Open Access

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

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:

{ ( b ( u ) ) t = ( g ( u ) u ) + f ( x , u , | u | 2 , t ) in  D × ( 0 , T ) , u n + γ u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) > 0 in  D ¯ ,

where D R N ( N 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 solutionblow-up solutionparabolic problemRobin boundary conditiongradient 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:
{ ( b ( u ) ) t = ( g ( u ) u ) + f ( x , u , q , t ) in  D × ( 0 , T ) , u n + γ u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) > 0 in  D ¯ ,
(1.1)

where q : = | u | 2 , D R N ( N 2 ) is a bounded domain with smooth boundary ∂D, / 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 D ¯ is the closure of D. Set R + : = ( 0 , + ) . We assume, throughout the paper, that b ( s ) is a C 3 ( R + ) function, b ( s ) > 0 for any s R + , g ( s ) is a positive C 2 ( R + ) function, f ( x , s , d , t ) is a nonnegative C 1 ( D ¯ × R + × R + ¯ × R + ) function, and u 0 ( x ) is a positive C 2 ( 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 D ¯ × [ 0 , T ) . Moreover, the regularity theorem [2] implies u ( x , t ) C 3 ( D × ( 0 , T ) ) C 2 ( D ¯ × [ 0 , T ) ) .

Many papers have studied the global and blow-up solutions of parabolic problems with a gradient term (see, for instance, [313]). 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 [1420] and the references cited therein). Some special cases of problem (1.1) have been treated already. Zhang [21] dealt with the following problem:
{ u t = ( g ( u ) u ) + f ( u ) in  D × ( 0 , T ) , u n + γ u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) > 0 in  D ¯ ,
where D R N ( N 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:
{ ( b ( u ) ) t = Δ u + f ( u ) in  D × ( 0 , T ) , u n + γ u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) > 0 in  D ¯ ,
where D R N ( N 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:
{ ( b ( u ) ) t = ( g ( u ) u ) + f ( u ) in  D × ( 0 , T ) , u n + γ u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) > 0 in  D ¯ ,

where D R N ( N 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 = | u | 2 , it seems that the methods of [2123] are not applicable for problem (1.1). In this paper, by constructing completely different auxiliary functions with those in [2123] 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 [2123].

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:
  1. (i)
    for any s R + ,
    ( s b ( s ) ) 0 , s b ( s ) ( s b ( s ) ) 0 , ( g ( s ) b ( s ) ) 0 , [ 1 g ( s ) ( g ( s ) b ( s ) ) + 1 b ( s ) ] + 1 g ( g ( s ) b ( s ) ) + 1 b ( s ) 0 ;
    (2.1)
     
  2. (ii)
    for any ( x , s , d , t ) D × R + × R + ¯ × R + ,
    f t ( x , s , d , t ) 0 , f d ( x , s , d , t ) [ ( 1 b ( s ) ) + 1 b ( s ) ] 0 , ( f ( x , s , d , t ) b ( s ) g ( s ) ) s f ( x , s , d , t ) b ( s ) g ( s ) 0 ;
    (2.2)
     
  3. (iii)
    m 0 + b ( s ) e s d s = + , m 0 : = min D ¯ u 0 ( x ) ;
    (2.3)
     
  4. (iv)
    α : = max D ¯ ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 > 0 , q 0 : = | u 0 | 2 .
    (2.4)
     
Then the solution u to problem (1.1) must be a global solution and
u ( x , t ) H 1 ( α t + H ( u 0 ( x , t ) ) ) , ( x , t ) D ¯ × R + ¯ ,
(2.5)
where
H ( z ) : = m 0 z b ( s ) e s d s , z 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
P = b u t u + b u t α e u u ,
(2.8)
Δ P = b u t | u | 2 + 2 b u u t + b u t Δ u + b Δ u t α e u | u | 2 α e u Δ u ,
(2.9)
and
P t = b ( u t ) 2 + b ( u t ) t α e u u t = b ( u t ) 2 + b ( g b Δ u + g b | u | 2 + f b ) t α e u u t = b ( u t ) 2 + ( g b g b ) u t Δ u + g Δ u t + ( g b g b ) u t | u | 2 + ( 2 g + 2 f q ) u u t + ( f u b f b α e u ) u t + f t .
(2.10)
It follows from (2.9) and (2.10) that
g b Δ P P t = ( b g b + b g b g ) u t | u | 2 + ( 2 b g b 2 g 2 f q ) u u t + ( 2 b g b g ) u t Δ u α g b e u | u | 2 α g b e u Δ u b ( u t ) 2 + ( b f b f u + α e u ) u t f t .
(2.11)
By (1.1), we have
Δ u = b g u t g g | u | 2 f g .
(2.12)
Substitute (2.12) into (2.11), to get
g b Δ P P t = ( b g b b g b g + ( g ) 2 g ) u t | u | 2 + ( 2 b g b 2 g 2 f q ) u u t ( b ) 2 g ( g b ) ( u t ) 2 + ( f g g b f b f u ) u t + ( α g b e u α g b e u ) | u | 2 + α f b e u f t .
(2.13)
With (2.8), we have
u t = 1 b P b b u t u + α e u b u .
(2.14)
Next, we substitute (2.14) into (2.13) to obtain
g b Δ P + [ 2 ( g b ) + 2 f q b ] u P P t = ( b g b + b g b g + ( g ) 2 g 2 ( b ) 2 g ( b ) 2 + 2 b f q b ) u t | u | 2 + ( 2 α b g ( b ) 2 e u α g b e u α g b e u 2 α f q b e u ) | u | 2 ( b ) 2 g ( g b ) ( u t ) 2 + ( f g g b f b f u ) u t + α f b e u f t .
(2.15)
In view of (2.7), we have
u t = 1 b P + α e u b .
(2.16)
Substituting (2.16) into (2.15), we get
g b Δ P + [ 2 ( g b ) + 2 f q b ] u P + { [ g ( 1 g ( g b ) ) + 2 f q ( 1 b ) ] | u | 2 + g ( b ) 2 ( f b g ) u } P P t = α e u { g [ ( 1 g ( g b ) + 1 b ) + 1 g ( g b ) + 1 b ] + 2 f q [ ( 1 b ) + 1 b ] } | u | 2 ( b ) 2 g ( g b ) ( u t ) 2 α g e u ( b ) 2 [ ( f b g ) u f b g ] f t .
(2.17)
The assumptions (2.1) and (2.2) guarantee that the right-hand side of (2.17) is nonnegative, i.e.,
g b Δ P + [ 2 ( g b ) + 2 f q b ] u P + { [ g ( 1 g ( g b ) ) + 2 f q ( 1 b ) ] | u | 2 + g ( b ) 2 ( f b g ) u } P P t 0 in  D × ( 0 , T ) .
(2.18)
By applying the maximum principle [24], it follows from (2.18) that P can attain its nonnegative maximum only for D ¯ × { 0 } or D × ( 0 , T ) . For D ¯ × { 0 } , by (2.4), we have
max D ¯ P ( x , 0 ) = max D ¯ { b ( u 0 ) ( u 0 ) t α e u 0 } = max D ¯ { ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) α e u 0 } = max D ¯ { e u 0 [ ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 α ] } = 0 .
We claim that P cannot take a positive maximum at any point ( x , t ) D × ( 0 , T ) . In fact, suppose that P takes a positive maximum at a point ( x 0 , t 0 ) D × ( 0 , T ) , then
P ( x 0 , t 0 ) > 0 and P n | ( x 0 , t 0 ) > 0 .
(2.19)
With (1.1) and (2.16), we have
P n = b u t u n + b u t n α e u u n = γ b u u t + b ( u 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 ) ( 1 b P + α 1 b e u ) + γ α u e u = γ ( u b ) b P + γ α e u u b ( u b ) b on  D × ( 0 , T ) .
(2.20)
Next, by using the fact that ( s b ( s ) ) 0 , s b ( s ) ( s b ( s ) ) 0 for any s R + , it follows from (2.20) that
P n | ( x 0 , t 0 ) 0 ,
which contradicts with inequality (2.19). Thus we know that the maximum of P in D ¯ × [ 0 , T ) is zero, i.e.,
P 0 in  D ¯ × [ 0 , T ) ,
and
b ( u ) e u u t α .
(2.21)
For each fixed x D ¯ , integration of (2.21) from 0 to t yields
0 t b ( u ) e u u t d t = u 0 ( x ) u ( x , t ) b ( s ) e s d s 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 T u ( x , t ) = + .
Passing to the limit as t T in (2.22) yields
u 0 ( x ) + b ( s ) e s d s α T
and
m 0 + b ( s ) e s d s = m 0 u 0 ( x ) b ( s ) e s d s + u 0 ( x ) + b ( s ) e s d s m 0 u 0 ( x ) b ( s ) e s d s + α T < + ,
which contradicts with assumption (2.3). This shows that u is global. Moreover, it follows from (2.22) that
u 0 ( x ) u ( x , t ) b ( s ) e s d s = m 0 u ( x , t ) b ( s ) e s d s m 0 u 0 ( x ) b ( s ) e s d s = H ( u ( x , t ) ) H ( u 0 ( x ) ) α t .
Since H is an increasing function, we have
u ( x , t ) 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:
  1. (i)
    for any s R + ,
    ( s b ( s ) ) 0 , s b ( s ) ( s b ( s ) ) 0 , ( g ( s ) b ( s ) ) 0 , [ 1 g ( s ) ( g ( s ) b ( s ) ) + 1 b ( s ) ] + 1 g ( g ( s ) b ( s ) ) + 1 b ( s ) 0 ;
    (3.1)
     
  2. (ii)
    for any ( x , s , d , t ) D × R + × R + ¯ × R + ,
    f t ( x , s , d , t ) 0 , f d ( x , s , d , t ) [ ( 1 b ( s ) ) + 1 b ( s ) ] 0 , ( f ( x , s , d , t ) b ( s ) g ( s ) ) s f ( x , s , d , t ) b ( s ) g ( s ) 0 ;
    (3.2)
     
  3. (iii)
    M 0 + b ( s ) e s d s < + , M 0 : = max D ¯ u 0 ( x ) ;
    (3.3)
     
  4. (iv)
    β : = min D ¯ ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 > 0 , q 0 : = | u 0 | 2 .
    (3.4)
     
Then the solution u of problem (1.1) must blow up in finite time T, and
T 1 β M 0 + b ( s ) e s d s ,
(3.5)
u ( x , t ) G 1 ( β ( T t ) ) , ( x , t ) D ¯ × [ 0 , T ) ,
(3.6)
where
G ( z ) : = z + 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
g b Δ Q + [ 2 ( g b ) + 2 f q b ] u Q + { [ g ( 1 g ( g b ) ) + 2 f q ( 1 b ) ] | u | 2 + g ( b ) 2 ( f b g ) u } Q Q t = β e u { g [ ( 1 g ( g b ) + 1 b ) + 1 g ( g b ) + 1 b ] + 2 f q [ ( 1 b ) + 1 b ] } | u | 2 ( b ) 2 g ( g b ) ( u t ) 2 β g e u ( b ) 2 [ ( f b g ) u f b g ] f t .
(3.9)
Assumptions (3.1) and (3.2) imply that the right-hand side in equality (3.9) is nonpositive, i.e.,
g b Δ Q + [ 2 ( g b ) + 2 f q b ] u Q + { [ g ( 1 g ( g b ) ) + 2 f q ( 1 b ) ] | u | 2 + g ( b ) 2 ( f b g ) u } Q Q t 0 in  D × ( 0 , T ) .
(3.10)
With (3.4), we have
min D ¯ Q ( x , 0 ) = min D ¯ { b ( u 0 ) ( u 0 ) t β e u 0 } = min D ¯ { ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) β e u 0 } = min D ¯ { e u 0 [ ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 β ] } = 0 .
(3.11)
Substituting P and α with Q and β in (2.20), respectively, we have
Q n = γ ( u b ) b Q + γ β e u u b ( u b ) b on  D × ( 0 , T ) .
(3.12)
Combining (3.10)-(3.12) with the fact that ( s b ( s ) ) 0 , s b ( s ) ( s b ( s ) ) 0 for any s R + , and applying the maximum principles again, it follows that the minimum of Q in D ¯ × [ 0 , T ) is zero. Thus
Q 0 in  D ¯ × [ 0 , T ) ,
and
b ( u ) e u u t β .
(3.13)
At the point x D ¯ , where u 0 ( x ) = M 0 , integrate (3.13) over [ 0 , t ] to get
0 t b ( u ) e u u t d t = M 0 u ( x , t ) b ( s ) e s d s 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
M 0 + b ( s ) e s d s M 0 u ( x , t ) b ( s ) e s d s β t .
(3.15)
Letting t + in (3.15), we have
M 0 + b ( s ) e s d s = + ,
which contradicts with assumption (3.3). This shows that u must blow up in finite time t = T . Furthermore, letting t T in (3.14), we get
T 1 β M 0 + b ( s ) e s d s .
By integrating inequality (3.13) over [ t , s ] ( 0 < t < s < T ), for each fixed x, we obtain
G ( u ( x , t ) ) G ( u ( x , t ) ) G ( u ( x , s ) ) = u ( x , t ) + b ( s ) e s d s u ( x , s ) + b ( s ) e s d s = u ( x , t ) u ( x , s ) b ( s ) e s d s = t s b ( u ) e u u t d t β ( s t ) .
Hence, by letting s T , we have
G ( u ( x , t ) ) β ( T t ) .
Since G is a decreasing function, we obtain
u ( x , t ) G 1 ( β ( T t ) ) .

The proof is complete. □

4 Applications

When b ( u ) u and f ( x , u , q , t ) f ( u ) , the results stated in Theorem 3.1 are valid. When g ( u ) 1 and f ( x , u , q , t ) f ( u ) or f ( x , u , q , t ) 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 [2123].

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:
{ u t = Δ u + 2 + u 1 + u | u | 2 + e u ( e u + e q ) 1 + u ( e t + | x | 2 ) in  D × ( 0 , T ) , u n + 2 u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = 2 | x | 2 in  D ¯ ,
where q = | 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:
{ ( u e u ) t = ( ( 1 + u ) e u u ) + ( e u + e q ) ( e t + | x | 2 ) in  D × ( 0 , T ) , u n + 2 u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = 2 | x | 2 in  D ¯ .
Now
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 .
In order to determine the constant α, we assume
s : = | x | 2 ,
then 0 s 1 and
α = max D ¯ ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 = max D ¯ { 32 | x | 2 4 | x | 4 18 + ( 1 + | x | 2 ) [ exp ( 4 + 2 | x | 2 ) + exp ( 2 + 5 | x | 2 ) ] } = max 0 s 1 { 32 s 4 s 2 18 + ( 1 + s ) [ exp ( 4 + 2 s ) + exp ( 2 + 5 s ) ] } = 50.4417 .
It is easy to check that (2.1)-(2.3) hold. By Theorem 2.1, u must be a global solution, and
u ( x , t ) H 1 ( α t + H ( u 0 ( x ) ) ) = 1 + 50.4417 t + ( 1 + u 0 ( x ) ) 2 = 1 + 50.4417 t + ( 3 | x | 2 ) 2 .
Example 4.2 Let u be a solution of the following problem:
{ u t = Δ u 1 u ( 1 + u ) | u | 2 + u ( e u e q ) 1 + u ( 6 + t | x | 2 ) in  D × ( 0 , T ) , u n + 2 u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = 2 | x | 2 in  D ¯ ,
where q = | 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:
{ ( u + ln u ) t = ( ( 1 + 1 u ) u ) + ( e u e q ) ( 6 + t | x | 2 ) in  D × ( 0 , T ) , u n + 2 u = 0 on  D × ( 0 , T ) , u ( x , 0 ) = 2 | x | 2 in  D ¯ .
Now we have
b ( u ) = u + ln u , g ( u ) = 1 + 1 u , f ( x , u , q , t ) = ( e u e q ) ( 6 + t | x | 2 ) , u 0 ( x ) = 2 | x | 2 , γ = 2 .
By setting
s : = | x | 2 ,
we have 0 s 1 and
β = min D ¯ ( g ( u 0 ) u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 = min D ¯ { 6 | x | 4 + 26 | x | 2 36 ( 2 | x | 2 ) 2 exp ( 2 | x | 2 ) + 6 [ 1 exp ( 3 | x | 2 2 ) ] } = min 0 s 1 { 6 s 2 + 26 s 36 ( 2 s ) 2 exp ( 2 s ) + 6 [ 1 exp ( 3 s 2 ) ] } = 0.0735 .
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
T 1 β M 0 + b ( s ) e s d s = 1 0.0735 2 + ( 1 + 1 s ) 1 e s d s = 2.5066 , u ( x , t ) G 1 ( β ( T t ) ) = G 1 ( 0.0735 ( T t ) ) ,
where
G ( z ) = z + b ( s ) e s d s = z + ( 1 + 1 s ) 1 e s d s , z 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

References

  1. Amann H: Quasilinear parabolic systems under nonlinear boundary conditions. Arch. Ration. Mech. Anal. 1986, 92: 153-192.MathSciNetView ArticleMATHGoogle Scholar
  2. Sperb RP: Maximum Principles and Their Applications. Academic Press, New York; 1981.MATHGoogle Scholar
  3. Ding JT, Guo BZ: Global existence and blow-up solutions for quasilinear reaction-diffusion equations with a gradient term. Appl. Math. Lett. 2011, 24: 936-942. 10.1016/j.aml.2010.12.052MathSciNetView ArticleMATHGoogle Scholar
  4. Tersenov A: The preventive effect of the convection and of the diffusion in the blow-up phenomenon for parabolic equations. Ann. Inst. Henri Poincaré, Anal. Non Linéaire 2004, 21: 533-541. 10.1016/j.anihpc.2003.10.001MathSciNetView ArticleMATHGoogle Scholar
  5. Zheng SN, Wang W: Effects of reactive gradient term in a multi-nonlinear parabolic problem. J. Differ. Equ. 2009, 247: 1980-1992. 10.1016/j.jde.2009.07.005View ArticleMATHMathSciNetGoogle Scholar
  6. Payne LE, Song JC: Lower bounds for blow-up time in a nonlinear parabolic problem. J. Math. Anal. Appl. 2009, 354: 394-396. 10.1016/j.jmaa.2009.01.010MathSciNetView ArticleMATHGoogle Scholar
  7. Chen SH: Global existence and blowup of solutions for a parabolic equation with a gradient term. Proc. Am. Math. Soc. 2001, 129: 975-981. 10.1090/S0002-9939-00-05666-5View ArticleMATHMathSciNetGoogle Scholar
  8. Chen SH: Global existence and blowup for quasilinear parabolic equations not in divergence form. J. Math. Anal. Appl. 2013, 401: 298-306. 10.1016/j.jmaa.2012.12.028MathSciNetView ArticleMATHGoogle Scholar
  9. Chipot M, Weissler FB: Some blowup results for a nonlinear parabolic equation with a gradient term. SIAM J. Math. Anal. 1989, 20: 886-907. 10.1137/0520060MathSciNetView ArticleMATHGoogle Scholar
  10. Fila M: Remarks on blow up for a nonlinear parabolic equation with a gradient term. Proc. Am. Math. Soc. 1991, 111: 795-801. 10.1090/S0002-9939-1991-1052569-9MathSciNetView ArticleMATHGoogle Scholar
  11. Souplet P, Weissler FB: Poincaré’s inequality and global solutions of a nonlinear parabolic equation. Ann. Inst. Henri Poincaré, Anal. Non Linéaire 1999, 16: 335-371. 10.1016/S0294-1449(99)80017-1MathSciNetView ArticleMATHGoogle Scholar
  12. Souplet P: Recent results and open problems on parabolic equations with gradient nonlinearities. Electron. J. Differ. Equ. 2001, 2001: 1-19.MathSciNetMATHGoogle Scholar
  13. Souplet P: Finite time blow-up for a non-linear parabolic equation with a gradient term and applications. Math. Methods Appl. Sci. 1996, 19: 1317-1333. 10.1002/(SICI)1099-1476(19961110)19:16<1317::AID-MMA835>3.0.CO;2-MMathSciNetView ArticleMATHGoogle Scholar
  14. Friedman A, Mcleod B: Blow-up of positive solutions of semilinear heat equations. Indiana Univ. Math. J. 1985, 34: 425-447. 10.1512/iumj.1985.34.34025MathSciNetView ArticleMATHGoogle Scholar
  15. Enache C: Blow-up phenomena for a class of quasilinear parabolic problems under Robin boundary condition. Appl. Math. Lett. 2011, 24: 288-292. 10.1016/j.aml.2010.10.006MathSciNetView ArticleMATHGoogle Scholar
  16. Payne LE, Schaefer PW: Blow-up in parabolic problems under Robin boundary conditions. Appl. Anal. 2008, 87: 699-707. 10.1080/00036810802189662MathSciNetView ArticleMATHGoogle Scholar
  17. Rault JF: The Fujita phenomenon in exterior domains under the Robin boundary conditions. C. R. Math. Acad. Sci. Paris 2011, 349: 1059-1061. 10.1016/j.crma.2011.09.006MathSciNetView ArticleMATHGoogle Scholar
  18. Li YF, Liu Y, Xiao SZ: Blow-up phenomena for some nonlinear parabolic problems under Robin boundary conditions. Math. Comput. Model. 2011, 54: 3065-3069. 10.1016/j.mcm.2011.07.034MathSciNetView ArticleMATHGoogle Scholar
  19. Liu Y, Luo SG, Ye YH: Blow-up phenomena for a parabolic problem with a gradient nonlinearity under nonlinear boundary conditions. Comput. Math. Appl. 2013, 65: 1194-1199. 10.1016/j.camwa.2013.02.014MathSciNetView ArticleMATHGoogle Scholar
  20. Li YF, Liu Y, Lin CH: Blow-up phenomena for some nonlinear parabolic problems under mixed boundary conditions. Nonlinear Anal., Real World Appl. 2010, 11: 3815-3823. 10.1016/j.nonrwa.2010.02.011MathSciNetView ArticleMATHGoogle Scholar
  21. Zhang LL: Blow-up of solutions for a class of nonlinear parabolic equations. Z. Anal. Anwend. 2006, 25: 479-486.MathSciNetMATHGoogle Scholar
  22. Zhang HL: Blow-up solutions and global solutions for nonlinear parabolic problems. Nonlinear Anal. TMA 2008, 69: 4567-4574. 10.1016/j.na.2007.11.013View ArticleMATHMathSciNetGoogle Scholar
  23. Ding JT: Global and blow-up solutions for nonlinear parabolic equations with Robin boundary conditions. Comput. Math. Appl. 2013, 65: 1808-1822. 10.1016/j.camwa.2013.03.013MathSciNetView ArticleGoogle Scholar
  24. Protter MH, Weinberger HF: Maximum Principles in Differential Equations. Prentice-Hall, Englewood Cliffs; 1967.MATHGoogle Scholar

Copyright

© Ding; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.