一类具有对数非线性项的四阶抛物型方程解的全局渐近性 Global Asymptotic Behavior of Solutions for a Class of Fourth-Order Logarithmic Nonlinear Parabolic Equations

doi:10.12677/AAM.2020.911236

Advances in Applied Mathematics
Vol. 09 No. 11 ( 2020 ), Article ID: 38828 , 10 pages
10.12677/AAM.2020.911236

一类具有对数非线性项的四阶抛物型方程解的全局渐近性

彭迪，石鹏

●How to Cite this Article

贵州民族大学，数据与信息工程学院，贵州贵阳

收稿日期：2020年11月2日；录用日期：2020年11月19日；发布日期：2020年11月26日

摘要

本文研究了一类具有对数非线性项的四阶抛物型方程的初值问题。通过运用势阱法及构造相应的能量泛函，证明了低初始能量条件下方程解的渐近性和爆破性，并给出了爆破时间的下界估计。

关键词

四阶抛物型方程，指数衰退，爆破时间，对数非线性

Global Asymptotic Behavior of Solutions for a Class of Fourth-Order Logarithmic Nonlinear Parabolic Equations

Di Peng, Peng Shi

School of Data and Information Engineering, Guizhou Minzu University, Guiyang Guizhou

Received: Nov. 2^nd, 2020; accepted: Nov. 19^th, 2020; published: Nov. 26^th, 2020

ABSTRACT

In this paper, the initial value problem for a class of fourth-order parabolic equations with logarithmic nonlinear terms is studied. By applying the potential well method and constructing the corresponding energy general function, the asymptotic and bursting properties of the solutions are proved for low initial capacities, and lower bound estimates of the bursting time are given.

Keywords:Fourth Order Parabolic Equation, Exponential Decay, Blow-Up Time, Logarithmic Nonlinearity

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

本文研究如下具有对数非线性项的四阶抛物型方程：

${\begin{array}{l} u_{t} + Δ^{2} u - d i v ({| \nabla u |}^{p - 2}) = {| u |}^{p - 2} u \log | u |, & (x, t) \in Ω \times (0, T), \\ u = \frac{\partial u}{\partial υ} = 0, & (x, t) \in Ω \times (0, T), \\ u (x, 0) = u_{0} (x), & x \in Ω . \end{array}$ (1)

其中 $Ω \in R^{N} (N \geq 1)$ 是有光滑边界的有界域， $T \in (0, + \infty]$ ， $υ$ 是 $\partial Ω$ 上的外法向量并且参数p满足下面的条件： $Ω \in R^{N} (N \geq 1)$

${\begin{array}{l} 2 < p < \infty, & N = 1, 2, 3, 4, \\ 2 < p < \frac{2 N}{N - 4}, & N \geq 5. \end{array}$ (2)

具有对数源的四阶抛物型微分方程近年来在工程技术、生物数学、材料科学及物理学中有重要应用( [1] [2] [3] )。Sattinger [4]，Payne、Sattinger [5] 首先提出并建立了势阱法，此后势阱法成为研究非线性发展方程初边值问题解的全局存在性重要工具，之后被Liu、Zhao [6] 进一步完善。近年来，具有对数源的偏微分方程被许多作者广泛研究( [7] [8] )，特别的，通过运用修正的势阱法和对数Sobolev不等式，Liao、Li [8] 研究了(1)在不同初始能量条件下解的性质，此外，本文借鉴了( [9] [10] )中的一些方法。

在本文中，我们应用文献 [8] 的一些记号，定义 $(\cdot, \cdot)$ 为 $L^{2} (Ω)$ 中的内积，

$(u, v) = \int_{Ω} u (x) v (x) d x, \forall u, v \in L^{2} ( Ω )$

${‖ \cdot ‖}_{p}$ 和 ${‖ \cdot ‖}_{H_{0}^{2} (Ω)}$ 分别为 $L^{p} (Ω)$ 和 $H_{0}^{2} (Ω)$ 的范数，即：

$\begin{array}{l} {‖ u ‖}_{p} = {\begin{array}{l} {(\int_{Ω} {| u (x) |}^{p} d x)}^{\frac{1}{p}}, & 1 \leq p < \infty, \\ ess \sup_{x \in Ω} | u (x) |, & p = \infty, \end{array} \forall u \in L^{p} (Ω) . \\ {‖ u ‖}_{H_{0}^{2} (Ω)} = \sqrt{{‖ u ‖}_{2}^{2} + {‖ \nabla u ‖}_{2}^{2} + {‖ Δ u ‖}_{2}^{2}}, \forall u \in H_{0}^{2} (Ω) . \end{array}$ (3)

由 [8] 知 $H_{0}^{2} (Ω)$ 是一个内积为 ${(\cdot, \cdot)}_{H_{0}^{2} (Ω)}$ 的Hilbert空间，内积定义如下

${(u, v)}_{H_{0}^{2} (Ω)} = (Δ u, Δ v), \forall u, v \in H_{0}^{2} (Ω),$

由上式我们可知| ${‖ Δ (\cdot) ‖}_{2}$ 是 ${‖ \cdot ‖}_{H_{0}^{2} (Ω)}$ 的等价范数。因此在下面的证明中，我们记 $H_{0}^{2} (Ω)$ 中的范数为：

${‖ u ‖}_{H_{0}^{2} (Ω)} = {‖ Δ u ‖}_{2}, \forall u \in H_{0}^{2} (Ω) .$ (4)

对任意 $u \in H_{0}^{2} (Ω)$ ，定义泛函：

$J (u) = \frac{1}{2} {‖ Δ u ‖}_{2}^{2} + \frac{1}{p} {‖ \nabla u ‖}_{p}^{p} - \frac{1}{p} \int_{Ω} {| u |}^{p} \log | u | d x + \frac{1}{p^{2}} {‖ u ‖}_{p}^{p},$ (5)

$I (u) = {‖ Δ u ‖}_{2}^{2} + {‖ \nabla u ‖}_{p}^{p} - \int_{Ω} {| u |}^{p} \log | u | d x .$ (6)

由p满足条件(2)知函数 $J (u)$ 和 $I (u)$ 在 $H_{0}^{2} (Ω)$ 上有定义且连续，并满足下面的关系式：

$J (u) = \frac{1}{p} I (u) + \frac{1}{p^{2}} {‖ u ‖}_{p}^{p} + \frac{p - 2}{2 p} {‖ Δ u ‖}_{2}^{2} .$ (7)

定义(1)的Nehari流形

$N = {u \in H_{0}^{2} (Ω) | I (u) = 0, {‖ Δ u ‖}_{2} \neq 0} .$ (8)

令

$W = {u \in H_{0}^{2} (Ω) | I (u) > 0, J (u) < d} \cup {0},$ (9)

$V = {u \in H_{0}^{2} (Ω) | I (u) < 0, J (u) < d},$ (10)

分别为(1)对应的位势阱和阱外集合。

其中，

$d = \inf_{u \in N} J (u)$ (11)

为势阱W的深度。

首先，我们回忆问题(1)弱解的定义：

定义1. 设函数 $u = u (x, t) \in L^{\infty} (0, T; H_{0}^{2} (Ω))$ 且 $u_{t} \in L^{2} (0, T; L^{2} (Ω))$ ，称函数u为问题(1)的弱解。若 $u (x, 0) = u_{0} (x)$ 且满足：

$(u_{t}, ϕ) + (Δ u, Δ ϕ) + ({| \nabla u |}^{p - 2} \nabla u, \nabla ϕ) = ({| u |}^{p - 2} u \log | u |, ϕ), a . e ., t > 0, \forall ϕ \in H_{0}^{2} (Ω),$ (12)

和

$\int_{0}^{t} {‖ u_{τ} ‖}_{2}^{2} d τ + J (u) = J (u_{0}) (\cdot, \cdot), a . e . t > 0.$ (13)

若对任意 $T > 0$ ，上述式子成立，则称u是问题(1)的一个全局弱解。

当 $J (u_{0}) < d$ 时，文献 [8] 给出了如下结论：

定理1. ( [8] 定理3.3) 设p满足条件(2)， $u_{0} \in H_{0}^{2} (Ω)$ 。若 $J (u_{0}) < d$ ， $I (u_{0}) > 0$ ，则问题(1)有一个全局弱解 $u \in L^{\infty} (0, T; H_{0}^{2} (Ω))$ ， $u_{t} \in L^{2} (0, T; L^{2} (Ω))$ 且 $u (t) \in W$ 对任意 $0 \leq t < T$ 成立。此外，假如u是问题(1)在满足条件 $0 < J (u_{0}) < d$ ， $I (u_{0}) > 0$ 下的一个全局弱解，则有

${‖ u ‖}_{W_{0}^{1, p}}^{2} \leq {‖ u_{0} ‖}_{W_{2}^{1, p}}^{2} {(\frac{2}{2 + ω (p - 2) t})}^{2 / (p - 2)}, \forall t \geq 0.$

定理2. ( [8] 定理3.4) 设p满足条件(2)， $u_{0} \in H_{0}^{2} (Ω)$ 。如果u是问题(1)在 $J (u_{0}) < d$ ， $I (u_{0}) < 0$ 的

一个弱解，则存在有限时间 $T^{*} > 0$ 使得u满足 $\lim_{t \to T^{*}} \int_{0}^{t} {‖ u ‖}_{2}^{2} d τ = + \infty$ ，且 $T^{*}$ 满足

$T^{*} \leq t^{*} + \frac{M^{*}}{(p / 2 - 1) M^{'} ( t * )}$

其中

$M (t) = \int_{0}^{t} {‖ u ‖}_{2}^{2} d τ, \forall t \geq 0.$

$t^{*} : = \max {\frac{2 p J (u (0))}{\frac{p - 2}{2 S^{2}} C_{*}}, {(\frac{p {‖ u_{0} ‖}_{2}^{2}}{\frac{p - 2}{4 S^{2}} \frac{C_{*}}{2}})}^{1 / 2}}$

通过对上述结论的分析，我们有如下两个问题：

(i) 问题1只考虑了u按 ${‖ u ‖}_{W_{0}^{1, p}}$ 的渐近性，能否求出u在 ${‖ \cdot ‖}_{H_{0}^{2} (Ω)}$ 下，即 ${‖ Δ u ‖}_{2}$ 的渐近性？(ii) 问题2只考虑了爆破时间T的上界，能否求出爆破时间T的下界？

本文主要回答上述两个问题。

首先考虑下面的特征值问题：

${\begin{array}{l} Δ^{2} u = λ u & if x \in Ω, \\ u = \frac{\partial u}{\partial n} = 0 & if x \in \partial Ω . \end{array}$ (14)

由 [11] 知(14)的第一特征值 $λ_{1} > 0$ 是正的，记为

$λ_{1} = \inf_{0 \neq u \in H_{0}^{2} (Ω)} \frac{{‖ Δ u ‖}_{2}^{2}}{{‖ u ‖}_{2}^{2}},$ (15)

我们有如下结论。

定理3. 设p满足条件(2)且 $u \in H_{0}^{2} (Ω)$ 。如果 $0 < J (u_{0}) < d_{0}$ 且 $I (u_{0}) > 0$ ，则问题(1)的解 $u (t)$ 关于t全局存在，且满足：

${‖ Δ u ‖}_{2}^{2} \leq \frac{2 p}{p - 2} [J (u_{0}) + {‖ u_{0} ‖}_{2}^{2}] \exp [- \frac{β λ_{1} (p - 2)}{λ_{1} (p - 2) + 2 p} t]$

定理4. 设p满足条件(2)且 $u \in H_{0}^{2} (Ω)$ 。如果 $0 < J (u_{0}) < d_{0}$ 且 $I (u_{0}) < 0$ ，则问题(1)的弱解在有限时

间T爆破，即 $\lim_{t \to T} \int_{0}^{t} {‖ u ‖}_{2}^{2} d τ = + \infty$ ，且

$T > \frac{{‖ u_{0} ‖}_{2}^{2 - 2 Γ}}{2^{Γ} \tilde{C} (Γ - 1)} .$

其中

$Γ = \frac{θ (p + ρ_{1})}{2 - (1 - θ) (p + ρ_{1})} > 1,$

$\tilde{C} = {(\frac{{\hat{C}}^{p + ρ_{1}}}{ρ_{1}^{\frac{(1 - θ) (p + ρ_{1})}{2}}})}^{\frac{2}{2 - (1 - θ) (p + ρ_{1})}},$

$\hat{C} = \sup_{H_{0}^{2} (Ω) ∖ {0}} \frac{{‖ u ‖}_{p + ρ_{1}}}{{‖ Δ u ‖}_{2}^{1 - θ} {‖ u ‖}_{2}^{θ}} \in (0, \infty),$

$θ = \frac{2^{*} \cdot 2 - 2 (p + ρ_{1})}{2^{*} (p + ρ_{1}) - 2 (p + ρ_{1})} \in (0, 1),$

$2^{*} = \frac{2 N}{N - 4} .$

注：上述式中的 $θ, \hat{C}, Γ$ 是有意义的，且满足 $θ \in (0, 1), Γ > 1$ 。事实上，1. 由于 $ρ_{1} \in (0, 2^{*} - p)$ ， $p \in (2, 2^{*})$ ，我们有

$p + ρ_{1} < 2^{*},$

故 $θ > 0$ 。另外，由 $p + ρ_{1} > 2$ ，知 $2 \cdot 2^{*} < 2^{*} (p + ρ_{1})$ ，即 $2 \cdot 2^{*} - 2 (p + ρ_{1}) < 2^{*} (p + ρ_{1}) - 2 (p + ρ_{1})$ ，因此 $θ < 1$ 。

2. 由插值不等式及 $2 < p + ρ_{1} < 2^{*}$ 知存在常数C使得

${‖ u ‖}_{p + ρ_{1}} \leq C {‖ u ‖}_{2^{*}}^{1 - θ} {‖ u ‖}_{2}^{θ},$

由 $H_{0}^{2} (Ω)$ 可嵌入 $L^{2^{*}} (Ω)$ ，则存在常数 $S_{2^{*}}$ 满足

$S_{2^{*}} = \sup_{0 \neq u \in H_{0}^{2} (Ω)} \frac{{‖ u ‖}_{2^{*}}}{{‖ Δ u ‖}_{2}} .$

则

${‖ u ‖}_{p + ρ_{1}} \leq C S_{2^{*}}^{1 - θ} {‖ Δ u ‖}_{2}^{1 - θ} {‖ u ‖}_{2}^{θ} .$

其中 $θ$ 满足

$\frac{1}{p + ρ_{1}} = \frac{1 - θ}{2^{*}} + \frac{θ}{2} .$

因此， $\begin{array}{l} \hat{C} = \end{array} C S_{2^{*}}^{1 - θ}$ 是有意义的。

3. 我们证明 $Γ > 1$ ，事实上，由 $Γ$ 和 $θ$ 的定义，直接计算，我们有

$Γ = \frac{2 (2^{*} - p - ρ_{1})}{2 (2^{*} - 2) - 2^{*} (p + ρ_{1} - 2)}$ (16)

又因 $ρ_{1} < 2^{*} - p$ ，我们有

$2^{*} (p + ρ_{1} - 2) < 2 (2^{*} - 2),$

再由 $p + ρ_{1} < 2^{*}$ 和(16)，有

$\begin{array}{l} Γ > 1 \Leftrightarrow 2 (2^{*} - p - ρ_{1}) > 2 (2^{*} - 2) - 2^{*} (p + ρ_{1} - 2) \\ \Leftrightarrow 2^{*} (p + ρ_{1} - 2) > 2 (2^{*} - 2) - 2 (2^{*} - p - ρ_{1}) = 2 (p + ρ_{1} - 2) . \end{array}$ (17)

2. 主要结论的证明

定理3的证明：因 $J (u_{0}) < d$ ， $I (u_{0}) > 0$ 。由定理2，知 $u (t)$ 是问题(1)的一个全局弱解，且 $u (t) \in W$ 对所有 $0 \leq t < \infty$ 成立。

首先，我们考虑 ${‖ Δ u ‖}_{2}^{2}$ 关于时间t的渐近性。由(9)，则

$I (u (t)) \geq 0.$ (18)

则由(13)，(7)，(18)和 $p > 2$ 有

$\begin{matrix} J (u_{0}) \geq J (u (t)) \\ = \frac{1}{p} I (u (t)) + \frac{1}{p^{2}} {‖ u (t) ‖}_{p}^{p} + \frac{p - 2}{2 p} {‖ Δ u (t) ‖}_{2}^{2} \\ \geq \frac{p - 2}{2 p} {‖ Δ u (t) ‖}_{2}^{2}, \end{matrix}$ (19)

由 $\log | u | < \frac{{| u |}^{ρ_{1}}}{ρ_{1}}$ 和(19)知

$\begin{array}{l} \int_{x \in Ω} {| u (t) |}^{p} \log | u (t) | d x \\ = \int_{{x \in Ω : | u (t) | \geq 1}} {| u (t) |}^{p} \log | u (t) | d x + \int_{{x \in Ω : | u (t) | \leq 1}} {| u (t) |}^{p} \log | u (t) | d x \\ \leq \frac{e^{- 1}}{ρ_{1}} \int_{{x \in Ω}} {| u (t) |}^{p + ρ_{1}} d x \leq \frac{e^{- 1}}{ρ_{1}} {‖ u (t) ‖}_{p + ρ 1}^{p + ρ_{1}} \\ = \frac{e^{- 1}}{ρ_{1}} {‖ u (t) ‖}_{p + ρ_{1}}^{p + ρ_{1} - 2} {‖ u (t) ‖}_{p + ρ_{1}}^{2} \leq \frac{e^{- 1}}{ρ_{1}} S_{p + ρ_{1}}^{2} {‖ u (t) ‖}_{p + ρ_{1}}^{p + ρ_{1} - 2} {‖ Δ u (t) ‖}_{2}^{2} \\ \leq \frac{e^{- 1}}{ρ_{1}} S_{p + ρ_{1}}^{p + ρ_{1}} {[\frac{2 p}{p - 2} J (u_{0})]}^{\frac{p + ρ_{1} - 2}{2}} {‖ Δ u (t) ‖}_{2}^{2} . \end{array}$ (20)

其中 $S_{p + ρ_{1}}$ 是 $H_{0}^{2} (Ω)$ 到 $L^{p + ρ_{1}} (Ω)$ 的最佳嵌入常数，且参数 $ρ_{1}$ 满足 $0 < ρ_{1} < \frac{2 N}{N - 4} - p$ 。

在(12)式中 $ϕ$ 取成u，由 $I (u)$ 的定义有如下等式：

$\frac{d}{d t} {‖ u (t) ‖}_{2}^{2} = - 2 ({‖ Δ u (t) ‖}_{2}^{2} + {‖ \nabla u (t) ‖}_{p}^{p} - \int_{x \in Ω} {| u (t) |}^{p} \log | u (t) | d x) = - 2 I (u (t))$ (21)

由上述 $I (u)$ 的定义及(20)则

$\begin{matrix} I (u (t)) = {‖ Δ u (t) ‖}_{2}^{2} + {‖ \nabla u (t) ‖}_{p}^{p} - \int_{Ω} {| u (t) |}^{p} \log | u (t) | d x \\ \geq {‖ Δ u (t) ‖}_{2}^{2} - \frac{e^{- 1}}{ρ_{1}} {‖ u (t) ‖}_{p + ρ_{1}}^{p + ρ_{1}} \\ \geq [1 - \frac{e^{- 1}}{ρ_{1}} S_{p + ρ_{1}}^{p + ρ_{1}} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p + ρ_{1} - 2}{2}}] {‖ Δ u (t) ‖}_{2}^{2}, \end{matrix}$ (22)

作辅助函数

$K (t) : = J (u (t)) + {‖ u (t) ‖}_{2}^{2},$ (23)

则由(15)和(19)，有

$K (t) \leq J (u (t)) + \frac{1}{λ_{1}} {‖ Δ u ‖}_{2}^{2} \leq [1 + \frac{2 p}{λ_{1} (p - 2)}] J (u (t)),$ (24)

此外，由(13)和(21)有

$\frac{d}{d t} K (t) = - {‖ u_{t} ‖}_{2}^{2} - 2 I (u (t)),$

类似于(20)，则

$\begin{matrix} {‖ u (t) ‖}_{p}^{p} = {‖ u (t) ‖}_{p}^{p - 2} {‖ u (t) ‖}_{p}^{2} \\ \leq S_{p}^{2} {‖ u (t) ‖}_{p}^{p - 2} {‖ Δ u ‖}_{2}^{2} \\ \leq S_{p}^{p} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p - 2}{2}} {‖ Δ u ‖}_{2}^{2}, \end{matrix}$ (25)

此处 $S_{p}$ 是 $H_{0}^{2} (Ω)$ 到 $L^{p} (Ω)$ 的最佳嵌入常数。由上式及(15)，(22)，下式成立

$\begin{matrix} \frac{d}{d t} K (t) = - {‖ u^{'} (t) ‖}_{2}^{2} - 2 I (u (t)) - β J (u (t)) + \frac{β}{p} I (u (t)) + \frac{β}{p^{2}} {‖ u (t) ‖}_{p}^{p} + \frac{(p - 2) β}{2 p} {‖ Δ u (t) ‖}_{2}^{2} \\ \leq (- 2 + \frac{β}{p}) I (u (t)) - β J (u (t)) + [\frac{(p - 2) β}{2 p} + \frac{β}{p^{2}} S_{H_{0}^{2} (Ω)}^{p} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p - 2}{2}}] {‖ Δ u (t) ‖}_{2}^{2} \\ \leq (- 2 + \frac{β}{p}) I (u (t)) - β J (u (t)) + [\frac{(p - 2) β}{2 p} + \frac{β}{p^{2}} S_{H_{0}^{2} (Ω)}^{p} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p - 2}{2}}] \\ . {[1 - \frac{e^{- 1}}{ρ_{1}} S_{p + ρ_{1}}^{p + ρ_{1}} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p + ρ_{1} - 2}{2}}]}^{- 1} I (u (t)) \\ = - β J (u (t)) + A I (u (t)) . \end{matrix}$ (26)

其中

$A = - 2 + [\frac{(p - 2) β}{2 p} + \frac{β}{p^{2}} S_{H_{0}^{2} (Ω)}^{p} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p - 2}{2}}] {[1 - \frac{e^{- 1}}{ρ_{1}} S_{p + ρ_{1}}^{p + ρ_{1}} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p + ρ_{1} - 2}{2}}]}^{- 1} + \frac{β}{p} .$

令 $β$ 满足

$β = \frac{2 p}{[\frac{p - 2}{2 p} + \frac{1}{p^{2}} S_{H_{0}^{2} (Ω)}^{p} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p - 2}{2}}] {[1 - \frac{e^{- 1}}{ρ_{1}} S_{p + ρ_{1}}^{p + ρ_{1}} {(\frac{2 p}{p - 2} J (u_{0}))}^{\frac{p + ρ_{1} - 2}{2}}]}^{- 1} + 1}$

则 $A = 0$ ，且由(24)知

$\begin{matrix} \frac{d}{d t} K (t) \leq - β J (u (t)) \\ \leq - β {[1 + \frac{2 p}{λ_{1} (p - 2)}]}^{- 1} K (t) \\ = - \frac{β λ_{1} (p - 2)}{λ_{1} (p - 2) + 2 p} K (t), \end{matrix}$ (27)

又由(23)，有

$\begin{matrix} J (u (t)) + {‖ u (t) ‖}_{2}^{2} = K (t) \leq K (0) \exp [- \frac{β λ_{1} (p - 2)}{λ_{1} (p - 2) + 2 p} t] \\ = [J (u_{0}) + {‖ u_{0} ‖}_{2}^{2}] \exp [- \frac{β λ_{1} (p - 2)}{λ_{1} (p - 2) + 2 p} t] . \end{matrix}$ (28)

则由(19)和(28)，可得

$\begin{matrix} {‖ Δ u ‖}_{2}^{2} \leq \frac{2 p}{p - 2} J (u (t)) \\ \leq \frac{2 p}{p - 2} [J (u_{0}) + {‖ u_{0} ‖}_{2}^{2}] \exp [- \frac{β λ_{1} (p - 2)}{λ_{1} (p - 2) + 2 p} t] . \end{matrix}$ (29)

定理4的证明：由定理2知存在有限时间 $T > 0$ ，使得问题(1)的弱解u在有限时间T爆破。下面，我们求T的下界。首先，我们定义如下一个函数

$L (t) : = \frac{1}{2} {‖ u (t) ‖}_{2}^{2},$

则有

$\lim_{t \to T} L (t) = \infty$ (30)

又根据(19)，则有

$\frac{1}{2} \frac{d}{d t} {‖ u (t) ‖}_{2}^{2} = - I (u (t)) = - {‖ Δ u (t) ‖}_{2}^{2} - {‖ \nabla u (t) ‖}_{p}^{p} + \int_{Ω} {| u (t) |}^{p} \log | u (t) | d x .$ (31)

由定理2，知 $I (u (t)) < 0$ 。因此，由不等式 $\log | u (t) | < \frac{{| u (t) |}^{ρ_{1}}}{ρ_{1}}$ 对任意 $ρ_{1} > 0$ 成立，下式

$\begin{matrix} {‖ u (t) ‖}_{p + ρ_{1}}^{p + ρ_{1}} \leq {\hat{C}}^{p + ρ_{1}} {‖ Δ u (t) ‖}_{2}^{(1 - θ) (p + ρ_{1})} {‖ u (t) ‖}_{2}^{θ (p + ρ_{1})} \\ = {\hat{C}}^{p + ρ_{1}} {({‖ Δ u (t) ‖}_{2}^{2})}^{\frac{(1 - θ) (p + ρ_{1})}{2}} \cdot {‖ u (t) ‖}_{2}^{θ (p + ρ_{1})} \\ \leq {\hat{C}}^{p + ρ_{1}} {({‖ Δ u (t) ‖}_{2}^{2} + {‖ \nabla u (t) ‖}_{p}^{p})}^{\frac{(1 - θ) (p + ρ_{1})}{2}} \cdot {‖ u (t) ‖}_{2}^{θ (p + ρ_{1})} \\ < {\hat{C}}^{p + ρ_{1}} {(\int_{Ω} {| u (t) |}^{p} \log | u (t) | d x)}^{\frac{(1 - θ) (p + ρ_{1})}{2}} \cdot {‖ u (t) ‖}_{2}^{θ (p + ρ_{1})} \\ < \frac{{\hat{C}}^{p + ρ_{1}}}{ρ_{1}^{\frac{(1 - θ) (p + ρ_{1})}{2}}} {({‖ u ‖}_{p + ρ_{1}}^{p + ρ_{1}})}^{\frac{(1 - θ) (p + ρ_{1})}{2}} \cdot {‖ u (t) ‖}_{2}^{θ (p + ρ_{1})} . \end{matrix}$ (32)

成立，其中 $ρ_{1} \in (0, \frac{2 N}{N - 4} - p)$ 。

又由于 $0 < ρ_{1} < 2^{*} - p$ ， $2 < p < 2^{*}$ 和 $θ = \frac{2^{*} \cdot 2 - 2 (p + ρ_{1})}{2^{*} (p + ρ_{1}) - 2 (p + ρ_{1})} \in (0, 1)$ ，故 $\frac{(1 - θ) (p + ρ_{1})}{2} < 1$ 。

因此，由(32)，有

${‖ u (t) ‖}_{p + ρ_{1}}^{p + ρ_{1}} < {(\frac{{\hat{C}}^{p + ρ_{1}}}{ρ_{1}^{\frac{(1 - θ) (p + ρ_{1})}{2}}})}^{\frac{2}{2 - (1 - θ) (p + ρ_{1})}} {({‖ u (t) ‖}_{2}^{2})}^{\frac{θ (p + ρ_{1})}{2 - (1 - θ) (p + ρ_{1})}}$ (33)

此外，由注记

$Γ = \frac{θ (p + ρ_{1})}{2 - (1 - θ) (p + ρ_{1})} > 1.$

联立(31)和(33)有

$\begin{matrix} L^{'} (t) = - {‖ Δ u (t) ‖}_{2}^{2} - {‖ \nabla u (t) ‖}_{p}^{p} + \int_{Ω} {| u (t) |}^{p} \log | u (t) | d x \\ \leq \int_{Ω} {| u (t) |}^{p} \log | u (t) | d x \leq \frac{1}{ε} {‖ u (t) ‖}_{p + ρ_{1}}^{p + ρ_{1}} \\ < \tilde{C} {({‖ u (t) ‖}_{2}^{2})}^{Γ} = 2^{Γ} \tilde{C} {(L (t))}^{Γ}, \end{matrix}$ (34)

其中 $\tilde{C} = {(\frac{{\hat{C}}^{p + ρ_{1}}}{ρ_{1}^{\frac{(1 - θ) (p + ρ_{1})}{2}}})}^{\frac{2}{2 - (1 - θ) (p + ρ_{1})}}$ 。

下面我们用反证法证明对任意 $t \in [0, T)$ ， $L (t) > 0$ 。假设存在一个 $t_{1} \geq 0$ 使得 ${‖ u (t_{1}) ‖}_{2}^{2} = 0$ ，则由(33)有 ${‖ u (t) ‖}_{p + ρ_{1}}^{p + ρ_{1}} < 0$ ，矛盾。则由(34)有

$\frac{L^{'} (t)}{{(L (t))}^{Γ}} < 2^{Γ} \tilde{C}$ (35)

对上式从0到t积分，有

${(L (0))}^{1 - Γ} - {(L (t))}^{1 - Γ} < 2^{Γ} \tilde{C} (Γ - 1) t$ (36)

让(37)中 $t \to T$ 并且运用(32)我们可得下面结论

$T > \frac{{(L (0))}^{1 - Γ}}{2^{Γ} \tilde{C} (Γ - 1)} = \frac{{‖ u_{0} ‖}_{2}^{2 - 2 Γ}}{2^{Γ} \tilde{C} (Γ - 1)}$ 。

致谢

作者对同行评阅人的意见和建议表示深深感谢。

文章引用

彭迪,石鹏. 一类具有对数非线性项的四阶抛物型方程解的全局渐近性
Global Asymptotic Behavior of Solutions for a Class of Fourth-Order Logarithmic Nonlinear Parabolic Equations[J]. 应用数学进展, 2020, 09(11): 2036-2045. https://doi.org/10.12677/AAM.2020.911236

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