Global existence and blow-up for a class of nonlinear reaction diffusion problems

Abstract

This paper deals with the global existence and blow-up of the solution for a class of nonlinear reaction diffusion problems. The purpose of this paper is to establish conditions on the data to guarantee the blow-up of the solution at some finite time, and conditions to ensure that the solution remains global. In addition, an upper bound for the ‘blow-up time’, an upper estimate of the ‘blow-up rate’, and an upper estimate of the global solution are also specified. Finally, as applications of the obtained results, some examples are presented.

MSC: 35K57, 35K55, 35B05.

1 Introduction

The global existence and blow-up for nonlinear reaction diffusion equations have been widely studied in recent years (see, for instance, [1]–[8]). In this paper, we consider the following problem:

$$\{\begin{array}{cc}{(g(x,u))}_t=\mathrm{\nabla }\cdot (a(u)b(x)\mathrm{\nabla }u)+f(u)\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=u_0(x)>0\hfill & \text{in }\overline{D},\hfill \end{array}$$

(1.1)

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D, $\overline{D}$ is the closure of D, $\partial /\partial n$ is the outward normal derivative on ∂D, T is the maximal existence time of u. Set ${\mathbb{R}}^{+}:=(0,+\mathrm{\infty })$. We assume, throughout this paper, that $a(s)$ is a positive $C^2({\mathbb{R}}^{+})$ function, $b(x)$ is a positive $C^1(\overline{D})$ function, $f(s)$ is a positive $C^2({\mathbb{R}}^{+})$ function, $g(x,s)$ is a $C^2(\overline{D}\times {\mathbb{R}}^{+})$ function, $g_s(x,s)>0$ for any $(x,s)\in \overline{D}\times {\mathbb{R}}^{+}$, and $u_0(x)$ is a positive $C^2(\overline{D})$ function. Under the above assumptions, it is well known from the classical parabolic equation theory [5] and maximum principle [9] that there exists a unique local positive solution for problem (1.1). Moreover, by the regularity theorem [10], $u(x,t)\in C^3(D\times (0,T))\cap C^2(\overline{D}\times [0,T))$.

Many authors discussed the global existence and blow-up for nonlinear reaction diffusion equations with Neumann boundary conditions and obtained a lot of interesting results [11]–[24]. Some special cases of (1.1) have been studied already. Lair and Oxley [25] investigated the following problem:

$$\{\begin{array}{cc}{u}_t=\mathrm{\nabla }\cdot (a(u)\mathrm{\nabla }u)+f(u)\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=u_0(x)>0\hfill & \text{in }\overline{D},\hfill \end{array}$$

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D. The necessary and sufficient conditions characterized by functions a and f were given for the global existence and blow-up solution. Zhang [26] dealt with the following problem:

$$\{\begin{array}{cc}{(g(u))}_t=\mathrm{\Delta }u+f(u)\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=u_0(x)>0\hfill & \text{in }\overline{D},\hfill \end{array}$$

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D. The sufficient conditions were obtained there for the existence of global and blow-up solutions. Gao et al.[27] considered the following problem:

$$\{\begin{array}{cc}{(g(u))}_t=\mathrm{\nabla }\cdot (a(u)\mathrm{\nabla }u)+f(u)\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=u_0(x)>0\hfill & \text{in }\overline{D},\hfill \end{array}$$

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D. The sufficient conditions were developed for the existence of global and blow-up solutions. Meanwhile, the upper estimate of the global solution, the upper bound of the ‘blow-up time’, and the upper estimate of the ‘blow-up rate’ were also given.

In this paper, we study reaction diffusion problem (1.1). Note that $f(u)$, $g(x,u)$ and $a(u)b(x)$ are nonlinear reaction, nonlinear diffusion and nonlinear convection, respectively. Since the diffusion function $g(x,u)$ depends not only on the concentration variable u but also on the space variable x, it seems that the methods of [26], [27] are not applicable for the problem (1.1). In this paper, by constructing completely different auxiliary functions from those in [26], [27] and technically using maximum principles, we obtain the conditions on the data to guarantee the blow-up of the solution at some finite time, and conditions to ensure that the solution remains global. In addition, an upper bound for the ‘blow-up time’, an upper estimate of the ‘blow-up rate’, and an upper estimate of the global solution are also given. Our results extend and supplement those obtained in [26], [27].

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

2 Blow-up solution

In this section we establish sufficient conditions on the data of the problem (1.1) to produce a blow-up of the solution $u(x,t)$ at some finite time T and under these conditions we derive an explicit upper bound for T and an explicit upper estimate of the ‘blow-up rate’. The main result of this section is formulated in the following theorem.

Theorem 2.1

Let$u(x,t)$be a solution of the problem (1.1). Assume that the data of the problem (1.1) satisfies the following conditions:

• (i) for any $(s,t)\in D\times {\mathbb{R}}^{+}$,

  $$\begin{array}{r}{a}^{\prime }(s)-a(s)\le 0,{\left(\frac{g_s(x,s)}{a(s)}\right)}_s\le 0,\\ {\left(\frac{a(s)f(s)-{(a(s)f(s))}^{\prime }}{a(s)}\right)}^{\prime }-\frac{a(s)f(s)-{(a(s)f(s))}^{\prime }}{a(s)}\le 0;\end{array}$$

  (2.1)

• (ii) the constant

  $$\alpha =\underset{\overline{D}}{min}\left\{\frac{{\mathrm{e}}^{u_0}}{f(u_0)g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]\right\}>0;$$

  (2.2)

• (iii) the integration

  $${\int }_{M_0}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s<+\mathrm{\infty },M_0=\underset{\overline{D}}{max}{u}_0(x).$$

  (2.3)

Then $u(x,t)$ must blow up in a finite time T and

$$T\le \frac{1}{\alpha }{\int }_{M_0}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s$$

(2.4)

as well as

$$u(x,t)\le {\mathrm{\Phi }}^{-1}(\alpha (T-t)),$$

(2.5)

where

$$\mathrm{\Phi }(z)={\int }_z^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s,z>0,$$

(2.6)

and${\mathrm{\Phi }}^{-1}$is the inverse function of Φ.

Proof

Consider the auxiliary function

$$Q(x,t)={\mathrm{e}}^u{u}_t-\alpha f(u).$$

(2.7)

Now we have

$$\mathrm{\nabla }Q={\mathrm{e}}^u{u}_t\mathrm{\nabla }u+{\mathrm{e}}^u\mathrm{\nabla }{u}_t-\alpha f^{\prime }\mathrm{\nabla }u,$$

(2.8)

$$\mathrm{\Delta }Q={\mathrm{e}}^u{u}_t{|\mathrm{\nabla }u|}^2+2{\mathrm{e}}^u\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t+{\mathrm{e}}^u{u}_t\mathrm{\Delta }u+{\mathrm{e}}^u\mathrm{\Delta }{u}_t-\alpha f^{″}{|\mathrm{\nabla }u|}^2-\alpha f^{\prime }\mathrm{\Delta }u,$$

(2.9)

and

$$\begin{array}{rl}{Q}_t=& {\mathrm{e}}^u{(u_t)}^2+{\mathrm{e}}^u{(u_t)}_t-\alpha f^{\prime }{u}_t\\ =& {\mathrm{e}}^u{(u_t)}^2+{\mathrm{e}}^u{(\frac{ab}{g_u}\mathrm{\Delta }u+\frac{a^{\prime }b}{g_u}{|\mathrm{\nabla }u|}^2+\frac{a}{g_u}\mathrm{\nabla }b\cdot \mathrm{\nabla }u+\frac{f}{g_u})}_t-\alpha f^{\prime }{u}_t\\ =& {\mathrm{e}}^u{(u_t)}^2+(\frac{a^{\prime }b}{g_u}-\frac{ab{g}_{uu}}{g_u^2}){\mathrm{e}}^u{u}_t\mathrm{\Delta }u+\frac{ab}{g_u}{\mathrm{e}}^u\mathrm{\Delta }{u}_t+(\frac{a^{″}b}{g_u}-\frac{a^{\prime }b{g}_{uu}}{g_u^2}){\mathrm{e}}^u{u}_t{|\mathrm{\nabla }u|}^2\\ +2\frac{a^{\prime }b}{g_u}{\mathrm{e}}^u(\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t)+(\frac{a^{\prime }}{g_u}-\frac{a{g}_{uu}}{g_u^2}){\mathrm{e}}^u{u}_t(\mathrm{\nabla }b\cdot \mathrm{\nabla }u)+\frac{a}{g_u}{\mathrm{e}}^u(\mathrm{\nabla }b\cdot \mathrm{\nabla }{u}_t)\\ +[(\frac{f^{\prime }}{g_u}-\frac{f{g}_{uu}}{g_u^2}){\mathrm{e}}^u-\alpha f^{\prime }]u_t.\end{array}$$

(2.10)

It follows from (2.9) and (2.10) that

$$\begin{array}{rl}\frac{ab}{g_u}\mathrm{\Delta }Q-Q_t=& (\frac{a^{\prime }b{g}_{uu}}{g_u^2}-\frac{a^{″}b}{g_u}+\frac{ab}{g_u}){\mathrm{e}}^u{u}_t{|\mathrm{\nabla }u|}^2+(2\frac{ab}{g_u}-2\frac{a^{\prime }b}{g_u}){\mathrm{e}}^u(\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t)\\ +(\frac{ab{g}_{uu}}{g_u^2}-\frac{a^{\prime }b}{g_u}+\frac{ab}{g_u}){\mathrm{e}}^u{u}_t\mathrm{\Delta }u-\alpha \frac{ab{f}^{″}}{g_u}{|\mathrm{\nabla }u|}^2-\alpha \frac{ab{f}^{\prime }}{g_u}\mathrm{\Delta }u-{\mathrm{e}}^u{(u_t)}^2\\ +(\frac{a{g}_{uu}}{g_u^2}-\frac{a^{\prime }}{g_u}){\mathrm{e}}^u{u}_t(\mathrm{\nabla }b\cdot \mathrm{\nabla }u)-\frac{a}{g_u}{\mathrm{e}}^u(\mathrm{\nabla }b\cdot \mathrm{\nabla }{u}_t)\\ +[(\frac{f{g}_{uu}}{g_u^2}-\frac{f^{\prime }}{g_u}){\mathrm{e}}^u+\alpha f^{\prime }]u_t.\end{array}$$

(2.11)

By the first equation of (1.1), we have

$$\mathrm{\Delta }u=\frac{g_u}{ab}{u}_t-\frac{a^{\prime }}{a}{|\mathrm{\nabla }u|}^2-\frac{1}{b}\mathrm{\nabla }b\cdot \mathrm{\nabla }u-\frac{f}{ab}.$$

(2.12)

Substitute (2.12) into (2.11) to obtain

$$\begin{array}{r}\frac{ab}{g_u}\mathrm{\Delta }Q-Q_t\\ =(\frac{{(a^{\prime })}^{2}b}{a{g}_u}-\frac{a^{\prime }b}{g_u}-\frac{a^{″}b}{g_u}+\frac{ab}{g_u}){\mathrm{e}}^u{u}_t{|\mathrm{\nabla }u|}^2+(2\frac{ab}{g_u}-2\frac{a^{\prime }b}{g_u}){\mathrm{e}}^u(\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t)\\ +\frac{a}{g_u}{\left(\frac{g_u}{a}\right)}_u{\mathrm{e}}^u{(u_t)}^2-\frac{a}{g_u}{\mathrm{e}}^u{u}_t(\mathrm{\nabla }b\cdot \mathrm{\nabla }u)+\alpha \frac{a{f}^{\prime }}{g_u}(\mathrm{\nabla }b\cdot \mathrm{\nabla }u)+(\frac{a^{\prime }f}{a{g}_u}-\frac{f}{g_u}-\frac{f^{\prime }}{g_u}){\mathrm{e}}^u{u}_t\\ +(\alpha \frac{a^{\prime }b{f}^{\prime }}{g_u}-\alpha \frac{ab{f}^{″}}{g_u}){|\mathrm{\nabla }u|}^2-\frac{a}{g_u}{\mathrm{e}}^u(\mathrm{\nabla }b\cdot \mathrm{\nabla }{u}_t)+\alpha \frac{f{f}^{\prime }}{g_u}.\end{array}$$

(2.13)

It follows from (2.8) that

$$\mathrm{\nabla }{u}_t={\mathrm{e}}^{-u}\mathrm{\nabla }Q-u_t\mathrm{\nabla }u+\alpha {\mathrm{e}}^{-u}{f}^{\prime }\mathrm{\nabla }u.$$

(2.14)

Substituting (2.14) into (2.13), we get

$$\begin{array}{r}\frac{ab}{g_u}\mathrm{\Delta }Q+(2\frac{b}{g_u}(a^{\prime }-a)\mathrm{\nabla }u+\frac{a}{g_u}\mathrm{\nabla }b)\cdot \mathrm{\nabla }Q-Q_t\\ =(\frac{{(a^{\prime })}^{2}b}{a{g}_u}+\frac{a^{\prime }b}{g_u}-\frac{a^{″}b}{g_u}-\frac{ab}{g_u}){\mathrm{e}}^u{u}_t{|\mathrm{\nabla }u|}^2+(2\alpha \frac{ab{f}^{\prime }}{g_u}-\alpha \frac{a^{\prime }b{f}^{\prime }}{g_u}-\alpha \frac{ab{f}^{″}}{g_u}){|\mathrm{\nabla }u|}^2\\ +\frac{a}{g_u}{\left(\frac{g_u}{a}\right)}_u{\mathrm{e}}^u{(u_t)}^2+(\frac{a^{\prime }f}{a{g}_u}-\frac{f}{g_u}-\frac{f^{\prime }}{g_u}){\mathrm{e}}^u{u}_t+\alpha \frac{f{f}^{\prime }}{g_u}.\end{array}$$

(2.15)

With (2.7), we have

$$u_t={\mathrm{e}}^{-u}Q+\alpha {\mathrm{e}}^{-u}f.$$

(2.16)

Substitute (2.16) into (2.15) to get

$$\begin{array}{r}\frac{ab}{g_u}\mathrm{\Delta }Q+(2\frac{b}{g_u}(a^{\prime }-a)\mathrm{\nabla }u+\frac{a}{g_u}\mathrm{\nabla }b)\cdot \mathrm{\nabla }Q\\ +\{\frac{ab}{g_u}[{\left(\frac{a^{\prime }}{a}\right)}^{\prime }-\frac{a^{\prime }}{a}+1]{|\mathrm{\nabla }u|}^2+\frac{a}{g_u}[{\left(\frac{f}{a}\right)}^{\prime }+\frac{f}{a}]\}Q-Q_t\\ =\alpha \frac{ab}{g_u}[{\left(\frac{af-{(af)}^{\prime }}{a}\right)}^{\prime }-\frac{af-{(af)}^{\prime }}{a}]{|\mathrm{\nabla }u|}^2\\ +\frac{a}{g_u}{\left(\frac{g_u}{a}\right)}_u{\mathrm{e}}^u{(u_t)}^2+\alpha \frac{f^2}{a{g}_u}(a^{\prime }-a).\end{array}$$

(2.17)

The assumptions (2.1) ensure that the right-hand side of (2.17) is nonpositive; that is,

$$\begin{array}{r}\frac{ab}{g_u}\mathrm{\Delta }Q+(2\frac{b}{g_u}(a^{\prime }-a)\mathrm{\nabla }u+\frac{a}{g_u}\mathrm{\nabla }b)\cdot \mathrm{\nabla }Q\\ +\{\frac{ab}{g_u}[{\left(\frac{a^{\prime }}{a}\right)}^{\prime }-\frac{a^{\prime }}{a}+1]{|\mathrm{\nabla }u|}^2+\frac{a}{g_u}[{\left(\frac{f}{a}\right)}^{\prime }+\frac{f}{a}]\}Q-Q_t\\ \le 0\text{in }D\times (0,T).\end{array}$$

(2.18)

Now, by (1.1), we have

$$\frac{\partial Q}{\partial n}={\mathrm{e}}^u\frac{\partial u_t}{\partial n}+{\mathrm{e}}^u{u}_t\frac{\partial u}{\partial n}-\alpha f^{\prime }\frac{\partial u}{\partial n}={\mathrm{e}}^u{\left(\frac{\partial u}{\partial n}\right)}_t=0\text{on }\partial D\times (0,T).$$

(2.19)

Furthermore, it follows from (2.2) that

$$\begin{array}{rl}\underset{\overline{D}}{min}Q(x,0)& =\underset{\overline{D}}{min}\{\frac{{\mathrm{e}}^{u_0}}{g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]-\alpha f(u_0)\}\\ =\underset{\overline{D}}{min}\{f(u_0)[\frac{{\mathrm{e}}^{u_0}}{f(u_0)g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]-\alpha ]\}\\ =0.\end{array}$$

(2.20)

Combining (2.18)-(2.20) and applying the maximum principle [9], we find that the minimum of Q in $\overline{D}\times [0,T)$ is zero. Hence,

$$Q\ge 0\text{in }\overline{D}\times [0,T);$$

that is,

$$\frac{{\mathrm{e}}^u}{f(u)}{u}_t\ge \alpha .$$

(2.21)

At the point $x_0\in \overline{D}$, where $u_0(x_0)=M_0$, integrating (2.21) over $[0,t]$, we get

$${\int }_0^t\frac{{\mathrm{e}}^u}{f(u)}{u}_t\mathrm{d}t={\int }_{M_0}^{u(x_0,t)}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s\ge \alpha t.$$

By the assumption (2.3), we know that $u(x,t)$ must blow up in finite time $t=T$, moreover,

$$T\le \frac{1}{\alpha }{\int }_{M_0}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s.$$

For each fixed x, integrating the inequality (2.21) over $[t,s]$ ($0<t<s<T$), we obtain

$$\begin{array}{rcl}\mathrm{\Phi }(u(x,t))& \ge & \mathrm{\Phi }(u(x,t))-\mathrm{\Phi }(u(x,s))={\int }_{u(x,t)}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s-{\int }_{u(x,s)}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s\\ =& {\int }_{u(x,t)}^{u(x,s)}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s={\int }_t^s\frac{{\mathrm{e}}^u}{f(u)}{u}_t\mathrm{d}t\ge \alpha (s-t).\end{array}$$

Letting $s\to T$, we have

$$\mathrm{\Phi }(u(x,t))\ge \alpha (T-t),$$

which implies

$$u(x,t)\le {\mathrm{\Phi }}^{-1}(\alpha (T-t)).$$

The proof is complete. □

3 Global solution

In this section we establish sufficient conditions on the data of the problem (1.1) in order to ensure that the solution has global existence. Under these conditions, we derive an explicit upper estimate of the global solution. The main results of this section are the following theorem.

Theorem 3.1

Let$u(x,t)$be a solution of the problem (1.1). Assume that the data of the problem (1.1) satisfies the following conditions:

• (i) for any $(s,t)\in D\times {\mathbb{R}}^{+}$,

  $$\begin{array}{r}{a}^{\prime }(s)-a(s)\ge 0,{\left(\frac{g_s(x,s)}{a(s)}\right)}_s\ge 0,\\ {\left(\frac{a(s)f(s)-{(a(s)f(s))}^{\prime }}{a(s)}\right)}^{\prime }-\frac{a(s)f(s)-{(a(s)f(s))}^{\prime }}{a(s)}\ge 0;\end{array}$$

  (3.1)

• (ii) the constant

  $$\beta =\underset{\overline{D}}{max}\left\{\frac{{\mathrm{e}}^{u_0}}{f(u_0)g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]\right\}>0;$$

  (3.2)

• (iii) the integration

  $${\int }_{m_0}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s<+\mathrm{\infty },m_0:=\underset{\overline{D}}{min}{u}_0(x).$$

  (3.3)

Then $u(x,t)$ must be a global solution and

$$u(x,t)\le \mathrm{\Psi }(\beta t+\mathrm{\Psi }(u_0(x))),$$

(3.4)

where

$$\mathrm{\Psi }(z)={\int }_{m_0}^z\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s,z\ge m_0,$$

(3.5)

and${\mathrm{\Psi }}^{-1}$is the inverse function of Ψ.

Proof

Consider an auxiliary function

$$P(x,t)={\mathrm{e}}^u{u}_t-\beta f(u).$$

(3.6)

In (2.17), by replacing Q and α by P and β, respectively, we have

$$\begin{array}{r}\frac{ab}{g_u}\mathrm{\Delta }P+(2\frac{b}{g_u}(a^{\prime }-a)\mathrm{\nabla }u+\frac{a}{g_u}\mathrm{\nabla }b)\cdot \mathrm{\nabla }P\\ +\{\frac{ab}{g_u}[{\left(\frac{a^{\prime }}{a}\right)}^{\prime }-\frac{a^{\prime }}{a}+1]{|\mathrm{\nabla }u|}^2+\frac{a}{g_u}[{\left(\frac{f}{a}\right)}^{\prime }+\frac{f}{a}]\}P-P_t\\ =\beta \frac{ab}{g_u}[{\left(\frac{af-{(af)}^{\prime }}{a}\right)}^{\prime }-\frac{af-{(af)}^{\prime }}{a}]{|\mathrm{\nabla }u|}^2\\ +\frac{a}{g_u}{\left(\frac{g_u}{a}\right)}_u{\mathrm{e}}^u{(u_t)}^2+\beta \frac{f^2}{a{g}_u}(a^{\prime }-a).\end{array}$$

(3.7)

It follows from (3.1) that the right-hand side of (3.7) is nonnegative; that is,

$$\begin{array}{r}\frac{ab}{g_u}\mathrm{\Delta }P+(2\frac{b}{g_u}(a^{\prime }-a)\mathrm{\nabla }u+\frac{a}{g_u}\mathrm{\nabla }b)\cdot \mathrm{\nabla }P\\ +\{\frac{ab}{g_u}[{\left(\frac{a^{\prime }}{a}\right)}^{\prime }-\frac{a^{\prime }}{a}+1]{|\mathrm{\nabla }u|}^2+\frac{a}{g_u}[{\left(\frac{f}{a}\right)}^{\prime }+\frac{f}{a}]\}P-P_t\ge 0\text{in }D\times (0,T).\end{array}$$

(3.8)

With (1.1), we have

$$\frac{\partial P}{\partial n}={\mathrm{e}}^u\frac{\partial u_t}{\partial n}+{\mathrm{e}}^u{u}_t\frac{\partial u}{\partial n}-\beta f^{\prime }\frac{\partial u}{\partial n}={\mathrm{e}}^u{\left(\frac{\partial u}{\partial n}\right)}_t=0\text{on }\partial D\times (0,T).$$

(3.9)

It follows from (3.2) that

$$\begin{array}{rl}\underset{\overline{D}}{max}P(x,0)& =\underset{\overline{D}}{max}\{\frac{{\mathrm{e}}^{u_0}}{g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]-\beta f(u_0)\}\\ =\underset{\overline{D}}{max}\{f(u_0)[\frac{{\mathrm{e}}^{u_0}}{f(u_0)g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]-\beta ]\}\\ =0.\end{array}$$

(3.10)

Combining (3.8)-(3.10) and applying the maximum principle, we know that the maximum of P in $\overline{D}\times [0,T)$ is zero; that is,

$$P(x,t)\le 0\text{in }\overline{D}\times [0,T).$$

(3.11)

From (3.11), we get

$$\frac{{\mathrm{e}}^u}{f(u)}{u}_t\le \beta .$$

(3.12)

For each fixed $x\in \overline{D}$, by integrating (3.12) over $[0,t]$, we have

$$\frac{1}{\beta }{\int }_0^t\frac{{\mathrm{e}}^u}{f(u)}{u}_t\mathrm{d}t=\frac{1}{\beta }{\int }_{u_0(x)}^{u(x,t)}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s\le t.$$

(3.13)

It follows from (3.13) and (3.3) that $u(x,t)$ must be a global solution. Furthermore, by (3.13), we have

$$\mathrm{\Psi }(u(x,t))-\mathrm{\Psi }(u_0(x))={\int }_{m_0}^{u(x,t)}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s-{\int }_{m_0}^{u_0(x)}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s={\int }_{u_0(x)}^{u(x,t)}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s\le \beta t.$$

Hence,

$$u(x,t)\le {\mathrm{\Psi }}^{-1}(\beta t+\mathrm{\Psi }(u_0(x))).$$

The proof is complete. □

4 Applications

When $g(x,u)\equiv g(u)$, $a(u)\equiv 1$, and $b(x)\equiv 1$ or $g(x,u)\equiv g(u)$ and $b(x)\equiv 1$, the conclusions of Theorems 2.1 and 3.1 are valid. In this sense, our results extend and supplement the results of [26], [27].

In what follows, we present several examples to demonstrate the applications of the obtained results.

Example 4.1

Let $u(x,t)$ be a solution of the following problem:

$$\{\begin{array}{cc}{u}_t=\mathrm{\Delta }u+{|\mathrm{\nabla }u|}^2+\frac{2}{1+{|x|}^2}{\sum }_{i=1}^3{x}_i\frac{\partial u}{\partial x_i}+\frac{3{\mathrm{e}}^u}{1+{|x|}^2}\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=2-{(1-{|x|}^2)}^2\hfill & \text{in }\overline{D},\hfill \end{array}$$

where $D=\{x=(x_1,x_2,x_3)|{|x|}^2=x_1^2+x_2^2+x_3^2<1\}$ is the unit ball of ${\mathbb{R}}^3$. The above problem may be turned into the following problem:

$$\{\begin{array}{cc}{({\mathrm{e}}^u(1+{|x|}^2))}_t=\mathrm{\nabla }\cdot ({\mathrm{e}}^u(1+{|x|}^2)\mathrm{\nabla }u)+3{\mathrm{e}}^{2u}\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=2-{(1-{|x|}^2)}^2\hfill & \text{in }\overline{D}.\hfill \end{array}$$

Now

$$\begin{array}{r}a(u)={\mathrm{e}}^u,b(x)=1+{|x|}^2,f(u)=3{\mathrm{e}}^{2u},\\ g(x,u)={\mathrm{e}}^u(1+{|x|}^2),u_0(x)=2-{(1-{|x|}^2)}^2.\end{array}$$

By setting

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

we have $0\le s\le 1$ and

$$\begin{array}{rcl}\alpha & =& \underset{\overline{D}}{min}\left\{\frac{{\mathrm{e}}^{u_0}}{f(u_0)g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]\right\}\\ =& \underset{\overline{D}}{min}\left\{\frac{12+16{|x|}^2-44{|x|}^4-16{|x|}^6+16{|x|}^8+3exp[2-{(1-{|x|}^2)}^2]}{3(1+{|x|}^2)exp[2-{(1-{|x|}^2)}^2]}\right\}\\ =& \underset{0\le s\le 1}{min}\left\{\frac{12+16s-44s^2-16s^3+16s^4+3exp[2-{(1-s)}^2]}{3(1+s)exp[2-{(1-s)}^2]}\right\}\\ =& 0.1391.\end{array}$$

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

$$T\le \frac{1}{\alpha }{\int }_{M_0}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^s}{f(s)}\mathrm{d}s=\frac{1}{0.1391}{\int }_2^{+\mathrm{\infty }}\frac{1}{3{\mathrm{e}}^s}\mathrm{d}s=0.3243$$

as well as

$$u(x,t)\le {\mathrm{\Phi }}^{-1}(\alpha (T-t))=ln\frac{1}{0.4173(T-t)}.$$

Example 4.2

Let $u(x,t)$ be a solution of the following problem:

$$\{\begin{array}{cc}{u}_t=\mathrm{\Delta }u+2{|\mathrm{\nabla }u|}^2+\frac{2}{1+{|x|}^2}{\sum }_{i=1}^3{x}_i\frac{\partial u}{\partial x_i}+\frac{{\mathrm{e}}^{-\frac{2}{3}u}}{2(1+{|x|}^2)}\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=1+{(1-{|x|}^2)}^2\hfill & \text{in }\overline{D},\hfill \end{array}$$

where $D=\{x=(x_1,x_2,x_3)|{|x|}^2=x_1^2+x_2^2+x_3^2<1\}$ is the unit ball of ${\mathbb{R}}^3$. The above problem can be transformed into the following problem:

$$\{\begin{array}{cc}{({\mathrm{e}}^{2u}(1+{|x|}^2))}_t=\mathrm{\nabla }\cdot (2{\mathrm{e}}^{2u}(1+{|x|}^2)\mathrm{\nabla }u)+{\mathrm{e}}^{\frac{u}{2}}\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=1+{(1-{|x|}^2)}^2\hfill & \text{in }\overline{D}.\hfill \end{array}$$

Now we have

$$\begin{array}{r}a(u)=2{\mathrm{e}}^{2u},b(x)=1+{|x|}^2,f(u)={\mathrm{e}}^{\frac{u}{2}},\\ g(x,u)={\mathrm{e}}^{2u}(1+{|x|}^2),u_0(x)=1+{(1-{|x|}^2)}^2.\end{array}$$

In order to determine the constant β, we assume

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

then $0\le s\le 1$ and

$$\begin{array}{rl}\beta =& \underset{\overline{D}}{max}\left\{\frac{{\mathrm{e}}^{u_0}}{f(u_0)g_u(x,u_0)}[\mathrm{\nabla }\cdot (a(u_0)b(x)\mathrm{\nabla }{u}_0)+f(u_0)]\right\}\\ =& \underset{\overline{D}}{max}\left\{\right(exp[\frac{1}{2}+\frac{1}{2}{(1-{|x|}^2)}^2]\\ \times [-24+64{|x|}^2-8{|x|}^4-64{|x|}^6+64{|x|}^8+exp(-\frac{3}{2}-\frac{3}{2}{(1-{|x|}^2)}^2)])\\ /(1+{|x|}^2)\}\\ =& \underset{0\le s\le 1}{max}\left\{\frac{exp[\frac{1}{2}+\frac{1}{2}{(1-s)}^2][-24+64s-8s^2-64s^3+64s^8+exp(-\frac{3}{2}-\frac{3}{2}{(1-s)}^2)]}{1+s}\right\}\\ =& 26.5635.\end{array}$$

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

$$\begin{array}{rl}u(x,t)& \le {\mathrm{\Psi }}^{-1}(\beta t+\mathrm{\Psi }(u_0(x)))=2ln[13.2818t+exp(\frac{1}{2}{u}_0(x))]\\ =2ln[13.2818t+exp(\frac{1}{2}+\frac{1}{2}{(1-{|x|}^2)}^2)].\end{array}$$

Author’s contributions

All results belong to Juntang Ding.