Open Access

L estimates of solutions for the quasilinear parabolic equation with nonlinear gradient term and L1 data

Boundary Value Problems20122012:19

DOI: 10.1186/1687-2770-2012-19

Received: 10 August 2011

Accepted: 15 February 2012

Published: 15 February 2012

Abstract

In this article, we study the quasilinear parabolic problem

u t - div ( u m u ) + u u β - 2 u q = u u α - 2 u p + g ( u ) , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω ; u ( x , t ) = 0 , x Ω , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ1_HTML.gif
(0.1)

where Ω is a bounded domain in N , m > 0 and g(u) satisfies |g(u)| ≤ K1|u|1+νwith 0 ≤ ν < m. By the Moser's technique, we prove that if α, β > 1, 0 ≤ p < q, 1 ≤ q < m + 2, p + α < q + β, there exists a weak solution u ( t ) L ( [ 0 , ) , L 1 ) L loc ( ( 0 , ) W 0 1 , m + 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq1_HTML.gif for all u0 L1(Ω). Furthermore, if 2qm + 2, we derive the L estimate for u(t). The asymptotic behavior of global weak solution u(t) for small initial data u0 L2(Ω) also be established if p + α > max{m + 2, q + β}.

2000 Mathematics Subject Classification: 35K20; 35K59; 35K65.

Keywords

quasilinear parabolic equation L estimates asymptotic behavior of solution

1 Introduction

In this article, we are concerned with the initial boundary value problem of the quasilinear parabolic equation with nonlinear gradient term
u t - div ( u m u ) + u u β - 2 u q = u u α - 2 u p + g ( u ) , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω , u ( x , t ) = 0 , x Ω , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ2_HTML.gif
(1.1)

where Ω is a bounded domain in N with smooth boundary Ω and m > 0, α, β > 1, 0 ≤ p < q, 1 ≤ q < m + 2.

Recently, Andreu et al. in [1] considered the following quasilinear parabolic problem
u t - Δ u + u u β - 2 u q = u u α - 2 u p , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω , u ( x , t ) = 0 , x Ω , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ3_HTML.gif
(1.2)
where α, β > 1, 0 ≤ p < q ≤ 2, p + α < q + β and u0 L1(Ω). By the so-called stability theorem with the initial data, they proved that there exists a generalized solution u(t) C([0, T], L1) for (1.2), in which u(t) satisfies A k ( u ) L 2 ( [ 0 , T ] , W 0 1 , 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq2_HTML.gif and
Ω J k ( u ( t ) - ϕ ( t ) ) d x + 0 t Ω ( u A k ( u - ϕ ) + u u β - 2 u q A k ( u - ϕ ) ) d x d s = 0 t Ω ( u u α - 2 u p A k ( u - ϕ ) - A k ( u - ϕ ) ϕ s ) d x d s + Ω J k ( u 0 - ϕ ( 0 ) ) d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ4_HTML.gif
(1.3)
for t [0, T] and ϕ L 2 ( [ 0 , T ] , W 0 1 , 2 ) L ( Q T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq3_HTML.gif, where Q T = Ω × (0, T], and for any k > 0,
A k ( u ) = - k u - k , u - k u k , k u k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ5_HTML.gif
(1.4)

J k (u) is the primitive of A k (u) such that J k (0) = 0. The problem similar to (1.2) has also been extensively considered, see [26] and the references therein. It is an interesting problem to prove the existence of global solution u(t) of (1.2) or (1.1) and to derive the L estimate for u(t) and u(t).

Porzio in [7] also investigated the solution of Leray-Lions type problem
u t = div ( a ( x , t , u , u ) ) , ( x , t ) Ω × ( 0 , + ) , u ( x , 0 ) = u 0 ( x ) , x Ω , u ( x , t ) = 0 , ( x , t ) Ω × ( 0 , + ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ6_HTML.gif
(1.5)
where a(x, t, s, ξ ) is a Carathé odory function satisfying the following structure condition
a ( x , t , s , ξ ) ξ θ ξ m , f o r ( x , t , s , ξ ) Ω × + × 1 × N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ7_HTML.gif
(1.6)
with θ > 0 and u0 L q (Ω), q ≥ 1. By the integral inequalities method, Porzio derived the L decay estimate of the form
u ( t ) L ( Ω ) C u 0 L q ( Ω ) α t - λ , t > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ8_HTML.gif
(1.7)

with C = C(N, q, m, θ), α = mq(N(m - 2) + mq)-1, λ = N(N(m - 2) + mq)-1.

In this article, we will consider the global existence of solution u(t) of (1.1) with u0 L1(Ω) and give the L estimates for u(t) under the similar condition in [1]. More specially, we will study the behavior of solution u(t) as t → 0+. Obviously, if m = 0 and g ≡ 0, problem (1.1) is reduced to (1.2). We remark that the methods used in our article are different from that of [1]. In L estimates, we use an improved Morser's technique as in [810]. Since the equation in (1.1) contains the nonlinear gradient term u|u|α-2|u| p and u|u|β-2|u| q , it is difficult to derive L estimates for u(t) and u(t).

This article is organized as follows. In Section 2, we state the main results and present some Lemmas which will be used later. In Section 3, we use these Lemmas to derive L estimates of u(t). Also the proof of the main results will be given in Section 3. The L estimates of u(t) are considered in Section 4. The asymptotic behavior of solution for the small initial data u0(x) is investigated in Section 5.

2 Preliminaries and main results

Let Ω be a bounded domain in N with smooth boundary Ω and · r , ·1,rdenote the Sobolev space L r (Ω) and W1,r(Ω) norms, respectively, 1 ≤ r ≤ ∞. We often drop the letter Ω in these notations.

Let us state our precise assumptions on the parameters p, q, α, β and the function g(u).

(H1) the parameters α, β > 1, 0 ≤ p < q < m + 2 < N, p + α < q + β and q(α - 1) ≥ p(β - 1),

(H2) the function g(u) C1 and K1 ≥ 0 and 0 ≤ ν < max{q + β - 2, m], such that
g ( u ) K 1 u 1 + ν , u 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equa_HTML.gif

(H3) the initial data u0 L1(Ω),

(H4) 2q ≤ 2 + m, α, β < 2 + m(1 + 1/N)/2,

(H5) the mean curvature of H(x) of Ω at x is non-positive with respect to the outward normal.

Remark 2.1 The assumptions (H1) and (H3) are similar to as in [1].

Definition 2.2 A measurable function u(t) = u(x, t) on Ω × [0, ∞) is said to be a global weak solution of the problem (1.1) if u(t) is in the class
C ( [ 0 , ) , L 1 ) L loc ( ( 0 , ) , W 0 1 , m + 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equb_HTML.gif
and u u β - 2 u q , u u α - 2 u p L loc 1 ( [ 0 , ) × Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq4_HTML.gif, and for any ϕ = ϕ ( x , t ) C 1 ( [ 0 , ) , C 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq5_HTML.gif the equality
0 T Ω - u ϕ t + u m u ϕ + u u β - 2 u q ϕ d x d t = Ω ( u 0 ( x ) ϕ ( x , 0 ) - u ( x , T ) ϕ ( x , T ) ) d x + 0 T Ω ( u u α - 2 u p + g ( u ) ) ϕ d x d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ9_HTML.gif
(2.1)

is valid for any T > 0.

Remark 2.3 In [1], the concept of generalized solution for (1.2) was introduced. A similar concept can be found in [7, 11]. By the definition, we know that weak solution is the generalized solution. Conversely, a generalized solution is not necessarily weak solution.

Our main results read as follows.

Theorem 2.4 Assume (H1)-(H3). Then the problem (1.1) admits a global weak solution u(t) which satisfies
u ( t ) L ( [ 0 , ) , L 1 ) C ( [ 0 , ) , L 1 ) L loc ( ( 0 , ) , W 0 1 , m + 2 ) , u t L loc 2 ( ( 0 , ) , L 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ10_HTML.gif
(2.2)
and the estimates
u ( t ) C 0 t - λ , 0 < t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ11_HTML.gif
(2.3)
Furthermore, if (H4) is satisfied, the solution u(t) has the following estimates
0 T s 1 + r u t ( s ) 2 2 d s C 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ12_HTML.gif
(2.4)
u ( t ) m + 2 C 0 t - ( 1 + λ ) / ( m + 2 ) , 0 < t T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ13_HTML.gif
(2.5)

with r > λ = N(mN + m + 2)-1 and C0 = C0(T, u01).

Theorem 2.5 Assume (H1)-(H5). Then the solution u(t) of (1.1) has the following L gradient estimate
u ( t ) C 0 t - σ , 0 < t T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ14_HTML.gif
(2.6)

with σ = (2 + 2λ + N)(mN + 2m + 4)-1 and C0 = C0(T, u01).

Remark 2.6 The estimates (2.3) and (2.6) give the behavior of u(t) and u(t) as

Theorem 2.7 Assume the parameters α, β > 1, γ ≥ 0, 0 ≤ q < m + 2 < N and p < m + 2 < p + α, α ≤ (m + 2 - p)(1 + 2N-1).

Then, d0 > 0, such that u0 L2(Ω) with u02 < d0, the initial boundary value problem
u t - div ( u m u ) + γ u u β - 2 u q = u α - 2 u u p , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , x Ω , u ( x , t ) = 0 , x Ω , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ15_HTML.gif
(2.7)
admits a solution u ( t ) L ( [ 0 , ) , L 2 ) W 0 1 , m + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq6_HTML.gif, which satisfies
u ( t ) 2 C ( 1 + t ) - 1 / m , t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ16_HTML.gif
(2.8)

where C = C(u02).

Theorem 2.8 Assume the parameters γ > 0, α, β > 1, 1 ≤ p < q < m + 2 < N and τ = N(μ - q)(q + β) ≤ 2(q2 + ) with μ = ( - )/(q - p) > q + β.

Then, d0 > 0, such that u0 L2 with u02 < d0, the initial boundary value problem
u t - div ( u m u ) + γ u u β - 2 u q = u α - 2 u u p , x Ω , t > 0 u ( x , 0 ) = u 0 ( x ) , x Ω ; u ( x , t ) = 0 , x Ω , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ17_HTML.gif
(2.9)
admits a solution u ( t ) L ( [ 0 , ) , L 2 ) W 0 1 , m + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq6_HTML.gif which satisfies
u ( t ) 2 C ( 1 + t ) - 1 / ( q + β - 2 ) , t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ18_HTML.gif
(2.10)

where C = C(u02).

To obtain the above results, we will need the following Lemmas.

Lemma 2.9 (Gagliardo-Nirenberg type inequality) Let β ≥ 0, N > p ≥ 1, q ≥ 1 + β and 1 ≤ rqpN(1 + β)/(N - p). Then for |u| β u W1,p(Ω), we have
u q C 0 1 / ( β + 1 ) u r 1 - θ u β u 1 , p θ / ( β + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equc_HTML.gif

with θ = (1 + β)(r-1 - q-1)/(N-1 - p-1 + (1 + β)r-1), where the constant C0 depends only on p, N.

The Proof of Lemma 2.9 can be obtained from the well-known Gagliardo-Nirenberg-Sobolev inequality and the interpolation inequality and is omitted here.

Lemma 2.10 [10] Let y(t) be a nonnegative differentiable function on (0, T] satisfying
y ( t ) + A t λ θ - 1 y 1 + θ ( t ) B t - k y ( t ) + C t - δ , 0 < t T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equd_HTML.gif
with A, θ > 0, λθ ≥ 1, B, C ≥ 0, k ≤ 1. Then, we have
y ( t ) A - 1 / θ ( 2 λ + 2 B T 1 - k ) 1 / θ t - λ + 2 C ( λ + B T 1 - k ) - 1 t 1 - δ , 0 < t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Eque_HTML.gif

3 Lestimate for u(t)

In this section, we derive a priori estimates of the assumed solutions u(t) and give a proof of Theorem 2.4. The solutions are in fact given as limits of smooth solutions of appropriate approximate equations and we may assume for our estimates that the solutions under consideration are sufficiently smooth.

Let u 0 , i C 0 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq7_HTML.gif and u0,iu0 in L1(Ω) as i → ∞. For i = 1, 2, ..., we consider the approximate problem of (1.1)
u t - div ( u 2 + i - 1 ) m 2 u + u u β - 2 u q = u u α - 2 u p + g ( u ) , x Ω , t > 0 , u ( x , 0 ) = u 0 , i ( x ) , x Ω , u ( x , t ) = 0 , x Ω , t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ19_HTML.gif
(3.1)

The problem (3.1) is a standard quasilinear parabolic equation and admits a unique smooth solution u i (t)(see Chapter 6 in [12]). We will derive estimates for u i (t). For the simplicity of notation, we write u instead of u i and u k for |u|k- 1u where k > 0. Also, let C, C j be generic constants independent of k, i, n changeable from line to line.

Lemma 3.1 Let (H1)-(H3) hold. Suppose that u(t) is the solution of (3.1), then u(t) L([0, ∞), L1).

Proof Let n = 1, 2, ..., and
f n ( s ) = 1 , 1 n s n s ( 2 - n s ) , 0 s 1 n - n s ( 2 + n s ) , - 1 n s 0 - 1 , s < - 1 n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equf_HTML.gif

It is obvious that f n (s) is odd and continuously differentiable in 1. Furthermore, f n ( s ) 1 , f n ( s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq8_HTML.gif and f n (s) → sign(s) uniformly in 1.

Multiplying the equation in (3.1) by f n (u) and integrating on Ω, we get
Ω f n ( u ) u t d x + Ω u m + 2 f n ( u ) d x + Ω u u β - 2 f n ( u ) u q d x Ω u u α - 2 f n ( u ) u p d x + Ω u u β - 2 f n ( u ) u q d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ20_HTML.gif
(3.2)
and the application of the Young inequality gives
Ω u u α - 2 f n ( u ) u p d x 1 4 Ω u u β - 2 f n ( u ) u q d x + C 1 Ω u μ - 1 d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ21_HTML.gif
(3.3)

where μ = ( - )(q - p)-1 ≥ 1, i.e q(α - 1) ≥ p(β - 1).

In order to get the estimate for the third term of left-hand side in (3.2), we denote
F n ( u ) = 0 u ( s s β - 2 f n ( s ) ) 1 / q d s , u 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equg_HTML.gif
It is easy to verify that F n (u) is odd in 1. Then, we obtain from the Sobolev inequality that
1 4 Ω u u β - 2 f n ( u ) u q d x = 1 4 Ω F n ( u ) q d x λ 0 Ω F n ( u ) q d x = λ 0 Ω n F n ( u ) q d x + λ 0 Ω n c F n ( u ) q d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ22_HTML.gif
(3.4)
with some λ0 > 0 and
Ω n = { x Ω | u ( x , t ) n - 1 } , Ω n c = Ω \ Ω n , n = 1 , 2 , . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equh_HTML.gif
We note that |F n (u)| q n-(q+β- 1)in Ω n c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq9_HTML.gif and
Ω n c F n ( u ) q d x n - ( q + β - 1 ) Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equi_HTML.gif
On the other hand, we have |u(x, t)| ≥ n-1 in Ω n and
F n ( u ) n - 1 u ( s s β - 2 f n ( s ) ) 1 / q d s q q + β - 1 u q + β - 1 q - n - q + β - 1 q in Ω n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equj_HTML.gif
This implies that there exists λ1 > 0, such that
λ 0 Ω n F n ( u ) q d x λ 1 Ω n u q + β - 1 d x - λ 1 Ω n - ( q + β - 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ23_HTML.gif
(3.5)
Then it follows from (3.4)-(3.5) that
1 4 Ω u u β - 2 f n ( u ) u q d x λ 1 Ω u q + β - 1 d x - C 2 n - ( q + β - 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ24_HTML.gif
(3.6)

with some C2 > 0.

Similarly, we have from the assumption (H2) and the Young inequality that
Ω g ( u ) f n ( u ) d x K 1 Ω u 1 + ν f n ( u ) d x K 1 Ω u 1 + ν d x λ 1 2 Ω n u q + β - 1 d x + C 2 ( 1 + n - 1 - ν ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ25_HTML.gif
(3.7)
Furthermore, the assumption μ < q + β implies that
C 1 Ω n u μ - 1 d x λ 1 2 Ω n u q + β - 1 d x + C 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ26_HTML.gif
(3.8)
Then (3.2)-(3.3) and (3.6)-(3.8) give that
Ω f n ( u ) u t d x + 1 2 Ω u u β - 2 f n ( u ) u q d x C 3 1 + n - 1 - ν + n - ( q + β - 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ27_HTML.gif
(3.9)
Letting n → ∞ in (3.9) yields
d d t u ( t ) 1 + 1 2 Ω u β - 1 u q d x C 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ28_HTML.gif
(3.10)
Note that
Ω u β - 1 u q d x = q q + β - 1 q Ω u 1 + β - 1 q q d x 2 λ 2 u 1 q + β - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equk_HTML.gif
with some λ2 > 0. Then (3.10) becomes
d d t u ( t ) 1 + λ 2 u ( t ) 1 q + β - 1 C 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ29_HTML.gif
(3.11)

This gives that u(t) L([0, ∞), L1) if u0 L1.

Remark 3.2 The differential inequality (3.10) implies that the solution u i (t) of (3.1) satisfies
0 T Ω u i β - 1 u i q d x d t C 0 for i = 1 , 2 , . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ30_HTML.gif
(3.12)

withC0 = C0(T, u01).

Lemma 3.3 Assume (H1)-(H4). Then, for any T > 0, the solution u(t) of (3.1) also satisfies the following estimates:
u ( t ) C 0 t - λ , 0 < t T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ31_HTML.gif
(3.13)

where λ = N(mN + m + 2)-1, C0 = C0(T, u01).

Proof Multiplying the equation in (3.1) by uk-1, k ≥ 2, we have
1 k d d t u ( t ) k k + ( k - 1 ) m + 2 k + 2 m + 2 u k + m m + 2 m + 2 m + 2 + Ω u β + k - 2 u q d x Ω u α + k - 2 u p d x + K 1 Ω u ν + k d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ32_HTML.gif
(3.14)
It follows from the Hö lder and Sobolev inequalities that
K 1 Ω u ν + k d x C u k θ 1 u 1 θ 2 u s θ 3 C u k θ 1 u k + m m + 2 m + 2 ( m + 2 ) θ 3 k + m k - 1 2 m + 2 k + 2 m + 2 u k + m m + 2 m + 2 m + 2 + C k σ u k k , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equl_HTML.gif

in which θ1 = k λ(m - ν + (m + 2)N-1), θ2 = ν λ(m + 2)N-1, θ3 = ν λ(k + m), σ = ν λ, s = N(k + m)(N - m - 2)-1.

Note that
Ω u α + k - 2 u p d x 1 4 Ω u β + k - 2 u q d x + C Ω u μ + k - 2 d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equm_HTML.gif
and
1 2 Ω u β + k - 2 u q d x C 1 k - q Ω u q + β + k - 2 q q d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equn_HTML.gif

with some C1 independent of k and μ = ( - )(q - p)-1 < q + β.

Without loss of generality, we assume k > 3 - μ. Similarly, we derive
C Ω u μ + k - 2 d x C u k - 2 μ 1 u 1 μ 2 u k * μ 3 C ξ 1 μ 2 u k μ 1 u k * μ 3 C u k μ 1 u q k / q q q μ 3 / q k A k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equo_HTML.gif
with ξ1 = supt≥0u(t)1 and
μ 1 = λ 0 ( k - 2 ) ( q + β - μ + q N - 1 ) , μ 2 = λ 0 μ q N - 1 , μ 3 = λ 0 μ q k , λ 0 = ( q + β + q / N ) - 1 , k * = q k N ( N - q ) - 1 , q k = q + β + k - 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equp_HTML.gif
Then, for any η > 0,
A k C η u q k / q q q + C η - θ / θ u k μ 1 θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ33_HTML.gif
(3.15)

with μ λ0θ = 1, (1 - μ λ0)θ' = 1.

Note that μ1θ' < k. Let η = C 1 2 C k - q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq10_HTML.gif. Then it follows from (3.15) that
A k C 1 2 k - q u q k / q q q + C k γ ( u k k + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ34_HTML.gif
(3.16)
with γ = qθ'θ-1 = λ0/(1 - μ λ0). Then, (3.14) becomes
1 k d d t u k k + k - 1 2 m + 2 k + 2 m + 2 u k + m m + 2 m + 2 m + 2 + C 1 2 k - q u q k / q q q C k σ 0 ( u k k + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equq_HTML.gif
or
d d t u k k + C 1 k - m u k + m m + 2 m + 2 m + 2 C k 1 + σ 0 ( u k k + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ35_HTML.gif
(3.17)

with σ0 = max{σ, γ} = max{ν λ, γ}.

Now we employ an improved Moser's technique as in [8, 9]. Let {k n } be a sequence defined by k1 = 1, kn = Rn-2(R - m - 1) + m(R - 1)-1(n = 2, 3, ...) with R > max{m + 1, m + 4 - μ} such that k n ≥ 3 - μ(n ≥ 2). Obviously, k n → ∞ as n → ∞.

By Lemma 2.9, we have
u ( t ) k n C 0 m + 2 m + k n u ( t ) k n - 1 1 - θ n u m + k n m + 2 m + 2 θ n ( m + 2 ) m + k n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ36_HTML.gif
(3.18)

with θ n = R N ( 1 - k n - 1 k n - 1 ) ( m + 2 + N ( R - 1 ) ) - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq11_HTML.gif.

Then, inserting (3.18) into (3.17) (k = k n ), we find that
d d t u ( t ) k n k n + C 1 C 0 - m + 2 θ n k n - m u ( t ) k n - 1 ( 1 - 1 / θ n ) ( m + k n ) u ( t ) k n ( m + k n ) / θ n C k n 1 + σ 0 ( u ( t ) k n k n + 1 ) , 0 < t T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ37_HTML.gif
(3.19)
or
d d t u ( t ) k n k n + C 1 C 0 - m + 2 θ n k n - m u ( t ) k n - 1 m - β n u ( t ) k n k n + β n C k n 1 + σ 0 ( u ( t ) k n k n + 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ38_HTML.gif
(3.20)
where β n = ( m + k n ) θ n - 1 - k n , n = 2 , 3 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq12_HTML.gif. It is easy to see that
θ n θ 0 = N ( R - 1 ) m + 2 + N ( R - 1 ) , β n k n - 1 m + 2 N ( R - 1 ) , as n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equr_HTML.gif
Denote
y n ( t ) = u ( t ) k n k n , 0 < t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equs_HTML.gif
Then (3.20) can be rewritten as follows
y n ( t ) + C 1 C - m + 2 θ n k n - m ( y n ( t ) ) 1 + β n / k n u ( t ) k n - 1 m - β n C k n 1 + σ 0 ( y n ( t ) + 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ39_HTML.gif
(3.21)
We claim that there exist a bounded sequence {ξ n } and a convergent sequence {λ n }, such that
u ( t ) k n ξ n t - λ n , 0 < t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ40_HTML.gif
(3.22)
Indeed, by Lemma 3.1, the estimate (3.22) holds for n = 1 if we take λ1 = 0, ξ1 = supt≥0u(t)1. If (3.22) is true for n - 1, then we have from (3.21) and (3.22) that
y n ( t ) + C 1 C - m + 2 θ n k n - m ( ξ n - 1 ) m - β n t Λ n τ n - 1 y n 1 + τ n ( t ) C k n 1 + σ 0 ( y n ( t ) + 1 ) , 0 t T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ41_HTML.gif
(3.23)
where
τ n = β n k n , Λ n = k n λ n , λ n = 1 + λ n - 1 ( β n - m ) β n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equt_HTML.gif
Applying Lemma 2.10 to (3.23), we have
y n ( t ) C 1 C - m + 2 θ n k n - m ξ n - 1 m - β n - 1 / τ n ( 2 k n λ n + 2 C T k n 1 + σ 0 ) 1 / τ n t - k n λ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ42_HTML.gif
(3.24)
This implies that for t (0, T),
u ( t ) k n C 1 C - m + 2 θ n k n - m ξ n - 1 m - β n - 1 / β n ( 2 k n λ n + 2 C T k n 1 + σ 0 ) 1 / β n t - λ n ξ n t - λ n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ43_HTML.gif
(3.25)
where
ξ n = ξ n - 1 C 1 C - m + 2 θ n k n - m - 1 / β n ( 2 k n λ n + 4 C T k n 1 + σ 0 ) 1 / β n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ44_HTML.gif
(3.26)

in which the fact k n ~ β n as n → ∞ has been used.

It is not difficult to show that {ξ n } is bounded. Furthermore, by Lemma 4 in [9], we have
1 + λ n - 1 ( β n - m ) β n λ = N m + 2 + m N , as n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equu_HTML.gif

Letting n → ∞ in (3.22) implies that (3.13) and we finish the Proof of Lemma 3.3.

Lemma 3.4. Let (H1)-(H4) hold. Then, the solution u(t) of (3.1) has the following estimates
0 T s 1 + r u t ( s ) 2 2 d s C 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ45_HTML.gif
(3.27)
and
u ( t ) m + 2 C 0 t - ( 1 + λ ) / ( m + 2 ) , 0 < t T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ46_HTML.gif
(3.28)

with r > λ = N(mN + m + 2)-1, C0 = C0(T, u01).

Proof We first choose r > λ and η(t) C[0, ∞) C1(0, ∞) such that η(t) = t r when t [0, 1]; η(t) = 2, when t ≥ 2 and η(t), η'(t) ≥ 0 in [0, ∞). Multiplying the equation in (3.1) by η(t)u, we have
1 2 η ( t ) u ( t ) 2 2 + 0 t η ( s ) u ( s ) m + 2 m + 2 d s + 0 t Ω u β u q η ( s ) d x d s 1 2 0 t η ( s ) u ( s ) 2 2 d s + 0 t Ω u α u p η ( s ) d x d s + K 1 0 t Ω u 2 + ν η ( s ) d x d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ47_HTML.gif
(3.29)
Note that
0 t Ω u α u p η ( s ) d x d s 1 2 0 t Ω u β u q η ( s ) d x d s + C 0 t Ω u μ η ( s ) d x d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equv_HTML.gif
Hence, we have
1 2 η ( t ) u ( t ) 2 2 + 0 t η ( s ) u ( s ) m + 2 m + 2 d s + 1 2 0 t Ω u β u q η ( s ) d x d s 1 2 0 t η ( s ) u ( s ) 2 2 d s + C 0 t Ω u μ η ( s ) d x d s + K 1 0 t Ω u 2 + ν η ( s ) d x d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ48_HTML.gif
(3.30)
By Lemma 3.1 and the estimate (3.13), we get
0 t η ( s ) u ( s ) 2 2 d s C 0 t s r - 1 u ( t ) 1 u ( t ) d s C t r - λ , 0 t < T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ49_HTML.gif
(3.31)
Since μ < q + β, we have from Sobolev inequality that
C 0 t Ω u μ η ( s ) d x d s 1 4 0 t Ω u β u q η ( s ) d x d s + C 0 t η ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ50_HTML.gif
(3.32)
Similarly, we have from 2 + ν < q + β that
K 1 0 t Ω u 2 + ν η ( s ) d x d s 1 4 0 t Ω u β u q η ( s ) d x d s + C 0 t η ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ51_HTML.gif
(3.33)
Therefore, it follows from (3.30)-(3.33) that
0 t Ω u m + 2 η ( s ) d x d s C t r - λ , 0 t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ52_HTML.gif
(3.34)
Next, let G ( u ) = 0 u g ( s ) d s , u 1 , ρ ( t ) = 0 t η ( s ) d s , t ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq13_HTML.gif. Furthermore, multiplying the equation in (3.1) by ρ(t)u t yields
ρ ( t ) u t ( t ) 2 2 + 1 m + 2 d d t Ω ρ ( t ) ( u 2 + i - 1 ) m + 2 2 d x + ρ ( t ) Ω G ( u ) d x ρ ( t ) m + 2 Ω ( u 2 + i - 1 ) m + 2 2 d x + d d t Ω ρ ( t ) G ( u ) d x + Ω ρ ( t ) u β - 1 u t u q d x + Ω ρ ( t ) u α - 1 u t u p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ53_HTML.gif
(3.35)
By the assumption p < q and the Cauchy inequality, we deduce
Ω u β - 1 u t u q d x 1 4 u t ( t ) 2 2 + C Ω u 2 ( β - 1 ) u 2 q d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ54_HTML.gif
(3.36)
and
Ω u α - 1 u t u p d x 1 4 u t ( t ) 2 2 + C Ω u 2 ( α - 1 ) u 2 p d x 1 4 u t ( t ) 2 2 + C Ω u 2 ( β - 1 ) u 2 q d x + C Ω u 2 ( μ - 1 ) d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ55_HTML.gif
(3.37)
and
Ω G ( u ) d x C 1 Ω u 2 + ν d x C h 2 + ν ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ56_HTML.gif
(3.38)

with h(t) = u(t).

Now, it follows from (H4) and (3.35)-(3.38) that
1 2 0 t ρ ( s ) u t ( s ) 2 2 d s + 1 m + 2 ρ ( t ) u ( t ) m + 2 m + 2 1 2 0 t η ( s ) u ( s ) m + 2 m + 2 d s + C ρ ( t ) h 2 + ν ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ57_HTML.gif
(3.39)
or
1 2 0 t ρ ( s ) u t ( s ) 2 2 d s + ρ ( t ) m + 2 u ( t ) m + 2 m + 2 C 0 t r - λ + C 0 0 t ρ ( s ) h 2 ( β - 1 ) ( s ) u ( s ) m + 2 m + 2 d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ58_HTML.gif
(3.40)

where C0 = C0(T, u01) and the fact 2 + λ ≥ 2(μ - 1)λ has been used.

Since the function h2(β-1)(t) L1([0, T]), the application of the Gronwall inequality to (3.40) gives
0 t ρ ( s ) u t ( s ) 2 2 d s + ρ ( t ) u m + 2 m + 2 C 0 t r - λ , 0 < t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ59_HTML.gif
(3.41)
Hence,
u m + 2 C 0 t - ( 1 + λ ) / ( m + 2 ) , 0 < t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ60_HTML.gif
(3.42)

and the Proof of Lemma 3.4 is completed.

Proof of Theorem 2.4 Noticing that the estimate constant C0 in (3.12)-(3.13) and (3.27)-(3.28) is independent of i, we have from the standard compact argument as in [1, 13, 14] that there exists a subsequence (still denoted by u i ) and a function u L s ( [ 0 , T ] , W 0 1 , s ( Ω ) ) , ( 1 s m + 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq14_HTML.gif satisfying
u i u weakly in L s ( [ 0 , T ] , W 0 1 , s ( Ω ) ) , u i u in L s ( Q T ) and a .e . in Q T , u i β - 1 u i q u β - 1 u q in L 1 ( Q T ) , u i α - 1 u i p u α - 1 u p in L 1 ( Q T ) , u i u in C ( [ 0 , T ] ; L 1 ( Ω ) ) , u i t u t weakly in L loc 2 ( 0 , T ; L 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ61_HTML.gif
(3.43)
Since A i ( u i ) = - div ( ( u i 2 + i - 1 ) m 2 u i ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq15_HTML.gif is bounded in ( W 0 1 , m + 2 ) * = W 0 - 1 , m + 2 m + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq16_HTML.gif, we see further that
A i ( u i ) χ weakly * in L l o c ( 0 , T ; ( W 0 1 , m + 2 ) * ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ62_HTML.gif
(3.44)

for some χ L loc ( 0 , T ] , ( W 0 1 , m + 2 ) * ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq17_HTML.gif. As the Proof of Theorem 1 in [9], we have χ = A(u) = -div((u m u).

Then, the function u is a global weak solution of (1.1). Furthermore, it follows from Lemma 3.4 that u(t) satisfies the estimate (2.4)-(2.5). The Proof of Theorem 2.4 is now completed.

4 Lestimate for u(t)

In this section, we use an argument similar to that in [9, 10, 15] and give the Proof of Theorem 2.5. Hence, we only consider the estimate of u for the smooth solution u(t) of (3.1). As above, let C, C j be the generic constants independent of k and i. Denote
D 2 u 2 = i , j = 1 N u i j 2 , u i j = 2 u x i x j . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equw_HTML.gif
Multiplying (3.1) by -div(|u|k-2u), km + 2 and integrating by parts, we have
1 k d d t u ( t ) k k + Ω u k + m - 2 D 2 u 2 d x + k - 2 4 Ω u k + m - 4 ( u 2 ) 2 d x - ( N - 1 ) Ω H ( x ) u k + m d S = Ω u u β - 2 u q div ( u k - 2 u ) d x - Ω u u p u α - 2 div ( u k - 2 u ) d x + Ω g ( u ) div ( u k - 2 u ) d x I + I I + I I I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ63_HTML.gif
(4.1)
Since
div ( u k - 2 u ) = u k - 2 Δ u + k - 2 2 u k - 4 u ( u 2 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ64_HTML.gif
(4.2)
we have
| div( | u | k 2 u ) | ( k 1 ) | u | k 2 | D 2 u | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ65_HTML.gif
(4.3)
and
I ( k - 1 ) Ω u β - 1 u q + k - 2 D 2 u d x = ( k - 1 ) Ω u k + m - 2 2 D 2 u u k + 2 q - m - 2 2 u β - 1 d x 1 4 Ω u k + m - 2 D 2 u 2 d x + C 0 k 2 Ω u k + 2 q - m - 2 u 2 ( β - 1 ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ66_HTML.gif
(4.4)
Similarly, we obtain the following estimates
I I 1 4 Ω u m + k - 2 D 2 u 2 d x + C 0 k 2 Ω u k + 2 p - m - 2 u 2 ( α - 1 ) d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ67_HTML.gif
(4.5)
and
I I I = Ω g ( u ) div ( u k - 2 u ) d x = - Ω g ( u ) u k d x K 1 Ω y ν u k d x C h ν ( t ) u ( t ) k k , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ68_HTML.gif
(4.6)

where h(t) = u(t)Ct.

Moreover, we assume that 2qm + 2, 2pm + 2, then (4.1) becomes
1 k d d t u k k + 1 2 Ω u k + m - 2 D 2 u 2 d x + k - 2 4 Ω u k + m - 4 ( u 2 ) 2 d x - ( N - 1 ) Ω H ( x ) u k + m d S C 0 k 2 Ω u k + 2 q - m - 2 u 2 ( β - 1 ) + u k + 2 p - m - 2 u 2 ( α - 1 ) d x + C h ν ( t ) u ( t ) k k C 0 k 2 h 1 ( t ) 1 + u ( t ) k k , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ69_HTML.gif
(4.7)

where h1(t) = max{h2(α-1)(t), h2(β-1)(t), h ν (t)}. Since α , β < 2 + m 2 1 + 1 N , ν < m + 2 + m N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq18_HTML.gif, we get h1(t) L1([0.T]) for any T > 0.

If H(x) ≤ 0 on Ω and N > 1, then by an argument of elliptic eigenvalue problem in [15], there exists λ1 > 0, such that
v 2 2 - ( N - 1 ) Ω v 2 H ( x ) d S λ 1 v 1 , 2 2 , v W 1 , 2 ( Ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ70_HTML.gif
(4.8)
Hence, by (4.7) and (4.8), we see that there exists C1 and C2 such that
d d t u ( t ) k k + C 1 u ( t ) k + m 2 1 , 2 2 C k 3 h 1 ( t ) ( 1 + u ( t ) k k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ71_HTML.gif
(4.9)
Let k1 = m + 2, R > m + 1, k n = Rn-2(R-1-m) + m (R-1)-1, θ n = R N ( 1 - k n - 1 k n - 1 ) ( R ( N - 1 ) + 2 ) - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq19_HTML.gif, n = 2, 3,.... Then, the application of Lemma 2.9 gives
u k n C 2 k n + m u k n - 1 1 - θ n u k n + m 2 1 , 2 2 θ n k n + m . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ72_HTML.gif
(4.10)
Inserting this into (4.9)(k = k n ), we get
d d t u k n k n + C 1 C - 2 / θ n u ( t ) k n - 1 ( k n + m ) ( 1 - 1 / θ n ) u ( t ) k n ( k n + m ) / θ n C 2 k n 3 h 1 ( t ) ( 1 + u ( t ) k n k n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ73_HTML.gif
(4.11)
By (3.28), we take y1 = max{1, C0}, z1 = (1 + λ)/(m + 2). As the Proof of Lemma 3.3, we can show that there exist bounded sequences y n and z n such that
u ( t ) k n y n t - z n , 0 < t T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ74_HTML.gif
(4.12)

in which z n σ = (2 + 2λ + N)(mN + 2m + 4)-1. Letting n → ∞ in (4.12), we have the estimate (2.6). This completes the Proof of Theorem 2.5.

5 Asymptotic behavior of solution

In this section, we will prove that the problem (1.1) admits a global solution if the initial data u0(x) is small under the assumptions of Theorems 2.7 and 2.8. Also, we derive the asymptotic behavior of solution u(t).

Proof of Theorem 2.7 The existence of solution for (1.1) in small u0 can be obtained by a similar argument as the Proof of Theorem 2.4. So, it is sufficient to derive the estimate (2.8).

Multiplying the equation in (2.7) by u and integrating over Ω, we obtain
1 2 d d t u ( t ) 2 2 + C 1 u ( t ) m + 2 m + 2 Ω u α u p d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ75_HTML.gif
(5.1)

with C 1 = m + 2 4 m + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq20_HTML.gif.

Since p < m + 2 < p + α, it follows from Lemma 2.9 that
Ω u α u p d x u ( t ) m + 2 p u s α C 0 u m + 2 p u r α ( 1 - θ ) u m + 2 α θ C 0 u ( t ) m + 2 m + 2 u ( t ) r p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ76_HTML.gif
(5.2)
with
s = α ( m + 2 ) m + 2 - p , θ = 1 r - 1 s 1 N + 1 r - 1 m + 2 - 1 , r = N p 1 m + 2 - p , p 1 = p + α - m - 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equx_HTML.gif
The assumption on α shows that r ≤ 2. Then, (5.1) can be rewritten as
1 2 d d t u ( t ) 2 2 + u ( t ) m + 2 m + 2 ( C 1 - C 0 u ( t ) 2 p 1 ) 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ77_HTML.gif
(5.3)
By the Sobolev embedding theorem,
u ( t ) m + 2 m + 2 C 2 u ( t ) m + 2 m + 2 C 2 u ( t ) 2 m + 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ78_HTML.gif
(5.4)
we obtain from (5.3) and (5.4) that d0 > 0, λ0 > 0, such that u02 < d0 and
ϕ ( t ) + λ 0 ϕ 1 + m / 2 ( t ) 0 , t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ79_HTML.gif
(5.5)
with ϕ ( t ) = u ( t ) 2 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq21_HTML.gif. This implies that
u ( t ) 2 C ( 1 + t ) - 1 / m , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ80_HTML.gif
(5.6)

where the constant C depends only u02. This completes the Proof of Theorem 2.7.

Proof of Theorem 2.8 Multiplying the equation in (2.9) by u and integrating over Ω, we obtain
1 2 d d t u ( t ) 2 2 + γ Ω u β u q d x Ω u α u p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ81_HTML.gif
(5.7)
Since p < q, q + β < p + α, it follows from the Hö lder inequality that
Ω u α u p d x u 1 + β / q q p u μ μ ( 1 - p / q ) C 1 u 1 + β / q q p u τ μ 1 ( 1 - p / q ) u 1 + β / q q μ 2 ( 1 - p / q ) C 1 u 1 + β / q q q u τ μ 1 ( 1 - p / q ) C 1 u 1 + β / q q q u 2 μ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ82_HTML.gif
(5.8)

with μ2 = q, μ1 = μ - q, μ3 = μ1(1 - p/q) and τ = N(μ- q)(q + β)(q2 + )-1 ≤ 2.

Then (5.7) becomes
d d t u ( t ) 2 2 + u 1 + β / q q q ( C 0 - C 1 u ( t ) 2 μ 3 ) 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ83_HTML.gif
(5.9)
This implies that d0 > 0, λ1 > 0, such that u02 < d0 and
ϕ ( t ) + λ 1 ϕ ( q + β ) / 2 ( t ) 0 , t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ84_HTML.gif
(5.10)
with ϕ ( t ) = u ( t ) 2 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_IEq21_HTML.gif. This implies that
u ( t ) 2 C ( 1 + t ) - 1 / ( q + β - 2 ) , t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-19/MediaObjects/13661_2011_Article_120_Equ85_HTML.gif
(5.11)

This is the estimate (2.10) and we finish the Proof of Theorem 2.8.

Declarations

Acknowledgements

The authors wish to express their gratitude to the referees for useful comments and suggestions.

Authors’ Affiliations

(1)
College of Science, Hohai University

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