1 Introduction

In this work, we consider the following non-local reaction diffusion problem with time dependent coefficient under the Dirichlet boundary condition

{ u t = Δ u + Ω u p d x k ( t ) u q , x = ( x 1 , x 2 , , x N ) Ω , t ( 0 , t ) , u ( x , t ) = 0 , x Ω , t ( 0 , t ) , u ( x , 0 ) = u 0 ( x ) 0 , x Ω ,
(1.1)

where Ω R N is a bounded domain with a smooth boundary Ω, Δ is the Laplace operator, and t is the possible blow-up time. By the maximum principle, it follows that u(x,t)0 in the time interval of existence. The coefficient k(t) is assumed to be nonnegative. The particular case of k=const of problem (1.1) has already been investigated by many authors, in [1, 2], they studied the question of blow-up for the solution, and in [35], they derived lower bounds for blow-up time under different boundary conditions. To deal with problem (1.1) with time dependent coefficient, we make the assumption on the parameters p and q, that is, p=q>1.

The motivation of this article comes from the work of Payne and Philippin in [6], where they investigated the blow-up phenomena of the solution of the following problem

{ u t = Δ u + k ( t ) f ( u ) , x = ( x 1 , , x N ) Ω , t ( 0 , t ) , u ( x , t ) = 0 , x Ω , u ( x , 0 ) = u 0 ( x ) , x Ω ,
(1.2)

where Ω is a bounded sufficiently smooth domain in R N , N2, and the coefficient k(t) is assumed nonnegative or strictly positive depending on the situation.

In next, we employ a method used by Kaplan in [7] to obtain a condition, which leads to blow-up at some finite time and also leads to an upper bound for the blow-up time. In Section 3, we derive the condition on the data of problem (1.1) sufficient to guarantee the global existence of u(x,t).

2 Conditions for blow-up in finite time t

Let λ 1 be the first eigenvalue, and let ϕ 1 be the associated eigenfunction of the Dirichlet-Laplace operator defined as

Δ ϕ 1 = λ 1 ϕ 1 , ϕ 1 >0,xΩ; ϕ 1 =0,xΩ,
(2.1)
Ω ϕ 1 dx=1.
(2.2)

Let the auxiliary function η(t)

η(t):= ( | Ω | k ( t ) ) 1 q 1 Ω u ϕ 1 dx
(2.3)

defined in (0, t ), where u(x,t) is the solution of (1.1) and q>1.

We assume that for all t(0, t ),

|Ω|>k(t)>0, k ( t ) | Ω | k ( t ) β, max x Ω ϕ 1 |Ω|1,
(2.4)

for some constant β, and

γ:= λ 1 β q 1 .
(2.5)

We deduce from (2.4) and (2.5) that

η ( t ) β q 1 η ( t ) + ( | Ω | k ( t ) ) 1 q 1 Ω ϕ 1 [ Δ u + Ω u q d x k ( t ) u q ] d x = γ η ( t ) + ( | Ω | k ( t ) ) 1 q 1 ( Ω u q d x k ( t ) Ω ϕ 1 u q d x ) γ η ( t ) + ( | Ω | k ( t ) ) 1 q 1 ( ( 1 max x Ω ϕ 1 k ( t ) ) Ω ϕ 1 u q d x ) γ η ( t ) + ( | Ω | k ( t ) ) q q 1 Ω ϕ 1 u q d x .
(2.6)

Furthermore, using (2.2) and Hölder’s inequality, we get

Ω ϕ 1 udx ( Ω ϕ 1 u q d x ) 1 q .
(2.7)

Combining (2.6) and (2.7), we get

η (t)γη(t)+ η q (t),t ( 0 , t ) .
(2.8)

By integrating (2.8), we get

( η ( t ) ) 1 q Θ(t):= { ( ( η ( 0 ) ) 1 q 1 γ ) e γ ( q 1 ) t + 1 γ , γ 0 , ( η ( 0 ) ) 1 q + ( 1 q ) t , γ = 0 .
(2.9)

If Θ( T 1 )=0 for some T 1 >0, then η(t) blows up at time t < T 1 . This result is summarized in the following theorem.

Theorem 1 Let u(x,t) be the solution of problem (1.1). Then the auxiliary function η(t) defined in (2.3) blows up at time t < T 1 with

T 1 := { 1 γ ( q 1 ) log ( 1 γ ( ( η ( 0 ) ) 1 q 1 γ ) ) if  0 < γ ( η ( 0 ) ) 1 q < 1 , 1 ( q 1 ) ( η ( 0 ) ) q 1 if  γ 0 .
(2.10)

3 Condition for global existence

In this section, our argument makes use of the following Sobolev-type inequality

( Ω v 6 d x ) 1 / 4 Γ ( Ω | v | 2 d x ) 3 / 4 ,Γ= 2.3 3 / 4 π ,
(3.1)

valid in R 3 for a nonnegative function v that vanishes on Ω. In this section, our results restricted to R 3 for proof of (3.1), see [8].

We consider the auxiliary function σ(t) defined as

σ(t):= M 1 ( | Ω | k ( t ) ) 2 n Ω u 2 n ( p 1 ) dx,t ( 0 , t ) ,
(3.2)

with

M:= ( | Ω | k ( 0 ) ) 2 n Ω u 0 2 n ( q 1 ) dx,
(3.3)

we assume that for all t(0, t ),

|Ω|>k(t)>0, k ( t ) | Ω | k ( t ) <β,
(3.4)

for some constant β. In (3.2)-(3.3), n is subjected to restrictions

n(q1)1,n> 3 4 .
(3.5)

For convenience, we set

v(x,t)= u n ( q 1 ) ,
(3.6)

and compute

σ ( t ) = 2 n k ( t ) | Ω | k ( t ) σ ( t ) + 2 n ( q 1 ) M 1 ( | Ω | k ( t ) ) 2 n × Ω u 2 n ( q 1 ) 1 [ Δ u + Ω u q d x k ( t ) u q ] d x
(3.7)

with

Ω u 2 n ( q 1 ) 1 Δudx= 2 n ( q 1 ) 1 n 2 ( q 1 ) 2 Ω |v | 2 dx,
(3.8)

due to (3.4), we obtain

σ ( t ) 2 n β σ ( t ) 2 [ 2 n ( q 1 ) 1 ] n ( q 1 ) ( | Ω | k ( t ) ) 2 n M 1 Ω | v | 2 d x + 2 n ( q 1 ) M 1 ( | Ω | k ( t ) ) 2 n [ | Ω | Ω v 2 + 1 n d x k ( t ) Ω v 2 + 1 n d x ] = 2 n β σ ( t ) 2 [ 2 n ( q 1 ) 1 ] n ( q 1 ) ( | Ω | k ( t ) ) 2 n M 1 Ω | v | 2 d x + 2 n ( q 1 ) M 1 ( | Ω | k ( t ) ) 2 n + 1 Ω v 2 + 1 n d x .
(3.9)

By using Hölder’s inequality,

Ω v 2 + 1 n dx ( Ω v 2 d x ) ( 4 n 1 ) / 4 n ( Ω v 6 d x ) 1 / 4 n ,
(3.10)

and Sobolev-type inequality (3.1), we obtain

( | Ω | k ( t ) ) 2 n + 1 Ω v 2 + 1 n d x ( | Ω | k ( t ) ) 2 n + 1 ( Ω v 2 d x ) ( 4 n 1 ) / 4 n ( Ω | v | 2 d x ) 3 / 4 n Γ 1 / n = Γ 1 / n M ( 4 n 1 ) / 4 n σ ( 4 n 1 ) / 4 n ( ( | Ω | k ( t ) ) 2 n + 1 Ω | v | 2 d x ) 3 / 4 n ,
(3.11)

where Γ is defined in (3.1). Joining (3.11) and (3.9), we obtain

σ ( t ) 2 n β σ ( t ) 2 [ 2 n ( q 1 ) 1 ] n ( q 1 ) M 1 ( | Ω | k ( t ) ) 2 n Ω | v | 2 d x + 2 n ( q 1 ) Γ 1 / n M 1 / 2 n σ ( 4 n 1 ) / 4 n ( M 1 ( | Ω | k ( t ) ) 2 n Ω | v | 2 d x ) 3 / 4 n = 2 n β σ ( t ) + 2 n ( λ 1 M 1 ( | Ω | k ( t ) ) 2 n Ω | v | 2 d x ) 3 / 4 n × { λ 3 / 4 n ( q 1 ) Γ 1 / n σ ( 4 n 1 ) / 4 n M 1 / 2 n 2 [ 2 n ( q 1 ) 1 ] n 2 ( q 1 ) × λ ( M 1 ( | Ω | k ( t ) ) 2 n λ 1 Ω | v | 2 d x ) ( 4 n 3 ) / 4 n }
(3.12)

with arbitrary λ0. Choosing λ:= λ 1 , the first eigenvalue of problem (2.1), we have

Ω |v | 2 dx λ 1 Ω v 2 dx,
(3.13)

by the Rayleigh principle. By using (3.13) in the last factor of (3.12), we obtain

σ ( t ) 2 n β σ ( t ) + 2 n ( λ 1 1 M 1 ( | Ω | k ( t ) ) 2 n Ω | v | 2 d x ) 3 / 4 n × { λ 1 3 / 4 n ( q 1 ) Γ 1 / n σ ( t ) ( 4 n 1 ) / 4 n M 1 / 2 n 2 [ 2 n ( q 1 ) 1 ] n 2 ( q 1 ) λ 1 σ ( t ) ( 4 n 3 ) / 4 n } = 2 n β σ ( t ) + 2 n σ ( t ) 3 / 4 n × σ ( t ) ( 4 n 3 ) / 4 n { ω σ ( t ) 1 / 2 n ( μ + β ) } ,
(3.14)

with

ω= λ 1 3 / 4 n (q1) Γ 1 / n M 1 / 2 n ,μ= 2 [ 2 n ( q 1 ) 1 ] n 2 ( q 1 ) λ 1 β.
(3.15)

Suppose that β is small enough to satisfy the condition

μ>0,
(3.16)

and that initial data is small enough to satisfy the condition

ωμ<0.
(3.17)

Then either ω ( σ ( t ) ) 1 / 2 n μ remains negative for all time, or there exists a first time t 0 such that

ω ( σ ( t 0 ) ) 1 / 2 n μ=0.
(3.18)

Then we obtain the differential inequality

σ (t)2nσ(t) { ω ( σ ( t ) ) 1 / 2 n μ } 0,t(0, t 0 ).
(3.19)

Integrating this differential inequality, we obtain

σ(t) { ( 1 ω μ ) e μ t + ω μ } 2 n ,t>0.
(3.20)

This result is summarized in the next theorem.

Theorem 2 Let Ω be a bounded domain in R 3 , and assume that the data of problem (1.1) satisfy conditions (3.4), (3.16), (3.17). Then the auxiliary function σ(t) defined in (3.2) satisfies (3.20), and u(x,t) exists for all time t>0.