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

Juntang Ding

doi:10.1186/1687-2770-2013-237

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

Juntang Ding¹Email author

Boundary Value Problems20132013:237

https://doi.org/10.1186/1687-2770-2013-237

Received: 25 July 2013

Accepted: 25 September 2013

Published: 8 November 2013

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{matrix} {(b (u))}_{t} = \nabla \cdot (g (u) \nabla u) + f (x, u, {| \nabla u |}^{2}, t) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + γ u = 0 & on \partial D \times (0, T), \\ u (x, 0) = u_{0} (x) > 0 & in \bar{D}, \end{matrix}

where $D \subset R^{N}$ ( $N \geq 2$ ) 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.

Keywords

global solution blow-up solution parabolic problem Robin boundary condition gradient term

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{matrix} {(b (u))}_{t} = \nabla \cdot (g (u) \nabla u) + f (x, u, q, t) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + γ u = 0 & on \partial D \times (0, T), \\ u (x, 0) = u_{0} (x) > 0 & in \bar{D}, \end{matrix}

(1.1)

where $q : = {| \nabla u |}^{2}$ , $D \subset R^{N}$ ( $N \geq 2$ ) 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 $\bar{D}$ is the closure of D. Set $R^{+} : = (0, + \infty)$ . We assume, throughout the paper, that $b (s)$ is a $C^{3} (R^{+})$ function, $b^{'} (s) > 0$ for any $s \in R^{+}$ , $g (s)$ is a positive $C^{2} (R^{+})$ function, $f (x, s, d, t)$ is a nonnegative $C^{1} (\bar{D} \times R^{+} \times \bar{R^{+}} \times R^{+})$ function, and $u_{0} (x)$ is a positive $C^{2} (\bar{D})$ 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 $\bar{D} \times [0, T)$ . Moreover, the regularity theorem [2] implies $u (x, t) \in C^{3} (D \times (0, T)) \cap C^{2} (\bar{D} \times [0, T))$ .

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{matrix} u_{t} = \nabla \cdot (g (u) \nabla u) + f (u) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + γ u = 0 & on \partial D \times (0, T), \\ u (x, 0) = u_{0} (x) > 0 & in \bar{D}, \end{matrix}

where

D \subset R^{N}

(

N \geq 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{matrix} {(b (u))}_{t} = Δ u + f (u) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + γ u = 0 & on \partial D \times (0, T), \\ u (x, 0) = u_{0} (x) > 0 & in \bar{D}, \end{matrix}

where

D \subset R^{N}

(

N \geq 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{matrix} {(b (u))}_{t} = \nabla \cdot (g (u) \nabla u) + f (u) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + γ u = 0 & on \partial D \times (0, T), \\ u (x, 0) = u_{0} (x) > 0 & in \bar{D}, \end{matrix}

where $D \subset R^{N}$ ( $N \geq 2$ ) 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 $q = {| \nabla u |}^{2}$ , 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 $s \in R^{+}$ ,
$\begin{aligned} {(s b^{'} (s))}^{'} \geq 0, s b^{'} (s) - {(s b^{'} (s))}^{'} \leq 0, {(\frac{g (s)}{b^{'} (s)})}^{'} \leq 0, \\ {[\frac{1}{g (s)} {(\frac{g (s)}{b^{'} (s)})}^{'} + \frac{1}{b^{'} (s)}]}^{'} + \frac{1}{g} {(\frac{g (s)}{b^{'} (s)})}^{'} + \frac{1}{b^{'} (s)} \leq 0; \end{aligned}$

(2.1)
(ii)

for any $(x, s, d, t) \in D \times R^{+} \times \bar{R^{+}} \times R^{+}$ ,
$\begin{aligned} f_{t} (x, s, d, t) \leq 0, f_{d} (x, s, d, t) [{(\frac{1}{b^{'} (s)})}^{'} + \frac{1}{b^{'} (s)}] \leq 0, \\ {(\frac{f (x, s, d, t) b^{'} (s)}{g (s)})}_{s} - \frac{f (x, s, d, t) b^{'} (s)}{g (s)} \leq 0; \end{aligned}$

(2.2)
(iii)

$\int_{m_{0}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s = + \infty, m_{0} : = min_{\bar{D}} u_{0} (x);$

(2.3)
(iv)

$α : = max_{\bar{D}} \frac{\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0)}{e^{u_{0}}} > 0, q_{0} : = {| \nabla u_{0} |}^{2} .$

(2.4)

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

u (x, t) \leq H^{- 1} (α t + H (u_{0} (x, t))), (x, t) \in \bar{D} \times \bar{R^{+}},

(2.5)

where

H (z) : = \int_{m_{0}}^{z} \frac{b^{'} (s)}{e^{s}} d s, z \geq m_{0},

(2.6)

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

Proof Consider the auxiliary function

P (x, t) : = b^{'} (u) u_{t} - α e^{u} .

(2.7)

Now we have

\nabla P = b^{″} u_{t} \nabla u + b^{'} \nabla u_{t} - α e^{u} \nabla u,

(2.8)

Δ P = b^{‴} u_{t} {| \nabla u |}^{2} + 2 b^{″} \nabla u \cdot \nabla u_{t} + b^{″} u_{t} Δ u + b^{'} Δ u_{t} - α e^{u} {| \nabla u |}^{2} - α e^{u} Δ u,

(2.9)

and

\begin{aligned} P_{t} = & b^{″} {(u_{t})}^{2} + b^{'} {(u_{t})}_{t} - α e^{u} u_{t} \\ = & b^{″} {(u_{t})}^{2} + b^{'} {(\frac{g}{b^{'}} Δ u + \frac{g^{'}}{b^{'}} {| \nabla u |}^{2} + \frac{f}{b^{'}})}_{t} - α e^{u} u_{t} \\ = & b^{″} {(u_{t})}^{2} + (g^{'} - \frac{b^{″} g}{b^{'}}) u_{t} Δ u + g Δ u_{t} + (g^{″} - \frac{b^{″} g^{'}}{b^{'}}) u_{t} {| \nabla u |}^{2} \\ + (2 g^{'} + 2 f_{q}) \nabla u \cdot \nabla u_{t} + (f_{u} - \frac{b^{″} f}{b^{'}} - α e^{u}) u_{t} + f_{t} . \end{aligned}

(2.10)

It follows from (2.9) and (2.10) that

\begin{aligned} \frac{g}{b^{'}} Δ P - P_{t} = & (\frac{b^{‴} g}{b^{'}} + \frac{b^{″} g^{'}}{b^{'}} - g^{″}) u_{t} {| \nabla u |}^{2} + (2 \frac{b^{″} g}{b^{'}} - 2 g^{'} - 2 f_{q}) \nabla u \cdot \nabla u_{t} \\ + (2 \frac{b^{″} g}{b^{'}} - g^{'}) u_{t} Δ u - α \frac{g}{b^{'}} e^{u} {| \nabla u |}^{2} - α \frac{g}{b^{'}} e^{u} Δ u - b^{″} {(u_{t})}^{2} \\ + (\frac{b^{″} f}{b^{'}} - f_{u} + α e^{u}) u_{t} - f_{t} . \end{aligned}

(2.11)

By (1.1), we have

Δ u = \frac{b^{'}}{g} u_{t} - \frac{g^{'}}{g} {| \nabla u |}^{2} - \frac{f}{g} .

(2.12)

Substitute (2.12) into (2.11), to get

\begin{aligned} \frac{g}{b^{'}} Δ P - P_{t} = & (\frac{b^{‴} g}{b^{'}} - \frac{b^{″} g^{'}}{b^{'}} - g^{″} + \frac{{(g^{'})}^{2}}{g}) u_{t} {| \nabla u |}^{2} + (2 \frac{b^{″} g}{b^{'}} - 2 g^{'} - 2 f_{q}) \nabla u \cdot \nabla u_{t} \\ - \frac{{(b^{'})}^{2}}{g} {(\frac{g}{b^{'}})}^{'} {(u_{t})}^{2} + (\frac{f g^{'}}{g} - \frac{b^{″} f}{b^{'}} - f_{u}) u_{t} + (α \frac{g^{'}}{b^{'}} e^{u} - α \frac{g}{b^{'}} e^{u}) {| \nabla u |}^{2} \\ + α \frac{f}{b^{'}} e^{u} - f_{t} . \end{aligned}

(2.13)

With (2.8), we have

\nabla u_{t} = \frac{1}{b^{'}} \nabla P - \frac{b^{″}}{b^{'}} u_{t} \nabla u + α \frac{e^{u}}{b^{'}} \nabla u .

(2.14)

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

\begin{aligned} \frac{g}{b^{'}} Δ P + [2 {(\frac{g}{b^{'}})}^{'} + 2 \frac{f_{q}}{b^{'}}] \nabla u \cdot \nabla P - P_{t} \\ = (\frac{b^{‴} g}{b^{'}} + \frac{b^{″} g^{'}}{b^{'}} - g^{″} + \frac{{(g^{'})}^{2}}{g} - 2 \frac{{(b^{″})}^{2} g}{{(b^{'})}^{2}} + 2 \frac{b^{″} f_{q}}{b^{'}}) u_{t} {| \nabla u |}^{2} \\ + (2 α \frac{b^{″} g}{{(b^{'})}^{2}} e^{u} - α \frac{g^{'}}{b^{'}} e^{u} - α \frac{g}{b^{'}} e^{u} - 2 α \frac{f_{q}}{b^{'}} e^{u}) {| \nabla u |}^{2} \\ - \frac{{(b^{'})}^{2}}{g} {(\frac{g}{b^{'}})}^{'} {(u_{t})}^{2} + (\frac{f g^{'}}{g} - \frac{b^{″} f}{b} - f_{u}) u_{t} + α \frac{f}{b^{'}} e^{u} - f_{t} . \end{aligned}

(2.15)

In view of (2.7), we have

u_{t} = \frac{1}{b^{'}} P + α \frac{e^{u}}{b^{'}} .

(2.16)

Substituting (2.16) into (2.15), we get

\begin{aligned} \frac{g}{b^{'}} Δ P + [2 {(\frac{g}{b^{'}})}^{'} + 2 \frac{f_{q}}{b^{'}}] \nabla u \cdot \nabla P \\ + {[g {(\frac{1}{g} {(\frac{g}{b^{'}})}^{'})}^{'} + 2 f_{q} {(\frac{1}{b^{'}})}^{'}] {| \nabla u |}^{2} + \frac{g}{{(b^{'})}^{2}} {(\frac{f b^{'}}{g})}_{u}} P - P_{t} \\ = - α e^{u} {g [{(\frac{1}{g} {(\frac{g}{b^{'}})}^{'} + \frac{1}{b^{'}})}^{'} + \frac{1}{g} {(\frac{g}{b^{'}})}^{'} + \frac{1}{b^{'}}] + 2 f_{q} [{(\frac{1}{b^{'}})}^{'} + \frac{1}{b^{'}}]} {| \nabla u |}^{2} \\ - \frac{{(b^{'})}^{2}}{g} {(\frac{g}{b^{'}})}^{'} {(u_{t})}^{2} - α \frac{g e^{u}}{{(b^{'})}^{2}} [{(\frac{f b^{'}}{g})}_{u} - \frac{f b^{'}}{g}] - f_{t} . \end{aligned}

(2.17)

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

\begin{aligned} \frac{g}{b^{'}} Δ P + [2 {(\frac{g}{b^{'}})}^{'} + 2 \frac{f_{q}}{b^{'}}] \nabla u \cdot \nabla P \\ + {[g {(\frac{1}{g} {(\frac{g}{b^{'}})}^{'})}^{'} + 2 f_{q} {(\frac{1}{b^{'}})}^{'}] {| \nabla u |}^{2} + \frac{g}{{(b^{'})}^{2}} {(\frac{f b^{'}}{g})}_{u}} P - P_{t} \\ \geq 0 in D \times (0, T) . \end{aligned}

(2.18)

By applying the maximum principle [24], it follows from (2.18) that P can attain its nonnegative maximum only for

\bar{D} \times {0}

or

\partial D \times (0, T)

. For

\bar{D} \times {0}

, by (2.4), we have

\begin{aligned} max_{\bar{D}} P (x, 0) & = max_{\bar{D}} {b^{'} (u_{0}) {(u_{0})}_{t} - α e^{u_{0}}} = max_{\bar{D}} {\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0) - α e^{u_{0}}} \\ = max_{\bar{D}} {e^{u_{0}} [\frac{\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0)}{e^{u_{0}}} - α]} = 0 . \end{aligned}

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 and \frac{\partial P}{\partial n} |_{(x_{0}, t_{0})} > 0 .

(2.19)

With (1.1) and (2.16), we have

\begin{aligned} \frac{\partial P}{\partial n} & = b^{″} u_{t} \frac{\partial u}{\partial n} + b^{'} \frac{\partial u_{t}}{\partial n} - α e^{u} \frac{\partial u}{\partial n} = - γ b^{″} u u_{t} + b^{'} {(\frac{\partial u}{\partial n})}_{t} + γ α u e^{u} \\ = - γ b^{″} u u_{t} + b^{'} {(- γ u)}_{t} + γ α u e^{u} = - γ {(u b^{'})}^{'} u_{t} + γ α u e^{u} \\ = - γ {(u b^{'})}^{'} (\frac{1}{b^{'}} P + α \frac{1}{b^{'}} e^{u}) + γ α u e^{u} \\ = - γ \frac{{(u b^{'})}^{'}}{b^{'}} P + γ α e^{u} \frac{u b^{'} - {(u b^{'})}^{'}}{b^{'}} on \partial D \times (0, T) . \end{aligned}

(2.20)

Next, by using the fact that

{(s b^{'} (s))}^{'} \geq 0

,

s b^{'} (s) - {(s b^{'} (s))}^{'} \leq 0

for any

s \in R^{+}

, it follows from (2.20) that

\frac{\partial P}{\partial n} |_{(x_{0}, t_{0})} \leq 0,

which contradicts with inequality (2.19). Thus we know that the maximum of P in

\bar{D} \times [0, T)

is zero, i.e.,

P \leq 0 in \bar{D} \times [0, T),

and

\frac{b^{'} (u)}{e^{u}} u_{t} \leq α .

(2.21)

For each fixed

x \in \bar{D}

, integration of (2.21) from 0 to t yields

\int_{0}^{t} \frac{b^{'} (u)}{e^{u}} u_{t} d t = \int_{u_{0} (x)}^{u (x, t)} \frac{b^{'} (s)}{e^{s}} d s \leq α t,

(2.22)

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

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

Passing to the limit as

t \to T^{-}

in (2.22) yields

\int_{u_{0} (x)}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s \leq α T

and

\int_{m_{0}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s = \int_{m_{0}}^{u_{0} (x)} \frac{b^{'} (s)}{e^{s}} d s + \int_{u_{0} (x)}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s \leq \int_{m_{0}}^{u_{0} (x)} \frac{b^{'} (s)}{e^{s}} d s + α T < + \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^{'} (s)}{e^{s}} d s = \int_{m_{0}}^{u (x, t)} \frac{b^{'} (s)}{e^{s}} d s - \int_{m_{0}}^{u_{0} (x)} \frac{b^{'} (s)}{e^{s}} d s = H (u (x, t)) - H (u_{0} (x)) \leq α t .

Since H is an increasing function, we have

u (x, t) \leq H^{- 1} (α 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 $s \in R^{+}$ ,
$\begin{aligned} {(s b^{'} (s))}^{'} \geq 0, s b^{'} (s) - {(s b^{'} (s))}^{'} \geq 0, {(\frac{g (s)}{b^{'} (s)})}^{'} \geq 0, \\ {[\frac{1}{g (s)} {(\frac{g (s)}{b^{'} (s)})}^{'} + \frac{1}{b^{'} (s)}]}^{'} + \frac{1}{g} {(\frac{g (s)}{b^{'} (s)})}^{'} + \frac{1}{b^{'} (s)} \geq 0; \end{aligned}$

(3.1)
(ii)

for any $(x, s, d, t) \in D \times R^{+} \times \bar{R^{+}} \times R^{+}$ ,
$\begin{aligned} f_{t} (x, s, d, t) \geq 0, f_{d} (x, s, d, t) [{(\frac{1}{b^{'} (s)})}^{'} + \frac{1}{b^{'} (s)}] \geq 0, \\ {(\frac{f (x, s, d, t) b^{'} (s)}{g (s)})}_{s} - \frac{f (x, s, d, t) b^{'} (s)}{g (s)} \geq 0; \end{aligned}$

(3.2)
(iii)

$\int_{M_{0}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s < + \infty, M_{0} : = max_{\bar{D}} u_{0} (x);$

(3.3)
(iv)

$β : = min_{\bar{D}} \frac{\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0)}{e^{u_{0}}} > 0, q_{0} : = {| \nabla u_{0} |}^{2} .$

(3.4)

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

T \leq \frac{1}{β} \int_{M_{0}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s,

(3.5)

u (x, t) \leq G^{- 1} (β (T - t)), (x, t) \in \bar{D} \times [0, T),

(3.6)

where

G (z) : = \int_{z}^{+ \infty} \frac{b^{'} (s)}{e^{s}} 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^{'} (u) u_{t} - β e^{u} .

(3.8)

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

\begin{aligned} \frac{g}{b^{'}} Δ Q + [2 {(\frac{g}{b^{'}})}^{'} + 2 \frac{f_{q}}{b^{'}}] \nabla u \cdot \nabla Q \\ + {[g {(\frac{1}{g} {(\frac{g}{b^{'}})}^{'})}^{'} + 2 f_{q} {(\frac{1}{b^{'}})}^{'}] {| \nabla u |}^{2} + \frac{g}{{(b^{'})}^{2}} {(\frac{f b^{'}}{g})}_{u}} Q - Q_{t} \\ = - β e^{u} {g [{(\frac{1}{g} {(\frac{g}{b^{'}})}^{'} + \frac{1}{b^{'}})}^{'} + \frac{1}{g} {(\frac{g}{b^{'}})}^{'} + \frac{1}{b^{'}}] + 2 f_{q} [{(\frac{1}{b^{'}})}^{'} + \frac{1}{b^{'}}]} {| \nabla u |}^{2} \\ - \frac{{(b^{'})}^{2}}{g} {(\frac{g}{b^{'}})}^{'} {(u_{t})}^{2} - β \frac{g e^{u}}{{(b^{'})}^{2}} [{(\frac{f b^{'}}{g})}_{u} - \frac{f b^{'}}{g}] - f_{t} . \end{aligned}

(3.9)

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

\begin{aligned} \frac{g}{b^{'}} Δ Q + [2 {(\frac{g}{b^{'}})}^{'} + 2 \frac{f_{q}}{b^{'}}] \nabla u \cdot \nabla Q \\ + {[g {(\frac{1}{g} {(\frac{g}{b^{'}})}^{'})}^{'} + 2 f_{q} {(\frac{1}{b^{'}})}^{'}] {| \nabla u |}^{2} + \frac{g}{{(b^{'})}^{2}} {(\frac{f b^{'}}{g})}_{u}} Q - Q_{t} \\ \leq 0 in D \times (0, T) . \end{aligned}

(3.10)

With (3.4), we have

\begin{aligned} min_{\bar{D}} Q (x, 0) & = min_{\bar{D}} {b^{'} (u_{0}) {(u_{0})}_{t} - β e^{u_{0}}} = min_{\bar{D}} {\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0) - β e^{u_{0}}} \\ = min_{\bar{D}} {e^{u_{0}} [\frac{\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0)}{e^{u_{0}}} - β]} = 0 . \end{aligned}

(3.11)

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

\frac{\partial Q}{\partial n} = - γ \frac{{(u b^{'})}^{'}}{b^{'}} Q + γ β e^{u} \frac{u b^{'} - {(u b^{'})}^{'}}{b^{'}} on \partial D \times (0, T) .

(3.12)

Combining (3.10)-(3.12) with the fact that

{(s b^{'} (s))}^{'} \geq 0

,

s b^{'} (s) - {(s b^{'} (s))}^{'} \geq 0

for any

s \in R^{+}

, and applying the maximum principles again, it follows that the minimum of Q in

\bar{D} \times [0, T)

is zero. Thus

Q \geq 0 in \bar{D} \times [0, T),

and

\frac{b^{'} (u)}{e^{u}} u_{t} \geq β .

(3.13)

At the point

x^{*} \in \bar{D}

, where

u_{0} (x^{*}) = M_{0}

, integrate (3.13) over

[0, t]

to get

\int_{0}^{t} \frac{b^{'} (u)}{e^{u}} u_{t} d t = \int_{M_{0}}^{u (x^{*}, t)} \frac{b^{'} (s)}{e^{s}} d s \geq β 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}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s \geq \int_{M_{0}}^{u (x^{*}, t)} \frac{b^{'} (s)}{e^{s}} d s \geq β t .

(3.15)

Letting

t \to + \infty

in (3.15), we have

\int_{M_{0}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s = + \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 \leq \frac{1}{β} \int_{M_{0}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s .

By integrating inequality (3.13) over

[t, s]

(

0 < t < s < T

), for each fixed x, we obtain

\begin{aligned} G (u (x, t)) & \geq G (u (x, t)) - G (u (x, s)) = \int_{u (x, t)}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s - \int_{u (x, s)}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s \\ = \int_{u (x, t)}^{u (x, s)} \frac{b^{'} (s)}{e^{s}} d s = \int_{t}^{s} \frac{b^{'} (u)}{e^{u}} u_{t} d t \geq β (s - t) . \end{aligned}

Hence, by letting

s \to T

, we have

G (u (x, t)) \geq β (T - t) .

Since G is a decreasing function, we obtain

u (x, t) \leq G^{- 1} (β (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{matrix} u_{t} = Δ u + \frac{2 + u}{1 + u} {| \nabla u |}^{2} + \frac{e^{- u} (e^{- u} + e^{q})}{1 + u} (e^{- t} + {| x |}^{2}) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + 2 u = 0 & on \partial D \times (0, T), \\ u (x, 0) = 2 - {| x |}^{2} & in \bar{D}, \end{matrix}

where

q = {| \nabla u |}^{2}

,

D = {x = (x_{1}, x_{2}, x_{3}) ∣ {| x |}^{2} < 1}

is the unit ball of

R^{3}

. The above problem can be transformed into the following problem:

{\begin{matrix} {(u e^{u})}_{t} = \nabla \cdot ((1 + u) e^{u} \nabla u) + (e^{- u} + e^{q}) (e^{- t} + {| x |}^{2}) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + 2 u = 0 & on \partial D \times (0, T), \\ u (x, 0) = 2 - {| x |}^{2} & in \bar{D} . \end{matrix}

Now

\begin{aligned} b (u) = u e^{u}, g (u) = (1 + u) e^{u}, f (x, u, q, t) = (e^{- u} + e^{q}) (e^{- t} + {| x |}^{2}), \\ u_{0} (x) = 2 - {| x |}^{2}, γ = 2 . \end{aligned}

In order to determine the constant α, we assume

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

then

0 \leq s \leq 1

and

\begin{aligned} α & = max_{\bar{D}} \frac{\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0)}{e^{u_{0}}} \\ = max_{\bar{D}} {32 {| x |}^{2} - 4 {| x |}^{4} - 18 + (1 + {| x |}^{2}) [exp (- 4 + 2 {| x |}^{2}) + exp (- 2 + 5 {| x |}^{2})]} \\ = max_{0 \leq s \leq 1} {32 s - 4 s^{2} - 18 + (1 + s) [exp (- 4 + 2 s) + exp (- 2 + 5 s)]} \\ = 50.4417 . \end{aligned}

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

\begin{aligned} u (x, t) & \leq H^{- 1} (α t + H (u_{0} (x))) = - 1 + \sqrt{50.4417 t + {(1 + u_{0} (x))}^{2}} \\ = - 1 + \sqrt{50.4417 t + {(3 - {| x |}^{2})}^{2}} . \end{aligned}

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

{\begin{matrix} u_{t} = Δ u - \frac{1}{u (1 + u)} {| \nabla u |}^{2} + \frac{u (e^{u} - e^{- q})}{1 + u} (6 + t {| x |}^{2}) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + 2 u = 0 & on \partial D \times (0, T), \\ u (x, 0) = 2 - {| x |}^{2} & in \bar{D}, \end{matrix}

where

q = {| \nabla u |}^{2}

,

D = {x = (x_{1}, x_{2}, x_{3}) ∣ {| x |}^{2} < 1}

is the unit ball of

R^{3}

. The above problem may be turned into the following problem:

{\begin{matrix} {(u + ln u)}_{t} = \nabla \cdot ((1 + \frac{1}{u}) \nabla u) + (e^{u} - e^{- q}) (6 + t {| x |}^{2}) & in D \times (0, T), \\ \frac{\partial u}{\partial n} + 2 u = 0 & on \partial D \times (0, T), \\ u (x, 0) = 2 - {| x |}^{2} & in \bar{D} . \end{matrix}

Now we have

\begin{aligned} b (u) = u + ln u, g (u) = 1 + \frac{1}{u}, f (x, u, q, t) = (e^{u} - e^{- q}) (6 + t {| x |}^{2}), \\ u_{0} (x) = 2 - {| x |}^{2}, γ = 2 . \end{aligned}

By setting

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

we have

0 \leq s \leq 1

and

\begin{aligned} β & = min_{\bar{D}} \frac{\nabla \cdot (g (u_{0}) \nabla u_{0}) + f (x, u_{0}, q_{0}, 0)}{e^{u_{0}}} \\ = min_{\bar{D}} {\frac{- 6 {| x |}^{4} + 26 {| x |}^{2} - 36}{{(2 - {| x |}^{2})}^{2} exp (2 - {| x |}^{2})} + 6 [1 - exp (- 3 {| x |}^{2} - 2)]} \\ = min_{0 \leq s \leq 1} {\frac{- 6 s^{2} + 26 s - 36}{{(2 - s)}^{2} exp (2 - s)} + 6 [1 - exp (- 3 s - 2)]} \\ = 0.0735 . \end{aligned}

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{matrix} T \leq \frac{1}{β} \int_{M_{0}}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s = \frac{1}{0.0735} \int_{2}^{+ \infty} (1 + \frac{1}{s}) \frac{1}{e^{s}} d s = 2.5066, \\ u (x, t) \leq G^{- 1} (β (T - t)) = G^{- 1} (0.0735 (T - t)), \end{matrix}

where

G (z) = \int_{z}^{+ \infty} \frac{b^{'} (s)}{e^{s}} d s = \int_{z}^{+ \infty} (1 + \frac{1}{s}) \frac{1}{e^{s}} d s, z \geq 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.

Declarations

Acknowledgements

This work was supported by the National Natural Science Foundation of China (Nos. 61074048 and 61174082) and the Research Project Supported by Shanxi Scholarship Council of China (Nos. 2011-011 and 2012-011).

Authors’ Affiliations

(1)

School of Mathematical Sciences, Shanxi University

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