Global and blow-up solutions for quasilinear parabolic equations with a gradient term and nonlinear boundary flux

Abstract

This work is concerned with positive classical solutions for a quasilinear parabolic equation with a gradient term and nonlinear boundary flux. We find sufficient conditions for the existence of global and blow-up solutions. Moreover, 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 given. Finally, some application examples are presented.

MSC:35R45, 35K65, 34A12.

1 Introduction

In this paper, we consider the quasilinear parabolic equation with a gradient term

$${(g(u))}_t=\mathrm{\nabla }\cdot (a(u)b(x)c(t)\mathrm{\nabla }u)+f(x,u,q,t)\text{in }D\times (0,T),$$

(1.1)

subject to the nonlinear boundary flux and initial conditions

$$\frac{\partial u}{\partial n}=h(x,t)r(u)\text{on }\partial D\times (0,T),$$

(1.2)

$$u(x,0)=u_0(x)\text{in }\overline{D}.$$

(1.3)

Here $D\subset {\mathbb{R}}^N$ ($N\ge 1$) is a bounded domain with a smooth boundary ∂D, $\overline{D}$ is the closure of D, $q={|\mathrm{\nabla }u|}^2$, n is the outer normal vector and T is the maximum existence time of $u(x,t)$. $a(u)b(x)c(t)$, $f(x,u,q,t)$ and $h(x,t)r(u)$ are nonlinear diffusion coefficient, reaction term and boundary flux, respectively. Let ${\mathbb{R}}^{+}=(0,+\mathrm{\infty })$, $\overline{{\mathbb{R}}^{+}}=[0,+\mathrm{\infty })$, and suppose that the function $g(s)\in C^2({\mathbb{R}}^{+})$, $g^{\prime }(s)>0$ for any $s>0$, $a(s)\in C^2({\mathbb{R}}^{+})$, $b(x)\in C^1(\overline{D})$, $c(t)\in C^1({\mathbb{R}}^{+})$, $f(x,u,q,t)\in C^1(\overline{D}\times {\mathbb{R}}^{+}\times \overline{{\mathbb{R}}^{+}}\times \overline{{\mathbb{R}}^{+}})$ is a nonnegative function, $h(x,t)\in C^1(\overline{D}\times (0,T))$, $r(s)\in C^2({\mathbb{R}}^{+})$ is a positive function, and the positive function $u_0(x)\in C^2(\overline{D})$ satisfies the compatibility conditions. Under these assumptions, the classical parabolic equation theory [[1], Section 3] ensures that there exists a unique classical solution $u(x,t)$ to problem (1.1)-(1.3) for some $T>0$, and the solution is positive over $\overline{D}\times [0,T)$. Moreover, by the regularity theorem [[2], Chapter 3], we know $u\in C^3(D\times (0,T))\cap C^2(\overline{D}\times (0,T))$.

Equation (1.1) describes the diffusion of concentration of some Newtonian fluids through porous media or the density of some biological species in many physical phenomena and combustion theories (see [3, 4]). The nonlinear Neumann boundary value condition (1.2) can be physically interpreted as the nonlinear radial law (see, e.g., [5, 6]).

In recent years the questions like blow-up and global solvability for nonlinear evolution equations have been investigated extensively by many authors. In particular, for the parabolic equations with a gradient term, we refer to [7–12]etc. For example, Souplet and Weissler [7] studied the semilinear parabolic equation

$$u_t=\mathrm{\Delta }u+f(u,\mathrm{\nabla }u)\text{in }D\times (0,T),$$

subject to the homogeneous Dirichlet boundary condition. By using the comparison principle and constructing a self-similar lower solution, they obtained sufficient conditions for global existence and blow-up solutions. Andreu [8] used a similar method to study the quasilinear parabolic equation

$$u_t=\mathrm{\Delta }{u}^m+f(u,\mathrm{\nabla }{u}^m)\text{in }D\times (0,T).$$

Chen [9] considered the following semilinear parabolic equation:

$$u_t=\mathrm{\Delta }u+f(u)+g(u){|\mathrm{\nabla }u|}^2\text{in }D\times (0,T),$$

with the homogeneous Dirichlet boundary condition. By estimating the integral of ratio of one solution to the other, the author proved both global existence and blow-up results. Then he used the same method to study a more generalized equation with a gradient term, see [10].

For the nonlinear parabolic equations with Neumann boundary conditions, Lair and Oxley [11] considered the quasilinear parabolic equation without a gradient term

$$u_t=\mathrm{\nabla }\cdot (a(u)\mathrm{\nabla }u)+f(u)\text{in }D\times (0,T),$$

subject to the homogeneous Neumann boundary conditions, and they obtained the necessary and sufficient conditions for the global existence and blow-up solution by the approximation method. Recently, Ding and Gao [12] investigated an initial boundary value problem of the quasilinear parabolic equation with a gradient term

$${(g(u))}_t=\mathrm{\Delta }u+f(x,u,{|\mathrm{\nabla }u|}^2,t)\text{in }D\times (0,T),$$

subject to boundary flux $\frac{\partial u}{\partial n}=r(u)$, and they obtained sufficient conditions for the global existence and blow-up solution, the upper estimate of global solution and blow-up time.

Motivated by the above works, we construct an appropriate auxiliary function and use the Hopf maximum principle to study problem (1.1)-(1.3). The aim of this paper is to obtain sufficient conditions for the existence of blow-up and global solution, an upper bound for the ‘blow-up time’, an upper estimate of the ‘blow-up rate’ and an upper estimate of the global solution and then to give some examples.

2 Main results and proof

We now state and prove the main results of this paper. Firstly, we give sufficient conditions of the existence of a blow-up solution of problem (1.1)-(1.3).

Theorem 1 Let $u\in C^3(D\times (0,T))\cap C^2(\overline{D}\times (0,T))$ be a solution of problem (1.1)-(1.3). Assume that the following conditions hold:

• (1) For any $(x,s,q,t)\in \overline{D}\times \mathbb{R}\times {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$,

  $$a(s)>0,b(x)>0,c(t)>0,r(s)>0,h(x,t)\ge 0;$$

  (2.1)

• (2) For any $(x,s,q,t)\in \overline{D}\times {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$,

  $$\begin{array}{r}{a}^{\prime }(s)\ge 0,h_t(x,t)\ge 0,f_q\ge 0,{\left(\frac{a(s)}{g^{\prime }(s)}\right)}^{\prime }\ge 0,r^{\prime }(s)\ge \frac{a^{\prime }(s)}{a(s)}r(s),\\ r^{″}(s)\ge \frac{a^{\prime }(s)}{a(s)}{r}^{\prime }(s),\end{array}$$

  (2.2)
  $$\begin{array}{r}{c}^{\prime }(t)\ge 0,g^{\prime }(s)>0,f_t(x,s,q,t)\ge \frac{c^{\prime }(t)}{c(t)}f(x,s,q,t),\\ f_s(x,s,q,t)\ge \frac{r^{\prime }(s)}{r(s)}f(x,s,q,t);\end{array}$$

  (2.3)

• (3) For any $x\in \{x\mid f(x,u_0,q_0,0)=0,x\in \overline{D}\}$,

  $$\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)\ge 0;$$

  (2.4)

• (4) The constant

  $$\beta =\underset{D_1}{min}\left\{\frac{a(u_0)}{g^{\prime }(u_0)r(u_0)}[\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)]\right\}>0,$$

  (2.5)

where $D_1=\{x\mid f(x,u_0,q_0,0)\ne 0,x\in \overline{D}\}\ne \varphi$, $q_0={|\mathrm{\nabla }{u}_0|}^2$;

• (5) The integration

  $${\int }_{M_0}^{+\mathrm{\infty }}\frac{a(s)}{r(s)}ds<+\mathrm{\infty },\mathit{\text{where }}{M}_0=\underset{\overline{D}}{max}{u}_0(x);$$

  (2.6)

then the solution $u(x,t)$ of system (1.1)-(1.3) must blow up in finite time T and

$$T\le \frac{1}{\beta }{\int }_{M_0}^{+\mathrm{\infty }}\frac{a(s)}{r(s)}ds,$$

(2.7)

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

(2.8)

where $\mathrm{\Phi }(z)={\int }_z^{+\mathrm{\infty }}\frac{a(s)}{r(s)}ds$, $z>0$, and ${\mathrm{\Phi }}^{-1}$ is the inverse function of Φ.

Proof Consider the auxiliary function

$$\mathrm{\Psi }=-\frac{1}{r(u)}{u}_t+\beta \frac{1}{a(u)}.$$

(2.9)

We find that

$$\mathrm{\nabla }\mathrm{\Psi }=\frac{r^{\prime }}{r^2}{u}_t\mathrm{\nabla }u-\frac{1}{r}\mathrm{\nabla }{u}_t-\beta \frac{a^{\prime }}{a^2}\mathrm{\nabla }u,$$

(2.10)

$$\begin{array}{rl}\mathrm{\Delta }\mathrm{\Psi }=& (\frac{r^{″}}{r^2}-2\frac{{(r^{\prime })}^2}{r^3})q{u}_t+2\frac{r^{\prime }}{r^2}\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t+\frac{r^{\prime }}{r^2}{u}_t\mathrm{\Delta }u-\frac{1}{r}\mathrm{\Delta }{u}_t\\ -\beta (\frac{a^{″}}{a^2}-2\frac{{(a^{\prime })}^2}{a^3})q-\beta \frac{(a^{\prime })}{a^2}\mathrm{\Delta }u,\end{array}$$

(2.11)

and

$$\begin{array}{rl}{\mathrm{\Psi }}_t=& \frac{r^{\prime }}{r^2}{(u_t)}^2-\frac{1}{r}{(u_t)}_t-\beta \frac{(a^{\prime })}{a^2}{u}_t\\ =& \frac{r^{\prime }}{r^2}{(u_t)}^2-\frac{1}{r}{\left[\frac{1}{g^{\prime }}(abc\mathrm{\Delta }u+a^{\prime }bcq+ac\mathrm{\nabla }b\cdot \mathrm{\nabla }u+f)\right]}_t-\beta \frac{(a^{\prime })}{a^2}{u}_t\\ =& \frac{r^{\prime }}{r^2}{(u_t)}^2-\beta \frac{(a^{\prime })}{a^2}{u}_t-\frac{1}{g^{\prime }r}(a^{\prime }bc{u}_t\mathrm{\Delta }u+ab{c}^{\prime }\mathrm{\Delta }u+abc\mathrm{\Delta }{u}_t+a^{″}bcq{u}_t\\ +a^{\prime }b{c}^{\prime }q+2a^{\prime }bc\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t\\ +a^{\prime }c{u}_t\mathrm{\nabla }b\cdot \mathrm{\nabla }u+a{c}^{\prime }\mathrm{\nabla }b\cdot \mathrm{\nabla }u+ac\mathrm{\nabla }b\cdot \mathrm{\nabla }{u}_t+2f_q\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t+f_t+f_u{u}_t)\\ +\frac{g^{″}}{{(g^{\prime })}^{2}r}(abc\mathrm{\Delta }u+a^{\prime }bcq+ac\mathrm{\nabla }b\cdot \mathrm{\nabla }u+f)u_t.\end{array}$$

(2.12)

Hence, from (2.11) and (2.12) we have

$$\begin{array}{r}\frac{abc}{g^{\prime }}\mathrm{\Delta }\mathrm{\Psi }-{\mathrm{\Psi }}_t\\ =(\frac{abc}{g^{\prime }}\frac{r^{″}}{r^2}-2\frac{abc}{g^{\prime }}\frac{{(r^{\prime })}^2}{r^3}+\frac{a^{″}bc}{g^{\prime }}\frac{1}{r}-\frac{a^{\prime }bc}{r}\frac{g^{″}}{{(g^{\prime })}^2})q{u}_t\\ +(2\frac{abc}{g^{\prime }}\frac{r^{\prime }}{r^2}+2\frac{abc}{g^{\prime }}\frac{1}{r}+2\frac{f_q}{g^{\prime }}\frac{1}{r})\mathrm{\nabla }u\cdot \mathrm{\nabla }{u}_t\\ +(\frac{abc}{g^{\prime }}\frac{r^{\prime }}{r^2}+\frac{abc}{g^{\prime }}\frac{1}{r}-\frac{abc}{r}\frac{g^{″}}{{(g^{\prime })}^2})u_t\mathrm{\Delta }u+(\frac{a^{\prime }b{c}^{\prime }}{g^{\prime }}\frac{1}{r}-\beta \frac{abc}{g^{\prime }}\frac{a^{″}}{a^2}+2\beta \frac{abc}{g^{\prime }}\frac{{(a^{\prime })}^2}{a^3})q\\ +(\frac{a^{\prime }b{c}^{\prime }}{g^{\prime }}\frac{1}{r}-\beta \frac{abc}{g^{\prime }}\frac{a^{\prime }}{a^2})\mathrm{\Delta }u-\frac{r^{\prime }}{r^2}{(u_t)}^2+\beta \frac{a^{\prime }}{a^2}{u}_t+\frac{a^{\prime }c}{g^{\prime }r}{u}_t\mathrm{\nabla }b\cdot \mathrm{\nabla }u+\frac{a{c}^{\prime }}{g^{\prime }r}\mathrm{\nabla }b\cdot \mathrm{\nabla }u\\ +\frac{f_t}{g^{\prime }r}+\frac{f_u}{g^{\prime }r}{u}_t-\frac{ac}{r}\frac{g^{″}}{{(g^{\prime })}^2}{u}_t\mathrm{\nabla }b\cdot \mathrm{\nabla }u-\frac{f}{r}\frac{g^{″}}{{(g^{\prime })}^2}{u}_t.\end{array}$$

(2.13)

Using (2.10) leads to

$$\mathrm{\nabla }{u}_t=-r\mathrm{\nabla }\mathrm{\Psi }-\beta \frac{a^{\prime }r}{a^2}\mathrm{\nabla }u+\frac{r^{\prime }}{r}{u}_t\mathrm{\nabla }u.$$

(2.14)

Now substituting (2.14) into (2.13) yields

$$\begin{array}{r}\frac{abc}{g^{\prime }}\mathrm{\Delta }\mathrm{\Psi }+(\frac{ac}{g^{\prime }}\mathrm{\nabla }b+2\frac{f_q}{g^{\prime }}\mathrm{\nabla }u+2\frac{bc}{g^{\prime }}\frac{{(ar)}^{\prime }}{r}\mathrm{\nabla }u)\mathrm{\nabla }\mathrm{\Psi }-{\mathrm{\Psi }}_t\\ =(\frac{abc}{g^{\prime }}\frac{r^{″}}{r^2}+\frac{a^{″}bc}{g^{\prime }}\frac{1}{r}+2\frac{a^{\prime }bc}{g^{\prime }}\frac{r^{\prime }}{r^2}+2\frac{f_q}{g^{\prime }}\frac{r^{\prime }}{r}-\frac{a^{\prime }bc}{r}\frac{g^{″}}{{(g^{\prime })}^2})q{u}_t\\ +(\frac{ab{c}^{\prime }}{g^{\prime }}\frac{1}{r}-\beta \frac{abc}{g^{\prime }}\frac{a^{\prime }}{a^2})\mathrm{\Delta }u\\ +(\frac{abc}{g^{\prime }}\frac{r^{\prime }}{r^2}+\frac{abc}{g^{\prime }}\frac{1}{r}-\frac{abc}{r}\frac{g^{″}}{{(g^{\prime })}^2})u_t\mathrm{\Delta }u+(\frac{a^{\prime }c}{g^{\prime }r}+\frac{ac}{g^{\prime }}\frac{r^{\prime }}{r^2}-\frac{ac}{r}\frac{g^{″}}{{(g^{\prime })}^2})u_t\mathrm{\nabla }b\cdot \mathrm{\nabla }u\\ +(\frac{a^{\prime }b{c}^{\prime }}{g^{\prime }}\frac{1}{r}-\beta \frac{abc}{g^{\prime }}\frac{a^{″}}{a^2}-2\beta \frac{abc}{g^{\prime }}\frac{a^{\prime }}{a^2}\frac{r^{\prime }}{r}-2\beta \frac{f_q}{g^{\prime }}\frac{a^{\prime }r}{a})q+(\beta \frac{a^{\prime }}{a^2}+\frac{f_u}{g^{\prime }r}-\frac{f}{r}\frac{g^{″}}{{(g^{\prime })}^2})u_t\\ -\frac{r^{\prime }}{r^2}{(u_t)}^2+(\frac{a{c}^{\prime }}{g^{\prime }r}-\beta \frac{ac}{g^{\prime }}\frac{a^{\prime }}{a^2})\mathrm{\nabla }b\cdot \mathrm{\nabla }u+\frac{f_t}{g^{\prime }r}.\end{array}$$

(2.15)

In fact, from (1.1) we see that

$$\mathrm{\Delta }u=\frac{1}{abc}(g^{\prime }{u}_t-a^{\prime }bcq-ac\mathrm{\nabla }b\cdot \mathrm{\nabla }u-f).$$

(2.16)

Thus combining (2.15) and (2.16), we arrive at

$$\begin{array}{r}\frac{abc}{g^{\prime }}\mathrm{\Delta }\mathrm{\Psi }+(\frac{ac}{g^{\prime }}\mathrm{\nabla }b+2\frac{f_q}{g^{\prime }}\mathrm{\nabla }u+2\frac{bc}{g^{\prime }}\frac{{(ar)}^{\prime }}{r}\mathrm{\nabla }u)\mathrm{\nabla }\mathrm{\Psi }-{\mathrm{\Psi }}_t\\ =(\frac{abc}{g^{\prime }}\frac{r^{″}}{r^2}+\frac{a^{″}bc}{g^{\prime }}\frac{1}{r}+\frac{a^{\prime }bc}{g^{\prime }}\frac{r^{\prime }}{r^2}+2\frac{f_q}{g^{\prime }}\frac{r^{\prime }}{r}-\frac{{(a^{\prime })}^{2}bc}{a{g}^{\prime }}\frac{1}{r})q{u}_t\\ +(\beta \frac{{(a^{\prime })}^{2}b{c}^{\prime }}{a^2{g}^{\prime }}-\beta \frac{a^{″}bc}{a{g}^{\prime }}-2\beta \frac{a^{\prime }bc}{a{g}^{\prime }}\frac{1}{r}-2\beta \frac{a^{\prime }r}{a^2}\frac{f_q}{g^{\prime }})q\\ +(\frac{c^{\prime }}{c}\frac{1}{r}-\frac{f}{g^{\prime }}\frac{r^{\prime }}{r^2}-\frac{a^{\prime }}{a}\frac{f}{g^{\prime }}\frac{1}{r}+\frac{f_u}{g^{\prime }r})u_t+\frac{f_t}{g^{\prime }r}-\frac{c^{\prime }}{c}\frac{f}{g^{\prime }r}\\ +\beta \frac{a^{\prime }}{a^2}\frac{f}{g^{\prime }}+(\frac{a^{\prime }}{a}\frac{1}{r}-\frac{1}{r}\frac{g^{″}}{{(g^{\prime })}^2}){(u_t)}^2.\end{array}$$

(2.17)

In view of (2.9), we have

$$u_t=-r\mathrm{\Psi }+\beta \frac{r}{a}.$$

(2.18)

If we substitute (2.18) into (2.17), then it is easy to obtain

$$\begin{array}{r}\frac{abc}{g^{\prime }}\mathrm{\Delta }\mathrm{\Psi }+(\frac{ac}{g^{\prime }}\mathrm{\nabla }b+2\frac{f_q}{g^{\prime }}\mathrm{\nabla }u+2\frac{bc}{g^{\prime }}\frac{{(ar)}^{\prime }}{r}\mathrm{\nabla }u)\mathrm{\nabla }\mathrm{\Psi }\\ +\{[{\left(\frac{a^{\prime }r}{a}\right)}^{\prime }+r^{″}]q\frac{abc}{g^{\prime }r}+2\frac{f_q}{g^{\prime }}\frac{r^{\prime }}{r}q+\frac{ar}{g^{\prime }}{\left(\frac{f}{ar}\right)}_u+\frac{c^{\prime }}{c}\}\mathrm{\Psi }-{\mathrm{\Psi }}_t\\ =\beta \frac{bc}{g^{\prime }}(\frac{r^{″}}{r}-\frac{a^{\prime }}{a}\frac{r^{\prime }}{r})q+2\beta \frac{f_q}{g^{\prime }}(\frac{r^{\prime }}{a}-\frac{a^{\prime }r}{a^2})q+\beta \frac{c^{\prime }}{ac}+(\frac{a^{\prime }}{a}\frac{1}{r}-\frac{1}{r}\frac{g^{″}}{g^{\prime }}){(u_t)}^2\\ +\frac{1}{g^{\prime }r}(f_t-\frac{c^{\prime }}{c}f)+\beta \frac{1}{a{g}^{\prime }}(f_u-f\frac{r^{\prime }}{r})\\ =\beta \frac{abc}{g^{\prime }r}{\left(\frac{r^{\prime }}{a}\right)}^{\prime }q+2\beta \frac{f_q}{g^{\prime }}{\left(\frac{r}{a}\right)}^{\prime }q+\beta \frac{c^{\prime }}{ac}+\frac{g^{\prime }}{ar}{\left(\frac{a}{g^{\prime }}\right)}^{\prime }{(u_t)}^2\\ +\frac{c}{g^{\prime }r}{\left(\frac{f}{c}\right)}_t+\beta \frac{r}{a{g}^{\prime }}{\left(\frac{f}{r}\right)}_u.\end{array}$$

(2.19)

From assumptions (2.1)-(2.3), it follows that the right-hand side of (2.19) is nonnegative, i.e.,

$$\begin{array}{r}\frac{abc}{g^{\prime }}\mathrm{\Delta }\mathrm{\Psi }+(\frac{ac}{g^{\prime }}\mathrm{\nabla }b+2\frac{f_q}{g^{\prime }}\mathrm{\nabla }u+2\frac{bc}{g^{\prime }}\frac{{(ar)}^{\prime }}{r}\mathrm{\nabla }u)\mathrm{\nabla }\mathrm{\Psi }\\ +\{[{\left(\frac{a^{\prime }r}{a}\right)}^{\prime }+r^{″}]q\frac{abc}{g^{\prime }r}+2\frac{f_q}{g^{\prime }}\frac{r^{\prime }}{r}q+\frac{ar}{g^{\prime }}{\left(\frac{f}{ar}\right)}_u+\frac{c^{\prime }}{c}\}\mathrm{\Psi }-{\mathrm{\Psi }}_t\ge 0.\end{array}$$

(2.20)

Then from (2.4) and (2.5) we have

$$\begin{array}{rl}\underset{\overline{D}}{max}\mathrm{\Psi }(x,0)& =\underset{\overline{D}}{max}\{-\frac{1}{g^{\prime }(u_0)r(u_0)}[\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)]+\beta \frac{1}{a(u_0)}\}\\ \le 0.\end{array}$$

(2.21)

And as we can see, an explicit calculation

$$\begin{array}{rl}\frac{\partial \mathrm{\Psi }}{\partial n}& =\frac{r^{\prime }}{r^2}{u}_t\frac{\partial u}{\partial n}-\frac{1}{r}\frac{\partial u_t}{\partial n}-\beta \frac{a^{\prime }}{a^2}\frac{\partial u}{\partial n}=\frac{r^{\prime }}{r^2}h{u}_t-\frac{1}{r}{(hr)}_t-\beta \frac{a^{\prime }}{a^2}hr\\ =\frac{r^{\prime }}{r^2}h{u}_t-h_t-\frac{r^{\prime }}{r}h{u}_t-\beta \frac{a^{\prime }}{a^2}hr=-h_t-\beta \frac{a^{\prime }}{a^2}hr\le 0\end{array}$$

(2.22)

holds on $\partial D\times (0,T)$. Thus, by combining (2.20)-(2.22) and using the Hopf maximum principle, we find that the maximum of Ψ on $\partial D\times (0,T)$ is 0, i.e.,

$$\mathrm{\Psi }\le 0\text{on }\partial D\times (0,T),$$

and by (2.9), it gives

$$\frac{a(u)}{r(u)}{u}_t\ge \beta .$$

(2.23)

Integrating (2.23) over $[0,t]$ at the point $x_0\in \overline{D}$, where $u_0(x_0)=M_0$, yields

$$\frac{1}{\beta }{\int }_{M_0}^{u(x_0,t)}\frac{a(s)}{r(s)}ds\ge t.$$

(2.24)

This together with assumption (2.6) shows that $u(x,t)$ must blow up in finite time T; moreover,

$$T\le \frac{1}{\beta }{\int }_{M_0}^{+\mathrm{\infty }}\frac{a(s)}{r(s)}ds.$$

(2.25)

For each fixed x, integrating inequality (2.23) over $[t,s]$ ($0<t<s<T$) leads to

$$\mathrm{\Phi }(u(x,t))\ge \mathrm{\Phi }(u(x,t))-\mathrm{\Phi }(u(x,s))={\int }_{u(x,t)}^{u(x,s)}\frac{a(s)}{r(s)}ds\ge \beta (s-t).$$

If we let $s\to T$, then formally

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

therefore

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

The proof is completed. □

The result on the global solution is stated as Theorem 2 below.

Theorem 2 Let $u\in C^3(D\times (0,T))\cap C^2(\overline{D}\times (0,T))$ be a solution of problem (1.1)-(1.3). Assume that the following conditions hold:

• (1) For any $(x,s,q,t)\in \overline{D}\times \mathbb{R}\times {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$,

  $$a(s)>0,b(x)>0,c(t)>0,r(s)>0,h(x,t)\ge 0;$$

  (2.26)

• (2) For any $(x,s,q,t)\in \overline{D}\times {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$,

  $$\begin{array}{r}{a}^{\prime }(s)\le 0,h_t(x,t)\le 0,f_q\le 0,{\left(\frac{a(s)}{g^{\prime }(s)}\right)}^{\prime }\le 0,\\ r^{\prime }(s)\ge \frac{a^{\prime }(s)}{a(s)}r(s),r^{″}(s)\le \frac{a^{\prime }(s)}{a(s)}{r}^{\prime }(s),\end{array}$$

  (2.27)
  $$\begin{array}{r}{c}^{\prime }(t)\le 0,g^{\prime }(s)>0,f_t(x,s,q,t)\le \frac{c^{\prime }(t)}{c(t)}f(x,s,q,t),\\ f_s(x,s,q,t)\le \frac{r^{\prime }(s)}{r(s)}f(x,s,q,t);\end{array}$$

  (2.28)

• (3) For any $x\in \{x\mid f(x,u_0,q_0,0)=0,x\in \overline{D}\}$,

  $$\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)\ge 0;$$

  (2.29)

• (4) The constant

  $$\alpha =\underset{D_1}{max}\left\{\frac{a(u_0)}{g^{\prime }(u_0)r(u_0)}[\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)]\right\}>0,$$

  (2.30)

where $D_1=\{x\mid f(x,u_0,q_0,0)=0,x\in \overline{D}\}\ne \varphi$, $q_0={|\mathrm{\nabla }{u}_0|}^2$;

• (5) The integration

  $${\int }_{m_0}^{+\mathrm{\infty }}\frac{a(s)}{r(s)}ds<+\mathrm{\infty },\mathit{\text{where }}{m}_0=\underset{\overline{D}}{min}{u}_0(x);$$

  (2.31)

then the solution $u(x,t)$ of system (1.1)-(1.3) must be a global solution and

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

(2.32)

where $\mathrm{\Psi }(z)={\int }_{m_0}^z\frac{a(s)}{r(s)}ds$, $z>0$, and ${\mathrm{\Psi }}^{-1}$ is the inverse function of Ψ.

Proof Consider the auxiliary function

$$\mathrm{\Phi }=-\frac{1}{r(u)}{u}_t+\alpha \frac{1}{a(u)}.$$

(2.33)

We first replace Ψ and β in (2.20) with Φ and α, respectively, and under assumptions (2.26)-(2.28), we get

$$\begin{array}{r}\frac{abc}{g^{\prime }}\mathrm{\Delta }\mathrm{\Phi }-(\frac{ac}{g^{\prime }}\mathrm{\nabla }b+2\frac{f_q}{g^{\prime }}\mathrm{\nabla }u+2\frac{bc}{g^{\prime }}\frac{{(ar)}^{\prime }}{r}\mathrm{\nabla }u)\mathrm{\nabla }\mathrm{\Phi }\\ +\{[{\left(\frac{a^{\prime }r}{a}\right)}^{\prime }+r^{″}]q\frac{abc}{g^{\prime }r}+2\frac{f_q}{g^{\prime }}\frac{r^{\prime }}{r}q+\frac{ar}{g^{\prime }}{\left(\frac{f}{ar}\right)}_u+\frac{c^{\prime }}{c}\}\mathrm{\Phi }-{\mathrm{\Phi }}_t\le 0.\end{array}$$

(2.34)

In fact, from (2.29) and (2.30) we can see that

$$\begin{array}{rl}\underset{\overline{D}}{min}\mathrm{\Phi }(x,0)& =\underset{\overline{D}}{min}\{-\frac{1}{g^{\prime }(u_0)r(u_0)}[\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)]+\alpha \frac{1}{a(u_0)}\}\\ \ge 0.\end{array}$$

(2.35)

Also, on $\partial D\times (0,T)$, it gives

$$\frac{\partial \mathrm{\Phi }}{\partial n}=-h_t-\alpha \frac{a^{\prime }}{a^2}hr\ge 0.$$

(2.36)

By combining (2.34)-(2.36) and using the Hopf maximum principle, we find that the minimum of Φ on $\partial D\times (0,T)$ is 0, i.e.,

$$\mathrm{\Phi }\ge 0\text{in }\partial D\times (0,T),$$

and by (2.33), we can see that

$$\frac{a(u)}{r(u)}{u}_t\le \alpha .$$

(2.37)

For each fixed x, integrating (2.37) over $[0,t]$ yields

$$\frac{1}{\alpha }{\int }_{u_0(x)}^{u(x,t)}\frac{a(s)}{r(s)}ds\le t.$$

(2.38)

This together with assumption (2.31) shows that $u(x,t)$ must be a global solution; moreover,

$$\mathrm{\Psi }(u(x,t))-\mathrm{\Psi }(u_0(x))={\int }_{u_0(x)}^{u(x,t)}\frac{a(s)}{r(s)}ds\le \alpha t,$$

therefore

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

The proof is completed. □

3 Applications

In what follows, we present several examples to demonstrate the applications of Theorems 1 and 2.

Example 1 Let u be a solution of

$$\begin{array}{c}{\left(e^{2u}\right)}_t=\mathrm{\nabla }\cdot \left(e^{3u}(1+\sum _{i=1}^3{x}_i^2)e^t\mathrm{\nabla }u\right)+(1+\sum _{i=1}^3{x}_i^2)e^{4u}q{e}^t\text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=2(1+t\sum _{i=1}^3{x}_i^4)e^{4u}\text{on }\partial D\times (0,T),\hfill \\ u(x,0)=u_0(x)=1+e^4\sum _{i=1}^3{x}_i^2\text{in }\overline{D},\hfill \end{array}$$

where $D=\{x=(x_1,x_2,x_3)\mid {\sum }_{i=1}^3{x}_i^2<1\}$, then we have

$$\begin{array}{c}g(u)=e^{2u},a(u)=e^{3u},b(x)=1+\sum _{i=1}^3{x}_i^2,c(t)=e^t,\hfill \\ f(x,u,q,t)=(1+\sum _{i=1}^3{x}_i^2)e^{4u}q{e}^t,h(x,t)=2(1+t\sum _{i=1}^3{x}_i^4),r(u)=e^{4u}.\hfill \end{array}$$

It is easy to verify that (2.1)-(2.4) hold. By (2.5), we find

$$\begin{array}{rl}\beta & =\underset{D_1}{min}\left\{\frac{a(u_0)}{g^{\prime }(u_0)r(u_0)}[\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)]\right\}\\ =\underset{1\le u_0<1+e^4}{min}\left\{\frac{1}{2}[3u_0{|\mathrm{\nabla }{u}_0|}^2+{|\mathrm{\nabla }{u}_0|}^2+u_0\mathrm{\Delta }{u}_0+e^{u_0}{u}_0{|\mathrm{\nabla }{u}_0|}^2]\right\}=3e^4.\end{array}$$

It follows from Theorem 1 that $u(x,t)$ must blow up in finite time T and

$$T\le \frac{1}{\beta }{\int }_{M_0}^{+\mathrm{\infty }}\frac{a(s)}{r(s)}ds=\frac{1}{\beta }{\int }_2^{+\mathrm{\infty }}\frac{e^{3s}}{e^{4s}}ds=\frac{1}{3}{e}^{-6},$$

and

$$u(x,t)\le {\mathrm{\Phi }}^{-1}(\beta (T-t))=ln\left[\frac{1}{3e^4}{(T-t)}^{-1}\right].$$

Example 2 Let u be a solution of

$$\begin{array}{c}{(u\sqrt{u})}_t=\mathrm{\nabla }\cdot \left(\frac{1}{\sqrt{u}}(1+\sum _{i=1}^3{x}_i^2)\frac{1}{1+t}\mathrm{\nabla }u\right)+(1+\sum _{i=1}^3{x}_i^2)\frac{1-q}{1+t}\sqrt{u}\text{in }D\times (0,T),\hfill \\ \frac{\partial u}{\partial n}=\sqrt{2}{(1+t\sum _{i=1}^3{x}_i^4)}^{-1}\sqrt{u}\text{in }\partial D\times (0,T),\hfill \\ u(x,0)=u_0(x)=1+\sum _{i=1}^3{x}_i^2\text{in }\overline{D},\hfill \end{array}$$

where $D=\{x=(x_1,x_2,x_3)\mid {\sum }_{i=1}^3{x}_i^2<1\}$, then we have

$$\begin{array}{c}g(u)=u\sqrt{u},a(u)=\frac{1}{\sqrt{u}},b(x)=(1+\sum _{i=1}^3{x}_i^2),c(t)=\frac{1}{1+t},\hfill \\ f(x,u,q,t)=(1+\sum _{i=1}^3{x}_i^2)\frac{1-q}{1+t}\sqrt{u},h(x,t)=\sqrt{2}{(1+t\sum _{i=1}^3{x}_i^4)}^{-1},r(u)=\sqrt{u}.\hfill \end{array}$$

It is easy to verify that (2.26)-(2.29) hold. By (2.30), we find

$$\begin{array}{rl}\alpha & =\underset{D_1}{max}\left\{\frac{a(u_0)}{g^{\prime }(u_0)r(u_0)}[\mathrm{\nabla }(a(u_0)b(x)c(0)\mathrm{\nabla }{u}_0)+f(x,u_0,q_0,0)]\right\}\\ =\underset{1\le u_0<2}{max}\left\{\frac{1}{3}[-u_0^{-2}{|\mathrm{\nabla }{u}_0|}^2+2u_0^{-1}\mathrm{\Delta }{u}_0+2(1-{|\mathrm{\nabla }{u}_0|}^{-2})]\right\}=\frac{14}{3}.\end{array}$$

It follows from Theorem 2 that $u(x,t)$ must be a global solution and

$$u(x,t)\le {\mathrm{\Psi }}^{-1}(\alpha t+\mathrm{\Psi }(u_0(x)))=exp(\alpha t+ln{u}_0)=u_{0}exp\left(\frac{14}{3}t\right).$$