本文研究如下带有双阻尼和记忆项的$ \sigma $-发展方程的柯西问题

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}+\left(-\Delta\right)^{\sigma}u+u_{t}+\left(-\Delta\right)^{\sigma}u_{t}=\int_{0}^{t}(t-s)^{-\gamma}|u|^{p}ds, \qquad t>0, \; x \in \mathbb{R}^{n}, \\ &u(x, 0)=u_{0}, \quad u_{t}(x, 0)=u_{1}, \qquad\; \; \; \; \; \; \; \; \; \qquad\qquad\qquad\qquad x \in \mathbb{R}^{n}, \\ \end{aligned} \right. \end{equation} $

(1.1)

其中$ \sigma\geq1 $为整数, $ \gamma \in (0, 1), \; p>1 $, $ (-\Delta)^{\sigma}f=\mathcal{F}^{-1}(|\xi|^{2\sigma}\mathcal{F}(f)(\xi)), \; \xi\in\mathbb{R}^{n}, \; |\xi|=(\xi^{2}_{1}+\cdots+\xi^{2}_{n})^{\frac{1}{2}}, \; \mathcal{F} $为傅里叶变换.

近年来, 对于带有阻尼项的$ \sigma $-发展方程的柯西问题已有很多研究. 考虑单阻尼情形, 文[1–4] 对带有阻尼项$ u_{t} $的问题进行研究, 得到线性问题解的衰减估计, 解的整体存在性以及渐近形态等结论. 带有阻尼项$ (-\Delta)^{\delta} u_{t}(\delta>0) $的研究见文[5–7], 也得到了该问题下关于解的衰减估计以及整体存在性等结论.

对于带有双阻尼的$ \sigma $-发展方程, 当$ \sigma=1 $时, 文[8] 考虑如下情形

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}-\Delta u+u_{t}-\Delta u_{t}=0, \qquad\qquad t>0, x \in \mathbb{R}^{n}, \\ &u(x, 0)=u_{0}, \quad u_{t}(x, 0)=u_{1}, \; \; \qquad x \in \mathbb{R}^{n}.\\ \end{aligned} \right. \end{equation} $

(1.2)

当初值属于加权$ L^{1} $空间时, 作者得到了问题(1.2) 的解在$ L^{2} $空间下的渐近形态, 当$ t\rightarrow \infty $时, 有$ u(t, x)\sim(P_{0}+P_{1})G_{t}(x) $, 其中$ P_{j}:=\int_{\mathbb{R}^{n}}u_{j}(x)dx(j=0, 1), \; G_{t}(x):=\frac{1}{(\sqrt{4\pi t})^{n}}e^{-\frac{|x|^{2}}{4t}} $为高斯核函数. 作者指出, 与只带某一单阻尼情形时解的渐近形态相比较, 此时阻尼项$ u_{t} $的影响更为显著.

对于如下带有双阻尼以及非线性项的半线性波动方程

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}-\Delta u+u_{t}-\Delta u_{t}=F(u, u_{t}, \nabla u), \qquad t>0, x \in \mathbb{R}^{n}, \\ &u(x, 0)=u_{0}, \quad u_{t}(x, 0)=u_{1}, \; \; \; \; \; \qquad\qquad x \in \mathbb{R}^{n}, \\ \end{aligned} \right. \end{equation} $

(1.3)

其中$ F(u)=|\partial_{t}^{i}\partial_{x}^{j}u|^{p}(i, j=0, 1) $. 文[9] 中, 作者得到当空间维数$ n\geq1 $时, 问题(1.3) 线性问题解的衰减估计, 并利用不动点定理证明了问题(1.3) 带有上面三类非线性项时小初值问题解的整体存在性, 解整体存在时得到指数p的范围分别为:

$ \begin{equation*} 1+\frac{2}{n}<p<1+\frac{2}{n-2}, \; 1<p<\frac{2}{n}, \; p>1+\frac{1}{n+1}. \end{equation*} $

当$ F(u)=|u|^{p} $时, 文[10] 证明了当空间维度$ n\leq4 $时小初值问题解的整体存在性以及爆破, 得到临界指数$ p_{crit}=1+\frac{2}{n} $, 这与Fujita指数一致. 此外, 文[11] 研究了问题(1.3) 初值的正则性和指数p容许范围之间的关系. 若$ n \geq 1, \; m\in [1, 2), \; s_{1}, \; s_{2} $为正实数, 当初值具有下述三种正则性:

$ \begin{equation} \nonumber \begin{aligned} &(i)\; (u_{0}, u_{1}) \in (H^{s_{1}}\cap L^{m})\times (L^{2}\cap L^{m});\\ &(ii)\; s_{1}\geq s_{2}>\frac{n}{2}, \; (u_{0}, u_{1}) \in (H^{s_{1}}\cap L^{m})\times (H^{s_{2}}\cap L^{m});\\ &(iii)\; s_{2}\leq s_{1}<\frac{n}{2}, \; (u_{0}, u_{1}) \in (H^{s_{1}}\cap L^{m})\times (H^{s_{2}}\cap L^{m});\\ \end{aligned} \end{equation} $

作者分别得到相应情形下小初值问题解的整体存在性.

当$ \sigma\geq1 $时, 文[12] 研究了如下带有双阻尼的半线性$ \sigma $-发展方程

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}+(-\Delta)^{\sigma} u+u_{t}+(-\Delta)^{\sigma} u_{t}=F(u), \qquad \; \; \; t>0, x \in \mathbb{R}^{n}, \\ &u(x, 0)=u_{0}, \quad u_{t}(x, 0)=u_{1}, \; \; \; \; \qquad\qquad\qquad x \in \mathbb{R}^{n}.\\ \end{aligned} \right. \end{equation} $

(1.4)

作者讨论了当$ F(u)=|u|^{p} $时的情形, 得到问题(1.4)对应线性问题解的衰减估计, 证明了小初值问题解的整体存在性以及爆破, 得到临界指数$ p_{crit}=1+\frac{2\sigma}{n}. $此外, 当$ F(u)=|u|^{p} $时, 问题(1.4) 解的衰减估计与其对应线性问题解的衰减估计一致, 并未出现衰减损失.

考虑方程带记忆项情形. 对于如下问题

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}-\Delta u+ u_{t}=\int_{0}^{t}(t-s)^{-\gamma}|u|^{p}ds, \; \; \; \; \; \quad t>0, x \in \mathbb{R}^{n}, \\ &u(x, 0)=u_{0}, \quad u_{t}(x, 0)=u_{1}, \; \; \; \qquad\qquad\; x \in \mathbb{R}^{n}, \\ \end{aligned} \right. \end{equation} $

(1.5)

其中$ \gamma \in (0, 1), \; p>1 $. 文[13] 研究了当空间维数$ n\leq5 $时, 小初值问题解的整体存在性. 文[14] 得到空间维数$ n\leq3 $时解的整体存在性以及任意空间维数下解的爆破, 得

$ \begin{equation} \nonumber p_{crit}=1+\frac{2(2-\gamma)}{n-2(1-\gamma)}. \end{equation} $

对于如下问题

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}-\Delta u+\mu (-\Delta)^{\sigma} u_{t}=\int_{0}^{t}(t-s)^{-\gamma}|u|^{p}ds, \qquad t>0, x \in \mathbb{R}^{n}, \\ &u(x, 0)=u_{0}, \quad u_{t}(x, 0)=u_{1}, \qquad\qquad\qquad\qquad\; x \in \mathbb{R}^{n}, \\ \end{aligned} \right. \end{equation} $

(1.6)

其中$ \mu>0, \gamma \in (0, 1), p>1, \sigma \in (0, 1) $. 当空间维数$ n\geq 1 $且初值满足$ \int_{\mathbb{R}^{n}}(u_{1}+\mu(-\Delta)^{\sigma}u_{0})dx>0 $时, 文[15] 得到问题(1.6) 的爆破结果. 文[16] 考虑空间维数$ n\geq 2, \; \sigma=\frac{1}{2} $情形, 得到问题(1.6) 小初值问题下的临界指数$ p_{crit}=\max\left\{1+\frac{3-\gamma}{n-2+\gamma}, \; \gamma^{-1}\right\}. $

在已知的关于方程同时带有阻尼项与记忆项的研究中, 大多考虑单阻尼情形, 而对于双阻尼情形得到的结论较少, 且就记忆项而言, 它的出现可能对解的一些性质产生影响. 受到这一启发, 本文在文[12] 工作的基础上, 考虑双阻尼情形下, 当问题(1.1) 带记忆项时对指数p的范围以及衰减估计的影响, 并得到解的爆破以及生命跨度上界的估计. 文章结构如下: 第二节给出相关引理. 主要结论将在第三节中给出, 第四节给出解的整体存在性, 爆破以及生命跨度估计的证明.

本文有如下记号:

(1) $ f\lesssim g $表示存在一个常数$ c>0 $使得$ f\leq cg $;

(2) $ L^{p}(0<p<\infty) $表示Lebesgue空间;

(3) $ H^{s, p}(\mathbb{R}^{n}) $表示非齐次Sobolev空间, 即

$ \qquad H^{s, p}(\mathbb{R}^{n}):=\left\{f \in L^{p}(\mathbb{R}^{n}):\| f\|_{H^{s, p}(\mathbb{R}^{n})}=\|\mathcal{F}^{-1}[(1+|\xi|^{2})^{\frac{s}{2}}\hat{f}]\|_{L^{p}(\mathbb{R}^{n})}<\infty, \; s\geq 0 \right\} $;

(4) $ f\in AC[0, T] $表示f在$ [0, T] $内绝对连续.

首先给出分数阶积分和分数阶导数的一些性质, 见 [17, 18]. 若$ f\in L^{1}(0, T), \; T>0, \; \alpha\in(0, 1) $, 则$ \alpha $阶Riemann-Liouville分数阶积分定义如下

$ \begin{eqnarray*} J^{\alpha}_{0|t}f(t):=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}f(s)ds, \qquad t>0, \\ J^{\alpha}_{t|T}f(t):=\frac{1}{\Gamma(\alpha)}\int_{t}^{T}(s-t)^{\alpha-1}f(s)ds, \qquad t<T, \end{eqnarray*} $

其中$ \Gamma(s) $为Gamma函数.

若$ f\in AC[0, T], \; T>0, \; \alpha\in(0, 1) $, 则$ \alpha $阶Riemann-Liouville型分数阶导数定义如下

$ \begin{equation} \nonumber D^{\alpha}_{0|t}f(t):=\partial_{t}J^{1-\alpha}_{0|t}f(t)=\frac{1}{\Gamma(1-\alpha)}\partial_{t}\int_{0}^{t}(t-s)^{-\alpha}f(s)ds, \qquad t>0. \end{equation} $

$ \begin{equation} \nonumber D^{\alpha}_{t|T}f(t):=-\partial_{t}J^{1-\alpha}_{t|T}f(t)=-\frac{1}{\Gamma(1-\alpha)}\partial_{t}\int_{t}^{T}(s-t)^{-\alpha}f(s)ds, \qquad t<T. \end{equation} $

若对于$ t \in [0, T], \alpha \in (0, 1), f, g \in C[0, T] $有$ D^{\alpha}_{0|t}f(t), D^{\alpha}_{t|T}g(t) $存在且连续, 则由分部积分公式可得

$ \begin{equation} \nonumber \int_{0}^{T}(D^{\alpha}_{0|t}f)(t)g(t)dt=\int_{0}^{T}f(t)(D^{\alpha}_{t|T}g)(t)dt. \end{equation} $

且对于任意的$ f\in AC^{n+1}[0, T] $, 以及整数$ n\geq 0, $有$ (-1)^{n}\partial^{n}_{t}D^{\alpha}_{t|T}f=D^{n+\alpha}_{t|T}f, $其中$ AC^{n+1}[0, T]:=\left\{f:[0, T] \rightarrow \mathbb{R}, \; \partial^{n}_{t}f\in AC[0, T]\right\} $, $ \partial^{n}_{t} $是关于t的n阶导数. 此外, 对于所有的$ 1 \leq q \leq \infty $, 有$ D^{\alpha}_{0|t}J^{\alpha}_{0|t}=Id_{L^{q}(0, T)} $在$ [0, T] $上几乎处处成立.

下面给出线性问题解的衰减估计.

引理2.1 ^[12]   问题(1.1) 对应线性问题的解满足如下$ (L^{1} \cap L^{2}-L^{2}) $估计:

$ \begin{equation} \|\partial_{t}^{j}(-\Delta)^{\frac{a}{2}}u(t, \cdot)\|_{L^{2}}\lesssim(1+t)^{-\frac{n}{4\sigma}-\frac{a}{2\sigma}-j}(\|u_{0}\|_{L^{1}\cap H^{a}}+\|u_{1}\|_{L^{1}\cap H^{[a+2(j-1)\sigma]^{+}}}), \end{equation} $

(2.1)

其中$ a\geq 0, \; n\geq 1, \; [\cdot]^{+}=\max\left\{0, \cdot\right\}, \; j=0, 1 $.

下面给出一些需要用到的引理.

引理2.2 ^[19]   若$ 1<q < \infty, \; \sigma > 0.\ $则对于任意的$ y \in H^{\sigma}(\mathbb{R}^{n}), $有如下分数阶G-N不等式

$ \begin{equation} \|y\|_{L^{q}(\mathbb{R}^{n})}\lesssim\|(-\Delta)^{\frac{\sigma}{2}}y\|^{\theta_{q}}_{L^{2}(\mathbb{R}^{n})}\|y\|^{1-\theta_{q}}_{L^{2}(\mathbb{R}^{n})}, \end{equation} $

(2.2)

其中$ \theta_{q}=\frac{n}{\sigma}(\frac{1}{2}-\frac{1}{q})\in [0, 1]. $

引理2.3 ^[14]   若$ \alpha \in \mathbb{R}, \; \beta>1, \; \gamma \in (0, 1), $则

$ \begin{equation} \nonumber \int_{0}^{t}(1+t-s)^{-\alpha}\int_{0}^{s}(s-\tau)^{-\gamma}(1+\tau)^{-\beta}d\tau ds \lesssim \left\{\begin{aligned} &(1+t)^{-\gamma}, \; \qquad \qquad\quad\; \alpha>1, \\ &(1+t)^{-\gamma}\log(2+t) , \quad\; \alpha=1, \\ &(1+t)^{1-\alpha-\gamma} , \; \qquad\qquad \alpha<1. \\ \end{aligned} \right. \end{equation} $

引理2.4 ^[17]   令$ T>0, $函数$ \omega:[0, T]\rightarrow \mathbb{R} $定义如下

$ \begin{equation} \nonumber \omega(t) = \left\{\begin{aligned} & (1-\frac{t}{T}), \qquad t\in[0, T], \\ & 0, \qquad \qquad \quad t>T. \end{aligned} \right. \end{equation} $

若$ 0<\alpha<1, \; \beta\gg 1, \; m\geq0, $则对所有的$ t\in [0, T] $, 有

$ \begin{equation} D^{m+a}_{t|T}\omega^{\beta}(t)=\frac{\Gamma(\beta+1)}{\Gamma(\beta+1-m-\alpha)}T^{-(m+\alpha)}\omega^{\beta-\alpha-m}(t). \end{equation} $

(2.3)

下面给出弱解的定义.

定义2.1   令$ p>1, \; T>0 $. 若$ (u_{0}, u_{1})\in L^{2}\times L^{2}, \; u\in L^{p}([0, T], L^{2p})\cap L^{1}([0, T], L^{2}) $且对于任意的测试函数$ \phi(t, x)\in C^{2}([0, T], L^{2})\cap C^{1}([0, T], H^{2\sigma} \cap L^{2})\cap C([0, T], H^{2\sigma}), \; \phi(T, \cdot)=0, \; \phi_{t}(T, \cdot)=0, $有如下等式成立

$ \begin{equation} \begin{aligned} &\Gamma(\alpha)\int_{0}^{T}\int_{\mathbb{R}^{n}}J^{\alpha}_{0|t}(|u|^{p})\phi(t, x)dxdt+\int_{\mathbb{R}^{n}}u_{0}(x)(\phi(0, x)+(-\Delta)^{\sigma}\phi(0, x)-\phi_{t}(0, x))dx\\ &+\int_{\mathbb{R}^{n}}u_{1}(x)\phi(0, x)dx=\int_{0}^{T}\int_{\mathbb{R}^{n}}u(\phi_{tt}(t, x)-(-\Delta)^{\sigma}\phi_{t}(t, x)+(-\Delta)^{\sigma}\phi(t, x)-\phi_{t}(t, x))dxdt, \end{aligned} \end{equation} $

(2.4)

则称u是$ (1.1) $的局部弱解. 若$ T=\infty $, 则称u是问题$ (1.1) $的整体弱解.

下面给出本文的主要结论.

定理3.1    当$ \sigma \geq \frac{n}{2} $时, 若$ \gamma \in (1-\frac{n}{4\sigma}, 1) $,

$ \begin{equation} p>1+\frac{2\sigma(2-\gamma)}{n-2\sigma(1-\gamma)}, \end{equation} $

(3.1)

则存在常数$ \varepsilon>0 $使得对任意的小初值$ (u_{0}, u_{1})\in\mathcal{A}:=(L^{1}\cap H^{\sigma})\times(L^{1}\cap L^{2}) $, $ \|(u_{0}, u_{1})\|_{\mathcal{A}}\leq \varepsilon, $问题(1.1) 都存在一个唯一的整体解$ u \in C([0, \infty), H^{\sigma})\cap C^{1}([0, \infty), L^{2}) $且满足如下估计

$ \begin{equation} \|u\|_{L^{2}} \lesssim (1+t)^{-\frac{n}{4\sigma}-\gamma+1}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \end{equation} $

(3.2)

$ \begin{equation} \||D|^{\sigma}u\|_{L^{2}} \lesssim \left\{\begin{aligned} (1+t)^{-\frac{n}{4\sigma}-\gamma+\frac{1}{2}}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \quad \sigma > \frac{n}{2}, \\ (1+t)^{-\gamma}\log(2+t)\|(u_{0}, u_{1})\|_{\mathcal{A}} , \quad \sigma=\frac{n}{2}, \\ \end{aligned} \right. \end{equation} $

(3.3)

$ \begin{equation} \|\partial_{t}u\|_{L^{2}}\lesssim (1+t)^{-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}. \end{equation} $

(3.4)

定理3.2    假设初值$ u_{0}=0, \; u_{1}\in L^{1}\cap L^{2} $且满足

$ \begin{equation} \int_{\mathbb{R}^{n}}u_{1}(x)dx>0. \end{equation} $

(3.5)

若$ \sigma \geq \frac{n}{2}, \; \gamma \in (1-\frac{n}{4\sigma}, 1) $,

$ \begin{equation} 1<p \leq 1+\frac{2\sigma(2-\gamma)}{n-2\sigma(1-\gamma)}, \end{equation} $

(3.6)

则问题(1.1) 不存在整体解. 且对于任意小的常数$ \varepsilon>0 $, 当初值$ u_{0}=0, \; u_{1} $变为$ \varepsilon u_{1} $时, 可得到生命跨度的估计

$ \begin{equation} \nonumber T_{\varepsilon} \leq C\varepsilon^{-\frac{1}{k}}, \end{equation} $

其中$ C>0 $,

$ \begin{equation*} k=\frac{(2-\gamma)p}{p-1}-\frac{2\sigma+n}{2\sigma}>0. \end{equation*} $

注3.1    事实上, $ F(t, u):=\Gamma(1-\gamma)J_{0|t}^{1-\gamma}(|u|^{p}), \; \lim_{\gamma\rightarrow 1}\Gamma(1-\gamma)F(t, u)=|u|^{p} $, 即当$ \gamma\rightarrow 1 $时, 可以得到非线性记忆项与幂次非线性项之间的关系. 综合定理3.1和定理3.2, 可得当$ \sigma \geq \frac{n}{2}, \; \gamma \in (1-\frac{n}{4\sigma}, 1) $时

$ \begin{equation*} p_{crit}=1+\frac{2\sigma(2-\gamma)}{n-2\sigma(1-\gamma)}, \end{equation*} $

令$ \gamma \rightarrow 1 $有$ p_{crit} \rightarrow 1+\frac{2\sigma}{n} $, 这与文[12] 中非线性项为幂次项时得到的临界指数一致.

定理3.3    当$ \sigma \in [\frac{n}{4}, \frac{n}{2}) $时, 若$ \sigma \in (\frac{n}{4}, \frac{n}{2}) $,

$ \begin{equation} \nonumber \gamma\in \Bigg(\max \left\{1-\frac{n}{4\sigma}, \frac{n}{\sigma}-\frac{2\sigma}{n}-\frac{\frac{n^{2}}{\sigma^{2}}-\frac{2n}{\sigma}}{4}+\frac{\frac{n^{3}}{\sigma^{3}}-\frac{8n^{2}}{\sigma^{2}}+\frac{12n}{\sigma}}{\frac{8n}{\sigma}-16}\right\}, \frac{3}{2}-\frac{n}{4\sigma}\Bigg), \end{equation} $

$ \begin{equation} \frac{1+\frac{n}{\sigma}-\frac{n^{2}}{4\sigma^{2}}}{\frac{3n}{4\sigma}+\gamma-1-\frac{n^{2}}{8\sigma^{2}}}<p<\frac{n}{n-2\sigma}; \end{equation} $

(3.7)

若$ \sigma=\frac{n}{4}, \; \gamma=\frac{1}{2}, \; p=2 $, 则存在常数$ \varepsilon>0 $使得对任意的小初值$ (u_{0}, u_{1})\in\mathcal{A}:=(L^{1}\cap H^{\sigma})\times(L^{1}\cap L^{2}) $, $ \|(u_{0}, u_{1})\|_{\mathcal{A}}\leq \varepsilon, $问题(1.1) 都存在一个唯一的整体解$ u \in C([0, \infty), H^{\sigma})\cap C^{1}([0, \infty), L^{2}) $且满足如下估计

$ \begin{equation} \|u\|_{L^{2}} \lesssim \left\{\begin{aligned} &(1+t)^{-\frac{n}{4\sigma}-\gamma+1}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \qquad\; \; \sigma\in (\frac{n}{4}, \frac{n}{2}), \\ &(1+t)^{-\gamma}\log(2+t)\|(u_{0}, u_{1})\|_{\mathcal{A}} , \quad \sigma=\frac{n}{4}, \\ \end{aligned} \right. \end{equation} $

(3.8)

$ \begin{equation} \||D|^{\sigma}u\|_{L^{2}}\lesssim (1+t)^{-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \end{equation} $

(3.9)

$ \begin{equation} \|\partial_{t}u\|_{L^{2}}\lesssim (1+t)^{-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}. \end{equation} $

(3.10)

本节在证明定理之前, 首先给出解的表达式以及在证明解的整体存在性过程中所需要的解空间的定义.

利用Duhamel原理, 可得问题(1.1) 的解, 形式如下

$ \begin{equation} u(t, x)=u^{ln}(t, x)+u^{nl}(t, x), \end{equation} $

(4.1)

其中

$ \begin{equation} \nonumber u^{ln}(t, x)=K_{0}\ast_{x}u_{0}(x)+K_{1}\ast_{x}u_{1}(x), \end{equation} $

$ \begin{equation} \nonumber u^{nl}(t, x)=\int_{0}^{t}K_{1}(t-\tau, x)\ast _{x}\Big(\int_{0}^{\tau}(\tau-s)^{-\gamma}|u(s, x)|^{p}\Big)dsd\tau, \end{equation} $

$ K_{i}(i=0, 1) $的定义与 ^[12]中一致. 为了证明定理3.1和定理3.3, 引入解空间$ X(t) $. 对任意的$ t>0 $, 以及$ \eta>0 $, 空间$ X(t) $定义如下

$ \begin{equation} X(t):=\left\{u \mid u\in C([0, t], H^{\sigma})\cap C^{1}([0, t], L^{2}), \; \|u\|_{X(t)}<\eta\right\}. \end{equation} $

(4.2)

定义算子$ N:u\in X(t) \rightarrow Nu , $其中

$ \begin{equation} Nu(t, x)=u^{ln}(t, x)+u^{nl}(t, x). \end{equation} $

(4.3)

问题(1.1) 的整体解即为算子N的不动点. 因此, 为了得到$ X(t) $中解的全局存在性和唯一性, 对于任意的$ u, \; v\in X(t) $需要证明以下两项估计

$ \begin{equation} \|Nu\|_{X(t)}\lesssim \|(u_{0}, u_{1})\|_{\mathcal{A}}+\|u\|^{p}_{X(t)}, \end{equation} $

(4.4)

$ \begin{equation} \|Nu-Nv\|_{X(t)}\lesssim \|u-v\|_{X(t)}(\|u\|^{p-1}_{X(t)}+\|v\|^{p-1}_{X(t)}). \end{equation} $

(4.5)

再应用Banach不动点定理, 可得小初值问题解的全局存在性结果.

证明定理3.1    当$ \sigma=\frac{n}{2} $时, 定义空间范数为

$ \begin{equation} \nonumber \begin{aligned} \|u\|_{X(t)}=&\sup\limits_{0\leq \tau \leq t}((1+\tau)^{-\frac{1}{2}+\gamma}\|u(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\gamma}\log^{-1}(2+\tau)\||D|^{\sigma}u(\tau, \cdot)\|_{L^{2}}\\ &\quad+(1+\tau)^{\gamma}\|\partial_{t}u(\tau, \cdot)\|_{L^{2}}). \end{aligned} \end{equation} $

当$ \sigma>\frac{n}{2} $时, 定义空间范数为

$ \begin{equation} \nonumber \|u\|_{X(t)}=\sup\limits_{0\leq \tau \leq t}((1+\tau)^{\frac{n}{4\sigma}+\gamma-1}\|u(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\frac{n}{4\sigma}+\gamma-\frac{1}{2}}\||D|^{\sigma}u(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\gamma}\|\partial_{t}u(\tau, \cdot)\|_{L^{2}}). \end{equation} $

下面只证明$ \sigma>\frac{n}{2} $时情形, $ \sigma=\frac{n}{2} $时证明与之类似.

由线性估计, 有

$ \begin{equation} \begin{aligned} \|u^{ln}\|_{X(t)}=&\sup\limits_{0\leq \tau \leq t}((1+\tau)^{\frac{n}{4\sigma}+\gamma-1}\|u^{ln}(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\frac{n}{4\sigma}+\gamma-\frac{1}{2}}\||D|^{\sigma}u^{ln}(\tau, \cdot)\|_{L^{2}}\\ &+(1+\tau)^{\gamma}\|\partial_{t}u^{ln}(\tau, \cdot)\|_{L^{2}})\\ \lesssim& \|(u_{0}, u_{1})\|_{\mathcal{A}} . \end{aligned} \end{equation} $

(4.6)

为了证明式(4.4), 接下来证明

$ \begin{equation} \|u^{nl}(t, x)\|_{X(t)}\lesssim \|u\|^{p}_{X(t)}. \end{equation} $

(4.7)

由引理2.1可得如下估计

$ \begin{equation} \|u^{nl}\|_{L^{2}}\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}}\int_{0}^{\tau}(\tau-s)^{-\gamma}\|u^{p}(\tau, \cdot)\|_{L^{1}\cap L^{2}}dsd\tau, \end{equation} $

(4.8)

$ \begin{equation} \||D|^{\sigma}u^{nl}\|_{L^{2}}\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-\frac{1}{2}}\int_{0}^{\tau}(\tau-s)^{-\gamma}\|u^{p}(\tau, \cdot)\|_{L^{1}\cap L^{2}}dsd\tau, \end{equation} $

(4.9)

$ \begin{equation} \|\partial_{t}u^{nl}\|_{L^{2}}\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-1}\int_{0}^{\tau}(\tau-s)^{-\gamma}\|u^{p}(\tau, \cdot)\|_{L^{1}\cap L^{2}}dsd\tau. \end{equation} $

(4.10)

由引理2.2中G-N不等式可以估计如下范数

$ \begin{equation} \nonumber \begin{aligned} \|u\|^{p}_{L^{sp}}&\lesssim \|u\|^{(1-\theta)p}_{L^{2}}\|(-\Delta)^{\frac{\sigma}{2}}u\|^{\theta p}_{L^{2}}\\ &\lesssim(1+\tau)^{(-\frac{n}{4\sigma}-\gamma+\frac{1}{2})\theta p}\|u\|^{\theta p}_{X(t)}(1+\tau)^{(-\frac{n}{4\sigma}-\gamma+1)(p-\theta p)}\|u\|^{(p-\theta p)}_{X(t)}\\ &\lesssim(1+\tau)^{-\frac{n}{2\sigma}(p-\frac{1}{s})-(\gamma-1)p}\|u\|^{p}_{X(t)}, \end{aligned} \end{equation} $

其中$ s=1, 2.\; \theta\in [0, 1] $且

$ \begin{equation} \nonumber \theta=\frac{n}{\sigma}(\frac{1}{2}-\frac{1}{sp}). \end{equation} $

从而得

$ \begin{equation} \begin{aligned} \|u^{p}\|_{L^{1}\cap L^{2}}\lesssim& \|u\|^{p}_{L^{p}}+\|u\|^{p}_{L^{2p}}\\ \lesssim& (1+\tau)^{-\frac{n}{2\sigma}(p-1)-(\gamma-1)p}\|u\|^{p}_{X(t)}+(1+\tau)^{-\frac{n}{2\sigma}(p-\frac{1}{2})-(\gamma-1)p}\|u\|^{p}_{X(t)}. \end{aligned} \end{equation} $

(4.11)

令$ \beta_{\gamma}(n, \sigma, p):=\frac{n}{2\sigma}(p-1)+(\gamma-1)p, \; \beta^{'}_{\gamma}(n, \sigma, p):=\frac{n}{2\sigma}(p-\frac{1}{2})+(\gamma-1)p $, 将式(4.11) 代入式(4.8) (4.9) 和(4.10) 有

$ \begin{equation} \begin{aligned} \|u^{nl}\|_{L^{2}}\lesssim& \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}}\int_{0}^{\tau}(\tau-s)^{-\gamma}(1+s)^{-\beta_{\gamma}(n, \sigma, p)}dsd\tau \|u\|^{p}_{X(t)} \\ &+\int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}}\int_{0}^{\tau}(\tau-s)^{-\gamma}(1+s)^{-\beta^{'}_{\gamma}(n, \sigma, p)}dsd\tau \|u\|^{p}_{X(t)}. \end{aligned} \end{equation} $

(4.12)

$ \begin{equation} \begin{aligned} \||D|^{\sigma}u^{nl}\|_{L^{2}}\lesssim& \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-\frac{1}{2}}\int_{0}^{\tau}(\tau-s)^{-\gamma}(1+s)^{-\beta_{\gamma}(n, \sigma, p)}dsd\tau \|u\|^{p}_{X(t)} \\ &+\int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-\frac{1}{2}}\int_{0}^{\tau}(\tau-s)^{-\gamma}(1+s)^{-\beta^{'}_{\gamma}(n, \sigma, p)}dsd\tau \|u\|^{p}_{X(t)} . \end{aligned} \end{equation} $

(4.13)

$ \begin{equation} \begin{aligned} \|\partial_{t}u^{nl}\|_{L^{2}}\lesssim& \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-1}\int_{0}^{\tau}(\tau-s)^{-\gamma}(1+s)^{-\beta_{\gamma}(n, \sigma, p)}dsd\tau \|u\|^{p}_{X(t)} \\ &+\int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-1}\int_{0}^{\tau}(\tau-s)^{-\gamma}(1+s)^{-\beta^{'}_{\gamma}(n, \sigma, p)}dsd\tau \|u\|^{p}_{X(t)}. \end{aligned} \end{equation} $

(4.14)

因此, 当$ \beta_{\gamma}(n, \sigma, p)>1, \; \beta^{'}_{\gamma}(n, \sigma, p)>1, \; \gamma\in (1-\frac{n}{4\sigma}, 1) $时, 可得$ p>1+\frac{2\sigma(2-\gamma)}{n-2\sigma(1-\gamma)} $, 此时运用引理2.3可得

$ \begin{equation} \nonumber \begin{aligned} \|u^{nl}\|_{L^{2}}&\lesssim (1+t)^{1-\gamma-\frac{n}{4\sigma}}\|u\|^{p}_{X(t)}, \end{aligned} \end{equation} $

$ \begin{equation} \nonumber \begin{aligned} \||D|^{\sigma}u^{nl}\|_{L^{2}}&\lesssim (1+t)^{\frac{1}{2}-\gamma-\frac{n}{4\sigma}}\|u\|^{p}_{X(t)} , \end{aligned} \end{equation} $

$ \begin{equation} \nonumber \begin{aligned} \|\partial_{t}u^{nl}\|_{L^{2}}&\lesssim (1+t)^{-\gamma}\|u\|^{p}_{X(t)} . \end{aligned} \end{equation} $

从而

$ \begin{equation} \nonumber \begin{aligned} (1+t)^{-1+\gamma+\frac{n}{4\sigma}}\|u^{nl}\|_{L^{2}}&\lesssim \|u\|^{p}_{X(t)} , \end{aligned} \end{equation} $

$ \begin{equation} \nonumber \begin{aligned} (1+t)^{-\frac{1}{2}+\gamma+\frac{n}{4\sigma}}\||D|^{\sigma}u^{nl}\|_{L^{2}}&\lesssim \|u\|^{p}_{X(t)} , \end{aligned} \end{equation} $

$ \begin{equation} \nonumber \begin{aligned} (1+t)^{\gamma}\|u^{nl}\|_{L^{2}}&\lesssim \|u\|^{p}_{X(t)} . \end{aligned} \end{equation} $

即得到

$ \begin{equation} \|u^{nl}\|_{X(t)}\lesssim \|u\|^{p}_{X(t)}. \end{equation} $

(4.15)

结合式(4.6) (4.15) 得

$ \begin{equation} \|Nu\|_{X(t)}\lesssim \|(u_{0}, u_{1})\|_{\mathcal{A}}+\|u\|^{p}_{X(t)}. \end{equation} $

(4.16)

下面证明式(4.5). 对于任意的$ u, \; v \in X(t), $有

$ \begin{equation} \nonumber Nu-Nv=u^{nl}-v^{nl}. \end{equation} $

由引理2.1有

$ \begin{equation} \begin{aligned} \|u^{nl}-v^{nl}\|_{L^{2}}&\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}}\int_{0}^{\tau}(\tau-s)^{-\gamma}\||u(\tau, \cdot)|^{p}-|v(\tau, \cdot)|^{p}\|_{L^{1}\cap L^{2}}dsd\tau, \\ \end{aligned} \end{equation} $

(4.17)

$ \begin{equation} \begin{aligned} \||D|^{\sigma}u^{nl}-|D|^{\sigma}v^{nl}\|_{L^{2}}&\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-\frac{1}{2}}\int_{0}^{\tau}(\tau-s)^{-\gamma}\||u(\tau, \cdot)|^{p}-|v(\tau, \cdot)|^{p}\|_{L^{1}\cap L^{2}}dsd\tau, \end{aligned} \end{equation} $

(4.18)

$ \begin{equation} \begin{aligned} \|\partial_{t}u^{nl}-\partial_{t}v^{nl}\|_{L^{2}}&\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-1}\int_{0}^{\tau}(\tau-s)^{-\gamma}\||u(\tau, \cdot)|^{p}-|v(\tau, \cdot)|^{p}\|_{L^{1}\cap L^{2}}dsd\tau. \\ \end{aligned} \end{equation} $

(4.19)

由Hölder不等式, 当$ r\geq1 $时有

$ \begin{equation} \nonumber \||u(\tau, \cdot)|^{p}-|v(\tau, \cdot)|^{p}\|_{L^{r}}\leq \||u(\tau, \cdot)|-|v(\tau, \cdot)|\|_{L^{rp}}(\|u(\tau, \cdot)\|^{p-1}_{L^{rp}}+\|v(\tau, \cdot)\|^{p-1}_{L^{rp}}). \end{equation} $

由引理2.2中G-N不等式得

$ \begin{equation} \begin{aligned} \||u(\tau, \cdot)|^{p}-|v(\tau, \cdot)|^{p}\|_{L^{1}\cap L^{2}}&\lesssim \|u-v\|_{L^{p}}(\|u\|^{p-1}_{L^{p}}+\|v\|^{p-1}_{L^{p}})+\|u-v\|_{L^{2p}}(\|u\|^{p-1}_{L^{2p}}+\|v\|^{p-1}_{L^{2p}})\\ &\lesssim(1+\tau)^{-\frac{n}{2\sigma}(p-1)-(\gamma-1)p}\|u-v\|_{X(t)}(\|u\|^{p-1}_{X(t)}+\|v\|^{p-1}_{X(t)})\\ &+(1+\tau)^{-\frac{n}{2\sigma}(p-\frac{1}{2})-(\gamma-1)p}\|u-v\|_{X(t)}(\|u\|^{p-1}_{X(t)}+\|v\|^{p-1}_{X(t)}).\\ \end{aligned} \end{equation} $

(4.20)

将(4.20) 带入(4.17), (4.18) 以及(4.19) 中, 令$ \beta_{\gamma}(n, \sigma, p)>1, \; \beta^{'}_{\gamma}(n, \sigma, p)>1, \; \gamma\in (1-\frac{n}{4\sigma}, 1) $, 利用引理2.3可得

$ \begin{equation} \|Nu-Nv\|_{X(t)}\lesssim \|u-v\|_{X(t)}(\|u\|^{p-1}_{X(t)}+\|v\|^{p-1}_{X(t)}). \end{equation} $

(4.21)

综上, 即证得(4.4)与(4.5), 再运用不动点定理则可得到在$ \sigma>\frac{n}{2} $情形下问题(1.1) 整体解的存在唯一性, 且有如下衰减估计

$ \begin{equation} \nonumber \|u\|_{L^{2}} \lesssim (1+t)^{-\frac{n}{4\sigma}-\gamma+1}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \end{equation} $

$ \begin{equation} \nonumber \||D|^{\sigma}u\|_{L^{2}} \lesssim (1+t)^{-\frac{n}{4\sigma}-\gamma+\frac{1}{2}}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \end{equation} $

$ \begin{equation} \nonumber \|\partial_{t}u\|_{L^{2}} \lesssim (1+t)^{-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}. \end{equation} $

与上述证明类似, 也可得$ \sigma=\frac{n}{2} $时整体解的存在唯一性和衰减估计

$ \begin{equation} \nonumber \|u\|_{L^{2}} \lesssim (1+t)^{\frac{1}{2}-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \end{equation} $

$ \begin{equation} \nonumber \||D|^{\sigma}u\|_{L^{2}} \lesssim (1+t)^{-\gamma}\log(2+t)\|(u_{0}, u_{1})\|_{\mathcal{A}}, \end{equation} $

$ \begin{equation} \nonumber \|\partial_{t}u\|_{L^{2}} \lesssim (1+t)^{-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}. \end{equation} $

下面证明$ \sigma\geq \frac{n}{2} $时解的爆破.

证明定理3.2    首先假设u是问题(1.1) 的整体解. 令$ \Psi\in C^{\infty}_{c}(\mathbb{R}^{n}) $为球对称函数, 满足:

(1) supp$ \Psi=B_{1} $,

(2) $ 0\leq \Psi \leq 1 $, 且对于任意的$ x\in B_{\frac{1}{2}}, \; \Psi(x)=1 $,

(3) 若$ |x_{1}|\leq|x_{2}| $, 则$ \Psi(x_{1})\geq \Psi(x_{2}) $, 其中$ B_{R}=\left\{x\in \mathbb{R}^{n}:|x|\leq R\right\} $.

$ \omega(t) $定义如下

$ \begin{equation} \nonumber \omega(t) = \left\{\begin{aligned} & 1-\frac{t}{T}, \qquad t\in[0, T], \\ & 0, \qquad \qquad \quad t>T, \end{aligned} \right. \end{equation} $

其中supp$ \omega=[0, T] $, 对于任意的$ \beta>k\geq 0, \; \omega^{\beta}(t)\in C^{k}_{c}([0, \infty)) $.

对任意的$ R\geq 1 $, 令$ \Psi_{R}(x):=\Psi(\frac{x}{R}), \; \Phi_{R}(t, x)=\omega^{\beta}(t)\Psi_{R}(x), $引入函数$ \phi(t, x):=D^{\alpha}_{t|T}\Phi_{R}(t, x)=D^{\alpha}_{t|T}\omega^{\beta}(t)\Psi_{R}(x) $, 则supp$ \phi\subset [0, T]\times B_{R} $.

令$ \alpha:=1-\gamma, \; \beta>(\alpha+2)p^{\prime} $, 其中$ p^{\prime} $是p的共轭指数. 由引理2.4得

$ \begin{equation} \nonumber \begin{aligned} &D^{\alpha}_{t|T}\Phi_{R}(t, x)=\Psi_{R}(x)D^{\alpha}_{t|T}\omega^{\beta}(t)=C\Psi_{R}(x)T^{-\alpha}\omega^{\beta-\alpha}(t), \\ &\partial_{t}D^{\alpha}_{t|T}\Phi_{R}(t, x)=-\Psi_{R}(x)D^{\alpha+1}_{t|T}\omega^{\beta}(t)=-C\Psi_{R}(x)T^{-\alpha-1}\omega^{\beta-\alpha-1}(t), \\ &\partial^{2}_{t}D^{\alpha}_{t|T}\Phi_{R}(t, x)=\Psi_{R}(x)D^{\alpha+2}_{t|T}\omega^{\beta}(t)=C\Psi_{R}(x)T^{-\alpha-2}\omega^{\beta-\alpha-2}(t), \end{aligned} \end{equation} $

其中C为常数. 由于$ \sigma $是整数, 有$ (-\Delta)^{\sigma}\Psi_{R}(x)=R^{-2\sigma}(-\Delta)^{\sigma}\Psi(\frac{x}{R}) $.

由弱解的定义(2.4) 有

$ \begin{equation} \nonumber \begin{aligned} &\Gamma(\alpha)\int_{0}^{T}\int_{B_{R}}J^{\alpha}_{0|t}(|u|^{p})\phi(t, x)dxdt+\int_{B_{R}}u_{1}(x)\phi(0, x)dx \\ =&\int_{0}^{T}\int_{B_{R}}u(\phi_{tt}(t, x)-(-\Delta)^{\sigma}\phi_{t}(t, x)+(-\Delta)^{\sigma}\phi(t, x)-\phi_{t}(t, x))dxdt. \end{aligned} \end{equation} $

令$ I_{R}:=\Gamma(\alpha)\int_{0}^{T}\int_{B_{R}}J^{\alpha}_{0|t}(|u|^{p})\phi(t, x)dxdt=\Gamma(\alpha)\int_{0}^{T}\int_{B_{R}}|u|^{p}\Phi_{R}dxdt $. 下面进行估计.

$ \begin{equation} \begin{aligned} \quad \int_{0}^{T}\int_{B_{R}}u\phi_{tt}dxdt&=\int_{0}^{T}\int_{B_{R}}u\Phi^{\frac{1}{p}}_{R}\Phi^{-\frac{1}{p}}_{R}\phi_{tt}dxdt \\ &\leq \epsilon\int_{0}^{T}\int_{B_{R}}u^{p}\Phi_{R}dxdt+C_{\epsilon}\int_{0}^{T}\int_{B_{R}}\Phi^{-\frac{1}{p-1}}_{R}\phi^{p^{\prime}}_{tt}dxdt\\ &\leq \epsilon I_{R}+C_{\epsilon}T^{-(\alpha+2)p^{\prime}}\int_{0}^{T}\int_{B_{R}}\omega^{-\frac{\beta}{p-1}}(t)\Psi^{-\frac{1}{p-1}}_{R}\Psi^{p^{\prime}}_{R}\omega^{(\beta-\alpha-2)p^{\prime}}(t)dxdt\\ &\leq \epsilon I_{R}+CT^{-(\alpha+2)p^{\prime}+1}R^{n}. \end{aligned} \end{equation} $

(4.22)

$ \begin{equation} \begin{aligned} \int_{0}^{T}\int_{B_{R}}u(-\Delta)^{\sigma}\phi dxdt&=\int_{0}^{T}\int_{B_{R}}u\Phi^{\frac{1}{p}}_{R}\Phi^{-\frac{1}{p}}_{R}(-\Delta)^{\sigma}\phi dxdt \\ &\leq \epsilon I_{R}+C_{\epsilon}T^{-\alpha p^{\prime}}\int_{0}^{T}\int_{B_{R}} R^{-2\sigma p^{\prime}}w^{\beta-\alpha p^{\prime}}(t) (\Psi(\frac{x}{R}))^{-\frac{1}{p-1}}\\ &\quad\times((-\Delta)^{\sigma}\Psi(\frac{x}{R}))^{p^{\prime}}dxdt\\ &\leq \epsilon I_{R}+CT^{-\alpha p^{\prime}+1}R^{n-2\sigma p^{\prime}}. \end{aligned} \end{equation} $

(4.23)

$ \begin{equation} \begin{aligned} -\int_{0}^{T}\int_{B_{R}}u(-\Delta)^{\sigma}\phi_{t} dxdt&=\int_{0}^{T}\int_{B_{R}}u(-\Delta)^{\sigma}D^{\alpha+1}_{t|T}\Phi_{R}dxdt\\ &=\int_{0}^{T}\int_{B_{R}}u\Phi^{\frac{1}{p}}_{R}\Phi^{-\frac{1}{p}}_{R}((-\Delta)^{\sigma}D^{\alpha+1}_{t|T}\Phi_{R}) dxdt \\ &\leq \epsilon I_{R}+C_{\epsilon}T^{-(\alpha+1) p^{\prime}}R^{-2\sigma p^{\prime}}\int_{0}^{T}\int_{B_{R}} w^{\beta-(\alpha+1) p^{\prime}}(t) (\Psi(\frac{x}{R}))^{-\frac{p^{\prime}}{p}}\\ &\quad\times((-\Delta)^{\sigma}\Psi(\frac{x}{R}))^{p^{\prime}}dxdt\\ &\leq \epsilon I_{R}+CT^{1-(\alpha+1) p^{\prime}}R^{n-2\sigma p^{\prime}}. \end{aligned} \end{equation} $

(4.24)

$ \begin{equation} \begin{aligned} \; -\int_{0}^{T}\int_{B_{R}}u\phi_{t} dxdt&=\int_{0}^{T}\int_{B_{R}}uD^{\alpha+1}_{t|T}\Phi_{R}dxdt\\ &\leq \epsilon I_{R}+C_{\epsilon}T^{-(\alpha+1) p^{\prime}} \int_{0}^{T}\int_{B_{R}} w^{(\beta-\alpha-1) p^{\prime}}(t) (\Psi(\frac{x}{R}))^{p^{\prime}} \Phi^{-\frac{1}{p-1}}_{R}dxdt\\ &\leq \epsilon I_{R}+CT^{1-(\alpha+1) p^{\prime}}R^{n}. \end{aligned} \end{equation} $

(4.25)

综合(4.22), (4.23), (4.24), (4.25) 有

$ \begin{equation} I_{R}+\int_{B_{R}}u_{1}(x)\phi(0, x)dx \lesssim 4\epsilon I_{R}+T^{1-\alpha p^{\prime}}R^{n}(T^{-2p^{\prime}}+R^{-2\sigma p^{\prime}}+T^{-p^{\prime}}R^{-2\sigma p^{'}}+T^{-p^{\prime}}), \end{equation} $

(4.26)

令$ \epsilon \in (0, \frac{1}{4}) $, 由(3.5) 以及$ 0\leq \Psi \leq 1 $, 得

$ \begin{equation} \nonumber I_{R} \lesssim T^{1-\alpha p^{\prime}}R^{n}(T^{-2p^{\prime}}+R^{-2\sigma p^{\prime}}+T^{-p^{\prime}}R^{-2\sigma p^{\prime}}+T^{-p^{\prime}}), \end{equation} $

取$ R=T^{\frac{1}{2\sigma}} $, 得

$ \begin{equation} \nonumber I_{R} \lesssim T^{1-(\alpha+1) p^{\prime}+\frac{n}{2\sigma}}, \end{equation} $

事实上, 当$ n>2\sigma\alpha $, 即$ \gamma>1-\frac{n}{2\sigma} $时, $ 1-(\alpha+1) p^{\prime}+\frac{n}{2\sigma}<0 $当且仅当$ p<1+\frac{2\sigma(2-\gamma)}{n-2\sigma(1-\gamma)} $. 由Beppo Levi单调收敛定理, 由于$ T\rightarrow \infty , \; \Phi_{T^{\frac{1}{2\sigma}}}\rightarrow 1 $, 可得

$ \begin{equation} \nonumber \lim\limits_{T\rightarrow \infty}I_{T^{\frac{1}{2\sigma}}}=0, \end{equation} $

因此可得$ u\equiv 0 $, 这与(3.5) 矛盾.

考虑$ \bar{p}:=1+\frac{2\sigma(2-\gamma)}{n-2\sigma(1-\gamma)} $临界情形, 令$ R=T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}}, \; K>1 $. 则由(4.26) 有

$ \begin{equation} \nonumber (1-4\epsilon)\int_{0}^{T}\int_{B_{T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}}}}|u|^{p}\Phi_{R}dxdt\leq \bar{C}K^{-\frac{n}{2\sigma}}(T^{-\bar{p}^{\prime}}+K^{\bar{p}^{\prime}}+T^{-\bar{p}^{\prime}}K^{\bar{p}^{\prime}}+1). \end{equation} $

因此, 令$ T\rightarrow \infty $, 对于某些$ C(K)>0 $有

$ \begin{equation} \int_{0}^{\infty}\int_{\mathbb{R}^{n}}|u|^{p}dxdt=\lim\limits_{T\rightarrow \infty}\int_{0}^{T}\int_{B_{T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}}}}|u|^{p}\Phi_{T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}}}dxdt\leq C(K). \end{equation} $

(4.27)

由于对于任意的正整数$ \theta $, 当$ |x|\leq \frac{R}{2} $时, $ (-\Delta)^{\theta}\Psi_{R}(x)=0.\ $与非临界情形证明类似, 当$ R=T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}} $时, 有

$ \begin{equation} \int_{0}^{T}\int_{\mathbb{R}^{n}}u\phi_{tt}dxdt\leq \epsilon\int_{0}^{T}\int_{\mathbb{R}^{n}}u^{\bar{p}}\Phi_{R}dxdt+C_{\epsilon}K^{-\frac{n}{2\sigma}}T^{-\bar{p}^{\prime}}, \end{equation} $

(4.28)

$ \begin{equation} \int_{0}^{T}\int_{\mathbb{R}^{n}}u\phi_{t}dxdt\leq \epsilon\int_{0}^{T}\int_{\mathbb{R}^{n}}u^{\bar{p}}\Phi_{R}dxdt+C_{\epsilon}K^{-\frac{n}{2\sigma}}, \end{equation} $

(4.29)

以及

$ \begin{equation} \nonumber \begin{aligned} \int_{0}^{T}\int_{\mathbb{R}^{n}}u(-\Delta)^{\sigma}\phi dxdt&=\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}u(-\Delta)^{\sigma}\phi dxdt\\ &\leq (\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}((-\Delta)^{\sigma}\phi)^{\bar{p}^{\prime}}\Phi^{-\frac{1}{\bar{p}-1}}_{R}dxdt)^{\frac{1}{\bar{p}^{\prime}}} \\ &\quad\times (\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}u^{\bar{p}}\Phi_{R}dxdt)^{\frac{1}{\bar{p}}}, \end{aligned} \end{equation} $

其中

$ \begin{equation} \int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}((-\Delta)^{\sigma}\phi)^{\bar{p}^{\prime}}\Phi^{-\frac{1}{\bar{p}-1}}_{R}dxdt\lesssim K^{-\frac{n}{2\sigma}+\bar{p}^{\prime}}. \end{equation} $

(4.30)

同理有

$ \begin{equation} \nonumber \begin{aligned} \int_{0}^{T}\int_{\mathbb{R}^{n}}u(-\Delta)^{\sigma}\phi_{t} dxdt&=\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}u(-\Delta)^{\sigma}\phi_{t} dxdt\\ &\leq (\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}((-\Delta)^{\sigma}\phi_{t})^{\bar{p}^{\prime}}\Phi^{-\frac{1}{\bar{p}-1}}_{R}dxdt)^{\frac{1}{\bar{p}^{\prime}}} \\ &\quad\times (\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}u^{\bar{p}}\Phi_{R}dxdt)^{\frac{1}{\bar{p}}}, \end{aligned} \end{equation} $

其中

$ \begin{equation} \int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}((-\Delta)^{\sigma}\phi_{t})^{\bar{p}^{\prime}}\Phi^{-\frac{1}{\bar{p}-1}}_{R}dxdt\lesssim K^{-\frac{n}{2\sigma}+\bar{p}^{\prime}}T^{-\bar{p}^{\prime}}. \end{equation} $

(4.31)

由(4.28), (4.20), (4.30), (4.31) 有

$ \begin{equation} \begin{aligned} (1-2\epsilon)\int_{0}^{T}\int_{\mathbb{R}^{n}}u^{\bar{p}}\Phi_{R}dxdt &\leq CK^{-\frac{n}{2\sigma}}(T^{-\bar{p}^{\prime}}+1)+K^{-\frac{n}{2\sigma\bar{p}^{\prime}}+1}(1+T^{-1})\\ &\quad\times(\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}u^{\bar{p}}\Phi_{R}dxdt)^{\frac{1}{\bar{p}}}. \end{aligned} \end{equation} $

(4.32)

由(4.27) 有$ u\in L^{p} $, 从而对于任意固定的$ K>0 $, 有

$ \begin{equation} \nonumber \lim\limits_{T\rightarrow \infty}\int_{0}^{T}\int_{|x|>(T^{\frac{1}{2\sigma}}K^{-\frac{1}{2\sigma}})/2}u^{\bar{p}}\Phi_{R}dxdt=0, \end{equation} $

从而当$ T\rightarrow \infty $, 由(4.32) 可得

$ \begin{equation} \int_{0}^{\infty}\int_{\mathbb{R}^{n}}u^{\bar{p}}dxdt\leq CK^{-\frac{n}{2\sigma}}. \end{equation} $

(4.33)

当$ K\rightarrow \infty $时有$ u\equiv 0 $, 这与(3.5) 矛盾. 综上, 爆破结论得证.

当$ u_{1}\in L^{1}\cap L^{2} $, 初值给定为$ (0, \varepsilon u_{1}) $时, 重复上述爆破证明步骤, 对于$ (\alpha+1)p^{\prime}>1+\frac{n}{2\sigma} $, 令$ k=(\alpha+1)p^{\prime}-1-\frac{n}{2\sigma} $, 由于

$ \begin{equation} \nonumber I_{R}+C\varepsilon\lesssim C\tau^{-(\alpha+1)p^{\prime}+1+\frac{n}{2\sigma}}=C\tau^{-k}, \\ \end{equation} $

则有

$ \begin{equation} \nonumber \begin{aligned} \varepsilon \leq C\tau^{-k}, \quad \tau\leq C\varepsilon^{-\frac{1}{k}}, \end{aligned} \end{equation} $

从而得到生命跨度估计

$ \begin{equation} \nonumber T_{\varepsilon}\leq C\varepsilon^{-\frac{1}{k}}, \end{equation} $

其中$ C>0 $,

$ \begin{equation} \nonumber k=\frac{(2-\gamma)p}{p-1}-\frac{2\sigma+n}{2\sigma}>0. \end{equation} $

证明定理3.3    当$ \sigma=\frac{n}{4} $时, 定义空间范数为

$ \begin{equation} \nonumber \|u\|_{X(t)}=\sup\limits_{0\leq \tau \leq t}((1+\tau)^{\gamma}\log^{-1}(2+\tau)\|u(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\gamma}\||D|^{\sigma}u(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\gamma}\|\partial_{t}u(\tau, \cdot)\|_{L^{2}}). \end{equation} $

当$ \sigma \in (\frac{n}{4}, \frac{n}{2}) $时, 定义空间范数为

$ \begin{equation} \nonumber \|u\|_{X(t)}=\sup\limits_{0\leq \tau \leq t}((1+\tau)^{\frac{n}{4\sigma}+\gamma-1}\|u(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\gamma}\||D|^{\sigma}u(\tau, \cdot)\|_{L^{2}}+(1+\tau)^{\gamma}\|\partial_{t}u(\tau, \cdot)\|_{L^{2}}). \end{equation} $

由线性问题解的衰减估计, 显然$ u^{ln} $属于$ X(t) $. 式(4.7) 的证明方法步骤与定理3.1中的证明类似, 只是在使用引理2.2时有不同. 当$ \sigma \in (\frac{n}{4}, \frac{n}{2}) $时, 此时G-N不等式变为

$ \begin{equation} \nonumber \begin{aligned} \|u\|^{p}_{L^{sp}}&\lesssim (1+\tau)^{-\frac{n}{2\sigma}(p-\frac{1}{s})-(\gamma-1)p+(\frac{n}{4\sigma}-\frac{1}{2})\frac{n}{\sigma}(\frac{p}{2}-\frac{1}{s})}\|u\|^{p}_{X(t)}, \end{aligned} \end{equation} $

其中$ s=1, 2.\; \theta\in [0, 1] $且

$ \begin{equation} \nonumber \theta=\frac{n}{\sigma}(\frac{1}{2}-\frac{1}{sp}). \end{equation} $

从而得

$ \begin{equation} \begin{aligned} \|u^{p}\|_{L^{1}\cap L^{2}}&\lesssim \|u\|^{p}_{L^{p}}+\|u\|^{p}_{L^{2p}}\\ &\lesssim (1+\tau)^{-\frac{n}{2\sigma}(p-1)-(\gamma-1)p+ (\frac{n}{4\sigma}-\frac{1}{2})\frac{n}{\sigma}(\frac{p}{2}-1)}\|u\|^{p}_{X(t)}\\ &\quad+(1+\tau)^{-\frac{n}{2\sigma}(p-\frac{1}{2})-(\gamma-1)p+(\frac{n}{4\sigma}-\frac{1}{2})\frac{n}{\sigma}(\frac{p}{2}-\frac{1}{2})}\|u\|^{p}_{X(t)}. \end{aligned} \end{equation} $

(4.34)

此时令$ \beta_{\gamma}(n, \sigma, p):=\frac{n}{2\sigma}(p-1)+(\gamma-1)p-(\frac{n}{4\sigma}-\frac{1}{2})\frac{n}{\sigma}(\frac{p}{2}-1), \; \beta^{'}_{\gamma}(n, \sigma, p):=\frac{n}{2\sigma}(p-\frac{1}{2})+(\gamma-1)p-(\frac{n}{4\sigma}-\frac{1}{2})\frac{n}{\sigma}(\frac{p}{2}-\frac{1}{2}) $, 相应地, 由引理2.1可得如下估计

$ \begin{equation} \|u^{nl}\|_{L^{2}}\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}}\int_{0}^{\tau}(\tau-s)^{-\gamma}\|u^{p}(\tau, \cdot)\|_{L^{1}\cap L^{2}}dsd\tau, \end{equation} $

(4.35)

$ \begin{equation} \||D|^{\sigma}u^{nl}\|_{L^{2}}\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-\frac{1}{2}}\int_{0}^{\tau}(\tau-s)^{-\gamma}\|u^{p}(\tau, \cdot)\|_{L^{1}\cap L^{2}}dsd\tau, \end{equation} $

(4.36)

$ \begin{equation} \|\partial_{t}u^{nl}\|_{L^{2}}\lesssim \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4\sigma}-1}\int_{0}^{\tau}(\tau-s)^{-\gamma}\|u^{p}(\tau, \cdot)\|_{L^{1}\cap L^{2}}dsd\tau. \end{equation} $

(4.37)

令$ \beta_{\gamma}(n, \sigma, p)>1, \; \beta^{'}_{\gamma}(n, \sigma, p)>1 $, 即得

$ \begin{equation} p>\frac{1+\frac{n}{\sigma}-\frac{n^{2}}{4\sigma^{2}}}{\frac{3n}{4\sigma}+\gamma-1-\frac{n^{2}}{8\sigma^{2}}}, \end{equation} $

(4.38)

由上述$ \theta $取值范围, 当$ \sigma \in (\frac{n}{4}, \frac{n}{2}) $时有

$ \begin{equation} 2\leq p\leq\frac{n}{n-2\sigma}. \end{equation} $

(4.39)

要让(4.38) 同时满足(4.39), 可得$ \gamma $范围为

$ \begin{equation} \nonumber \gamma\in \Bigg(\max \left\{1-\frac{n}{4\sigma}, \frac{n}{\sigma}-\frac{2\sigma}{n}-\frac{\frac{n^{2}}{\sigma^{2}}-\frac{2n}{\sigma}}{4}+\frac{\frac{n^{3}}{\sigma^{3}}-\frac{8n^{2}}{\sigma^{2}}+\frac{12n}{\sigma}}{\frac{8n}{\sigma}-16}\right\}, \frac{3}{2}-\frac{n}{4\sigma}\Bigg). \end{equation} $

此时将(4.34) 带入(4.35), (4.36) 和(4.37), 再利用引理2.3, 可得如下估计

$ \begin{equation} \begin{aligned} \|u^{nl}\|_{L^{2}}&\lesssim (1+t)^{1-\gamma-\frac{n}{4\sigma}}\|u\|^{p}_{X(t)}, \end{aligned} \end{equation} $

(4.40)

$ \begin{equation} \begin{aligned} \||D|^{\sigma}u^{nl}\|_{L^{2}}&\lesssim (1+t)^{-\gamma}\|u\|^{p}_{X(t)}, \end{aligned} \end{equation} $

(4.41)

$ \begin{equation} \begin{aligned} \|\partial_{t}u^{nl}\|_{L^{2}}&\lesssim (1+t)^{-\gamma}\|u\|^{p}_{X(t)}. \end{aligned} \end{equation} $

(4.42)

从而可得

$ \begin{equation} \nonumber \|u^{nl}\|_{X(t)}\lesssim \|u\|^{p}_{X(t)}. \end{equation} $

式(4.5) 的证明与定理3.1中证明类似, 得到(4.4) (4.5) 后再运用不动点定理可得解的整体存在唯一性. $ \sigma=\frac{n}{4} $时的证明也与定理3.1中证明类似. 得到$ \sigma\in [\frac{n}{4}, \frac{n}{2}) $时衰减估计如下

$ \begin{equation} \nonumber \|u\|_{L^{2}} \lesssim \left\{\begin{aligned} &(1+t)^{-\frac{n}{4\sigma}-\gamma+1}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \qquad\; \; \; \sigma\in (\frac{n}{4}, \frac{n}{2}), \\ &(1+t)^{-\gamma}\log(2+t)\|(u_{0}, u_{1})\|_{\mathcal{A}} , \quad \sigma=\frac{n}{4}, \\ \end{aligned} \right. \end{equation} $

$ \begin{equation} \nonumber \||D|^{\sigma}u\|_{L^{2}}\lesssim (1+t)^{-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}, \end{equation} $

$ \begin{equation} \nonumber \|\partial_{t}u\|_{L^{2}}\lesssim (1+t)^{-\gamma}\|(u_{0}, u_{1})\|_{\mathcal{A}}. \end{equation} $

注4.1    定理$ 3.3 $中$ \gamma $范围为

$ \begin{equation} \nonumber \gamma\in \Bigg(\max \left\{1-\frac{n}{4\sigma}, \frac{n}{\sigma}-\frac{2\sigma}{n}-\frac{\frac{n^{2}}{\sigma^{2}}-\frac{2n}{\sigma}}{4}+\frac{\frac{n^{3}}{\sigma^{3}}-\frac{8n^{2}}{\sigma^{2}}+\frac{12n}{\sigma}}{\frac{8n}{\sigma}-16}\right\}, \frac{3}{2}-\frac{n}{4\sigma}\Bigg). \end{equation} $

通过计算可知, 在$ \sigma \in (\frac{n}{4}, \frac{n}{2}) $上$ \gamma $的范围非空.