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Blow-up problems for a compressible reactive gas model

Abstract

This paper investigates a compressible reactive gas model with homogeneous Dirichlet boundary conditions. Under the parameters and the initial data satisfying some conditions, we prove that the solutions have global blow-up, and the blow-up rate is uniform in all compact subsets of the domain. Moreover, the blow-up rates of | u ( t ) | and | v ( t ) | are precisely determined.

MSC:35K05, 35K55, 35D55.

1 Introduction and main results

In this paper, we investigate blow-up and the blow-up rate of nonnegative solutions for the following degenerate reaction-diffusion system with nonlocal sources:

{ u t = ( u m u ) + a u p 1 v B , α 1 p 2 , ( x , t ) B × ( 0 , T ) , v t = ( v n v ) + b v q 1 u B , α 2 q 2 , ( x , t ) B × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) , v ( x , 0 ) = v 0 ( x ) , x B , u ( x , t ) = v ( x , t ) = 0 , ( x , t ) B × ( 0 , T ) ,
(1.1)

where B=B(0,R) R N (N1) is a ball centered at the origin with the radius R R + , a,b>0, exponents p 2 , q 2 , α 1 , α 2 1, m,n, p 1 , q 1 >0, and T< is the maximal existence time of a solution, B , α α = B | | α dx.

The system (1.1) models such as heat propagations in a two-components combustible mixture gases [1]. This problem is worth studying because of the applications to heat and mass transport processes (see [2, 3]). In addition, there exist interesting interactions among the multi-nonlinearities described by these exponents in the problem (1.1).

In the past decades, many physical phenomena have been formulated into nonlocal mathematical models and studied by many authors. Here, we will recall some of those results concerning the first initial boundary value problem.

At first, the global solutions and blow-up problems for a single parabolic equation with nonlocal nonlinearity sources had been studied extensively, see [410] and references therein. As a typical example, in [4] Souplet considered the equation with spatial integral term

u t =Δu+g ( Ω f ( u ( x , t ) ) d x )
(1.2)

and the equation with both local and nonlocal terms

u t =Δu+ Ω f ( u ( x , t ) ) dx+h ( u ( x , t ) ) .
(1.3)

These two equations are related to some ignition models for compressible reactive gases. The author introduced a method to investigate the profile of blow-up solutions of (1.2) and (1.3) and observed the asymptotic blow-up behaviors of the solutions. In addition, an important model in the theory of nuclear reactor dynamics can be described by the following equation with the space-time integral term:

u t =Δu+f ( 0 t Ω g ( u ( y , s ) ) β ( y ) d y d s ) .
(1.4)

The blow-up of its solutions was studied by Pao [5], Guo and Su [6].

In 2003, Li and Xie [7] considered the following problem:

v τ =Δ v m +a v p 1 Ω v q 1 dx.
(1.5)

By introducing some transformations u= v m , t=mτ (1.5) takes the form

u t = u p ( Δ u + a u r Ω u s d x ) .
(1.6)

Then they proved that the solution of (1.6) blows up in finite time for large initial data and obtained the blow-up rate. Recently, Liu et al. in [8] investigated the blow-up rate of solutions to diffusion equation (1.6). Their approach was based on sub- and super-solution methods which were very different from those previously used in the study of the blow-up rate. They proved, by using the maximum principle, that the solutions have global blow-up, and the rate of blow-up is uniform in all compact subsets of the domain. Here the global blow-up means that there exists 0<T<+ such that

lim t T | u ( , t ) | =or lim t T | v ( , t ) | =for all xΩ.

Secondly, we should point out that in the case of m=n=0, the system (1.1) becomes a semilinear system. To our knowledge, there do not seem to be any results in the literature on blow-up problems of these types. But other related works of the semilinear case have been deeply investigated by many authors, e.g., see [11, 12], and the authors of this paper in [13] studied the system

u t =Δu+a(x) u p 1 (x,t) v q 1 (0,t), v t =Δv+b(x) v p 2 (x,t) u q 2 (0,t),

where the simultaneous and non-simultaneous blow-up criteria were obtained by using the fundamental solution of the heat equation. On the other hand, there are many known results concerning the global solutions and blow-up problems for the parabolic system with local nonlinearities, localized nonlinearities and nonlinear boundary conditions, see [1417] and references therein. In particular, Ling and Wang in [18] considered the following degenerate parabolic system:

u t =Δ u m + v p 1 u α p 2 , v t =Δ v n + u q 1 v β q 2

in a bounded domain Ω, with the help of the super- and sub-solution methods, the critical exponent of the system was determined. Motivated by the above works, under the following conditions:

0<m< p 1 <1,0<n< q 1 <1, p 2 q 2 >(1 p 1 )(1 q 1 ),
(1.7)

we consider a more general degenerate parabolic system (1.1) which includes the problems considered in [7, 8] and [17] as special cases. Employing the ideas in [7, 8], we describe the blow-up rate of the radially symmetric solutions to (1.1). Here we discuss the blow-up of radially symmetric solutions as well as derive their blow-up rate. Moreover, we get the accurate coefficient of the blow-up rate. For the related discussion on a radially symmetric solution, we refer the readers to [19] and references therein.

In this paper, we always assume that the initial data ( u 0 , v 0 )V (V is defined by (1.8)) and satisfies the following (H1)-(H3) or (H4):

(H1) u 0 (x), v 0 (x) C 2 + α (B)C( B ¯ ), α(0,1).

(H2) u 0 (x), v 0 (x)>0 in B, u 0 (x)= v 0 (x)=0, u 0 ν , v 0 ν <0 on ∂B.

(H3) u 0 (x), v 0 (x) are radially symmetric, u 0 (r), v 0 (r)0 for r(0,R), r=|x|.

Denote the set of initial data, depending only on the radial variable in the spherical coordinate system of R N :

V= { ( u 0 ( r ) , v 0 ( r ) ) | Φ 1 ( r ) > 0 , Φ 2 ( r ) > 0 } ,
(1.8)

where

Φ 1 ( r ) = m ( u 0 ( r ) ) 2 u 0 m 1 ( r ) + u 0 m ( r ) ( u 0 ( r ) + N 1 r u 0 ( r ) ) + a u 0 p 1 ( r ) v 0 ( r ) B , α 1 p 2 , Φ 2 ( r ) = n ( v 0 ( r ) ) 2 v 0 n 1 ( r ) + v 0 n ( r ) ( v 0 ( r ) + N 1 r v 0 ( r ) ) + b v 0 q 1 ( r ) u 0 ( r ) B , α 2 q 2 .

It is noted that the set V is not empty. For example, for the simplest case N=R=1 and a=b=1, for any constant exponents m, n and p i , q i , α i , i=1,2, there exist positive constants a 1 , a 2 such that ( u 0 , v 0 )V with u 0 (r)= a 1 /2 a 1 r 2 /2, v 0 (r)= a 2 /2 a 2 r 2 /2, r[0,1).

(H4) Let δ 0 , k 1 , k 2 be positive constants (will be given in Section 3), and there exists a constant δ> δ 0 such that

{ Δ u 20 + a ( 1 + m p 1 ) v 20 B , μ 1 σ 1 δ u 20 k 1 + 1 r 1 0 , Δ v 20 + b ( 1 + n q 1 ) u 20 B , μ 2 σ 2 δ v 20 k 2 + 1 r 2 0 ,
(1.9)

here u 20 , v 20 and σ i , μ i , r i are defined by (2.9) and (2.6).

Then, our main results read as follows in detail.

Theorem 1 Assume that ( u 0 , v 0 )V and satisfies (H 1)-(H 3). If ρ 2 > 1 a ( 1 + m p 1 ) 1 b ( 1 + n q 1 ) , then the positive solution (u,v) of (1.1) blows up in finite time, where ρ is defined by (2.12).

Theorem 2 Under the assumptions of Theorem  1, if Δ u 0 ,Δ v 0 0 on B ¯ and ( u 0 , v 0 ) satisfies (H 4), then the following statements hold uniformly on any compact subset of B:

lim t T u 1 p 1 ( x , t ) ( 1 p 1 ) G ˜ 1 ( t ) =a, lim t T v 1 q 1 ( x , t ) ( 1 q 1 ) G ˜ 2 ( t ) =b,
(1.10)

where G ˜ 1 (t) and G ˜ 2 (t) are defined by (3.1).

Theorem 3 Under the assumptions of Theorem  2, if (1 q 1 )(1+m p 1 )< p 2 ( q 2 m) and (1 p 1 )(1+n q 1 )< q 2 ( p 2 n), then

(1.11)
(1.12)

uniformly on compact subsets of B, where

θ 1 = ( σ 1 μ 1 + σ 1 β 2 ( σ 2 μ 2 σ 1 μ 1 ) ) 1 k 1 , θ 1 = ( σ 2 μ 2 + σ 2 β 1 ( σ 1 μ 1 σ 2 μ 2 ) ) 1 k 2 .

This paper is organized as follows. The result pertaining to blow-up of a solution in finite time is presented in Section 2, while results regarding the blow-up rates are established in Section 3. Some discussions are given in Section 4.

2 Proof of Theorem 1

In this section, we will discuss the blow-up of the solution to (1.1) and prove Theorem 1. By a simple computation, we have

(2.1)
(2.2)

Since 1+m p 1 >0, m< p 1 , from (2.2), we can derive the inequality

u m p 1 Δu 1 1 + m p 1 Δ u 1 + m p 1 .
(2.3)

Moreover, by (1.1), (2.1) and (2.3), we have

1 1 p 1 u 1 p 1 t = u p 1 u t = u p 1 ( m u m 1 | u | 2 + u m Δ u + a u p 1 v B , α 1 p 2 ) 1 1 + m p 1 Δ u 1 + m p 1 + a v B , α 1 p 2 .

Thus,

1 + m p 1 1 p 1 ( u 1 + m p 1 ) 1 p 1 1 + m p 1 t Δ u 1 + m p 1 +a(1+m p 1 ) v B , α 1 p 2 .
(2.4)

Similarly,

1 + n q 1 1 q 1 ( v 1 + n q 1 ) 1 q 1 1 + n q 1 t Δ v 1 + n q 1 +b(1+n q 1 ) u B , α 2 q 2 .
(2.5)

Denote u 1 = u 1 + m p 1 , v 1 = v 1 + n q 1 and

{ r 1 = m 1 + m p 1 , σ 1 = p 2 1 + n q 1 , μ 1 = α 1 1 + n q 1 , r 2 = n 1 + n q 1 , σ 2 = q 2 1 + m p 1 , μ 2 = α 2 1 + m p 1 .
(2.6)

Then 0< r 1 , r 2 <1, σ 1 , σ 2 , μ 1 , μ 2 >1 and u 1 , v 1 satisfy

{ u 1 t u 1 r 1 ( Δ u 1 + a ( 1 + m p 1 ) v 1 B , μ 1 σ 1 ) , v 1 t v 1 r 2 ( Δ v 1 + b ( 1 + n q 1 ) u 1 B , μ 2 σ 2 ) .
(2.7)

Consider now the following problem:

{ u 2 t = u 2 r 1 ( Δ u 2 + a ( 1 + m p 1 ) v 2 B , μ 1 σ 1 ) , ( x , t ) B × ( 0 , T ) , v 2 t = v 2 r 2 ( Δ v 2 + a ( 1 + n q 1 ) u 2 B , μ 2 σ 2 ) , ( x , t ) B × ( 0 , T ) , u 2 ( x , t ) = v 2 ( x , t ) = 0 , ( x , t ) B × ( 0 , T ) , u 2 ( x , 0 ) = u 20 ( x ) , v 2 ( x , 0 ) = v 20 ( x ) , x B ,
(2.8)

where

{ v 2 B , μ 1 σ 1 = v B , α 1 p 2 , u 2 B , μ 2 σ 2 = u B , α 2 q 2 , u 20 ( x ) = ( u 0 ( x ) ) 1 + m p 1 , v 20 ( x ) = ( v 0 ( x ) ) 1 + n q 1 .
(2.9)

Since u 0 (x), v 0 (x) satisfy (H1)-(H2), then (2.8) has a unique classical solution ( u 2 , v 2 ) (see [20]). In the meantime, by the comparison principle, we observe

u 2 (x,t) u 1 + m p 1 (x,t), v 2 (x,t) v 1 + n q 1 (x,t),(x,t)B×(0,T).
(2.10)

Let G be a bounded domain of R N . Consider the problem

{ ω t = d ω r 0 ( Δ ω + a 0 G ω d x ) , x G , t > 0 , ω ( x , t ) = c , x G , t > 0 , ω ( x , 0 ) = c , x G ,
(2.11)

where 0< r 0 <1 and d, a 0 ,c>0 are some constants. By the standard method (see [3]), we can show that (2.11) has a unique classical solution ω(x,t) and ω(x,t)c. Denote by φ 0 (x) the unique positive solution of the linear elliptic problem

Δ φ 0 (x)=1,xG; φ 0 (x)=0,xG.

Set ρ 0 = G φ 0 (x)dx, then we have:

Lemma 1 If ρ 0 >1/ a 0 , then the positive solution ω(x,t) of (2.11) blows up in finite time.

Proof Set H(t)= G ω 1 r 0 φ 0 dx, then

1 1 r 0 H ( t ) = d ( G Δ ω φ 0 d x + a 0 G ω d x G φ 0 d x ) d ( a 0 ρ 0 1 ) G ω d x d ( a 0 ρ 0 1 ) ( G ω φ 0 d x ) / M ,

where M= max x G ¯ φ 0 (x). Let z= ω 1 r 0 , then

G z t (x,t) φ 0 (x)dxd(1 r 0 )( a 0 ρ 0 1) ( G z 1 / ( 1 r 0 ) φ 0 d x ) /M.

Since 1/(1 r 0 )>1, from Jensen’s inequality, it follows that

G z t (x,t) φ 0 (x)dxd(1 r 0 )( a 0 ρ 0 1) ( ρ 0 ) r 0 / ( 1 r 0 ) ( G z φ 0 d x ) 1 / ( 1 r 0 ) /M.

That is H (t) C 0 H 1 / ( 1 r 0 ) (t). In view of H(0)>0, it follows that there exists T< such that lim t T H(t)=+, and hence ω(x,t) blows up in finite time. □

Let φ(x) be the unique positive solution of the following linear elliptic problem:

Δφ(x)=1,xB,φ(x)=0,xB

and

ρ=min { ρ 01 = φ B , μ 1 σ 1 , ρ 02 = φ B , μ 2 σ 2 } .
(2.12)

Lemma 2 If ρ 2 > 1 a ( 1 + m p 1 ) 1 b ( 1 + n q 1 ) , then for the solution ( u 2 , v 2 ) of (2.8), there exists a sufficiently small constant ε>0 such that

εφ(x) u 2 (x,t),εφ(x) v 2 (x,t)

for all (x,t) B ¯ ×[0,T).

Proof From (H1) and (H2) we see that there exists a sufficiently small constant ε>0 such that

εφ(x) u 20 (x),εφ(x) v 20 (x),x B ¯
(2.13)

and

a(1+m p 1 )ρ> ε 1 σ 1 ε σ 2 1 > ( b ( 1 + n q 1 ) ρ ) 1 .
(2.14)

Let s 1 (x,t)=εφ(x), s 2 (x,t)=εφ(x), then we have by (2.14)

{ s 1 t s 1 r 1 ( Δ s 1 + a ( 1 + m p 1 ) s 2 B , μ 1 σ 1 ) s 1 r 1 ( ε a ( 1 + m p 1 ) ε σ 1 ρ ) 0 , s 2 t s 2 r 2 ( Δ s 2 + b ( 1 + n q 1 ) s 1 B , μ 2 σ 2 ) 0 , x B , 0 < t < T , s 1 ( x , t ) = s 2 ( x , t ) = 0 , x B , 0 < t < T .
(2.15)

Thus it follows from (2.13) and (2.15) that ( s 1 , s 2 ) is a sub-solution of (2.8). Hence, (εφ,εφ)( u 2 , v 2 ) by the comparison principle. □

Lemma 3 The solution ( u 2 , v 2 ) of (2.8) blows up in finite time if ρ 2 > 1 a ( 1 + m p 1 ) 1 b ( 1 + n q 1 ) and u 0 , v 0 satisfy (H 1)-(H 3).

Proof In view of ρ 2 > 1 a ( 1 + m p 1 ) 1 b ( 1 + n q 1 ) , we can choose a smooth sub-ball B 1 B such that

ρ 1 2 > 1 a ( 1 + m p 1 ) 1 b ( 1 + n q 1 ) ,

where ρ 1 =min{ ρ 11 = φ 1 B 1 , μ 1 σ 1 , ρ 12 = φ 1 B 1 , μ 2 σ 2 } and φ 1 (x)>0 satisfies

Δ φ 1 (x)=1,x B 1 ; φ 1 (x)=0,x B 1 .

On the other hand, there exists a sufficiently small ε 0 >0 such that

B 1 φ 1 (x)dx ε 0 φ 1 B 1 , μ 1 σ 1 , B 1 φ 1 (x)dx ε 0 φ 1 B 1 , μ 2 σ 2 .
(2.16)

Let η=ε min B ¯ 1 φ, here ε is determined by Lemma 2. Then η>0 and

u 2 (x,t)η, v 2 (x,t)η,(x,t) B ¯ 1 ×(0,T)

by Lemma 2. Therefore, ( u 2 , v 2 ) in B 1 ×(0,T) satisfies

{ u 2 t = u 2 r 1 ( Δ u 2 + a ( 1 + m p 1 ) v 2 B , μ 1 σ 1 ) u 2 t u 2 r 1 ( Δ u 2 + a ( 1 + m p 1 ) v 2 B 1 , μ 1 σ 1 ) , v 2 t = v 2 r 2 ( Δ v 2 + b ( 1 + n q 1 ) u 2 B , μ 2 σ 2 ) v 2 r 2 ( Δ v 2 + b ( 1 + n q 1 ) u 2 B 1 , μ 2 σ 2 ) , u 2 ( x , t ) η , v 2 ( x , t ) η , ( x , t ) B 1 × ( 0 , T ) , u 2 ( x , 0 ) = u 20 ( x ) η , v 2 ( x , 0 ) = v 20 ( x ) η , x B 1 .
(2.17)

Now, consider the following system:

{ u 3 t = u 3 r 1 ( Δ u 3 + a ( 1 + m p 1 ) v 3 B 1 , μ 1 σ 1 ) , x B 1 , t > 0 , v 3 t = v 3 r 2 ( Δ v 3 + b ( 1 + n q 1 ) u 3 B 1 , μ 2 σ 2 ) , x B 1 , t > 0 , u 3 ( x , t ) = v 3 ( x , t ) = η , x B 1 , t > 0 , u 3 ( x , 0 ) = v 3 ( x , 0 ) = η , x B 1 .
(2.18)

Similarly, we can show that there exists a nonnegative classical solution ( u 3 , v 3 ) of (2.18) for (x,t) B 1 ×(0, T ), where T denotes the maximal existence time. The standard comparison principle for a parabolic system implies that T T and

u 2 (x,t) u 3 (x,t), v 2 (x,t) v 3 (x,t),(x,t) B ¯ 1 ×(0,T).
(2.19)

Therefore, it suffices to show that ( u 3 , v 3 ) blows up in finite time, because if so, its upper bound ( u 2 , v 2 ) does exist up to a finite time T.

Since the initial data (η,η) is a sub-solution of (2.18), the standard super-solution and sub-solution methods assert that u 3 t 0, v 3 t 0, which implies that

Δ u 3 +a(1+m p 1 ) v 3 B 1 , μ 1 σ 1 0,Δ v 3 +b(1+n q 1 ) u 3 B 1 , μ 2 σ 2 0.

Hence u 3 , v 3 η for (x,t) B ¯ 1 ×[0, T ). Thus, ( u 3 , v 3 ) satisfies

{ u 3 t η r 1 r u 3 r ( Δ u 3 + a ( 1 + m p 1 ) v 3 B 1 , μ 1 σ 1 ) , ( x , t ) B 1 × ( 0 , T ) , v 3 t η r 2 r v 3 r ( Δ v 3 + b ( 1 + n q 1 ) u 3 B 1 , μ 2 σ 2 ) , ( x , t ) B 1 × ( 0 , T )
(2.20)

with the corresponding initial and boundary conditions and 0<r<min{ r 1 , r 2 }.

Since ρ 1 2 > 1 a ( 1 + m p 1 ) 1 b ( 1 + n q 1 ) , there exist positive constants l 1 , l 2 with l 1 , l 2 >1, and l such that

{ a ( 1 + m p 1 ) ρ 1 > l 1 l 2 > 1 b ( 1 + n q 1 ) ρ 1 , ρ 1 > 1 l > l 1 | B 1 | μ 1 1 μ 1 ε 0 a ( 1 + m p 1 ) l 2 σ 1 , ρ 1 > 1 l > l 2 | B 1 | μ 2 1 μ 2 ε 0 b ( 1 + n q 1 ) l 1 σ 2 .
(2.21)

Let

ω 1 (x,t)= l 1 ω(x,t), ω 2 (x,t)= l 2 ω(x,t),

where ω(x,t) is a unique positive solution of (2.11) with

d=min { η r 1 r , η r 2 r } , r 0 =r, a 0 = l ε 0 ,c=min { 1 l 1 , 1 l 2 } η,G= B 1 .

From (2.21) and Lemma 1, we know that ω(x,t) blows up in finite time T 0 <. Moreover, ω t 0, that is, Δω+ a 0 B 1 ωdx0, since the initial data is a sub-solution of (2.11). In addition, from σ 1 , σ 2 >1 and Hölder’s inequality, we have

{ B 1 ω d x | B 1 | μ 1 1 μ 1 ( B 1 ω μ 1 d x ) 1 μ 1 | B 1 | μ 1 1 μ 1 ω B 1 , μ 1 σ 1 , B 1 ω d x | B 1 | μ 2 1 μ 2 ( B 1 ω μ 2 d x ) 1 μ 2 | B 1 | μ 2 1 μ 2 ω B 1 , μ 2 σ 2 .
(2.22)

Thus, a series of computations yields

{ ω 1 t η r 1 r ω 1 r ( Δ ω 1 + a ( 1 + m p 1 ) ω 2 B 1 , μ 1 σ 1 ) = l 1 d ω r ( Δ ω + l ε 0 B 1 ω d x ) l 1 η r 1 r ( l 1 ω ) r ( Δ ω + a ( 1 + m p 1 ) l 2 σ 1 l 1 ω B 1 , μ 1 σ 1 ) l 1 d ω r ( Δ ω + l ε 0 | B 1 | μ 1 1 μ 1 ω B 1 , μ 1 σ 1 ) l 1 η r 1 r ( l 1 ω ) r ( Δ ω + a ( 1 + m p 1 ) l 2 σ 1 l 1 ω B 1 , μ 1 σ 1 ) 0 , ω 2 t η r 2 r ω 2 r ( Δ ω 2 + b ( 1 + n q 1 ) ω 1 B 1 , μ 2 σ 2 ) 0 , x B 1 , 0 < t < T 0 , ω 1 ( x , t ) = l 1 c η , ω 2 ( x , t ) = l 2 c η , x B 1 , 0 t < T 0 , ω 1 ( x , 0 ) = l 1 c η , ω 2 ( x , 0 ) = l 2 c η , x B 1 .
(2.23)

It follows from (2.20), (2.23) and the comparison principle that ( ω 1 , ω 2 )( u 3 , v 3 ). Hence ( u 3 , v 3 ) blows up in finite time, and so does the solution ( u 2 , v 2 ) of (2.8) from (2.19). The proof now is completed. □

Considering Lemma 3 and (2.10), we directly obtain the results of Theorem 1.

3 Proofs of Theorems 2 and 3

In this section, we assume that the solution (u,v) of (1.1) blows up in finite time T and will prove Theorems 2 and 3. We use c or C to denote the generic constant depending only on the structural data of the problem, and it may be different even in the same formula.

For the problem (1.1), denote

g ˜ 1 ( t ) = v B , α 1 p 2 , g ˜ 2 ( t ) = u B , α 2 q 2 , G ˜ 1 ( t ) = 0 t g ˜ 1 ( s ) d s , G ˜ 2 ( t ) = 0 t g ˜ 2 ( s ) d s .
(3.1)

Then we have

Lemma 4 Suppose that u 0 , v 0 satisfy (H 1)-(H 3), then we have

lim t T G ˜ i (t)= lim t T sup g ˜ i (t)=,i=1,2.

Proof According to the hypotheses, we know that u(0,t)u(x,t), v(0,t)v(x,t), (x,t)B×(0,T). Let

U ˜ (t)= max x B ¯ u(x,t)=u(0,t), V ˜ (t)= max x B ¯ v(x,t)=v(0,t).
(3.2)

Then, U ˜ (t), V ˜ (t) are Lipschitz continuous (see [21]) and U ˜ =u(0,t)=0, V ˜ =v(0,t)=0. Since ( u 0 , v 0 ) is radially symmetric and non-increasing in r=|x|, (u,v) is also a radially symmetric and non-increasing function, i.e., u r (r,t), v r (r,t)0 with r=|x|. Thus, u(x,t) and v(x,t) always reach their maxima at x=0, which means that Δu(0,t),Δv(0,t)0 for any 0<t<T, i.e., Δ U ˜ ,Δ V ˜ 0 for any 0<t<T. Therefore, it follows from (2.1) and (1.1) that

U ˜ (t)a U ˜ p 1 (t) g ˜ 1 (t), V ˜ (t)b V ˜ q 1 (t) g ˜ 2 (t).
(3.3)

Integrating (3.3) over (0,t), we obtain

{ 1 1 p 1 U ˜ 1 p 1 ( t ) a G ˜ 1 ( t ) + 1 1 p 1 U ˜ 1 p 1 ( 0 ) , 1 1 q 1 V ˜ 1 q 1 ( t ) b G ˜ 2 ( t ) + 1 1 q 1 V ˜ 1 q 1 ( 0 ) .
(3.4)

From lim t T U ˜ (t)= lim t T V ˜ (t)= and 0< U ˜ (0), V ˜ (0)<, we get

lim t T G ˜ i (t)= lim t T sup g ˜ i (t)=,i=1,2.

 □

Next, we first give some auxiliary lemmas about the solutions of (2.8), which will be used in the proofs of theorems. Similar to (3.2), we let

U(t)= max x B ¯ u 2 (x,t),V(t)= max x B ¯ v 2 (x,t).
(3.5)

By (2.8), we see that U(t) and V(t) satisfy

U t a ( 1 + m p 1 ) | B | σ 1 μ 1 U r 1 V σ 1 , V t b ( 1 + n q 1 ) | B | σ 2 μ 2 V r 2 U σ 2 , a.e.  t ( 0 , T ) .
(3.6)

Let β 1 =1 r 1 + σ 2 , β 2 =1 r 2 + σ 1 , then β 1 , β 2 >0. By Young’s inequality, we have

( U β 1 + V β 2 ) t ( β 1 a ( 1 + m p 1 ) | B | σ 1 μ 1 + β 2 b ( 1 + n q 1 ) | B | σ 2 μ 2 ) U β 1 σ 2 β 1 V β 2 σ 1 β 2 C ( U β 1 + V β 2 ) σ 2 β 1 + σ 1 β 2 .

Integrating the above inequality over (t,T), we obtain

U β 1 + V β 2 C ( T t ) β 1 β 2 d ,
(3.7)

where d= σ 1 σ 2 (1 r 1 )(1 r 2 )>0 by (1.7).

Lemma 5 Suppose that u 0 , v 0 satisfy (H 1)-(H 3) and the solution ( u 2 , v 2 ) of (2.8) blows up in finite time T. Then, we have

lim t T sup g i (t)= lim t T G i (t)=,i=1,2,

where

g 1 ( t ) = v 2 B , μ 1 σ 1 , g 2 ( t ) = u 2 B , μ 2 σ 2 , G 1 ( t ) = 0 t g 1 ( s ) d s , G 2 ( t ) = 0 t g 2 ( s ) d s .
(3.8)

Proof Let U(t),V(t) be as (3.5), then from (2.8), we have

U ( t ) a ( 1 + m p 1 ) U r 1 ( t ) g 1 ( t ) , V ( t ) b ( 1 + n q 1 ) V r 2 ( t ) g 2 ( t ) , a.e.  t [ 0 , T ) .
(3.9)

Integrating (3.9) over (0,t), we obtain

{ 1 1 r 1 U 1 r 1 ( t ) a ( 1 + m p 1 ) G 1 ( t ) + 1 1 r 1 U 1 r 1 ( 0 ) , 1 1 r 2 V 1 r 2 ( t ) b ( 1 + n q 1 ) G 2 ( t ) + 1 1 r 2 V 1 r 2 ( 0 ) .
(3.10)

Similar to the proofs of G ˜ i (t) and g ˜ i (t), we have

lim t T G i (t)= lim t T sup g i (t)=,i=1,2.

 □

Lemma 6 Suppose that u 0 , v 0 satisfy (H 1)-(H 4). Then, we have

u 2 t δ u 2 k 1 + 1 0, v 2 t δ v 2 k 2 + 1 0,(x,t)B×(0,T),
(3.11)

here k 1 =d/ β 2 , k 2 =d/ β 1 .

Proof Set J 1 (x,t)= u 2 t δ u 2 k 1 + 1 , J 2 (x,t)= u 2 t δ u 2 k 2 + 1 . Then,

lim x B J 1 (x,t)0, lim x B J 2 (x,t)0; J 1 (x,0)0, J 2 (x,0)0,xB.

A series of computations yields J 1 t = u 2 t t δ( k 1 +1) u 2 k 1 u 2 t and

u 2 t t = r 1 u 2 1 ( J 1 2 + 2 δ u 2 k 1 + 1 J 1 + δ 2 u 2 2 k 1 + 2 ) + u 2 r 1 Δ J 1 + δ ( k 1 + 1 ) k 1 u 2 k 1 1 + r 1 | u 2 | 2 + δ ( k 1 + 1 ) u 2 k 1 + r 1 Δ u 2 + a ( 1 + m p 1 ) σ 1 u 2 r 1 v 2 μ 1 σ 1 μ 1 B v 2 μ 1 1 ( J 2 + δ v 2 k 2 + 1 ) d x = u 2 r 1 Δ J 1 + ( 2 r 1 δ u 2 k 1 + δ ( k 1 + 1 ) u 2 k 1 ) J 1 + r 1 u 2 1 J 1 2 + δ ( k 1 + 1 ) k 1 u 2 k 1 1 + r 1 | u 2 | 2 + ( r 1 δ 2 + ( k 1 + 1 ) δ 2 ) u 2 2 k 1 + 1 ( k 1 + 1 ) a ( 1 + m p 1 ) δ u 2 k 1 + r 1 v 2 μ 1 σ 1 + a ( 1 + m p 1 ) σ 1 u 2 r 1 v 1 μ 1 σ 1 μ 1 B v 2 μ 1 1 J 2 d x + a ( 1 + m p 1 ) σ 1 δ u 2 r 1 v 2 μ 1 σ 1 μ 1 v 2 μ 1 + k 2 μ 1 + k 2 .

From the condition (1.7), it is easy to calculate that k 1 +1> r 1 . Then, it entails

(3.12)

By the Hölder inequality, for any 0<θ<1, it follows that

v 2 μ 1 σ 1 = v 2 μ 1 ( σ 1 μ 1 ) θ v 2 μ 1 σ 1 ( σ 1 μ 1 ) θ v 2 μ 1 ( σ 1 μ 1 ) θ v 2 μ 1 + k 2 σ 1 ( σ 1 μ 1 ) θ | B | k 2 [ σ 1 ( σ 1 μ 1 ) θ ] μ 1 ( μ 1 + k 2 ) .

Furthermore, by Young’s inequality, for any ε>0 and l 1 , l 2 >1 satisfying 1/ l 1 +1/ l 2 =1, the following inequality holds:

u 2 k 1 v 2 μ 1 σ 1 u 2 k 1 v 2 μ 1 ( σ 1 μ 1 ) θ v 2 μ 1 + k 2 σ 1 ( σ 1 μ 1 ) θ | B | k 2 [ σ 1 ( σ 1 μ 1 ) θ ] μ 1 ( μ 1 + k 2 ) | B | k 2 [ σ 1 ( σ 1 μ 1 ) θ ] μ 1 ( μ 1 + k 2 ) ( ( ε u 2 k 1 ) l 1 l 1 + 1 l 2 ( 1 ε v 2 μ 1 ( σ 1 μ 1 ) θ v 2 μ 1 + k 2 σ 1 ( σ 1 μ 1 ) θ ) l 2 ) .
(3.13)

Now, we take

l 1 = k 2 + σ 1 k 2 , l 2 = k 2 + σ 1 σ 1 ,θ= σ 1 k 2 + σ 1 ,ε= ( k 1 + 1 k 2 + σ 1 | B | k 2 σ 1 μ 1 ( k 2 + σ 1 ) ) σ 1 k 2 + σ 1 .

Therefore, by (3.12) and (3.13), it follows that

J 1 t u 2 r 1 Δ J 1 2 r 1 δ u 2 k 1 J 1 a ( 1 + m p 1 ) σ 1 u 2 r 1 v 2 μ 1 σ 1 μ 1 B v 2 μ 1 1 J 2 d x r 1 δ ( δ δ 1 ) u 2 2 k 1 + 2 0 ,

where

δ 1 = a ( 1 + m p 1 ) k 2 r 1 | B | σ 1 μ 1 ( k 1 + 1 k 2 + σ 1 ) σ 1 k 2 + 1 .

We can determine a number δ 2 in the similar way. Let δ 0 =max{ δ 1 , δ 2 }, similar to the above, one has

J 2 t v 2 r 2 Δ J 2 2 r 2 δ v 2 k 2 J 2 b(1+n q 1 ) σ 2 v 2 r 2 u 2 μ 2 σ 2 μ 2 B u 2 μ 2 1 J 1 dx0.

By the comparison principle of Lemma 1 in [20], we have J 1 , J 2 0. This completes the proof. □

Lemma 7 Suppose that u 0 , v 0 satisfy (H 1)-(H 4), then there exist positive constants c and C such that

{ c max x B ¯ u 2 ( x , t ) ( T t ) 1 / k 1 C , c max x B ¯ v 2 ( x , t ) ( T t ) 1 / k 2 C .
(3.14)

Proof It follows from (3.11) that

U t δ U k 1 + 1 , V t δ V k 2 + 1 ,t(0,T).
(3.15)

Combining with (3.6), we can obtain

U k 1 + 1 r 1 a ( 1 + m p 1 ) δ | B | σ 1 μ 1 V σ 1 , V k 2 + 1 r 2 b ( 1 + n q 1 ) δ | B | σ 2 μ 2 U σ 2 .
(3.16)

The direct computation yields k 1 +1 r 1 = σ 1 β 1 / β 2 , k 2 +1 r 1 = σ 2 β 2 / β 1 . It follows from (3.16) that

U β 1 ( a ( 1 + m p 1 ) δ ) β 2 σ 1 | B | β 2 μ 1 V β 2 , V β 2 ( b ( 1 + n q 1 ) δ ) β 1 σ 2 | B | β 1 μ 2 U β 1 .
(3.17)

Therefore, combining (3.17) with (3.7) gives

cU(t) ( T t ) 1 / k 1 ,cV(t) ( T t ) 1 / k 2 .

Integrating (3.15) from t to T, we end the proof. □

Lemma 8 Suppose that u 0 , v 0 satisfy (H 1)-(H 4) and Δ u 0 ,Δ v 0 0, then

lim t T u 2 1 r 1 ( x , t ) ( 1 r 1 ) G 1 ( t ) =a(1+m p 1 ), lim t T v 2 1 r 2 ( x , t ) ( 1 r 2 ) G 2 ( t ) =b(1+n q 1 )
(3.18)

uniformly on compact subsets of B.

Proof Here we consider the first eigenvalue problem

Δϕ(x)= λ 1 ϕ(x),xB;ϕ(x)=0,xB.

Normalize ϕ(x) as ϕ(x)>0 in B and B ϕ(x)dx=1. Define

z(x,t)=a(1+m p 1 ) G 1 (t) 1 1 r 1 u 2 1 r 1 (x,t),γ(t)= B z(y,t)ϕ(y)dy.

A series of computations yields

γ ( t ) = B ( a ( 1 + m p 1 ) g 1 ( t ) u 2 r 1 u 2 t ) ϕ ( y ) d y = B Δ u 2 ( y , t ) ϕ ( y ) d y = λ 1 B u 2 ( y , t ) ϕ ( y ) d y = λ 1 ( 1 r 1 ) 1 1 r 1 B ( a ( 1 + m p 1 ) G 1 ( t ) z ( y , t ) ) 1 1 r 1 ϕ ( y ) d y λ 1 ( 1 r 1 ) 1 1 r 1 B ( a ( 1 + m p 1 ) G 1 ( t ) + z ( y , t ) ) 1 1 r 1 ϕ ( y ) d y C ( G 1 1 1 r 1 ( t ) + B ( z ( y , t ) ) 1 1 r 1 ϕ ( y ) d y ) ,

where z =max{z,0}. By (3.10), we know that inf B z(x,t)C, which means z (x,t)C. Then

γ (t)C G 1 1 1 r 1 (t)+C.
(3.19)

Integrating (3.19) from 0 to t yields

γ(t)C ( 1 + 0 t G 1 1 1 r 1 ( s ) d s ) .

That is

B | z ( y , t ) | ϕ(y)dyC ( 1 + 0 t G 1 1 1 r 1 ( s ) d s ) .
(3.20)

Denote B ϱ ={yB:ϱ|y|<R}. Since Δz0, using Lemma 4.5 in [4], we obtain

sup B ϱ z(x,t) C ϱ N + 1 ( 1 + 0 t G 1 1 1 r 1 ( s ) d s ) .
(3.21)

It follows from (3.21) and (3.10) that

C G 1 ( t ) a(1+m p 1 ) u 2 1 r 1 ( 1 r 1 ) G 1 ( t ) C ( 1 + 0 t G 1 1 1 r 1 ( s ) d s ) G 1 ( t )
(3.22)

for any x B ϱ . On the other hand, we know from (3.10), (3.14) and Δ u 0 ,Δ v 0 0 that

c ( T t ) ( 1 r 1 ) ( 1 r 2 + σ 1 ) d G 1 (t)C.

Therefore,

{ c ( T t ) 1 r 2 + σ 1 d G 1 1 1 r 1 ( t ) C , c ( T t ) ( 1 r 1 ) ( 1 r 2 + σ 1 ) d + 1 G 1 ( t ) C .

Noting that

1 r 2 + σ 1 d < ( 1 r 1 ) ( 1 r 2 + σ 1 ) d +11 r 2 < σ 1 ( σ 2 r 1 ).

Then

lim t T 0 t G 1 1 1 r 1 ( s ) d s G 1 ( t ) = lim t T G 1 1 1 r 1 ( t ) G 1 ( t ) =0.

Thus

lim t T u 2 1 r 1 ( x , t ) ( 1 r 1 ) G 1 ( t ) =a(1+m p 1 ).

Similarly,

lim t T v 2 1 r 2 ( x , t ) ( 1 r 2 ) G 2 ( t ) =b(1+n q 1 ).

 □

Proof of Theorem 2 According to u 2 u 1 + m p 1 , it follows from (2.9), (3.1), (3.8) and (3.18) that

lim t T inf u 1 p 1 ( x , t ) ( 1 p 1 ) G ˜ 1 ( t ) lim t T u 2 1 r 1 ( x , t ) ( 1 r 1 ) G 1 ( t ) 1 1 + m p 1 =a.
(3.23)

On the other hand, from (3.4), we estimate

lim t T sup U ˜ 1 p 1 ( t ) ( 1 p 1 ) G ˜ 1 ( t ) a.
(3.24)

Combining (3.23) with (3.24), we obtain

lim t T u 1 p 1 ( x , t ) ( 1 p 1 ) G ˜ 1 ( t ) =a.

Similarly,

lim t T v 1 q 1 ( x , t ) ( 1 q 1 ) G ˜ 2 ( t ) =b.

This completes the proof of the theorem. □

Proof of Theorem 3 By Theorem 2, we have that, as tT,

{ G ˜ 1 ( t ) = v B , α 1 p 2 | B | p 2 α 1 ( b ( 1 q 1 ) ) p 2 1 q 1 G ˜ 2 p 2 1 q 1 ( t ) , G ˜ 2 ( t ) = u B , α 2 q 2 | B | q 2 α 2 ( a ( 1 p 1 ) ) q 2 1 p 1 G ˜ 1 q 2 1 p 1 ( t ) ,

where the notation uv means that lim t T u(t)/v(t)=1. Hence, we obtain

d G ˜ 1 d G ˜ 2 | B | p 2 α 1 q 2 α 2 ( a ( 1 p 1 ) ) q 2 1 p 1 ( b ( 1 q 1 ) ) p 2 1 q 1 G ˜ 1 q 2 1 p 1 G ˜ 2 p 2 1 q 1 .

A series of computations yields

{ G ˜ 1 ( t ) | B | θ 1 ( 1 p 1 ) d β 2 ( 1 p 1 ) d a ( 1 p 1 ) ( 1 + m p 1 ) ( β 2 a ( 1 + m p 1 ) ) ( 1 p 1 ) ( 1 q 1 ) d ( β 1 b ( 1 + n q 1 ) ) p 2 ( 1 p 1 ) d ( T t ) β 2 ( 1 p 1 ) d , G ˜ 2 ( t ) | B | θ 2 ( 1 q 1 ) d β 1 ( 1 q 1 ) d b ( 1 q 1 ) ( 1 + n q 1 ) ( β 1 b ( 1 + n q 1 ) ) ( 1 p 1 ) ( 1 q 1 ) d ( β 2 a ( 1 + m p 1 ) ) q 2 ( 1 q 1 ) d ( T t ) β 1 ( 1 q 1 ) d .

Combining with Lemma 7, we obtain the results of Theorem 3 immediately. □

4 Discussions

The results in this paper show the interactions among the multi-nonlinearities in the reaction-diffusion system (1.1). Roughly speaking, either large exponents m, n, large coupling exponents p 2 , q 2 or large constants a, b benefit from the occurrence of the finite blow-up. For example, to make a finite blow-up to the problem (1.1), for fixed m, p 1 , p 2 , α 1 and n, q 1 , q 2 , α 2 , constants a and b should be properly large such that the following inequality

ρ 2 > 1 a ( 1 + m p 1 ) 1 b ( 1 + n q 1 )

holds.

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Acknowledgements

The authors are supported by National Natural Science Foundation of China and they would like to express their many thanks to the editor and reviewers for their constructive suggestions to improve the previous version of this paper. This work is supported by the NNSF of China (11071100).

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Ling, Z., Wang, Z. Blow-up problems for a compressible reactive gas model. Bound Value Probl 2012, 101 (2012). https://doi.org/10.1186/1687-2770-2012-101

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