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

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。

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.

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：

$\{\begin{array}{ll}{u}_t+{\Delta }^{2}u-div\left({\left|\nabla u\right|}^{p-2}\right)={\left|u\right|}^{p-2}u\log \left|u\right|,\hfill & \left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ u=\frac{\partial u}{\partial \upsilon }=0,\hfill & \left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & x\in \Omega .\hfill \end{array}$ (1)

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

$\{\begin{array}{ll}2<p<\infty ,\hfill & N=1,2,3,4,\hfill \\ 2<p<\frac{2N}{N-4},\hfill & N\ge 5.\hfill \end{array}$ (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 $\left(\cdot ,\cdot \right)$ 为 $L^2\left(\Omega \right)$ 中的内积，

$\left(u,v\right)={\displaystyle {\int }_{\Omega }u}\left(x\right)v\left(x\right)\text{d}x,\forall u,v\in L^2(\; \Omega \; )$

${\Vert \cdot \Vert }_p$ 和 ${\Vert \cdot \Vert }_{H_0^2\left(\Omega \right)}$ 分别为 $L^p\left(\Omega \right)$ 和 $H_0^2\left(\Omega \right)$ 的范数，即：

$\begin{array}{l}{\Vert u\Vert }_p=\{\begin{array}{ll}{\left({\displaystyle {\int }_{\Omega }{\left|u\left(x\right)\right|}^p\text{d}x}\right)}^{\frac{1}{p}},\hfill & 1\le p<\infty ,\hfill \\ \text{ess}\underset{x\in \Omega }{\sup }\left|u\left(x\right)\right|,\hfill & p=\infty ,\hfill \end{array}\forall u\in L^p\left(\Omega \right).\\ {\Vert u\Vert }_{H_0^2\left(\Omega \right)}=\sqrt{{\Vert u\Vert }_2^2+{\Vert \nabla u\Vert }_2^2+{\Vert \Delta u\Vert }_2^2},\forall u\in H_0^2\left(\Omega \right).\end{array}$ (3)

由 [8] 知 $H_0^2\left(\Omega \right)$ 是一个内积为 ${\left(\cdot ,\cdot \right)}_{H_0^2\left(\Omega \right)}$ 的Hilbert空间，内积定义如下

${\left(u,v\right)}_{H_0^2\left(\Omega \right)}=\left(\Delta u,\Delta v\right),\forall u,v\in H_0^2\left(\Omega \right),$

由上式我们可知| ${\Vert \Delta (\cdot )\Vert }_2$ 是 ${\Vert \cdot \Vert }_{H_0^2\left(\Omega \right)}$ 的等价范数。因此在下面的证明中，我们记 $H_0^2\left(\Omega \right)$ 中的范数为：

${\Vert u\Vert }_{H_0^2\left(\Omega \right)}={\Vert \Delta u\Vert }_2,\forall u\in H_0^2\left(\Omega \right).$ (4)

对任意 $u\in H_0^2\left(\Omega \right)$，定义泛函：

$J\left(u\right)=\frac{1}{2}{\Vert \Delta u\Vert }_2^2+\frac{1}{p}{\Vert \nabla u\Vert }_p^p-\frac{1}{p}{\displaystyle {\int }_{\Omega }{\left|u\right|}^p\log \left|u\right|\text{d}x}+\frac{1}{p^2}{\Vert u\Vert }_p^p,$ (5)

$I\left(u\right)={\Vert \Delta u\Vert }_2^2+{\Vert \nabla u\Vert }_p^p-{\displaystyle {\int }_{\Omega }{\left|u\right|}^p}\log \left|u\right|\text{d}x.$ (6)

由p满足条件(2)知函数 $J\left(u\right)$ 和 $I\left(u\right)$ 在 $H_0^2\left(\Omega \right)$ 上有定义且连续，并满足下面的关系式：

$J\left(u\right)=\frac{1}{p}I\left(u\right)+\frac{1}{p^2}{\Vert u\Vert }_p^p+\frac{p-2}{2p}{\Vert \Delta u\Vert }_2^2.$ (7)

定义(1)的Nehari流形

$N=\left\{u\in H_0^2\left(\Omega \right)|I\left(u\right)=0,{\Vert \Delta u\Vert }_2\ne 0\right\}.$ (8)

令

$W=\left\{u\in H_0^2\left(\Omega \right)|I\left(u\right)>0,J\left(u\right)<d\right\}\cup \left\{0\right\},$ (9)

$V=\left\{u\in H_0^2\left(\Omega \right)|I\left(u\right)<0,J\left(u\right)<d\right\},$ (10)

分别为(1)对应的位势阱和阱外集合。

其中，

$d=\underset{u\in N}{\inf }J\left(u\right)$ (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 $u=u\left(x,t\right)\in L^{\infty }\left(0,T;H_0^2\left(\Omega \right)\right)$ 且 $u_t\in L^2\left(0,T;L^2\left(\Omega \right)\right)$，称函数u为问题(1)的弱解。若 $u\left(x,0\right)=u_0\left(x\right)$ 且满足：

$\left(u_t,\varphi \right)+\left(\Delta u,\Delta \varphi \right)+\left({\left|\nabla u\right|}^{p-2}\nabla u,\nabla \varphi \right)=\left({\left|u\right|}^{p-2}u\log \left|u\right|,\varphi \right),a.e.,t>0,\forall \varphi \in H_0^2\left(\Omega \right),$ (12)

和

${\displaystyle {\int }_0^t{\Vert u_{\tau }\Vert }_2^2\text{d}\tau }+J\left(u\right)=J\left(u_0\right)\left(\cdot ,\cdot \right)\text{,}a\text{.}e.t>0.$ (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 $J\left(u_0\right)<d$ 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)， $u_0\in H_0^2\left(\Omega \right)$。若 $J\left(u_0\right)<d$， $I\left(u_0\right)>0$，则问题(1)有一个全局弱解 $u\in L^{\infty }\left(0,T;H_0^2\left(\Omega \right)\right)$， $u_t\in L^2\left(0,T;L^2\left(\Omega \right)\right)$ 且 $u\left(t\right)\in W$ 对任意 $0\le t<T$ 成立。此外，假如u是问题(1)在满足条件 $0<J\left(u_0\right)<d$， $I\left(u_0\right)>0$ 下的一个全局弱解，则有

${\Vert u\Vert }_{W_0^{1,p}}^2\le {\Vert u_0\Vert }_{W_2^{1,p}}^2{\left(\frac{2}{2+\omega \left(p-2\right)t}\right)}^{2/\left(p-2\right)},\forall t\ge 0.$

定理2. ( [8] 定理3.4) 设p满足条件(2)， $u_0\in H_0^2\left(\Omega \right)$。如果u是问题(1)在 $J\left(u_0\right)<d$， $I\left(u_0\right)<0$ 的

一个弱解，则存在有限时间 $T^{\ast }>0$ 使得u满足 $\underset{t\to T^{\ast }}{\lim }{\displaystyle {\int }_0^t{\Vert u\Vert }_2^2\text{d}\tau }=+\infty$，且 $T^{*}$ 满足

$T^{\ast }\le t^{\ast }+\frac{M^{\ast }}{\left(p/2-1\right)M^{\prime }(\; t\; \ast \; )}$

其中

$M\left(t\right)={\displaystyle {\int }_0^t{\Vert u\Vert }_2^2\text{d}\tau },\forall t\ge 0.$

$t^{\ast }:=\max \left\{\frac{2pJ\left(u\left(0\right)\right)}{\frac{p-2}{2S^2}{C}_{\ast }},{\left(\frac{p{\Vert u_0\Vert }_2^2}{\frac{p-2}{4S^2}\frac{C_{\ast }}{2}}\right)}^{1/2}\right\}$

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 ${\Vert u\Vert }_{W_0^{1,p}}$ 的渐近性，能否求出u在 ${\Vert \cdot \Vert }_{H_0^2\left(\Omega \right)}$ 下，即 ${\Vert \Delta u\Vert }_2$ 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：

$\{\begin{array}{ll}{\Delta }^{2}u=\lambda u\hfill & \text{if}x\in \Omega ,\hfill \\ u=\frac{\partial u}{\partial n}=0\hfill & \text{if}x\in \partial \Omega .\hfill \end{array}$ (14)

由 [11] 知(14)的第一特征值 ${\lambda }_1>0$ 是正的，记为

${\lambda }_1=\underset{0\ne u\in H_0^2\left(\Omega \right)}{\inf }\frac{{\Vert \Delta u\Vert }_2^2}{{\Vert u\Vert }_2^2},$ (15)

我们有如下结论。

定理3. 设p满足条件(2)且 $u\in H_0^2\left(\Omega \right)$。如果 $0<J\left(u_0\right)<d_0$ 且 $I\left(u_0\right)>0$，则问题(1)的解 $u\left(t\right)$ 关于t全局存在，且满足：

${\Vert \text{Δ}u\Vert }_2^2\le \frac{2p}{p-2}\left[J\left(u_0\right)+{\Vert u_0\Vert }_2^2\right]\exp \left[-\frac{\beta {\lambda }_1\left(p-2\right)}{{\lambda }_1\left(p-2\right)+2p}t\right]$

定理4. 设p满足条件(2)且 $u\in H_0^2\left(\Omega \right)$。如果 $0<J\left(u_0\right)<d_0$ 且 $I\left(u_0\right)<0$，则问题(1)的弱解在有限时

间T爆破，即 $\underset{t\to T}{\lim }{\displaystyle {\int }_0^t{\Vert u\Vert }_2^2\text{d}\tau }=+\infty$，且

$T>\frac{{\Vert u_0\Vert }_2^{2-2\Gamma }}{2^{\Gamma }\tilde{C}\left(\Gamma -1\right)}.$

其中

$\Gamma =\frac{\theta \left(p+{\rho }_1\right)}{2-\left(1-\theta \right)\left(p+{\rho }_1\right)}>1,$

$\tilde{C}={\left(\frac{{\hat{C}}^{p+{\rho }_1}}{{\rho }_1^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}}\right)}^{\frac{2}{2-\left(1-\theta \right)\left(p+{\rho }_1\right)}},$

$\hat{C}=\underset{H_0^2\left(\Omega \right)\setminus \left\{0\right\}}{\sup }\frac{{\Vert u\Vert }_{p+{\rho }_1}}{{\Vert \Delta u\Vert }_2^{1-\theta }{\Vert u\Vert }_2^{\theta }}\in \left(0,\infty \right),$

$\theta =\frac{2^{\ast }\cdot 2-2\left(p+{\rho }_1\right)}{2^{\ast }\left(p+{\rho }_1\right)-2\left(p+{\rho }_1\right)}\in \left(0,1\right),$

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

注：上述式中的 $\theta ,\hat{C},\Gamma$ 是有意义的，且满足 $\theta \in \left(0,1\right),\Gamma >1$。事实上，1. 由于 ${\rho }_1\in \left(0,2^{\ast }-p\right)$， $p\in \left(2,2^{\ast }\right)$，我们有

$p+{\rho }_1<2^{\ast },$

故 $\theta >0$。另外，由 $p+{\rho }_1>2$，知 $2\cdot 2^{\ast }<2^{\ast }\left(p+{\rho }_1\right)$，即 $2\cdot 2^{\ast }-2\left(p+{\rho }_1\right)<2^{\ast }\left(p+{\rho }_1\right)-2\left(p+{\rho }_1\right)$，因此 $\theta <1$。

2. 由插值不等式及 $2<p+{\rho }_1<2^{\ast }$ 知存在常数C使得

${\Vert u\Vert }_{p+{\rho }_1}\le C{\Vert u\Vert }_{2^{\ast }}^{1-\theta }{\Vert u\Vert }_2^{\theta },$

由 $H_0^2\left(\Omega \right)$ 可嵌入 $L^{2^{\ast }}\left(\Omega \right)$，则存在常数 $S_{2^{\ast }}$ 满足

$S_{2^{\ast }}=\underset{0\ne u\in H_0^2\left(\Omega \right)}{\sup }\frac{{\Vert u\Vert }_{2^{\ast }}}{{\Vert \Delta u\Vert }_2}.$

则

${\Vert u\Vert }_{p+{\rho }_1}\le C{S}_{2^{\ast }}^{1-\theta }{\Vert \Delta u\Vert }_2^{1-\theta }{\Vert u\Vert }_2^{\theta }.$

其中 $\theta$ 满足

$\frac{1}{p+{\rho }_1}=\frac{1-\theta }{2^{\ast }}+\frac{\theta }{2}.$

因此， $\begin{array}{l}\hat{C}=\hfill \end{array}C{S}_{2^{\ast }}^{1-\theta }$ 是有意义的。

3. 我们证明 $\Gamma >1$，事实上，由 $\Gamma$ 和 $\theta$ 的定义，直接计算，我们有

$\Gamma =\frac{2\left(2^{\ast }-p-{\rho }_1\right)}{2\left(2^{\ast }-2\right)-2^{\ast }\left(p+{\rho }_1-2\right)}$ (16)

又因 ${\rho }_1<2^{\ast }-p$，我们有

$2^{\ast }\left(p+{\rho }_1-2\right)<2\left(2^{\ast }-2\right),$

再由 $p+{\rho }_1<2^{\ast }$ 和(16)，有

$\begin{array}{l}\Gamma >1\iff 2\left(2^{\ast }-p-{\rho }_1\right)>2\left(2^{\ast }-2\right)-2^{\ast }\left(p+{\rho }_1-2\right)\\ \iff 2^{\ast }\left(p+{\rho }_1-2\right)>2\left(2^{\ast }-2\right)-2\left(2^{\ast }-p-{\rho }_1\right)=2\left(p+{\rho }_1-2\right).\end{array}$ (17)

2. 主要结论的证明

定理3的证明：因 $J\left(u_0\right)<d$， $I\left(u_0\right)>0$。由定理2，知 $u\left(t\right)$ 是问题(1)的一个全局弱解，且 $u\left(t\right)\in W$ 对所有 $0\le t<\infty$ 成立。

首先，我们考虑 ${\Vert \Delta u\Vert }_2^2$ 关于时间t的渐近性。由(9)，则

$I\left(u\left(t\right)\right)\ge 0.$ (18)

则由(13)，(7)，(18)和 $p>2$ 有

$\begin{array}{c}J\left(u_0\right)\ge J\left(u\left(t\right)\right)\\ =\frac{1}{p}I\left(u\left(t\right)\right)+\frac{1}{p^2}{\Vert u\left(t\right)\Vert }_p^p+\frac{p-2}{2p}{\Vert \Delta u\left(t\right)\Vert }_2^2\\ \ge \frac{p-2}{2p}{\Vert \Delta u\left(t\right)\Vert }_2^2,\end{array}$ (19)

由 $\log \left|u\right|<\frac{{\left|u\right|}^{{\rho }_1}}{{\rho }_1}$ 和(19)知

$\begin{array}{l}{\displaystyle {\int }_{x\in \Omega }{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}\\ ={\displaystyle {\int }_{\left\{x\in \Omega :\left|u\left(t\right)\right|\ge 1\right\}}{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}+{\displaystyle {\int }_{\left\{x\in \Omega :\left|u\left(t\right)\right|\le 1\right\}}{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}\\ \le \frac{e^{-1}}{{\rho }_1}{\displaystyle {\int }_{\left\{x\in \Omega \right\}}{\left|u\left(t\right)\right|}^{p+{\rho }_1}dx}\le \frac{e^{-1}}{{\rho }_1}{\Vert u\left(t\right)\Vert }_{p+\rho 1}^{p+{\rho }_1}\\ =\frac{e^{-1}}{{\rho }_1}{\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^{p+{\rho }_1-2}{\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^2\le \frac{e^{-1}}{{\rho }_1}{S}_{p+{\rho }_1}^2{\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^{p+{\rho }_1-2}{\Vert \Delta u\left(t\right)\Vert }_2^2\\ \le \frac{e^{-1}}{{\rho }_1}{S}_{p+{\rho }_1}^{p+{\rho }_1}{\left[\frac{2p}{p-2}J\left(u_0\right)\right]}^{\frac{p+{\rho }_1-2}{2}}{\Vert \Delta u\left(t\right)\Vert }_2^2.\end{array}$ (20)

其中 $S_{p+{\rho }_1}$ 是 $H_0^2\left(\Omega \right)$ 到 $L^{p+{\rho }_1}\left(\Omega \right)$ 的最佳嵌入常数，且参数 ${\rho }_1$ 满足 $0<{\rho }_1<\frac{2N}{N-4}-p$。

在(12)式中 $\varphi$ 取成u，由 $I\left(u\right)$ 的定义有如下等式：

$\frac{\text{d}}{\text{d}t}{\Vert u\left(t\right)\Vert }_2^2=-2\left({\Vert \Delta u\left(t\right)\Vert }_2^2+{\Vert \nabla u\left(t\right)\Vert }_p^p-{\displaystyle {\int }_{x\in \Omega }{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}\right)=-2I\left(u\left(t\right)\right)$ (21)

由上述 $I\left(u\right)$ 的定义及(20)则

$\begin{array}{c}I\left(u\left(t\right)\right)={\Vert \Delta u\left(t\right)\Vert }_2^2+{\Vert \nabla u\left(t\right)\Vert }_p^p-{\displaystyle {\int }_{\Omega }{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}\\ \ge {\Vert \Delta u\left(t\right)\Vert }_2^2-\frac{e^{-1}}{{\rho }_1}{\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^{p+{\rho }_1}\\ \ge \left[1-\frac{e^{-1}}{{\rho }_1}{S}_{p+{\rho }_1}^{p+{\rho }_1}{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p+{\rho }_1-2}{2}}\right]{\Vert \Delta u\left(t\right)\Vert }_2^2,\end{array}$ (22)

作辅助函数

$K\left(t\right):=J\left(u\left(t\right)\right)+{\Vert u\left(t\right)\Vert }_2^2,$ (23)

则由(15)和(19)，有

$K\left(t\right)\le J\left(u\left(t\right)\right)+\frac{1}{{\lambda }_1}{\Vert \Delta u\Vert }_2^2\le \left[1+\frac{2p}{{\lambda }_1\left(p-2\right)}\right]J\left(u\left(t\right)\right),$ (24)

此外，由(13)和(21)有

$\frac{\text{d}}{\text{d}t}K\left(t\right)=-{\Vert u_t\Vert }_2^2-2I\left(u\left(t\right)\right),$

类似于(20)，则

$\begin{array}{c}{\Vert u\left(t\right)\Vert }_p^p={\Vert u\left(t\right)\Vert }_p^{p-2}{\Vert u\left(t\right)\Vert }_p^2\\ \le S_p^2{\Vert u\left(t\right)\Vert }_p^{p-2}{\Vert \Delta u\Vert }_2^2\\ \le S_p^p{\left(\frac{2p}{p-2}J(u_0)\right)}^{\frac{p-2}{2}}{\Vert \Delta u\Vert }_2^2,\end{array}$ (25)

此处 $S_p$ 是 $H_0^2\left(\Omega \right)$ 到 $L^p\left(\Omega \right)$ 的最佳嵌入常数。由上式及(15)，(22)，下式成立

$\begin{array}{c}\frac{\text{d}}{\text{d}t}K\left(t\right)=-{\Vert u^{\prime }\left(t\right)\Vert }_2^2-2I\left(u\left(t\right)\right)-\beta J\left(u\left(t\right)\right)+\frac{\beta }{p}I\left(u\left(t\right)\right)+\frac{\beta }{p^2}{\Vert u\left(t\right)\Vert }_p^p+\frac{\left(p-2\right)\beta }{2p}{\Vert \Delta u\left(t\right)\Vert }_2^2\\ \le \left(-2+\frac{\beta }{p}\right)I\left(u\left(t\right)\right)-\beta J\left(u\left(t\right)\right)+\left[\frac{\left(p-2\right)\beta }{2p}+\frac{\beta }{p^2}{S}_{H_0^2\left(\Omega \right)}^p{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p-2}{2}}\right]{\Vert \Delta u\left(t\right)\Vert }_2^2\\ \le \left(-2+\frac{\beta }{p}\right)I\left(u\left(t\right)\right)-\beta J\left(u\left(t\right)\right)+\left[\frac{\left(p-2\right)\beta }{2p}+\frac{\beta }{p^2}{S}_{H_0^2\left(\Omega \right)}^p{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p-2}{2}}\right]\\ .{\left[1-\frac{e^{-1}}{{\rho }_1}{S}_{p+{\rho }_1}^{p+{\rho }_1}{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p+{\rho }_1-2}{2}}\right]}^{-1}I\left(u\left(t\right)\right)\\ =-\beta J\left(u\left(t\right)\right)+AI\left(u\left(t\right)\right).\end{array}$ (26)

其中

$A=-2+\left[\frac{\left(p-2\right)\beta }{2p}+\frac{\beta }{p^2}{S}_{H_0^2\left(\Omega \right)}^p{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p-2}{2}}\right]{\left[1-\frac{e^{-1}}{{\rho }_1}{S}_{p+{\rho }_1}^{p+{\rho }_1}{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p+{\rho }_1-2}{2}}\right]}^{-1}+\frac{\beta }{p}.$

令 $\beta$ 满足

$\beta =\frac{2p}{\left[\frac{p-2}{2p}+\frac{1}{p^2}{S}_{H_0^2\left(\Omega \right)}^p{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p-2}{2}}\right]{\left[1-\frac{e^{-1}}{{\rho }_1}{S}_{p+{\rho }_1}^{p+{\rho }_1}{\left(\frac{2p}{p-2}J\left(u_0\right)\right)}^{\frac{p+{\rho }_1-2}{2}}\right]}^{-1}+1}$

则 $A=0$，且由(24)知

$\begin{array}{c}\frac{\text{d}}{\text{d}t}K\left(t\right)\le -\beta J\left(u\left(t\right)\right)\\ \le -\beta {\left[1+\frac{2p}{{\lambda }_1\left(p-2\right)}\right]}^{-1}K\left(t\right)\\ =-\frac{\beta {\lambda }_1\left(p-2\right)}{{\lambda }_1\left(p-2\right)+2p}K\left(t\right),\end{array}$ (27)

又由(23)，有

$\begin{array}{c}J\left(u\left(t\right)\right)+{\Vert u\left(t\right)\Vert }_2^2=K\left(t\right)\le K\left(0\right)\exp \left[-\frac{\beta {\lambda }_1\left(p-2\right)}{{\lambda }_1\left(p-2\right)+2p}t\right]\\ =\left[J\left(u_0\right)+{\Vert u_0\Vert }_2^2\right]\exp \left[-\frac{\beta {\lambda }_1\left(p-2\right)}{{\lambda }_1\left(p-2\right)+2p}t\right].\end{array}$ (28)

则由(19)和(28)，可得

$\begin{array}{c}{\Vert \Delta u\Vert }_2^2\le \frac{2p}{p-2}J\left(u\left(t\right)\right)\\ \le \frac{2p}{p-2}\left[J\left(u_0\right)+{\Vert u_0\Vert }_2^2\right]\exp \left[-\frac{\beta {\lambda }_1\left(p-2\right)}{{\lambda }_1\left(p-2\right)+2p}t\right].\end{array}$ (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

$L\left(t\right):=\frac{1}{2}{\Vert u\left(t\right)\Vert }_2^2,$

则有

$\underset{t\to T}{\lim }L\left(t\right)=\infty$ (30)

又根据(19)，则有

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert u\left(t\right)\Vert }_2^2=-I\left(u\left(t\right)\right)=-{\Vert \Delta u\left(t\right)\Vert }_2^2-{\Vert \nabla u\left(t\right)\Vert }_p^p+{\displaystyle {\int }_{\Omega }{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}.$ (31)

由定理2，知 $I\left(u\left(t\right)\right)<0$。因此，由不等式 $\log \left|u\left(t\right)\right|<\frac{{\left|u\left(t\right)\right|}^{{\rho }_1}}{{\rho }_1}$ 对任意 ${\rho }_1>0$ 成立，下式

$\begin{array}{c}{\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^{p+{\rho }_1}\le {\hat{C}}^{p+{\rho }_1}{\Vert \Delta u\left(t\right)\Vert }_2^{\left(1-\theta \right)\left(p+{\rho }_1\right)}{\Vert u\left(t\right)\Vert }_2^{\theta \left(p+{\rho }_1\right)}\\ ={\hat{C}}^{p+{\rho }_1}{\left({\Vert \Delta u\left(t\right)\Vert }_2^2\right)}^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}\cdot {\Vert u\left(t\right)\Vert }_2^{\theta \left(p+{\rho }_1\right)}\\ \le {\hat{C}}^{p+{\rho }_1}{\left({\Vert \Delta u\left(t\right)\Vert }_2^2+{\Vert \nabla u\left(t\right)\Vert }_p^p\right)}^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}\cdot {\Vert u\left(t\right)\Vert }_2^{\theta \left(p+{\rho }_1\right)}\\ <{\hat{C}}^{p+{\rho }_1}{\left({\displaystyle {\int }_{\Omega }{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}\right)}^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}\cdot {\Vert u\left(t\right)\Vert }_2^{\theta \left(p+{\rho }_1\right)}\\ <\frac{{\hat{C}}^{p+{\rho }_1}}{{\rho }_1^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}}{\left({\Vert u\Vert }_{p+{\rho }_1}^{p+{\rho }_1}\right)}^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}\cdot {\Vert u\left(t\right)\Vert }_2^{\theta \left(p+{\rho }_1\right)}.\end{array}$ (32)

成立，其中 ${\rho }_1\in \left(0,\frac{2N}{N-4}-p\right)$。

又由于 $0<{\rho }_1<2^{\ast }-p$， $2<p<2^{\ast }$ 和 $\theta =\frac{2^{\ast }\cdot 2-2\left(p+{\rho }_1\right)}{2^{\ast }\left(p+{\rho }_1\right)-2\left(p+{\rho }_1\right)}\in \left(0,1\right)$，故 $\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}<1$。

因此，由(32)，有

${\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^{p+{\rho }_1}<{\left(\frac{{\hat{C}}^{p+{\rho }_1}}{{\rho }_1^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}}\right)}^{\frac{2}{2-\left(1-\theta \right)\left(p+{\rho }_1\right)}}{\left({\Vert u\left(t\right)\Vert }_2^2\right)}^{\frac{\theta \left(p+{\rho }_1\right)}{2-\left(1-\theta \right)\left(p+{\rho }_1\right)}}$ (33)

此外，由注记

$\Gamma =\frac{\theta \left(p+{\rho }_1\right)}{2-\left(1-\theta \right)\left(p+{\rho }_1\right)}>1.$

联立(31)和(33)有

$\begin{array}{c}{L}^{\prime }\left(t\right)=-{\Vert \Delta u\left(t\right)\Vert }_2^2-{\Vert \nabla u\left(t\right)\Vert }_p^p+{\displaystyle {\int }_{\Omega }{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}\\ \le {\displaystyle {\int }_{\Omega }{\left|u\left(t\right)\right|}^p\log \left|u\left(t\right)\right|\text{d}x}\le \frac{1}{\epsilon }{\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^{p+{\rho }_1}\\ <\tilde{C}{\left({\Vert u\left(t\right)\Vert }_2^2\right)}^{\Gamma }=2^{\Gamma }\tilde{C}{\left(L\left(t\right)\right)}^{\Gamma },\end{array}$ (34)

其中 $\tilde{C}={\left(\frac{{\hat{C}}^{p+{\rho }_1}}{{\rho }_1^{\frac{\left(1-\theta \right)\left(p+{\rho }_1\right)}{2}}}\right)}^{\frac{2}{2-\left(1-\theta \right)\left(p+{\rho }_1\right)}}$。

下面我们用反证法证明对任意 $t\in \left[0,T\right)$， $L\left(t\right)>0$。假设存在一个 $t_1\ge 0$ 使得 ${\Vert u\left(t_1\right)\Vert }_2^2=0$，则由(33)有 ${\Vert u\left(t\right)\Vert }_{p+{\rho }_1}^{p+{\rho }_1}<0$，矛盾。则由(34)有

$\frac{L^{\prime }\left(t\right)}{{\left(L\left(t\right)\right)}^{\Gamma }}<2^{\Gamma }\tilde{C}$ (35)

对上式从0到t积分，有

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

让(37)中 $t\to T$ 并且运用(32)我们可得下面结论

$T>\frac{{\left(L\left(0\right)\right)}^{1-\Gamma }}{2^{\Gamma }\tilde{C}\left(\Gamma -1\right)}=\frac{{\Vert u_0\Vert }_2^{2-2\Gamma }}{2^{\Gamma }\tilde{C}\left(\Gamma -1\right)}$。

致谢

作者对同行评阅人的意见和建议表示深深感谢。

参考文献