带有强阻尼时滞项的<i>m</i>-Laplacian型波方程解的爆破

高云龙; 林荣瑞; 佘连兵; 李爱静

doi:10.6054/j.jscnun.2021015

带有强阻尼时滞项的m-Laplacian型波方程解的爆破

六盘水师范学院数学与计算机科学学院, 六盘水 553004

基金项目:

国家自然科学基金项目 11571283

贵州省教育厅自然基金项目黔教合KY字[2019]139

贵州省教育厅自然基金项目黔教合KY字[2019]143)

六盘水师范学院校级项目 LPSSYKYJJ201801

六盘水师范学院校级项目 LPSSKJTD201907

详细信息

通讯作者:
高云龙, Email: gyl0813101x@163.com

中图分类号: O175.2
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出版历程
- 收稿日期: 2020-08-03
- 网络出版日期: 2021-03-23
- 刊出日期: 2021-02-24

Blow-up of Solutions to the m-Laplacian Type Wave Equation with Strong Delay Terms

School of Mathematics and Information Engineering, Liupanshui Normal University, Liupanshui 553004, China

摘要

摘要: 研究带有强阻尼时滞项的m-Laplacian型波方程: ${{u}_{tt}}-{{\Delta }_{m}}u-\Delta u+g*\Delta u-{{\mu }_{1}}\Delta {{u}_{t}}\left( x, t \right)-{{\mu }_{2}}\Delta {{u}_{t}}\left( x, t-\tau \right)={{\left| u \right|}^{p-2}}u$ 解的爆破：当初始能量0 < E(0) < E₁时, 利用能量函数构造凹函数L₁(t), 得到微分不等式 $\frac{\text{d}{{L}_{1}}\left( t \right)}{\text{d}t}\ge {{\xi }_{0}}L_{1}^{1+\nu }\left( t \right)\ \left( {{\xi }_{0}}>0, \nu >0, t\ge 0 \right)$ , 在(0, t)上对此微分不等式积分, 从而可知存在有限时间T^*>0, 使得当时间t趋于T^*时, 该m-Laplacian型波方程的解爆破; 当初始能量E(0) < 0时, 构造凹函数L₂(t), 通过同样的方法得到该方程的解存在有限时间爆破.
- m-Laplacian /
- 强阻尼时滞项 /
- 爆破
Abstract: Blow-up of solutions to the m-Laplacian type wave equation with strong delay was studied: ${{u}_{tt}}-{{\Delta }_{m}}u-\Delta u+g*\Delta u-{{\mu }_{1}}\Delta {{u}_{t}}\left( x, t \right)-{{\mu }_{2}}\Delta {{u}_{t}}\left( x, t-\tau \right)={{\left| u \right|}^{p-2}}u$ . When the initial energy 0 < E(0) < E₁, the concave function L₁(t) was constructed with the energy function, and the differential inequality $\frac{\text{d}{{L}_{1}}\left( t \right)}{\text{d}t}\ge {{\xi }_{0}}L_{1}^{1+\nu }\left( t \right)\ \left( {{\xi }_{0}}>0, \nu >0, t\ge 0 \right)$ was obtained. Then, the differential inequality was integrated in (0, t), and it was proved that there was a finite time T^*>0, so that when the time t was tended to T^*, the m-Laplacian type wave equation underwent blow-up of solutions. When the initial energy E(0) < 0, a concave function L₂(t) was also constructed. With the same method, it was found that the solutions to the equation had a finite-time blow-up.
- m-Laplacian /
- strong delay terms /
- blow-up

HTML全文

含有m-Laplacian项的拟线性波方程源于非线性Voigt模型, 描述的是粘弹性材料杆的纵向振动, 特别是可描述受外力作用的Ludwick材料的振动^[1-2]. 近年来, 学者们对带有p-Laplacian项的偏微分方程解的性态作了深入研究, 如：PEI等^[3]研究了p-Laplacian型波方程u_tt-Δ_pu-Δu_t=f(u), 其中2 < p < 3且f(u)的指数为超临界情形, 证明了该方程的整体弱解的存在以及当初始能量为负值时其解存在有限时间爆破; NAKAO^[4]在适当假设下, 证明了p-Laplacian项为div{σ(|▽u|²)▽u}的偏微分方程的能量解具有多项式衰减性; RAPOSO等^[5]考虑了带有记忆项和p-Laplacian项但不含非线性源项和强阻尼项的波方程, 主要通过Galerkin方法证明该方程的整体解存在且能量函数也具有多项式衰减性. 此外, 时滞项是系统不稳定的一个重要因素, 含有时滞项的方程受到越来越多学者的关注, 如: MESSAOUDI等^[6]研究了带有强阻尼时滞项的波方程u_tt-Δu-μ₁Δu_t-μ₂Δu_t(t- τ)=0, 当假设|μ₂|≤μ₁时证明方程解的适定性以及能量函数具有指数衰减性; FENG^[7]在一定条件下利用能量扰动法得到方程 ${u_{tt}} - \Delta u + \int_0^t {g\left( {t - s} \right)} \;\Delta u\left( {x, s} \right){\rm{d}}s - {\mu _1}\Delta {u_t} - {\mu _2}\Delta {u_t}\left( {t - \tau } \right) = 0$ 解的指数衰减性; MOHAMMAD和MUHAMMAD^[8]在文献[7]的基础上研究带有非线性源项|u|^p-1u的柯西问题, 证明在初始能量为负时, 解存在有限时间爆破. 更多解的爆破研究可参考文献[9-12].

基于文献[3, 6-8]的研究, 本文考虑可描述弹性杆纵波振动的带有强阻尼时滞项的粘弹性方程：

$\left\{\begin{array}{l} u_{t t}-\Delta_{m} u-\Delta u+g * \Delta u-\mu_{1} \Delta u_{t}(x, t)-\mu_{2} \Delta u_{t}(x, t-\tau)= \\ \begin{array}{ll} \ \ \ \ |u|^{p-2} u \ \ \ \ \ \ (x \in \varOmega, t>0), \\ u(x, t)=0 \ \ \ \ (x \in \partial \varOmega, t \geqslant 0), \\ u_{t}(x, t-\tau)=f_{0}(x, t-\tau) \quad(x \in \varOmega, t \in(0, \tau)), \\ u(x, 0)=u_{0}(x), u_{t}(x, 0)=u_{1}(x) \quad(x \in \varOmega), \end{array} \end{array}\right.$

(1)

其中，u(x, t)表示振动的位移, g*Δu表示粘弹性弱阻尼, -Δu_t表示弹性体的内部阻尼, Ω是 $\mathbb{R}$ ⁿ (n≥1)上带有光滑边界∂Ω的有界区域, g(·): $\mathbb{R}$ ⁺→ $\mathbb{R}$ ⁺是正函数, m、μ₁、μ₂、p为常数,

$\Delta_{m} u=\operatorname{div}\left(|\nabla u|^{m-2} \nabla u\right), (g * \Delta u)(t)=\int_{0}^{t} g(t-s) \Delta u(s) \mathrm{d} s.$

本文主要利用凹性方法, 证明当初始能量0 < E(0) < E₁和E(0) < 0时, 方程组(1)的解都存在有限时间的爆破, 并给出爆破时间T^*的上界估计.

1. 预备知识

本文用到Sobolev空间W₀^{1, m}(Ω) (m≥2)和赋范线性空间L^p(Ω)及其范数‖ · ‖_p. 特别地, L²(Ω)的范数表示为‖ · ‖. 正常数K>0为紧嵌入W₀^{1, m}(Ω)↪L^p(Ω)的最佳嵌入常数. 对方程组(1)中涉及到的函数g(·)和正常数m、p假设如下：

(A1) g: $\mathbb{R}$ ⁺→ $\mathbb{R}$ ⁺是非增可微函数, 满足: 1- $\int_{0}^{+\infty }{g\left( s \right)\text{d}s}={{l}_{0}}>0$ , g′(t)≤0, t≥0.

(A2) 当n>m时, m < p≤ $\frac{mn}{n-m}$ ; 当n≤m时, m < p < +∞.

(A3) 常数p和函数g(·) 满足: $\int_{0}^{+\infty }{g\left( s \right)}\text{d}s\le \frac{p\left( p-2 \right)}{{{\left( p-1 \right)}^{2}}}$ .

类似文献[6], 引进新变量

$z(x, \rho, t)=u_{t}(x, t-\rho \tau)((x, \rho, t) \in \varOmega \times(0, 1) \times(0, +\infty)),$

则有

$\tau z(x, \rho, t)+z_{\rho}(x, \rho, t)=0 \quad((x, \rho, t) \in \varOmega \times(0, 1) \times(0, +\infty)).$

从而方程组(1)等价于：

$\left\{\begin{array}{l} u_{t t}-\Delta_{m} u-\Delta u+g * \Delta u-\mu_{1} \Delta u_{t}(x, t)-\mu_{2} \Delta z(x, 1, t)= \\ \quad|u|^{p-2} u \quad((x, t) \in \varOmega \times(0, +\infty)), \\ \tau z(x, \rho, t)+z_{\rho}(x, \rho, t)=0 \\ \quad((x, \rho, t) \in \varOmega \times(0, 1) \times(0, +\infty)), \\ u(x, t)=0 \quad((x, t) \in \partial \varOmega \times[0, +\infty)), \\ z(x, \rho, 0)=f_{0}(x, -\rho \tau)\quad((x, \rho) \in \varOmega \times(0, 1)), \\ u(x, 0)=u_{0}(x), u_{t}(x, 0)=u_{1}(x)\quad(x \in \varOmega) . \end{array}\right.$

(2)

定义方程组(2)的修正能量函数为

$\begin{array}{l} E(t):=\frac{1}{2}\left\|u_{t}\right\|^{2}+\frac{1}{m}\|\nabla u\|_{m}^{m}+ \\ \quad \frac{1}{2}\left(1-\int_{0}^{t} g(s) \mathrm{d} s\right)\|\nabla u\|^{2}+\frac{1}{2}(g \circ \nabla u)(t)+ \\ \quad \frac{\xi}{2} \int_{\varOmega} \int_{0}^{1}|\nabla z(x, \rho, t)|^{2} \mathrm{~d} \rho \mathrm{d} x-\frac{1}{p}\|u\|_{p}^{p} \quad(t \geqslant 0), \end{array}$

(3)

其中

$\begin{array}{c} (g \circ \varphi)(t)=\int_{0}^{t} g(t-s)\|\varphi(t)-\varphi(s)\|^{2} \mathrm{~d} s, \\ \tau\left|\mu_{2}\right|<\xi<\tau\left(2 \mu_{1}-\left|\mu_{2}\right|\right), \mu_{1}>\left|\mu_{2}\right|. \end{array}$

下面给出方程组(1)的弱解定义.

定义 1 任给定初始值(u₀, u₁)∈W₀^{1, m}(Ω)×L²(Ω), 称 $\mathcal{H}$ =(u, u_t)∈C( $\mathbb{R}$ ⁺, W₀^{1, m}(Ω)×L²(Ω))为方程组(1)的弱解, 若满足: $\mathcal{H}$ (0)=(u₀, u₁), 且对任意w∈H₀¹(Ω)，有

$\begin{array}{l} \left(u_{t t}, w\right)+\left(|\nabla u|^{m-2} \nabla u, \nabla w\right)+(\nabla u, \nabla w)- \\ \ \ \ \ \ \ \ \ \int_{0}^{t} g(t-s)(\nabla u(s), \nabla w) \mathrm{d} s+\mu_{1}\left(\nabla u_{t}, \nabla w\right)+ \\ \ \ \ \ \ \ \ \ \mu_{2}\left(\nabla u_{t}(t-\tau), \nabla w\right)=\left(|u|^{p-2} u, w\right). \end{array}$

类似文献[5]、[6]、[10]的证明, 可直接得到方程组(2)的局部解的存在唯一性：

定理 1 假设(A1)、(A2)成立, 若初始值满足(u₀, u₁, f₀)∈W₀^{1, m}(Ω)×L²(Ω)×H¹(Ω×(- τ, 0)), 则方程组(2)存在唯一局部解, 且存在足够小的T>0, 有

$\begin{array}{c} u \in C\left([0, T) ; W_{0}^{1, m}(\varOmega)\right), \\ u_{t} \in C\left([0, T) ; H^{1}(\varOmega)\right) \cap L^{2}([0, T) \times \varOmega), \\ z \in C\left([0, T) ; H^{1}(\varOmega \times(0, 1))\right). \end{array}$

引理 1 设u是方程组(2)的解, 则存在正常数C₁>0, 使得

$E^{\prime}(t) \leqslant-C_{1}\left(\left\|\nabla u_{t}\right\|^{2}+\|\nabla z(x, 1, t)\|^{2}\right)<0.$

证明用u_t乘以方程组(2)中的第1个方程, 并在Ω上积分, 得

$\begin{array}{l} \frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{2}\left\|u_{t}\right\|^{2}+\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{1}{2}\left(1-\int_{0}^{t} g(s) \mathrm{d} s\right)\|\nabla u\|^{2}+\right. \\ \quad\; \; \left.\frac{1}{2}(g \circ \nabla u)(t)-\frac{1}{p}\|u\|_{p}^{p}\right]=\frac{1}{2}\left(g^{\prime} \circ \nabla u\right)(t)- \\ \quad\; \; \frac{1}{2} g(t)\|\nabla u\|^{2}-\mu_{1}\left\|\nabla u_{t}\right\|^{2}- \\ \quad\; \; \mu_{2} \int_{\varOmega} \nabla z(x, 1, t) \cdot \nabla u_{t}(x, t) \mathrm{d} x. \end{array}$

(4)

用 $-\frac{\xi }{\tau }\Delta z$ 乘以方程组(2)的第2个方程, 并在Ω×(0, 1)上积分，得

$\begin{array}{l} \frac{\xi}{2} \frac{\mathrm{d}}{\mathrm{d} t} \int_{\varOmega} \int_{0}^{1}|\nabla z(x, \rho, t)|^{2} \mathrm{~d} \rho \mathrm{d} x+ \\ \quad\; \;\frac{\xi}{2 \tau}\left(\|\nabla z(x, 1, t)\|^{2}-\left\|\nabla u_{t}\right\|^{2}\right)=0. \end{array}$

(5)

由Young不等式^[13], 可得

$\begin{array}{l} -\mu_{2} \int_{\varOmega} \nabla z(x, 1, t) \cdot \nabla u_{t}(x, t) \mathrm{d} x \leqslant \\ \quad\; \;\frac{\left|\mu_{2}\right|}{2}\|\nabla z(x, 1, t)\|^{2}+\frac{\left|\mu_{2}\right|}{2}\left\|\nabla u_{t}\right\|^{2}. \end{array}$

(6)

由式(4)~(6), 可得

$\begin{array}{l} E^{\prime}(t) \leqslant \frac{1}{2}\left(g^{\prime} \circ \nabla u\right)(t)-\frac{1}{2} g(t)\|\nabla u\|^{2}+\\ \quad\; \;\left(\frac{\xi}{2 \tau}+\frac{\left|\mu_{2}\right|}{2}-\mu_{1}\right)\left\|\nabla u_{t}\right\|^{2}+\left(\frac{\left|\mu_{2}\right|}{2}-\frac{\xi}{2 \tau}\right)\|\nabla z(x, 1, t)\|^{2}. \end{array}$

(7)

因此, 令 $0<{{C}_{1}}\le \min \left\{ {{\mu }_{1}}-\frac{\left| {{\mu }_{2}} \right|}{2}-\frac{\xi }{2\tau }, \frac{\xi }{2\tau }-\frac{\left| {{\mu }_{2}} \right|}{2} \right\}$ , 由式(7)和假设(A1)可知结论成立.

2. 解的爆破

下面考虑方程组(2)的初始能量为正值和负值时解的爆破情况. 首先, 引入3个正常数：

$K_{1}=\max \{1, K\}, \zeta_{1}=\frac{K_{1}^{p m /(m-p)}}{m}, E_{1}=\left(1-\frac{m}{p}\right) \zeta_{1}.$

引理 2 若假设(A1)、(A2)成立, 且满足

$E(0)<E_{1}, \zeta_{1}<\frac{1}{m}\left\|\nabla u_{0}\right\|_{m}^{m}+\frac{l}{2}\left\|\nabla u_{0}\right\|^{2} \leqslant \frac{K_{1}^{-m}}{m},$

(8)

则存在正常数ζ₂>ζ₁, 使得

$\frac{1}{m}\|\nabla u(t)\|_{m}^{m}+\frac{l}{2}\|\nabla u(t)\|^{2} \geqslant \zeta_{2} \quad(t \geqslant 0),$

(9)

$\|u(t)\|_{p}^{p} \geqslant\left(m^{\frac{1}{m}} K_{1}\right)^{p} \zeta_{2}^{p / m} \quad(t \geqslant 0).$

(10)

证明由式(3)和假设(A1)、(A2), 可得

$\begin{array}{l} E(t) \geqslant \frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-\frac{1}{p}\|u\|_{p}^{p} \geqslant \\ \quad\quad\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-\frac{K_{1}^{p}}{p}\|\nabla u\|_{m}^{p} \geqslant \\ \quad\quad\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)- \\ \quad\quad\frac{\left(m^{\frac{1}{m}} K_{1}\right)^{p}}{p}\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)^{\frac{p}{m}}= \\ \quad\quad\zeta(t)-\frac{\left(m^{\frac{1}{m}} K_{1}\right)^{p}}{p} \zeta^{\frac{p}{m}}(t):=f(\zeta(t)). \end{array}$

(11)

其中ζ(t)= $\frac{1}{m}\left\| \nabla u \right\|_{m}^{m}+\frac{l}{2}{{\left\| \nabla u \right\|}^{2}}$ . 易知f(ζ(t))在ζ₁>0处有最大值E₁, 且有

$f^{\prime}(\zeta)=1-\frac{\left(m^{\frac{1}{m}} K_{1}\right)^{p}}{m} \zeta^{\frac{p-m}{m}}.$

(12)

当ζ ∈(0, ζ₁) 时, f′(ζ)>0, 所以f(ζ)在(0, ζ₁)上严格单调递增; 当ζ∈(ζ₁, +∞)时, f′(ζ) < 0, 所以f(ζ)在(ζ₁, +∞)上严格单调递减. 又由于E(0) < E₁=f(ζ₁), 则存在正常数ζ₂ ∈(ζ₁, +∞), 使得f(ζ₂)=E(0). 由式(11)可知f(ζ(0))≤E(0)=f(ζ₂), 所以ζ(0)≥ζ₂.

假设存在t₁>0, 有 $\frac{1}{m}\left\| \nabla u\left( {{t}_{1}} \right) \right\|_{m}^{m}+\frac{l}{2}{{\left\| \nabla u\left( {{t}_{1}} \right) \right\|}^{2}}<{{\zeta }_{2}}$ . 进一步地, 由式(11)和f(ζ)的单调性可得E(t₂)≥ f(ζ(t₂))>f(ζ₂)=E(0), 在t ∈(0, T)上与E(t) < E(0)矛盾, 故式(9)成立.

再一次利用式(11), 可得

$\begin{aligned} &\frac{1}{p}\|u\|_{p}^{p} \geqslant \frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-E(t) \geqslant\\ &\ \ \ \ \frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-E(0) \geqslant \zeta_{2}-E(0)=\\ &\ \ \ \ \frac{\left(m^{\frac{1}{m}} K\right)^{p}}{p} \zeta_{2}^{p / m}. \end{aligned}$

(13)

因此, 式(10)成立.

定理 2 在引理2的假设下, 进一步假设(A3)成立, 则方程组(2)的解存在有限时间爆破.

证明设函数

$L_{1}(t)=G_{1}^{1-\alpha}(t)+\varepsilon \int_{\varOmega} u u_{t} \mathrm{~d} x+\frac{\varepsilon \mu_{1}}{2}\|\nabla u\|^{2} \quad(t \in[0, T)),$

(14)

其中, ε>0为足够小的正实数, 且

$0<\alpha \leqslant \min \left\{\frac{m-2}{p}, \frac{m-2}{2 m}\right\},$

(15)

$G_{1}(t):=E_{2}-E(t), E_{2} \in\left(E(0), E_{1}\right).$

(16)

下面将证明存在C>0, 使L₁(t)满足微分不等式

$L_{1}^{\prime}(t) \geqslant C L_{1}^{q}(t) \quad(q>1).$

首先, 由式(3)、(16)和引理1, 得

$\begin{array}{l} 0<G_{1}(0)<G_{1}(t) \leqslant\\ \ \ \ \ E_{2}-\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)+\frac{1}{p}\|u\|_{p}^{p} \quad(t \geqslant 0). \end{array}$

由引理2, 有

$E_{2}-\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right) \leqslant E_{2}-\zeta_{2} \leqslant E_{1}-\zeta_{1}=-\frac{m}{p} \zeta_{1}.$

因此

$0<G_{1}(0)<G_{1}(t) \leqslant \frac{1}{p}\|u\|_{p}^{p}.$

(17)

对L₁(t)关于时间t求导, 有

$\begin{array}{l} L_{1}^{\prime}(t)=(1-\alpha) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\varepsilon\left\|u_{t}\right\|^{2}+ \\ \ \ \ \ \ \ \ \ \varepsilon \int_{\varOmega} u u_{t t} \mathrm{~d} x+\varepsilon \mu_{1} \int_{\varOmega} \nabla u \cdot \nabla u_{t} \mathrm{~d} x= \\ \ \ \ \ \ \ \ \ (1-\alpha) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\varepsilon\left\|u_{t}\right\|^{2}-\varepsilon\|\nabla u\|_{m}^{m}- \\ \ \ \ \ \ \ \ \ \varepsilon\|\nabla u\|^{2}+\varepsilon \int_{\varOmega} \nabla u(t) \int_{0}^{t} g(t-s) \nabla u(s) \mathrm{d} s \mathrm{~d} x- \\ \ \ \ \ \ \ \ \ \varepsilon \mu_{2} \int_{\varOmega} \nabla u(x, t) \cdot \nabla z(x, 1, t) \mathrm{d} x+\varepsilon\|u\|_{p}^{p}. \end{array}$

(18)

由Young不等式, 有

$\begin{array}{l} \int_{\varOmega} \nabla u(t) \int_{0}^{t} g(t-s) \nabla u(s) \mathrm{d} s \mathrm{~d} x= \\ \ \ \ \ \ \ \ \ \int_{0}^{t} g(t-s) \int_{\varOmega} \nabla u(t) \cdot(\nabla u(s)-\nabla u(t)) \mathrm{d} x \mathrm{~d} s+ \\ \ \ \ \ \ \ \ \ \int_{0}^{t} g(s) \mathrm{d} s\|\nabla u\|^{2} \geqslant-\delta_{1}(g \circ \nabla u)(t)+\\ \ \ \ \ \ \ \ \ \left(1-\frac{1}{4 \delta_{1}}\right) \int_{0}^{t} g(s) \mathrm{d} s\|\nabla u\|^{2}, \end{array}$

(19)

$\begin{array}{l} -\mu_{2} \int_{\varOmega} \nabla u(x, t) \cdot \nabla z(x, 1, t) \mathrm{d} x \geqslant\\ \quad-\frac{\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2}-\left|\mu_{2}\right| \delta_{2}\|\nabla z(x, 1, t)\|^{2}. \end{array}$

(20)

取0<a<1, δ₁=p(1-a)/2, 将式(19)、(20)代入式(18), 可得

$\begin{array}{l} L_{1}^{\prime}(t) \geqslant(1-\alpha) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\varepsilon p(1-a) G_{1}(t)+\\ \quad\varepsilon\left(1+\frac{p(1-a)}{2}\right)\left\|u_{t}\right\|^{2}+\varepsilon\left[\frac{p(1-a)}{2}-1+\right.\\ \quad\left.\left(1-\frac{1}{2 p(1-a)}-\frac{p(1-a)}{2}\right) \int_{0}^{t} g(s) \mathrm{d} s\right]\|\nabla u\|^{2}+\\ \quad\varepsilon a\|u\|_{p}^{p}+\frac{\varepsilon p(1-a) \xi}{2} \int_{\varOmega} \int_{0}^{1}|\nabla z(x, \rho, t)|^{2} \mathrm{~d} \rho \mathrm{d} x+\\ \quad\varepsilon\left(\frac{p(1-a)}{m}-1\right)\|\nabla u\|_{m}^{m}-\frac{\varepsilon\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2}-\\ \quad\varepsilon\left|\mu_{2}\right| \delta_{2}\|\nabla z(x, 1, t)\|^{2}-\varepsilon p(1-a) E_{2}. \end{array}$

(21)

下面对式(21)中一些项进行估计. 令δ₂=MG₁^-α(t) (M>0), 由引理1可得

$\begin{array}{l} -\varepsilon \mid \mu_{2} \mid \delta_{2}\|\nabla z(x, 1, t)\|^{2}=\\ \quad -\varepsilon M\left|\mu_{2}\right| G_{1}^{-\alpha}(t)\|\nabla z(x, 1, t)\|^{2} \geqslant \\ \quad \frac{\varepsilon M\left|\mu_{2}\right|}{C_{1}} E^{\prime}(t) G_{1}^{-\alpha}(t)=-\frac{\varepsilon M\left|\mu_{2}\right|}{C_{1}} G_{1}^{-\alpha}(t) G_{1}^{\prime}(t). \end{array}$

(22)

利用Hölder不等式^[14]和嵌入W₀^{1, m}(Ω)↪L^p(Ω), 结合式(17), 有

$\begin{array}{l} -\frac{\varepsilon\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2}=-\frac{\varepsilon\left|\mu_{2}\right|}{4 M} G_{1}^{\alpha}(t)\|\nabla u\|^{2} \geqslant \\ \quad -\frac{\varepsilon\left|\mu_{2}\right||\varOmega|^{\frac{m-2}{m}}}{4 M p^{\alpha}}\|u\|_{p}^{p \alpha}\|\nabla u\|_{m}^{2} \geqslant \\ \quad -\frac{\varepsilon\left|\mu_{2}\right||\varOmega|^{\frac{m-2}{m}} K^{p \alpha}}{4 M p^{\alpha}}\|\nabla u\|_{m}^{p \alpha+2} . \end{array}$

(23)

进一步地, 由 ${{z}^{\nu }}\le z+1\le \left( 1+\frac{1}{b} \right)\left( z+b \right)$ (z≥0, 0 < ν≤1, b≥0)和式(23), 可得

$-\frac{\varepsilon\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2} \geqslant-\frac{\varepsilon C_{2}}{M}\left(\|\nabla u\|_{m}^{m}+G_{1}(t)\right),$

(24)

其中 ${C_2}: = \frac{{\left| {{\mu _2}} \right|{{\left| \mathit{\Omega } \right|}^{\frac{{m - 2}}{m}}}{K^{p\alpha }}\left( {1 + {G_1}\left( 0 \right)} \right)}}{{4{p^\alpha }{G_1}\left( 0 \right)}}$ .

由假设(A3)和p>m, 可得

$\begin{array}{l} \frac{p(1-a)}{2}-1+\left(1-\frac{1}{2 p(1-a)}-\frac{p(1-a)}{2}\right) \int_{0}^{t} g(s) \mathrm{d} s>0, \\ \frac{p(1-a)}{m}-1 \leqslant p(1-a)-2+\left(2-\frac{1}{p(1-a)}-p(1-a)\right)(1-l). \end{array}$

因此, 令a>0充分小, 取适当的E₂ ∈(E(0), E₁), 并结合引理2可得

$\begin{array}{l} \varepsilon\left[\frac{p(1-a)}{2}-1+\left(1-\frac{1}{2 p(1-a)}-\frac{p(1-a)}{2}\right) \int_{0}^{t} g(s) \mathrm{d} s\right] \times \\ \quad \|\nabla u\|^{2}+\varepsilon\left(\frac{p(1-a)}{m}-1\right)\|\nabla u\|_{m}^{m}-\varepsilon p(1-a) E_{2} \geqslant \\ \quad \varepsilon(p(1-a)-m)\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)- \\ \quad \varepsilon p(1-a) E_{2} \geqslant \varepsilon C_{3}\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right), \end{array}$

(25)

其中C₃=p(1-a)-m- $\frac{{p\left( {1 - a} \right){E_2}}}{{{\zeta _2}}}$ ≥0.

将式(22)~(25)代入式(21), 可得

$\begin{aligned} L_{1}^{\prime}(t) & \geqslant\left(1-\alpha-\frac{\varepsilon M\left|\mu_{2}\right|}{C_{1}}\right) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\\ & \varepsilon\left(p(1-a)-\frac{C_{2}}{M}\right) G_{1}(t)+\varepsilon\left(1+\frac{p(1-a)}{2}\right)\left\|u_{t}\right\|^{2}+\\ & \varepsilon\left(\frac{C_{3}}{m}-\frac{C_{2}}{M}\right)\|\nabla u\|_{m}^{m}+\frac{\varepsilon l C_{3}}{2}\|\nabla u\|^{2}+\varepsilon a\|u\|_{p}^{p} . \end{aligned}$

(26)

在式(26)中, 当a>0充分小、M>0足够大时, 可使p(1-a)- $\frac{{{C_2}}}{M}$ ≥0, 且 $\frac{{{C_3}}}{m} - \frac{{{C_2}}}{M}$ ≥0. 进一步地, 令ε>0充分小, 可使1-α- $\frac{{\varepsilon M\left| {{\mu _2}} \right|}}{{{C_1}}}$ ≥0. 故存在正常数C₄>0, 使得

$\begin{array}{l} L_{1}^{\prime}(t) \geqslant C_{4}\left(\left\|u_{t}\right\|^{2}+\|\nabla u\|_{m}^{m}+\|\nabla u\|^{2}+\|u\|_{p}^{p}+G_{1}(t)\right) \\ \quad\quad(t \geqslant 0) . \end{array}$

(27)

另一方面, 由式(14)可得

$L_{1}^{1 /(1-\alpha)}(t) \geqslant C_{5}\left(G_{1}(t)+\left(\int_{\varOmega} u u_{t} \mathrm{~d} x\right)^{\frac{1}{1-\alpha}}+\|\nabla u\|^{\frac{2}{1-\alpha}}\right).$

(28)

利用式(15)、(17), 结合Young不等式、Poincaré不等式^[13], 得

$\left(\int_{\varOmega} u u_{t} \mathrm{~d} x\right)^{\frac{1}{1-\alpha}} \leqslant C_{6}\left(G_{1}(t)+\|\nabla u\|_{m}^{m}+\left\|u_{t}\right\|^{2}\right) .$

(29)

再由式(15)、(17), 有

$\begin{array}{l}\|\nabla u\|^{\frac{2}{1-\alpha}}=\left(\|\nabla u\|^{m}\right)^{\frac{2}{m(1-\alpha)}} \leqslant\\ \quad\left(1+\frac{1}{G_{1}(0)}\right)\left(G_{1}(t)+\|\nabla u\|_{m}^{m}\right) \leqslant \\ \quad\left(1+\frac{1}{G_{1}(0)}\right)\left(G_{1}(t)+|\varOmega|^{\frac{m-2}{2}}\|\nabla u\|_{m}^{m}\right). \end{array}$

(30)

可得

$L_{1}^{1 /(1-\alpha)}(t) \leqslant C_{6}\left(G_{1}(t)+\|\nabla u\|_{m}^{m}+\left\|u_{t}\right\|^{2}\right) .$

(31)

最后, 综合式(27)、(31)可知：存在C>0, 使得

$L_{1}^{\prime}(t) \geqslant C L_{1}^{1 /(1-\alpha)}(t) \quad(t \geqslant 0).$

(32)

对式(32)在(0, t)上积分, 有

$L_{1}^{1 /(1-\alpha)}(t) \geqslant \frac{1}{L_{1}^{-\alpha /(1-\alpha)}(0)-\frac{C \alpha}{1-\alpha} t},$

所以, 存在T^*≤ $\frac{{1 - \alpha }}{{C\alpha L_1^{\alpha /\left( {1 - \alpha } \right)}\left( 0 \right)}}$ , 当t趋于T^*时, L₁(t)趋于无穷, 即方程组(2)的解爆破.

定理 3 若假设(A1)~(A3)成立, 且初始能量E(0) < 0时, 则方程组(2)的解同样存在有限时间爆破.

证明设函数

$L_{2}(t)=G_{2}^{1-\beta}(t)+\varepsilon \int_{\varOmega} u u_{t} \mathrm{~d} x+\frac{\varepsilon \mu_{1}}{2}\|\nabla u\|^{2} \quad(t \in[0, T)),$

(33)

其中, ε>0为足够小的正实数, 0 < β≤min $\left\{ {\frac{{m - 2}}{p}, \frac{{m - 2}}{{2m}}} \right\}$ , 且G₂(t)=-E(t).

由式(3)、(16)和引理1, 可得0 < G₂(0) < G₂(t)≤ $\frac{1}{p}\left\| u \right\|_p^p$ (t≥0). 参照定理2的证明方法(只需令E₂=0), 定理得证.

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