本文研究如下具有强阻尼和对数非线性源项的p-Laplacian型波动方程的初边值问题

$ \begin{align} \left\{ \begin{aligned} &{u_{tt}} + {\Delta _p}u - \Delta {u_t} = |u{|^{p - 2}}u\ln |u|,(x,t) \in \Omega \times (0,T),\\ &u(x,0) = {u_0}(x),{u_t}(x,0) = {u_1}(x),x \in \Omega ,\\ &u(x,t) = 0,(x,t) \in \partial \Omega \times [0,T], \end{aligned} \right. \end{align} $

(1.1)

其中$ \Omega \subset {R^n} $$ \left( {n \ge 1} \right) $是一个具有光滑边界$ \partial \Omega $的有界区域, $ {\Delta _p}u = - div( {\left| {\nabla u} \right|^{p - 2}}\nabla u) $, $ {u_0}(x) \in {W^{1,p}}(\Omega ) $, $ {u_1} \in {L^2}(\Omega ) $, 指数$ p $满足

$ (H)\qquad2 < p < + \infty ,n \le p;\quad2 < p \le \frac{{np}}{{n - p}},n > p. $

非线性波动方程可用来描述物理学和其他应用科学中的许多现象, 如粘弹性力学和量子力学理论^{[1, 2]}, 目前关于其解的性质已有许多文献, 例如[3–5]. 考虑如下p-Laplacian型波动方程的初边值问题

$ \begin{align} \left\{ \begin{aligned} &{u_{tt}} + {\Delta _p}u = |u{|^{r - 2}}u,(x,t) \in \Omega \times (0,T),\\ &u(x,0) = {u_0}(x),{u_t}(x,0) = {u_1}(x),x \in \Omega ,\\ &u(x,t) = 0,(x,t) \in \partial \Omega \times [0,T], \end{aligned} \right. \end{align} $

(1.2)

由势阱法可定义方程(1.2)的势阱深度Tsutsumi ^[8]得到了在Ibrahim和Lyaghfouri^[12]得到了解的Ye^[13]等人通过势阱法及对数Sobolev不等式证明了方程整体解的存在性, 并用凸方法得到了解的$ L^2 $范数在有限时刻爆破.

对于问题

$ \begin{align} \left\{ \begin{aligned} &{u_{tt}} + {\Delta _p}u - \Delta {u_t} = f(u),(x,t) \in \Omega \times (0,T),\\ &u(x,0) = {u_0}(x),{u_t}(x,0) = {u_1}(x),x \in \Omega ,\\ &u(x,t) = 0,(x,t) \in \partial \Omega \times [0,T], \end{aligned} \right. \end{align} $

(1.3)

其中$ f \in {C^1} $, $ \left| {f(u)} \right| \le c{\left| u \right|^r} $, $ \left| u \right| \ge 1 $, $ 1 \leq r<\frac{5 p}{2(3-p)} $, 定义方程(1.3)的能量泛函为

$ E(t):=\frac{1}{2}\left\|u_{t}(t)\right\|_{2}^{2}+\frac{1}{p}\|\nabla u(t)\|_{p}^{p}-\displaystyle{\int}_{\Omega} F(u(t)) d x, $

其中Pei^[14]研究了在Chen^[15]等人使用Galerkin方法证明了解的存在性, 得到了解以多项式衰减的估计. 若源项为对数非线性项, 当Di^[16]等人利用势阱法考虑了全局解的存在唯一性, 解的衰减估计以及解爆破的上下界. Ma^[17]得到了解以指数衰减的估计和解在无穷时刻爆破.若源项为非线性项, 当Zu^[18]在$ 3 $维情况下对上述结果进行推广得出在不同的初始能量下弱解在有限时刻爆破的上界及下界.

受文[16]的启发, 本文进一步考虑$ p>2 $时解的存在性, 衰减估计和爆破情况, 并结合文献[18]中得到解的衰减估计的方法, 对数Sobolev不等式, 势阱理论和Sobolev嵌入定理, 使用Galerkin方法得到本文结论.

为了说明本文的主要结果，下面引入能量泛函

$ \begin{gather} E(t): = E(u,{u_t}) = \frac{1}{2}{\left\| {{u_t}} \right\|^2} + \frac{1}{p}\left\| {\nabla u} \right\|_p^p + \frac{1}{{{p^2}}}\left\| u \right\|_p^p - \frac{1}{p}\displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx, \end{gather} $

(1.4)

$ \begin{gather} E(0) = \frac{1}{2}{\left\| {{u_1}} \right\|^2} + \frac{1}{p}\left\| {\nabla {u_0}} \right\|_p^p + \frac{1}{{{p^2}}}\left\| {{u_0}} \right\|_p^p - \frac{1}{p}\displaystyle{\int\limits_\Omega} {{{\left| {{u_0}} \right|}^p}\ln } \left| {{u_0}} \right|dx, \end{gather} $

(1.5)

$ \begin{gather} J(u) = \frac{1}{p}\left\| {\nabla u} \right\|_p^p + \frac{1}{{{p^2}}}\left\| u \right\|_p^p - \frac{1}{p}\displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx, \end{gather} $

(1.6)

$ \begin{gather} I(u) = \left\| {\nabla u} \right\|_p^p - \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx, \end{gather} $

(1.7)

下面的定理为本文的主要结论.

定理1.1    假设$ (H) $成立, $ u_{0} \in W_{0}^{1, p}(\Omega) $, $ u_{1} \in L^{2}(\Omega) $, $ 0 < E(0) < d $, $ I(u_0) >0 $, 那么问题(1.1)存在唯一整体解$ u(x,t) $满足$ u \in {L^\infty }(0,\infty ;W_0^{1,p}) $, $ {u_t} \in {L^2}(0,\infty ;W_0^{1,2}) $.

定理1.2    假设$ {u_0} $, $ {u_1} $, $ p $满足定理1.1, $ u(x,t) $是问题(1.1)的解, 则存在一个正常数$ C = C({u_0},{u_1}) $, 使得能量泛函$ E(t) $满足以下多项式衰减估计

$ \begin{align} \begin{aligned} E(t) \le C{(1 + t)^{ - p/ (p - 2)}},\quad\forall t \ge 0. \end{aligned} \end{align} $

(1.8)

定理1.3    假设$ (H) $成立, $ u_{0} \in W_{0}^{1, p}(\Omega) $, $ u_{1} \in L^{2}(\Omega) $, $ \int_{\Omega} u_{0} u_{1} dx>0 $, $ 0 < E(0) < d $, $ I(u_0) <0 $, 则问题(1.1)的弱解在有限时间$ {T^*} $内爆破, 即

$ \begin{align} \begin{aligned} \mathop {\lim }\limits_{t \to {T^{ * - }}} \left( {{{\left\| u \right\|}^2} + \displaystyle{\int\limits_0^t} {{{\left\| {\nabla u} \right\|}^2}} dt} \right) = + \infty . \end{aligned} \end{align} $

(1.9)

且时间$ {T^*} $的上限如下

$ \begin{align} \begin{aligned} T^{*} \leq \frac{2 b {\eta ^2}+2\left\|u_{0}\right\|_{2}^{2}}{(p-2) b\eta+(p-2) \int_{\Omega} u_{0} u_{1} d x-2\left\|\nabla u_{0}\right\|_{2}^{2}}, \end{aligned} \end{align} $

(1.10)

其中$ b = d - E(0) > 0 $, $ \eta > \max \left\{ {0,\frac{{2{{\left\| {{u_0}} \right\|}^2} - (p - 2) {\int_{\Omega} u_{0} u_{1} dx}}}{{b(p - 2)}}} \right\} $.

首先, 我们介绍一些将在整篇论文中使用的定义和引理.

为了方便, 本文中所提到的$ C $是一个不依赖于未知函数的正常数, 在不同地方其具体值可能不同. 我们定义这些符号$ W_0^{1,p}(\Omega ) = W_0^{1,p}, {L^2}(\Omega ) = {L^2}, {\left\| u \right\|_{{L^p}(\Omega )}} = {\left\| u \right\|_p}, {\left\| u \right\|_{{L^2}(\Omega )}} = \left\| u \right\|, {(u,v) = \displaystyle{\int}_\Omega {uv} dx,u,v \in W_0^{1,p}.} $

为了证明方程解的整体存在性，还需定义泛函

$ \begin{gather} \mathcal {N} = \{ u \in W_0^{1,p}\backslash \{ 0\} |I(u) = 0,{\left\| {\nabla u} \right\|_p} \ne 0\} , \end{gather} $

(2.1)

根据(1.6), (1.7),

$ \begin{align} J(u) = \frac{1}{p}I(u) + \frac{1}{{{p^2}}}\left\| u \right\|_p^p, \end{align} $

(2.2)

定义集合

$ \begin{gather} W = \{ u \in W_0^{1,p}|J(u) < d,I(u) > 0\} \cup \{ 0\} , \end{gather} $

(2.3)

$ \begin{gather} V = \{ u \in W_0^{1,p}|J(u) < d,I(u) < 0\}, \end{gather} $

(2.4)

以及定义势阱深度

$ \begin{align} d = \inf \{ \mathop {\sup }\limits_{\lambda > 0} J(\lambda u)|u \in W_0^{1,p}\backslash \{ 0\},{\left\| {\nabla u} \right\|_p} \ne 0\} = \mathop {\inf }\limits_{u \in \mathcal{N}} J(u). \end{align} $

(2.5)

在本文中, 我们考虑问题(1.1)的弱解$ u(x,t) $, 其定义如下. 在不造成混淆的情况下, 我们有时会用$ u(t) $来表示$ u(x,t) $.

定义2.1 (弱解)    若$ u(0) = {u_0} \in W_0^{1,p} $, $ {u_t}(0) = {u_1} \in {L^2} $, 且$ u \in {L^\infty }(0,T ;W_0^{1,p} $), $ {u_t} \in {L^2}(0,T ;W_0^{1,2}) $, 满足$ \left( {{u_t}(t),v} \right) + \displaystyle{\int}_0^t {\left( {|\nabla u{|^{p - 2}}\nabla u,\nabla v} \right)} dt + \left( {\nabla u,\nabla v} \right) = \displaystyle{\int\limits_0^t} {\left( {|u{|^{p - 2}}u\ln |u|,v} \right)} dt + \left( {{u_1},v} \right) + \left( {\nabla {u_0},\nabla v} \right), \forall v \in W_0^{1,p}. $则称$ u(x,t) $是问题(1.1)在$ \Omega \times \left( {0,T} \right) $的弱解.

下面给出需要在定理证明过程中用到的引理.

引理2.4    ^[19] 设$ p $满足$ 1 \le p < n $, 对一切$ u \in H_0^1 $, 有

$ \begin{align} \begin{aligned} {\left\| u \right\|_q} \le C{\left\| {\nabla u} \right\|_p}, \end{aligned} \end{align} $

(2.6)

其中$ q $满足如下关系

$ \begin{equation} \begin{cases} 2 \le q < + \infty ,&\text{如果} 2 \le n \le p;\\ 2 \le q < \frac{{np}}{{n - p}},&\text{如果} 2 < p < n. \end{cases} \notag \end{equation} $

引理2.5    ^[20][$ {L^p} $-对数Sobolev不等式]若$ u \in H_0^1 $, 有

$ \begin{align} \begin{aligned} \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx \le \left\| u \right\|_p^p\ln {\left\| u \right\|_p} + \frac{{(p - 2){a^2}}}{{4\pi }}\left\| u \right\|_p^p + \frac{{{a^2}}}{{2\pi }}\left\| {\nabla u} \right\|_p^p - \frac{n}{p}(1 + \ln a)\left\| u \right\|_p^p, \end{aligned} \end{align} $

(2.7)

其中$ a >0 $是一个常数.

引理2.6    ^[21]假设$ E(t) $是一非增长函数, 常数$ q \ge 0 $, $ \gamma > 0 $, 满足

$ \begin{align} \begin{aligned} \int\limits_S^\infty {{E^{q + 1}}(t)} dt \le {\gamma ^{ - 1}}{E^q}(0)E(S),\quad\forall S \ge 0. \end{aligned} \end{align} $

(2.8)

则有

$ \begin{gather} E(t) \le E(0){(\frac{{1 + q}}{{1 + q\gamma t}})^{1/q}},\quad\forall t \ge 0,q > 0, \end{gather} $

(2.9)

$ \begin{gather} E(t) \le E(0){e^{1 - \gamma t}},\quad\forall t \ge 0,q = 0. \end{gather} $

(2.10)

引理2.7    ^[22] 假设$ \rho $是一个正常数, 则有

$ \begin{gather} {\phi ^p}\log \phi \le \frac{{{e^{ - 1}}}}{\rho }{\phi ^{p + \rho }},\quad\forall \phi \ge 1, \end{gather} $

(2.11)

$ \begin{gather} \left| {{\phi ^p}\log \phi } \right| \le {(e\rho )^{ - 1}},\quad\forall 0 < \phi < 1. \end{gather} $

(2.12)

引理2.8    ^[23] 假设$ G(t) $是一个正的二次可微函数, 对$ \forall t > 0 $, 满足

$ \begin{equation} G(t){G^{''}}(t) - (1 + \alpha ){({G^\prime}(t))^2} \ge 0, \end{equation} $

(2.13)

其中$ \alpha > 0 $. 若$ G(t) > 0 $, $ {G^\prime}(t) > 0 $, 则$ \exists {T^ * } \le \frac{{G(0)}}{{\alpha {G^\prime}(0)}} $, 使得

$ \begin{equation} \mathop {\lim }\limits_{t \to {T^{ * - }}} G(t) = \infty . \end{equation} $

(2.14)

同时, 给出势阱深度$ d $和势阱的一些性质.

引理2.9    若$ \forall u \in W_0^{1,p} $, $ {\left\| u \right\|_p} \ne 0 $, 有

(1) $ \mathop {\lim }\limits_{\lambda \to {0^ + }} J(\lambda u) = 0 $, $ \mathop {\lim }\limits_{\lambda \to + \infty } J(\lambda u) = - \infty $；

(2) $ \exists {\lambda ^ * } > 0 $, 使得$ \frac{d}{{d\lambda }}J(\lambda u){|_{\lambda = {\lambda ^ * }}} = 0 $；且$ J(\lambda u) $在$ 0 < \lambda < {\lambda ^ * } $单调递增, 在$ {\lambda ^ * } < \lambda < + \infty $单调递减；

(3) 当$ 0 < \lambda < {\lambda ^ * } $时, $ I(\lambda u) > 0 $, 而$ {\lambda ^ * } < \lambda < + \infty $时, $ I(\lambda u) < 0 $, 且$ I({\lambda ^ * }u) = 0 $.

证     (1)由$ J(\lambda u) $的定义, 对$ \forall u \in W_0^{1,p} $, 有

$ J(\lambda u) = \frac{{{\lambda ^p}}}{p}\left\| {\nabla u} \right\|_p^p - \frac{{{\lambda ^p}}}{p}\displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx - \frac{{{\lambda ^p}}}{p}\left\| u \right\|_p^p\ln \lambda + \frac{{{\lambda ^p}}}{{{p^2}}}\left\| u \right\|_p^p. $

则$ (1) $成立.

(2) 对$ J(\lambda u) $求导, 可得

$ \frac{d}{{d\lambda }}J(\lambda u) = {\lambda ^{p - 1}}(\left\| {\nabla u} \right\|_p^p - \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx - \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| \lambda \right|dx). $

令$ \frac{d}{{d\lambda }}J(\lambda u) = g(\lambda ) $, 有

$ \begin{equation} \begin{split} \frac{d}{{d\lambda }}g(\lambda ) = (p - 1){\lambda ^{p - 2}}(\left\| {\nabla u} \right\|_p^p - \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx - \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| \lambda \right|dx) - {\lambda ^{p - 2}}\left\| u \right\|_p^p. \end{split}\nonumber \end{equation} $

若$ \frac{d}{{d\lambda }}J(\lambda u){|_{\lambda = {\lambda ^ * }}} = 0 $, 则

$ {\lambda ^ * } = \exp (\frac{{\left\| {\nabla u} \right\|_p^p - \int\limits_\Omega {{{\left| u \right|}^p}\ln } \left| u \right|dx}}{{\left\| u \right\|_p^p}}). $

且有$ \frac{d}{{d\lambda }}g(\lambda ){|_{\lambda = {\lambda ^ * }}} = - {\lambda ^{p - 2}}\left\| u \right\|_p^p < 0. $满足$ J(\lambda u) $在$ 0 < \lambda < {\lambda ^ * } $单调递增, 在$ {\lambda ^ * } < \lambda < + \infty $单调递减.

(3) 由$ I(\lambda u) $的定义, 对$ \forall u \in W_0^{1,p} $, 有

$ I(\lambda u) = {\lambda ^p}(\left\| {\nabla u} \right\|_p^p - \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| u \right|dx - \displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln } \left| \lambda \right|dx). $

因此, $ I({\lambda ^ * }u) = {\lambda ^ * }\frac{d}{{d{\lambda ^ * }}}J({\lambda ^ * }u) = 0 $.

引理2.10    假设$ u \in W_0^{1,p} $, $ {\left\| {\nabla u} \right\|_p} \ne 0 $, 则有$ d \ge M $, 其中$ M=\frac{1}{p^{2}}(2 \pi)^{\frac{n}{2}} e^{\frac{2(n+p)-p^{2}}{2}} $.

证     由引理2.9和(1.6), 可得

$ \sup\limits_{\lambda \geq 0} J(\lambda u) = J({\lambda ^ * }u) = \frac{1}{{{p^2}}}\left\| {{\lambda ^ * }u} \right\|_p^p, $

由引理2.5, 令$ a = \sqrt {2\pi } $, 可知

$ \begin{equation} \begin{split} I(\lambda u) = {\lambda ^p}\left\| {\nabla u} \right\|_p^p - \displaystyle{\int\limits_\Omega} {{{\left| {\lambda u} \right|}^p}\ln } \left| {\lambda u} \right|dx \le \left\| {\lambda u} \right\|_p^p(\frac{n}{p}\ln \sqrt {2\pi } e - \ln {\left\| {\lambda u} \right\|_p} - \frac{{p - 2}}{2}). \end{split}\nonumber \end{equation} $

又由$ I({\lambda ^ * }u) = 0 $可知$ (\frac{n}{p}\ln \sqrt {2\pi } e - \ln {\left\| {{\lambda ^ * }u} \right\|_p} - \frac{{p - 2}}{2}) \le 0, $从而得到$ \left\|{\lambda ^ * }u\right\|_{p}^{p} \geq(2 \pi)^{\frac{n}{2}}e^{\frac{2(n+p)-p^{2}}{2}}. $

因此，根据$ d $的定义，可得$ d \ge M > 0 $.

在这一节中, 我们将证明弱解的全局存在性和唯一性.

证明定理1.1  证明解的整体存在性分为三个步骤. 首先, 第一步, 构造近似解. 假设$ \left\{ {{w_j}} \right\}_{j = 1}^\infty $为$ W_0^{1,p} $中的一组基, 且$ {w_j} $是满足狄利克雷边界条件的Laplace算子的特征函数, 即

$ \begin{align} \left\{ \begin{aligned} &- \Delta {w_j} = \lambda_j {w_j},x \in \Omega ,\\ &{w_j} = 0,x \in \partial \Omega. \end{aligned} \right. \end{align} $

(3.1)

同时$ \left\{ {{w_j}} \right\}_{j = 1}^\infty $在$ {L^2} $中是标准正交的, $ {E_v} $是由$ \left\{ {{w_1},{w_2}, \cdots ,{w_j}} \right\} $张成的线性子空间. 接下来, 构造问题(1.1)的近似解$ {u_m}(x,t) = \sum\limits_{i = 1}^m {{g_{jm}}(t)} {w_j},{g_{jm}}(t) \in {C^2}[0,T],\forall T > 0, $其中未知函数$ {g_{jm}}(t) $满足以下关系

$ \begin{equation} \begin{split} &\left( {{u_{mt}}(t),{w_j}} \right) + \displaystyle{\int\limits_0^t} {\left( {|\nabla {u_m}(t){|^{p - 2}}\nabla {u_m}(t),\nabla {w_j}} \right)} dt + \left( {\nabla {u_m},\nabla {w_j}} \right)\\ =& \displaystyle{\int\limits_0^t} {\left( {|{u_m}{|^{p - 2}}{u_m}\ln |{u_m}|,{w_j}} \right)} dt + \left( {{u_{1m}},{w_j}} \right) + \left( {\nabla {u_{0m}},\nabla {w_j}} \right). \end{split} \end{equation} $

(3.2)

其初值条件为: $ {u_m}(0) = {u_{0m}} $, $ {u_{mt}}(0) = {u_{1m}}. $由于$ W_0^{1,p} $在$ {L^2} $中稠密, 则初值条件满足

$ \begin{equation} \begin{split} {u_{0m}} = \sum\limits_{i = 1}^m {{g_{jm}}(0)} {w_j} = \sum\limits_{i = 1}^m {{\alpha _{jm}}(t)} {w_j} \to {u_0}(x) \text{在}W_0^{1,p} \text{中},m \to \infty, \end{split} \end{equation} $

(3.3)

$ \begin{equation} \begin{split} {u_{1m}} = \sum\limits_{i = 1}^m {{g_{jmt}}(0)} {w_j} = \sum\limits_{i = 1}^m {{\beta _{jm}}(t)} {w_j} \to {u_1}(x) \text{在 }{L^2}\text{中},m \to \infty. \end{split} \end{equation} $

(3.4)

因此, 根据常微分方程中的Picard迭代法, 问题$ (3.2) $-$ (3.4) $在$ \left[ {0,t_m} \right] $, $ 0 < t_m < T $中存在一个局部解$ {u_m}(t) $.

第二步, 为了证明解的整体存在性，需要对构造的近似解$ {u_m}(x,t) $进行先验估计. 先对(3.2)关于$ t $求导, 然后在其两边同时乘以$ g_{jmt} $, 得到

$ \begin{align} \left( {{u_{mtt}},{u_{mt}}} \right) + \left( {|\nabla {u_m}{|^{p - 2}}\nabla {u_m},\nabla {u_{mt}}} \right) + \left( {\nabla {u_m},\nabla {u_{mt}}} \right) = \left( {|{u_m}{|^{p - 2}}{u_m}\ln |{u_m}|,{u_{mt}}} \right). \end{align} $

(3.5)

对(3.5)关于$ j $从$ 1 $到$ m $求和, 并在$ [0,t] $上进行积分, 得到

$ \begin{align} \begin{aligned} E({u_m},{u_{mt}}) + \displaystyle{\int\limits_0^t} {{{\left\| {\nabla {u_{mt}}} \right\|}^2}d\tau } = E({u_{0m}},{u_{1m}}) < M \le d. \end{aligned} \end{align} $

(3.6)

接下来, 证明

$ \begin{align} \begin{aligned} ({u_m},{u_{mt}}) \in W,\quad\forall t \ge 0. \end{aligned} \end{align} $

(3.7)

假设(3.7)不成立, 即存在最小时间$ {t^*} $, 使得$ ({u_m},{u_{mt}}) \in \partial W $, 即有

$ \begin{align} \begin{aligned} E({u_m}({t^ * }),{u_{mt}}({t^ * })) = d, \end{aligned} \end{align} $

(3.8)

或者

$ \begin{align} \begin{aligned} I({u_m}({t^ * })) = 0. \end{aligned} \end{align} $

(3.9)

由(3.6)可以知道, (3.8)显然不成立. 若(3.9)成立, 根据$ d $的定义, 可以得到$ E({u_m}({t^ * }),{u_{mt}}({t^ * })) > J({u_m}({t^ * })) \ge d $, 这也是不成立的. 因此, 由(3.6)和(3.7)可以得到

$ \begin{align} \begin{aligned} \frac{1}{2}{\left\| {{u_{mt}}} \right\|^2} + \frac{1}{{{p^2}}}\left\| {{u_m}} \right\|_p^p + \displaystyle{\int\limits_0^t} {{{\left\| {\nabla {u_{mt}}} \right\|}^2}d\tau } < E({u_{0m}},{u_{1m}}) < M \le d. \end{aligned} \end{align} $

(3.10)

则有

$ \begin{gather} {\left\| {{u_{mt}}} \right\|^2} < 2M, \end{gather} $

(3.11)

$ \begin{gather} \left\| {{u_m}} \right\|_p^p < M{p^2}, \end{gather} $

(3.12)

$ \begin{gather} \displaystyle{\int\limits_0^t} {{{\left\| {\nabla {u_{mt}}} \right\|}^2}d\tau } < M. \end{gather} $

(3.13)

根据(1.6), (1.7), 引理2.5, 令$ a = \sqrt \pi $, 整理可得

$ \begin{align} \left\| {\nabla {u_m}} \right\|_p^p &= 2I({u_m}) + 2\displaystyle{\int\limits_\Omega} {{{\left| {{u_m}} \right|}^p}\ln } \left| {{u_m}} \right|dx - \left\| {\nabla {u_m}} \right\|_p^p \\ &= 2pJ({u_m}) - \left\| {\nabla {u_m}} \right\|_p^p - \frac{2}{p}\left\| {{u_m}} \right\|_p^p + 2\displaystyle{\int\limits_\Omega} {{{\left| {{u_m}} \right|}^p}\ln } \left| {{u_m}} \right|dx \\ &\le 2pJ({u_m}) - \frac{2}{p}\left\| {{u_m}} \right\|_p^p + \frac{{p - 2}}{2}\left\| {{u_m}} \right\|_p^p - \frac{n}{p}(\ln \pi {e^2})\left\| {{u_m}} \right\|_p^p + 2\ln {\left\| {{u_m}} \right\|_p}\left\| {{u_m}} \right\|_p^p \\ &\le 2pM + \frac{{(p - 2){p^2}}}{2}M + 2pM\ln (M{p^2}) \\ &\le {C_M}. \end{align} $

(3.14)

根据$ \left( {{\Delta _p}u,v} \right) = \left( {|\nabla u(t){|^{p - 2}}\nabla u(t),\nabla v} \right) $, 由(3.14)和Hölder不等式, 可得

$ \begin{gather} {\left\| {{\Delta _p}u} \right\|_{{W^{ - 1,p^\prime}}}} \le \left\| {\nabla u} \right\|_p^{p - 1} \le C_M^{p - 1}. \end{gather} $

(3.15)

第三步, 对近似解$ {{u_m}} $取极限. 由估计$ (3.11)-(3.15) $可知, 存在一个函数列$ \left\{ {{u_m}} \right\} $的子序列(仍记为$ \left\{ {{u_m}} \right\} $)和函数$ u $, 使得

$ \begin{gather} u_{m} \stackrel{*}{\rightharpoonup}u \text {在}L^{\infty}\left(0,\infty ; W_0^{1,p}\right)\text{中}, \end{gather} $

(3.16)

$ \begin{gather} u_{mt} \stackrel{}{\rightharpoonup} u \text{在}L^{2}\left(0, \infty ; W_0^{1,2}\right)\text{中}, \end{gather} $

(3.17)

$ \begin{gather} \Delta _ { p } u _ { m } \stackrel{*}{\rightharpoonup} \chi \text{在}L^{\infty}\left(0,\infty ; {W^{ - 1,p^\prime}}\right)\text{中}. \end{gather} $

(3.18)

其中$ \chi = {\Delta _p}u $(证明见[13]). 利用Aubin–Lions–Simon^[24]引理和上式可知

$ \begin{gather} {u_m} \to u \text{在}C([0,\infty);{L^2})\text{中}, \end{gather} $

(3.19)

$ \begin{gather} |{u_m}{|^{p - 2}}{u_m}\ln |{u_m}|\xrightarrow{a.e.}|u{|^{p - 2}}u\ln |u|,Q = [0,\infty) \times \Omega . \end{gather} $

(3.20)

因此, 令$ {\Omega _1} = \left\{ {x \in \Omega |\left| {{u_m}(x,t)} \right| \le 1} \right\},{\Omega _2} = \left\{ {x \in \Omega |\left| {{u_m}(x,t)} \right| \ge 1} \right\} $, 通过引理2.4, 引理2.7和(3.14), 得到

$ \begin{align} \displaystyle{\int}_\Omega {{{\left| {{\phi ^{(m)}}} \right|}^{{p^\prime }}}} dx &= \displaystyle{\int}_{x \in \Omega_1 } {{{\left| {{\phi ^{(m)}}} \right|}^{{p^\prime }}}} dx + \displaystyle{\int}_{x \in \Omega_2} {{{\left| {{\phi ^{(m)}}} \right|}^{{p^\prime }}}} dx \\ &\le {(e(p - 1))^{ - {p^\prime }}}|\Omega | + {(e\mu )^{ - {p^\prime }}}\displaystyle{\int}_{x \in \Omega_2} {{{\left| {{u_m}} \right|}^{(p - 1 + \mu ){p^\prime }}}} dx \\ &\le {(e(p - 1))^{ - {p^\prime }}}|\Omega | + {(e\mu )^{ - {p^\prime }}}C^{(p - 1 + \mu ){p^\prime }}\left\| {\nabla {u_m}} \right\|_2^{(p - 1 + \mu ){p^\prime }} \\ &\le {(e(p - 1))^{ - {p^\prime }}}|\Omega | + {(e\mu )^{ - {p^\prime }}}C^{(p - 1 + \mu ){p^\prime }}\left( {\frac{{2pM}}{{p - 2}}} \right)\frac{{(p - 1 + \mu ){p^\prime }}}{2}. \end{align} $

(3.21)

其中$ {\phi ^{(m)}} = |{u_m}{|^{p - 2}}{u_m}\ln |{u_m}| $，$ {p^\prime } = \frac{p}{{p - 1}} $. 由(3.20), (3.21)有

$ \begin{gather} |{u_m}{|^{p - 2}}{u_m}\ln |{u_m}| \stackrel{*}{\rightharpoonup} |u{|^{p - 2}}u\ln |u|\text{在}{L^\infty }(0,\infty;{L^{p^\prime }})\text{中} . \end{gather} $

(3.22)

令$ m \to \infty $, 根据(3.2)得到

$ \begin{equation} \begin{split} &\left( {{u_t}(t),w} \right) + \displaystyle{\int\limits_0^t} {\left( {|\nabla u(t){|^{p - 2}}\nabla u(t),\nabla w} \right)} dt + \left( {\nabla u,\nabla w} \right)\\ =& \displaystyle{\int\limits_0^t} {\left( {|u{|^{p - 2}}u\ln |u|,w} \right)} dt + \left( {{u_1},w} \right) + \left( {\nabla {u_0}, \nabla w} \right),\forall w \in W_0^{1,p}. \end{split}\nonumber \end{equation} $

且由(3.3)和(3.4)知道$ u $满足初值条件. 因此$ u $是问题(1.1)的全局弱解.

接下来, 证明解的唯一性. 假设$ u,v $是问题(1.1)的两个解, 令$ W(t) = {u_t}(t) -{v_t}(t) $, 且$ u(0) = v(0),{u_t}(0) = {v_t}(0) $, $ z = u - v $, 则由问题(1.1)可得

$ \begin{gather} {W_t} + {\Delta _p}u - {\Delta _p}v - \Delta W = f(u) - f(v). \end{gather} $

(3.23)

其中$ f(u) = {\left| u \right|^{p - 2}}u\ln \left| u \right|,u \in R $. 接下来, 将方程(3.23)与$ W $做内积, 并关于$ t $积分, 整理化简得到

$ \begin{align*} &{\left\| {W(t)} \right\|^2} + 2\displaystyle{\int\limits_0^t} {{{\left\| {\nabla W(s)} \right\|}^2}ds} + 2\displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {({{\left| {\nabla u} \right|}^{p - 2}}\nabla u - {{\left| {\nabla v} \right|}^{p - 2}}\nabla v)\nabla W} dxd\tau } \notag\\ =& \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {({{\left| u \right|}^{p - 2}}u\ln \left| u \right| - {{\left| v \right|}^{p - 2}}v\ln \left| v \right|)W} dxd\tau }\notag \\ =& \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {((p - 1)({{\left| {u + \theta v} \right|}^{p - 2}}\ln \left| {u + \theta v} \right|) + {{\left| {u + \theta v} \right|}^{p - 2}})zW} dxd\tau } \notag\\ \end{align*} $

$ \begin{align} =& \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {(p - 1)({{\left| {u + \theta v} \right|}^{p - 2}}\ln \left| {u + \theta v} \right|)zW} dxd\tau } + \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {({{\left| {u + \theta v} \right|}^{p - 2}})zW} dxd\tau } \\ =& {I_1} + {I_2}. \end{align} $

(3.24)

其中$ 0 < \theta < 1 $. 由Poincaré不等式, 令$ {U_\varepsilon } = \varepsilon u + (1 - \varepsilon )v $, $ 0 \le \varepsilon \le 1 $, 可以得到

$ \begin{gather} \left| {\nabla {U_\varepsilon }(\tau )} \right| \le \left| {\nabla u(\tau )} \right| + \left| {\nabla v(\tau )} \right|, \end{gather} $

(3.25)

$ \begin{gather} \left| {\nabla z(\tau )} \right| \le \int\limits_0^\tau {\left| {\nabla W(s)} \right|} ds. \end{gather} $

(3.26)

因此利用Hölder不等式和Sobolev不等式可将(3.24)右边第一项化简为

$ \begin{align} {I_1}& = \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {(p - 1)({{\left| {u + \theta v} \right|}^{p - 2}}\ln \left| {u + \theta v} \right|)zW} dxd\tau } \\ &\le (p - 1)\displaystyle{\int\limits_0^t} {(\displaystyle{\int\limits_\Omega} {{{\left| {{{\left| {u + \theta v} \right|}^{p - 2}}\ln \left| {u + \theta v} \right|} \right|}^n})dx{)^{\frac{1}{n}}}{{\left\| z \right\|}_{\frac{{2n}}{{n - 2}}}}\left\| W \right\|} d\tau }. \end{align} $

(3.27)

且由引理2.7, 可以得到

$ \begin{equation} \begin{split} \displaystyle{\int\limits_\Omega} {{{\left| {{{\left| {u + v\theta v} \right|}^{p - 2}}\ln \left| {u + \theta v} \right|} \right|}^n})dx} \le (e{(p - 2)^{ - n}})\left| \Omega \right| + {C(e\mu )^{ - n}}{(\frac{{2pM}}{{p - 2}})^{\frac{{(\mu + p - 2)n}}{2}}}. \end{split} \end{equation} $

(3.28)

其中$ \mu > 0 $. 将(3.28)代入(3.27)可得

$ \begin{equation} \begin{split} {I_1} \le C\displaystyle{\int\limits_0^t} {{{\left\| {\nabla W(s)} v\right\|}^2}ds} . \end{split} \end{equation} $

(3.29)

同理, (3.24)右边第二项可化简为

$ \begin{align} {I_2}& = \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {({{\left| {u + \theta v} \right|}^{p - 2}})zW} dxd\tau } \le \displaystyle{\int\limits_0^t} {\left\| {u + \theta v} \right\|_{n(p - 2)}^{p - 2}{{\left\| z \right\|}_{\frac{{2n}}{{n - 2}}}}\left\| W \right\|d\tau } \\& \le C\displaystyle{\int\limits_0^t} {{{\left\| {\nabla u + \theta \nabla v} \right\|}^{p - 2}}\left\| {\nabla z} \right\|\left\| W \right\|d\tau } \le C\displaystyle{\int\limits_0^t} {{{\left\| {\nabla W(s)} \right\|}^2}ds} . \end{align} $

(3.30)

(3.24)左边第三项利用Hölder不等式可化简为

$ \begin{align} \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {\left| {({{\left| {\nabla u} \right|}^{p - 2}}\nabla u - {{\left| {\nabla v} \right|}^{p - 2}}\nabla v)\nabla W} \right|} dxd\tau } &\le \displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {\left| {\int\limits_0^1 {\frac{d}{{d\varepsilon }}({{\left| {\nabla {U_\varepsilon }} \right|}^{p - 2}}\nabla {U_\varepsilon })} d\varepsilon} \right|\left| {\nabla W} \right|} dxd\tau } \\ &\le (p - 1)\displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {\int\limits_0^1 {{{\left| {\nabla {U_\varepsilon }} \right|}^{p - 2}}{\nabla z(\tau )}} \left| {\nabla W} \right|} d\varepsilon dxd\tau } = {I_3}. \end{align} $

(3.31)

因此, 可得

$ \begin{align} &{I_3} \le C\displaystyle{\int\limits_0^t} {\displaystyle{\int\limits_\Omega} {\int\limits_0^\tau {({{\left| {\nabla u(\tau )} \right|}^{p - 2}} + {{\left| {\nabla v(\tau )} \right|}^{p - 2}})\left| {\nabla W(\tau )} \right|} \left| {\nabla W(s)} \right|} dxdsd\tau } \\ &\le C\displaystyle{\int\limits_0^t} {\int\limits_0^\tau {({{\left\| {\nabla u(\tau )} \right\|}_p^{p - 2}} + {{\left\| {\nabla v(\tau )} \right\|}_p^{p - 2}})\left\| {\nabla W(s)} \right\|} \left\| {\nabla W(\tau )} \right\|dxdsd\tau } \\ &\le C\displaystyle{\int\limits_0^t} {\int\limits_0^\tau {\left\| {\nabla W(s)} \right\|} \left\| {\nabla W(\tau )} \right\|dsd\tau } \le C{(\displaystyle{\int\limits_0^t} {\left\| {\nabla W(s)} \right\|ds} )^2} \\&\le Ct\displaystyle{\int\limits_0^t} {{{\left\| {\nabla W(s)} \right\|}^2}ds} . \end{align} $

(3.32)

由(3.29), (3.30)和(3.32)可得

$ \begin{equation} \begin{split} {\left\| {W(t)} \right\|^2} + 2\displaystyle{\int\limits_0^t} {{{\left\| {\nabla W(s)} \right\|}^2}ds} \le {C_t}\displaystyle{\int\limits_0^t} {({{\left\| {\nabla W(s)} \right\|}^2})ds} . \end{split} \end{equation} $

(3.33)

因此, 由(2.5)可知, $ \exists {T_1} > 0 $, 使得

$ \begin{equation} \begin{split} W(t) = 0,\quad 0 \le t \le {T_1}. \end{split} \end{equation} $

(3.34)

即$ u = v $, 则问题(1.1)存在唯一的整体弱解.

在本节中, 通过给定一些适当的条件, 建立了问题(1.1)的能量的多项式衰减估计.

证明定理1.2     根据方程(1.1), 可以得到

$ \begin{equation} \begin{split} \frac{d}{{dt}}E(t) + {\left\| {\nabla {u_t}} \right\|^2} = 0,\quad\forall t \ge 0. \end{split} \end{equation} $

(4.1)

即$ E(t) $在$ [0,\infty ) $上是一个非递增函数. 假设$ q = (p - 2)/p > 0 $, 将方程(1.1)两端与$ {E^q}(t)u(t) $做内积, 得到

$ \begin{align} \displaystyle{\int}_S^T {{E^q}} (t)\displaystyle{\int}_\Omega u \left( {{u_{tt}} - {\Delta _p}u - \Delta {u_t} - {{\left| u \right|}^{p - 2}}u\ln \left| u \right|} \right)dxdt = 0, \quad \forall T>S \geq 0. \end{align} $

(4.2)

将等式(4.2)中的每一项分别展开, 有

$ \begin{align*} &\displaystyle{\int}_S^T {{E^q}} (t)\left( {u,{u_{tt}}} \right)dt= \left. {{E^q}(t)\left( {u,{u_t}} \right)} \right|_S^T \notag \\ &- \displaystyle{\int}_S^T {\left( {q{E^{q - 1}}(t){E^\prime }(t)\left( {u,{u_t}} \right) + {E^q}(t){{\left\| {{u_t}(t)} \right\|}^2}} \right)} dt,\notag \\ &- \displaystyle{\int}_S^T {{E^q}} (t)\left( {u,{\Delta _p}u} \right)dt = \displaystyle{\int}_S^T {{E^q}} (t)\left\| {\nabla u(t)} \right\|_p^pdt ,\notag\\ \end{align*} $

$ \begin{align} &- \displaystyle{\int}_S^T {{E^q}} (t)\left( {u,\Delta {u_t}} \right)dt = \displaystyle{\int}_S^T {{E^q}} (t)\left( {\nabla u,\nabla {u_t}} \right)dt, \\ &- \displaystyle{\int}_S^T {{E^q}} (t)\left( {u,{{\left| u \right|}^{p - 2}}u\ln \left| u \right|} \right)dt = \displaystyle{\int}_S^T {{E^q}} (t)\displaystyle{\int\limits_\Omega} {{{\left| u \right|}^p}\ln \left| u \right|} dt . \end{align} $

(4.3)

由(4.2)可得

$ \begin{align} p\displaystyle{\int}_S^T {{E^{q + 1}}(t)} dt =& \left. { - {E^q}(t)\left( {u,{u_t}} \right)} \right|_S^T + q\displaystyle{\int}_S^T {{E^{q - 1}}(t){E^\prime }(t)\left( {u,{u_t}} \right)} dt\\ &+ (1 + \frac{p}{2})\displaystyle{\int}_S^T {{E^q}(t){{\left\| {{u_t}(t)} \right\|}^2}} dt - \displaystyle{\int}_S^T {{E^q}} (t)\left( {\nabla u,\nabla {u_t}} \right)dt \\ &+ \frac{1}{p}\displaystyle{\int}_S^T {{E^q}} (t)\left\| {u(t)} \right\|_p^pdt = {G_1} + {G_2} + {G_3} + {G_4} + {G_5} . \end{align} $

(4.4)

根据引理2.4, (1.4)和(4.1),

$ \begin{equation} \begin{split} \left\| {\nabla {u_t}(t)} \right\| \le {( - {E^\prime }(t))^{1/2}},{\left\| {\nabla u(t)} \right\|_p} \le p{E^{1/p}}(t),\quad\forall t \ge 0. \end{split} \end{equation} $

(4.5)

$ \begin{equation} \begin{split} \left| {{E^q}(t)\left( {u,{u_t}} \right)} \right| \le {E^q}(t)\left\| {u(t)} \right\|\left\| {{u_t}(t)} \right\| \le C{E^q}(t){\left\| {\nabla u(t)} \right\|_p}\left\| {{u_t}(t)} \right\| \le C{E^{q + 1 + 1/p}}(t). \end{split} \end{equation} $

(4.6)

因为$ E(t) $是一个非增长函数, 则$ {G_1} $可化简为

$ \begin{equation} \begin{split} \left| {{G_1}} \right| \le C{E^{q + 1 + 1/p}}(S),\quad\forall T \ge S \ge 0. \end{split} \end{equation} $

(4.7)

同理, 可把$ {G_2} $, $ {G_3} $, $ {G_4} $, $ {G_5} $化简,

$ \begin{equation} \begin{split} &\left| {{G_2}} \right| \\ \le& C\displaystyle{\int}_S^T {{E^{q - 1}}} (t)\left| {{E^\prime }(t)} \right|\left\| {u(t)} \right\|\left\| {{u_t}(t)} \right\|dt \le C\displaystyle{\int}_S^T {{E^{q + 1/p}}} (t)\left| {{E^\prime }(t)} \right|dt \le C{E^{q + 1 + 1/p}}(S). \end{split} \end{equation} $

(4.8)

$ \begin{equation} \begin{split} \left| {{G_3}} \right| \le C\displaystyle{\int}_S^T {{E^q}(t){{\left\| {\nabla {u_t}(t)} \right\|}^2}} dt = C\displaystyle{\int}_S^T {{E^q}(t)( - {E^\prime }(t))} dt \le C{E^{q + 1}}(S). \end{split} \end{equation} $

(4.9)

$ \begin{align} \left| {{G_4}} \right| &\le C\displaystyle{\int}_S^T {{E^q}} (t)\left\| {\nabla u(t)} \right\|\left\| {\nabla {u_t}(t)} \right\|dt \le C\displaystyle{\int}_S^T {{E^{q + 1/p}}} (t){( - {E^\prime }(t))^{1/2}}dt \\ &\le \displaystyle{\int}_S^T {{E^{q + 1}}} (t)dt + C\displaystyle{\int}_S^T {{E^{q + 2/p - 1}}(t)( - {E^\prime }(t))} dt \le \displaystyle{\int}_S^T {{E^{q + 1}}} (t)dt + C{E^{q + 2/p}}(S). \end{align} $

(4.10)

$ \begin{equation} \begin{split} \left| {{G_5}} \right| \le C\displaystyle{\int}_S^T {{E^q}} (t)\left\| {\nabla u(t)} \right\|_p^pdt \le C{E^{q + 1}}(S). \end{split} \end{equation} $

(4.11)

于是由$ (4.4) $和$ {G_2} $, $ {G_3} $, $ {G_4} $, $ {G_5} $的估计, 可得

$ \begin{equation} \begin{split} \displaystyle{\int}_S^T {{E^{q + 1}}} (t)dt &\le C({E^{q + 1 + 1/p}}(S) + {E^{q + 1}}(S) + {E^{q + 2/p}}(S))\\ &\le CE(S)({E^{q + 1/p}}(S) + {E^q}(S) + {E^{q + 2/p - 1}}(S))\\ &\le CE(S){E^q}(0)({E^{1/p}}(0) + 1 + {E^{2/p - 1}}(0)). \end{split} \end{equation} $

(4.12)

令$ T \to \infty $，对$ \forall T \ge S \ge 0 $, 可以得到

$ \begin{equation} \begin{split} \displaystyle{\int}_S^\infty {{E^{q + 1}}} (t)dt \le {\gamma ^{ v- 1}}E(S){E^q}(0),\quad\forall S \ge 0. \end{split} \end{equation} $

(4.13)

因此, 根据引理2.6, 可得

$ \begin{equation} \begin{split} E(t) \le E(0){\left( {\frac{{1 + q}}{{1 + q\gamma vt}}} \right)^{1/q}} \le CE(0){(1 + t)^{ - p/(p - 2)}}. \end{split} \end{equation} $

(4.14)

于是问题(1.1)的能量是以多项式衰减的.

在本节中, 我们考虑了$ ({u_0},{u_1}) \in V $的弱解在有限时间内爆破的结果, 并给出了问题(1.1)爆破时间的上限.

引理5.11    假设$ ({u_0},{u_1}) \in V $，$ E(0) < d $, 则对$ \forall t \in [0,T) $，$ (u,{u_t}) \in V $, 且满足

$ \begin{equation} \begin{split} \left\| u \right\|_p^p \ge {p^2}d,\quad \forall t \in [0,T). \end{split} \end{equation} $

(5.1)

证     假设在$ t = {t_0} $时, $ (u,{u_t}) \notin V $. 即存在一列$ \{ {t_n}\} $, 当$ {t_n} \to t_0^ - $, 有$ I(u({t_n})) < 0 $, $ E({t_n}) < d $. 利用$ {\left\| \cdot \right\|_{W_0^{1,p}}} $的弱下半连续性, 可得

$ \begin{equation} \begin{split} I(u({t_0})) < \mathop {\lim }\limits_{n \to \infty } \inf I(u({t_n})) < 0,\quad E({t_0}) < \mathop {\lim }\limits_{n \to \infty } \inf E({t_n}) < d. \end{split} \end{equation} $

(5.2)

由于$ (u({t_0}),{u_t}({t_0})) \notin V $, 则有$ I(u({t_0})) = 0 $或者$ E(u({t_0})) > d $. 但$ E(u({t_0})) > d $与(5.2)矛盾; 且若$ I(u({t_0})) = 0 $, 根据$ d $的定义, 得到$ E(u({t_0})) > d $, 与(5.2)矛盾. 即$ (u,{u_t}) \in V $. 因此, 根据引理2.9和$ (u,{u_t}) \in V $,

$ \begin{equation} \begin{split} d \le J({\lambda ^ * }u) < J(u) < \frac{1}{{{p^2}}}\left\| u \right\|_p^p. \end{split} \end{equation} $

(5.3)

证明定理1.3    这个定理将利用$ Levine $的凹性论证理论来证明. 假设$ u $是问题(1.1)的全局弱解, 则$ {T^ * } = \infty $. 对任一$ T > 0 $, $ b > 0 $, $ \eta > 0 $, 定义$ G(t) $

$ \begin{equation} \begin{split} G(t) = {\left\| u \right\|^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla u} \right\|}^2}} d\tau + (T - t){\left\| {\nabla {u_0}} \right\|^2} + b{\left( {t + \eta } \right)^2}. \end{split} \end{equation} $

(5.4)

首先, $ G(t) $关于$ t $求一阶导, 有

$ \begin{equation} \begin{split} {G^\prime }(t) = 2\displaystyle{\int}_\Omega u {u_t}dx + 2\displaystyle{\int}_0^t {\displaystyle{\int}_\Omega \nabla } u \cdot \nabla {u_t}dxd\tau + 2b\left( {t + \eta } \right) . \end{split} \end{equation} $

(5.5)

$ G(t) $关于$ t $求二阶导, 有

$ \begin{align*} {G^{\prime \prime }}(t) =& 2{\left\| {{u_t}} \right\|^2} + 2\displaystyle{\int}_\Omega {{u_{tt}}} udx - 2\displaystyle{\int}_\Omega u \Delta {u_t}dx + 2b \notag \\ \end{align*} $

$ \begin{align} =& 2{\left\| {{u_t}} \right\|^2} + 2\displaystyle{\int}_\Omega u \left[ { - {\Delta _p}u + |u{|^{p - 2}}u\ln |u|} \right]dx + 2b \\ =& 2{\left\| {{u_t}} \right\|^2} - 2\left\| {\nabla u} \right\|_p^p + 2\displaystyle{\int}_\Omega | u{|^p}\ln |u|dx + 2b \\ =& 2{\left\| {{u_t}} \right\|^2} - 2I(u) + 2b. \end{align} $

(5.6)

化简$ {G^{\prime \prime }}(t) $可得

$ \begin{equation} \begin{split} {G^{\prime \prime }}(t) =& 2{\left\| {{u_t}} \right\|^2} - (2pE(t) - p{\left\| {{u_t}} \right\|^2} - \frac{2}{p}\left\| u \right\|_p^p) + 2b\\ \ge& (p + 2){\left\| {{u_t}} \right\|^2} - 2pE(0) + \frac{2}{p}\left\| u \right\|_p^p + 2p\displaystyle{\int\limits_0^t} {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau + 2b\\ \ge& (p + 2){\left\| {{u_t}} \right\|^2} + 2p(d - E(0)) + 2p\displaystyle{\int\limits_0^t} {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau + 2b. \end{split} \end{equation} $

(5.7)

由$ b $的定义, $ p>2 $, 可得

$ \begin{equation} \begin{split} {G^{\prime \prime }}(t) \ge (p + 2)\left[ {\left\| {{u_t}} \right\|_2^2 + \displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau + b} \right]. \end{split} \end{equation} $

(5.8)

根据(5.5), 有

$ \begin{align} {\left( {{G^\prime }(t)} \right)^2} =& 4{\left[ {\left( {u,{u_t}} \right) + \displaystyle{\int}_0^t {\displaystyle{\int}_\Omega \nabla } u \cdot \nabla {u_t}dxd\tau + b\left( {t + \eta } \right)} \right]^2} \\ =& 4{\left( {u,{u_t}} \right)^2} + 8\left( {u,{u_t}} \right)\displaystyle{\int}_0^t {\displaystyle{\int}_\Omega \nabla } u \cdot \nabla {u_t}dxd\tau + 8b\left( {u,{u_t}} \right)(t + \eta ) \\ &+ 4{\left( {\displaystyle{\int}_0^t {\displaystyle{\int}_\Omega \nabla } u \cdot \nabla {u_t}dxd\tau } \right)^2} + 8b(t + \eta )\displaystyle{\int}_0^t {\displaystyle{\int}_\Omega \nabla } u \cdot \nabla {u_t}dxd\tau + 4{b^2}{(t + \eta )^2}. \end{align} $

(5.9)

由Cauchy-Schwarz不等式和Young不等式, 得到$ {\left( {{G^\prime }(t)} \right)^2} $的估计为

$ \begin{align} &{\left( {{G^\prime }(t)} \right)^2}\\ \le& 4({\left\| u \right\|^2}{\left\| {{u_t}} \right\|^2} + 2\left\| u \right\|\left\| {{u_t}} \right\|{\left( {\displaystyle{\int}_0^t {{{\left\| {\nabla u} \right\|}^2}} \;{\rm{d}}\tau } \right)^{\frac{1}{2}}}{\left( {\displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau } \right)^{\frac{1}{2}}} + 2b(t + \eta )\left\| u \right\|\left\| {{u_t}} \right\| \\ &+ \displaystyle{\int}_0^t {{{\left\| {\nabla u} \right\|}^2}} \;{\rm{d}}\tau \displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau + 2b(t + \eta ){\left( {\displaystyle{\int}_0^t {{{\left\| {\nabla u} \right\|}^2}} \;{\rm{d}}\tau } \right)^{\frac{1}{2}}}{\left( {\displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau } \right)^{\frac{1}{2}}} + {b^2}{(t + \eta )^2}) \\ \le& 4({\left\| u \right\|^2}\left( {{{\left\| {{u_t}} \right\|}^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau } \right) + \displaystyle{\int}_0^t {{{\left\| {\nabla u} \right\|}^2}} \;{\rm{d}}\tau \left( {{{\left\| {{u_t}} \right\|}^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau } \right) \\ &+ b\left( {{{\left\| u \right\|}^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla u} \right\|}^2}} \;{\rm{d}}\tau } \right) + b{(t + \eta )^2}\left( {{{\left\| {{u_t}} \right\|}^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau } \right) + {b^2}{(t + \eta )^2}) \\ =& 4\left( {{{\left\| u \right\|}^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla u} \right\|}^2}} \;{\rm{d}}\tau + b{{(t + \eta )}^2}} \right)\left( {{{\left\| {{u_t}} \right\|}^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau + b} \right). \end{align} $

(5.10)

根据(5.4), (5.8)和(5.10), 计算化简得到

$ \begin{eqnarray} \begin{aligned} G(t){G^{\prime \prime }}(t) - \frac{{p + 2}}{4}{\left( {{G^\prime }(t)} \right)^2}\ge& (p + 2)(T - t){\left\| {\nabla {u_0}} \right\|^2}\left[ {{{\left\| {{u_t}} \right\|}^2} + \displaystyle{\int}_0^t {{{\left\| {\nabla {u_\tau }} \right\|}^2}} \;{\rm{d}}\tau + b} \right] \ge 0,\\ & t \in [0,T]. \end{aligned} \end{eqnarray} $

(5.11)

由$ \eta $的定义, 则有

$ \begin{gather} G(0) = \left\| {{u_0}} \right\|_2^2 + T{\left\| {\nabla {u_0}} \right\|^2} + b{\eta ^2} > 0, \end{gather} $

(5.12)

$ \begin{gather} {G^\prime }(0) = 2\left( {{u_0},{u_1}} \right) + 2b\eta > 0 . \end{gather} $

(5.13)

因此, 根据引理2.8, $ \exists {T^ * } > 0 $, 满足

$ \begin{equation} \begin{split} {T^*} \le \frac{{4G(0)}}{{(p - 2){G^\prime }(0)}} = \frac{{2b{\eta ^2} + 2{{\left\| {{u_0}} \right\|}^2} + 2T{{\left\| {\nabla {u_0}} \right\|}^2}}}{{(p - 2)(b\eta + ({u_0},{u_1}))}}. \end{split} \end{equation} $

(5.14)

即有

$ \begin{equation} \begin{split} T^{*} \leq T(\eta) = \frac{2 b {\eta ^2}+2\left\|u_{0}\right\|_{2}^{2}}{(p-2) b\eta+(p-2) \int_{\Omega} u_{0} u_{1} d x-2\left\|\nabla u_{0}\right\|_{2}^{2}}. \end{split} \end{equation} $

(5.15)

且$ \mathop {\lim }\limits_{t \to {T^{ * - }}} G(t) = \infty $, 与假设矛盾. 故定理1.3得证.