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

Abstract

In this article, we study the quasilinear parabolic problem

$$\left\{\begin{array}{c}{u}_t-\mathsf{\text{div}}\left({\left|\nabla u\right|}^m\nabla u\right)+u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q=u{\left|u\right|}^{\alpha -2}{\left|\nabla u\right|}^p+g\left(u\right),x\in \Omega ,t>0,\hfill \\ u\left(x,0\right)=u_0\left(x\right),x\in \Omega ;u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0,\hfill \end{array}\right.$$

(0.1)

where Ω is a bounded domain in ℝ^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 $u\left(t\right)\in L^{\infty }\left(\left[0,\infty \right),L^1\right)\cap L_{\mathsf{\text{loc}}}^{\infty }\left(\left(0,\infty \right)W_0^{1,m+2}\right)$ for all u₀ ∈ L¹(Ω). Furthermore, if 2q ≤ m + 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.

1 Introduction

In this article, we are concerned with the initial boundary value problem of the quasilinear parabolic equation with nonlinear gradient term

$$\left\{\begin{array}{c}{u}_t-\mathsf{\text{div}}\left({\left|\nabla u\right|}^m\nabla u\right)+u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q=u{\left|u\right|}^{\alpha -2}{\left|\nabla u\right|}^p+g\left(u\right),x\in \Omega ,t>0,\hfill \\ u\left(x,0\right)=u_0\left(x\right),x\in \Omega ,u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0,\hfill \end{array}\right.$$

(1.1)

where Ω is a bounded domain in ℝ^Nwith 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

$$\left\{\begin{array}{c}{u}_t-\Delta u+u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q=u{\left|u\right|}^{\alpha -2}{\left|\nabla u\right|}^p,x\in \Omega ,t>0,\hfill \\ u\left(x,0\right)=u_0\left(x\right),x\in \Omega ,u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0,\hfill \end{array}\right.$$

(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 $A_k\left(u\right)\in L^2\left(\left[0,T\right],W_0^{1,2}\right)$ and

$$\begin{array}{c}\underset{\Omega }{\int }{J}_k\left(u\left(t\right)-\varphi \left(t\right)\right)\mathsf{\text{d}}x+\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }\left(\nabla u\cdot \nabla A_k\left(u-\varphi \right)+u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q{A}_k\left(u-\varphi \right)\right)\mathsf{\text{d}}x\mathsf{\text{d}}s\\ =\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }\left(u{\left|u\right|}^{\alpha -2}{\left|\nabla u\right|}^p{A}_k\left(u-\varphi \right)-A_k\left(u-\varphi \right){\varphi }_s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s+\underset{\Omega }{\int }{J}_k\left(u_0-\varphi \left(0\right)\right)\mathsf{\text{d}}x\end{array}$$

(1.3)

for ∀t ∈ [0, T] and $\forall \varphi \in L^2\left(\left[0,T\right],W_0^{1,2}\right)\cap L^{\infty }\left(Q_T\right)$, where Q _T = Ω × (0, T], and for any k > 0,

$$A_k\left(u\right)=\left\{\begin{array}{cc}-k\hfill & \hfill u\le -k,\hfill \\ u\hfill & \hfill -k\le u\le k,\hfill \\ k\hfill & \hfill u\ge k.\hfill \end{array}\right.$$

(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 [2–6] 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

$$\left\{\begin{array}{cc}{u}_t=\mathsf{\text{div}}\left(a\left(x,t,u,\nabla u\right)\right),\hfill & \hfill \left(x,t\right)\in \Omega \times \left(0,+\infty \right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & \hfill x\in \Omega ,\hfill \\ u\left(x,t\right)=0,\hfill & \hfill \left(x,t\right)\in \partial \Omega \times \left(0,+\infty \right),\hfill \end{array}\right.$$

(1.5)

where a(x, t, s, ξ ) is a Carathé odory function satisfying the following structure condition

$$a\left(x,t,s,\xi \right)\xi \ge \theta {\left|\xi \right|}^m,for\forall \left(x,t,s,\xi \right)\in \Omega \times {ℝ}^{+}\times {ℝ}^1\times {ℝ}^N$$

(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

$${∥u\left(t\right)∥}_{L^{\infty }\left(\Omega \right)}\le C{∥u_0∥}_{L^q\left(\Omega \right)}^{\alpha }{t}^{-\lambda },t>0$$

(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 [8–10]. Since the equation in (1.1) contains the nonlinear gradient term u|u|^α-2|∇u|^pand 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 ℝ^Nwith smooth boundary ∂ Ω and ∥·∥ _r , ∥·∥_1,rdenote the Sobolev space L^r(Ω) and W^1,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

$$\left|g\left(u\right)\right|\le K_1{\left|u\right|}^{1+\nu },\forall u\in {ℝ}^1,$$

(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

$$C\left(\left[0,\infty \right),L^1\right)\cap L_{\mathsf{\text{loc}}}^{\infty }\left(\left(0,\infty \right),W_0^{1,m+2}\right)$$

and $u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q,u{\left|u\right|}^{\alpha -2}{\left|\nabla u\right|}^p\in L_{\mathsf{\text{loc}}}^1\left(\left[0,\infty \right)\times \Omega \right)$, and for any $\varphi =\varphi \left(x,t\right)\in C^1\left(\left[0,\infty \right),C_0^1\left(\Omega \right)\right)$ the equality

$$\begin{array}{c}\underset{0}{\overset{T}{\int }}\underset{\Omega }{\int }\left\{-u{\varphi }_t+{\left|\nabla u\right|}^m\nabla u\nabla \varphi +u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q\varphi \right\}\mathsf{\text{d}}x\mathsf{\text{d}}t\\ =\underset{\Omega }{\int }\left(u_0\left(x\right)\varphi \left(x,0\right)-u\left(x,T\right)\varphi \left(x,T\right)\right)\mathsf{\text{d}}x+\underset{0}{\overset{T}{\int }}\underset{\Omega }{\int }\left(u{\left|u\right|}^{\alpha -2}{\left|\nabla u\right|}^p+g\left(u\right)\right)\varphi \mathsf{\text{d}}x\mathsf{\text{d}}t\end{array}$$

(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

$$u\left(t\right)\in L^{\infty }\left(\left[0,\infty \right),L^1\right)\cap C\left(\left[0,\infty \right),L^1\right)\cap L_{\mathsf{\text{loc}}}^{\infty }\left(\left(0,\infty \right),W_0^{1,m+2}\right),u_t\in L_{\mathsf{\text{loc}}}^2\left(\left(0,\infty \right),L^2\right)$$

(2.2)

and the estimates

$${∥u\left(t\right)∥}_{\infty }\le C_0{t}^{-\lambda },0<t\le T.$$

(2.3)

Furthermore, if (H₄) is satisfied, the solution u(t) has the following estimates

$$\underset{0}{\overset{T}{\int }}{s}^{1+r}{∥u_t\left(s\right)∥}_2^2\mathsf{\text{d}}s\le C_0,$$

(2.4)

$${∥\nabla u\left(t\right)∥}_{m+2}\le C_0{t}^{-\left(1+\lambda \right)/\left(m+2\right)},0<t\le T,$$

(2.5)

with r > λ = N(mN + m + 2)^-1 and C₀ = C₀(T, ∥u₀∥₁).

Theorem 2.5 Assume (H₁)-(H₅). Then the solution u(t) of (1.1) has the following L^∞ gradient estimate

$${∥\nabla u\left(t\right)∥}_{\infty }\le C_0{t}^{-\sigma },0<t\le T,$$

(2.6)

with σ = (2 + 2λ + N)(mN + 2m + 4)^-1 and C₀ = C₀(T, ∥u₀∥₁).

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 ∥u₀∥₂ < d₀, the initial boundary value problem

$$\left\{\begin{array}{c}{u}_t-\mathsf{\text{div}}\left({\left|\nabla u\right|}^m\nabla u\right)+\gamma u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q={\left|u\right|}^{\alpha -2}u{\left|\nabla u\right|}^p,x\in \Omega ,t>0,\hfill \\ u\left(x,0\right)=u_0\left(x\right),x\in \Omega ,u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0,\hfill \end{array}\right.$$

(2.7)

admits a solution $u\left(t\right)\in L^{\infty }\left(\left[0,\infty \right),L^2\right)\cap W_0^{1,m+2}$, which satisfies

$${∥u\left(t\right)∥}_2\le C{\left(1+t\right)}^{-1/m},t\ge 0.$$

(2.8)

where C = C(∥u₀∥₂).

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 + Nβ) with μ = (qα - pβ)/(q - p) > q + β.

Then, ∃d₀ > 0, such that u₀ ∈ L² with ∥u₀∥₂ < d₀, the initial boundary value problem

$$\left\{\begin{array}{c}{u}_t-\mathsf{\text{div}}\left({\left|\nabla u\right|}^m\nabla u\right)+\gamma u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q={\left|u\right|}^{\alpha -2}u{\left|\nabla u\right|}^p,x\in \Omega ,t>0\hfill \\ u\left(x,0\right)=u_0\left(x\right),x\in \Omega ;u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0,\hfill \end{array}\right.$$

(2.9)

admits a solution $u\left(t\right)\in L^{\infty }\left(\left[0,\infty \right),L^2\right)\cap W_0^{1,m+2}$ which satisfies

$${∥u\left(t\right)∥}_2\le C{\left(1+t\right)}^{-1/\left(q+\beta -2\right)},t\ge 0.$$

(2.10)

where C = C(∥u₀∥₂).

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 ≤ r ≤ q ≤ pN(1 + β)/(N - p). Then for |u|^βu ∈ W^1,p(Ω), we have

$${∥u∥}_q\le C_0^{1/\left(\beta +1\right)}{∥u∥}_r^{1-\theta }{∥{\left|u\right|}^{\beta }u∥}_{1,p}^{\theta /\left(\beta +1\right)}$$

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

$$y^{\prime }\left(t\right)+A{t}^{\lambda \theta -1}{y}^{1+\theta }\left(t\right)\le B{t}^{-k}y\left(t\right)+C{t}^{-\delta },0<t\le T$$

with A, θ > 0, λθ ≥ 1, B, C ≥ 0, k ≤ 1. Then, we have

$$y\left(t\right)\le A^{-1/\theta }{\left(2\lambda +2B{T}^{1-k}\right)}^{1/\theta }{t}^{-\lambda }+2C{\left(\lambda +B{T}^{1-k}\right)}^{-1}{t}^{1-\delta },0<t\le T.$$

3 L^∞estimate 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 $u_{0,i}\in C_0^2\left(\Omega \right)$ and u_0,i→ u₀ in L¹(Ω) as i → ∞. For i = 1, 2, ..., we consider the approximate problem of (1.1)

$$\left\{\begin{array}{c}{u}_t-\mathsf{\text{div}}\left({\left({\left|\nabla u\right|}^2+i^{-1}\right)}^{\frac{m}{2}}\nabla u\right)+u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q=u{\left|u\right|}^{\alpha -2}{\left|\nabla u\right|}^p+g\left(u\right),x\in \Omega ,t>0,\hfill \\ u\left(x,0\right)=u_{0,i}\left(x\right),x\in \Omega ,u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0.\hfill \end{array}\right.$$

(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^kfor |u|^{k- 1}u 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

$$f_n\left(s\right)=\left\{\begin{array}{cc}1,\hfill & \hfill \frac{1}{n}\le s\hfill \\ ns\left(2-ns\right),\hfill & \hfill 0\le s\le \frac{1}{n}\hfill \\ -ns\left(2+ns\right),\hfill & \hfill -\frac{1}{n}\le s\le 0\hfill \\ -1,\hfill & \hfill s<-\frac{1}{n}.\hfill \end{array}\right.$$

It is obvious that f _n (s) is odd and continuously differentiable in ℝ¹. Furthermore, $\left|f_n\left(s\right)\right|\le 1,f_n^{\prime }\left(s\right)\ge 0$ and f _n (s) → sign(s) uniformly in ℝ¹.

Multiplying the equation in (3.1) by f _n (u) and integrating on Ω, we get

$$\begin{array}{c}\underset{\Omega }{\int }{f}_n\left(u\right)u_t\mathsf{\text{d}}x+\underset{\Omega }{\int }{\left|\nabla u\right|}^{m+2}{f^{\prime }}_n\left(u\right)\mathsf{\text{d}}x+\underset{\Omega }{\int }u{\left|u\right|}^{\beta -2}{f}_n\left(u\right){\left|\nabla u\right|}^q\mathsf{\text{d}}x\\ \le \underset{\Omega }{\int }u{\left|u\right|}^{\alpha -2}{f}_n\left(u\right){\left|\nabla u\right|}^p\mathsf{\text{d}}x+\underset{\Omega }{\int }u{\left|u\right|}^{\beta -2}{f}_n\left(u\right){\left|\nabla u\right|}^q\mathsf{\text{d}}x\end{array}$$

(3.2)

and the application of the Young inequality gives

$$\underset{\Omega }{\int }u{\left|u\right|}^{\alpha -2}{f}_n\left(u\right){\left|\nabla u\right|}^p\mathsf{\text{d}}x\le \frac{1}{4}\underset{\Omega }{\int }u{\left|u\right|}^{\beta -2}{f}_n\left(u\right){\left|\nabla u\right|}^q\mathsf{\text{d}}x+C_1\underset{\Omega }{\int }{\left|u\right|}^{\mu -1}\mathsf{\text{d}}x,$$

(3.3)

where μ = (qα - pβ)(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

$$F_n\left(u\right)=\underset{0}{\overset{u}{\int }}{\left(s{\left|s\right|}^{\beta -2}{f}_n\left(s\right)\right)}^{1/q}\mathsf{\text{d}}s,u\in {ℝ}^1.$$

It is easy to verify that F _n (u) is odd in ℝ¹. Then, we obtain from the Sobolev inequality that

$$\begin{array}{ll}\hfill \frac{1}{4}\underset{\Omega }{\int }u{\left|u\right|}^{\beta -2}{f}_n\left(u\right){\left|\nabla u\right|}^q\mathsf{\text{d}}x& =\frac{1}{4}\underset{\Omega }{\int }{\left|\nabla F_n\left(u\right)\right|}^q\mathsf{\text{d}}x\\ \ge {\lambda }_0\underset{\Omega }{\int }{\left|F_n\left(u\right)\right|}^q\mathsf{\text{d}}x={\lambda }_0\underset{{\Omega }_n}{\int }{\left|F_n\left(u\right)\right|}^q\mathsf{\text{d}}x+{\lambda }_0\underset{{\Omega }_n^c}{\int }{\left|F_n\left(u\right)\right|}^q\mathsf{\text{d}}x\end{array}$$

(3.4)

with some λ₀ > 0 and

$${\Omega }_n=\left\{x\in \Omega |\left|u\left(x,t\right)\right|\ge n^{-1}\right\},{\Omega }_n^c=\Omega \backslash {\Omega }_n,n=1,2,\dots .$$

We note that |F _n (u)|^q≤ n^{-(q+β- 1)}in ${\Omega }_n^c$ and

$$\underset{{\Omega }_n^c}{\int }{\left|F_n\left(u\right)\right|}^q\mathsf{\text{d}}x\le n^{-\left(q+\beta -1\right)}\left|\Omega \right|.$$

On the other hand, we have |u(x, t)| ≥ n^-1 in Ω _n and

$$\left|F_n\left(u\right)\right|\ge \underset{n-1}{\overset{\left|u\right|}{\int }}{\left(s{\left|s\right|}^{\beta -2}{f}_n\left(s\right)\right)}^{1/q}\mathsf{\text{d}}s\ge \frac{q}{q+\beta -1}\left({\left|u\right|}^{\frac{q+\beta -1}{q}}-n^{-\frac{q+\beta -1}{q}}\right)\mathsf{\text{in}}{\Omega }_n.$$

This implies that there exists λ₁ > 0, such that

$${\lambda }_0\underset{{\Omega }_n}{\int }{\left|F_n\left(u\right)\right|}^q\mathsf{\text{d}}x\ge {\lambda }_1\underset{{\Omega }_n}{\int }{\left|u\right|}^{q+\beta -1}\mathsf{\text{d}}x-{\lambda }_1\left|\Omega \right|n^{-\left(q+\beta -1\right)}$$

(3.5)

Then it follows from (3.4)-(3.5) that

$$\frac{1}{4}\underset{\Omega }{\int }u{\left|u\right|}^{\beta -2}{f}_n\left(u\right){\left|\nabla u\right|}^q\mathsf{\text{d}}x\ge {\lambda }_1\underset{\Omega }{\int }{\left|u\right|}^{q+\beta -1}\mathsf{\text{d}}x-C_2{n}^{-\left(q+\beta -1\right)}$$

(3.6)

with some C₂ > 0.

Similarly, we have from the assumption (H₂) and the Young inequality that

$$\begin{array}{ll}\hfill \underset{\Omega }{\int }\left|g\left(u\right)f_n\left(u\right)\right|\mathsf{\text{d}}x& \le K_1\underset{\Omega }{\int }{\left|u\right|}^{1+\nu }\left|f_n\left(u\right)\right|\mathsf{\text{d}}x\\ \le K_1\underset{\Omega }{\int }{\left|u\right|}^{1+\nu }\mathsf{\text{d}}x\le \frac{{\lambda }_1}{2}\underset{{\Omega }_n}{\int }{\left|u\right|}^{q+\beta -1}\mathsf{\text{d}}x+C_2\left(1+n^{-1-\nu }\right).\end{array}$$

(3.7)

Furthermore, the assumption μ < q + β implies that

$$C_1\underset{{\Omega }_n}{\int }{\left|u\right|}^{\mu -1}\mathsf{\text{d}}x\le \frac{{\lambda }_1}{2}\underset{{\Omega }_n}{\int }{\left|u\right|}^{q+\beta -1}\mathsf{\text{d}}x+C_2.$$

(3.8)

Then (3.2)-(3.3) and (3.6)-(3.8) give that

$$\underset{\Omega }{\int }{f}_n\left(u\right)u_t\mathsf{\text{d}}x+\frac{1}{2}\underset{\Omega }{\int }u{\left|u\right|}^{\beta -2}{f}_n\left(u\right){\left|\nabla u\right|}^q\mathsf{\text{d}}x\le C_3\left(1+n^{-1-\nu }+n^{-\left(q+\beta -1\right)}\right).$$

(3.9)

Letting n → ∞ in (3.9) yields

$$\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_1+\frac{1}{2}\underset{\Omega }{\int }{\left|u\right|}^{\beta -1}{\left|\nabla u\right|}^q\mathsf{\text{d}}x\le C_3.$$

(3.10)

Note that

$$\underset{\Omega }{\int }{\left|u\right|}^{\beta -1}{\left|\nabla u\right|}^q\mathsf{\text{d}}x={\left(\frac{q}{q+\beta -1}\right)}^q\underset{\Omega }{\int }{\left|\nabla u^{1+\frac{\beta -1}{q}}\right|}^q\mathsf{\text{d}}x\ge 2{\lambda }_2\ge {∥u∥}_1^{q+\beta -1}$$

with some λ₂ > 0. Then (3.10) becomes

$$\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_1+{\lambda }_2{∥u\left(t\right)∥}_1^{q+\beta -1}\le C_3.$$

(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

$$\underset{0}{\overset{T}{\int }}\underset{\Omega }{\int }{\left|u_i\right|}^{\beta -1}{\left|\nabla u_i\right|}^q\mathsf{\text{d}}x\mathsf{\text{d}}t\le C_0\mathsf{\text{for}}i=1,2,\dots .$$

(3.12)

withC₀ = C₀(T, ∥u₀∥₁).

Lemma 3.3 Assume (H₁)-(H₄). Then, for any T > 0, the solution u(t) of (3.1) also satisfies the following estimates:

$${∥u\left(t\right)∥}_{\infty }\le C_0{t}^{-\lambda },0<t\le T,$$

(3.13)

where λ = N(mN + m + 2)^-1, C₀ = C₀(T, ∥u₀∥₁).

Proof Multiplying the equation in (3.1) by u^k-1, k ≥ 2, we have

$$\begin{array}{c}\frac{1}{k}\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_k^k+\left(k-1\right){\left(\frac{m+2}{k+2}\right)}^{m+2}{∥\nabla u^{\frac{k+m}{m+2}}∥}_{m+2}^{m+2}+\underset{\Omega }{\int }{\left|u\right|}^{\beta +k-2}{\left|\nabla u\right|}^q\mathsf{\text{d}}x\\ \le \underset{\Omega }{\int }{\left|u\right|}^{\alpha +k-2}{\left|\nabla u\right|}^p\mathsf{\text{d}}x+K_1\underset{\Omega }{\int }{\left|u\right|}^{\nu +k}\mathsf{\text{d}}x.\end{array}$$

(3.14)

It follows from the Hö lder and Sobolev inequalities that

$$\begin{array}{c}{K}_1\underset{\Omega }{\int }{\left|u\right|}^{\nu +k}\mathsf{\text{d}}x\le C{∥u∥}_k^{{\theta }_1}{∥u∥}_1^{{\theta }_2}{∥u∥}_s^{{\theta }_3}\le C{∥u∥}_k^{{\theta }_1}{∥\nabla u^{\frac{k+m}{m+2}}∥}_{m+2}^{\frac{\left(m+2\right){\theta }_3}{k+m}}\\ \le \frac{k-1}{2}{\left(\frac{m+2}{k+2}\right)}^{m+2}{∥\nabla u^{\frac{k+m}{m+2}}∥}_{m+2}^{m+2}+C{k}^{\sigma }{∥u∥}_k^k,\end{array}$$

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

$$\underset{\Omega }{\int }{\left|u\right|}^{\alpha +k-2}{\left|\nabla u\right|}^p\mathsf{\text{d}}x\le \frac{1}{4}\underset{\Omega }{\int }{\left|u\right|}^{\beta +k-2}{\left|\nabla u\right|}^q\mathsf{\text{d}}x+C\underset{\Omega }{\int }{\left|u\right|}^{\mu +k-2}\mathsf{\text{d}}x$$

and

$$\frac{1}{2}\underset{\Omega }{\int }{\left|u\right|}^{\beta +k-2}{\left|\nabla u\right|}^q\mathsf{\text{d}}x\ge C_1{k}^{-q}\underset{\Omega }{\int }{\left|\nabla u^{\frac{q+\beta +k-2}{q}}\right|}^q\mathsf{\text{d}}x$$

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

Without loss of generality, we assume k > 3 - μ. Similarly, we derive

$$\begin{array}{ll}\hfill C\underset{\Omega }{\int }{\left|u\right|}^{\mu +k-2}\mathsf{\text{d}}x& \le C{∥u∥}_{k-2}^{{\mu }_1}{∥u∥}_1^{{\mu }_2}{∥u∥}_{k^{*}}^{{\mu }_3}\le C{\xi }_1^{{\mu }_2}{∥u∥}_k^{{\mu }_1}{∥u∥}_{k^{*}}^{{\mu }_3}\\ \le C{∥u∥}_k^{{\mu }_1}{∥\nabla u^{q_k/q}∥}_q^{q{\mu }_3/q_k}\equiv A_k\end{array}$$

with ξ₁ = sup_t≥0∥u(t)∥₁ and

$$\begin{array}{c}{\mu }_1={\lambda }_0\left(k-2\right)\left(q+\beta -\mu +q{N}^{-1}\right),{\mu }_2={\lambda }_0\mu q{N}^{-1},{\mu }_3={\lambda }_0\mu q_k,\\ {\lambda }_0={\left(q+\beta +q/N\right)}^{-1},k^{*}=q_{k}N{\left(N-q\right)}^{-1},q_k=q+\beta +k-2.\end{array}$$

Then, for any η > 0,

$$A_k\le C\eta {∥\nabla u^{q_k/q}∥}_q^q+C{\eta }^{-{\theta }^{\prime }/\theta }{∥u∥}_k^{{\mu }_1{\theta }^{\prime }}$$

(3.15)

with μ λ₀θ = 1, (1 - μ λ₀)θ' = 1.

Note that μ₁θ' < k. Let $\eta =\frac{C_1}{2C}{k}^{-q}$. Then it follows from (3.15) that

$$A_k\le \frac{C_1}{2}{k}^{-q}{∥\nabla u^{q_k/q}∥}_q^q+C{k}^{\gamma }\left({∥u∥}_k^k+1\right)$$

(3.16)

with γ = qθ'θ^-1 = qμ λ₀/(1 - μ λ₀). Then, (3.14) becomes

$$\frac{1}{k}\frac{d}{\mathsf{\text{d}}t}{∥u∥}_k^k+\frac{k-1}{2}{\left(\frac{m+2}{k+2}\right)}^{m+2}{∥\nabla u^{\frac{k+m}{m+2}}∥}_{m+2}^{m+2}+\frac{C_1}{2}{k}^{-q}{∥\nabla u^{q_k/q}∥}_q^q\le C{k}^{{\sigma }_0}\left({∥u∥}_k^k+1\right)$$

or

$$\frac{d}{\mathsf{\text{d}}t}{∥u∥}_k^k+C_1{k}^{-m}{∥\nabla u^{\frac{k+m}{m+2}}∥}_{m+2}^{m+2}\le C{k}^{1+{\sigma }_0}\left({∥u∥}_k^k+1\right)$$

(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, k_n = R^n-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

$${∥u\left(t\right)∥}_{k_n}\le C_0^{\frac{m+2}{m+k_n}}{∥u\left(t\right)∥}_{k_{n-1}}^{1-{\theta }_n}{∥\nabla u^{\frac{m+k_n}{m+2}}∥}_{m+2}^{\frac{{\theta }_n\left(m+2\right)}{m+k_n}}$$

(3.18)

with ${\theta }_n=RN\left(1-k_{n-1}{k}_n^{-1}\right){\left(m+2+N\left(R-1\right)\right)}^{-1}$.

Then, inserting (3.18) into (3.17) (k = k _n ), we find that

$$\begin{array}{c}\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_{k_n}^{k_n}+C_1{C}_0^{-\frac{m+2}{{\theta }_n}}{k}_n^{-m}{∥u\left(t\right)∥}_{k_{n-1}}^{\left(1-1/{\theta }_n\right)\left(m+k_n\right)}{∥u\left(t\right)∥}_{k_n}^{\left(m+k_n\right)/{\theta }_n}\\ \le C{k}_n^{1+{\sigma }_0}\left({∥u\left(t\right)∥}_{k_n}^{k_n}+1\right),0<t\le T,\end{array}$$

(3.19)

or

$$\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_{k_n}^{k_n}+C_1{C}_0^{-\frac{m+2}{{\theta }_n}}{k}_n^{-m}{∥u\left(t\right)∥}_{k_{n-1}}^{m-{\beta }_n}{∥u\left(t\right)∥}_{k_n}^{k_n+{\beta }_n}\le C{k}_n^{1+{\sigma }_0}\left({∥u\left(t\right)∥}_{k_n}^{k_n}+1\right),$$

(3.20)

where ${\beta }_n=\left(m+k_n\right){\theta }_n^{-1}-k_n,n=2,3,\dots$. It is easy to see that

$${\theta }_n\to {\theta }_0=\frac{N\left(R-1\right)}{m+2+N\left(R-1\right)},{\beta }_n{k}_n^{-1}\to \frac{m+2}{N\left(R-1\right)},\mathsf{\text{as}}n\to \infty .$$

Denote

$$y_n\left(t\right)={∥u\left(t\right)∥}_{k_n}^{k_n},0<t\le T.$$

Then (3.20) can be rewritten as follows

$$y_n^{\prime }\left(t\right)+C_1{C}^{-\frac{m+2}{{\theta }_n}}{k}_n^{-m}{\left(y_n\left(t\right)\right)}^{1+{\beta }_n/k_n}{∥u\left(t\right)∥}_{k_{n-1}}^{m-{\beta }_n}\le C{k}_n^{1+{\sigma }_0}\left(y_n\left(t\right)+1\right).$$

(3.21)

We claim that there exist a bounded sequence {ξ _n } and a convergent sequence {λ _n }, such that

$${∥u\left(t\right)∥}_{k_n}\le {\xi }_n{t}^{-{\lambda }_n},0<t\le T.$$

(3.22)

Indeed, by Lemma 3.1, the estimate (3.22) holds for n = 1 if we take λ₁ = 0, ξ₁ = sup_t≥0∥u(t)∥₁. If (3.22) is true for n - 1, then we have from (3.21) and (3.22) that

$$y_n^{\prime }\left(t\right)+C_1{C}^{-\frac{m+2}{{\theta }_n}}{k}_n^{-m}{\left({\xi }_{n-1}\right)}^{m-{\beta }_n}{t}^{{\Lambda }_n{\tau }_n-1}{y}_n^{1+{\tau }_n}\left(t\right)\le C{k}_n^{1+{\sigma }_0}\left(y_n\left(t\right)+1\right),0\le t\le T,$$

(3.23)

where

$${\tau }_n=\frac{{\beta }_n}{k_n},{\Lambda }_n=k_n{\lambda }_n,{\lambda }_n=\frac{1+{\lambda }_{n-1}\left({\beta }_n-m\right)}{{\beta }_n}.$$

Applying Lemma 2.10 to (3.23), we have

$$y_n\left(t\right)\le {\left(C_1{C}^{-\frac{m+2}{{\theta }_n}}{k}_n^{-m}{\xi }_{n-1}^{m-{\beta }_n}\right)}^{-1/{\tau }_n}{\left(2k_n{\lambda }_n+2CT{k}_n^{1+{\sigma }_0}\right)}^{1/{\tau }_n}{t}^{-k_n{\lambda }_n}.$$

(3.24)

This implies that for t ∈ (0, T),

$${∥u\left(t\right)∥}_{k_n}\le {\left(C_1{C}^{-\frac{m+2}{{\theta }_n}}{k}_n^{-m}{\xi }_{n-1}^{m-{\beta }_n}\right)}^{-1/{\beta }_n}{\left(2k_n{\lambda }_n+2CT{k}_n^{1+{\sigma }_0}\right)}^{1/{\beta }_n}{t}^{-{\lambda }_n}\le {\xi }_n{t}^{-{\lambda }_n},$$

(3.25)

where

$${\xi }_n={\xi }_{n-1}{\left(C_1{C}^{-\frac{m+2}{{\theta }_n}}{k}_n^{-m}\right)}^{-1/{\beta }_n}{\left(2k_n{\lambda }_n+4CT{k}_n^{1+{\sigma }_0}\right)}^{1/{\beta }_n},$$

(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

$$\frac{1+{\lambda }_{n-1}\left({\beta }_n-m\right)}{{\beta }_n}\to \lambda =\frac{N}{m+2+mN},\mathsf{\text{as}}n\to \infty .$$

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

$$\underset{0}{\overset{T}{\int }}{s}^{1+r}{∥u_t\left(s\right)∥}_2^2\mathsf{\text{d}}s\le C_0$$

(3.27)

and

$${∥\nabla u\left(t\right)∥}_{m+2}\le C_0{t}^{-\left(1+\lambda \right)/\left(m+2\right)},0<t\le T,$$

(3.28)

with r > λ = N(mN + m + 2)^-1, C₀ = C₀(T, ∥u₀∥₁).

Proof We first choose r > λ and η(t) ∈ C[0, ∞) ⋂ C¹(0, ∞) such that η(t) = t^rwhen 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

$$\begin{array}{c}\frac{1}{2}\eta \left(t\right){∥u\left(t\right)∥}_2^2+\underset{0}{\overset{t}{\int }}\eta \left(s\right){∥\nabla u\left(s\right)∥}_{m+2}^{m+2}\mathsf{\text{d}}s+\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\beta }{\left|\nabla u\right|}^q\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s\\ \le \frac{1}{2}\underset{0}{\overset{t}{\int }}{\eta }^{\prime }\left(s\right){∥u\left(s\right)∥}_2^2\mathsf{\text{d}}s+\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\alpha }{\left|\nabla u\right|}^p\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s+K_1\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{2+\nu }\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s.\end{array}$$

(3.29)

Note that

$$\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\alpha }{\left|\nabla u\right|}^p\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s\le \frac{1}{2}\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\beta }{\left|\nabla u\right|}^q\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s+C\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\mu }\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s.$$

Hence, we have

$$\begin{array}{c}\frac{1}{2}\eta \left(t\right){∥u\left(t\right)∥}_2^2+\underset{0}{\overset{t}{\int }}\eta \left(s\right){∥\nabla u\left(s\right)∥}_{m+2}^{m+2}\mathsf{\text{d}}s+\frac{1}{2}\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\beta }{\left|\nabla u\right|}^q\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s\\ \le \frac{1}{2}\underset{0}{\overset{t}{\int }}{\eta }^{\prime }\left(s\right){∥u\left(s\right)∥}_2^2\mathsf{\text{d}}s+C\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\mu }\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s+K_1\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{2+\nu }\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s.\end{array}$$

(3.30)

By Lemma 3.1 and the estimate (3.13), we get

$$\underset{0}{\overset{t}{\int }}{\eta }^{\prime }\left(s\right){∥u\left(s\right)∥}_2^2\mathsf{\text{d}}s\le C\underset{0}{\overset{t}{\int }}{s}^{r-1}{∥u\left(t\right)∥}_1{∥u\left(t\right)∥}_{\infty }\mathsf{\text{d}}s\le C{t}^{r-\lambda },0\le t<T.$$

(3.31)

Since μ < q + β, we have from Sobolev inequality that

$$C\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\mu }\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s\le \frac{1}{4}\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\beta }{\left|\nabla u\right|}^q\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s+C\underset{0}{\overset{t}{\int }}\eta \left(s\right)\mathsf{\text{d}}s.$$

(3.32)

Similarly, we have from 2 + ν < q + β that

$$K_1\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{2+\nu }\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s\le \frac{1}{4}\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|u\right|}^{\beta }{\left|\nabla u\right|}^q\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s+C\underset{0}{\overset{t}{\int }}\eta \left(s\right)\mathsf{\text{d}}s.$$

(3.33)

Therefore, it follows from (3.30)-(3.33) that

$$\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }{\left|\nabla u\right|}^{m+2}\eta \left(s\right)\mathsf{\text{d}}x\mathsf{\text{d}}s\le C{t}^{r-\lambda },0\le t\le T.$$

(3.34)

Next, let $G\left(u\right)={\int }_0^{u}g\left(s\right)\mathsf{\text{d}}s,u\in {ℝ}^1,\rho \left(t\right)={\int }_0^t\eta \left(s\right)\mathsf{\text{d}}s,t\in \left(0,\infty \right)$. Furthermore, multiplying the equation in (3.1) by ρ(t)u _t yields

$$\begin{array}{l}\rho \left(t\right){∥u_t\left(t\right)∥}_2^2+\frac{1}{m+2}\frac{d}{\mathsf{\text{d}}t}\underset{\Omega }{\int }\rho \left(t\right){\left({\left|\nabla u\right|}^2+i^{-1}\right)}^{\frac{m+2}{2}}\mathsf{\text{d}}x+{\rho }^{\prime }\left(t\right)\underset{\Omega }{\int }G\left(u\right)\mathsf{\text{d}}x\\ \le \frac{{\rho }^{\prime }\left(t\right)}{m+2}\underset{\Omega }{\int }{\left({\left|\nabla u\right|}^2+i^{-1}\right)}^{\frac{m+2}{2}}\mathsf{\text{d}}x+\frac{d}{\mathsf{\text{d}}t}\underset{\Omega }{\int }\rho \left(t\right)G\left(u\right)\mathsf{\text{d}}x\\ +\underset{\Omega }{\int }\rho \left(t\right){\left|u\right|}^{\beta -1}\left|u_t\right|{\left|\nabla u\right|}^q\mathsf{\text{d}}x+\underset{\Omega }{\int }\rho \left(t\right){\left|u\right|}^{\alpha -1}\left|u_t\right|{\left|\nabla u\right|}^p\mathsf{\text{d}}x.\end{array}$$

(3.35)

By the assumption p < q and the Cauchy inequality, we deduce

$$\underset{\Omega }{\int }{\left|u\right|}^{\beta -1}\left|u_t\right|{\left|\nabla u\right|}^q\mathsf{\text{d}}x\le \frac{1}{4}{∥u_t\left(t\right)∥}_2^2+C\underset{\Omega }{\int }{\left|u\right|}^{2\left(\beta -1\right)}{\left|\nabla u\right|}^{2q}\mathsf{\text{d}}x$$

(3.36)

and

$$\begin{array}{ll}\hfill \underset{\Omega }{\int }{\left|u\right|}^{\alpha -1}\left|u_t\right|{\left|\nabla u\right|}^p\mathsf{\text{d}}x& \le \frac{1}{4}{∥u_t\left(t\right)∥}_2^2+C\underset{\Omega }{\int }{\left|u\right|}^{2\left(\alpha -1\right)}{\left|\nabla u\right|}^{2p}\mathsf{\text{d}}x\\ \le \frac{1}{4}{∥u_t\left(t\right)∥}_2^2+C\underset{\Omega }{\int }{\left|u\right|}^{2\left(\beta -1\right)}{\left|\nabla u\right|}^{2q}\mathsf{\text{d}}x+C\underset{\Omega }{\int }{\left|u\right|}^{2\left(\mu -1\right)}\mathsf{\text{d}}x\end{array}$$

(3.37)

and

$$\underset{\Omega }{\int }\left|G\left(u\right)\right|\mathsf{\text{d}}x\le C_1{\underset{\Omega }{\int }\left|u\right|}^{2+\nu }\mathsf{\text{d}}x\le C{h}^{2+\nu }\left(t\right)$$

(3.38)

with h(t) = ∥u(t)∥_∞.

Now, it follows from (H₄) and (3.35)-(3.38) that

$$\begin{array}{l}\frac{1}{2}\underset{0}{\overset{t}{\int }}\rho \left(s\right){∥u_t\left(s\right)∥}_2^{2}ds+\frac{1}{m+2}\rho \left(t\right){∥\nabla u\left(t\right)∥}_{m+2}^{m+2}\le \frac{1}{2}\underset{0}{\overset{t}{\int }}\eta \left(s\right){∥\nabla u\left(s\right)∥}_{m+2}^{m+2}\mathsf{\text{d}}s+C\rho \left(t\right)h^{2+\nu }\left(t\right)\end{array}$$

(3.39)

or

$$\begin{array}{c}\frac{1}{2}\underset{0}{\overset{t}{\int }}\rho \left(s\right){∥u_t\left(s\right)∥}_2^2\mathsf{\text{d}}s+\frac{\rho \left(t\right)}{m+2}{∥\nabla u\left(t\right)∥}_{m+2}^{m+2}\\ \le C_0{t}^{r-\lambda }+C_0\underset{0}{\overset{t}{\int }}\rho \left(s\right)h^{2\left(\beta -1\right)}\left(s\right){∥\nabla u\left(s\right)∥}_{m+2}^{m+2}\mathsf{\text{d}}s\end{array}$$

(3.40)

where C₀ = C₀(T, ∥u₀∥₁) and the fact 2 + λ ≥ 2(μ - 1)λ has been used.

Since the function h^2(β-1)(t) ∈ L¹([0, T]), the application of the Gronwall inequality to (3.40) gives

$$\underset{0}{\overset{t}{\int }}\rho \left(s\right){∥u_t\left(s\right)∥}_2^2\mathsf{\text{d}}s+\rho \left(t\right){∥\nabla u∥}_{m+2}^{m+2}\le C_0{t}^{r-\lambda },0<t\le T.$$

(3.41)

Hence,

$${∥\nabla u∥}_{\mathsf{\text{m}}+2}\le C_0{t}^{-\left(1+\lambda \right)/\left(m+2\right)},0<t\le T.$$

(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 $u\in L^s\left(\left[0,T\right],W_0^{1,s}\left(\Omega \right)\right),\left(1\le s\le m+2\right)$ satisfying

$$\begin{array}{c}{u}_i\rightharpoonup u\mathsf{\text{weakly}}\mathsf{\text{in}}{L}^s\left(\left[0,T\right],W_0^{1,s}\left(\Omega \right)\right),\\ u_i\to u\mathsf{\text{in}}{L}^s\left(Q_T\right)\mathsf{\text{and}}\mathsf{\text{a}}\mathsf{\text{.e}}.\mathsf{\text{in}}{Q}_T,\\ {\left|u_i\right|}^{\beta -1}{\left|\nabla u_i\right|}^q\to {\left|u\right|}^{\beta -1}{\left|\nabla u\right|}^q\mathsf{\text{in}}{L}^1\left(Q_T\right),\\ {\left|u_i\right|}^{\alpha -1}{\left|\nabla u_i\right|}^p\to {\left|u\right|}^{\alpha -1}{\left|\nabla u\right|}^p\mathsf{\text{in}}{L}^1\left(Q_T\right),\\ u_i\to u\mathsf{\text{in}}C\left(\left[0,T\right];L^1\left(\Omega \right)\right),\\ \frac{\partial u_i}{\partial t}\to \frac{\partial u}{\partial t}\mathsf{\text{weakly}}\mathsf{\text{in}}{L}_{\mathsf{\text{loc}}}^2\left(0,T;L^2\right).\end{array}$$

(3.43)

Since $A_i\left(u_i\right)=-\mathsf{\text{div}}\left({\left({\left|\nabla u_i\right|}^2+i^{-1}\right)}^{\frac{m}{2}}\nabla u_i\right)$ is bounded in ${\left(W_0^{1,m+2}\right)}^{*}=W_0^{-1,\frac{m+2}{m+1}}$, we see further that

$$A_i\left(u_i\right)\to \chi \mathsf{\text{weakly}}*\mathsf{\text{in}}{L}_{loc}^{\infty }\left(0,T;\left(W_0^{1,m+2}\right)*\right)$$

(3.44)

for some $\chi \in L_{\text{loc}}^{\infty }(0,T],(W_0^{1,m+2})*)$. 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 L^∞estimate 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

$${\left|D^{2}u\right|}^2=\sum _{i,j=1}^N{u}_{ij}^2,u_{ij}=\frac{{\partial }^{2}u}{\partial x_i\partial x_j}.$$

Multiplying (3.1) by -div(|∇u|^k-2∇u), k ≥ m + 2 and integrating by parts, we have

$$\begin{array}{c}\frac{1}{k}\frac{d}{\mathsf{\text{d}}t}\left({∥\nabla u\left(t\right)∥}_k^k\right)+\underset{\Omega }{\int }{\left|\nabla u\right|}^{k+m-2}{\left|D^{2}u\right|}^2\mathsf{\text{d}}x+\frac{k-2}{4}\underset{\Omega }{\int }{\left|\nabla u\right|}^{k+m-4}{\left|\nabla \left({\left|\nabla u\right|}^2\right)\right|}^2\mathsf{\text{d}}x\\ -\left(N-1\right)\underset{\partial \Omega }{\int }H\left(x\right){\left|\nabla u\right|}^{k+m}\mathsf{\text{d}}S\\ =\underset{\Omega }{\int }u{\left|u\right|}^{\beta -2}{\left|\nabla u\right|}^q\mathsf{\text{div}}\left({\left|\nabla u\right|}^{k-2}\nabla u\right)\mathsf{\text{d}}x-\underset{\Omega }{\int }{u\left|\nabla u\right|}^p{\left|u\right|}^{\alpha -2}\mathsf{\text{div}}\left({\left|\nabla u\right|}^{k-2}\nabla u\right)\mathsf{\text{d}}x\\ +\underset{\Omega }{\int }g\left(u\right)\mathsf{\text{div}}\left({\left|\nabla u\right|}^{k-2}\nabla u\right)\mathsf{\text{d}}x\equiv I+II+III.\end{array}$$

(4.1)

Since

$$\mathsf{\text{div}}\left({\left|\nabla u\right|}^{k-2}\nabla u\right)={\left|\nabla u\right|}^{k-2}\Delta u+\frac{k-2}{2}{\left|\nabla u\right|}^{k-4}\nabla u\nabla \left({\left|\nabla u\right|}^2\right),$$

(4.2)

we have

$$\left|\text{div(}{\left|\nabla u\right|}^{k-2}\nabla u)\right|\le (k-1){\left|\nabla u\right|}^{k-2}\left|D^{2}u\right|$$

(4.3)

and

$$\begin{array}{ll}\hfill \left|I\right|& \le \left(k-1\right)\underset{\Omega }{\int }{\left|u\right|}^{\beta -1}{\left|\nabla u\right|}^{q+k-2}\left|D^{2}u\right|\mathsf{\text{d}}x\\ =\left(k-1\right)\underset{\Omega }{\int }{\left|\nabla u\right|}^{\frac{k+m-2}{2}}\left|D^{2}u\right|{\left|\nabla u\right|}^{\frac{k+2q-m-2}{2}}{\left|u\right|}^{\beta -1}\mathsf{\text{d}}x\\ \le \frac{1}{4}\underset{\Omega }{\int }{\left|\nabla u\right|}^{k+m-2}{\left|D^{2}u\right|}^2\mathsf{\text{d}}x+C_0{k}^2\underset{\Omega }{\int }{\left|\nabla u\right|}^{k+2q-m-2}{\left|u\right|}^{2\left(\beta -1\right)}\mathsf{\text{d}}x.\end{array}$$

(4.4)

Similarly, we obtain the following estimates

$$\left|II\right|\le \frac{1}{4}\underset{\Omega }{\int }{\left|\nabla u\right|}^{m+k-2}{\left|D^{2}u\right|}^2\mathsf{\text{d}}x+C_0{k}^2\underset{\Omega }{\int }{\left|\nabla u\right|}^{k+2p-m-2}{\left|u\right|}^{2\left(\alpha -1\right)}\mathsf{\text{d}}x$$

(4.5)

and

$$\begin{array}{ll}\hfill III& =\underset{\Omega }{\int }g\left(u\right)\mathsf{\text{div}}\left({\left|\nabla u\right|}^{k-2}\nabla u\right)\mathsf{\text{d}}x=-\underset{\Omega }{\int }{g}^{\prime }\left(u\right){\left|\nabla u\right|}^k\mathsf{\text{d}}x\\ \le K_1\underset{\Omega }{\int }{\left|y\right|}^{\nu }{\left|\nabla u\right|}^k\mathsf{\text{d}}x\le C{h}^{\nu }\left(t\right){∥\nabla u\left(t\right)∥}_k^k,\end{array}$$

(4.6)

where h(t) = ∥u(t)∥_∞ ≤ Ct^-λ.

Moreover, we assume that 2q ≤ m + 2, 2p ≤ m + 2, then (4.1) becomes

$$\begin{array}{c}\frac{1}{k}\frac{d}{\mathsf{\text{d}}t}\left({∥\nabla u∥}_k^k\right)+\frac{1}{2}\underset{\Omega }{\int }{\left|\nabla u\right|}^{k+m-2}{\left|D^{2}u\right|}^2\mathsf{\text{d}}x+\frac{k-2}{4}\underset{\Omega }{\int }{\left|\nabla u\right|}^{k+m-4}{\left|\nabla \left({\left|\nabla u\right|}^2\right)\right|}^2\mathsf{\text{d}}x\\ -\left(N-1\right)\underset{\partial \Omega }{\int }H\left(x\right){\left|\nabla u\right|}^{k+m}\mathsf{\text{d}}S\\ \le C_0{k}^2\underset{\Omega }{\int }\left({\left|\nabla u\right|}^{k+2q-m-2}{\left|u\right|}^{2\left(\beta -1\right)}+{\left|\nabla u\right|}^{k+2p-m-2}{\left|u\right|}^{2\left(\alpha -1\right)}\right)\mathsf{\text{d}}x+C{h}^{\nu }\left(t\right){∥\nabla u\left(t\right)∥}_k^k\\ \le C_0{k}^2{h}_1\left(t\right)\left(1+{∥\nabla u\left(t\right)∥}_k^k\right),\end{array}$$

(4.7)

where h₁(t) = max{h^2(α-1)(t), h^2(β-1)(t), h^ν(t)}. Since $\alpha ,\beta <2+\frac{m}{2}\left(1+\frac{1}{N}\right),\nu <m+2+\frac{m}{N}$, 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

$${∥\nabla v∥}_2^2-\left(N-1\right)\underset{\partial \Omega }{\int }{v}^{2}H\left(x\right)\mathsf{\text{d}}S\ge {\lambda }_1{∥v∥}_{1,2}^2,\forall v\in W^{1,2}\left(\Omega \right).$$

(4.8)

Hence, by (4.7) and (4.8), we see that there exists C₁ and C₂ such that

$$\frac{d}{\mathsf{\text{d}}t}\left({∥\nabla u\left(t\right)∥}_k^k\right)+C_1{∥{\left|\nabla u\left(t\right)\right|}^{\frac{k+m}{2}}∥}_{1,2}^2\le C{k}^3{h}_1\left(t\right)\left(1+{∥\nabla u\left(t\right)∥}_k^k\right).$$

(4.9)

Let k₁ = m + 2, R > m + 1, k _n = R^n-2(R-1-m) + m (R-1)^-1, ${\theta }_n=RN\left(1-k_{n-1}{k}_n^{-1}\right){\left(R\left(N-1\right)+2\right)}^{-1}$, n = 2, 3,.... Then, the application of Lemma 2.9 gives

$${∥\nabla u∥}_{k_n}\le C^{\frac{2}{k_n+m}}{∥\nabla u∥}_{k_{n-1}}^{1-{\theta }_n}{∥{\left|\nabla u\right|}^{\frac{k_n+m}{2}}∥}_{1,2}^{\frac{2{\theta }_n}{k_{n+m}}}.$$

(4.10)

Inserting this into (4.9)(k = k _n ), we get

$$\begin{array}{c}\frac{d}{\mathsf{\text{d}}t}\left({∥\nabla u∥}_{k_n}^{k_n}\right)+C_1{C}^{-2/{\theta }_n}{∥\nabla u\left(t\right)∥}_{k_{n-1}}^{\left(k_n+m\right)\left(1-1/{\theta }_n\right)}{∥\nabla u\left(t\right)∥}_{k_n}^{\left(k_n+m\right)/{\theta }_n}\\ \le C_2{k}_n^3{h}_1\left(t\right)\left(1+{∥\nabla u\left(t\right)∥}_{k_n}^{k_n}\right).\end{array}$$

(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

$${∥\nabla u\left(t\right)∥}_{k_n}\le y_n{t}^{-z_n},0<t\le T,$$

(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

$$\frac{1}{2}\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_2^2+C_1{∥\nabla u\left(t\right)∥}_{m+2}^{m+2}\le \underset{\Omega }{\int }{\left|u\right|}^{\alpha }{\left|\nabla u\right|}^p\mathsf{\text{d}}x$$

(5.1)

with $C_1={\left(\frac{m+2}{4}\right)}^{m+2}$.

Since p < m + 2 < p + α, it follows from Lemma 2.9 that

$$\begin{array}{ll}\hfill \underset{\Omega }{\int }{\left|u\right|}^{\alpha }{\left|\nabla u\right|}^p\mathsf{\text{d}}x& \le {∥\nabla u\left(t\right)∥}_{m+2}^p{∥u∥}_s^{\alpha }\le C_0{∥\nabla u∥}_{m+2}^p{∥u∥}_r^{\alpha \left(1-\theta \right)}{∥\nabla u∥}_{m+2}^{\alpha \theta }\\ \le C_0{∥\nabla u\left(t\right)∥}_{m+2}^{m+2}{∥u\left(t\right)∥}_r^{p1}\end{array}$$

(5.2)

with

$$s=\frac{\alpha \left(m+2\right)}{m+2-p},\theta =\left(\frac{1}{r}-\frac{1}{s}\right){\left(\frac{1}{N}+\frac{1}{r}-\frac{1}{m+2}\right)}^{-1},r=\frac{N{p}_1}{m+2-p},p_1=p+\alpha -m-2.$$

The assumption on α shows that r ≤ 2. Then, (5.1) can be rewritten as

$$\frac{1}{2}\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_2^2+{∥\nabla u\left(t\right)∥}_{m+2}^{m+2}\left(C_1-C_0{∥u\left(t\right)∥}_2^{p_1}\right)\le 0.$$

(5.3)

By the Sobolev embedding theorem,

$${∥\nabla u\left(t\right)∥}_{m+2}^{m+2}\ge C_2{∥u\left(t\right)∥}_{m+2}^{m+2}\ge C_2{∥u\left(t\right)∥}_2^{m+2},$$

(5.4)

we obtain from (5.3) and (5.4) that ∃d₀ > 0, λ₀ > 0, such that ∥u₀∥₂ < d₀ and

$${\varphi }^{\prime }\left(t\right)+{\lambda }_0{\varphi }^{1+m/2}\left(t\right)\le 0,t\ge 0$$

(5.5)

with $\varphi \left(t\right)={∥u\left(t\right)∥}_2^2$. This implies that

$${∥u\left(t\right)∥}_2\le C{\left(1+t\right)}^{-1/m},t\ge 0,$$

(5.6)

where the constant C depends only ∥u₀∥₂. 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

$$\frac{1}{2}\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_2^2+\gamma \underset{\Omega }{\int }{\left|u\right|}^{\beta }{\left|\nabla u\right|}^q\mathsf{\text{d}}x\le \underset{\Omega }{\int }{\left|u\right|}^{\alpha }{\left|\nabla u\right|}^p\mathsf{\text{d}}x.$$

(5.7)

Since p < q, q + β < p + α, it follows from the Hö lder inequality that

$$\begin{array}{ll}\hfill \underset{\Omega }{\int }{\left|u\right|}^{\alpha }{\left|\nabla u\right|}^p\mathsf{\text{d}}x& \le {∥\nabla u^{1+\beta /q}∥}_q^p{∥u∥}_{\mu }^{\mu \left(1-p/q\right)}\\ \le C_1{∥\nabla u^{1+\beta /q}∥}_q^p{∥u∥}_{\tau }^{{\mu }_1\left(1-p/q\right)}{∥\nabla u^{1+\beta /q}∥}_q^{{\mu }_2\left(1-p/q\right)}\\ \le C_1{∥\nabla u^{1+\beta /q}∥}_q^q{∥u∥}_{\tau }^{{\mu }_1\left(1-p/q\right)}\le C_1{∥\nabla u^{1+\beta /q}∥}_q^q{∥u∥}_2^{{\mu }_3}\end{array}$$

(5.8)

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

Then (5.7) becomes

$$\frac{d}{\mathsf{\text{d}}t}{∥u\left(t\right)∥}_2^2+{∥\nabla u^{1+\beta /q}∥}_q^q\left(C_0-C_1{∥u\left(t\right)∥}_2^{{\mu }_3}\right)\le 0.$$

(5.9)

This implies that ∃d₀ > 0, λ₁ > 0, such that ∥u₀∥₂ < d₀ and

$${\varphi }^{\prime }\left(t\right)+{\lambda }_1{\varphi }^{\left(q+\beta \right)/2}\left(t\right)\le 0,t\ge 0$$

(5.10)

with $\varphi \left(t\right)={∥u\left(t\right)∥}_2^2$. This implies that

$${∥u\left(t\right)∥}_2\le C{\left(1+t\right)}^{-1/\left(q+\beta -2\right)},t\ge 0.$$

(5.11)

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