Global and blow-up solutions for nonlinear parabolic problems with a gradient term under Robin boundary conditions

Abstract

In this paper, we study the global and blow-up solutions of the following nonlinear parabolic problems with a gradient term under Robin boundary conditions:

$$\{\begin{array}{cc}{(b(u))}_t=\mathrm{\nabla }\cdot (g(u)\mathrm{\nabla }u)+f(x,u,{|\mathrm{\nabla }u|}^2,t)\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}+\gamma u=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=u_0(x)>0\hfill & \text{in }\overline{D},\hfill \end{array}$$

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D. By constructing auxiliary functions and using maximum principles, the sufficient conditions for the existence of a global solution, an upper estimate of the global solution, the sufficient conditions for the existence of a blow-up solution, an upper bound for ‘blow-up time’, and an upper estimate of ‘blow-up rate’ are specified under some appropriate assumptions on the functions f, g, b and initial value $u_0$.

MSC:35K55, 35B05, 35K57.

1 Introduction

In this paper, we study the global and blow-up solutions of the following nonlinear parabolic problems with a gradient term under Robin boundary conditions:

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

(1.1)

where $q:={|\mathrm{\nabla }u|}^2$, $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D, $\partial /\partial n$ represents the outward normal derivative on ∂D, γ is a positive constant, $u_0$ is the initial value, T is the maximal existence time of u, and $\overline{D}$ is the closure of D. Set ${\mathbb{R}}^{+}:=(0,+\mathrm{\infty })$. We assume, throughout the paper, that $b(s)$ is a $C^3({\mathbb{R}}^{+})$ function, $b^{\prime }(s)>0$ for any $s\in {\mathbb{R}}^{+}$, $g(s)$ is a positive $C^2({\mathbb{R}}^{+})$ function, $f(x,s,d,t)$ is a nonnegative $C^1(\overline{D}\times {\mathbb{R}}^{+}\times \overline{{\mathbb{R}}^{+}}\times {\mathbb{R}}^{+})$ function, and $u_0(x)$ is a positive $C^2(\overline{D})$ function. Under the above assumptions, the classical theory [1] of parabolic equation assures that there exists a unique classical solution $u(x,t)$ with some $T>0$ for problem (1.1) and the solution is positive over $\overline{D}\times [0,T)$. Moreover, the regularity theorem [2] implies $u(x,t)\in C^3(D\times (0,T))\cap C^2(\overline{D}\times [0,T))$.

Many papers have studied the global and blow-up solutions of parabolic problems with a gradient term (see, for instance, [3–13]). Some authors have discussed the global and blow-up solutions of parabolic problems under Robin boundary conditions and have got a lot of meaningful results (see [14–20] and the references cited therein). Some special cases of problem (1.1) have been treated already. Zhang [21] dealt with the following problem:

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

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D. By constructing auxiliary functions and using maximum principles, the sufficient conditions characterized by functions f, g and $u_0$ were given for the existence of a blow-up solution. Zhang [22] investigated the following problem:

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

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D. By constructing some auxiliary functions and using maximum principles, the sufficient conditions were obtained there for the existence of global and blow-up solutions. Meanwhile, the upper estimate of a global solution, the upper bound of ‘blow-up time’ and the upper estimate of ‘blow-up rate’ were also given. Ding [21] considered the following problem:

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

where $D\subset {\mathbb{R}}^N$ ($N\ge 2$) is a bounded domain with smooth boundary ∂D. By constructing some appropriate auxiliary functions and using a first-order differential inequality technique, the sufficient conditions were obtained for the existence of global and blow-up solutions. For the blow-up solution, an upper and a lower bound on blow-up time were also given.

In this paper, we study problem (1.1). Since the function $f(x,u,q,t)$ contains a gradient term $q={|\mathrm{\nabla }u|}^2$, it seems that the methods of [21–23] are not applicable for problem (1.1). In this paper, by constructing completely different auxiliary functions with those in [21–23] and technically using maximum principles, we obtain some existence theorems of a global solution, an upper estimate of the global solution, the existence theorems of a blow-up solution, an upper bound of ‘blow-up time’, and an upper estimates of ‘blow-up rate’. Our results extend and supplement those obtained [21–23].

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

2 Global solution

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

Theorem 2.1 Let u be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are satisfied:

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

  $$\begin{array}{r}{(s{b}^{\prime }(s))}^{\prime }\ge 0,s{b}^{\prime }(s)-{(s{b}^{\prime }(s))}^{\prime }\le 0,{\left(\frac{g(s)}{b^{\prime }(s)}\right)}^{\prime }\le 0,\\ {[\frac{1}{g(s)}{\left(\frac{g(s)}{b^{\prime }(s)}\right)}^{\prime }+\frac{1}{b^{\prime }(s)}]}^{\prime }+\frac{1}{g}{\left(\frac{g(s)}{b^{\prime }(s)}\right)}^{\prime }+\frac{1}{b^{\prime }(s)}\le 0;\end{array}$$

  (2.1)

• (ii) for any $(x,s,d,t)\in D\times {\mathbb{R}}^{+}\times \overline{{\mathbb{R}}^{+}}\times {\mathbb{R}}^{+}$,

  $$\begin{array}{r}{f}_t(x,s,d,t)\le 0,f_d(x,s,d,t)[{\left(\frac{1}{b^{\prime }(s)}\right)}^{\prime }+\frac{1}{b^{\prime }(s)}]\le 0,\\ {\left(\frac{f(x,s,d,t)b^{\prime }(s)}{g(s)}\right)}_s-\frac{f(x,s,d,t)b^{\prime }(s)}{g(s)}\le 0;\end{array}$$

  (2.2)
• (iii)
  $${\int }_{m_0}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s=+\mathrm{\infty },m_0:=\underset{\overline{D}}{min}{u}_0(x);$$

  (2.3)
• (iv)
  $$\alpha :=\underset{\overline{D}}{max}\frac{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)}{{\mathrm{e}}^{u_0}}>0,q_0:={|\mathrm{\nabla }{u}_0|}^2.$$

  (2.4)

Then the solution u to problem (1.1) must be a global solution and

$$u(x,t)\le H^{-1}(\alpha t+H(u_0(x,t))),(x,t)\in \overline{D}\times \overline{{\mathbb{R}}^{+}},$$

(2.5)

where

$$H(z):={\int }_{m_0}^z\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s,z\ge m_0,$$

(2.6)

and $H^{-1}$ is the inverse function of H.

Proof Consider the auxiliary function

$$P(x,t):=b^{\prime }(u)u_t-\alpha {\mathrm{e}}^u.$$

(2.7)

Now we have

$$\mathrm{\nabla }P=b^{″}{u}_t\mathrm{\nabla }u+b^{\prime }\mathrm{\nabla }{u}_t-\alpha {\mathrm{e}}^u\mathrm{\nabla }u,$$

(2.8)

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

(2.9)

and

$$\begin{array}{rl}{P}_t=& b^{″}{(u_t)}^2+b^{\prime }{(u_t)}_t-\alpha {\mathrm{e}}^u{u}_t\\ =& b^{″}{(u_t)}^2+b^{\prime }{(\frac{g}{b^{\prime }}\mathrm{\Delta }u+\frac{g^{\prime }}{b^{\prime }}{|\mathrm{\nabla }u|}^2+\frac{f}{b^{\prime }})}_t-\alpha {\mathrm{e}}^u{u}_t\\ =& b^{″}{(u_t)}^2+(g^{\prime }-\frac{b^{″}g}{b^{\prime }})u_t\mathrm{\Delta }u+g\mathrm{\Delta }{u}_t+(g^{″}-\frac{b^{″}{g}^{\prime }}{b^{\prime }})u_t{|\mathrm{\nabla }u|}^2\\ +(2g^{\prime }+2f_q)\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t+(f_u-\frac{b^{″}f}{b^{\prime }}-\alpha {\mathrm{e}}^u)u_t+f_t.\end{array}$$

(2.10)

It follows from (2.9) and (2.10) that

$$\begin{array}{rl}\frac{g}{b^{\prime }}\mathrm{\Delta }P-P_t=& (\frac{b^{‴}g}{b^{\prime }}+\frac{b^{″}{g}^{\prime }}{b^{\prime }}-g^{″})u_t{|\mathrm{\nabla }u|}^2+(2\frac{b^{″}g}{b^{\prime }}-2g^{\prime }-2f_q)\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t\\ +(2\frac{b^{″}g}{b^{\prime }}-g^{\prime })u_t\mathrm{\Delta }u-\alpha \frac{g}{b^{\prime }}{\mathrm{e}}^u{|\mathrm{\nabla }u|}^2-\alpha \frac{g}{b^{\prime }}{\mathrm{e}}^u\mathrm{\Delta }u-b^{″}{(u_t)}^2\\ +(\frac{b^{″}f}{b^{\prime }}-f_u+\alpha {\mathrm{e}}^u)u_t-f_t.\end{array}$$

(2.11)

By (1.1), we have

$$\mathrm{\Delta }u=\frac{b^{\prime }}{g}{u}_t-\frac{g^{\prime }}{g}{|\mathrm{\nabla }u|}^2-\frac{f}{g}.$$

(2.12)

Substitute (2.12) into (2.11), to get

$$\begin{array}{rl}\frac{g}{b^{\prime }}\mathrm{\Delta }P-P_t=& (\frac{b^{‴}g}{b^{\prime }}-\frac{b^{″}{g}^{\prime }}{b^{\prime }}-g^{″}+\frac{{(g^{\prime })}^2}{g})u_t{|\mathrm{\nabla }u|}^2+(2\frac{b^{″}g}{b^{\prime }}-2g^{\prime }-2f_q)\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t\\ -\frac{{(b^{\prime })}^2}{g}{\left(\frac{g}{b^{\prime }}\right)}^{\prime }{(u_t)}^2+(\frac{f{g}^{\prime }}{g}-\frac{b^{″}f}{b^{\prime }}-f_u)u_t+(\alpha \frac{g^{\prime }}{b^{\prime }}{\mathrm{e}}^u-\alpha \frac{g}{b^{\prime }}{\mathrm{e}}^u){|\mathrm{\nabla }u|}^2\\ +\alpha \frac{f}{b^{\prime }}{\mathrm{e}}^u-f_t.\end{array}$$

(2.13)

With (2.8), we have

$$\mathrm{\nabla }{u}_t=\frac{1}{b^{\prime }}\mathrm{\nabla }P-\frac{b^{″}}{b^{\prime }}{u}_t\mathrm{\nabla }u+\alpha \frac{{\mathrm{e}}^u}{b^{\prime }}\mathrm{\nabla }u.$$

(2.14)

Next, we substitute (2.14) into (2.13) to obtain

$$\begin{array}{r}\frac{g}{b^{\prime }}\mathrm{\Delta }P+[2{\left(\frac{g}{b^{\prime }}\right)}^{\prime }+2\frac{f_q}{b^{\prime }}]\mathrm{\nabla }u\cdot \mathrm{\nabla }P-P_t\\ =(\frac{b^{‴}g}{b^{\prime }}+\frac{b^{″}{g}^{\prime }}{b^{\prime }}-g^{″}+\frac{{(g^{\prime })}^2}{g}-2\frac{{(b^{″})}^{2}g}{{(b^{\prime })}^2}+2\frac{b^{″}{f}_q}{b^{\prime }})u_t{|\mathrm{\nabla }u|}^2\\ +(2\alpha \frac{b^{″}g}{{(b^{\prime })}^2}{\mathrm{e}}^u-\alpha \frac{g^{\prime }}{b^{\prime }}{\mathrm{e}}^u-\alpha \frac{g}{b^{\prime }}{\mathrm{e}}^u-2\alpha \frac{f_q}{b^{\prime }}{\mathrm{e}}^u){|\mathrm{\nabla }u|}^2\\ -\frac{{(b^{\prime })}^2}{g}{\left(\frac{g}{b^{\prime }}\right)}^{\prime }{(u_t)}^2+(\frac{f{g}^{\prime }}{g}-\frac{b^{″}f}{b}-f_u)u_t+\alpha \frac{f}{b^{\prime }}{\mathrm{e}}^u-f_t.\end{array}$$

(2.15)

In view of (2.7), we have

$$u_t=\frac{1}{b^{\prime }}P+\alpha \frac{{\mathrm{e}}^u}{b^{\prime }}.$$

(2.16)

Substituting (2.16) into (2.15), we get

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

(2.17)

The assumptions (2.1) and (2.2) guarantee that the right-hand side of (2.17) is nonnegative, i.e.,

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

(2.18)

By applying the maximum principle [24], it follows from (2.18) that P can attain its nonnegative maximum only for $\overline{D}\times \{0\}$ or $\partial D\times (0,T)$. For $\overline{D}\times \{0\}$, by (2.4), we have

$$\begin{array}{rl}\underset{\overline{D}}{max}P(x,0)& =\underset{\overline{D}}{max}\{b^{\prime }(u_0){(u_0)}_t-\alpha {\mathrm{e}}^{u_0}\}=\underset{\overline{D}}{max}\{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)-\alpha {\mathrm{e}}^{u_0}\}\\ =\underset{\overline{D}}{max}\left\{{\mathrm{e}}^{u_0}[\frac{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)}{{\mathrm{e}}^{u_0}}-\alpha ]\right\}=0.\end{array}$$

We claim that P cannot take a positive maximum at any point $(x,t)\in \partial D\times (0,T)$. In fact, suppose that P takes a positive maximum at a point $(x_0,t_0)\in \partial D\times (0,T)$, then

$$P(x_0,t_0)>0\text{and}\frac{\partial P}{\partial n}{|}_{(x_0,t_0)}>0.$$

(2.19)

With (1.1) and (2.16), we have

$$\begin{array}{rl}\frac{\partial P}{\partial n}& =b^{″}{u}_t\frac{\partial u}{\partial n}+b^{\prime }\frac{\partial u_t}{\partial n}-\alpha {\mathrm{e}}^u\frac{\partial u}{\partial n}=-\gamma b^{″}u{u}_t+b^{\prime }{\left(\frac{\partial u}{\partial n}\right)}_t+\gamma \alpha u{\mathrm{e}}^u\\ =-\gamma b^{″}u{u}_t+b^{\prime }{(-\gamma u)}_t+\gamma \alpha u{\mathrm{e}}^u=-\gamma {\left(u{b}^{\prime }\right)}^{\prime }{u}_t+\gamma \alpha u{\mathrm{e}}^u\\ =-\gamma {\left(u{b}^{\prime }\right)}^{\prime }(\frac{1}{b^{\prime }}P+\alpha \frac{1}{b^{\prime }}{\mathrm{e}}^u)+\gamma \alpha u{\mathrm{e}}^u\\ =-\gamma \frac{{(u{b}^{\prime })}^{\prime }}{b^{\prime }}P+\gamma \alpha {\mathrm{e}}^u\frac{u{b}^{\prime }-{(u{b}^{\prime })}^{\prime }}{b^{\prime }}\text{on }\partial D\times (0,T).\end{array}$$

(2.20)

Next, by using the fact that ${(s{b}^{\prime }(s))}^{\prime }\ge 0$, $s{b}^{\prime }(s)-{(s{b}^{\prime }(s))}^{\prime }\le 0$ for any $s\in {\mathbb{R}}^{+}$, it follows from (2.20) that

$$\frac{\partial P}{\partial n}{|}_{(x_0,t_0)}\le 0,$$

which contradicts with inequality (2.19). Thus we know that the maximum of P in $\overline{D}\times [0,T)$ is zero, i.e.,

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

and

$$\frac{b^{\prime }(u)}{{\mathrm{e}}^u}{u}_t\le \alpha .$$

(2.21)

For each fixed $x\in \overline{D}$, integration of (2.21) from 0 to t yields

$${\int }_0^t\frac{b^{\prime }(u)}{{\mathrm{e}}^u}{u}_t\mathrm{d}t={\int }_{u_0(x)}^{u(x,t)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s\le \alpha t,$$

(2.22)

which implies that u must be a global solution. Actually, if that u blows up at finite time T, then

$$\underset{t\to T^{-}}{lim}u(x,t)=+\mathrm{\infty }.$$

Passing to the limit as $t\to T^{-}$ in (2.22) yields

$${\int }_{u_0(x)}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s\le \alpha T$$

and

$${\int }_{m_0}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s={\int }_{m_0}^{u_0(x)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s+{\int }_{u_0(x)}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s\le {\int }_{m_0}^{u_0(x)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s+\alpha T<+\mathrm{\infty },$$

which contradicts with assumption (2.3). This shows that u is global. Moreover, it follows from (2.22) that

$${\int }_{u_0(x)}^{u(x,t)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s={\int }_{m_0}^{u(x,t)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s-{\int }_{m_0}^{u_0(x)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s=H(u(x,t))-H(u_0(x))\le \alpha t.$$

Since H is an increasing function, we have

$$u(x,t)\le H^{-1}(\alpha t+H(u_0(x))).$$

The proof is complete. □

3 Blow-up solution

The following theorem is the main result for the blow-up solution.

Theorem 3.1 Let u be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are fulfilled:

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

  $$\begin{array}{r}{(s{b}^{\prime }(s))}^{\prime }\ge 0,s{b}^{\prime }(s)-{(s{b}^{\prime }(s))}^{\prime }\ge 0,{\left(\frac{g(s)}{b^{\prime }(s)}\right)}^{\prime }\ge 0,\\ {[\frac{1}{g(s)}{\left(\frac{g(s)}{b^{\prime }(s)}\right)}^{\prime }+\frac{1}{b^{\prime }(s)}]}^{\prime }+\frac{1}{g}{\left(\frac{g(s)}{b^{\prime }(s)}\right)}^{\prime }+\frac{1}{b^{\prime }(s)}\ge 0;\end{array}$$

  (3.1)

• (ii) for any $(x,s,d,t)\in D\times {\mathbb{R}}^{+}\times \overline{{\mathbb{R}}^{+}}\times {\mathbb{R}}^{+}$,

  $$\begin{array}{r}{f}_t(x,s,d,t)\ge 0,f_d(x,s,d,t)[{\left(\frac{1}{b^{\prime }(s)}\right)}^{\prime }+\frac{1}{b^{\prime }(s)}]\ge 0,\\ {\left(\frac{f(x,s,d,t)b^{\prime }(s)}{g(s)}\right)}_s-\frac{f(x,s,d,t)b^{\prime }(s)}{g(s)}\ge 0;\end{array}$$

  (3.2)
• (iii)
  $${\int }_{M_0}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s<+\mathrm{\infty },M_0:=\underset{\overline{D}}{max}{u}_0(x);$$

  (3.3)
• (iv)
  $$\beta :=\underset{\overline{D}}{min}\frac{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)}{{\mathrm{e}}^{u_0}}>0,q_0:={|\mathrm{\nabla }{u}_0|}^2.$$

  (3.4)

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

$$T\le \frac{1}{\beta }{\int }_{M_0}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s,$$

(3.5)

$$u(x,t)\le G^{-1}(\beta (T-t)),(x,t)\in \overline{D}\times [0,T),$$

(3.6)

where

$$G(z):={\int }_z^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s,z>0,$$

(3.7)

and $G^{-1}$ is the inverse function of G.

Proof Construct the following auxiliary function:

$$Q(x,t):=b^{\prime }(u)u_t-\beta {\mathrm{e}}^u.$$

(3.8)

Replacing P and α with Q and β in (2.17), respectively, we get

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

(3.9)

Assumptions (3.1) and (3.2) imply that the right-hand side in equality (3.9) is nonpositive, i.e.,

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

(3.10)

With (3.4), we have

$$\begin{array}{rl}\underset{\overline{D}}{min}Q(x,0)& =\underset{\overline{D}}{min}\{b^{\prime }(u_0){(u_0)}_t-\beta {\mathrm{e}}^{u_0}\}=\underset{\overline{D}}{min}\{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)-\beta {\mathrm{e}}^{u_0}\}\\ =\underset{\overline{D}}{min}\left\{{\mathrm{e}}^{u_0}[\frac{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)}{{\mathrm{e}}^{u_0}}-\beta ]\right\}=0.\end{array}$$

(3.11)

Substituting P and α with Q and β in (2.20), respectively, we have

$$\frac{\partial Q}{\partial n}=-\gamma \frac{{(u{b}^{\prime })}^{\prime }}{b^{\prime }}Q+\gamma \beta {\mathrm{e}}^u\frac{u{b}^{\prime }-{(u{b}^{\prime })}^{\prime }}{b^{\prime }}\text{on }\partial D\times (0,T).$$

(3.12)

Combining (3.10)-(3.12) with the fact that ${(s{b}^{\prime }(s))}^{\prime }\ge 0$, $s{b}^{\prime }(s)-{(s{b}^{\prime }(s))}^{\prime }\ge 0$ for any $s\in {\mathbb{R}}^{+}$, and applying the maximum principles again, it follows that the minimum of Q in $\overline{D}\times [0,T)$ is zero. Thus

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

and

$$\frac{b^{\prime }(u)}{{\mathrm{e}}^u}{u}_t\ge \beta .$$

(3.13)

At the point $x^{\ast }\in \overline{D}$, where $u_0(x^{\ast })=M_0$, integrate (3.13) over $[0,t]$ to get

$${\int }_0^t\frac{b^{\prime }(u)}{{\mathrm{e}}^u}{u}_t\mathrm{d}t={\int }_{M_0}^{u(x^{\ast },t)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s\ge \beta t,$$

(3.14)

which implies that u must blow up in finite time. Actually, if u is a global solution of (1.1), then for any $t>0$, (3.14) shows

$${\int }_{M_0}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s\ge {\int }_{M_0}^{u(x^{\ast },t)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s\ge \beta t.$$

(3.15)

Letting $t\to +\mathrm{\infty }$ in (3.15), we have

$${\int }_{M_0}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s=+\mathrm{\infty },$$

which contradicts with assumption (3.3). This shows that u must blow up in finite time $t=T$. Furthermore, letting $t\to T$ in (3.14), we get

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

By integrating inequality (3.13) over $[t,s]$ ($0<t<s<T$), for each fixed x, we obtain

$$\begin{array}{rl}G(u(x,t))& \ge G(u(x,t))-G(u(x,s))={\int }_{u(x,t)}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s-{\int }_{u(x,s)}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s\\ ={\int }_{u(x,t)}^{u(x,s)}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s={\int }_t^s\frac{b^{\prime }(u)}{{\mathrm{e}}^u}{u}_t\mathrm{d}t\ge \beta (s-t).\end{array}$$

Hence, by letting $s\to T$, we have

$$G(u(x,t))\ge \beta (T-t).$$

Since G is a decreasing function, we obtain

$$u(x,t)\le G^{-1}(\beta (T-t)).$$

The proof is complete. □

4 Applications

When $b(u)\equiv u$ and $f(x,u,q,t)\equiv f(u)$, the results stated in Theorem 3.1 are valid. When $g(u)\equiv 1$ and $f(x,u,q,t)\equiv f(u)$ or $f(x,u,q,t)\equiv f(u)$, the conclusions of Theorems 2.1 and 3.1 still hold true. In this sense, our results extend and supplement the results of [21–23].

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

Example 4.1 Let u be a solution of the following problem:

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

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

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

Now

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

In order to determine the constant α, we assume

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

then $0\le s\le 1$ and

$$\begin{array}{rl}\alpha & =\underset{\overline{D}}{max}\frac{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)}{{\mathrm{e}}^{u_0}}\\ =\underset{\overline{D}}{max}\{32{|x|}^2-4{|x|}^4-18+(1+{|x|}^2)[exp(-4+2{|x|}^2)+exp(-2+5{|x|}^2)]\}\\ =\underset{0\le s\le 1}{max}\{32s-4s^2-18+(1+s)[exp(-4+2s)+exp(-2+5s)]\}\\ =50.4417.\end{array}$$

It is easy to check that (2.1)-(2.3) hold. By Theorem 2.1, u must be a global solution, and

$$\begin{array}{rl}u(x,t)& \le H^{-1}(\alpha t+H(u_0(x)))=-1+\sqrt{50.4417t+{(1+u_0(x))}^2}\\ =-1+\sqrt{50.4417t+{(3-{|x|}^2)}^2}.\end{array}$$

Example 4.2 Let u be a solution of the following problem:

$$\{\begin{array}{cc}{u}_t=\mathrm{\Delta }u-\frac{1}{u(1+u)}{|\mathrm{\nabla }u|}^2+\frac{u({\mathrm{e}}^u-{\mathrm{e}}^{-q})}{1+u}(6+t{|x|}^2)\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}+2u=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=2-{|x|}^2\hfill & \text{in }\overline{D},\hfill \end{array}$$

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

$$\{\begin{array}{cc}{(u+lnu)}_t=\mathrm{\nabla }\cdot ((1+\frac{1}{u})\mathrm{\nabla }u)+({\mathrm{e}}^u-{\mathrm{e}}^{-q})(6+t{|x|}^2)\hfill & \text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}+2u=0\hfill & \text{on }\partial D\times (0,T),\hfill \\ u(x,0)=2-{|x|}^2\hfill & \text{in }\overline{D}.\hfill \end{array}$$

Now we have

$$\begin{array}{r}b(u)=u+lnu,g(u)=1+\frac{1}{u},f(x,u,q,t)=({\mathrm{e}}^u-{\mathrm{e}}^{-q})(6+t{|x|}^2),\\ u_0(x)=2-{|x|}^2,\gamma =2.\end{array}$$

By setting

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

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

$$\begin{array}{rl}\beta & =\underset{\overline{D}}{min}\frac{\mathrm{\nabla }\cdot (g(u_0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)}{{\mathrm{e}}^{u_0}}\\ =\underset{\overline{D}}{min}\{\frac{-6{|x|}^4+26{|x|}^2-36}{{(2-{|x|}^2)}^{2}exp(2-{|x|}^2)}+6[1-exp(-3{|x|}^2-2)]\}\\ =\underset{0\le s\le 1}{min}\{\frac{-6s^2+26s-36}{{(2-s)}^{2}exp(2-s)}+6[1-exp(-3s-2)]\}\\ =0.0735.\end{array}$$

Again it is easy to check that (3.1)-(3.3) hold. By Theorem 3.1, u must blow up in finite time T, and

$$\begin{array}{c}T\le \frac{1}{\beta }{\int }_{M_0}^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s=\frac{1}{0.0735}{\int }_2^{+\mathrm{\infty }}(1+\frac{1}{s})\frac{1}{{\mathrm{e}}^s}\mathrm{d}s=2.5066,\hfill \\ u(x,t)\le G^{-1}(\beta (T-t))=G^{-1}(0.0735(T-t)),\hfill \end{array}$$

where

$$G(z)={\int }_z^{+\mathrm{\infty }}\frac{b^{\prime }(s)}{{\mathrm{e}}^s}\mathrm{d}s={\int }_z^{+\mathrm{\infty }}(1+\frac{1}{s})\frac{1}{{\mathrm{e}}^s}\mathrm{d}s,z\ge 0,$$

and $G^{-1}$ is the inverse function of G.

Remark 4.1 We can see from Example 4.1 that when the equation has a gradient term with exponential increase, the functions g and b increase exponentially to ensure that the solution of (1.1) blows up. It follows from Example 4.2 that when the equation has a gradient term with exponential decay, the appropriate assumptions on the functions g and b can guarantee the solution of (1.1) to be global.