本文研究以下具有多项式非线性项和粘性项的非线性抛物方程的初边值问题

$ \begin{equation} \begin{cases} u_{t}-{\Delta}u+{ \int_{0}^{t}{g(t-s){\Delta}u(x, s)ds}} = \vert u\vert^{p-2}u, &\text{在}{\;}{\Omega}{\times}(0, T)\text{中}, \\ u(x, t) = 0, &\text{在}{\;}{\partial}{\Omega}{\times}(0, T)\text{上}, \\ u(x, 0) = u_{0}(x), &\text{在}{\;}{\Omega}\text{中}, \\ \end{cases} \end{equation} $

(1.1)

其中$ g:\mathbb{R}^{+}\to {\mathbb{R}}^+ $是有界的$ C^1 $函数, $ u_{0}{\in}H_0^1({\Omega}), \ T{\in}(0, +{\infty}], \ {\Omega}{\subset}{\mathbb{R}}^n(n{\geq}1) $是具有光滑边界$ {\partial}{\Omega} $的有界区域.

材料学和力学中对粘弹性的应用非常广泛, 其中粘弹性依赖于温度和外力作用的时间. 所谓的粘弹性是指任何物体对其本身的形变历史都具有一个长期的记忆^[1]. 特别的, 当研究具有记忆材料的热传导时, 经典的Fourier定律被以下形式所取代

$ \begin{equation*} \label{1.6} q = -d\nabla u-\int_{-\infty}^{t}\nabla [g(x, t)u(x, t)]ds, \end{equation*} $

其中$ q $与单位长度的温差成正比, $ u $是温度, $ d $是热传导系数, 积分项表示材料中的记忆效应, $ q $不是线性地依赖于$ u $. 对于这类问题的研究引起了广泛的关注, 本文将研究该类问题$ (1.1) $的解的存在性和爆破性. 当方程$ (1.1) $中的$ g = 0 $时, 方程如下

$ u_{t}-{\Delta}u = |u|^{p-2}u, $

可以描述诸多化学反应、热传导过程和种群动力学. 针对这类问题在解的存在性方面可以参考文献[2-5]. 关于解的爆破性方面也得到了广泛的研究, 可参考文献[6-12].

当方程$ (1.1) $中的$ g\neq 0 $时, 可以描述电流变流体的扩散和记忆材料中的热传导(见文献[13, 14]). 由于粘弹性项的影响, 问题变得更加复杂, 为了解决相关问题, 需要对$ g $进行精确的估计和细致的观察. 近年来, 也有不少相关的研究. 在$ g $和$ p $满足合适的条件下, Messaoudi^{[15, 16]}给出了方程的强解在有限时刻内爆破的证明. 之后, Tian^[17]在此基础上建立了一个新的爆破准则, 利用微分不等式技巧给出了解的爆破时间的上界. Di等^[18]和Sun等^[19]研究了含有$ {\Delta}u_{t} $项的非线性拟抛物方程解的存在性和爆破性问题. 还有一些相关的研究, 可以参考文献[20-22]. 本文假设以下条件

条件A   $ g:\mathbb{R}^{+}\to {\mathbb{R}}^+ $是有界的$ C^1 $函数, 满足

$ \begin{equation*} g(s)\geq 0, \ g'(s)\leq 0, \ 1-\int_0^{+\infty}{g(s)ds} = l>0. \end{equation*} $

条件B   常数$ p $满足

$ \begin{equation*} \text{当}\ n = 1, 2\ \text{时}, p>2; \text{当}\ n\geq 3\ \text{时}, 2<p<\frac{2n}{n-1}. \end{equation*} $

定义1.1   (弱解)对任意$ v\in H_0^1(\Omega), t\in (0, T) $, 函数$ u(x, t) $满足$ u(x, t)\in L^{\infty}(0, +\infty; H_0^1(\Omega)) $, $ u_{t}(x, t)\in L^{2}(0, +\infty; H_0^1(\Omega)) $及以下方程

$ \begin{equation} \big(u_t(\cdot, t), v\big)_2+\big({\nabla{u}}(\cdot, t), {\nabla{v}}\big)_2-\big({\int_{0}^{t}{g(t-s){\nabla}u(x, s)ds}}, \nabla{v}\big)_2 = \big(\vert u(\cdot, t)\vert^{p-2}u(\cdot, t), v\big)_2, \end{equation} $

(1.2)

则称$ u = u(x, t) $是方程$ (1.1) $的弱解, 其中$ u(x, 0) = u_0(x)\in H_0^1(\Omega) $, $ (\cdot, \cdot)_2 $表示内积$ (\cdot, \cdot)_{L^2(\Omega)} $.

定义1.2   (最大存在时间)设$ u(x, t) $为方程$ (1.1) $的弱解. 我们定义$ u(x, t) $的最大存在时间$ T $如下:

$ (1) $若对任意$ 0\leq t<+\infty, u(x, t) $都存在, 则$ T = +\infty $.

$ (2) $若存在$ t_0\in(0, +\infty) $使得当$ 0\leq t<t_0 $时, $ u(x, t) $存在, 但在$ t = t_0 $处$ u(x, t) $不存在, 则$ T = t_0 $.

定义1.3   (有限时刻爆破)设$ u(x, t) $为方程$ (1.1) $的弱解. 如果$ u(x, t) $的最大存在时间$ T $是有限的, 且满足

$ \lim\limits_{t\rightarrow T^-} \Vert u(\cdot, t)\Vert_{L^2(\Omega)} = +\infty, $

则称$ u(x, t) $在有限时刻爆破.

为了说明本文的主要结果, 需要引入以下修正的能量泛函

$ \begin{equation} \begin{aligned} J(u) = &\frac{1}{2}\int_0^t{g(t-s)\Vert{\nabla}u(\cdot, t)-\nabla u(\cdot, s)\Vert _{L^2(\Omega)}^2ds}+\frac{1}{2}\big(1-\int_0^t{g(s)ds}\big)\Vert\nabla u(\cdot, t)\Vert_{L^2(\Omega)}^2\\ &-\frac{1}{p}\Vert u(x, t)\Vert_{L^{p}(\Omega)}^{p}, \end{aligned} \end{equation} $

(1.3)

$ \begin{equation} \begin{split} I(u) = &\int_0^t{g(t-s)\Vert{\nabla}u(\cdot, t)-\nabla u(\cdot, s)\Vert _{L^2(\Omega)}^2ds}+\big(1-\int_0^t{g(s)ds}\big)\Vert\nabla u(\cdot, t)\Vert_{L^2(\Omega)}^2\\ &-\Vert u(x, t)\Vert_{L^{p}(\Omega)}^{p}. \end{split} \end{equation} $

(1.4)

则有

$ \begin{equation} J(u) = \frac{1}{2}I(u)+\frac{p-2}{2p}\Vert u(x, t)\Vert_{L^{p}(\Omega)}^{p}. \end{equation} $

(1.5)

本文的主要结果如下.

定理1.1   若条件A、B成立, $ u_0\in H_0^1(\Omega), J(u_0)<d $且$ I(u_0)\geq 0 $. 则方程$ (1.1) $存在全局弱解$ u = u(x, t)\in L^{\infty}(0, +\infty;H_0^1(\Omega)), u_{t}(x, t )\in L^{2}(0, +\infty;H_0^1(\Omega)) $.

定理1.2   若条件A、B成立, $ u_0\in H_0^1(\Omega), J(u_0)<E_1, \int_0^{+\infty}{g(s)ds}<\frac{\varepsilon_0}{2p+\varepsilon_0}. $则方程$ (1.1) $的弱解将在有限时刻爆破, 并且有

$ t^\nu\leq K^\nu+A_0. $

这里$ E_1>0 $见$ (3.3) $式, $ 0<\varepsilon_0<p-2, $常数$ K $满足$ {\rm{supp}}(u_0)\subset B_{K}(0), $常数$ 0<\nu<1, $ $ A_0>0 $见$ (3.14) $式.

注1.1   在$ g $和$ p $具有相似但不同的限制条件下, Messaoudi^{[15, 16]}证明了具有正、负或消失的初始能量时(初始能量都满足$ J(u_0)<E_0, E_0>0 $) 该方程的解在有限时刻爆破. 在定理1.2中, 我们将这一能量范围进行了扩大, 得到一个新的更大的能量范围$ E_1, $即$ E_0<E_1, $具体证明见$ (3.5) $式.

注1.2   在定理1.2的条件下, 可以得出$ I(u_0)<0 $, 与定理1.1不存在冲突.

本文结构如下: 在第2节中, 利用位势井理论和Galerkin方法给出定理1.1的证明. 在第3节中, 给出相关引理和定理1.2的证明.

为了证明方程解的存在性, 需要定义

$ \begin{equation*} \mathcal{N} = \{u\in H_0^1(\Omega)\vert \;I(u) = 0, \int_{\Omega}\vert \nabla u\vert^2dx\neq 0\}, \end{equation*} $

$ \begin{equation*} d = \inf\limits_{u\in\mathcal{N}}J(u), \end{equation*} $

显然有$ d>0 $. 从而, 定义

$ \begin{equation*} \begin{aligned} W& = \{u\in H_0^1(\Omega)\vert \;I(u)>0, J(u)<d\}\cup\{0\}, \;\\ V& = \{u\in H_0^1(\Omega)\vert \;I(u)<0, J(u)<d\}. \end{aligned} \end{equation*} $

接下来给出能量恒等式.

引理2.1  

$ \begin{equation*} \begin{aligned} J(u_0) = &J(u(t))+\int_0^t\Vert u_\tau \Vert_{L^2(\Omega)}^2d\tau+\frac{1}{2}\int_0^t g(\tau)\Vert \nabla u(\tau) \Vert_{L^2(\Omega)}^2d\tau\\ &-\frac{1}{2}\int_0^t \int_0^{\tau} g'(\tau-s)\Vert \nabla u(\tau)-\nabla u(s) \Vert_{L^2(\Omega)}^2 dsd\tau. \end{aligned} \end{equation*} $

这里$ J(u(t)) = J(u), $见$ (1.3) $式. 事实上, 由于$ u = u(x, t) $是$ x, t $的函数, 由$ (1.3) $式可知, $ J(u) $是$ t $的函数, 为了突出这一点, 故用记号$ J(u(t)). $后文会用到类似的其他记号.

证明   结合(1.3)式, 定义

$ \begin{equation} \begin{aligned} E(t) = J(u) = &\frac{1}{2}\int_0^t{g(t-s)\Vert{\nabla}u(t)-\nabla u(s)\Vert _{L^2(\Omega)}^2ds}+\frac{1}{2}\big(1-\int_0^t{g(s)ds}\big)\Vert\nabla u(t)\Vert_{L^2(\Omega)}^2\\ &-\frac{1}{p}\Vert u(t)\Vert_{L^{p}(\Omega)}^{p}. \end{aligned} \end{equation} $

(2.1)

利用(1.2)式, 可得

$ \begin{equation*} \begin{aligned} \frac{d}{dt}E(t) = &\frac{1}{2}\int_0^t{g'(t-s)\Vert{\nabla}u(t)-\nabla u(s)\Vert _{L^2(\Omega)}^2ds}+\int_0^{t}g(t-s)\int_{\Omega}\nabla u_{t}(t)[\nabla u(t)-\nabla u(s)]dxds\\ &-\frac{1}{2}g(t)\Vert \nabla u(t)\Vert_{L^2(\Omega)}^2-\int_0^{t}g(t-s)\int_{\Omega}\nabla u_{t}(t)\nabla u(t)dxds\\ &+\int_{\Omega}\nabla u_{t}(t)\cdot\nabla u(t)dx-\int_{\Omega}u_{t}(t)u^{p-1}(t)dx\\ = &\frac{1}{2}\int_0^t{g'(t-s)\Vert{\nabla}u(t)-\nabla u(s)\Vert _{L^2(\Omega)}^2ds}-\frac{1}{2}g(t)\Vert \nabla u(t)\Vert_{L^2(\Omega)}^2+\int_{\Omega}\nabla u_{t}(t)\nabla u(t)dx\\ &-\int_0^{t}g(t-s)\int_{\Omega}\nabla u_{t}(t)\nabla u(s)dxds-\int_{\Omega}u_{t}(t)u^{p-1}(t) dx\\ = &\frac{1}{2}\int_0^t{g'(t-s)\Vert{\nabla}u(t)-\nabla u(s)\Vert _{L^2(\Omega)}^2ds}-\frac{1}{2}g(t)\Vert \nabla u(t)\Vert_{L^2(\Omega)}^2-\int_{\Omega}u_{t}^2(t)dx. \end{aligned} \end{equation*} $

上式两端在$ [0, t] $上进行积分, 可得

$ \begin{equation*} \begin{aligned} J(u(t))-J(u_0) = &\frac{1}{2}\int_0^t \int_0^{\tau} g'(\tau-s)\Vert \nabla u(\tau)-\nabla u(s) \Vert_{L^2(\Omega)}^2 dsd\tau-\frac{1}{2}\int_0^t g(\tau)\Vert \nabla u(\tau) \Vert_{L^2(\Omega)}^2d\tau\\ &-\int_0^t\Vert u_\tau(\tau) \Vert_{L^2(\Omega)}^2d\tau, \end{aligned} \end{equation*} $

从而引理2.1得证.

现在, 给出方程解的存在性的证明.

证明定理1.1   在$ J(u_0)<d $, $ I(u_0)\geq 0 $时, 不难得到$ J(u_0)\geq 0 $. 考虑若$ J(u_0) = 0 $且$ I(u_0)\geq 0, $则$ u_0 = 0 $, 这是一个平凡解. 若$ J(u_0)>0 $, $ I(u_0) = 0 $, 则有$ \int_\Omega |\nabla u_0|^2dx\neq0 $, 进而$ J(u_0)\geq d, $这与$ J(u_0)<d $相矛盾. 所以我们只需考虑$ 0<J(u_0)<d $, $ I(u_0)>0 $的情况.

下面利用Galerkin方法构造问题(1.1) 的近似解$ u_m(x, t) $. 我们选择$ \{\omega_j(x)\} $作为$ H_0^1(\Omega) $上的一组正交基, 令

$ \begin{equation*} u_m(x, t) = \sum\limits_{j = 1}^{m}h^j_m(t)\omega_j(x), \ {\rm {}}\ m = 1, 2, \cdots, \end{equation*} $

且对于$ k = 1, 2, \cdots, m $, 满足

$ \begin{equation} (u_{mt}, \omega_k)_2+({\nabla{u}}_m, {\nabla{\omega}}_k)_2-\big(\int_{0}^{t}g(t-s){\nabla}u_m(x, s)ds, \nabla{\omega}_k\big)_2 = (|u_m|^{p-2}u_{m} , \omega_k)_2, \\ \end{equation} $

(2.2)

$ \begin{equation} u_m(x, 0) = \sum\limits_{j = 1}^{m}a^j_m\omega_j(x)\to u_0, \ \text{在}\ H_0^1(\Omega) \text{中}. \end{equation} $

(2.3)

同时, 当$ m $足够大, 且$ t\in [0, T) $时($ T $是$ u_m(x, t) $的最大存在时间), 能量恒等式如下

$ \begin{equation} \begin{aligned} J(u_m(0)) = &J(u_m(t))+\int_0^t\Vert u_{m\tau}\Vert_{L^2(\Omega)}^2 d\tau+\frac{1}{2}\int_0^t g(\tau)\Vert \nabla u_m(\tau)\Vert_{L^2(\Omega)}^2 d\tau\\ &-\frac{1}{2}\int_0^t \int_0^{\tau}g'(\tau-s)\Vert \nabla u_m(\tau)-\nabla u_m(s)\Vert_{L^2(\Omega)}^2 ds d\tau<d. \end{aligned} \end{equation} $

(2.4)

现在证明对于足够大的$ m $和任意的$ 0\leq t<T $, 有$ u_m(x, t)\in W $. 假设该命题不成立, 则存在$ t_0\in (0, T) $使得$ u_m(x, t_0)\in \partial W $, 从而

$ \begin{equation*} I(u_m(t_0)) = 0, \;\Vert \nabla u_m(t_0)\Vert_{L^2(\Omega)}\neq 0, \ {\rm{\text{或者}}}\ J(u_m(t_0)) = d. \end{equation*} $

结合(2.4)式, 可知$ J(u_m(t_0)) = d $不成立. 又若$ I(u_m(t_0)) = 0, \;\Vert \nabla u_m(t_0)\Vert_{L^2(\Omega)}\neq 0 $, 根据$ d $的定义, 有$ J(u_m(t_0))\geq d $, 这与$ (2.4) $式相矛盾. 从而可得, 对足够大的$ m $和任意的$ 0\leq t<T $, 有$ u_m(x, t)\in W $. 进而, 以$ u_m(x, T) $为初始能量, 重复上述讨论, 则对于足够大的$ m $和任意的$ 0\leq t<+\infty $, 都有$ u_m(x, t)\in W $.

根据

$ \begin{equation*} \begin{aligned} J(u_{m}(t)) = &\frac{1}{p}I(u_{m}(t))+\frac{p-2}{2p}\int_0^{t}g(t-s)\Vert\nabla u_{m}(t)-\nabla u_{m}(s)\Vert_{L^2(\Omega)}^2ds\\ &+(1-\int_0^{t}g(s)ds)\frac{p-2}{2p}\Vert\nabla u_{m}(t)\Vert_{L^2(\Omega)}^2<d, \end{aligned} \end{equation*} $

可知

$ \begin{equation*} l\frac{p-2}{2p}\Vert\nabla u_{m}(t)\Vert_{L^2(\Omega)}^2\leq (1-\int_0^{t}g(s)ds)\frac{p-2}{2p}\Vert\nabla u_{m}(t)\Vert_{L^2(\Omega)}^2<d, \end{equation*} $

则有

$ \begin{equation} \Vert\nabla u_{m}(t)\Vert_{L^2(\Omega)}^2\leq \frac{2p}{(p-2)l}d. \end{equation} $

(2.5)

由(2.4)式可以得到

$ \begin{equation} \int_0^t\Vert u_{m\tau}\Vert_{L^2(\Omega)}^2 d\tau <d. \end{equation} $

(2.6)

另一方面, 通过计算可得如下结果

$ \begin{equation} \begin{aligned} \int_{\Omega}(u_{m}|u_{m}|^{p-2})^{\frac{p}{p-1}}dx = \Vert u_{m}\Vert_{L^{p}(\Omega)}^{p} \leq C_{*}^{p}\Vert\nabla u_{m}\Vert_{L^{2}(\Omega)}^{p} \leq C_{*}^{p}(\frac{2pd}{(p-2)l})^{\frac{p}{2}}\leq C_{d}. \end{aligned} \end{equation} $

(2.7)

这里$ C_{*} $是Sobolev空间$ H_0^1(\Omega) \hookrightarrow L^{p}(\Omega) $的最佳嵌入常数.

这里用$ \mathop{\longrightarrow}\limits_{ }^{w^*} $表示弱星收敛. 根据(2.5)-(2.7)式, 存在一个子序列$ \{u_m\} $和$ u $使得, 当$ m\longrightarrow\infty $, 有

$ \begin{equation*} \begin{aligned} &u_m\mathop{\longrightarrow}\limits_{ }^{w^*} u \text{在} L^{\infty}(0, \infty;H_0^1(\Omega)) \text{中}\text{且} a.e. \text{在} \Omega\times [0, +\infty) \text{中}, \\ &u_{mt}\mathop{\longrightarrow}\limits_{ }^{w^*} u_t \text{在} L^2(0, \infty;H_0^1(\Omega)) \text{中}, \\ &|u_m|^{p-2}u_m\mathop{\longrightarrow}\limits_{ }^{w^*} |u|^{p-2}u \text{在} L^{\infty}(0, \infty;L^{\frac{p}{p-1}}(\Omega)) \text{中}. \end{aligned} \end{equation*} $

因此, 在(2.2) 式中, 当$ k $固定, $ m\longrightarrow +\infty $时, 有

$(u_t, \omega_k)_2+({\nabla{u}}, {\nabla{\omega}}_k)_2-(\int_{0}^{t}g(t-s){\nabla}u(x, s)ds, \nabla{\omega}_k)_2 = (|u|^{p-2}u , \omega_k)_2$,

进而对任意$ v\in H_0^1(\Omega) $, 有

$ (u_t, v)_2+({\nabla{u}}, \nabla{v})_2-(\int_{0}^{t}g(t-s){\nabla}u(x, s)ds, \nabla{v})_2 = (|u|^{p-2}u, , v)_2, \;t>0. $

同时, 从(2.3)式中, 可以得到$ u(x, 0) = u_0(x) \in H_0^1(\Omega) $. 因此, $ u $是问题(1.1)的全局弱解. 定理1.1证毕.

在证明解的爆破性之前, 先介绍一个必需的引理. 已知能量函数(见$ (2.1) $式)

$ \begin{equation} \begin{aligned} E(t)& = \frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^2(\Omega)}^2ds+\frac{1}{2}(1-\int_0^{t}g(s)ds)\Vert\nabla u(t)\Vert_{L^2(\Omega)}^2-\frac{1}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ &\geq \frac{1}{2}\big(l\Vert\nabla u(t)\Vert_{L^2(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^2(\Omega)}^2ds\big)-\frac{1}{p}B^{p}l^{\frac{p}{2}}\Vert\nabla u\Vert_{L^2(\Omega)}^{p}\\ &\geq \frac{1}{2}\big(l\Vert\nabla u(t)\Vert_{L^2(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^2(\Omega)}^2ds\big)\\ &\ \ \ \ -\frac{B^{p}}{p}\big(l\Vert\nabla u(t)\Vert_{L^2(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^2(\Omega)}^2ds\big)^{\frac{p}{2}}\\ & = \frac{1}{2}\lambda^2-\frac{B^{p}}{p}\lambda^{p}: = G(\lambda), \end{aligned} \end{equation} $

(3.1)

这里$ B = C_{*}/l $ ($ C_{*} $是Sobolev空间$ H_0^1(\Omega) \hookrightarrow L^{p}(\Omega) $的最佳嵌入常数),

$ \begin{equation*} \lambda = \lambda(t) = \big(l\Vert\nabla u(t)\Vert_{L^2(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^2(\Omega)}^2ds\big)^{\frac{1}{2}}. \end{equation*} $

则有

$ \begin{equation*} G'(\lambda) = \lambda-B^{p}\lambda^{p-1}, \end{equation*} $

且$ G(\lambda) $在$ 0<\lambda<\lambda_0 $时单调递增, 在$ \lambda>\lambda_0 $时单调递减, 在$ \lambda_0 = (B^{p})^{-\frac{1}{p-2}} $处取得最大值, 最大值为$ G_{max} = (\frac{1}{2}-\frac{1}{p})B^{\frac{-2p}{p-2}} = (1-\frac{2}{p})\omega_0 $, 其中$ \omega_0 = \frac{1}{2}\lambda_0^2 $. 令$ E_0 = (\frac{1}{2}-\frac{1}{p})B^{\frac{-2p}{p-2}} = (1-\frac{2}{p})\omega_0. $有

$ \begin{equation} \begin{aligned} \frac{1}{p}\Vert u(t)\Vert_{L^{p}(\Omega)}^{p}&\geq \frac{1}{2}\big(l\Vert\nabla u(t)\Vert_{L^2(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^2(\Omega)}^2ds\big)-E(t). \end{aligned} \end{equation} $

(3.2)

设$ u(x, t) $是问题(1.1)的弱解, 令

$ \begin{equation} \begin{aligned} \omega_1 = \inf\limits_{t\in R_0^{+}}\frac{1}{2}\lambda^2(t)\geq 0, \ \omega_2 = \inf\limits_{t\in R_0^{+}}\frac{1}{p}\Vert u(t)\Vert_{L^{p}(\Omega)}^{p}, \ {E}_1 = (\frac{p}{2}-1)\omega_2. \end{aligned} \end{equation} $

(3.3)

引理3.1   设$ u(x, t) $是方程$ (1.1) $的弱解. 若$ J(u_0)<{E}_1 $, 则对任意的$ t\in[0, T) $, 都有$ \omega_2>0, \lambda(t)>\lambda_0, E(t)<{E}_1, E_0<E_1 $.

证明   由引理2.1和$ (3.2) $式可知, 对$ \forall t\in [0, T) $

$ E(t)<J(u_0)<{E}_1, \ \frac{1}{p}\Vert u(t)\Vert_{L^{p}(\Omega)}^{p}\geq \frac{1}{2}\lambda^2(t)-J(u_0)\geq \omega_1-J(u_0), $

$ \omega_2\geq \omega_1-J(u_0)>\omega_1-(\frac{p}{2}-1)\omega_2, $

于是有

$ \begin{equation} \omega_2>\frac{2}{p}\omega_1\geq0. \end{equation} $

(3.4)

由Sobolev嵌入定理, 可得

$ \frac{B^{p}}{p}\big(l\Vert\nabla u(t)\Vert_{L^{2}(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\big)^{\frac{p}{2}}\geq\frac{1}{p}\Vert u(t)\Vert_{L^{p}(\Omega)}^{p}. $

则有

$ \begin{equation*} \begin{aligned} &(\frac{p}{2}-1)\frac{B^{p}}{p}\big(l\Vert\nabla u(t)\Vert_{L^{2}(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\big)^{\frac{p}{2}}\\ \geq&(\frac{p}{2}-1)\frac{1}{p}\Vert u(t)\Vert_{L^{p}(\Omega)}^{p}\geq(\frac{p}{2}-1)\omega_2> E(t)\\ \geq&\frac{1}{2}\big(l\Vert\nabla u(t)\Vert_{L^{2}(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\big)\\ &-\frac{B^{p}}{p}\big(l\Vert\nabla u(t)\Vert_{L^{2}(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\big)^{\frac{p}{2}}, \end{aligned} \end{equation*} $

因此, 可以得到

$ \lambda(t) = \big(l\Vert\nabla u(t)\Vert_{L^{2}(\Omega)}^2+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\big)^{\frac{1}{2}}> B^{-\frac{p}{p-2}} = \lambda_0. $

再结合$ (3.3) $式, 可以得到

$ \begin{equation} E_1 = (\frac{p}{2}-1)\omega_2>\frac{2}{p}(\frac{p}{2}-1)\inf\limits_{t\in R_0^{+}}\frac{1}{2}\lambda^2(t)>(\frac{1}{2}-\frac{1}{p})\inf\limits_{t\in R_0^{+}}\frac{1}{2}\lambda^2(t)>(\frac{1}{2}-\frac{1}{p})\lambda_0^2 = E_0. \end{equation} $

(3.5)

引理3.1得证.

接下来给出方程解的爆破性的证明及爆破时间的上界.

证明定理1.2   由$ J(u_0)<{E}_1 $, 定义$ H(t) = E_2-E(t), $其中$ E_2\in([J(u_0)]^{+}, {E}_1) $且$ E_2>0 $, 则有

$ H'(t) = -E'(t)\geq 0, \ H(t)\geq H(0) = E_2-J(u_0)>0. $

由$ E_2 $的定义, 可以得出$ {E}_1-E(t)>H(t). $又由$ -E(t)\leq\frac{1}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}, $可推断

$ H(t)<{E}_1+\frac{1}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p} = (\frac{p}{2}-1)\omega_2+\frac{1}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\leq\frac{1}{2}\Vert u\Vert_{L^{p}(\Omega)}^{p}. $

所以

$ \begin{equation} 0<H(0)<H(t)<\frac{1}{2}\Vert u\Vert_{L^{p}(\Omega)}^{p}. \end{equation} $

(3.6)

结合$ \omega_2>0, 0<E_2<{E}_1, $取足够小的$ \varepsilon_0>0 $, 使得

$ \begin{equation} 0<\varepsilon_0\omega_2\leq(p-2)\omega_2-2E_2, \end{equation} $

(3.7)

这意味着$ \varepsilon_0< p-2. $接下来, 定义加权函数

$ \begin{equation} \mathcal{K}(t) = (K+t)^{A}H^{1-\alpha}(t)+ M(t), \end{equation} $

(3.8)

其中$ M(t) = \frac{1}{2}\Vert u\Vert_{L^{2}(\Omega)}^2 $, 常数$ 0<A<1 $, 并且常数$ K $满足$ {\rm{supp}}(u_0)\subset B_{K}(0) $. 根据方程$ (1.1) $和柯西不等式可知

$ \begin{equation*} \begin{aligned} &M'(t)\\ = &\int_{\Omega}uu_{t}dx\\ = &-\Vert \nabla u\Vert_{L^{2}(\Omega)}^2+\int_{\Omega}\int_0^{t}g(t-s)\nabla u(t)\cdot\nabla u(s)dsdx+\int_{\Omega}\vert u\vert^{p-2}u^2dx\\ \geq& -\Vert \nabla u\Vert_{L^{2}(\Omega)}^2+\frac{1}{2}\int_0^{t}g(s)ds\Vert \nabla u\Vert_{L^{2}(\Omega)}^2-\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds+\Vert u\Vert_{L^{p}(\Omega)}^{p}. \end{aligned} \end{equation*} $

利用$ (3.1) $式和$ (3.7) $式, 不等式右边加上一项$ -(p-\varepsilon_0)E(t)+(p-\varepsilon_0)E(t) $, 可得

$ \begin{equation} \begin{aligned} &M'(t)\\ \geq&(p-\varepsilon_0-2)\big[\frac{1}{2}(1-\int_0^{t}g(s)ds)\Vert \nabla u\Vert_{L^{2}(\Omega)}^2+\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\big]\\ &+\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds-\frac{1}{2}\int_0^{t}g(s)ds\Vert \nabla u\Vert_{L^{2}(\Omega)}^2\\ &-(p-\varepsilon_0)E(t)+\Vert u\Vert_{L^{p}(\Omega)}^{p}-\frac{p-\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ \geq&(p-\varepsilon_0-2)\big[\frac{1}{2}(1-\int_0^{t}g(s)ds)\Vert \nabla u\Vert_{L^{2}(\Omega)}^2+\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds-E(t)\big]\\ & +\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds-\frac{1}{2}\int_0^{t}g(s)ds\Vert \nabla u\Vert_{L^{2}(\Omega)}^2-2E(t)+\frac{\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ \geq &(p-\varepsilon_0-2)\omega_2+\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds-\frac{1}{2}\int_0^{t}g(s)ds\Vert \nabla u\Vert_{L^{2}(\Omega)}^2\\ & -2E_2+2H(t)+\frac{\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ \geq&(p-2)\omega_2-(p-2)\omega_2+2E_2+\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\\ &-\frac{1}{2}\int_0^{t}g(s)ds\Vert \nabla u\Vert_{L^{2}(\Omega)}^2-2E_2+2H(t)+\frac{\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ = &\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds-\frac{1}{2}\int_0^{t}g(s)ds\Vert \nabla u\Vert_{L^{2}(\Omega)}^2+2H(t)+\frac{\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}. \end{aligned} \end{equation} $

(3.9)

通过$ (3.8) $式和$ (3.9) $式, 可以得到

$ \begin{equation*} \begin{aligned} \mathcal{K}'(t)& = (1-\alpha)(K+t)^{A}H^{-\alpha}(t)H'(t)+ M'(t)\\ &\geq (1-\alpha)(K+t)^{A}H^{-\alpha}(t)H'(t)+\frac{1}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\\ &\ \ \ \ -\frac{1}{2}\int_0^{t}g(s)ds\Vert \nabla u\Vert_{L^{2}(\Omega)}^2+2 H(t)+\frac{\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ &\geq (1-\alpha)(K+t)^{A}H^{-\alpha}(t)H'(t)+(2+\gamma)H(t)-\gamma E_2+\frac{\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}-\frac{\gamma}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ &\ \ \ \ +\frac{1+\gamma}{2}\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds+(\frac{\gamma}{2}-\frac{1+\gamma}{2}\int_0^{t}g(s)ds)\Vert \nabla u\Vert_{L^{2}(\Omega)}^2. \end{aligned} \end{equation*} $

这里$ \gamma>0 $. 根据$ \omega_2 $和$ {E}_1 $的定义, 则$ (3.6) $式变形为如下形式

$ \begin{equation*} \begin{aligned} E_2<{E}_1 = (\frac{p}{2}-1)\omega_2\leq(\frac{p}{2}-1)\frac{1}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p} = \frac{p-2}{2p}\Vert u\Vert_{L^{p}(\Omega)}^{p}. \end{aligned} \end{equation*} $

现在, 选定$ \gamma = \frac{\varepsilon_0}{2p} $, 可以得到

$ \begin{equation*} \begin{aligned} \frac{\varepsilon_0}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}-\frac{\gamma}{p}\Vert u\Vert_{L^{p}(\Omega)}^{p}-\gamma E_2&>(\frac{\varepsilon_0}{p}-\frac{\gamma}{p}-\gamma\frac{p-2}{2p})\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ & = (\frac{\varepsilon_0}{p}-\frac{\gamma}{2})\Vert u\Vert_{L^{p}(\Omega)}^{p} = \frac{3\varepsilon_0}{4p}\Vert u\Vert_{L^{p}(\Omega)}^{p}> 0. \end{aligned} \end{equation*} $

因此

$ \begin{equation} \begin{aligned} \mathcal{K}'(t)&\geq(1-\alpha)(K+t)^{A}H^{-\alpha}(t)H'(t)+(\frac{1}{2}+\frac{\varepsilon_0}{4p})\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\\ &\ \ \ \ +(2+\frac{\varepsilon_0}{2p})H(t)+\frac{3\varepsilon_0}{4p}\Vert u\Vert_{L^{p}(\Omega)}^{p}+\big[\frac{\varepsilon_0}{4p}-(\frac{1}{2}+\frac{\varepsilon_0}{4p})\int_0^{t}g(s)ds\big]\Vert \nabla u\Vert_{L^{2}(\Omega)}^2\\ &\geq (1-\alpha)(K+t)^{A}H^{-\alpha}(t)H'(t)+(2+\frac{\varepsilon_0}{2p})H(t)+\frac{3\varepsilon_0}{4p}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ &\ \ \ \ +(\frac{1}{2}+\frac{\varepsilon_0}{4p})\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds\; \; \; \big({\text{这里}}\; \int_0^{\infty}g(s)ds\leq\frac{\varepsilon_0}{2p+\varepsilon_0}\big)\\ &\geq\mu\big(H(t)+\int_0^{t}g(t-s)\Vert\nabla u(t)-\nabla u(s)\Vert_{L^{2}(\Omega)}^2ds+\Vert u\Vert_{L^{p}(\Omega)}^{p}\big)\; \; \; \big({\text{取}}\; \alpha<1\big)\\ &\geq\mu\big(H(t)+\Vert u\Vert_{L^{p}(\Omega)}^{p}\big)>0, \\ \end{aligned} \end{equation} $

(3.10)

其中$ \mu = \min\big\{2+\frac{\varepsilon_0}{2p}, \frac{1}{2}+\frac{\varepsilon_0}{4p}, \frac{3\varepsilon_0}{4p}\big\}>0 $, 且有$ \mathcal{K}(0)\geq K^{A}H^{1-\alpha}(0)+\frac{1}{2}\Vert u_0\Vert_{L^{2}(\Omega)}^2>0. $从而有$ \mathcal{K}(t)\geq\mathcal{K}(0)>0, \forall\; t\geq 0. $另一方面, 可以得到

$ \begin{equation} \begin{aligned} \big(\frac{1}{2}\int_{\Omega}u^2dx\big)^{\frac{1}{1-\alpha}}&\leq(\frac{1}{2})^{\frac{1}{1-\alpha}}\Vert u\Vert_{L^{2}(\Omega)}^{\frac{2}{1-\alpha}} \leq c_1(K+t)^{\frac{n(p-2)}{p(1-\alpha)}}\Vert u\Vert_{L^{p}(\Omega)}^{\frac{2}{1-\alpha}}\\ &\leq c_1(K+t)^{\frac{n(p-2)}{p(1-\alpha)}}\Vert u\Vert_{L^{p}(\Omega)}^{p}\Vert u\Vert_{L^{p}(\Omega)}^{\frac{2}{1-\alpha}-p} \leq c_1c_2(K+t)^{\frac{n(p-2)}{p(1-\alpha)}}\Vert u\Vert_{L^{p}(\Omega)}^{p}\\ & = c_3(K+t)^{\frac{n(p-2)}{p(1-\alpha)}}\Vert u\Vert_{L^{p}(\Omega)}^{p}, \end{aligned} \end{equation} $

(3.11)

这里$ c_2 = (2H(0))^{\frac{2}{p(1-\alpha)}-1}, c_3 = c_1c_2 $, 并且选取合适的$ \alpha $使得$ \frac{2}{1-\alpha}\leq p $.

再根据$ (3.8) $和$ (3.11) $式, 有

$ \begin{equation} \begin{aligned} \mathcal{K}^{\frac{1}{1-\alpha}}(t)&\leq 2^{\frac{1}{1-\alpha}}\left[(K+t)^{\frac{A}{1-\alpha}}H(t)+\big(\frac{1}{2}\int_{\Omega}u^2dx\big)^{\frac{1}{1-\alpha}}\right]\\ &\leq c_4(K+t)^{\frac{A_1}{1-\alpha}}[H(t)+\Vert u\Vert_{L^{p}(\Omega)}^{p}], \end{aligned} \end{equation} $

(3.12)

其中$ c_4 = \max\big\{2^{\frac{1}{1-\alpha}}, c_32^{\frac{1}{1-\alpha}}\big\} $, $ A_1 = \max\{A, \frac{n(p-2)}{p}\big\} $. 结合$ (3.10) $式和$ (3.12) $式, 进而有

$ \begin{equation} \begin{aligned} \mathcal{K}'(t)&\geq c_5\mathcal{K}^{\frac{1}{1-\alpha}}(t)(K+t)^{-\frac{A_1}{1-\alpha}}, \end{aligned} \end{equation} $

(3.13)

其中$ c_5 = \frac{\mu}{c_4} $. 缩小$ \alpha $的范围, 使得$ 0<\alpha<1-A_1 $. 令$ 1+\beta = \frac{1}{1-\alpha} $, 对$ (3.13) $式在$ (0, t) $上进行积分, 可以得到$ \mathcal{K}^{-\beta}(t)-\mathcal{K}^{-\beta}(0)\leq -c_5\beta\int_0^{t}(K+s)^{-A_1(1+\beta)}ds, $最后变形得到

$ t^{1-A_1(1+\beta)}\leq K^{1-A_1(1+\beta)}+\frac{1-A_1(1+\beta)}{c_5\beta}\mathcal{K}^{-\beta}(0), $

即$ t^\nu\leq K^\nu+A_0, $其中

$ \begin{equation} \nu = 1-A_1(1+\beta)\; \mbox{且}\; 0<\nu<1, \; A_0 = \frac{1-A_1(1+\beta)}{c_5\beta}\mathcal{K}^{-\beta}(0)\; \mbox{且}\; A_0>0. \end{equation} $

(3.14)

定理1.2证毕.