2}}{a} \biggl(\frac{a}{g'} \biggr)'(u_{t})^{2} \\& \qquad {}-\beta\frac{a{\mathrm{e}}^{u}}{(g')^{2}} \biggl[ \biggl(\frac{fg'}{a} \biggr)_{u}-\frac{fg'}{a} \biggr]. \end{aligned}$$
(2.18)

By assumptions (2.1) and (2.2) the right-hand side of (2.18) is nonpositive, that is,

$$\begin{aligned}& \frac{ab}{g'}\Delta\Psi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Psi \\& \quad {}+ \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} + \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Psi- \Psi _{t}\leq0 \quad \mbox{in } D\times(0,T). \end{aligned}$$
(2.19)

Now by (2.4) we have

$$\begin{aligned} \min_{\overline{D}}\Psi(x,0) =&\min _{\overline{D}} \bigl\{ g'(u_{0}) (u_{0})_{t}-\beta{\mathrm{e}}^{u_{0}} \bigr\} \\ =&\min_{\overline{D}} \bigl\{ \nabla\cdot \bigl(a(u_{0})b(x) \nabla u_{0} \bigr)+f(x,u_{0})-\beta{\mathrm{e}}^{u_{0}} \bigr\} \\ =&\min_{\overline{D}} \biggl\{ {\mathrm{e}}^{u_{0}} \biggl[ \frac{\nabla\cdot(a(u_{0})b(x)\nabla u_{0})+f(x,u_{0})}{{\mathrm{e}}^{u_{0}}}-\beta \biggr] \biggr\} =0. \end{aligned}$$
(2.20)

It follows from (1.1) that

$$ \frac{\partial\Psi}{\partial n}=g''u_{t} \frac{\partial u}{\partial n} +g'\frac{\partial u_{t}}{\partial n} -\beta{\mathrm{e}}^{u} \frac{\partial u}{\partial n} =g' \biggl(\frac{\partial u}{\partial n} \biggr)_{t}=0 \quad \mbox{on } \partial D\times(0,T). $$
(2.21)

The assumptions concerning the functions a, b, f, g, and \(u_{0}\) in Section 1 imply that we can use maximum principles to (2.19)-(2.21). Combining (2.19)-(2.21) and applying maximum principles [23], it follows that the minimum of Ψ in \(\overline{D}\times[0,T)\) is zero. Thus, we have

$$\Psi\geq0 \quad \mbox{in } \overline{D}\times[0,T), $$

that is, the differential inequality

$$ \frac{g'(u)}{{\mathrm{e}}^{u}}u_{t}\geq\beta. $$
(2.22)

Suppose that \(x_{0}\in\overline{D}\) and \(u_{0}(x_{0})=M_{0}\). At the \(x_{0}\), integrate (2.22) over \([0,t]\) to get

$$ \int^{t}_{0}\frac{g'(u)}{{\mathrm{e}}^{u}}u_{t}\,{ \mathrm{d}}t = \int^{u(x_{0},t)}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s\geq \beta t, $$
(2.23)

which implies that u must blow up in finite time. Actually, if u is a global solution of (1.1), then for any \(t>0\), it follows from (2.23) that

$$ \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s \geq \int^{u(x_{0},t)}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s \geq \beta t. $$
(2.24)

Letting \(t\rightarrow+\infty\) in (2.24) yields

$$\int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s=+ \infty, $$

which contradicts with assumption (2.3). This shows that u must blow up in a finite time \(t=T\). Furthermore, letting \(t\rightarrow T\) in (2.23), we have

$$T\leq\frac{1}{\beta} \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s. $$

Integrating inequality (2.22) over \([t,s]\) (\(0< t< s< T\)) yields, for each fixed x, that

$$\begin{aligned} H \bigl(u(x,t) \bigr) \geq& H \bigl(u(x,t) \bigr) -H \bigl(u(x,s) \bigr) = \int^{+\infty}_{u(x,t)}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s- \int ^{+\infty}_{u(x,s)}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s \\ =& \int^{u(x,s)}_{u(x,t)}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s= \int^{s}_{t}\frac {g'(u)}{{\mathrm{e}}^{u}}u_{t}\,{ \mathrm{d}}t\geq\beta(s-t). \end{aligned}$$

Passing to the limit as \(s\rightarrow T^{-}\) gives

$$H\bigl(u(x,t)\bigr)\geq\beta(T-t). $$

Since H is a decreasing function, we have

$$u(x,t)\leq H^{-1} \bigl(\beta(T-t) \bigr). $$

The proof is complete. □

3 Global solution

The following theorem is the main result for the global solution.

Theorem 3.1

Let u be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are satisfied:

  1. (i)

    for any \(s\in\mathbb{R}^{+}\),

    $$ \biggl(\frac{a(s)}{g'(s)} \biggr)'\leq0,\quad \biggl[ \frac{1}{a(s)} \biggl(\frac{a(s)}{g'(s)} \biggr)' - \frac{1}{g'(s)} \biggr]'- \biggl[\frac{1}{a(s)} \biggl( \frac {a(s)}{g'(s)} \biggr)' -\frac{1}{g'(s)} \biggr]\leq0; $$
    (3.1)
  2. (ii)

    for any \((x,s)\in D\times\mathbb{R}^{+}\),

    $$ \biggl(\frac{f(x,s)g'(s)}{a(s)} \biggr)_{s} +\frac{f(x,s)g'(s)}{a(s)} \leq0; $$
    (3.2)
  3. (iii)
    $$ \int^{+\infty}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s=+ \infty,\qquad m_{0}:=\min_{\overline{D}}u_{0}(x); $$
    (3.3)
  4. (iv)
    $$ \alpha:=\max_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{-u_{0}}}>0. $$
    (3.4)

Then the solution u of (1.1) must be a global solution, and

$$ u(x,t)\leq G^{-1} \bigl(\alpha t+G\bigl(u_{0}(x,t) \bigr) \bigr), \quad \forall (x,t)\in\overline{D}\times\overline{\mathbb{R}}^{+}, $$
(3.5)

where

$$ G(z):= \int^{z}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s, \quad z\geq m_{0}, $$
(3.6)

and \(G^{-1}\) is the inverse function of G.

Proof

Construct the auxiliary function

$$ \Phi(x,t):=g'(u)u_{t}-\alpha{ \mathrm{e}}^{-u}. $$
(3.7)

By using the same reasoning process with that of (2.9)-(2.18), we have

$$\begin{aligned}& \frac{ab}{g'}\Delta\Phi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Phi \\& \qquad {} + \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} - \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Phi- \Phi_{t} \\& \quad = -\alpha ab{\mathrm{e}}^{-u} \biggl\{ \biggl[\frac{1}{a} \biggl(\frac {a}{g'} \biggr)'-\frac{1}{g'} \biggr]' - \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)'-\frac{1}{g'} \biggr] \biggr\} |\nabla u|^{2} -\frac{(g')^{2}}{a} \biggl(\frac{a}{g'} \biggr)'(u_{t})^{2} \\& \qquad {} -\alpha\frac{a{\mathrm{e}}^{-u}}{(g')^{2}} \biggl[ \biggl(\frac{fg'}{a} \biggr)_{u}+\frac{fg'}{a} \biggr]. \end{aligned}$$
(3.8)

From assumptions (3.1) and (3.2) we see that the right-hand side of (3.8) is nonnegative, that is,

$$\begin{aligned}& \frac{ab}{g'}\Delta\Phi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Phi \\& \quad {}+ \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} - \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Phi- \Phi_{t} \geq0 \quad \mbox{in } D\times(0,T). \end{aligned}$$
(3.9)

By (3.4) we have

$$\begin{aligned} \max_{\overline{D}}\Phi(x,0) =& \max _{\overline{D}} \bigl\{ g'(u_{0}) (u_{0})_{t}-\alpha{\mathrm{e}}^{-u_{0}} \bigr\} \\ =&\max_{\overline{D}} \bigl\{ \nabla\cdot \bigl(a(u_{0})b(x) \nabla u_{0} \bigr)+f(x,u_{0})-\alpha{\mathrm{e}}^{-u_{0}} \bigr\} \\ =&\max_{\overline{D}} \biggl\{ {\mathrm{e}}^{-u_{0}} \biggl[ \frac{\nabla\cdot(a(u_{0})b(x)\nabla u_{0})+f(x,u_{0})}{{\mathrm{e}}^{-u_{0}}}-\alpha \biggr] \biggr\} =0. \end{aligned}$$
(3.10)

Repeating the arguments for (2.21), we have

$$ \frac{\partial\Phi}{\partial n}=0 \quad \mbox{on } \partial D\times(0,T). $$
(3.11)

Combining (3.9)-(3.11) and applying the maximum principles again, we get that the maximum of Φ in \(\overline{D}\times[0,T)\) is zero. Hence, we have

$$\Phi\leq0 \quad \mbox{in } \overline{D}\times[0,T), $$

that is, the differential inequality

$$ \frac{g'(u)}{{\mathrm{e}}^{-u}}u_{t}\leq\alpha. $$
(3.12)

For each fixed \(x\in\overline{D}\), integrate (3.12) over \([0,t]\) to produce

$$ \int^{t}_{0}\frac{g'(u)}{{\mathrm{e}}^{-u}}u_{t}\,{ \mathrm{d}}t = \int^{u(x,t)}_{u_{0}(x)} \frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s \leq \alpha t, $$
(3.13)

which shows that u must be a global solution. In fact, suppose that u blows up at finite time T, that is,

$$\lim_{t\rightarrow T^{-}}u(x,t)=+\infty. $$

Passing to the limit as \(t\rightarrow T^{-}\) in (3.13) gives

$$\int^{+\infty}_{u_{0}(x)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s\leq \alpha T $$

and

$$\int^{+\infty}_{m_{0}} \frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s = \int^{u_{0}(x)}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s + \int^{+\infty}_{u_{0}(x)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s \leq \int^{u_{0}(x)}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s+ \alpha T< +\infty, $$

which is a contradiction. This shows that u is global. Moreover, (3.13) implies that

$$\int^{u(x,t)}_{u_{0}(x)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s = \int_{m_{0}}^{u(x,t)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s - \int^{u_{0}(x)}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s =G \bigl(u(x,t) \bigr)-G\bigl(u_{0}(x)\bigr) \leq\alpha t. $$

Since G is an increasing function, we have

$$u(x,t)\leq G^{-1} \bigl(\alpha t+G\bigl(u_{0}(x)\bigr) \bigr). $$

The proof is complete. □

4 Applications

When \(g(u)\equiv u\), \(b(x)\equiv1\), and \(f(x,u)\equiv f(u)\), problem (1.1) is problem (1.2) studied by Lair and Oxley [20]. When \(a(u)\equiv1\), \(b(x)\equiv1\), and \(f(x,u)\equiv f(u)\), problem (1.1) is problem (1.3) discussed by Zhang [21]. When \(b(x)\equiv1\) and \(f(x,u)\equiv f(u)\), problem (1.1) is problem (1.4) considered by Ding and Guo [22]. In these three cases, the conclusions of Theorems 2.1 and 3.1 still hold. In this sense, our results extend and supplement the results of [20–22].

In what follows, we present several examples to demonstrate applications of Theorems 2.1 and 3.1.

Example 4.1

Let u be a solution of the following problem:

$$\left \{ \textstyle\begin{array}{l@{\quad}l} (2{\mathrm{e}}^{\frac{u}{2}}+u )_{t} =\nabla\cdot ( (1+{\mathrm{e}}^{\frac{u}{2}} ) (1+\| x\|^{2} )\nabla u ) +7{\mathrm{e}}^{u}-\|x\|^{2} &\mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 &\mbox{on } \partial D\times(0,T), \\ u(x,0)=1+ (1-\|x\|^{2} )^{2} & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$

where \(D= \{x=(x_{1},x_{2},x_{3}) \mid \|x\| ^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}<1 \}\) is the unit ball of \(\mathbb{R}^{3}\). Now we have

$$\begin{aligned}& g(u)=2{\mathrm{e}}^{\frac{u}{2}}+u, \qquad a(u)=1+{\mathrm{e}}^{\frac{u}{2}}, \qquad b(x)=1+\|x\|^{2}, \\& f(x,u)=7{\mathrm{e}}^{u}-\|x\|^{2}, \qquad u_{0}(x)=1+ \bigl(1-\|x\|^{2} \bigr)^{2}. \end{aligned}$$

In order to determine the constant β, we assume that

$$s=\|x\|^{2}. $$

Then \(0\leq s\leq1\) and

$$\begin{aligned} \beta =&\min_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{u_{0}}} \\ =&\min_{\overline{D}} \bigl\{ \bigl({\mathrm{e}}^{-1-(1-\|x\|^{2})^{2}}+{ \mathrm{e}}^{-\frac{1}{2}-\frac{1}{2}(1-\|x\|^{2})^{2}} \bigr) \bigl(-12+28\|x\|^{2} \bigr) \\ &{}+8{\mathrm{e}}^{-\frac{1}{2}-\frac{1}{2} (1-\|x\| ^{2} )^{2}}\|x\|^{2} \bigl(1+\|x\|^{2} \bigr) \bigl(1-\|x\|^{2} \bigr)^{2} +7-\|x\|^{2}{ \mathrm{e}}^{-1- (1-\|x\|^{2} )^{2}}\bigr\} \\ =&\min_{0\leq s\leq1} \bigl\{ \bigl({\mathrm{e}}^{-1-(1-s)^{2}}+{ \mathrm{e}}^{-\frac{1}{2}-\frac{1}{2}(1-s)^{2}} \bigr) (-12+28s) \\ &{}+8{\mathrm{e}}^{-\frac{1}{2}-\frac{1}{2}(1-s)^{2}}s(1+s) (1-s)^{2}+7-s{ \mathrm{e}}^{-1-(1-s)^{2}}\bigr\} \\ =&0.9614. \end{aligned}$$

It is easy to check that (2.1)-(2.3) hold. By Theorem 2.1, u must blow up in a finite time T, and

$$\begin{aligned}& T\leq\frac{1}{\beta} \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s = \frac{1}{0.9614} \int^{+\infty}_{1}\frac{{\mathrm{e}}^{\frac {s}{2}}+1}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s =1.4025, \\& u(x,t)\leq H^{-1} \bigl(\beta(T-t) \bigr) =\ln\frac{1}{ (\sqrt{1+0.9614(T-t)}-1 )^{2}}. \end{aligned}$$

Example 4.2

Let u be a solution of the following problem:

$$\left \{ \textstyle\begin{array}{l@{\quad}l} (\ln ({\mathrm{e}}^{u}-1 )-u )_{t} =\nabla\cdot (\frac{1}{{\mathrm{e}}^{u}-1} (1+\|x\|^{2} )\nabla u ) +{\mathrm{e}}^{-u} (1+\|x\|^{2} ) &\mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 &\mbox{on } \partial D\times(0,T), \\ u(x,0)=1+ (1-\|x\|^{2} )^{2} & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$

where \(D= \{x=(x_{1},x_{2},x_{3}) \mid \|x\| ^{2}=x_{2}^{2}+x_{2}^{2}+x_{3}^{2}<1 \}\) is the unit ball of \(\mathbb{R}^{3}\). Now we have

$$\begin{aligned}& g(u)=\ln \bigl({\mathrm{e}}^{u}-1 \bigr)-u, \qquad a(u)= \frac{1}{{\mathrm{e}}^{u}-1}, \qquad b(x)=1+\|x\|^{2}, \\& f(x,u)={\mathrm{e}}^{-u} \bigl(1+\|x\|^{2} \bigr), \qquad u_{0}(x)=1+ \bigl(1-\| x\|^{2} \bigr)^{2}. \end{aligned}$$

By setting

$$s=\|x\|^{2}, $$

we have \(0\leq s\leq1\) and

$$\begin{aligned} \alpha =&\min_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{-u_{0}}} \\ =&\min_{\overline{D}} \biggl\{ \frac{1}{ ({\mathrm{e}}^{1+(1-\|x\|^{2})^{2}}-1 )^{2}} \bigl[ \bigl(-12+28 \|x\|^{2} \bigr){\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}} \bigl({\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}}-1 \bigr) \\ &{}-16\|x\|^{2} \bigl(1+\|x\|^{2} \bigr) \bigl(1-\|x \|^{2} \bigr)^{2}{\mathrm{e}}^{2+2 (1-\|x\|^{2} )^{2}} + \bigl(1+\|x \|^{2} \bigr) \bigl({\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}}-1 \bigr)^{2} \bigr]\biggr\} \\ =&\min_{0\leq s\leq1} \biggl\{ \frac{1}{ ({\mathrm{e}}^{1+(1-s)^{2}}-1 )^{2}} \bigl[ (-12+28s ){ \mathrm{e}}^{1+ (1-s )^{2}} \bigl({\mathrm{e}}^{1+ (1-s )^{2}}-1 \bigr) \\ &{}-16s (1+s ) (1-s )^{2}{\mathrm{e}}^{2+2 (1-s )^{2}} + (1+s ) \bigl({ \mathrm{e}}^{1+ (1-s )^{2}}-1 \bigr)^{2}\bigr]\biggr\} \\ =&27.3116. \end{aligned}$$

Again, it is easy to check that (3.1)-(3.3) hold. By Theorem 3.1, u must be a global solution, and

$$u(x,t)\leq G^{-1} \bigl(\alpha t+G\bigl(u_{0}(x)\bigr) \bigr) =\ln \bigl[1+{\mathrm{e}}^{27.3116t} \bigl({\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}}-1 \bigr) \bigr]. $$