1. Introduction

In this article, we consider the positive solution of the following degenerate parabolic equation

u t = f ( u ) ( Δ u + a Ω u ( x , t ) d x ) , x Ω , t > 0 , u ( x , t ) = Ω g ( x , y ) u l ( y , t ) d y , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω ,
(1.1)

where a, l > 0 and Ω is a bounded domain in RN (N ≥ 1) with smooth boundary ∂Ω.

There have been many articles dealing with properties of solutions to degenerate parabolic equations with homogeneous Dirichlet boundary condition (see [14] and references therein). For example, Deng et al. [5] studied the parabolic equation with nonlocal source

u t = f ( u ) ( Δ u + a Ω u d x ) ,
(1.2)

which is subjected to homogeneous Dirichlet boundary condition. It was proved that there exists no global positive solution if and only if 1 ( s f ( s ) ) d s < and Ω φ ( x ) d x > 1 a , where φ(x) is the unique positive solution of the linear elliptic problem

- Δ φ = 1 , x Ω ; φ ( x ) = 0 , x Ω .
(1.3)

However, there are some important phenomena formulated into parabolic equations which are coupled with nonlocal boundary conditions in mathematical modeling such as thermoelasticity theory (see [6, 7]). Friedman [8] studied the problem of nonlocal boundary conditions for linear parabolic equations of the type

u t A u = c ( x ) u , x Ω , t > 0 , u ( x , t ) = Ω K ( x , y ) u ( y , t ) d y , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω ,
(1.4)

with uniformly elliptic operator A = i , j = 1 n a i j ( x ) 2 x i x j + i = 1 n b i ( x ) x i + c ( x ) and c(x)≤ 0. It was proved that the unique solution of (1.4) tends to 0 monotonically and exponentially as t →+∞ provided that Ω φ ( x , y ) d y ρ < 1 , x Ω .

Parabolic equations with both nonlocal sources and nonlocal boundary conditions have been studied as well (see [912]). Lin and Liu [13] considered the problem of the form

u t = Δ u + Ω g ( u ) d x , x Ω , t > 0 , u ( x , t ) = Ω K ( x , y ) u ( y , t ) d y , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω .
(1.5)

They established local existence, global existence, and nonexistence of solutions, and discussed the blow-up properties of solutions.

Chen and Liu [14] considered the following nonlinear parabolic equation with a localized reaction source and a weighted nonlocal boundary condition

u t = f ( u ) ( Δ u + a u ( x 0 , t ) ) , x Ω , t > 0 , u ( x , t ) = Ω g ( x , y ) u ( y , t ) d y , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω .
(1.6)

Under certain conditions, they obtained blow-up criteria. Furthermore, they derived the uniform blow-up estimate for some special f(u).

In recent few years, reaction-diffusion problems coupled with nonlocal nonlinear boundary conditions have also been studied. Gladkov and Kim [15] considered the following problem for a single semilinear heat equation

u t = Δ u + c ( x , t ) u p , x Ω , t > 0 , u ( x , t ) = Ω φ ( x , y , t ) u l ( y , t ) d y , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω ,
(1.7)

where p, l > 0. They obtained some criteria for the existence of global solution as well as for the solution to blow-up in finite time.

For other works on parabolic equations and systems with nonlocal nonlinear boundary conditions, we refer readers to [1620] and the references therein.

Motivated by those of works above, we will study the problem (1.1) and want to understand how the function f(u) and the coefficient a, the weight function g(x, y) and the nonlinear term ul (y, t) in the boundary condition play substantial roles in determining blow-up or not of solutions.

In this article, we give the following hypotheses:

(H1) u 0 ( x ) C 2 + α ( Ω ) C ( Ω ¯ ) for α∈(0,1),u0(x) > 0 in Ω, u 0 ( x ) = Ω g ( x , y ) u 0 l ( y ) dy on ∂Ω.

(H2) g(x, y)≢0 is a nonnegative and continuous function defined for x Ω , y Ω ¯ .

(H3) fC([0,∞))∩C1(0,∞), f > 0, f' ≥ 0 in (0,∞).

The main results of this article are stated as follows.

Theorem 1.1. Assume that 0 < l ≤ 1 and Ω g ( x , y ) d y < 1 for all x∈∂Ω.

  1. (1)

    If a is sufficiently small, then the solution of (1.1) exists globally;

  2. (2)

    If a is sufficiently large, then the solution of (1.1) also exists globally provided that δ + 1 ( s f ( s ) ) d s = + for some δ > 0.

Theorem 1.2. Assume that l > 1 and Ω g ( x , y ) d y < 1 for all x∈∂Ω. Then the solution of (1.1) exists globally provided that a and u0(x) are sufficiently small. While the solution blows up in finite time if a,u0(x) are sufficiently large and δ + 1 ( s f ( s ) ) d s < + for some δ > 0.

Theorem 1.3. Assume that l > 1 and Ω g ( x , y ) d y 1 for all x∈∂Ω. If δ + 1 ( s f ( s ) ) d s < + for some δ > 0, then the solution of (1.1) blows up in finite time provided that u0(x) is large enough.

Theorem 1.4. If δ + 1 ( s f ( s ) ) d s < + for some δ > 0 and a > Ω φ ( x ) d x - 1 , where φ(x) is the solution of (1.3), then there exists no global positive solution of (1.1).

To describe conditions for blow-up of solutions, we need an additional assumption on the initial data u 0 .

(H4) There exists a constant ε > ε1 > 0 such that Δ u 0 + a Ω u 0 ( x ) d x ε u 0 , where ε1 will be given later.

Theorem 1.5. Assume u0(x) satisfies (H1), (H2), and (H4), Δu0 ≤ 0 in Ω holds, and let f(u) = up, 0 < p ≤ 1, l = 1, then the following limits converge uniformly on any compact subset of Ω:

  1. (1)

    If 0 < p < 1, lim t T * u ( x , t ) ( T * - t ) 1 p = ( a p Ω ) - 1 p .

  2. (2)

    If p = 1, lim t T * ln ( T * - t ) - 1 ln u ( x , t ) =1.

This article is organized as follows. In Section 2, we establish the comparison principle and the local existence. Some criteria regarding to global existence and finite time blow-up for problem (1.1) are given in Section 3. In Section 4, the global blow-up result and the blow-up rate estimate of blow-up solutions for the special case of f (u) = up , 0 < p ≤ 1 and l = 1 are obtained.

2. Comparison principle and local existence

First, we start with the definition of subsolution and supersolution of (1.1) and comparison principle. Let Q T = Ω × (0, T), S T = ∂Ω × (0, T), and Q ¯ T = Ω ¯ × [ 0 , T ) .

Definition 2.1. A function u(x, t) is called a subsolution of (1.1) on Q T , if u ¯ C 2 , 1 ( Q T ) C ( Q ¯ T ) satisfies

u ¯ t f ( u ¯ ) ( Δ u ¯ + a Ω u ¯ d x ) , x Ω , t > 0 , u ¯ ( x , t ) Ω g ( x , y ) u ¯ l ( y , t ) d y , x Ω , t > 0 , u ¯ ( x , 0 ) u 0 ( x ) , x Ω .
(2.1)

Similarly, a supersolution ū(x, t) of (1.1) is defined by the opposite inequalities.

A solution of problem (1.1) is a function which is both a subsolution and a supersolution of problem (1.1).

The following comparison principle plays a crucial role in our proofs which can be obtained by similar arguments as [10] and its proof is therefore omitted here.

Lemma 2.2. Suppose that w ( x , t ) C 2 , 1 ( Q T ) C ( Q ¯ T ) and satisfies

w t - d ( x , t ) Δ w c 1 ( x , t ) w + c 2 ( x , t ) Ω c 3 ( x , t ) w ( x , t ) d x , ( x , t ) Q T , w ( x , t ) c 4 ( x , t ) Ω c 5 ( x , y ) w l ( y , t ) d y , ( x , t ) S T ,
(2.2)

where d(x, t), c i (x, t)(i = 1,2,3,4) are bounded functions and d(x, t)≥ 0, c i (x, t)≥ 0 (i = 2,3,4) in Q T , c5 (x, y)≥ 0 for x∈∂Ω, y∈Ω and is not identically zero. Then, w(x, 0) > 0 for x Ω ¯ implies w(x, t) > 0 in Q T . Moreover, c5 (x, y) ≡ 0 or if c 4 ( x , t ) Ω c 5 ( x , y ) d y 1 on S T , then w(x, 0) ≥ 0 for x Ω ¯ implies w(x, t) ≥ 0 in Q T .

On the basis of the above lemmas, we obtain the following comparison principle of (1.1).

Lemma 2.3. Let u and v be nonnegative subsolution and supersolution of (1.1), respectively, with u(x, 0) ≤ v(x, 0) for x Ω ¯ . Then, uv in Q T if uη or vη for some small positive constant η holds.

Local in time existence of positive classical solutions of (1.1) can be obtained by using fixed point theorem [21], the representation formula and the contraction mapping principle as in [13]. By the above comparison principle, we get the uniqueness of solution to the problem. The proof is more or less standard, so is omitted here.

3. Global existence and blow-up in finite time

In this section, we will use super- and subsolution techniques to derive some conditions on the existence or nonexistence of global solution.

Proof of Theorem 1.1. (1) Let ψ(x) be the unique positive solution of the linear elliptic problem

- Δ ψ = ε 0 , x Ω , ψ ( x ) = Ω g ( x , y ) d y , x Ω ,
(3.1)

where ε0 is a positive constant such that 0 < ψ(x) < 1 (since Ω g ( x , y ) d y < 1 , there exists such ε0). Let K ¯ = max x Ω ¯ ψ ( x ) , K ¯ = min x Ω ¯ ψ ( x ) .

We define a function w(x, t) as following:

w ( x , t ) = M ψ ( x ) ,
(3.2)

where M ≥ 1 is a constant to be determined later. Then, we have

w | Ω = M Ω g ( x , y ) d y M Ω g ( x , y ) ψ l ( x ) d y = M 1 - l Ω g ( x , y ) w l ( y , t ) d y Ω g ( x , y ) w l ( y , t ) d y .
(3.3)

On the other hand, we have for x ∈ Ω, t > 0,,

w t - f ( w ) ( Δ w + a Ω w ( x , t ) d x ) f ( M ψ ( x ) ) M ( ε 0 - a Ω K ¯ ) .
(3.4)

We choose M = max { K ¯ - 1 max x Ω ¯ u 0 ( x ) , 1 } and set a 0 = ε 0 ( Ω K ¯ ) - 1 , then it is easy to verify that w(x, t) is a supersolution of (1.1) provided that aa0. By comparison principle, u(x, t) ≤ w(x, t), then u(x, t) exists globally.

  1. (2)

    Consider the following problem

    z ( t ) = b 1 f ( K ¯ z ( t ) ) z ( t ) , t > 0 , z ( 0 ) = z 0 ,
    (3.5)

where z 0 > max { K ¯ - 1 max x Ω ¯ u 0 ( x ) , 1 } , b1 is a positive constant to be fixed later. It follows from hypothesis (H3) and the theory of ordinary differential equation (ODE) that there exists a unique solution z (t) to problem (3.5) and z (t) is increasing. If δ + 1 ( s f ( s ) ) d s = + for some positive δ, we know that z (t) exists globally and z (t) ≥ z0.

Let v(x, t) = z (t) ψ (x), where ψ (x) is given by (3.1), then for x ∈ Ω, t > 0, we obtain

v t - f ( v ) ( Δ v + a Ω v ( x , t ) d x ) = z ( t ) ψ ( x ) - f ( z ( t ) ψ ( x ) ) ( z ( t ) Δ ψ ( x ) + a Ω z ( t ) ψ ( x ) d x ) z ( t ) K ¯ - f ( K ¯ z ( t ) ) z ( t ) ( a K ¯ Ω - ε 0 ) = f ( K ¯ z ( t ) ) z ( t ) ( b 1 K ¯ - ( a K ¯ Ω - ε 0 ) ) .
(3.6)

Set a 1 = ε 0 ( K ¯ Ω ) - 1 , if a is sufficiently large such that a > a1, then we can choose

b 1 = K ¯ - 1 ( a K ¯ Ω - ε 0 ) > 0 . Thus,

v t - f ( v ) ( Δ v + a Ω v ( x , t ) d x ) 0 .
(3.7)

On the other hand, for x ∈ ∂Ω, t > 0, we get

v ( x , t ) = z ( t ) Ω g ( x , y ) d y > z ( t ) Ω g ( x , y ) ψ l ( y ) d y > Ω g ( x , y ) z l ( t ) ψ l ( y ) d y = Ω g ( x , y ) v l ( y , t ) d y .
(3.8)

Here, we use the conclusions 0 < ψ (x) < 1 and z(t) > 1.

Also for x Ω ¯ , we have

v ( x , 0 ) = z ( 0 ) ψ ( x ) = z 0 ψ ( x ) z 0 K ¯ u 0 ( x ) .
(3.9)

And the inequalities (3.5)-(3.9) show that v(x, t) is a supersolution of (1.1). Again by using the comparison principle, we obtain the global existence of u(x, t). The proof is complete.

Proof of Theorem 1.2. The proof of global existence part is similar to the first case of Theorem 1.1. For any given positive constant M ≤ 1, w (x) = (x) is a supersolution of problem (1.1) provided that u0 (x) ≤ ψ (x) < 1 and a < ε 0 ( Ω K ¯ ) - 1 , so the solution of (1.1) exists globally by using the comparison principle.

To prove the bow-up result, we introduce the elliptic problem

- Δ φ ( x ) = 1 , x Ω ; φ ( x ) = Ω g ( x , y ) φ ( y ) d y , x Ω .

Under the hypothesis (H2) and Ω g ( x , y ) d y < 1 , we know that it exists a unique positive solution φ(x). Let K * = min x Ω ¯ φ ( x ) , K * = max x Ω ¯ φ ( x ) , and z(t) be the solution of the following ODE

z ( t ) = f ( K * z ( t ) ) z ( t ) , t > 0 , z ( 0 ) = z 1 > 1 .
(3.10)

Then, z(t) is increasing and z (t) ≥ z1. Due to the condition δ + 1 ( s f ( s ) ) d s < + for some positive constant δ, we know that z (t) of problem (3.10) blows up in finite time.

If a and u0 (x) are so large that a ( K * ) l - 1 ( K * + l ) Ω - 1 K * - l , u0 (x) ≥ z1 (K*) l , then we set v1(x, t) = z(t)φl (x). For x ∈ Ω, t > 0, we obtain

v 1 t - f ( v 1 ) ( Δ v 1 + a Ω v 1 ( x , t ) d x ) = z ( t ) φ l ( x ) - f ( z ( t ) φ l ( x ) ) z ( t ) ( l ( l - 1 ) φ l - 2 ( x ) φ 2 + l φ l - 1 ( x ) Δ φ ( x ) + a Ω φ l ( x ) d x ) z ( t ) ( K * ) l - f ( K * l z ( t ) ) z ( t ) ( a K * l Ω - l ( K * ) l - 1 ) 0 .
(3.11)

For x ∈ ∂Ω, t > 0, by Jensen's inequality, we get

v 1 ( x , t ) = z ( t ) φ l ( x ) = z ( t ) ( Ω g ( x , y ) φ ( y ) d y ) l z ( t ) ( Ω g ( x , y ) d y ) l - 1 ( Ω g ( x , y ) φ l ( y ) d y ) Ω g ( x , y ) z l ( t ) φ l ( y ) d y = Ω g ( x , y ) v 1 l ( y ) d y .
(3.12)

Also for x Ω ¯ , we have

v 1 ( x , 0 ) = z ( 0 ) φ l ( x ) = z 1 φ l ( x ) z 1 ( K * ) l u 0 ( x ) .
(3.13)

The inequalities (3.10)-(3.13) show that v1(x, t) is a subsolution of problem (1.1). Since v1(x, t) blows up in finite time, u(x, t) also blows up in finite time by comparison principle.

Proof of Theorem 1.3. Let z(t) be the solution of the following ODE

z ( t ) = b 2 f ( z ( t ) ) z ( t ) , t > 0 , z ( 0 ) = z 2 .
(3.14)

where 0 < b2 < a |Ω|. If u0(x) is large enough, we can set 1 < z 2 < min x Ω ¯ u 0 ( x ) . Then, z (t) is increasing and satisfies z (t) ≥ z2 > 1. Moreover, z (t) of problem (3.14) blows up in finite time.

Set s (x, t) = z (t), then we have for x ∈ Ω, t > 0,

s t - f ( s ) ( Δ s + a Ω s ( x , t ) d x ) = z ( t ) - a f ( z ( t ) ) Ω z ( t ) = ( b 2 - a Ω ) f ( z ( t ) ) z ( t ) < 0 .
(3.15)

For x ∈ ∂Ω, t > 0,

s ( x , t ) = z ( t ) Ω g ( x , y ) z ( t ) d y < Ω g ( x , y ) z l ( t ) d y = Ω g ( x , y ) s l ( y , t ) d y .
(3.16)
s ( x , 0 ) = z ( 0 ) = z 2 < u 0 ( x ) , x Ω ¯ .
(3.17)

From (3.14)-(3.17), we see that s (x, t) is a subsolution of (1.1). Hence, u (x, t) ≥ s (x, t) by comparison principle, which implies u (x, t) blows up in finite time. This completes the proof.

Proof of Theorem 1.4. Consider the following equation

v t = f ( v ) ( Δ v + a Ω v d x ) , x Ω , t > 0 , v ( x , t ) = 0 , x Ω , t > 0 , v ( x , 0 ) = v 0 ( x ) , x Ω ,
(3.18)

and let v (x, t) be the solution to problem (3.18). It is obvious that v (x, t) is a subsolution of (1.1). By Theorem 1 in [5], we can obtain the result immediately.

4. Blow-up rate estimate

Now, we consider problem (1.1) with f (u) = up , 0 < p ≤ 1 and l = 1, i.e.,

u t = u p ( Δ u + a Ω u ( x , t ) d x ) , x Ω , t > 0 , u ( x , t ) = Ω g ( x , y ) u ( y , t ) d y , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω ,
(4.1)

where Ω g ( x , y ) d y 1 for all x ∈ ∂Ω, and suppose that the solution of (4.1) blows up in finite time T*.

Set U ( t ) = max x Ω ¯ u ( x , t ) , then U(t) is Lipschitz continuous.

Lemma 4.1. Suppose that u0 satisfies (H1), (H2), and (H4), then there exists a positive constant c0 such that

U ( t ) c 0 ( T * - t ) - 1 p .
(4.2)

Proof. By the first equation in (4.1), we have (see [22])

U ( t ) a Ω U 1 + p ( t ) ,a.e.t ( 0 , T * ) ,
(4.3)

Hence,

- ( U - p ( t ) ) ap Ω .
(4.4)

Integrating (4.3) over (t, T*), we can get

U ( t ) ( a p Ω ) - 1 p ( T * - t ) - 1 p .
(4.5)

Setting c0 = (ap |Ω| p)-1/p, then we draw the conclusion.

Lemma 4.2. Under the conditions of Lemma 4.1, there exists a constant ε1, which will be given below, such that

u t - ε 1 u p + 1 0, ( x , t ) Ω× ( 0 , T * ) .
(4.6)

Proof. Let J(x, t) = u t -ε1up+1for (x, t) ∈Ω × (0, T*), a series of computations yields

J t - u p Δ J - 2 p ε 1 u p J - a u p Ω J d x = p u - 1 J 2 + ε 1 ( p + 1 ) p u 2 p - 1 u 2 + p ε 1 2 u 2 p + 1 + a ε 1 u p Ω u p + 1 d x - a ε 1 ( p + 1 ) u 2 p Ω u d x p ε 1 2 u 2 p + 1 + a ε 1 u p Ω u p + 1 d x - a ε 1 ( p + 1 ) u 2 p Ω u d x .
(4.7)

By virtue of Hölder inequality, we have

Ω u d x Ω p ( p + 1 ) ( Ω u p + 1 d x ) 1 ( p + 1 ) .

Furthermore, by Young's inequality, for any θ > 0, the following inequality holds

u 2 p Ω u d x Ω p ( p + 1 ) u p u p ( Ω u p + 1 d x ) 1 ( p + 1 ) Ω p ( p + 1 ) u p ( p ( p + 1 ) ( θ u p ) ( p + 1 ) p + 1 ( p + 1 ) θ - ( p + 1 ) Ω u p + 1 d x ) = ( 1 ( p + 1 ) ) Ω p ( p + 1 ) ( p θ ( p + 1 ) p u 2 p + 1 + θ - ( p + 1 ) u p Ω u p + 1 d x ) .
(4.8)

Using (4.8) and taking θ = Ω p ( p + 1 ) 2 , ε1 = a|Ω|, then (4.7) becomes

J t - u p ΔJ-2p ε 1 u p J-a u p Ω Jdxp ε 1 ( ε 1 - a Ω ) u 2 p + 1 =0,
(4.9)

Fix (x, t) ∈∂Ω × (0, T*), then we have

J ( x , t ) = u t - ε 1 u p + 1 = Ω g ( x , y ) u t ( y , t ) dy- ε 1 ( Ω g ( x , y ) u ( y , t ) d y ) p + 1 .

Since u t (y, t) = J(y, t) + ε1up+1(y, t), we have

Ω g ( x , y ) u t ( y , t ) d y - ε 1 Ω g ( x , y ) u ( y , t ) d y p + 1 = Ω g ( x , y ) J ( y , t ) d y + ε 1 Ω g ( x , y ) u p + 1 ( y , t ) d y - Ω g ( x , y ) u ( y , t ) d y p + 1 .

Noticing that p > 0, 0 < F ( x ) = Ω g ( x , y ) d y 1 , x Ω , we can apply Jensen's inequality to the last integral in the above inequality,

Ω g ( x , y ) u p + 1 ( y , t ) d y - ( Ω g ( x , y ) u ( y , t ) d y ) p + 1 F ( x ) Ω g ( x , y ) u ( y , t ) d y F ( x ) p + 1 - Ω g ( x , y ) u ( y , t ) d y p + 1 0 .

Hence, for (x, t) ∈∂Ω × (0, T*), we have

J ( x , t ) Ω g ( x , y ) J ( y , t ) dy.
(4.10)

On the other hand, (H4) implies that

J ( x , 0 ) > 0 , x Ω .
(4.11)

Owing to u(x, t) is a positive continuous function for ( x , t ) Ω ¯ × [ 0 , T * ) , it follows from (4.9)-(4.11) and Lemma 2.2 that J(x, t) ≥ 0 for ( x , t ) Ω ¯ × [ 0 , T * ) , i.e., u t ε1up+1. This completes the proof.

Integrating (4.6) from t to T*, we conclude that

u ( x , t ) c 2 ( T * - t ) - 1 p ,
(4.12)

where c2 = (ε1p)-1/pis a positive constant independent of t. Combining (4.2) with (4.12), we obtain the following result.

Theorem 4.3. Under the conditions of Lemma 4.1, if u (x, t) is the solution of (4.1) and blows up in finite time T*, then there exist positive constants c1, c2, such that

c 1 ( T * - t ) - 1 p max x Ω ¯ u ( x , t ) c 2 ( T * - t ) - 1 p .

Lemma 4.4. Assume that u0(x) satisfies (H1), (H2), and (H4), Δu0 ≤ 0 in Ω. u(x, t) is the solution of problem (4.1). Then, Δu ≤ 0 in any compact subsets of Ω × (0, T*).

The proof is similar to that of Lemma 1.1 in [14].

Denote

g ( t ) =a Ω udx,G ( t ) = 0 t g ( s ) ds.

Lemma 4.5. Under the conditions of Lemma 4.4, it holds that

lim t T * g ( t ) = , lim t T * G ( t ) = .
(4.13)

Proof. From Lemma 4.3, we have

u t u p g ( t ) ,a.e.t [ 0 , T * ) .
(4.14)

Integrating (4.14) over (0, t), we obtain

1 1 - p u 1 - p ( x , t ) 0 t g ( s ) ds+ 1 1 - p u 0 1 - p ( x ) ,0<p<1,
(4.15)
ln u ( x , t ) 0 t g ( s ) ds+ ln u 0 ( x ) ,p=1.
(4.16)

In view of lim t T * u ( x , t ) =, lim t T * G ( t ) =. Noting that u t ≥ 0 by the assumption of the initial function, then we see that g(t) is monotone nondecreasing. Therefore, lim t T * g ( t ) =.

Lemma 4. 6. Under the conditions of Lemma 4.4, then we have

( 1 ) lim t T * u 1 - p ( x , t ) ( 1 - p ) G ( t ) = lim t T * u ( , t ) 1 - p ( 1 - p ) G ( t ) =1,0<p<1,
(4.17)
( 2 ) lim t T * ln u ( x , t ) G ( t ) = lim t T * ln u ( , t ) G ( t ) =1,p=1,
(4.18)

uniformly on any compact subsets of Ω.

Proof. Let λ > 0 be the principal eigenvalue of -Δ in Ω with the null Dirichlet boundary condition, and ϕ(x) be the corresponding eigenfunction satisfying ϕ(x) > 0, Ω ϕ ( x ) dx=1.

In case of (1). Define z1(x, t) = G (t) - u1-p/(1-p), γ 1 ( t ) = Ω z 1 ( y , t ) ϕ ( y ) dy. A direct computation shows

γ 1 ( t ) = Ω ( g ( t ) - u - p ( y , t ) u t ( y , t ) ) ϕ ( y ) d y = - Ω Δ u ( y , t ) ϕ ( y ) d y  =  λ Ω ϕ ( y ) u ( y , t ) d y + Ω u ( ϕ n ) d S λ Ω ( G ( t ) - z 1 ( y , t ) ) 1 ( 1 - p ) ϕ ( y ) d y C 1 ( G 1 ( 1 - p ) ( t ) + Ω ( z 1 - ( y , t ) ) 1 ( 1 - p ) ϕ ( y ) d y ) ,

where z 1 - = max { - z 1 , 0 } and using the equality Ω ( ϕ n ) d S = - λ < 0 . From (4.15), we know that

inf Ω z 1 ( y , t ) - C ,
(4.19)

which means z 1 - C . Then,

γ 1 ( t ) C 2 G 1 ( 1 - p ) ( t ) + C 3 ,
(4.20)

Integrate (4.20) from 0 to t,

γ 1 ( t ) C 4 ( 1 + 0 t G 1 ( 1 - p ) ( s ) d s ) ,
(4.21)

Thus, (4.19) and (4.21) imply

Ω z 1 ( y , t ) ϕ ( y ) dy C 5 ( 1 + 0 t G 1 ( 1 - p ) ( s ) d s ) .

Define K ρ = {y∈ Ω: dist(y, ∂Ω)≥ ρ}. Since -Δz1 ≤ 0 in Ω × (0, T*). Using Lemma 4.5 in [1], we obtain

sup K ρ z 1 ( x , t ) C 6 ρ N + 1 ( 1 + 0 t G 1 ( 1 - p ) ( s ) d s ) .
(4.22)

It follows from (4.22) and (4.15) that

- k 1 G ( t ) 1- u 1 - p ( x , t ) ( 1 - p ) G ( t ) K 1 ρ N + 1 1 + 0 t G 1 ( 1 - p ) ( s ) d s G ( t ) ,
(4.23)

for any xK ρ and t ∈ (0,T*), where k1 and K1 are positive constants.

We know from Theorem 4.3 that

0 t G 1 ( 1 - p ) ( s ) ds C 7 0 t ( T * - s ) - 1 p ds.
(4.24)

In view of (4.15) and Theorem 4.3, it follows that

G ( t ) C 8 u 1 - p C 9 ( T * - t ) - ( 1 - p ) p .
(4.25)

From (4.23)-(4.25), we get

- k 1 G ( t ) 1- u 1 - p ( x , t ) ( 1 - p ) G ( t ) C 10 ρ N + 1 1 + 0 t ( T * - s ) - 1 p d s ( T * - t ) - ( 1 - p ) p .
(4.26)

It is obvious that

lim t T * 0 t ( T * - s ) - 1 p d s ( T * - t ) - ( 1 - p ) p =0.

Thus,

lim t T * u 1 - p ( x , t ) ( 1 - p ) G ( t ) = lim t T * u ( , t ) 1 - p ( 1 - p ) G ( t ) =1.

In case of (2). We define z2 (x, t) = G (t) - ln u(x, t), γ 2 ( t ) = Ω z 2 ( y , t ) ϕ ( y ) dy. Then,

γ 2 ( t ) = Ω ( g ( t ) - u - 1 ( y , t ) u t ( y , t ) ) ϕ ( y ) d y = - Ω Δ u ( y , t ) ϕ ( y ) d y  =  λ Ω ϕ ( y ) u ( y , t ) d y + Ω u ( ϕ n ) d S λ Ω exp { G ( t ) - z 2 ( y , t ) } ϕ ( y ) d y .

From (4.16), we know that

inf Ω z 2 ( y , t ) - C ,
(4.27)

Then,

γ 2 ( t ) C 11 exp { G ( t ) } ,
(4.28)

Integrate (4.28) from 0 to t yields

γ 2 ( t ) C 12 ( 1 + 0 t exp { G ( s ) } d s ) ,
(4.29)

Thus, (4.29) and (4.27) imply

Ω z 2 ( y , t ) ϕ ( y ) dy C 13 ( 1 + 0 t exp { G ( s ) } d s ) .

Define K ζ = {y∈ Ω: dist(y, ∂Ω)≥ ζ}. Since -Δz2 ≤ 0 in Ω × (0, T*), we obtain

sup K ζ z 2 ( x , t ) C 14 ζ N + 1 ( 1 + 0 t exp { G ( s ) } d s ) .
(4.30)

It follows from (4.30) and (4.27) that

- k 2 G ( t ) 1- ln u ( x , t ) G ( t ) K 2 ζ N + 1 1 + 0 t exp { G ( s ) } d s G ( t ) ,
(4.31)

for any xK ζ and t ∈ (0,T*).

By Theorem 4.3, we have

G ( t ) = 0 t g ( s ) d s a 0 t ( Ω U ( s ) d x ) d s a Ω ( ε 1 ) - 1 0 t ( T * - s ) - 1 d s ln ( T * - t ) - 1 + ln T * .
(4.32)

On the other hand, we know form (4.16) and Theorem 4.3 that

G ( t ) C 16 ln ( T * - t ) .
(4.33)

From (4.31)-(4.33), we get

- k 2 G ( t ) 1- ln u ( x , t ) G ( t ) C 17 ζ N + 1 1 + T * 0 t ( T * - s ) - 1 d s ln ( T * - t ) .

It is easy to derive

lim t T * 1 + T * 0 t ( T * - s ) - 1 d s ln ( T * - t ) =0.

Thus,

lim t T * ln u ( x , t ) G ( t ) = lim t T * ln u ( , t ) G ( t ) =1.

This completes the proof.

Proof of Theorem 1.5. Case 1: 0 < p < 1. Form (4.17), we have

u ( x , t ) ~ ( ( 1 - p ) G ( t ) ) 1 ( 1 - p ) as t T * ,

where the notation u ~ v means lim t T * u ( t ) v ( t ) =1.

Furthermore,

G ( t ) =g ( t ) =a Ω udx~a Ω ( ( 1 - p ) G ( t ) ) 1 ( 1 - p ) as t T * .
(4.34)

Integrating (4.34) over (t, T*) yields

G ( t ) ~ ( 1 - p ) - 1 ( a p Ω ( T * - t ) ) - ( 1 - p ) p as  t T * .
(4.35)

So, we can get our conclusion by using (4.17) and (4.35).

Case 2: p = 1. In this case, for any given σ: 0 < σ ≪ 1. By (4.18), there exists 0 < t0 < T* such that

( 1 - σ ) G ( t ) ln u ( x , t ) ( 1 + σ ) G ( t ) ,xΩ,t [ t 0 , T * ) .

Therefore,

a Ω exp { ( 1 - σ ) G ( t ) } G ( t ) =a Ω udxa Ω exp { ( 1 + σ ) G ( t ) } , t 0 t T * .
(4.36)

In view of the right-hand side of the (4.36), we have

exp { - ( 1 + σ ) G ( t ) } dG ( t ) a Ω dt, t 0 t T * .

Integrating the above inequality from t to T* yields that

exp { - ( 1 + σ ) G ( t ) } a ( 1 + σ ) Ω ( T * - t ) , t 0 t T * .

Namely,

G ( t ) ( - 1 ( 1 + σ ) ) ln [ a ( 1 + σ ) Ω ( T * - t ) ] , t 0 t T * .
(4.37)

Similar arguments to the left-hand side of (3.36) yield that

G ( t ) ( - 1 ( 1 - σ ) ) ln [ a ( 1 - σ ) Ω ( T * - t ) ] , t 0 t T * .
(4.38)

Consequently, (4.37) and (4.38) guarantee that for t0tT*,

( - 1 ( 1 + σ ) ) ln [ a ( 1 + σ ) Ω ( T * - t ) ] G ( t ) ( - 1 ( 1 - σ ) ) ln [ a ( 1 - σ ) Ω ( T * - t ) ] .
(4.39)

Letting σ → 0, we have

lim t T * G ( t ) ln ( T * - t ) - 1 =1,
(4.40)

because of lim t T * G ( t ) =. Due to ln u(x, t) ~ G(t) uniformly on any compact subset of Ω, the proof is complete.