一类具变指数源项的Klein-Gordon方程解的爆破

The Blow-Up of Solutions to the Klein-Gordon Equation with Variable Exponent Source Term

摘要: 本文研究了一类具有变指数源项和非线性阻尼项的Klein-Gordon方程在低初始能量条件下弱解的爆破性质，探讨解在有限时间内爆破的条件及爆破时间的上界估计。我们运用微分不等式技术，通过定义能量泛函和辅助函数，利用Sobolev嵌入不等式、Hölder不等式等工具，分析低初始能量下解的行为，推导解的爆破条件。在给定的变指数条件和松弛函数假设下，证明了当初始能量低于某一临界值时，弱解在有限时间内发生爆破，并获得了爆破时间的上界估计。变指数源项和非线性阻尼项的共同作用会导致解在低初始能量下有限时间爆破，研究结果为该类方程的解的定性分析提供了理论依据，拓展了变指数非线性偏微分方程的爆破理论。

Abstract: This paper studies the blow-up properties of weak solutions of a class of Klein-Gordon equations with variable exponential source term and nonlinear damping term under the condition of low initial energy, and explores the conditions for the blow-up of solutions within a finite time and the upper bound estimation of the blow-up time. We apply differential inequality techniques, and by defining energy functionals and auxiliary functions, and using tools such as Sobolev embedding inequalities and Hölder inequalities, we analyze the behavior of solutions under low initial energy and derive the blow-up conditions of the solutions. Under the given variable exponential condition and the assumption of the relaxation function, it is proved that when the initial energy is lower than a certain critical value, the weak solution blows up within a finite time, and the upper bound estimation of the blow-up time is obtained. The combined effect of the variable exponential source term and the nonlinear damping term can lead to the blow-up of the solution in a finite time at a low initial energy. The research results provide a theoretical basis for the qualitative analysis of the solutions of this type of equation and expand the blow-up theory of variable exponential nonlinear partial differential equations.

1. 引言

在本文中，主要研究了一类具变指数源项的Klein-Gordon方程的初边值问题

$\{\begin{array}{l}{u}_{tt}-\Delta u+u-div\left(a\left(x,t\right)\nabla u\right)+{\displaystyle {\int }_0^{t}g\left(t-\tau \right)\Delta u\left(\tau \right)\text{d}\tau }+{\left|u_t\right|}^{m\left(x\right)-2}{u}_t={\left|u\right|}^{p\left(x\right)-2}u,\left(x,t\right)\in \Omega \times \left(0,T\right),\\ u\left(x,t\right)=0,\left(x,t\right)\in \partial \Omega \times \left(0,T\right),\\ u\left(x,0\right)=u_0\left(x\right),u_t\left(x,0\right)=u_1\left(x\right),x\in \Omega ,\end{array}$ (1)

其中$\Omega \subset R^n\left(n\ge 1\right)$ 是具光滑边界$\partial \Omega$ 的有界区域，系数$a\left(x,t\right)$ 是连续函数，且满足

$0<a^{-}:=\text{ess}\underset{x\in \Omega }{\inf }a\left(x,t\right)\le a\left(x,t\right)\le a^{+}:=\text{ess}\underset{x\in \Omega }{\sup }a\left(x,t\right)<\infty ,$ (2)

指数$m\left(x\right)$ 和$p\left(x\right)$ 是两个连续函数，且满足

$2<m^{-}:=\text{ess}\underset{x\in \Omega }{\inf }m\left(x\right)\le m\left(x\right)\le m^{+}:=\text{ess}\underset{x\in \Omega }{\sup }m\left(x\right)<\frac{2n-2}{n-2},$ (3)

$2<p^{-}:=\text{ess}\underset{x\in \Omega }{\inf }p\left(x\right)\le p\left(x\right)\le p^{+}:=\text{ess}\underset{x\in \Omega }{\sup }p\left(x\right)<\frac{2n-2}{n-2},$ (4)

且有连续的对数模，即满足

$\forall x,y\in \Omega ,\left|x-y\right|<1,\left|m\left(x\right)-m\left(y\right)\right|+\left|p\left(x\right)-p\left(y\right)\right|\le \omega \left(\left|x-y\right|\right),$ (5)

其中$\underset{t\to 0^{+}}{\lim }\sup \omega \left(\tau \right)\ln \frac{1}{\tau }=C<\infty$ 。

对于松弛函数$g:R^{+}\to R^{+}$ 是非负有界的$C^1$ 函数，且满足

$g\left(0\right)>0,g^{\prime }\left(\tau \right)\le 0,1-{\displaystyle {\int }_0^{\infty }g\left(\tau \right)\text{d}\tau }=l>0,$ (6)

和

${\displaystyle {\int }_0^{\infty }g\left(\tau \right)\text{d}\tau }<\frac{\frac{p^{-}}{2}-1}{\frac{p^{-}}{2}-1+\frac{1}{2p^{-}}}.$ (7)

接着，对于$a$ 是$C^1\left(\overline{\Omega }\times \left[0,\infty \right)\right)$ 类的对称矩阵，因此存在常数$a^{+},a^{-}>0$ ，对于所有的$\xi \in {ℝ}^n$ ，有

$a^{-}{\left|\xi \right|}^2\le a\xi \cdot \xi \le a^{+}{\left|\xi \right|}^2$ (8)

和

$a_t\xi \cdot \xi \le 0,$ (9)

其中$a_t=\frac{\partial a\left(x,t\right)}{\partial t}$ 。

问题(1)的方程源于各种物理现象的建模，如粘弹性和系统控制粘弹性配置的纵向运动服从非线性Boltzmann模型，或电流变流体，粘弹性流体，多孔介质中的过滤过程，温度依赖黏度的流体，图像处理等，这些现象都涉及非标准增长条件，即非线性可变指数的方程。近年来，Klein-Gordon方程和非线性可变指数方程的相关问题深受国内外学者的广泛关注。Klein-Gordon方程在相对论量子力学和量子场论中被认为是一个非线性波动方程，它作为Schrödinger方程的相对论形式，描述了在自由状态下具有零自旋的粒子的运动，它的应用包括分析基本粒子的行为，研究晶体中位错的传播，以及研究约瑟夫森结中的电流流动等。关于这些问题的更多细节可以在以下的研究中找到。

孙丽丽等人[1]研究了一类变指数非线性双曲方程$u_{tt}=div\left(a\left(x,t\right)\nabla u\right)+b\left(x,t\right)u{\left|u\right|}^{p\left(x,t\right)-1}-c\left(x,t\right)u_t{\left|u_t\right|}^{q\left(x,t\right)-1}$ ，证明了初值在一定条件下解的爆破，得到了爆破时间的下界和上界。Antontsev等人[2]研究了一类具强阻尼和变指数源项的非线性板Petrovsky方程$u_{tt}+{\Delta }^{2}u-\Delta u_t+{\left|u_t\right|}^{p\left(x\right)-2}{u}_t={\left|u\right|}^{q\left(x\right)-2}u$ ，利用Banach压缩映射原理，在对变量指数$p\left(x\right)$ 和$q\left(x\right)$ 的适当假设下，得到了弱解的局部存在性，还证明了弱解的全局存在性和负初始能量时解在有限时间内爆破。Yιlmaz等人[3]研究了一类具阻尼项的变指数非线性Petrovsky方程$u_{tt}-\Delta u-\Delta u_{tt}+{\Delta }^{2}u-\Delta u_t+{\left|u_t\right|}^{p\left(x\right)-2}{u}_t={\left|u\right|}^{q\left(x\right)-2}u$ ，证明了负初始能量时解在有限时间内爆破。廖梦兰等人[4]研究了具有变指数非线性的四阶方程$u_{tt}+{\Delta }^{2}u-M\left({\Vert \nabla u\Vert }_2^2\right)\Delta u-\Delta u_t+{\left|u_t\right|}^{m\left(x\right)-2}{u}_t$ $={\left|u\right|}^{p\left(x\right)-2}u$ ，证明了解在任意高的初始能量下会发生爆破现象，也给出了爆破时间的上界。祝相宇[5]运用多种不等式技术研究了一类具变指数增长条件的基尔霍夫粘弹性方程${\left|u_t\right|}^{\rho }{u}_{tt}-M\left({\Vert \nabla u\Vert }_2^2\right)\Delta u+{\displaystyle {\int }_0^{t}g\left(t-\tau \right)\Delta u\left(\tau \right)\text{d}\tau }+{\left|u_t\right|}^{m\left(x\right)-2}{u}_t={\left|u\right|}^{p\left(x\right)-2}u$ 在不同初始能量条件下的爆破性质。Ouaoua等人[6]研究了具变指数高阶非线性波动方程$u_{tt}+{\left(-\Delta \right)}^{m}u+{\left|u_t\right|}^{r\left(x\right)-2}{u}_t={\left|u\right|}^{p\left(x\right)-2}u$ 在负初始能量时，通过使用Lyapunov函数，证明了解在有限时间内爆破。Baghaei [7]研究了一类具变指数伪抛物方程$u_t-\Delta u_t-\nabla \cdot \left({\left|\nabla u\right|}^{p\left(x\right)-2}\nabla u\right)+\omega {\left|u\right|}^{m\left(x\right)-2}{u}_t={\left|u\right|}^{p\left(x\right)-2}u$ 解的爆破性质。

对于更多的相关研究工作，感兴趣的读者可参见文献[8]-[10]。受上述文献启发，一个自然的问题是，对于类似的具有扩散项和粘弹性项的变指数非线性Klein-Gordon方程，其弱解的爆破性会有怎样的变化呢？这正是本文的主要研究方向。

本文的结构如下：第2节给出一些必要的符号说明和引理；第3节证明问题(1)的弱解在低初始能量时有限时间爆破。

2. 预备知识

首先，我们介绍Orlicz-Sobolev类型的Banach空间

$L^{p\left(x\right)}\left(\Omega \right)=\left\{u\left(x\right)|u\left(x\right)是实可测函数,{\rho }_{p\left(x\right)}\left(u\right)={\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x<\infty }\right\},$

赋予以下范数

${\Vert u\Vert }_{p\left(x\right)}=\inf \left\{\lambda >0|{\displaystyle {\int }_{\Omega }{\left|\frac{u\left(x\right)}{\lambda }\right|}^{p\left(x\right)}\text{d}x\le 1}\right\},$

另外，$L^{p\left(x\right)}\left(\Omega \right)$ 空间的对偶空间为$L^{q\left(x\right)}\left(\Omega \right)$ ，其中$\frac{1}{p\left(x\right)}+\frac{1}{q\left(x\right)}=1$ 。

变指数Sobolev空间$W^{1,p\left(x\right)}\left(\Omega \right)$ 定义为

$W^{1,p\left(x\right)}\left(\Omega \right)=\left\{u\in L^{p\left(x\right)}\left(\Omega \right);\left|\nabla u\right|\in L^{p\left(x\right)}\left(\Omega \right)\right\},$

赋予如下范数

${\Vert u\Vert }_{1,p\left(x\right)}={\Vert u\Vert }_{p\left(x\right)}+{\Vert \nabla u\Vert }_{p\left(x\right)},$

进一步，定义$W_0^{1,p\left(x\right)}\left(\Omega \right)$ 为$C_0^{\infty }\left(\Omega \right)$ 在范数${\Vert \cdot \Vert }_{1,p\left(x\right)}$ 下的完备化空间。下面我们给出一些必要的引理。

引理1 [11] (Hölder不等式)对于任意的$u\in L^{p\left(x\right)}\left(\Omega \right)$ 和$v\in L^{q\left(x\right)}\left(\Omega \right)$ 有

$\left|{\displaystyle {\int }_{\Omega }uv\text{d}x}\right|\le \left(\frac{1}{p^{-}}+\frac{1}{q^{-}}\right){\Vert u\Vert }_{p\left(x\right)}\left(\Omega \right){\Vert v\Vert }_{q\left(x\right)}\left(\Omega \right)\le 2{\Vert u\Vert }_{p\left(x\right)}{\Vert v\Vert }_{q\left(x\right)}.$

引理2 [12] 若$p\left(x\right),q\left(x\right)\in C_{+}\left(\overline{\Omega }\right)=\left\{h\in C\left(\overline{\Omega }\right):\underset{x\in \overline{\Omega }}{\min }h\left(x\right)>1\right\}$ 且满足

$q\left(x\right)<p^{*}\left(x\right)=\{\begin{array}{l}\frac{np\left(x\right)}{n-p\left(x\right)},如果p\left(x\right)<n;\hfill \\ +\infty ,如果p\left(x\right)\ge n,\hfill \end{array}$

则连续紧嵌入成立。

引理3 [12]对于任意的$u\in L^{p\left(x\right)}\left(\Omega \right)$ ，则

$\begin{array}{l}\left(1\right){\Vert u\Vert }_{p\left(x\right)}<1\left(=1;>1\right)\iff {\rho }_{p\left(x\right)}\left(u\right)<1\left(=1;>1\right);\\ \left(2\right){\Vert u\Vert }_{p\left(x\right)}>1\Rightarrow {\Vert u\Vert }_{p\left(x\right)}^{p^{-}}\le {\rho }_{p\left(x\right)}\left(u\right)\le {\Vert u\Vert }_{p\left(x\right)}^{p^{+}};\\ \left(3\right){\Vert u\Vert }_{p\left(x\right)}<1\Rightarrow {\Vert u\Vert }_{p\left(x\right)}^{p^{+}}\le {\rho }_{p\left(x\right)}\left(u\right)\le {\Vert u\Vert }_{p\left(x\right)}^{p^{-}};\end{array}$

定义1 (弱解的定义)令$T>0$ ，如果$u\in C\left(0,T;H_0^1\left(\Omega \right)\right)\cap C^1\left(0,T;L^2\left(\Omega \right)\right)$ ，$u_t\in L^{m\left(x\right)}\left(\Omega \times \left(0,T\right)\right)$ ，$u_{tt}\in L^{\infty }\left(0,T;H^{-1}\left(\Omega \right)\right)$ ，且对任意的$\omega \in H_0^1\left(\Omega \right)$ ，有下列成立：

$\begin{array}{l}\left(1\right)\langle u_{tt},\omega \rangle +\left(\nabla u,\nabla \omega \right)+\left(u,\omega \right)+\left(a\left(x,t\right)\nabla u,\nabla \omega \right)-\left({\displaystyle {\int }_0^{t}g\left(t-\tau \right)\nabla u\left(\tau \right)d\tau },\nabla \omega \right)+\left({\left|u_t\right|}^{m\left(x\right)-2}{u}_t,\omega \right)\\ =\left({\left|u\right|}^{p\left(x\right)-2}u,\omega \right),\\ \left(2\right)u\left(x,0\right)=u_0\left(x\right)\in H_0^1\left(\Omega \right),u_t\left(x,0\right)=u_1\left(x\right)\in L^2\left(\Omega \right),\end{array}$

则称$u\left(x,t\right)$ 是问题(1)在$\Omega \times \left(0,T\right)$ 上的弱解。

定理1 若初始条件$u_0\in H_0^1\left(\Omega \right),u_1\in L^2\left(\Omega \right)$ 。假设(2)~(9)式成立，则存在$T>0$ ，使得问题(1)在$\Omega \times \left[0,T\right]$ 中存在唯一的局部解$u\left(x,t\right)$ 。

定理1的证明可以通过使用压缩映射原理结合Galerkin方法得到，类似文献[13]中所示。在本论文中，作者省略具体过程。

定义2 设$u\left(x,t\right)$ 是问题(1)的弱解，将最大存在时间$T_{\max }$ 定义如下

$T_{\max }=\sup \left\{T>0|u\left(t\right)在\left[0,T\right]上存在\right\}.$

1) 如果$T_{\max }=+\infty$ ，则弱解$u\left(x,t\right)$ 是整体的；

2) 如果$T_{\max }<+\infty$ ，则弱解$u\left(x,t\right)$ 在有限时间内爆破，$T_{\max }$ 是爆破时间。

为了方便讨论问题，定义如下的能量泛函：

$\begin{array}{l}E\left(t\right)=\frac{1}{2}{\Vert u_t\Vert }_2^2+\frac{1}{2}\left(1-{\displaystyle {\int }_0^{t}g\left(\tau \right)\text{d}\tau }\right){\Vert \nabla u\Vert }_2^2+\frac{1}{2}\left(g\circ \nabla u\right)\left(t\right)\hfill \\ +\frac{1}{2}{\Vert u\Vert }_2^2+\frac{1}{2}{\displaystyle {\int }_{\Omega }a\left(x,t\right)\nabla u\cdot \nabla u\text{d}x}-{\displaystyle {\int }_{\Omega }\frac{1}{p\left(x\right)}{\left|u\right|}^{p\left(x\right)}\text{d}x,}\hfill \end{array}$ (10)

其中$g\circ \nabla u={\displaystyle {\int }_0^{t}g\left(t-\tau \right)}{\Vert \nabla u\left(t\right)-\nabla u\left(\tau \right)\Vert }_2^2\text{d}\tau$ 。

为了方便叙述，我们引入一些记号

$B_1=\max \left\{1,\frac{B}{l^{\frac{1}{2}}}\right\},{\lambda }_1={\left(B_1^2\right)}^{\frac{-2}{p^{-}-2}},E_1=\left(\frac{1}{2}-\frac{1}{p^{-}}\right){\left(B_1^2\right)}^{\frac{-p^{-}}{p^{-}-2}},$

其中，$B$ 是Sobolev嵌入的最佳常数。

引理4 若$u\left(x,t\right)$ 为问题(1)的弱解，则$E\left(t\right)$ 是关于时间$t$ 的非增函数。

证明 将问题(1)的第一个方程左右两边各乘以$u_t$ 并在$\Omega$ 上积分可得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left(\frac{1}{2}{\Vert u_t\Vert }_2^2+\frac{1}{2}\left(1-{\displaystyle {\int }_0^{t}g\left(\tau \right)\text{d}\tau }\right){\Vert \nabla u\Vert }_2^2+\frac{1}{2}\left(g\circ \nabla u\right)\left(t\right)+\frac{1}{2}{\Vert u\Vert }_2^2+\frac{1}{2}{\displaystyle {\int }_{\Omega }a\left(x,t\right)\nabla u\cdot \nabla u\text{d}x}\right)\\ =\frac{1}{2}{\displaystyle {\int }_{\Omega }{a}_t\left(x,t\right)\nabla u\cdot \nabla u\text{d}x}-{\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)}\text{d}x}+\frac{1}{2}\left(g^{\prime }\circ \nabla u\right)-\frac{1}{2}g\left(t\right){\Vert \nabla u\Vert }_2^2+\frac{\text{d}}{\text{d}t}\left({\displaystyle {\int }_{\Omega }\frac{1}{p\left(x\right)}{\left|u\right|}^{p\left(x\right)}\text{d}x}\right),\end{array}$

结合松弛函数$g$ 和对称矩阵$a$ 的定义，有

$E\left(t\right)-\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{a}_{\tau }\left(x,\tau \right)\nabla u\cdot \nabla u\text{d}x\text{d}\tau }}+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\left|u_{\tau }\right|}^{m\left(x\right)}\text{d}x\text{d}\tau }}-\frac{1}{2}{\displaystyle {\int }_0^t\left(g^{\prime }\circ \nabla u\right)\text{d}t}+\frac{1}{2}{\displaystyle {\int }_0^{t}g\left(\tau \right){\Vert \nabla u\Vert }_2^2\text{d}\tau }=E\left(0\right),$

将上式两端求导可知

$E^{\prime }\left(t\right)=\frac{1}{2}{\displaystyle {\int }_{\Omega }{a}_t\left(x,t\right)\nabla u\cdot \nabla u\text{d}x}-{\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)}\text{d}x}+\frac{1}{2}\left(g^{\prime }\circ \nabla u\right)-\frac{1}{2}g\left(t\right){\Vert \nabla u\Vert }_2^2,$ (11)

证明完成。

引理5 假设$u\left(x,t\right)$ 为问题(1)的弱解，若$u_0\in H_0^1\left(\Omega \right)$ ，$u_1\in L^2\left(\Omega \right)$ ，且满足(2)~(9)式，进一步假设$E\left(0\right)<E_1$ 和${\lambda }_1<\lambda \left(0\right)=B_1^{2}l{\Vert \nabla u_0\Vert }_2^2$ ，然后存在一个常数${\lambda }_2>{\lambda }_1$ ，这样

$B_1^{2}l{\Vert \nabla u\Vert }_2^2\ge {\lambda }_2,\forall t\ge 0.$

证明 利用(10)式、引理3，以及Sobolev嵌入不等式，我们发现

$\begin{array}{l}E\left(t\right)\ge \frac{1}{2}\left(1-{\displaystyle {\int }_0^{t}g\left(\tau \right)\text{d}\tau }\right){\Vert \nabla u\Vert }_2^2-\frac{1}{p^{-}}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\\ \ge \frac{l}{2}{\Vert \nabla u\Vert }_2^2-\frac{1}{p^{-}}\max \left\{{\Vert u\Vert }_{p\left(x\right)}^{p^{-}},{\Vert u\Vert }_{p\left(x\right)}^{p^{+}}\right\}\\ \ge \frac{1}{2}{\Vert \nabla u\Vert }_2^2-\frac{1}{p^{-}}\max \left\{{\left(B{\Vert \nabla u\Vert }_2\right)}^{p^{-}},{\left(B{\Vert \nabla u\Vert }_2\right)}^{p^{+}}\right\}\\ \ge \frac{l}{2}{\Vert \nabla u\Vert }_2^2-\frac{1}{p^{-}}\max \left\{{\left(B_1{l}^{\frac{1}{2}}{\Vert \nabla u\Vert }_2\right)}^{p^{-}},{\left(B_1{l}^{\frac{1}{2}}{\Vert \nabla u\Vert }_2\right)}^{p^{+}}\right\}\\ =\frac{1}{2B_1^2}\lambda -\frac{1}{p^{-}}\max \left\{{\lambda }^{\frac{p^{-}}{2}},{\lambda }^{\frac{p^{+}}{2}}\right\}:=G\left(\lambda \right),\end{array}$

其中$\lambda \left(t\right)=B_1^{2}l{\Vert \nabla u\Vert }_2^2$ 。通过直接分析$G\left(\lambda \right)$ 的性质，我们推导出$G\left(\lambda \right)$ 满足以下性质

$\begin{array}{l}{G}^{\prime }\left(\lambda \right)=\{\begin{array}{l}\frac{1}{2B_1^2}-\frac{p^{+}}{2p^{\_}}{\lambda }^{\frac{p^{+}-2}{2}}<0,\lambda >1,\hfill \\ \frac{1}{2B_1^2}-\frac{1}{2}{\lambda }^{\frac{p^{-}-2}{2}},0<\lambda <1,\hfill \end{array}\\ {G^{\prime }}_{+}\left(1\right)=\frac{1}{2B_1^2}-\frac{p^{+}}{2p^{\_}}<0,{G^{\prime }}_{-}\left(1\right)=\frac{1}{2B_1^2}-\frac{1}{2}<0,\\ G^{\prime }\left({\lambda }_1\right)=0,0<{\lambda }_1<1.\end{array}$

很容易证实，$G\left(\lambda \right)$ 在$\left(0,{\lambda }_1\right)$ 上严格递增，而在$\left({\lambda }_1,+\infty \right)$ 上严格递减，当$\lambda \to +\infty$ 时，$G\left(\lambda \right)\to -\infty$ ，令$G\left({\lambda }_1\right)=E_1$ 。由于$E\left(0\right)<E_1$ ，存在${\lambda }_2$ 和${\tilde{\lambda }}_2$ ，满足${\tilde{\lambda }}_2>{\lambda }_1>{\lambda }_2$ ，使得$G\left({\lambda }_2\right)=G\left({\tilde{\lambda }}_2\right)=E\left(0\right)$ 。设$\lambda \left(0\right)=B_1^{2}l{\Vert \nabla u_0\Vert }_2^2$ ，则

$G\left(\lambda \left(0\right)\right)\le E\left(0\right)=G\left({\lambda }_2\right)=G\left({\tilde{\lambda }}_2\right).$ (12)

1) 如果${\lambda }_1>\lambda \left(0\right)$ ，那么(12)式意味着$\lambda \left(0\right)\le {\tilde{\lambda }}_2$ 。利用反证法，存在某个$t_0>0$ ，使得$\lambda \left(t_0\right)>{\tilde{\lambda }}_2$ 。由于$\lambda \left(t\right)$ 的连续性，我们选择$t_0$ 使得${\tilde{\lambda }}_2<\lambda \left(t_0\right)<{\lambda }_1$ ，因此有$E\left(0\right)=G\left({\tilde{\lambda }}_2\right)<G\left(\lambda \left(t_0\right)\right)\le E\left(t_0\right)$ ，这与(11)式相矛盾。

2) 如果${\lambda }_1<\lambda \left(0\right)$ ，那么(12)式意味着$\lambda \left(0\right)\ge {\lambda }_2$ 。利用反证法，存在某个$t_0>0$ ，使得$\lambda \left(t_0\right)<{\lambda }_2$ 。由于$\lambda \left(t\right)$ 的连续性，我们选择$t_0$ 使得${\lambda }_1<\lambda \left(t_0\right)<{\lambda }_2$ ，因此有$E\left(0\right)=G\left({\lambda }_2\right)<G\left(\lambda \left(t_0\right)\right)\le E\left(t_0\right)$ ，这与(11)式相矛盾。

证明完成。

3. 主要结果

在本节中，我们利用一阶微分不等式技术证明了问题(1)的弱解在低初始能量下在有限时间内爆破。

引理6 假设满足引理5中的条件。对于$t\in \left[0,T\right)$ ，定义$H\left(t\right)=E_2-E\left(t\right)$ ，其中$E_2\in \left(E\left(0\right),E_1\right)$ 足够接近$E\left(0\right)$ ，有

$0<H\left(0\right)\le H\left(t\right)\le {\displaystyle {\int }_{\Omega }\frac{1}{p\left(x\right)}}{\left|u\right|}^{p\left(x\right)}\text{d}x\le \frac{1}{p^{-}}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}}\text{d}x.$

证明 由$H\left(t\right)$ 的定义得出$H^{\prime }\left(t\right)=-E^{\prime }\left(t\right)\ge 0$ ，暗示$H(t)$ 是一个非递减函数，因此

$H\left(t\right)\ge H\left(0\right)=E_2-E\left(0\right)>0,\forall t\in \left[0,T\right),$

根据(10)式和引理5可知

$\begin{array}{l}H\left(t\right)=E_2-E\left(t\right)\\ \le E_2-\frac{l}{2}{\Vert \nabla u\Vert }_2^2+{\displaystyle {\int }_{\Omega }\frac{1}{p\left(x\right)}{\left|u\right|}^{p\left(x\right)}\text{d}x}\\ \le E_1-\frac{{\lambda }_2}{2B_1^2}+\frac{1}{p^{-}}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\\ \le E_1-\frac{{\lambda }_1}{2B_1^2}+\frac{1}{p^{-}}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\\ \le \frac{1}{p^{-}}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}.\end{array}$

证明完成。

定理2 假设指数满足

$2<m^{-}\le m\left(x\right)\le m^{+}<p^{-}\le p\left(x\right)\le p^{+}\le \frac{2n-2}{n-2},$

和

$1-l={\displaystyle {\int }_0^{\infty }g\left(\tau \right)\text{d}\tau }<\frac{\frac{p^{-}}{2}-1}{\frac{p^{-}}{2}-1+\frac{1}{2p^{-}}},$

且初始能量$E\left(0\right)<E_1$ 和${\lambda }_1<{\lambda }_0=B_1^{2}l{\Vert \nabla u_0\Vert }_2^2$ ，则问题(1)的弱解在有限时间$T^{*}$ 内爆破且满足

$\underset{t\to T^{*-}}{\lim }\left({\Vert u_t\Vert }_2^2+{\Vert \nabla u\Vert }_2^2+{\Vert u\Vert }_2^2+{\Vert u\Vert }_{p^{+}}^{p^{+}}\right)=+\infty .$ (13)

证明 运用反证法，假设(13)式不成立，则对于任意的$t\in \left[0,T^{*}\right]$ 且$T^{*}<+\infty$ 有

${\Vert u_t\Vert }_2^2+{\Vert \nabla u\Vert }_2^2+{\Vert u\Vert }_2^2+{\Vert u\Vert }_{p^{+}}^{p^{+}}\le C_{*},$ (14)

其中，$C_{*}$ 是一个正常数。

构造辅助函数

$L\left(t\right)=H^{1-\alpha }\left(t\right)+\epsilon {\displaystyle {\int }_{\Omega }{u}_{t}u\text{d}x},$

其中$\epsilon >0$ 足够小且

$0<\alpha \le \min \left\{\frac{p^{-}-m^{+}}{p^{-}\left(m^{+}-1\right)},\frac{p^{-}-2}{2p^{-}}\right\}.$

下面的证明分为三步。

第一步：$L^{\prime }\left(t\right)$ 的估计。由$L\left(t\right)$ 的定义和$H^{\prime }\left(t\right)$ 的定义可知

$\begin{array}{l}{L}^{\prime }\left(t\right)=\left(1-\alpha \right)H^{-\alpha }\left(t\right)H^{\prime }\left(t\right)+\epsilon {\Vert u_t\Vert }_2^2+\epsilon {\displaystyle {\int }_{\Omega }{u}_{tt}u\text{d}x}\\ \ge \left(1-\alpha \right)H^{-\alpha }\left(t\right){\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)}\text{d}x}+\epsilon {\Vert u_t\Vert }_2^2-\epsilon \left(1+a^{+}\right){\Vert \nabla u\Vert }_2^2-\epsilon {\Vert u\Vert }_2^2\\ +\epsilon {\displaystyle {\int }_0^{t}g\left(t-\tau \right)}{\displaystyle {\int }_{\Omega }\nabla u\left(\tau \right)\nabla u\left(t\right)\text{d}x\text{d}\tau }-\epsilon {\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)-2}{u}_{t}u\text{d}x}+\epsilon {\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x},\end{array}$

由Hölder不等式和Young不等式，选择$0<{\epsilon }_1\le \frac{a^{-}{p}^{-}-2a^{+}}{a^{-}{p}^{-}}$ ，可得

$\begin{array}{l}\epsilon {\displaystyle {\int }_0^{t}g\left(t-\tau \right)}{\displaystyle {\int }_{\Omega }\nabla u\left(\tau \right)\nabla u\left(t\right)\text{d}x\text{d}\tau }\\ \ge -\epsilon \frac{p^{-}\left(1-{\epsilon }_1\right)}{2}\left(g\circ \nabla u\right)+\epsilon \left(1-\frac{1}{2p^{-}\left(1-{\epsilon }_1\right)}\right){\displaystyle {\int }_0^{t}g\left(\tau \right)\text{d}\tau {\Vert \nabla u\Vert }_2^2},\end{array}$

结合上面两个式子，易知

第1.1步：$\epsilon \frac{\frac{l}{2}\left(\frac{p^{-}\left(1-{\epsilon }_1\right)}{2}-1\right)-\frac{1}{2p^{-}\left(1-{\epsilon }_1\right)}\frac{1-l}{2}}{l}{\left(B_1^2\right)}^{\frac{-p^{-}}{p^{-}-2}}-\epsilon p^{-}\left(1-{\epsilon }_1\right)E_2$ 的估计。根据定理2中的条件，有

$E\left(0\right)<\frac{1}{2}\left(\frac{1}{2}-\frac{1}{p^{-}}\right)\left(1-\frac{1-l}{p^{-}\left(p^{-}-2\right)l}\right){\left(B_1^2\right)}^{\frac{-p^{-}}{p^{-}-2}}=\frac{\left(\frac{p^{-}}{2}-1\right)\frac{l}{2}-\frac{1}{2p^{-}}\frac{1-l}{2}}{l{p}^{-}}{\left(B_1^2\right)}^{\frac{-p^{-}}{p^{-}-2}}<E_1,$

这里我们取${\epsilon }_1>0$ 足够小，且$E_2\in \left(E\left(0\right),E_1\right)$ 足够接近$E\left(0\right)$ 则

$\begin{array}{l}\epsilon \frac{\frac{l}{2}\left(\frac{p^{-}\left(1-{\epsilon }_1\right)}{2}-1\right)-\frac{1}{2p^{-}\left(1-{\epsilon }_1\right)}\frac{1-l}{2}}{l}{\left(B_1^2\right)}^{\frac{-p^{-}}{p^{-}-2}}-\epsilon p^{-}\left(1-{\epsilon }_1\right)E_2\hfill \\ \begin{array}{l}\ge \epsilon \frac{\frac{l}{2}\left(\frac{p^{-}\left(1-{\epsilon }_1\right)}{2}-1\right)-\frac{1}{2p^{-}\left(1-{\epsilon }_1\right)}\frac{1-l}{2}}{l}{\left(B_1^2\right)}^{\frac{-p^{-}}{p^{-}-2}}\hfill \\ -\epsilon \left(1-{\epsilon }_1\right)\frac{\left(\frac{p^{-}}{2}-1\right)\frac{l}{2}-\frac{1}{2p^{-}}\frac{1-l}{2}}{l}{\left(B_1^2\right)}^{\frac{-p^{-}}{p^{-}-2}}\hfill \end{array}\hfill \\ \ge 0.\hfill \end{array}$

第1.2步：$-\epsilon {\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)-2}{u}_{t}u\text{d}x}$ 的估计。应用Young不等式${\epsilon }_2>1$ ，Sobolev嵌入不等式，由引理3和引理5，我们得到

$\begin{array}{l}\left|{\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)-2}{u}_{t}u\text{d}x}\right|\\ \le {\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)-1}{H}^{-\alpha \frac{m\left(x\right)-1}{m\left(x\right)}}\left(t\right)H^{\alpha \frac{m\left(x\right)-1}{m\left(x\right)}}\left(t\right)\left|u\right|\text{d}x}\\ \le {\epsilon }_2{H}^{-\alpha }\left(t\right){\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)}\text{d}x}+\frac{1}{{\epsilon }_2^{m^{-}-1}}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{m\left(x\right)}{H}^{\alpha \left(m\left(x\right)-1\right)}\left(t\right)\text{d}x}\\ \le {\epsilon }_2{H}^{-\alpha }\left(t\right){\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)}\text{d}x}+\frac{2C_1^{\alpha \left(m^{-}-m^{+}\right)}}{{\epsilon }_2^{m^{-}-1}}{H}^{\alpha \left(m^{+}-1\right)}\left(t\right){\displaystyle {\int }_{\Omega }{\left|u\right|}^{m\left(x\right)}\text{d}x}\\ \le {\epsilon }_2{H}^{-\alpha }\left(t\right){\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)}\text{d}x}+\frac{C_2}{{\epsilon }_2^{m^{-}-1}}{H}^{\alpha \left(m^{+}-1\right)}\left(t\right)\max \left\{{\Vert u\Vert }_{p\left(x\right)}^{m^{+}},{\Vert u\Vert }_{p\left(x\right)}^{m^{-}}\right\},\end{array}$

其中$C_1=\min \left\{H\left(0\right),1\right\}$ ，$C_2=2{\left(1+\left|\Omega \right|\right)}^{m^{+}}{C}_1^{\alpha \left(m^{-}-m^{+}\right)}$ 。接下来，我们有

${\Vert u\Vert }_{p\left(x\right)}^{m^{+}}\le \max \left\{{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{+}}},{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{-}}}\right\}\le \max \left\{{\left(p^{-}H\left(t\right)\right)}^{\frac{m^{+}}{p^{+}}-\frac{m^{+}}{p^{-}}},1\right\}{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{-}}},$

和

${\Vert u\Vert }_{p\left(x\right)}^{m^{-}}\le \max \left\{{\left(p^{-}H\left(t\right)\right)}^{\frac{m^{-}}{p^{+}}-\frac{m^{+}}{p^{-}}},{\left(p^{-}H\left(t\right)\right)}^{\frac{m^{-}-m^{+}}{p^{-}}}\right\}{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{-}}},$

这意味着

$\max \left\{{\Vert u\Vert }_{p\left(x\right)}^{m^{+}},{\Vert u\Vert }_{p\left(x\right)}^{m^{-}}\right\}\le C_3{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{-}}},$

其中，$C_3=2\min {\left\{p^{-}H\left(0\right),1\right\}}^{\frac{m^{-}}{p^{+}}-\frac{m^{+}}{p^{-}}}$ 。回顾$0<\alpha \le \frac{p^{-}-m^{+}}{p^{-}\left(m^{+}-1\right)}$ 和引理6，显然，

$\begin{array}{l}{H}^{\alpha \left(m^{+}-1\right)}\left(t\right)\max \left\{{\Vert u\Vert }_{p\left(x\right)}^{m^{+}},{\Vert u\Vert }_{p\left(x\right)}^{m^{-}}\right\}\hfill \\ \le C_3{H}^{\alpha \left(m^{+}-1\right)}\left(t\right){\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{-}}}\hfill \\ \le C_3\frac{H^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}\left(t\right)}{H^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}\left(0\right)}{H}^{1-\frac{m^{+}}{p^{-}}}\left(t\right)H^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}\left(0\right){\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{-}}}\hfill \\ \le C_3{\left(\frac{1}{p^{-}}\right)}^{1-\frac{m^{+}}{p^{-}}}{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{1-\frac{m^{+}}{p^{-}}}{H}^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}\left(0\right){\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{m^{+}}{p^{-}}}\hfill \\ \le C_3{\left(\frac{1}{p^{-}}\right)}^{1-\frac{m^{+}}{p^{-}}}{C}_1^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x},\hfill \end{array}$

由上述可知，

$\begin{array}{l}{L}^{\prime }\left(t\right)\ge \left(1-\alpha -\epsilon {\epsilon }_2\right)H^{-\alpha }\left(t\right){\displaystyle {\int }_{\Omega }{\left|u_t\right|}^{m\left(x\right)}\text{d}x}+\epsilon \left(1+\frac{p^{-}\left(1-{\epsilon }_1\right)}{2}\right){\Vert u_t\Vert }_2^2\\ +\epsilon \left(\frac{l}{2}\left(\frac{p^{-}\left(1-{\epsilon }_1\right)}{2}-1\right)-\frac{1}{2p^{-}\left(1-{\epsilon }_1\right)}\frac{1-l}{2}\right){\Vert \nabla u\Vert }_2^2+\epsilon \left(\frac{p^{-}\left(1-{\epsilon }_1\right)}{2}-1\right){\Vert u\Vert }_2^2\\ +\epsilon \left({\epsilon }_1-\frac{C_1^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}{C}_2{C}_3{\left(\frac{1}{p^{-}}\right)}^{1-\frac{m^{+}}{p^{-}}}}{{\epsilon }_2^{m^{-}-1}}\right){\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}+\epsilon p^{-}\left(1-{\epsilon }_1\right)H\left(t\right),\end{array}$

我们适当的常数${\epsilon }_2$ ，使

$\frac{C_1^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}{C}_2{C}_3{\left(\frac{1}{p^{-}}\right)}^{1-\frac{m^{+}}{p^{-}}}}{{\epsilon }_2^{m^{-}-1}}<{\epsilon }_1<\frac{p^{-}-2}{p^{-}},$

然后选择足够小的$\epsilon >0$ 使得$1-\alpha >\epsilon {\epsilon }_2$ 。因此，我们得到

$L^{\prime }\left(t\right)\ge M_1\left(H\left(t\right)+{\Vert u_t\Vert }_2^2+{\Vert \nabla u\Vert }_2^2+{\Vert u\Vert }_2^2+{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right),$ (15)

其中

$\begin{array}{l}{M}_1=\epsilon \min \{1+\frac{p^{-}\left(1-{\epsilon }_1\right)}{2},\frac{l}{2}\left(\frac{p^{-}\left(1-{\epsilon }_1\right)}{2}-1\right)-\frac{1}{2p^{-}\left(1-{\epsilon }_1\right)}\frac{1-l}{2},\begin{array}{c}\\ \\ \\ \end{array}\\ \frac{p^{-}\left(1-{\epsilon }_1\right)}{2}-1,{\epsilon }_1-\frac{C_1^{\alpha \left(m^{+}-1\right)+\frac{m^{+}}{p^{-}}-1}{C}_2{C}_3{\left(\frac{1}{p^{-}}\right)}^{1-\frac{m^{+}}{p^{-}}}}{{\epsilon }_2^{m^{-}-1}},p^{-}\left(1-{\epsilon }_1\right)\},\end{array}$

由(15)式表明$L\left(t\right)\ge L\left(0\right)$ 。因此，对于充分小的$\epsilon$ ，我们有

$L\left(0\right)=H^{1-\alpha }\left(0\right)+\epsilon {\displaystyle {\int }_{\Omega }{u}_1{u}_0\text{d}x}>0.$

第二步。$L\left(t\right)$ 的估计。应用Hölder不等式，Young不等式和Sobolev嵌入不等式，我们得到

$\begin{array}{l}{\left|{\displaystyle {\int }_{\Omega }{u}_{t}u\text{d}x}\right|}^{\frac{1}{1-\alpha }}\le {\left({\Vert u_t\Vert }_2{\Vert u\Vert }_2\right)}^{\frac{1}{1-\alpha }}\hfill \\ \le {\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}{\Vert u_t\Vert }_2^{\frac{1}{1-\alpha }}{\Vert u\Vert }_{p\left(x\right)}^{\frac{1}{1-\alpha }}\hfill \\ \le \frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\mu }{\Vert u_t\Vert }_2^{\frac{1}{1-\alpha }\mu }+\frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\nu }{\Vert u\Vert }_{p\left(x\right)}^{\frac{1}{1-\alpha }\nu },\hfill \end{array}$

其中$\frac{1}{\mu }+\frac{1}{\nu }=1$ 。选择$\mu =2\left(1-\alpha \right)>1$ ，然后$\nu =\frac{2\left(1-\alpha \right)}{2\left(1-\alpha \right)-1}$ ，进一步得到

${\left|{\displaystyle {\int }_{\Omega }{u}_{t}u\text{d}x}\right|}^{\frac{1}{1-\alpha }}\le \frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\mu }{\Vert u_t\Vert }_2^2+\frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\nu }{\Vert u\Vert }_{p\left(x\right)}^{\frac{2}{2\left(1-\alpha \right)-1}},$ (16)

由$0<\alpha <\frac{p^{-}-2}{2p^{-}}$ ，我们得到

$\begin{array}{l}{\Vert u\Vert }_{p\left(x\right)}^{\frac{2}{2\left(1-\alpha \right)-1}}\le \max \left\{{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{2}{p^{-}\left(2\left(1-\alpha \right)-1\right)}},{\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)}^{\frac{2}{p^{+}\left(2\left(1-\alpha \right)-1\right)}}\right\}\hfill \\ \le \left\{{\left(p^{-}H\left(t\right)\right)}^{\frac{2-p^{-}\left(2\left(1-\alpha \right)-1\right)}{p^{+}\left(2\left(1-\alpha \right)-1\right)}},{\left(p^{-}H\left(t\right)\right)}^{\frac{2-p^{+}\left(2\left(1-\alpha \right)-1\right)}{p^{+}\left(2\left(1-\alpha \right)-1\right)}}\right\}{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\hfill \\ \le C_4{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\hfill \end{array}$ (17)

其中$C_4=\min {\left\{\left(p^{-}H\left(0\right)\right),1\right\}}^{\frac{2-p^{+}\left(2\left(1-\alpha \right)-1\right)}{p^{+}\left(2\left(1-\alpha \right)-1\right)}}$ 。将(17)式代入(16)式，我们得到

${\left|{\displaystyle {\int }_{\Omega }{u}_{t}u\text{d}x}\right|}^{\frac{1}{1-\alpha }}\le \frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\mu }{\Vert u_t\Vert }_2^2+\frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\nu }{C}_4{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x.}$ (18)

由(17)式和不等式${\left(a_1+a_2+\cdots +a_l\right)}^k\le 2^{\left(k-1\right)\left(l-1\right)}\left(a_1^k+a_2^k+\cdots +a_l^k\right)$ ，其中$a_i\ge 0$ ，$k\ge 1$ 和$i=1,2,\cdots ,l$ ，我们得到

$\begin{array}{l}{L}^{\frac{1}{1-\alpha }}\left(t\right)={\left(H^{1-\alpha }\left(t\right)+\epsilon {\displaystyle {\int }_{\Omega }{u}_{t}u\text{d}x}\right)}^{\frac{1}{1-\alpha }}\hfill \\ \le 2^{\frac{\alpha }{1-\alpha }}\left(H\left(t\right)+{\left(\epsilon {\displaystyle {\int }_{\Omega }{u}_{t}u\text{d}x}\right)}^{\frac{1}{1-\alpha }}\right)\hfill \\ \le M_2\left(H\left(t\right)+{\Vert u_t\Vert }_2^2+{\Vert \nabla u\Vert }_2^2+{\Vert u\Vert }_2^2+{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)\hfill \end{array}$ (19)

其中

$M_2=2^{\frac{\alpha }{1-\alpha }}\max \left\{1,{\epsilon }^{\frac{1}{1-\alpha }}\frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\mu },{\epsilon }^{\frac{1}{1-\alpha }}\frac{{\left(1+\left|\Omega \right|\right)}^{\frac{1}{1-\alpha }}}{\nu }{C}_4\right\}.$

结合(15)式和(19)式，我们得到

$L^{\prime }\left(t\right)\ge \frac{M_1}{M_2}{L}^{\frac{1}{1-\alpha }}\left(t\right),\forall t\ge 0.$

上式在$\left(0,t\right)$ 上进行简单积分，得到

$L^{\frac{\alpha }{1-\alpha }}\left(t\right)\ge \frac{1}{L^{\frac{\alpha }{\alpha -1}}\left(0\right)-\frac{M_1}{M_2}\frac{\alpha }{1-\alpha }t},$

这表明$L\left(t\right)$ 在有限时间$T^{*}$ 处爆破，且满足

$T^{*}\le \frac{M_2}{M_1}\frac{1-\alpha }{\alpha }{L}^{\frac{\alpha }{1-\alpha }}\left(0\right),$

此外

$\underset{t\to T^{*-}}{\lim }\left(H\left(t\right)+{\Vert u_t\Vert }_2^2+{\Vert \nabla u\Vert }_2^2+{\Vert u\Vert }_2^2+{\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\right)=+\infty ,$

由此，我们很容易地得出结论

${\displaystyle {\int }_{\Omega }{\left|u\right|}^{p\left(x\right)}\text{d}x}\le {\displaystyle {\int }_{\left\{\left|u\right|\ge 1\right\}}{\left|u\right|}^{p^{+}}\text{d}x}+{\displaystyle {\int }_{\left\{\left|u\right|<1\right\}}{\left|u\right|}^{p^{-}}\text{d}x}\le {\Vert u\Vert }_{p^{+}}^{p^{+}}+\left|\Omega \right|,$

我们易知

$\underset{t\to T^{*-}}{\lim }\left({\Vert u_t\Vert }_2^2+{\Vert \nabla u\Vert }_2^2+{\Vert u\Vert }_2^2+{\Vert u\Vert }_{p^{+}}^{p^{+}}\right)=+\infty ,$

这与(14)式矛盾，因此问题(1)的解在有限时间内爆破。

致 谢

作者对编辑和审稿人提出的重要建议表示诚挚的感谢。

基金项目

吉林省科技发展计划项目资助(20240101307JC)。

NOTES

^*通讯作者。