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Article

Global Existence and Uniform Blow-Up to a Nonlocal Parabolic System with Nonlinear Boundary Conditions Arising in a Thermal Explosion Theory

School of Mathematics and Statistics, Shandong Normal University, Jinan 250358, China
*
Author to whom correspondence should be addressed.
Mathematics 2023, 11(9), 1993; https://doi.org/10.3390/math11091993
Submission received: 13 March 2023 / Revised: 15 April 2023 / Accepted: 19 April 2023 / Published: 23 April 2023

Abstract

:
This paper deals with a nonlinear nonlocal parabolic system with nonlinear heat-loss boundary conditions, which arise in the thermal explosion model. Firstly, we prove a comparison principle for some kinds of parabolic systems under nonlinear boundary conditions. Using this, we improve a new theorem of the sub-and-super solution. Secondly, based on the new sub-and-super solution theorem, the sufficient conditions that the solution exists and blows up uniformly in finite time are presented. Then, we generalize some of the lemmas related to uniform blow-up solutions, which are used to introduce the uniform blow-up profiles of solutions. Finally, we give several numerical simulations to illustrate the existence and uniform blow-up of solutions.

1. Introduction

In this paper, we discuss the following semi-linear parabolic system with nonlinear nonlocal sources and nonlinear boundary conditions:
u t = Δ u + k 1 Ω u m ( x , t ) v n ( x , t ) d x , ( x , t ) Q T , v t = Δ v + k 2 Ω u p ( x , t ) v q ( x , t ) d x , ( x , t ) Q T , n · u + g 1 ( u ) u = n · v + g 2 ( v ) v = 0 , ( x , t ) S T , u ( x , 0 ) = u 0 ( x ) , v ( x , 0 ) = v 0 ( x ) , x Ω ,
where  k 1 , k 2 , m , n , p , q  are positive constants,  Q T = Ω × ( 0 , T ] S T = Ω × ( 0 , T ] T > 0  and  Ω  is a bounded region in  R N  ( N > 1  is an integer) with a smooth boundary  Ω n  is an outward unit normal vector of  Ω g 1 , g 2 , u 0 , v 0  are nonnegative continuous functions satisfying some assumptions, and  | Ω |  denotes the Lebesgue measure of  Ω .
Our study is motivated by [1,2], in which the existence and blow-up of partial differential equations under nonlinear boundary conditions were researched. As stated in [1], in certain thermal explosion problems that require long induction times (times prior to ignition), such as the safe storage of energetic materials or nuclear waste, the Dirichlet boundary condition  u = v = 0  is no longer applicable because during this time the boundary of the reaction material will be preheated to a temperature significantly higher than the surrounding temperature. As a result, we need the heat-loss boundary conditions in (1) to describe the temperature distribution of the boundary. In recent years, many authors have investigated the reaction-diffusion system with Dirichlet boundary conditions as follows:
u t d 1 ( x , t ) Δ u = f ( u , v ) , ( x , t ) Ω × ( 0 , T ) , v t d 2 ( x , t ) Δ v = g ( u , v ) , ( x , t ) Ω × ( 0 , T ) , u ( x , t ) = v ( x , t ) = 0 , ( x , t ) Ω × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) , v ( x , 0 ) = v 0 ( x ) , x Ω ,
For details, see [3,4,5,6,7] and their references. In [3], the blow-up and global existence criteria of the nonlocal parabolic system
u t = f ( u ) Δ u + a Ω v d x , v t = g ( v ) Δ v + b Ω u d x
were obtained. In [4,7], the authors studied nonlocal nonlinear parabolic systems of the form
u t u s 1 Δ u = a u m Ω v n d x , v t v s 2 Δ v = b v p Ω u q d x
where  m , n , p , q , s 1 , s 2 > 0  are constants. They all gave some results on the existence of solutions, the blow-up solutions, and the blow-up rate or profile. In [5], Li et al. suggested that if one of  α > 1 , β > 1  or  p q > ( 1 α ) ( 1 β )  holds, the solution of (2) blows up in finite time for sufficiently large initial data, when
f ( u , v ) = Ω u α v p d x , g ( u , v ) = Ω u q v β d x
and  d i = 1 , i = 1 , 2 . Furthermore, they also obtained the blow-up profile of the solution
lim t T * u ( x , t ) ( T * t ) p β + 1 p q ( 1 α ) ( 1 β ) = q α + 1 p β + 1 p p q ( 1 α ) ( 1 β ) × p β + 1 | Ω | ( p q ( 1 α ) ( 1 β ) ) p β + 1 p q ( 1 α ) ( 1 β ) , lim t T * v ( x , t ) ( T * t ) q α + 1 p q ( 1 α ) ( 1 β ) = p β + 1 q α + 1 q p q ( 1 α ) ( 1 β ) × q α + 1 | Ω | ( p q ( 1 α ) ( 1 β ) ) q α + 1 p q ( 1 α ) ( 1 β ) ,
uniformly on any compact subset of  Ω , if  q > α 1 p > β 1 , and  p q > ( 1 α ) ( 1 β ) . Thereafter, in [8], Kong and Wang extended these results to parabolic systems with nonlocal boundaries
u = Ω φ ( x , y ) u ( y , t ) d y , v = Ω ψ ( x , y ) v ( y , t ) d y
by improving the methods used in [9,10].
In addition, there are some interesting results on the global well-posed theory related to our work that can be found in [11,12,13,14].
In this paper, we present the global existence, the uniform blow-up criteria, and the uniform blow-up profiles of system (1) with nonlinear heat-loss boundary conditions. To this end, we first give the following assumptions.
(H1)
u 0 v 0 C 2 + α ( Ω ) C 1 ( Ω ¯ ) ( 0 < α < 1 )  are nonnegative, bounded, and  u 0 n v 0 n < 0 ;
(H2)
g i > 0 , i = 1 , 2  satisfy the local Lipschitz condition;
(H3)
g i , i = 1 , 2  are increasing functions.
This paper is divided into five sections. In Section 2, we prove some new comparison principles for a modification of our system and using the obtained principles, we can obtain the local existence of (1) if our system has sub- and super-solutions as we define. In Section 3, we give some new definitions of uniform blow-up solutions and global solutions and present the global existence and uniform blow-up criteria of solutions to system (1), respectively, with the help of the methods used in [8,15]. In Section 4, we generalize several relevant lemmas from [9,10] and based on them give the blow-up profile of (1), which describes the asymptotic behavior of the solution near the blow-up time. In Section 5, we simulate numerically the global existence and uniform blow-up of solutions.

2. Local Existence

In this section, we intend to prove the local existence of solutions to (1) using the monotone iterative technique (see [16,17,18]) and the fixed-point theorem. To do this, we need to define the sub- and super-solutions, as well as to generalize the comparison principle from [2,8,15]. Here we define the sub- and super-solutions as follows.
Definition 1.
A pair of nonnegative functions  ( u ¯ ( x , t ) , v ¯ ( x , t ) )  is called a super-solution of (1) if  ( u ¯ , v ¯ ) ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) × ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) 0 < T +  and satisfies
u ¯ t Δ u ¯ + k 1 Ω u ¯ m ( x , t ) v ¯ n ( x , t ) d x , ( x , t ) Q T , v ¯ t Δ v ¯ + k 2 Ω u ¯ p ( x , t ) v ¯ q ( x , t ) d x , ( x , t ) Q T , n · u ¯ + g 1 ( u ¯ ) u ¯ 0 , n · v ¯ + g 1 ( v ¯ ) v ¯ 0 , ( x , t ) S T , u ¯ ( x , 0 ) u 0 ( x ) , v ¯ ( x , 0 ) v 0 ( x ) , x Ω .
Similarly, a sub-solution can be defined with reversed inequalities in (7).
Then, we prove the following comparison principle.
Lemma 1.
Suppose that  a i  and  b i  are the continuous and bounded functions on  Q ¯ T , respectively;  c i  and  d i  are the nonnegative and bounded functions in  Q T ; and  f i  and  h i  are the nonnegative functions and the positive increasing functions on  Ω ¯ , respectively. If  u , v C 2 , 1 ( Q T ) C 1 ( Q ¯ T )  satisfy
u t a 1 Δ u b 1 u + f 1 ( x ) Ω ( c 1 u + d 1 v ) d x , ( x , t ) Q T , v t a 2 Δ v b 2 v + f 2 ( x ) Ω ( c 2 u + d 2 v ) d x , ( x , t ) Q T , n · u + h 1 ( u ) u 0 , n · v + h 2 ( v ) v 0 , ( x , t ) S T , u ( x , 0 ) = u 0 ( x ) 0 , v ( x , 0 ) = v 0 ( x ) 0 , x Ω
then  u , v 0  in  Q T .
Proof. 
Firstly, we prove that if  u , v  satisfy
n · u + h 1 ( u ) u > 0 , n · v + h 2 ( v ) v > 0 , ( x , t ) S T , u ( x , 0 ) = u 0 ( x ) > 0 , v ( x , 0 ) = v 0 ( x ) > 0 , x Ω
then  u , v > 0  in  Q T .
Let  b ¯ i = max Q ¯ T b i i = 1 , 2  and  γ 1 > max { b ¯ 1 , b ¯ 2 } . Set  w = e γ 1 t u z = e γ 1 t v , then we have
w t a 1 Δ w + ( γ 1 b 1 ) w f 1 ( x ) Ω ( c 1 w + d 1 z ) d x , ( x , t ) Q T , z t a 2 Δ z + ( γ 1 b 2 ) z f 2 ( x ) Ω ( c 2 w + d 2 z ) d x , ( x , t ) Q T , n · w + h 1 ( u ) w > 0 , n · z + h 2 ( v ) z > 0 , ( x , t ) S T , w ( x , 0 ) = u 0 ( x ) > 0 , z ( x , 0 ) = v 0 ( x ) > 0 , x Ω .
By boundary conditions and Assumption (H1),  w , z > 0  on  S T  can be obtained. Considering initial conditions  w ( x , 0 ) > 0 z ( x , 0 ) > 0 , by continuity, there exists a  t > 0  such that  w ( x , t ) > 0 z ( x , t ) > 0  for all  ( x , t ) Ω × [ 0 , t ] . Let  A = { t < T : w , z > 0 in Ω × [ 0 , t ] }  and take  t ¯ = sup A , then we have  0 < t ¯ T . If  t ¯ = T , the conclusion is proven, and if  t ¯ < T , with no loss of generality we can suppose that there are some  x ¯ Ω ¯  such that  w ( x ¯ , t ¯ ) = 0 . Therefore, w takes the minimum on  Ω ¯ × ( 0 , t ¯ ] . According to the first inequality of (10), we obtain
w t a 1 Δ w ( b 1 γ 1 ) w , ( x , t ) Ω × ( 0 , t ¯ ] .
By the strong maximum principle (Theorem 1 in [19] Chap. 2),  w 0  can be obtained, which is a contradiction. Hence,  w , z > 0 ; that is  u , v > 0 .
Then, we prove that if  u , v  satisfy
n · u + h 1 ( u ) u 0 , n · v + h 2 ( v ) v 0 , ( x , t ) S T , u ( x , 0 ) = u 0 ( x ) 0 , v ( x , 0 ) = v 0 ( x ) 0 , x Ω
then  u , v 0 .
Let  b ¯ i = max Q T | b i | c ¯ i = max Q T ( c i + d i ) f ¯ i = max Ω f i i = 1 , 2  and
w ( x , t ) = u ( x , t ) + η e γ 2 t , z ( x , t ) = v ( x , t ) + η e γ 2 t ,
where  γ 2 max i = 1 , 2 ( b ¯ i + f ¯ i c ¯ i | Ω | ) η  is an arbitrary positive real number. Then, we know  w ( x , 0 ) = u 0 ( x ) + η > 0 z ( x , 0 ) = v 0 ( x ) + η > 0 . Due to the boundary conditions, we have
n · w + h 1 ( w ) w n · u + h 1 ( u ) u + η h 1 ( w ) e γ 2 t η h 1 ( w ) e γ 2 t > 0 , n · z + h 2 ( z ) z n · v + h 2 ( v ) v + η h 2 ( z ) e γ 2 t η h 2 ( z ) e γ 2 t > 0 ,
which holds in  S T . Further, in  Q T , we have
w t a 1 Δ w b 1 w f 1 Ω ( c 1 w + d 1 z ) d x ( γ 2 b 1 f 1 ( c 1 + d 1 ) | Ω | ) η e γ 2 t 0 , z t a 2 Δ z b 2 z f 2 Ω ( c 2 w + d 2 z ) d x ( γ 2 b 2 f 2 ( c 2 + d 2 ) | Ω | ) η e γ 2 t 0 .
Therefore, we know that  w , z > 0 , then, letting  η 0 + u , v 0  is obtained. The proof of Lemma 1 is completed. □
We give the following conclusion about the properties of the sub-solutions and super-solutions.
Lemma 2.
Suppose Assumptions (H1)–(H3) hold,  g 1 g 2  are derivable, and the nonnegative  ( u ̲ , v ̲ ) ( u ¯ , v ¯ ) ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) × ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) 0 < T +  are the sub- and super-solutions to system (1), respectively.  ( u , v ) ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) × ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) )  is a solution to problem (1). If Case (i)
n · u ¯ + g 1 ( u ¯ ) u ¯ > 0 , n · v ¯ + g 1 ( v ¯ ) v ¯ > 0 , ( x , t ) S T , u ¯ ( x , 0 ) > u 0 ( x ) , v ¯ ( x , 0 ) > v 0 ( x ) x Ω
is satisfied, we have  u ¯ > u v ¯ > v ; otherwise, if Case (ii)
n · u ̲ + g 1 ( u ̲ ) u ̲ < 0 , n · v ̲ + g 1 ( v ̲ ) v ̲ < 0 , ( x , t ) S T , u ̲ ( x , 0 ) < u 0 ( x ) , v ̲ ( x , 0 ) < v 0 ( x ) x Ω
is satisfied, we have  u > u ̲ v > v ̲ .
Proof. 
Case (i). We assume that (16) holds. Let  w = u ¯ u z = v ¯ v , and  ψ 1 ( s ) = g 1 ( s ) s + g 1 ( s ) ψ 2 ( s ) = g 2 ( s ) s + g 2 ( s ) ; obviously,  ψ 1 ψ 2 > 0 . According to (1), and using the mean value theorem, we have
w t Δ w k 1 Ω ( u ¯ m v ¯ n u m v n ) d x = k 1 Ω ( m ξ 1 m 1 ξ 2 n w + n ξ 1 m ξ 2 n 1 z ) d x , ( x , t ) Q T , z t Δ z k 2 Ω ( u ¯ p v ¯ q u p v q ) d x = k 2 Ω ( p ξ 1 p 1 ξ 2 q w + q ξ 1 p ξ 2 q 1 z ) d x , ( x , t ) Q T , n · w + ψ 1 ( ζ 1 ) w > 0 , n · z + ψ 2 ( ζ 2 ) z > 0 , ( x , t ) S T , w ( x , 0 ) > 0 , z ( x , 0 ) > 0 , x Ω .
where  ξ 1 ζ 1  are between u and  u ¯ ξ 2 ζ 2  are between v and  v ¯ . Similar to the proof of Lemma 1, let  w ˜ = e t w z ˜ = e t z , then, (18) can be transformed into
w ˜ t Δ w ˜ + w ˜ k 1 Ω ( m ξ 1 m 1 ξ 2 n w + n ξ 1 m ξ 2 n 1 z ) d x , ( x , t ) Q T , z ˜ t Δ z ˜ + z ˜ k 2 Ω ( p ξ 1 p 1 ξ 2 q w + q ξ 1 p ξ 2 q 1 z ) d x , ( x , t ) Q T , n · w ˜ + ψ 1 ( ζ 1 ) w ˜ > 0 , n · z ˜ + ψ 2 ( ζ 2 ) z ˜ > 0 , ( x , t ) S T , w ˜ ( x , 0 ) > 0 , z ˜ ( x , 0 ) > 0 , x Ω .
By boundary conditions and  ψ 1 ψ 2 > 0 , we have  w ˜ z ˜ > 0  on  S T . Since  w ˜ ( x , 0 ) z ˜ ( x , 0 ) > 0 , by continuity, there exists a  t > 0  such that  w ˜ z ˜ > 0  for all  ( x , t ) Ω × [ 0 , t ] . Let  A = { t < T : w ˜ , z ˜ > 0 in Ω × [ 0 , t ] }  and  t ¯ = sup A , then we have  0 < t T . When  t ¯ = T , the conclusion is obvious and we discuss the case  t ¯ < T . Supposing that there are some  x ¯ Ω ¯  such that  w ˜ ( x ¯ , t ¯ ) = 0 w ˜  takes the minimum on  Ω ¯ × ( 0 , t ¯ ] . According to the first inequality of (19), we obtain
w ˜ t Δ w ˜ w ˜ , ( x , t ) Ω × ( 0 , t ¯ ] .
By the strong maximum principle (Theorem 1 in [19] Chap. 2),  w ˜ 0  can be obtained, which is a contradiction. We have  w ˜ z ˜ > 0 ; that is, w z > 0 u ¯ > u  and  v ¯ > v  are proved.
Case (ii). The proof of Case (ii) is similar to that of Case (i), and we omit it. □
Based on Definition 1 and Lemma 1, we give the theorem describing the local existence of (1).
Theorem 1.
Suppose Assumptions (H1)–(H3) hold, the nonnegative  ( u ̲ , v ̲ ) ( u ¯ , v ¯ ) ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) × ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) 0 < T +  are the sub- and super-solutions to system (1), respectively, and  u ̲ u ¯ v ̲ v ¯ , then there exists a pair of functions  ( u , v )  satisfying  u ̲ u u ¯ v ̲ v v ¯ , which is a solution to system (1).
Proof. 
According to Assumptions (H1)(H2), we know that for any  u 1 u 2  given, since the function  g 1 ( u ) u  is Lipschitz continuous, there exists a fixed real number  L 1  such that
| g 1 ( u 1 ) u 1 ( g 1 ( u 2 ) u 2 ) | L 1 | u 1 u 2 | ,
thus
g 1 ( u 1 ) u 1 ( g 1 ( u 2 ) u 2 ) L 1 ( u 1 u 2 ) .
Let  G l 1 = g 1 ( u ) u + L 1 u  and similarly  G l 2 ( v ) = g 2 ( v ) v + L 2 v , which are increasing. Then, we consider the following auxiliary problem
w t Δ w + w = u + k 1 Ω u m v n d x , ( x , t ) Q T , z t Δ z + z = v + k 2 Ω u p v q d x , ( x , t ) Q T , n · w + L 1 w = G l 1 ( u ) , n · z + L 2 z = G l 2 ( v ) , ( x , t ) S T , w ( x , 0 ) = u 0 ( x ) , z ( x , 0 ) = v 0 ( x ) , x Ω ,
where  u , v C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) 0 < T +  are two nonnegative functions. The auxiliary problem (22) has Robin boundary conditions, which means that there exists a unique solution  ( w , z )  to it. Define nonlinear operator  T 1 : [ u ̲ , u ¯ ] [ u ̲ , u ¯ ]  and  T 2 : [ v ̲ , v ¯ ] [ v ̲ , v ¯ ]  such that  w = T 1 u  and  z = T 2 v . Then, construct the following sequences
w ˜ 1 = T 1 u ¯ , w ˜ 2 = T 1 w ˜ 1 , , w ˜ n = T 1 w ˜ n 1 , w ^ 1 = T 1 u ̲ , w ^ 2 = T 1 w ^ 1 , , w ^ n = T 1 w ^ n 1 ; z ˜ 1 = T 2 v ¯ , z ˜ 2 = T 2 z ˜ 1 , , z ˜ n = T 2 z ˜ n 1 , z ^ 1 = T 1 v ̲ , z ^ 2 = T 2 z ^ 1 , , z ^ n = T 2 z ^ n 1 .
Next, we will show that the operators  T 1  and  T 2  are increasing. For any  u ̲ u 1 u 2 u ¯  and  v ̲ v 1 v 2 v ¯ , let  w 1 = T 1 u 1 w 2 = T 1 u 2 , and  z 1 = T 2 v 1 z 2 = T 2 v 2 , and take  W = w 2 w 1 Z = z 2 z 1 . Then, we have
W t Δ W + W = k 1 Ω ( u 2 m v 2 n u 1 m v 1 n ) d x + ( u 2 u 1 ) 0 , ( x , t ) Q T , Z t Δ Z + Z = k 2 Ω ( u 2 p v 2 q u 1 p v 1 q ) d x + ( v 2 v 1 ) 0 , ( x , t ) Q T , n · W + L 1 W = G l 1 ( u 2 ) G l 1 ( u 1 ) 0 , ( x , t ) S T , n · Z + L 2 Z = G l 2 ( v 2 ) G l 2 ( v 1 ) 0 , ( x , t ) S T , W ( x , 0 ) = Z ( x , 0 ) = 0 , x Ω .
Applying Lemma 1 with  a i = 1 b i = 1 c i = d i = 0 h i = L i i = 1 , 2 , we know  W , Z 0 ; that is,  w 2 w 1 z 2 z 1 . Letting  W = u ¯ w ˜ 1  and  Z = v ¯ z ^ 1 , we can obtain
W t Δ W + W = k 1 Ω ( u ¯ m v ¯ n u ¯ m v ¯ n ) d x + ( u ¯ u ¯ ) = 0 , ( x , t ) Q T , Z t Δ Z + Z = 0 , ( x , t ) Q T , n · W + L 1 W G l 1 ( u ¯ ) G l 1 ( u ¯ ) = 0 , ( x , t ) S T , n · Z + L 2 Z 0 , ( x , t ) S T , W ( x , 0 ) = Z ( x , 0 ) = 0 , x Ω ,
thus,  u ¯ w ˜ 1  and  v ¯ z ˜ 1  can be deduced. Similarly,  w ^ 1 u ̲  and  z ^ 1 v ̲  can be deduced, too.
By mathematical induction on n, we have
u ̲ w ^ 1 w ^ 2 w ^ n w ˜ n w ˜ n 1 w ˜ 1 u ¯ , v ̲ z ^ 1 z ^ 2 z ^ n z ˜ n z ˜ n 1 z ˜ 1 v ¯ ,
which shows that the sequences  { w ˜ n } { w ^ n } { z ˜ n } { z ^ n }  are increasing and bounded. Therefore, limits
w ˜ 0 = lim n w ˜ n , w ^ 0 = lim n w ^ n , z ˜ 0 = lim n z ˜ n , z ^ 0 = lim n z ^ n
satisfying  w ˜ 0 = T 1 w ˜ 0 z ˜ 0 = T 2 z ˜ 0  and  w ^ 0 = T 1 w ^ 0 z ^ 0 = T 2 z ^ 0 , exist. Considering the compactness of the nonlinear operators  T 1 T 2  and  | Ω | < , we know that  ( w ˜ 0 , z ˜ 0 )  and  ( w ^ 0 , z ^ 0 )  are the solutions to problem (1). The local existence of the solutions to problem (1) is proved. □

3. Global Existence and Uniform Blow-Up Results

In this section, we will use the sub-and-super solution theorem and combine it with Lemma 2 to give the global existence and uniform blow-up conditions of system (1). Firstly, we define the uniform blow-up solution.
Definition 2.
The nonnegative solution  ( u , v )  to problem (1) blows up uniformly in finite time if there exists a positive real number  T <  such that
lim sup t T min x Ω ¯ ( u ( x , t ) + v ( x , t ) ) = + .
Further, the solution  ( u , v )  of problem (1) exists globally if for all  t ( 0 , + ) ,
max x Ω ¯ ( u ( x , t ) + v ( x , t ) ) < + .
Then, we refer to [8,15] to give Theorems 2 and 3, which are about global existence and blow-up, respectively. Further, the proofs are given using the sub-and-super solution theorem.
Theorem 2.
Suppose Assumptions (H1)–(H3) hold,  g 1 g 2  are derivable and
m < 1 , p < 1 , n q ( 1 m ) ( 1 p ) .
Then, the solution  ( u , v )  of this problem exists globally of system (1).
Proof. 
First, according to condition (30), we have
1 m n · 1 p q 1 ,
so there exist two positive constants  α 1 , β 1 > 1  such that
1 m n β 1 α 1 , 1 p q α 1 β 1 .
Due to the continuity of functions  g 1  and  g 2 , let  δ 0 = min { min g 1 , min g 2 }  and  c 0 > B δ 0 max { α 1 , β 1 } . We consider the following eigenvalue problem
Δ φ + λ φ = 0 , x Ω , φ = c 0 , x Ω .
The first eigenvalue of problem (33) is denoted by  λ . Take  B = max Ω ¯ | φ | γ 0 = max Ω ¯ φ  and  k ¯ 1 = γ 0 m + n k 1 | Ω | k ¯ 2 = γ 0 p + q k 2 | Ω | . Then, let  s ( t )  be the unique solution to the Cauchy problem
d s d t = λ s ( t ) + ( α 1 + β 1 2 ) c 0 2 B 2 s ( t ) + k ¯ 1 α 1 1 c 0 α 1 s m α 1 + n β 1 α 1 + 1 ( t ) + k ¯ 2 β 1 1 c 0 β 1 s p α 1 + q β 1 β 1 + 1 ( t ) , t > 0 , s ( 0 ) = 1 c 0 max max Ω ¯ u 0 ( x ) 1 α 1 , max Ω ¯ v 0 ( x ) 1 β 1 + 1 .
where  m α 1 + n β 1 α 1 + 1 p α 1 + q β 1 β 1 + 1 < 1  such that  s ( t )  exists globally. Set  u ¯ = s α 1 ( t ) φ α 1 ( x ) v ¯ = s β 1 ( t ) φ β 1 ( x ) ( x , t ) Ω × ( 0 , + ) . Then, we assert that  ( u ¯ , v ¯ )  is a super-solution to problem (1). In  Ω × ( 0 , + ) , we have
Δ u ¯ + k 1 Ω u ¯ m v ¯ n d x = s α 1 ( t ) Δ φ α 1 + k 1 s m α 1 + n β 1 ( t ) Ω φ m α 1 + n β 1 d x s α 1 ( t ) α 1 φ α 1 1 Δ φ + α 1 ( α 1 1 ) φ α 1 2 ( φ ) 2 + k ¯ 1 s m α 1 + n β 1 ( t ) = s α 1 ( t ) λ α 1 φ α 1 + α 1 ( α 1 1 ) φ α 1 2 ( φ ) 2 + k ¯ 1 s m α 1 + n β 1 ( t ) α 1 s α 1 1 ( t ) φ α 1 λ + ( α 1 1 ) c 0 2 B 2 + k ¯ 1 α 1 1 c 0 α 1 s m α 1 + n β 1 α 1 ( t ) s ( t ) α 1 s α 1 1 ( t ) s ( t ) φ α 1 = u ¯ t ,
similarly,
Δ v ¯ + k 2 Ω u ¯ p v ¯ q d x = s β 1 ( t ) Δ φ β 1 + k 2 s p α 1 + q β 1 ( t ) Ω φ p α 1 + q β 1 d x s β 1 ( t ) β 1 φ β 1 1 Δ φ + β 1 ( β 1 1 ) φ β 1 2 ( φ ) 2 + k ¯ 2 s p α 1 + q β 1 ( t ) = s β 1 ( t ) λ β 1 φ β 1 + β 1 ( β 1 1 ) φ β 1 2 ( φ ) 2 + k ¯ 2 s p α 1 + q β 1 ( t ) β 1 s β 1 1 ( t ) φ β 1 λ + ( β 1 1 ) c 0 2 B 2 + k ¯ 2 β 1 1 c 0 β 1 s p α 1 + q β 1 β 1 ( t ) s ( t ) β 1 s β 1 1 ( t ) s ( t ) φ β 1 = v ¯ t .
Since  φ = c 0  on  Ω , we have that
n · u ¯ + g 1 ( u ¯ ) u ¯ = φ α 1 1 s α 1 ( α 1 n · φ + c 0 g ( u ¯ ) ) > 0 ,
and
n · v ¯ + g 2 ( v ¯ ) v ¯ = φ β 1 1 s β 1 ( β 1 n · φ + c 0 g ( v ¯ ) ) > 0
hold in  Ω × ( 0 , + ) . Considering the initial values, we have
u ¯ ( x , 0 ) = s α 1 ( 0 ) φ α 1 > φ α 1 c 0 α 1 max Ω ¯ u 0 u 0 ,
and
v ¯ ( x , 0 ) = s β 1 ( 0 ) φ β 1 > φ β 1 c 0 β 1 max Ω ¯ v 0 v 0 .
Hence,  ( u ¯ , v ¯ )  is a super-solution to problem (1), and satisfies (16). Applying Lemma 2, the solution  ( u , v )  to problem (1) satisfies  u < u ¯ v < v ¯ . Since  u ¯ v ¯  exist globally, Theorem 2 is proven. □
Theorem 3.
Suppose Assumptions (H1)–(H3) hold and  g 1 g 2  are derivable.
Case (i). If one of the following conditions holds
( a ) m > 1 ; ( b ) q > 1 ; ( c ) n p > ( 1 m ) ( 1 q ) ,
and  u 0 ( x ) , v 0 ( x )  are sufficiently large, then the solution  ( u , v )  of problem (1) blows up uniformly in a finite time.
Case (ii). If (40) is satisfied and  u 0 ( x ) , v 0 ( x )  are sufficiently small, then the solution  ( u , v )  exists globally.
Proof. 
Case (i) (a). Assuming that  m > 1 , we consider the eigenvalue problem (33) with  c 0 = 0 , and for convenience, let  max Ω ¯ φ = 1 2 . Take  α 2 β 2 > 1  such that  p α 2 + q β 2 β 2 > 0  and set
γ = min { m α 2 + n β 2 α 2 + 1 , p α 2 + q β 2 β 2 + 1 } .
Obviously,  γ > 1 . Let  h ( t )  be the unique solution to the following Cauchy problem
d h d t = λ h ( t ) + min k 1 α 2 Ω φ m α 2 + n β 2 d x , k 2 β 2 Ω φ p α 2 + q β 2 d x h γ ( t ) , t > 0 , h ( 0 ) = h 0 ,
where  h 0 > 0  is sufficiently large such that the solution to (42) blows up in a finite time.
Let  u 0 ( x ) v 0 ( x )  be sufficiently large such that  min { ( min Ω ¯ u 0 ) 1 α 2 , ( min Ω ¯ v 0 ) 1 β 2 } > h 0  and  u ̲ = h α 2 ( t ) ( φ ( x ) + δ ) α 2 v ̲ = h β 2 ( t ) ( φ ( x ) + δ ) β 2 ( x , t ) Ω × ( 0 , T * ε ] , where  ε ( 0 , T * )  and  δ ( 0 , 1 2 )  are arbitrarily real numbers. Then, we will show that  ( u ̲ , v ̲ )  is a sub-solution to problem (1). In  Ω × ( 0 , T * ε ] , we yield
Δ u ̲ + k 1 Ω u ̲ m v ̲ n d x = h α 2 ( t ) Δ ( φ + δ ) α 2 + k 1 h m α 2 + n β 2 ( t ) Ω ( φ + δ ) m α 2 + n β 2 d x h α 2 ( t ) λ α 2 ( φ + δ ) α 2 + α 2 ( α 2 1 ) ( φ + δ ) α 2 2 ( φ ) 2 + k 1 h m α 2 + n β 2 ( t ) Ω φ m α 2 + n β 2 d x α 2 h α 2 1 ( t ) ( φ + δ ) α 2 λ + k 1 α 2 h m α 2 + n β 2 α 2 ( t ) Ω φ m α 2 + n β 2 d x h ( t ) α 2 h α 2 1 ( t ) h ( t ) ( φ + δ ) α 2 = u ̲ t ,
and
Δ v ̲ + k 2 Ω u ̲ p v ̲ q d x = h β 2 ( t ) Δ ( φ + δ ) β 2 + k 2 h p α 2 + q β 2 ( t ) Ω ( φ + δ ) p α 2 + q β 2 d x h β 2 ( t ) λ β 2 ( φ + δ ) β 2 + β 2 ( β 2 1 ) ( φ + δ ) β 2 2 ( φ ) 2 + k 2 h p α 2 + q β 2 ( t ) Ω φ p α 2 + q β 2 d x β 2 h β 2 1 ( t ) ( φ + δ ) β 2 λ + k 2 β 2 h p α 2 + q β 2 β 2 ( t ) Ω φ p α 2 + q β 2 d x h ( t ) β 2 h β 2 1 ( t ) h ( t ) ( φ + δ ) β 2 = v ̲ t .
On  Ω × ( 0 , T * ε ] , we yield
n · u ̲ + g 1 ( u ̲ ) u ̲ = δ α 2 1 h α 2 ( t ) ( α 2 n · φ + δ g 1 ( u ̲ ) ) < 0 , n · v ̲ + g 2 ( v ̲ ) v ̲ = δ β 2 1 h β 2 ( t ) ( β 2 n · φ + δ g 2 ( v ̲ ) ) < 0 ,
when  δ  is sufficiently close to 0. Since
u ̲ ( x , 0 ) = h 0 α 2 ( φ + δ ) α 2 < u 0 ( x ) , x Ω , v ̲ ( x , 0 ) = h 0 β 2 ( φ + δ ) β 2 < v 0 ( x ) , x Ω ,
we can obtain that  ( u ̲ , v ̲ )  is a sub-solution to problem (1) and satisfies (17). Applying Lemma 2, we know that  u > u ̲ v > v ̲  in  Ω × ( 0 , T * ε ] . Due to the arbitrary nature of  ε , letting  ε 0 + u u ̲ v v ̲  in  Ω × ( 0 , T * )  can be obtained.
Specifically, we have  u > u ̲ = δ α 2 h α 2 ( t )  and  v > v ̲ = δ β 2 h β 2 ( t )  on  Ω × ( 0 , T * ) . Hence, the solution  ( u , v )  blows up globally when  m > 1  and  u 0 ( x ) v 0 ( x )  are sufficiently large.
Case (i) (b). We omit the proof for the comparable case in which  q > 1 .
Case (i) (c). When  0 < m 1 0 < q 1  and  n p > ( 1 m ) ( 1 q ) , we know that
n 1 m 1 q p > 0 .
Take  ε 0 , n 1 m 1 q p  and  α 2 β 2 = 1 q p + ε , then there exist two positive constants  α 2 β 2 > 1  such that
1 q p < α 2 β 2 , 1 m n < β 2 α 2 ,
which means that  m α 2 + n β 2 α 2 + 1 > 1 p α 2 + q β 2 β 2 + 1 > 1 . When  γ = min { m α 2 + n β 2 α 2 + 1 > 1 p α 2 + q β 2 β 2 + 1 }  is used, and the solution  ( u , v )  also blows up, as shown in Case (i).
Case (ii). By the continuity of functions  g 1  and  g 2 , we have  min g 1 , min g 2 > 0 . Let positive constant  δ  be sufficiently small such that  δ < min { min g 1 , min g 2 }  and  e ( x )  be the unique solution to the elliptic problem
Δ e ( x ) = 1 , x Ω , n · e ( x ) + δ e ( x ) = 0 , x Ω
Then, there exists a positive constant M such that  0 e ( x ) M . Set
u ¯ ( x , t ) = α 3 ( 1 + e ( x ) ) , v ¯ ( x , t ) = β 3 ( 1 + e ( x ) ) ,
where
α 3 k 1 | Ω | α 3 m β 3 n ( 1 + M ) m + n , β 3 k 2 | Ω | α 3 p β 3 q ( 1 + M ) p + q .
When  x Ω , we have
n · u ¯ + g 1 ( u ¯ ) u ¯ = α 3 g 1 ( u ¯ ) + α 3 e ( x ) ( g 1 ( u ¯ ) δ ) > 0 , n · v ¯ + g 2 ( v ¯ ) u ¯ = β 3 g 2 ( v ¯ ) + β 3 e ( x ) ( g 2 ( v ¯ ) δ ) > 0 .
Taking the initial values of problem (1 u 0 ( x ) v 0 ( x )  are sufficiently small such that  u 0 ( x ) < u ¯ ( x , t ) v 0 ( x ) < v ¯ ( x , t ) , we have that  ( u ¯ , v ¯ )  is a super-solution to (1), and satisfies (16). Using Lemma 2, we obtain  u ¯ > u v ¯ > v . Thus, u and v exist globally. Theorem 3 is then fully proved. □

4. Uniform Blow-Up Profile of the Solution

In this section, according to the ideas of [10], we discuss the uniform blow-up profile of the solution to (1).
We denote
F 1 ( t ) = k 1 Ω u m v n d x , G 1 ( t ) = 0 t F 1 ( s ) d s ; F 2 ( t ) = k 2 Ω u p v q d x , G 2 ( t ) = 0 t F 2 ( s ) d s .
Lemma 3.
Assume that (H1)-(H3) hold,  Δ u 0 Δ v 0 < 0 , and  g 1 g 2  can be derived. Let  ( u , v ) ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) ) × ( C 2 , 1 ( Q T ) C 1 ( Q ¯ T ) )  be a solution to problem (1). If  u , v  blow up at the finite time  T *  simultaneously, then we have that for all  ε > 0 , there exists  C 1 > 0  such that
C 1 Δ u max sup s [ 0 , t ] F 1 ( s ) , C 1 , C 1 Δ v max sup s [ 0 , t ] F 2 ( s ) , C 1
in  Ω × [ T * 2 , T * ε ] .
Proof. 
We take the first equation in (1) as an example. Let  w = Δ u . Since  Δ F 1 ( t ) = 0 , we obtain
w t Δ w = 0 , ( x , t ) Ω × [ T * 2 , T * ε ] .
It is easy to see that w can obtain its maximum and minimum in  Ω × [ T * 2 , T * ε ] .
On the one hand, if there is a point  ( x 0 , t 0 ) x 0 Ω T * 2 t 0 T * ε  such that  w ( x 0 , t 0 ) = max Ω ¯ × [ T * 2 , T * ε ] w ( x , t ) , then according to the strong maximum principle of heat equations, we know that w is a constant when  t [ T * 2 , t 0 ] .
On the other hand, when  t > t 0 , w cannot obtain its maximum and minimum in  Ω × ( t 0 , T * ε ] . For  ( x , t ) Ω × ( t 0 , T * ε ] , set  ψ ( u ) = g 1 ( u ) u + g 1 ( u ) . Obviously,  ψ ( u ) > 0 . A straightforward computation yields
t u n = t ( g 1 ( u ) u ) = ψ ( u ) u t .
Further, according to  w = u t + F 1 ( t ) , we deduce
w n = u t n = ψ ( u ) u t .
Applying Theorem 6 in [20] Chap. 3, when w takes the minimum value, we have that
w n = ψ ( u ) ( F 1 ( t ) w ) < 0 .
Since u, v F 1 ( t ) > 0  and by continuity, there exists a constant  0 < C 1 max Ω ¯ u ( x , T * 2 )  such that
Δ u C 1 .
And, when w takes the maximum value, we have that
w n = ψ ( u ) ( F 1 ( t ) w ) > 0 .
So we can deduce that
Δ u sup s [ 0 , t ] F 1 ( s )
for  t [ T * 2 , T * ε ] . Combining (55) and (57), we have
C 1 Δ u max sup s [ 0 , t ] F 1 ( s ) , C 1 .
The proof of the inequality on  Δ v  is similar to the one on  Δ u , which is omitted. □
According to (50), we have
u t C 1 u t Δ u = F 1 ( t ) , v t C 1 v t Δ v = F 2 ( t ) .
Integrating Equation (59) from  T * 2  to t yields the following lemma.
Lemma 4.
Under the assumptions of Lemma 3, we have that for all  ε > 0
C 2 u ( x , t ) C 2 + G 1 ( t ) , C 2 v ( x , t ) C 2 + G 2 ( t )
where  C 2 , C 2 > 0  are constants and  t [ T * 2 , T * ε ] .
Proof. 
After performing the integration, we have
u ( x , t ) u ( x , T * 2 ) C 1 ( t T * 2 ) G 1 ( t ) G 1 ( T * 2 ) G 1 ( t ) , v ( x , t ) v ( x , T * 2 ) C 1 ( t T * 2 ) G 2 ( t ) G 2 ( T * 2 ) G 2 ( t ) ,
Taking  C 2 = max max x Ω u ( x , T * 2 ) + T * 2 C 1 , max x Ω v ( x , T * 2 ) + T * 2 C 1 ,
u ( x , t ) C 2 + G 1 ( t ) , v ( x , t ) C 2 + G 2 ( t )
are obtained. Since u v > 0  in  Ω ¯ × ( 0 , T * )  and due to their continuity, there exists a constant  C 2 > 0  such that u v C 2 . □
Remark 1.
An obvious conclusion deduced from Lemma 4 is that
lim t T * G 1 ( t ) = + , lim t T * G 2 ( t ) = + ,
when u, v blow up.
Set
H 1 ( t ) = 0 t G 1 ( s ) d s , H 2 ( t ) = 0 t G 1 ( s ) d s ,
where  t [ 0 , T ) .
We consider the eigenvalue problem (33) with  c 0 0 . Set  λ  as the first eigenvalue of (33), and  φ > 0  is normalized with  Ω φ d x = 1 . Define
K 1 ( t ) = Ω u φ n d S , K 2 ( t ) = Ω v φ n d S ,
where  t [ 0 , T ) . We have  K 1 ( t ) K 2 ( t ) < 0 .
Lemma 5.
Under the assumptions of Lemma 3, letting  x Ω ¯ , and any  ρ > 0  such that (i)  B ( x , ρ ) Ω , if  x Ω ; or (ii)  B ( y , ρ ) Ω = { x } B ( y , ρ ) Ω  where  y Ω  is fixed, if  x Ω ; and  C 3 > 0  is a constant, it holds that
sup x K ρ [ G 1 ( t ) u ( x , t ) ] C 3 ρ N ( 1 + H 1 ( t ) ) , sup x K ρ [ G 2 ( t ) v ( x , t ) ] C 3 ρ N ( 1 + H 2 ( t ) ) ,
in  Ω × [ T * 2 , T * ε ] , for all  ε > 0 , where  K ρ = B ( x , ρ ) ( or  B ( y , ρ ) ) .
Proof. 
Case (i). Let  z ( x , t ) = G 1 ( t ) u ( x , t ) c 0 = 0  in (33) and  β ( t ) = Ω z ( x , t ) φ ( x ) d x . By Green’s formula, we have
β ( t ) = Ω ( F 1 ( t ) u t ( x , t ) ) φ ( x ) d x = Ω φ ( x ) Δ u d x = Ω u ( x , t ) Δ φ d x + K 1 ( t ) = λ Ω u ( x , t ) φ ( x ) d x + K 1 ( t ) = λ β + λ G 1 ( t ) + K 1 ( t ) .
Solving this ODE, we have
β ( t ) = β ( 0 ) e λ t + 0 t e λ ( t s ) ( K 1 ( s ) + λ G 1 ( s ) ) β ( 0 ) e λ t + λ 0 t e λ ( t s ) G 1 ( s ) d s C ( 1 + H 1 ( t ) ) .
On the other hand, Lemma 4 implies
inf Ω z ( x , t ) C 2
when  t [ T * 2 , T * ε ]  for all  ε > 0 , which, combined with Equation (67), implies
Ω | z ( x , t ) | φ ( x ) d x C ( 1 + H 1 ( t ) ) .
Further, by Lemma 3, we have
Δ z C 1 0 .
Fixing  K ρ , whose center is noted as x, the function  z ( y , t ) y K ρ Ω  is a subharmonic implied by (70). Applying the mean-value inequality for subharmonic functions, it follows that
z ( x , t ) 1 | K ρ | K ρ z ( y , t ) d y 1 | K ρ | K ρ | z ( y , t ) | d y .
We know that  inf K ρ φ c 0 , where  c 0 > 0  is related to the distance from  Ω , which, together with (69) and (71), can imply that
z ( x , t ) C ρ N K ρ | z ( y , t ) | φ ( y ) d y C ρ N ( 1 + H 1 ( t ) ) .
Similarly, it follows that
G 2 ( t ) v ( x , t ) C ρ N ( 1 + H 2 ( t ) ) .
Selecting  C 3 max { C , C } , Case (i) can be proved.
Case (ii). Letting  c 0 > 0  in (33), considering the mean value theorem, (66) and (67) are transformed into
β ( t ) = λ β + λ G 1 ( t ) + K 1 ( t ) + c 0 Ω u n d S ,
and
β ( t ) = β ( 0 ) e λ t + 0 t e λ ( t s ) K 1 ( s ) + λ G 1 ( s ) + c 0 Ω u ( x , s ) n d S d s β ( 0 ) e λ t + 0 t e λ ( t s ) c 0 Ω ( Δ u ( x , s ) ) d x + λ G 1 ( s ) d s = β ( 0 ) e λ t + 0 t e λ ( t s ) c 0 | Ω | Δ u ( x 0 , s ) + λ G 1 ( s ) d s C 1 + H 1 ( t ) 0 t Δ u ( x 0 , s ) d s .
where  x 0 Ω , respectively. Similarly, since  inf K ρ c 0 > 0  and  z ( y , t ) y K ρ Ω ¯  is a subharmonic function. Lemma 3 deduces that  0 t Δ u d s  is bounded on  Ω × ( T * 2 , T * ε ]  for all  ε > 0 , so it follows that
z ( x , t ) C ρ N K ρ | z ( y , t ) | φ ( y ) d y C ρ N ( 1 + H 1 ( t ) ) ,
too. Choosing sufficiently large  C 3 , Lemma 5 is proved completely. □
Lemma 6.
Under the assumptions of Lemma 3, we have that
lim sup t T * min x Ω ¯ u ( x , t ) = + , lim sup t T * min x Ω ¯ v ( x , t ) = +
if and only if
lim t T * G 1 ( t ) = + , lim t T * G 2 ( t ) = + .
Further, if (77) or (78) is fulfilled, then
lim t T * u ( x , t ) G 1 ( t ) = lim t T * u ( t ) G 1 ( t ) = 1 , lim t T * v ( x , t ) G 2 ( t ) = lim t T * v ( t ) G 2 ( t ) = 1 ,
uniformly on  Ω ¯ .
Proof. 
(77) ⇒ (78) is a simple corollary of Lemma 4. Assuming (78) holds, we use the method in [10]. According to Lemma 5 Case (i), for any fixed  K ρ Ω , combining Lemma 4 and  G 1 ( t ) , G 2 ( t ) > 0  yields
C 2 G 1 ( t ) 1 u ( x , t ) G 1 ( t ) C 3 ρ N 1 + H 1 ( t ) G 1 ( t ) , C 2 G 2 ( t ) 1 v ( x , t ) G 2 ( t ) C 3 ρ N 1 + H 2 ( t ) G 2 ( t ) .
Since  G 1 G 2  are increasing, it follows that for all  ε > 0 ,
0 1 + H 1 ( t ) G 1 ( t ) 0 T * ε G 1 ( s ) d s G 1 ( t ) + ε , 0 1 + H 2 ( t ) G 2 ( t ) 0 T * ε G 2 ( s ) d s G 2 ( t ) + ε ,
which, together with  G 1 ( t ) G 2 ( t ) +  implies that  lim t T * H i ( t ) G i ( t ) = 0 , i = 1 , 2 .
lim t T * u ( x , t ) G 1 ( t ) = lim t T * v ( x , t ) G 2 ( t ) = 1
on  K ρ  is deduced.
Integrating the first two equations in (1) from 0 to t, we have
u ( x , t ) 0 t Δ u ( x , s ) d s u 0 ( x ) = G 1 ( t ) , ( x , t ) K ρ × [ T * 2 , T * ) v ( x , t ) 0 t Δ v ( x , s ) d s v 0 ( x ) = G 2 ( t ) , ( x , t ) K ρ × [ T * 2 , T * ) .
Since (82), we know that
lim t T * 0 t Δ u d s G 1 ( t ) = lim t T * 0 t Δ v d s G 2 ( t ) = 0 .
On the other hand, according to Lemma 4, it follows that  F 1 ( t ) F 2 ( t ) > 0 ; that is,
u t Δ u > 0 , v t Δ v > 0 .
Applying the maximum principle, we know that the solution  ( u , v )  can take the minimum in  Ω . For all  x Ω ¯ , combining Lemmas 4 and 5 Case (ii) and  G 1 ( t ) , G 2 ( t ) > 0 , there exists a fixed  K ρ Ω ¯  such that
C 2 G 1 ( t ) 1 u ( x , t ) G 1 ( t ) C 3 ρ N 1 + H 1 ( t ) 0 t Δ u ( x 01 , s ) d s G 1 ( t ) , C 2 G 2 ( t ) 1 v ( x , t ) G 2 ( t ) C 3 ρ N 1 + H 2 ( t ) 0 t Δ v ( x 02 , s ) d s G 2 ( t ) .
where  x 01 x 02 Ω . Since  lim t T * H i ( t ) G i ( t ) = 0 i = 1 , 2  and (84), (85) can imply (79) uniformly on  Ω ¯ . Further, we obtain
lim sup t T * min x Ω ¯ u ( x , t ) = lim sup t T * min x Ω u ( x , t ) = + , lim sup t T * min x Ω ¯ v ( x , t ) = lim sup t T * min x Ω v ( x , t ) = + .
The proof of Lemma 6 is completed. □
Using the above lemmas and corollaries, we can obtain the following theorems describing blow-up solutions by the same method as in [8,21]. Theorems 4 and 5 give the necessary and sufficient conditions that u, v blow up simultaneously. Further, Theorem 6 describes the blow-up profile of (1) when u, v blow up simultaneously.
Theorem 4.
Assuming (H1)–(H3) hold,  Δ u 0 Δ v 0 < 0 g 1 g 2  are derivable. Let  ( u , v )  be a classical solution to problem 1. If u and v blow up in a finite time  T *  simultaneously, then (i)  p m 1  and  n q 1 , or (ii)  p < m 1  and  n < q 1  must be satisfied.
Theorem 5.
Under the assumption of Theorem 4. Let  ( u , v )  be a classical blow-up solution to problem 1. If  p m 1 > 0  and  n q 1 > 0 , then u and v blow up simultaneously and uniformly.
Theorem 6.
Under the assumption of Theorem 4, the following results hold uniformly on  Ω ¯ :
(i) 
if (a)  n > q 1 p > m 1  or (b)  n < q 1 p < m 1 , then we have
lim t T * u ( x , t ) ( T * t ) θ = lim t T * u ( t ) ( T * t ) θ = k 1 | Ω | θ k 2 k 1 θ σ n n + 1 q θ , lim t T * v ( x , t ) ( T * t ) σ = lim t T * v ( t ) ( T * t ) σ = k 2 | Ω | σ k 1 k 2 σ θ p p + 1 m σ ,
where
θ = n + 1 q n p ( m 1 ) ( q 1 ) , σ = p + 1 m n p ( m 1 ) ( q 1 ) ;
(ii) 
if  p = m 1 > 0  and  n > q 1 > 0 , then we have
lim t T * u p ( x , t ) [ ln u ( x , t ) ] n n + 1 q ( T * t ) = ( p k 1 | Ω | ) 1 k 2 ( n + 1 q ) k 1 n n + 1 q , lim t T * v n + 1 q | ln ( T * t ) | 1 = k 2 k 1 n + 1 q p ;
(iii) 
if  p > m 1 > 0  and  n = q 1 > 0 , then we have
lim t T * u p + 1 m ( x , t ) | ln ( T * t ) | 1 = k 1 k 2 p + 1 m n , lim t T * v n ( x , t ) [ ln v ( x , t ) ] p p + 1 m ( T * t ) = ( n k 2 | Ω | ) 1 k 1 ( p + 1 m ) k 2 p p + 1 m ;
(iv) 
if  p = m 1 > 0  and  n = q 1 > 0 , then we have
lim t T * ln u ( x , t ) | ln ( T * t ) | = lim t T * ln u ( t ) | ln ( T * t ) | = k 1 p k 1 + n k 2 , lim t T * ln v ( x , t ) | ln ( T * t ) | = lim t T * ln v ( t ) | ln ( T * t ) | = k 2 p k 1 + n k 2 .
The proofs of Theorems 4–6 are very similar to [8]; we only need to pay attention to the constants  k 1 k 2 , so we only give the proof by taking Theorem 6 Case (i) as an example, and omit the others.
Proof of Theorem 6 Case (i).
We have
G 1 ( t ) = F 1 ( t ) = k 1 Ω u m ( x , t ) v n ( x , t ) d x k 1 | Ω | G 1 m ( t ) G 2 n ( t ) ,
G 2 ( t ) = F 2 ( t ) = k 2 Ω u p ( x , t ) v q ( x , t ) d x k 2 | Ω | G 1 p ( t ) G 2 q ( t ) ,
it follows that
k 2 G 1 p m ( t ) d G 1 k 1 G 2 n q ( t ) d G 2 .
When  n > q 1  and  p > m 1 , integrating (89) from 0 to t, yields  k 2 p + 1 m G 1 p + 1 m ( t ) k 1 n + 1 q G 2 n + 1 q ( t ) . That is,
k 2 ( n + 1 q ) G 1 p + 1 m ( t ) k 1 ( p + 1 m ) G 2 n + 1 q ( t ) .
While  n < q 1  and  p < m 1 , since  lim t T * G i = + i = 1 , 2 , integrating (89) from t to  T * , we have
k 2 ( n + 1 q ) G 1 p + 1 m ( t ) k 1 ( p + 1 m ) G 2 n + 1 q ( t ) ,
similarly. By (87) and (90), we find
G 1 ( t ) k 1 | Ω | k 2 ( n + 1 q ) k 1 ( p + 1 m ) n n + 1 q G 1 m + n ( p + 1 m ) n + 1 q ( t ) = k 1 | Ω | k 2 k 1 θ σ n n + 1 q G 1 m + n σ θ ( t ) .
Since  1 m n σ θ = 1 θ < 0  and  lim t T * G 1 ( t ) = + , integrating (92) from t to  T * , we obtain
G 1 ( t ) k 1 | Ω | θ k 2 k 1 θ σ n n + 1 q ( T * t ) θ .
According to (93) and Lemma 6, it follows that uniformly on  Ω ¯ ,
u ( x , t ) u ( t ) k 1 | Ω | θ k 2 k 1 θ σ n n + 1 q ( T * t ) θ ,
that is,
lim t T * u ( x , t ) ( T * t ) θ = lim t T * u ( t ) ( T * t ) θ = k 1 | Ω | θ k 2 k 1 θ σ n n + 1 q θ
holds uniformly on  Ω ¯ .
Combining (88) and (91), and applying the same proofs of  G 2  and v, we obtain that
lim t T * v ( x , t ) ( T * t ) σ = lim t T * v ( t ) ( T * t ) σ = k 2 | Ω | σ k 1 k 2 σ θ p p + 1 m σ
holds uniformly on  Ω ¯ . □

5. Numerical Simulations

In this section, we give numerical simulations for several specific cases to illustrate Theorems 2 and 3. All calculations are performed through Mathematica.
Firstly, letting  m = 0.2 n = 0.5 p = q = 0.3  and  T = 10 , we obtain problem (1) of the following form
u t = Δ u + Ω u 0.2 ( t , x , y ) v 0.5 ( t , x , y ) d σ , ( t , x , y ) ( 0 , 10 ) × Ω , v t = Δ v + Ω u 0.3 ( t , x , y ) v 0.3 ( t , x , y ) d σ , ( t , x , y ) ( 0 , 10 ) × Ω , n · u + u 2 = n · v + v 1.5 = 0 , ( t , x , y ) ( 0 , 10 ) × Ω , u ( 0 , x , y ) = v ( 0 , x , y ) = 1 1 2 ( x 2 + y 2 ) , ( x , y ) Ω ,
where  Ω = { ( x , y ) : x 2 + y 2 1 } . Because of the symmetry, we only need to simulate the values of its solution on  0 x 1  as shown below.
In this case,  g 1 ( u ) = u g 2 ( v ) = v 1 2  satisfies (H2) and (H3); since  u 0 ( x , y ) = v 0 ( x , y ) = 1 1 2 ( x 2 + y 2 ) > 0 ,
u 0 n = v 0 n = ( x , y ) · ( x , y ) = ( x 2 + y 2 ) = 1 < 0 ,
Further, (94) satisfies (H1) and  m = 0.2 < 1 p = 0.3 < 1 n q = 0.15 0.56 = ( 1 m ) ( 1 p ) . Therefore, (94) satisfies the conditions of Theorem 2, so its solution should exist globally, which can be illustrated by Figure 1.
Next, we present a numerical simulation to illustrate the blow-up profiles using Mathematica. As stated in Theorem 3, we consider problem (1) in the following form satisfying (40):
u t = Δ u + Ω u 1.2 ( t , x , y ) v 0.4 ( t , x , y ) d σ , ( t , x , y ) ( 0 , T * ) × Ω , v t = Δ v + Ω u 0.3 ( t , x , y ) v 1.3 ( t , x , y ) d σ , ( t , x , y ) ( 0 , T * ) × Ω , n · u + u 2 = n · v + v 1.5 = 0 , ( t , x , y ) ( 0 , T * ) × Ω , u ( 0 , x , y ) = v ( 0 , x , y ) = 1 1 12 ( x 4 + y 4 ) , ( x , y ) Ω ,
where  Ω = { ( x , y ) : x 2 + y 2 1 }  and  T *  is the blow-up time that will be given by the numerical simulation. In this case, the initial values satisfy
u 0 n = v 0 n = ( x , y ) · ( 1 3 x 3 , 1 3 y 3 ) = 1 3 ( x 4 + y 4 ) < 0
and
Δ u 0 = Δ v 0 = ( x 2 + y 2 ) < 0 .
According to Theorem 3, we know that the solution to (95) blows up in a finite time. Without loss of generality, we take as an example the value of function u when  t ( 0 , 0.979 ) x [ 0 , 1 ] y = 0 , as in Figure 2.
As expected, Since (95) satisfies  p m 1 > 0  and  n q 1 > 0 , the solution  ( u , v )  blows up at approximately  T * = 0.979 , simultaneously. We take m = 1.2, n = 0.5, p = 1.3, q = 0.3 and replace the initial value conditions with
u 0 ( x , y ) = v 0 ( x , y ) = 1 4 1 48 ( x 4 + y 4 ) ,
which does not satisfy that  u 0 ( x ) , v 0 ( x )  are sufficiently large in Theorem 3. We resolve the solution  ( u , v )  of problem (95) (see Figure 3).
Figure 3 shows that when the initial values are very small, the solution  ( u , v )  does not blow up even if (40) is satisfied. Then, we change the values of m, n, p and q in (95) and perform numerical simulations for the rest of the cases mentioned in Theorem 6. This is shown in Figure 4, Figure 5, Figure 6 and Figure 7.

Author Contributions

Methodology, W.M.; formal analysis, B.Y. All authors have read and agreed to the published version of the manuscript.

Funding

This work is supported by the National Natural Science Foundation of China (62073203) and the Fund of Natural Science of Shandong Province (ZR2018MA022).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. The solution  ( u , v )  in  { ( t , x , y ) : 0 < t < 10 , 0 x 1 , y = 0 } .
Figure 1. The solution  ( u , v )  in  { ( t , x , y ) : 0 < t < 10 , 0 x 1 , y = 0 } .
Mathematics 11 01993 g001
Figure 2. The solution  ( u , v )  in  { ( t , x , y ) : 0 < t < 0.979 , 0 x 1 , y = 0 } .
Figure 2. The solution  ( u , v )  in  { ( t , x , y ) : 0 < t < 0.979 , 0 x 1 , y = 0 } .
Mathematics 11 01993 g002
Figure 3. The solution  ( u , v )  in  { ( t , x , y ) : 0 < t < 5.02 , 0 x 1 , y = 0 } .
Figure 3. The solution  ( u , v )  in  { ( t , x , y ) : 0 < t < 5.02 , 0 x 1 , y = 0 } .
Mathematics 11 01993 g003
Figure 4. m = 1.4 , n = 0.4 , p = 0.3 , q = 1.5 .
Figure 4. m = 1.4 , n = 0.4 , p = 0.3 , q = 1.5 .
Mathematics 11 01993 g004
Figure 5. m = 2 , n = 0.4 , p = 1 , q = 1.3 .
Figure 5. m = 2 , n = 0.4 , p = 1 , q = 1.3 .
Mathematics 11 01993 g005
Figure 6. m = 1.2 , n = 0.8 , p = 0.3 , q = 1.8 .
Figure 6. m = 1.2 , n = 0.8 , p = 0.3 , q = 1.8 .
Mathematics 11 01993 g006
Figure 7. m = 2 , n = 0.8 , p = 1 , q = 1.8 .
Figure 7. m = 2 , n = 0.8 , p = 1 , q = 1.8 .
Mathematics 11 01993 g007
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Ma, W.; Yan, B. Global Existence and Uniform Blow-Up to a Nonlocal Parabolic System with Nonlinear Boundary Conditions Arising in a Thermal Explosion Theory. Mathematics 2023, 11, 1993. https://doi.org/10.3390/math11091993

AMA Style

Ma W, Yan B. Global Existence and Uniform Blow-Up to a Nonlocal Parabolic System with Nonlinear Boundary Conditions Arising in a Thermal Explosion Theory. Mathematics. 2023; 11(9):1993. https://doi.org/10.3390/math11091993

Chicago/Turabian Style

Ma, Wenyuan, and Baoqiang Yan. 2023. "Global Existence and Uniform Blow-Up to a Nonlocal Parabolic System with Nonlinear Boundary Conditions Arising in a Thermal Explosion Theory" Mathematics 11, no. 9: 1993. https://doi.org/10.3390/math11091993

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