本文考虑如下带有对数阻尼项波动方程的柯西问题

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}+Lu+\left( I+L \right) ^{-1}u_t=\left| u \right|^p, \; \; \; \; \; \; \; \; \; \; \left( t, x \right) \in \left( 0, \infty \right) \times\mathbb{R}^n, \\ &u(0, x)=u_{0}(x), \; \; \; \; \; u_{t}(0, x)=u_{1}(x), \; \; \; \; x\in\mathbb{R}^n, \end{aligned} \right. \end{equation} $

(1.1)

算子

$ \begin{equation} L=\log(I+(-\Delta)^{\sigma}), \sigma\in R, \end{equation} $

(1.2)

其通过如下Fourier逆变换定义

$ \begin{equation} (Lf)(x)= \mathcal{F}_{\xi \rightarrow x}^{-1}\left( \log \left( 1+|\xi |^{2\sigma} \right) \widehat{f}\left( \xi \right) \right) \left( x \right), \end{equation} $

(1.3)

这里

$ \begin{equation} \widehat{f}\left( \xi \right) =\displaystyle{\int}_{\mathbb{R}^n}{e^{-ix\cdot \xi}}f\left( x \right) dx, \; \; \; \xi \in\mathbb{R}^n, \end{equation} $

(1.4)

问题(1.1)可以看成某种带有结构阻尼机制的类波方程, 下面先回顾已有波动方程的研究结果. Strauss^[1]研究了如下经典波动方程的柯西问题

$ \begin{equation} \left\{ \begin{array}{l} u_{tt}-\varDelta u=|u|^p, \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; t>0, x\in\mathbb{R}^n, \\ u\left( 0, x \right) =u_0\left( x \right) , u_t\left( 0, x \right) =u_1\left( x \right) , \ x\in\mathbb{R}^n, \\ \end{array} \right. \end{equation} $

(1.5)

得到临界指数$ p_c $, $ p_c $为$ \left( n-1 \right) p^2-\left( n-1 \right) p-2=0, n\geqslant 2 $的正根. 若$ p>p_c $时, 问题(1.5)的解在小初值情形下整体存在, 若$ 1<p<p_c $问题(1.5)的解在有限时刻爆破. 文^[2]在超临界情形下证明问题(1.5)解的整体解存在, 文^{[3, 4]}在次临界情形下证明问题(1.5)的解在有限时刻爆破.

随后, 一些学者研究带阻尼项的波动方程^{[5, 6]}. 文^[5]研究如下半线性阻尼波动方程的柯西问题

$ \begin{equation} \left\{ \begin{aligned} &u_{tt}-\Delta u+u_t=|u|^p, \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; t> 0, x\in\mathbb{R}^n, \\ &u\left( 0, x \right) =u_0\left( x \right), \; \; u_t\left( 0, x \right) =u_1\left( x \right), \; \; \; \; x\in\mathbb{R}^n, \\ \end{aligned} \right. \end{equation} $

(1.6)

得到了临界指数

$ p_c=1+\frac{2}{n}. $

若初值足够小且具有紧支集, 非线性指数$ p $满足如下条件时

$ \left\{ \begin{array}{l} p_c<p<\infty , \ \; \left( n=1, 2 \right), \\ p_c<p<\frac{n}{n-2}, \ \left( n\geqslant3 \right), \\ \end{array} \right. $

问题$ (1.6) $的解整体存在. 文^[6]研究次临界时, 若$ \int_{R^n} u_0(x)dx>0 $, 问题(1.6)的解在有限时刻爆破.

近年来, 许多学者将问题(1.6)推广到带有分数阶结构阻尼的波动方程. 文^[7]研究如下分数阶波动方程的柯西问题

$ \begin{equation} \left\{ \begin{array}{l} u_{tt}+\left( -\bigtriangleup \right) ^{\sigma}u+\left( -\bigtriangleup \right) ^{\delta}u_t=|u|^p, \ \ \ \ t> 0, x\in\mathbb{R}^n , \\ \left( u, u_t \right) \left( 0, x \right) =\left( u_0, u_1 \right) \left( x \right), \; \; \; \; \; \; \; \; \; \; \; \; x\in\mathbb{R}^n, \end{array} \right. \end{equation} $

(1.7)

得到了临界指数

$ p_c=1+\frac{2\sigma}{\left( n-2\delta \right) _+}. $

在假设小初值情形下, 通过整体迭代法证明了当$ p>p_c $时, 问题$ (1.7) $的解整体存在, 利用检验函数法证明了$ 0<p<p_c $问题$ (1.7) $的解在有限时刻爆破.

Charao^[8]在波动方程中引入了对数型阻尼项, 研究如下具有对数阻尼项波动方程的柯西问题

$ \begin{equation} \begin{aligned} \left\{ \begin{array}{l} u_{tt}-\varDelta u+\log ( I+\left( -\varDelta \right) ^{\theta} ) u_t=0, \ \left( t, x \right) \in \left( 0, \infty \right) \times \mathbb{R}^n, \\ u\left( 0, x \right) =0, u_t\left( 0, x \right) =u_1\left( x \right) , \; \; \; \; \; \; \; x\in \mathbb{R}^n, \\ \end{array} \right. \end{aligned} \end{equation} $

(1.8)

得到了问题$ (1.8) $解的渐近性态, 以及解在$ L^2 $意义下的最优估计.

文^[9]将问题$ (1.8) $推广到带对数阻尼机制的对数型类波方程

$ \begin{equation} \begin{aligned} \left\{ \begin{array}{l} u_{tt}+Lu+\left( I+L \right) ^{-1}u_t=0, \; \; \; \; \; \; \; \; \; \; \; \; ( t, x ) \in \left( 0, \infty \right) \times\mathbb{R}^n, \\ u\left( 0, x \right) =u_0\left( x \right), u_t\left( 0, x \right) =u_1\left( x \right), \ \ \ \ x\in\mathbb{R}^n, \\ \end{array} \right. \end{aligned} \end{equation} $

(1.9)

其中$ L=\log(I-\Delta) $, 问题(1.9)可以看成是如下模型的推广

$ \begin{equation} \begin{aligned} u_{tt}-\varDelta u_{tt}-\varDelta u+\varDelta ^2u+u_t=0, \end{aligned} \end{equation} $

(1.10)

得到了其在$ L^2 $框架下解的渐近性态, 并应用它来研究关于$ L^2 $范数的解的最优衰减率.

基于以上文献的启发, 本文在问题$ (1.9) $基础上考虑非线性问题$ (1.1) $, 研究是否可以得到一个临界指数$ p_c $, 使得$ p>p_c $时问题$ (1.1) $的解整体存在, $ 0<p<p_c $时问题$ (1.1) $的解有限时刻爆破. 本文的难点在于合理处理对数阻尼项的估计. 本文结构安排如下: 第二节引入相关符号以及证明文章主要结论所需公式引理; 第三节证明问题$ (1.1) $解的整体存在性; 第四节证明问题$ (1.1) $解的爆破. 下面给出本文的主要结论.

定理1.1 假设$ m\in[1, 2) $, $ \max\{\sigma, \; n(\frac{1}{m}-\frac{1}{2})+2k\}\leqslant l <k+\sigma $. 若

$ \frac{2}{m}\leqslant p\leqslant \left\{ \begin{array}{l} \; \; \; \; \; \infty \ \; \; \; \; \; \; n\leqslant \frac{4k}{2-m}, \\ \frac{n-2\left( l-\sigma \right)}{n-2k}\; \; \; \; n>2k, \\ \end{array} \right. $

指数$ p $满足

$ p>1+\frac{2\sigma m}{n}, $

初值$ (u_0, u_1)\in \mathcal{A}= (L^m\cap \dot{H}^{k+l})\times (L^m\cap \dot{H}^{k+l-\sigma}) $, 且存在一个足够小的常数$ \varepsilon $使得

$ \lVert ( u_0, u_1) \rVert _{\mathcal{A}}=\lVert u_0 \rVert _{L^m}+\lVert u_0 \rVert _{\dot{H}^{k+l}}+\lVert u_1 \rVert _{L^m}+\lVert u_1 \rVert _{\dot{H}^{k+l-\sigma}} \leqslant \varepsilon, $

问题$ (1.1) $存在唯一的整体解

$ u\in C([0, \infty), \dot{H}^k), $

且有如下衰减估计

$ \lVert u \rVert _{ \dot{H}^{k}}\lesssim \left( 1+t \right)^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}(\frac{1}{m}-\frac{1}{2})} \lVert (u_0, u_1) \rVert _{\mathcal{A}} . $

定理1.2 假设$ \sigma\geqslant 1 $, $ m\in[1, 2) $, $ 0<k<\sigma $, 当$ n\leqslant2k, \; p>1 $或$ n>2k, p\leqslant \frac{n}{n-2k} $时, 设初值$ (u_0, u_1)\in(L^m\cap \ {L}^{2})\times (L^m\cap {L}^{2}) $, 满足$ \int_{\mathbb{R}^n} u_0(x)dx\geqslant 0 $且

$ \displaystyle{\int}_{\mathbb{R}^n} u_1(x)dx>0, m=1, $

或

$ u_1(x)\gtrsim|x|^{-\frac{n}{m}}(\log(1+|x|))^{-1}, \; |x|\rightarrow \infty, \; m\in(1, 2), $

若$ 1<p<1+\frac{2\sigma m}{n} $时, 问题$ (1.1) $的解在有限时刻爆破.

本文引入如下记号.

(1) 齐次Sobolev空间

$ \dot {H}^{s}_{p}:=\{f\in S^{'}:\lVert \mathcal{F}^{-1}( |\xi |^s\widehat{f} ) \rVert _{L^p}< \infty , s\geqslant 0\}. $

(2) $ f\lesssim g $表示存在一个常数$ C $使得$ f\leqslant Cg $.

下面通过引理2.1给出$ \rho(\xi) $的定义.

引理2.1 设$ \delta _0\in ( 0, 1 ) $, 令$ \varphi(\xi) =\log( 1+|\xi |^{2\sigma})( 1+\log( 1+|\xi |^{2\sigma} )) $, $ \psi(\xi ) =\dfrac{1}{1+\log ( 1+|\xi |^{2\sigma})} $, $ \phi (\xi ) =\sqrt{\log ( 1+| \xi|^{2\sigma})} $, 则

(1) 若$ |\xi|\leqslant \delta _0 $, 则$ \varphi ( \xi ) \leqslant \psi ( \xi ) , \varphi ( \xi ) \leqslant \phi ( \xi ). $

(2) 若$ | \xi |\geqslant \delta _0 $, 则$ \psi ( \xi) \leqslant \varphi ( \xi ) , \psi ( \xi ) \leqslant \phi ( \xi ). $

证 当$ r\geqslant0 $时, 设$ \theta ( r) =( 1+\log( 1+r^{2\sigma} ) ) \sqrt{\log ( 1+r^{2\sigma} )}-1. $

由于$ \theta(0)=-1 $和$ \theta(1)=(1+\log2)\sqrt{2}-1>0 $, $ \theta(r) $关于$ r $连续, 存在$ 0<\delta_{0}<1 $使得$ \theta(\delta_{0})=0 $.

当$ 0\leqslant r\leqslant\delta_{0} $时, 有$ (1+\log(1+r^{2\sigma}))\sqrt{\log(1+r^{2\sigma})}\leqslant1, $即$ \varphi ( r) \leqslant\phi (r)\leqslant\psi(r ). $

当$ r\geqslant \delta_{0} $时, 有$ (1+\log(1+r^{2\sigma}))\sqrt{\log(1+r^{2\sigma})}\geqslant 1, $即$ \varphi (r )\geqslant \phi (r )\geqslant \psi(r ). $综上, 证毕.

对于$ \delta_{0}>0 $, 当$ \xi\in{\mathbb{R}^n} $定义如下函数

$ \begin{equation} \rho ( \xi ) =\left\{ \begin{array}{l} \frac{1}{2}\log( 1+| \xi|^{2\sigma})( 1+\log( 1+|\xi |^{2\sigma} ) ), \; \; |\xi |\leqslant \delta _0, \\ \dfrac{1}{2( 1+\log( 1+| \xi |^{2\sigma}) )}, \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; | \xi |> \delta _0.\\ \end{array} \right. \end{equation} $

(2.1)

设$ G(t, \xi), \; H(t, \xi) $是下面线性问题的基本解

$ \begin{equation} \begin{aligned} \left\{ \begin{array}{l} \left( I+\log \left( I+\left( -\varDelta \right) ^{\sigma} \right) \right) G_{tt}+\log \left( I+\left( -\varDelta \right) ^{\sigma} \right) \left( I+\log \left( I+\left( -\varDelta \right) ^{\sigma} \right) \right) G+G_t=0, \\ G\left( 0, x \right) =\delta \left( x \right), \\ G_t\left( 0, x \right) =0.\\ \end{array} \right. \\ \left\{ \begin{array}{l} \left( I+\log \left( I+\left( -\varDelta \right) ^{\sigma} \right) \right) H_{tt}+\log \left( I+\left( -\varDelta \right) ^{\sigma} \right) \left( I+\log \left( I+\left( -\varDelta \right) ^{\sigma} \right) \right) H+H_t=0, \\ H\left( 0, x \right) =0, \\ H_t\left( 0, x \right) =\delta \left( x \right).\\ \end{array} \right. \end{aligned} \end{equation} $

(2.2)

其中$ \delta(x) $是Dirichlet函数. $ (2.2) $式在Fourier空间中的解可以写为

$ \begin{equation} \begin{aligned} \widehat{G}(t, \xi)=\mathcal{L}^{-1}[\dfrac{\lambda}{1+\log(1+|\xi|^{2\sigma})\lambda^2+\lambda+\log(1+|\xi|^{2\sigma})(1+\log(1+|\xi|^{2\sigma})))}], \\ \widehat{H}(t, \xi)=\mathcal{L}^{-1}[\dfrac{1}{1+\log(1+|\xi|^{2\sigma})\lambda^2+\lambda+\log(1+|\xi|^{2\sigma})(1+\log(1+|\xi|^{2\sigma})))}]. \end{aligned} \end{equation} $

(2.3)

(2.3)式由Laplace逆变换定义. 由于在此过程中并未考虑到分母的零点, 下面证明算子$ \widehat G(t, \xi), \; \widehat H(t, \xi) $存在.

引理2.2 $ \widehat G(t, \xi), \widehat H(t, \xi) $存在.

证 定义

$ F(\lambda)=(1+\log(1+|\xi|^{2\sigma}))\lambda^2+\lambda+\log(1+|\xi|^{2\sigma})(1+\log(1+|\xi|^{2\sigma}). $

为了证明$ \mathcal{L}^{-1}[{\lambda}/{F(\lambda)}] $存在, 我们需要考虑$ F(\lambda) $的零点. 若$ \lambda_{1}=\sigma_{1}+i\nu_1 $是$ F(\lambda) $的一个零点, 即$ \sigma_{1} $和$ \nu_1 $满足

$ \begin{equation} \begin{aligned} {ReF}(\lambda) &=({\sigma }_1^2-{\nu }_{1}^{2}) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) +{\sigma }_1+\log ( 1+| \xi |^{2\sigma}) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) =0, \\ {ImF}(\lambda) &=2{\sigma }_1{\nu }_1( 1+\log ( 1+| \xi |^{2\sigma} ) ) +{\nu }_1=0. \end{aligned} \end{equation} $

(2.4)

当$ |\xi|=0 $时, $ \sigma_{1}=\nu_{1}=0 $(显然), 当$ |\xi|\ne0 $时, $ \sigma_{1}<0 $. (反证法) 若$ \sigma _1\geqslant0 $, $ \nu_{1}=0. $

$ ReF(\lambda_{1})={\sigma }_1^2 ( 1+\log ( 1+| \xi |^{2\sigma} ) ) +{\sigma }_1+\log ( 1+| \xi |^{2\sigma}) ( 1+\log ( 1+| \xi |^{2\sigma} ) )\geqslant0, $

与$ ReF(\lambda_{1})=0 $矛盾. 若$ \sigma _1\geqslant0 $, $ \nu_{1}\ne0. $

$ ImF(\lambda_{1})=\nu_{1}(2\sigma_{1}(1+\log(1+|\xi|^{2\sigma}))+1)\ne0, $

与$ ImF(\lambda_{1})=0, $矛盾. 因此$ \sigma_{1}<0 $.

综上, 当$ |\xi|=0 $时, $ {\lambda}/{F(\lambda)} $在$ {\lambda \in \mathbb{C}, Re(\lambda)>0} $解析. 当$ |\xi|\ne0 $时, $ {\lambda}/{F(\lambda)} $在$ {\lambda \in \mathbb{C}, Re(\lambda)\geqslant0} $解析, 若$ \lambda=\sigma+iv, \; \sigma>max\{Re \lambda_s\} $, 其中$ \lambda_{s} $是$ F(\lambda) $所有奇异点. 有

$ \begin{align*} \mathcal{L}^{-1}\mathcal[\dfrac{\lambda}{F(\lambda)}]= \int_{\sigma -i\infty}^{\sigma +i\infty}{\dfrac{\lambda e^{\lambda t}}{F\left( \lambda \right)}} d \lambda =\int_{-\infty}^{+\infty}{\dfrac{i\left( \sigma +i\nu \right) e^{\left( \sigma +i\nu \right) t}}{F\left( \sigma +i\nu \right)}} d\nu =\int_{\left\{ \nu ;\left| \nu \right|\leqslant R \right\}}{+\int_{\left\{ \nu ;\left| \nu \right|>R \right\}}{=J_{1}+J_{2}}}, \end{align*} $

易知$ J_{1} $收敛, 下面我们考虑$ J_{2} $, 由

$ \dfrac{\lambda}{F(\lambda)}=\dfrac{1}{\lambda}-\dfrac{\log(1+|\xi|^{2\sigma})(1+\log(1+\left|\xi\right|^{2\sigma}))}{\lambda F(\lambda)}, $

知$ J_{2} $收敛, 所以$ \widehat G(t, \xi) $存在. 同理得$ \widehat H(t, \xi) $存在.

由Duhamel原理, 问题$ (1.1) $的解可以表示为

$ \begin{equation} u=G(t, x)\ast u_{0}(x)+H(t, x)\ast u_1(x)+\displaystyle{\int}_{0}^{t}H(t-\tau, x)\ast|u|^pd\tau, \end{equation} $

(2.5)

其线性问题的解为

$ \begin{equation} u^{ln}=G(t, x)\ast u_{0}(x)+H(t, x)\ast u_1(x). \end{equation} $

(2.6)

为得到算子$ \widehat G(t, \xi) $, $ \widehat H(t, \xi) $的估计, 先利用能量法得到问题$ (1.1) $的解在Fourier空间中的逐点估计. 下面先对其线性问题进行Fourier变换, 得

$ \begin{equation} \left\{ \begin{array}{l} \left( 1+\log \left( 1+|\xi |^{2\sigma} \right) \right) \widehat{u}_{tt}+\log \left( 1+|\xi |^{2\sigma} \right) \left( 1+\log \left( 1+|\xi |^{2\sigma} \right) \right) \widehat{u}+\widehat{u}_t=0, t>0, \xi\in\mathbb{R}^n, \\ \widehat{u}\left( 0, \xi \right) =\widehat{u}_0\left( \xi \right), \; \; \; \; \; \; \widehat{u}_t\left( 0, \xi \right) =\widehat{u}_1\left( \xi \right), \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \xi\in\mathbb{R}^n. \end{array} \right. \end{equation} $

(2.7)

引理2.3 若$ t>0, \; \xi \in R_{\xi}^n $, 假设$ u(t, x) $是问题$ (1.1) $的解, 则

$ \begin{equation} \begin{aligned} &\; \; \; \; \; (1+\log(1+|\xi|^{2\sigma}))|\widehat{u_{t}}(t, \xi)|^2+\log(1+|\xi|^{2\sigma})(1+\log(1+|\xi|^{2\sigma}))|\widehat{u}(t, \xi)|^2 \\ &\lesssim(1+\log(1+|\xi|^{2\sigma}))e^{-\frac{\rho(\xi)}{2}t}|\widehat{u_1}(t, \xi)|^2+\log(1+|\xi|^{2\sigma})(1+\log(1+|\xi|^{2\sigma}))e^{-\frac{\rho(\xi)}{2}}|\widehat{u}_0(\xi)|^2. \end{aligned} \end{equation} $

(2.8)

证 $ (2.7) $式两边同乘$ \overline{\widehat{u_t}} $得

$ \begin{equation} \dfrac{d}{dt}E^{'}(t, \xi)+| \widehat{u_{t}}|^2=0, \end{equation} $

(2.9)

其中

$ \begin{equation} E^{'}( t, \xi) =\frac{1}{2}(1+\log(1+|\xi|^{2\sigma}))|\widehat{u}_t|+\frac{1}{2}\log(1+|\xi |^{2\sigma})( 1+\log(1+|\xi|^{2\sigma}))|\widehat{u}|^2. \end{equation} $

(2.10)

$ (2.7) $式两边同时乘$ \rho(\xi) $得

$ \begin{equation} \rho ( \xi ) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) \widehat{u}_{tt}\overline{\widehat{u}}+\rho ( \xi ) \log ( 1+| \xi |^{2\sigma}) ( 1+\log ( 1+| \xi|^{2\sigma} ) ) | \widehat{u} |^2+\frac{1}{2}\rho ( \xi )\frac{d}{dt}| \widehat{u} |^2=0, \end{equation} $

(2.11)

取实部得

$ \begin{equation} \begin{aligned} &\; \; \; \; \; \; \frac{d}{dt}[ \rho ( \xi) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) \text{Re}( \widehat{u}_t\overline{\widehat{u}} ) +\frac{1}{2}\rho ( \xi )\log ( 1+| \xi |^{2\sigma} ) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) ] \\&\; \; \; \; \; \; \; \; \; \; \; +\rho ( \xi ) \log ( 1+| \xi |^{2\sigma} ) ( 1+\log ( 1+| \xi |^{2\sigma} ) )| \widehat{u} |^2\\ &=\rho ( \xi ) ( 1+\log ( 1+| \xi|^{2\sigma} ) ) | \widehat{u}_t |^2. \end{aligned} \end{equation} $

(2.12)

定义如下函数

$ \begin{equation} \begin{aligned} E( t, \xi ) &=E^{'}( t, \xi ) +\rho( \xi )( 1+\log ( 1+| \xi |^{2\sigma}) ) {Re}( \widehat{u}_t\overline{\widehat{u}} ) +\frac{1}{2}\rho ( \xi)| \widehat{u}|^2, \\ F(t, \xi) &=|\widehat{u}_t|^2+\rho(\xi)\log(1+|\xi|^{2\sigma})(1+\log(1+|\xi |^{2\sigma})) |\widehat{u}|^2, \\ R(t, \xi) &=\rho ( \xi ) ( 1+\log ( 1+| \xi|^{2\sigma} )) | \widehat{u}_t |^2. \end{aligned} \end{equation} $

(2.13)

得

$ \begin{equation} \frac{d}{dt}E( t, \xi ) +F( t, \xi ) =R( t, \xi ), \ \ \ \ \ t>0, \; \xi \in R_{\xi}^{n}. \end{equation} $

(2.14)

下面估计$ E(t, \xi) $

$ \begin{align*} E( t, \xi ) &\leqslant E^{'}( t, \xi ) +\rho ( \xi ) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) | \widehat{u_t}|| \widehat{u}|+\dfrac{1}{2}\rho ( \xi )| \widehat{u} |^2\\ & \leqslant E^{'}( t, \xi) +\dfrac{1}{2}(1+\log ( 1+| \xi |^{2\sigma} ))| \widehat{u_t} |^2+\frac{1}{2}\rho ( \xi ) ^2( 1+\log( 1+| \xi |^{2\sigma} ) )| \widehat{u}|^2+\dfrac{1}{2}\rho( \xi )| \widehat{u} |^2\\ &\leqslant 3E^{'}( t, \xi ), \\ E( t, \xi) &=\rho( \xi ) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) \text{Re}( \widehat{u}_t\overline{\widehat{u}} ) +\frac{\rho ( \xi )}{2}| \widehat{u} |^2\\ & \geqslant\frac{1}{4}( 1+\log ( 1+| \xi |^{2\sigma} ) ) | \widehat{u_t} |^2+\frac{1}{4}\log ( 1+| \xi |^{2\sigma} ) ( 1+\log ( 1+| \xi |^{2\sigma} ) ) | \widehat{u} |^2\\ &=\dfrac{1}{2}E^{'}( t, \xi ), \end{align*} $

得

$ \begin{equation} \frac{1}{2}E^{'}( t, \xi ) \leqslant E( t, \xi ) \leqslant 3E^{'}( t, \xi ), \end{equation} $

(2.15)

由(2.13)和(2.14)式得

$ \begin{equation} \begin{aligned} \dfrac{d}{dt}E(t, \xi)+\frac{1}{2}\rho(\xi)E(t, \xi) &=R(t, \xi)-F(t, \xi)+\dfrac{1}{2}\rho(\xi)E(t, \xi)\\ &\leqslant-\dfrac{1}{8}|\widehat{u}_{t}|^2-\dfrac{1}{4}\rho(\xi)\log(1+|\xi|^{2\sigma})(1+\log(1+|\xi|^{2\sigma}))|\widehat{u}|^2\\ &\leqslant0, \end{aligned} \end{equation} $

(2.16)

即

$ E(t, \xi)\lesssim E(0, \xi)e^{-\frac{\rho(\xi)}{2}t}, $

将$ (2.15) $式代入上式得

$ \begin{equation} E(t, \xi)\lesssim E^{'}(0, \xi)e^{-\frac{\rho(\xi)}{2}t}. \end{equation} $

(2.17)

将$ E(t, \xi), \; E^{'}(t, \xi) $代入$ (2.17) $式, 引理2.3得证.

引理2.4 基本解$ G(t, \xi), H(t, \xi) $满足

$ |\widehat G(t, \xi)|\lesssim e^{-c\rho(\xi)t}, \; |\widehat H(t, \xi)|\lesssim[\log(1+|\xi|^{2\sigma})]^{-\frac{1}{2}}e^{-c\rho(\xi)t}. $

证 在$ (2.8) $式中, $ u $满足$ (2.6) $式, 令$ \widehat{u}_{1}=0 $, 得

$ \begin{equation} \begin{aligned} \; \; \; \; |\widehat{G}_{t}(t, \xi)|^2+\log(1+|\xi|^{2\sigma})|\widehat{G}(t, \xi)|^2\lesssim\log(1+|\xi|^{2\sigma})e^{-\frac{\rho(\xi)}{2}t}, \\ \end{aligned} \end{equation} $

(2.18)

令$ \widehat{u}_{0}=0 $, 得

$ \begin{equation} \begin{aligned} \; \; \; |\widehat{H}_{t}(t, \xi)|^2+\log(1+|\xi|^{2\sigma})|\widehat{H}(t, \xi)|^2\lesssim e^{-\frac{\rho(\xi)}{2}t}. \end{aligned} \end{equation} $

(2.19)

引理2.5 假设$ k\geqslant0, \; l\geqslant 0, \; m\in [1, 2) $有

$ \begin{equation} \begin{aligned} \lVert D _{x}^{k}G\left( t \right)\ast \varphi \rVert _{L^2}&\lesssim ( 1+t )^{-\frac{n}{2\sigma}(\frac{1}{m}-\frac{1}{2})-\frac{k}{2\sigma}} \lVert \varphi \rVert _{L^m}+\left( 1+t \right) ^{-\frac{l}{2\sigma}}\lVert \varphi \rVert _{\dot{H}^{k+l}} , \\ \lVert D _{x}^{k}H\left( t \right) \ast \psi \rVert _{L^2}&\lesssim ( 1+t )^{-\frac{n}{2\sigma}(\frac{1}{m}-\frac{1}{2})-\frac{k}{2\sigma}} \lVert \psi \rVert _{L^m}+\left( 1+t \right) ^{-\frac{l}{2\sigma}}\lVert \psi \rVert_{\dot{H}^{k+l-\sigma}} . \end{aligned} \end{equation} $

(2.20)

证 当$ |\xi|\leqslant\delta_{0} $时, $ \rho( \xi ) |\gtrsim |\xi|^{2\sigma} $; 当$ |\xi|>\delta_{0} $时, $ \rho( \xi ) |\gtrsim|\xi|^{-2\sigma} $, 由引理2.4得

$ \begin{equation} \begin{aligned} \lVert D _{x}^{k}G\left( t \right) \ast \varphi \rVert _{L^2}^{2}&\lesssim \int_{\mathbb{R}^n}{|\xi |^{2k}e^{-c\rho \left( \xi \right) t}}|\hat{\varphi}\left( \xi \right) |^2d\xi \\&\lesssim \int_{\left\{ \xi ;|\xi |\leqslant \delta _0 \right\}}{|\xi |^{2k}e^{-c|\xi |^{2\sigma}t}}|\hat{\varphi}\left( \xi \right) |^2d\xi +\int_{\left\{ \xi ;|\xi |>\delta _0 \right\}}{|\xi |^{2k}e^{-c|\xi |^{-2\sigma}t}}|\hat{\varphi}\left( \xi \right) |^2d\xi\\& \lesssim \left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{2}{m}-1 \right) -\frac{k}{\sigma}}\lVert \varphi \rVert ^2_{L^m}+\left( 1+t \right) ^{-\frac{l}{\sigma}}\lVert \varphi \rVert _{\dot{H}^{k+l}}^2, \\ \lVert D _{x}^{k}H\left( t \right) \ast \psi \rVert _{L^2}^{2}&\lesssim \int_{\mathbb{R}^n}{|\xi |^{2k}\left[ \log \left( 1+|\xi |^{\sigma} \right) \right] ^{-1}e^{-c\rho \left( \xi \right) t}}|\hat{\psi}\left( \xi \right) |^2d\xi \\&\lesssim \int_{\left\{ \xi ;|\xi |\leqslant \delta _0 \right\}}{|\xi |^{2k}e^{-c|\xi |^{2\sigma}t}}|\hat{\psi}\left( \xi \right) |^2d\xi +\int_{\left\{ \xi ;|\xi |>\delta _0 \right\}}{|\xi |^{-\sigma}e^{-c|\xi |^{-2\sigma}t}}|\hat{\psi}\left( \xi \right) |^2d\xi \\& \lesssim \left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{2}{m}-1 \right) -\frac{k}{\sigma}}\lVert \psi \rVert _{L^m}^{2}+\left( 1+t \right) ^{-\frac{l}{\sigma}}\lVert \psi \rVert^2 _{\dot{H}^{k+l-\sigma}}, \end{aligned} \end{equation} $

(2.21)

其中$ k\geqslant0, \; l\geqslant 0, \; m\in [1, 2) $.

引理2.6 (G-N不等式) 假设$ p, \; p_0, \; p_1\in(1, \infty) $且$ k\in[0, s), \; s>0, $$ f\in L^{p_0}(\mathbb{R}^n)\cap\dot{H}^s_{p_1}(\mathbb{R}^n) $有

$ \lVert f\rVert _{\dot {H}^{k}_p(\mathbb{R}^n)}\lesssim \lVert f\rVert^{1-\beta}_{L^{p_0}(\mathbb{R}^n)}\lVert f\rVert^\beta_{\dot {H}^s_{p_1}(\mathbb{R}^n)}, $

其中$ \beta=\beta_{k, s}=\frac{\left( \frac{1}{p_0}-\frac{1}{p}+\frac{k}{n} \right)}{\left( \frac{1}{p_0}-\frac{1}{p_1}+\frac{s}{n} \right)} $且$ \beta \in \left[ \frac{k}{s}, 1 \right]. $

引理2.7 (Leibniz法则) 假设$ s>0, \; 1\leqslant r\leqslant \infty, \; 1<p_1, \; p_2, \; q_1, \; q_2\leqslant \infty $满足

$ \frac{1}{r}=\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{q_1}+\frac{1}{q_2}, $

当$ f\in \dot {H}_{p_1}^{s}(\mathbb{R}^n )\cap L^{q_1}(\mathbb{R}^n ) $, $ g\in \dot {H}_{p_2}^{s}(\mathbb{R}^n)\cap L^{q_2}(\mathbb{R}^n) $时, 有

$ \lVert fg \rVert _{\dot {H}_{p_2}^{s}(\mathbb{R}^n)}\lesssim \lVert f \rVert _{\dot {H}_{p_1}^{s}(\mathbb{R}^n)}\lVert g \rVert _{L^{p_2}(\mathbb{R}^n)}+\lVert f \rVert _{L^{q_1}(\mathbb{R}^n )}\lVert g \rVert _{\dot{H}_{q_2}^{s}(\mathbb{R}^n)}. $

引理2.8 (链式法则) 假设$ s>0, \; p>[s] $, 且$ 1<r, r_1, r_2<\infty $满足

$ \frac{1}{r}=\frac{p-1}{r_1}+\frac{1}{r_2}, $

当$ f\in \dot {H}_{r_2}^{s}(\mathbb{R}^n ) \cap L^{r_1}(\mathbb{R}^n ) $时, 有

$ \lVert f|f|^{p-1}\rVert_{\dot {H}_{r}^{s}(\mathbb{R}^n )}+\lVert |f|^p \rVert _{\dot{H}_{r}^{s}(\mathbb{R}^n)}\lesssim \lVert f \rVert _{L^{r_1}(\mathbb{R}^n )}^{p-1}\lVert f \rVert _{\dot {H}_{r_2}^{s}(\mathbb{R}^n )}. $

引理2.9^[5] 若$ \alpha, \beta\in \mathbb{R} $, 有

$ \displaystyle{\int}_0^t{\left( 1+t-\tau \right) ^{-\alpha}}\left( 1+\tau \right) ^{-\beta}d\tau \lesssim \left\{\begin{array}{l} \left( 1+t \right) ^{-\min \left\{ \alpha , \beta \right\}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \max \left\{ \alpha , \beta \right\} >1, \\ \left( 1+t \right) ^{-\min \left\{ \alpha , \beta \right\}}\log \left( e+t \right) \; \; \; \; \; \; \max \left\{ \alpha , \beta \right\} =1, \\ \left( 1+t \right) ^{1-\alpha -\beta}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \max \left\{ \alpha , \beta \right\} <1. \end{array} \right. $

定义空间

$ X(T)=\{u\in C([0, T], \; \dot{H}^k)\}, $

装备相应的范数

$ \lVert u \rVert _{X( T )}:=\underset{0\leqslant t\leqslant T}{\sup}\{(1+t)^{\frac{n}{2\sigma}(\frac{1}{m}-\frac{1}{2})}\lVert u( t, \cdot ) \rVert _{L^2}+(1+t)^{\frac{k}{2\sigma}+\frac{n}{2\sigma}(\frac{1}{m}-\frac{1}{2})}\lVert |D|^ku( t, \cdot) \rVert _{L^2}\}. $

定义算子$ N $

$ \begin{align*} Nu(t, x)&=G(t, x)\ast u_{0}(x)+H(t, x)\ast u_1(x)+\int_{0}^{t}H(t-\tau, x)\ast|u|^pd\tau\\&=u^{ln}(t, x)+u^{nl}(t, x), \end{align*} $

下证算子$ N $满足如下两个不等式

$ \begin{equation} \lVert Nu \rVert _{X( T )}\lesssim \lVert (u_0, u_1) \rVert _{\mathcal{A}}+\lVert u \rVert _{_{X\left( T \right)}}^{p}. \end{equation} $

(3.1)

$ \begin{equation} \lVert Nu-Nv \rVert _{X( T)}\lesssim \lVert u-v \rVert _{X( T) }( \lVert u \rVert _{_{X( T )}}^{p-1}+\lVert v \rVert_{_{X( T )}}^{p-1} ). \end{equation} $

(3.2)

由引理$ 2.5 $得

$ \begin{equation} \begin{aligned} \lVert u^{\ln}\left( t, x \right) \rVert _{L^2}&\lesssim \lVert G\left( t \right) \ast u_0 \rVert _{L^2}+\lVert H\left( t \right) \ast u_1 \rVert _{L^2}\\ &\lesssim \left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)} \lVert (u_0, u_1) \rVert _{\mathcal{A}}, \\ \lVert |D|^ku^{\ln}\left( t, x \right) \rVert _{L^2}&\lesssim \lVert G\left( t \right) \ast u_0 \rVert _{\dot{H}^k}+\lVert H\left( t \right) \ast u_1 \rVert _{\dot{H}^k}\\ &\lesssim \left( 1+t \right) ^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}( \frac{1}{m}-\frac{1}{2} )}\lVert (u_0, u_1) \rVert _{\mathcal{A}}, \end{aligned} \end{equation} $

(3.3)

其中$ l\geqslant n(\frac{1}{m}-\frac{1}{2})+2k $. 综上, 有

$ \begin{equation} \lVert u^{\ln}\left( t, x \right) \rVert _{X(T)}\lesssim \lVert (u_0, u_1) \rVert _{\mathcal{A}}. \end{equation} $

(3.4)

下证

$ \begin{equation} \lVert u^{nl} \rVert _{X\left( T \right)}\lesssim \lVert u \rVert ^{p}_{X\left( T \right)}. \end{equation} $

(3.5)

在$ [0, t/2] $用$ (L^2\cap L^m)-L^2 $估计和$ \left[t/2, t\right] $用$ L^2-L^2 $估计, 可得

$ \begin{equation} \begin{aligned} \lVert u^{nl}\left( t, \cdot \right) \rVert _{L^{2}}&\lesssim \int_0^t{\lVert H\left( t-\tau , \cdot \right) \ast \left| u\left( \tau , \cdot \right) \right|^p \rVert _{L^2}}d\tau\\ &\lesssim\int_0^{\frac{t}{2}}{\left( 1+t-\tau \right)}^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert |u\left( \tau , \cdot \right) |^p \rVert _{L^{m}}d\tau +\int_{\frac{t}{2}}^t\lVert |u\left( \tau , \cdot \right) |^p \rVert _{L^{2}}d\tau.\\ \end{aligned} \end{equation} $

(3.6)

由引理$ 2.6 $得

$ \begin{equation} \begin{aligned} \lVert |u\left( \tau , \cdot \right)|^p \rVert _{L^m}=\lVert u \rVert _{L^{mp}}^{p}\lesssim \lVert u \rVert _{L^2}^{\left( 1-\beta _1 \right) p}\lVert |D|^ku \rVert _{L^2}^{\beta _1p}\lesssim \left( 1+\tau \right) ^{-\frac{np}{2\sigma}(\frac{1}{m}-\frac{1}{mp})}\lVert u( \tau , \cdot ) \rVert _{X\left( T \right)}^{p}, \\ \lVert |u\left( \tau , \cdot \right) |^p \rVert _{L^2}=\lVert u\rVert _{L^{2p}}^{p}\lesssim \lVert u\rVert _{L^2}^{\left( 1-\beta _2 \right) p}\lVert |D|^ku \rVert _{L^2}^{\beta _2p}\lesssim \left( 1+\tau \right) ^{-\frac{np}{2\sigma}(\frac{1}{m}-\frac{1}{2p})}\lVert u( \tau , \cdot ) \rVert _{X\left( T \right)}^{p}. \end{aligned} \end{equation} $

(3.7)

其中$ \beta_{1}, \beta_{2} $满足

$ \begin{equation} \beta_1=\tfrac{n}{k}(\tfrac{1}{2}-\tfrac{1}{mp})\in[0, 1], \; \beta_2=\tfrac{n}{k}(\tfrac{1}{2}-\tfrac{1}{2p})\in[0, 1]. \end{equation} $

(3.8)

解得

$ \left\{ \begin{array}{l} \; \; \; \; \; p\geqslant \frac{2}{m} \; \; \; \; \; \; \; \; \; \; n\leqslant 2k, \\ \tfrac{2}{m}\leqslant p \leqslant \frac{n}{n-2k} \ \ \ \ \ n>2k.\\ \end{array} \right. $

若$ -\frac{np}{2\sigma}(\frac{1}{m}-\frac{1}{mp})+1<0 $, 即$ p>1+\frac{2\sigma m}{n} $有

$ \begin{equation} \begin{aligned} \int_0^{\frac{t}{2}}{\left( 1+t-\tau\right)}^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\left( 1+\tau \right) ^{-\frac{np}{2\sigma}\left( \frac{1}{m}-\frac{1}{mp} \right)}d\tau \lesssim\left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}, \end{aligned} \end{equation} $

(3.9)

若$ -\frac{np}{2\sigma}( \frac{1}{m}-\frac{1}{2p})+1<-\frac{n}{2\sigma}( \frac{1}{m}-\frac{1}{2}) $, 即$ p>1+\frac{2\sigma m}{n} $有

$ \begin{equation} \begin{aligned} \int_{\frac{t}{2}}^t{\left( 1+\tau \right) ^{-\frac{np}{2\sigma}\left( \frac{1}{m}-\frac{1}{2p} \right)}}d\tau\lesssim \left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}, \end{aligned} \end{equation} $

(3.10)

其中$ \tau \in [0, t/2], \; (1+t-\tau)\approx (1+t) $, $ \tau\in[t/2, t], \; (1+\tau)\approx(1+t) $. (3.9)和(3.10)式代入(3.6)式得

$ \begin{equation} \begin{aligned} \lVert u^{nl}\left( t, \cdot \right) \rVert _{L^{2}} &\lesssim \left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert u \rVert _{X\left( T \right)}^{p}. \end{aligned} \end{equation} $

(3.11)

下面对$ \lVert |D|^ku^{nl}\left( t, \cdot \right) \rVert _{L^2} $估计, 由(3.7)式得

$ \begin{equation} \begin{aligned} \lVert |D|^ku^{nl}\left( t, \cdot \right) \rVert _{L^2} &\lesssim \int_0^{\frac{t}{2}}{\left( 1+t-\tau \right)}^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert |u\left( \tau , \cdot \right) |^p \rVert _{L^m}d\tau\\ &\; \; \; \; \; +\int_{\frac{t}{2}}^t{\left( 1+t-\tau \right)}^{-\frac{k}{2\sigma}}\lVert |u\left( \tau , \cdot \right) |^p \rVert _{L^2}d\tau+\int_0^t{\left( 1+t-\tau \right)}^{-\frac{l}{2\sigma}}\lVert |u\left( \tau , \cdot \right)|^p \rVert _{\dot{H}^{k+l-\sigma}}d\tau \\ &\lesssim\int_0^{\frac{t}{2}}(1+t-\tau)^{-\frac{k}{2\sigma}-\frac{np}{2\sigma}(\frac{1}{m}-\frac{1}{2})}(1+\tau)^{-\frac{np}{2\sigma}(\frac{1}{m}-\frac{1}{mp})}\lVert u(\tau, \cdot) \rVert^p_{X(T)} d\tau \\ &\; \; \; \; \; +\int_{\frac{t}{2}}^t(1+t-\tau)^{-\frac{k}{2\sigma}}(1+\tau)^{-\frac{np}{2\sigma}(\frac{1}{m}-\frac{1}{2p})}\lVert u(\tau, \cdot) \rVert^p_{X(T)} d\tau\\ &\; \; \; \; \; +\int_0^t{\left( 1+t-\tau \right)}^{-\frac{l}{2\sigma}+\frac{k}{2\sigma}}\lVert |u(\tau, \cdot)|^p \rVert _{\dot{H}^{l-\sigma}}d\tau. \end{aligned} \end{equation} $

(3.12)

当$ \sigma\leqslant l<k+\sigma $, 由引理$ 2.6 $和引理$ 2.8 $得

$ \begin{equation} \begin{aligned} \lVert \left| u\left( \tau , \cdot \right) \right|^p \rVert _{\dot{H}^{l-\sigma}}&\lesssim \lVert u( \tau , \cdot ) \rVert _{L^{r_1}}^{p-1}\lVert u\left( \tau , \cdot \right) \rVert _{\dot{H}_{r_2}^{l-\sigma}}, \\ \lVert u( \tau , \cdot ) \rVert _{L^{r_1}}^{p-1} &\lesssim \lVert u( \tau , \cdot ) \rVert _{L^2}^{( 1-\beta _{r_1} ) ( p-1 )}\lVert |D|^ku( \tau , \cdot ) \rVert _{L^2}^{\beta _{r1}( p-1 )}\\ &\lesssim ( 1+\tau)^{-\frac{n(p-1)}{2\sigma}(\frac{1}{m}-\frac{1}{r_1})} \lVert u \rVert _{X( T )}^{p-1}, \\ \lVert u\left( \tau , \cdot \right) \rVert _{\dot{H}_{r_2}^{l-\sigma}} &\lesssim \lVert u( \tau , \cdot ) \rVert _{L^2}^{( 1-\beta _{r_2} )}\lVert |D|^ku( \tau , \cdot) \rVert _{L^2}^{\beta _{r2}}\\ &\lesssim(1+\tau)^{-\frac{n}{2\sigma}(\frac{1}{m}-\frac{1}{r_2}+\frac{l-\sigma}{n})}\lVert u \rVert _{X\left( T \right)}. \end{aligned} \end{equation} $

(3.13)

其中$ \frac{1}{2}=\frac{p-1}{r_1}+\frac{1}{r_2} $, $ \beta_{r_{1}} $, $ \beta_{r_{2}} $满足

$ \beta_{r_{1}}=\tfrac{n}{k}(\tfrac{1}{2}-\tfrac{1}{r_1})\in[0, 1], \; \beta_{r_{2}}=\tfrac{n}{k}(\tfrac{1}{2}-\tfrac{1}{r_2}+\tfrac{{l-\sigma}}{n})\in[\tfrac{{l-\sigma}}{k}, 1]. $

解得

$ \left\{ \begin{array}{l} \; \; \; \; \; \; \; p>1 \; \; \; \; \; \; \; \; \; \; \; \; \; n\leqslant 2k, \\ 1<p\leqslant\frac{n-2(l-\sigma)}{n-2k} \ \ \ \ \ n>2k.\\ \end{array} \right. $

由$ (3.13) $式得

$ \begin{equation} \begin{aligned} \lVert \left| u\left( \tau , \cdot \right) \right|^p \rVert _{\dot{H}^{l-\sigma}}\lesssim (1+\tau)^{-\frac{np}{2\sigma}(\frac{1}{m}-\frac{1}{2p})}\lVert u \rVert _{X\left( T \right)}^p. \end{aligned} \end{equation} $

(3.14)

若$ k<\sigma $, $ p>1+\frac{2\sigma m}{n} $则

$ \begin{equation} \begin{aligned} \int_0^{\frac{t}{2}}{\left( 1+t-\tau \right)}^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\left( 1+\tau \right) ^{-\frac{np}{2\sigma}\left( \frac{1}{m}-\frac{1}{mp} \right)} \lesssim\left( 1+t \right) ^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}, \\ \int_{\frac{t}{2}}^t{\left( 1+t-\tau \right)}^{-\frac{k}{2\sigma}}\left( 1+\tau \right) ^{-\frac{np}{2\sigma}\left( \frac{1}{m}-\frac{1}{2p} \right)}\lesssim \left( 1+t \right) ^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}. \end{aligned} \end{equation} $

(3.15)

其中$ \tau \in [0, t/2], \; (1+t-\tau)\approx (1+t) $, $ \tau\in[t/2, t], (1+\tau)\approx(1+t) $. 若$ l\geqslant n(\frac{1}{m}-\frac{1}{2})+2k $, 且$ p>1+\frac{2\sigma m}{n} $, 由引理$ 2.9 $得

$ \begin{equation} \begin{aligned} \int_0^t{\left( 1+t-\tau \right)}^{-\frac{l}{2\sigma}+\frac{k}{2\sigma}}\lVert |u|^p \rVert _{\dot{H}^{l-\sigma}}d\tau &\lesssim \int_0^t{\left( 1+t-\tau \right)}^{-\frac{l}{2\sigma}+\frac{k}{2\sigma}}\left( 1+\tau \right) ^{-\frac{np}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert u \rVert _{X\left( T \right)}^{p}\\ &\lesssim \left( 1+t \right) ^{-\min \left\{ \frac{l}{2\sigma}-\frac{k}{2\sigma}, \frac{np}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right) \right\}}\lVert u \rVert _{X\left( T \right)}^{p}\\ &\lesssim \left( 1+t \right) ^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right) }\lVert u \rVert _{X\left( T \right)}^{p}. \end{aligned} \end{equation} $

(3.16)

(3.15)和(3.16)式代入(3.12)式得

$ \begin{equation} \begin{aligned} \lVert |D|^ku^{nl}\left( t, \cdot \right) \rVert _{L^2}\lesssim \left( 1+t \right) ^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert u\rVert _{X\left( T \right)}^{p} \end{aligned}. \end{equation} $

(3.17)

由(3.4)和(3.5)式知(3.1)式成立. 下面估计$ \lVert u^{nl}-v^{nl} \rVert _{L^2} $.

$ \begin{equation} \begin{aligned} \lVert u^{nl}(t, \cdot)-v^{nl}(t, \cdot) \rVert _{L^2}&\lesssim \int_0^t{\lVert H\left( t \right) \ast \left( \left| u\left( \tau , \cdot \right) \right|^p-\left| v\left( \tau , \cdot \right) \right|^p \right) \rVert}_{L^2}d\tau\\ &\lesssim \int_0^{\frac{t}{2}}{\left( 1+t-\tau \right)}^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert |u\left( \tau , \cdot \right) |^p-|v\left( \tau , \cdot \right) |^p \rVert _{L^m}d\tau\\ & \; \; \; \; \; +\int_{\frac{t}{2}}^t{\lVert |u\left( \tau , \cdot \right) |^p-|v\left( \tau , \cdot \right) |^p \rVert _{L^2}}d\tau, \end{aligned} \end{equation} $

(3.18)

由H{ö}lder不等式和引理2.6得

$ \begin{equation} \begin{aligned} \lVert | u( \tau , \cdot )|^p-| v( \tau , \cdot ) |^p \rVert _{L^m}&\lesssim \lVert u( \tau , \cdot) -v( \tau , \cdot ) \rVert _{L^{mp}}( \lVert u( \tau , \cdot ) \rVert _{L^{mp}}^{p-1}+\lVert u( \tau , \cdot ) \rVert _{L^{mp}}^{p-1} ) \\ &\lesssim ( 1+\tau) ^{-\frac{np}{2\sigma}( \frac{1}{m}-\frac{1}{mp} )}\lVert u-v \rVert _{X( T )}( \lVert u \rVert _{X( T)}^{p-1}+\lVert v \rVert _{X( T )}^{p-1} ), \\ \lVert | u( \tau , \cdot )|^p-| v( \tau , \cdot ) |^p \rVert _{L^2}&\lesssim \lVert u( \tau , \cdot) -v( \tau , \cdot ) \rVert _{L^{mp}}( \lVert u( \tau , \cdot ) \rVert _{L^{mp}}^{p-1}+\lVert u( \tau , \cdot ) \rVert _{L^{mp}}^{p-1} ) \\ &\lesssim ( 1+\tau) ^{-\frac{np}{2\sigma}( \frac{1}{m}-\frac{1}{2p} )}\lVert u-v \rVert _{X( T )}( \lVert u \rVert _{X( T)}^{p-1}+\lVert v \rVert _{X( T )}^{p-1} ). \end{aligned} \end{equation} $

(3.19)

$ (3.19) $式代入$ (3.18) $式得

$ \begin{equation} \lVert u^{nl}-v^{nl} \rVert _{L^2}\lesssim \left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert u-v \rVert _{X\left( T \right)}\left( \lVert u \rVert _{X\left( T \right)}^{p-1}+\lVert v \rVert _{X\left( T \right)}^{p-1} \right). \end{equation} $

(3.20)

下面对$ \lVert |D|^k\left( u^{nl}-v^{nl} \right) \rVert _{L^2} $估计.

$ \begin{equation} \begin{aligned} \lVert |D|^k\left( u^{nl}-v^{nl} \right) \rVert _{L^2}&\lesssim \int_0^t{\lVert |D|^kH\left( t-\tau \right) \ast \left( \left| u\left( \tau , \cdot \right) \right|^p-\left| v\left( \tau , \cdot \right) \right|^p \right) \rVert _{L^2}}d\tau\\ & \lesssim \int_0^{\frac{t}{2}}{\left( 1+t-\tau \right) ^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}}\lVert \left( \left| u\left( \tau , \cdot \right) \right|^p-\left| v\left( \tau , \cdot \right) \right|^p \right) \rVert _{_{L^m}}d\tau\\ & \; \; \; \; \; + \int_{\frac{t}{2}}^t{\left( 1+t-\tau \right) ^{-\frac{k}{2\sigma}}}\lVert \left( \left| u\left( \tau , \cdot \right) \right|^p-\left| v\left( \tau , \cdot \right) \right|^p \right) \rVert _{_{L^2}}d\tau \\ &\; \; \; \; \; +\int_0^t{\left( 1+t-\tau \right) ^{-\frac{l}{2\sigma}+\frac{k}{2\sigma}}}\lVert \left( \left| u\left( \tau , \cdot \right) \right|^p-\left| v\left( \tau , \cdot \right) \right|^p \right) \rVert _{_{\dot{H}^{l-\sigma}}}d\tau. \end{aligned} \end{equation} $

(3.21)

引入积分表达式

$ \begin{equation} |u|^p-|v|^p=p\displaystyle{\int}_0^1{\left( u-v \right) \cdot G\left( \omega u+\left( 1-\omega \right) v \right) d\omega}. \end{equation} $

(3.22)

其中$ G=u|u|^{p-2} $. 由$ (3.22) $式, 引理$ 2.7 $和Minkowski不等式得

$ \begin{equation} \begin{aligned} \lVert |u\left( \tau , \cdot \right) |^p-|v\left( \tau , \cdot \right) |^p \rVert _{\dot{H}^{l-\sigma}}&\lesssim \int_0^1\lVert D^{l-\sigma}\left( \left( u-v \right) G\left( \omega u+\left( 1-\omega \right) v \right) \right) \rVert _{L^2}d\tau \\ &\lesssim \lVert u-v \rVert _{\dot{H}_{s_1}^{l-\sigma}}\int_0^1{\lVert G\left( \omega u+\left( 1-\omega \right) v \right) \rVert}_{L^{s_2}}d\omega\\ & \; \; \; \; \; +\lVert u-v \rVert _{L^{s_3}}\int_0^1{\lVert G\left( \omega u+\left( 1-\omega \right) v \right) \rVert _{\dot{H}_{s_4}^{l-\sigma}}}d\omega\\ &\lesssim \lVert u-v \rVert _{\dot{H}_{s_1}^{l-\sigma}}\left( \lVert u \rVert _{L^{s_2\left( p-1 \right)}}^{p-1}+\lVert v \rVert _{L^{s_2\left( p-1 \right)}}^{p-1} \right) \\ &\; \; \; \; \; +\lVert u-v \rVert _{L^{s_3}}\int_0^1{\lVert G\left( \omega u+\left( 1-\omega \right) v \right) \rVert _{\dot{H}_{s_4}^{l-\sigma}}}d\omega , \end{aligned} \end{equation} $

(3.23)

其中$ \frac{1}{2}=\frac{1}{s_1}+\frac{1}{s_2}=\frac{1}{s_3}+\frac{1}{s_4} $. 若$ \sigma\leqslant l<k+\sigma $, 由引理$ 2.6 $得

$ \begin{equation} \begin{aligned} \lVert u-v \rVert _{\dot {H}_{s_1}^{l-\sigma}}&\lesssim \lVert u-v \rVert _{L^2}^{( 1-\beta _3)}\lVert | D |^k( u-v) \rVert _{L^2}^{\beta _3}\\ &\lesssim ( 1+t ) ^{-\frac{n}{2\sigma}( \frac{1}{m}-\frac{1}{s_1} ) -\frac{l-\sigma}{2\sigma}}\lVert u-v \rVert _{X( T )}, \\ \lVert u \rVert _{L^{s_2\left( p-1 \right)}}^{p-1}&\lesssim \lVert u \rVert _{L^2}^{\left( 1-\beta _4 \right) \left( p-1 \right)}\lVert \left| D \right|^ku \rVert _{L^2}^{\beta _4\left( p-1 \right)}\\ &\lesssim \left( 1+t \right) ^{-\frac{n\left( p-1 \right)}{2\sigma}\left( \frac{1}{m}-\frac{1}{s_2(p-1)} \right)}\lVert u \rVert _{X\left( T \right)}^{p-1}, \\ \lVert v \rVert _{L^{s_2\left( p-1 \right)}}^{p-1}&\lesssim \lVert v \rVert _{L^2}^{\left( 1-\beta _4 \right) \left( p-1 \right)}\lVert \left| D \right|^kv \rVert _{L^2}^{\beta _4\left( p-1 \right)}\\ &\lesssim \left( 1+t \right) ^{-\frac{n\left( p-1 \right)}{2\sigma}\left( \frac{1}{m}-\frac{1}{s_2(p-1)} \right)}\lVert v \rVert _{X\left( T \right)}^{p-1}\\ \lVert u-v \rVert _{L^{s_3}}&\lesssim \lVert u-v \rVert _{L^2}^{( 1-\beta _5)}\lVert|D |^k( u-v)\rVert _{L^2}^{\beta _5}\\ &\lesssim ( 1+t) ^{-\frac{n}{2\sigma}( \frac{1}{m}-\frac{1}{s_3} )}\lVert u-v \rVert _{X( T )}. \end{aligned} \end{equation} $

(3.24)

其中$ \beta_3 $, $ \beta_4 $, $ \beta_5 $满足

$ \begin{align*} \beta_3=\frac{n}{k}(\frac{1}{2}-\frac{1}{s_1})+\frac{l-\sigma}{k}\in[\frac{l-\sigma}{k}, 1], \; \beta_4=\beta_5=\frac{n}{k}(\frac{1}{2}-\frac{1}{s_2})\in[0, 1]. \end{align*} $

$ \begin{equation} \begin{aligned} \int_0^1{\lVert G\left( \omega u+\left( 1-\omega \right) v \right) \rVert _{\dot {H}_{s_4}^{l-\sigma}}}d\omega &=\int_0^1{\lVert D^{l-\sigma}G\left( \omega u+\left( 1-\omega \right) v \right) \rVert _{L^{s_4}}}d\omega \\ &\lesssim \int_0^1{\lVert D^{l-\sigma}\left( \omega u+\left( 1-\omega \right) v \right) \rVert _{L^{s_5}}\cdot}\lVert \omega u+\left( 1-\omega \right) v \rVert _{L^{s_6}}^{p-2}d\omega\\ &\lesssim \int_0^1{\left( 1+t \right) ^{-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{s_5} \right) -\frac{l-\sigma}{2\sigma}}}\lVert \omega u+\left( 1-\omega \right) v \rVert _{X\left( T \right)}\\ &\; \; \; \; \; \times \left( 1+t \right) ^{-\frac{n\left( p-2 \right)}{2\sigma}\left( \frac{1}{m}-\frac{1}{s_6(p-2)} \right)}\lVert \omega u+\left( 1-\omega \right) v \rVert _{X\left( T \right)}^{p-2}d\omega\\ &\lesssim \left( 1+t \right) ^{-\frac{n\left( p-1 \right)}{2\sigma}\left( \frac{1}{m}-\frac{1}{s_4(p-1)} \right) -\frac{l-\sigma}{2\sigma}}\left( \lVert u \rVert _{X\left( T \right)}^{p-1}+\lVert v \rVert _{X\left( T \right)}^{p-1} \right), \end{aligned} \end{equation} $

(3.25)

其中$ \frac{1}{s_4}=\frac{1}{s_5}+\frac{p-2}{s_6} $. (3.24)和(3.25)式代入(3.23)式得

$ \begin{equation} \lVert \left| u\left( \tau , \cdot \right) \right|^p-\left| v\left( \tau , \cdot \right) \right|^p \rVert _{\dot{H}^{l-\sigma}}\lesssim \left( 1+\tau \right) ^{-\frac{np}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right)}\lVert u-v \rVert _{X\left( T \right)}\left( \lVert u \rVert _{X\left( T \right)}^{p-1}+\lVert v \rVert _{X\left( T \right)}^{p-1} \right), \end{equation} $

(3.26)

$ (3.26) $式代入$ (3.21) $式, 再由$ (3.15) $和$ (3.16) $式得

$ \begin{equation} \begin{aligned} \lVert |D|^k( u^{nl}-v^{nl} ) \rVert _{L^2} &\lesssim\lVert u-v \rVert _{X( T )}( \lVert u \rVert _{X\left( T\right)}^{p-1}-\lVert v \rVert _{X( T )}^{p-1} )\\ &\; \; \; \; (\int_0^{\frac{t}{2}}{( 1+t ) ^{-\frac{n}{2\sigma}( \frac{1}{m}-\frac{1}{s_5} ) -\frac{k}{2\sigma}}}( 1+t ) ^{-\frac{n}{2\sigma}( \frac{1}{m}-\frac{1}{mp} )} d\tau \\ & \; \; \; \; +\int_{\frac{t}{2}}^t{\left( 1+t \right) ^{-\frac{k}{2\sigma}}}( 1+t ) ^{-\frac{np}{2\sigma}( \frac{1}{m}-\frac{1}{2p} )}d\tau\\ &\; \; \; \; +\int_0^t{( 1+t) ^{-\frac{l}{2\sigma}+\frac{k}{2\sigma}}}( 1+t ) ^{-\frac{np}{2\sigma}( \frac{1}{m}-\frac{1}{2} )}d\tau )\\ & \lesssim \left( 1+t \right) ^{-\frac{k}{2\sigma}-\frac{n}{2\sigma}\left( \frac{1}{m}-\frac{1}{2} \right) }\lVert u-v \rVert _{X\left( T \right)}\left( \lVert u \rVert _{X\left( T \right)}^{p-1}+\lVert v \rVert _{X\left( T \right)}^{p-1} \right).\\ \end{aligned} \end{equation} $

(3.27)

由$ (3.20) $式和(3.27)式知(3.2)式成立, 再由Banach不动点定理知问题$ (1.1) $在小初值情形下整体解存在. 定理1.1证毕.

为证明定理1.2, 先给出如下引理.

引理4.1^[10] 假设$ \left< x \right> =\left( 1+|x|^2 \right) ^{\frac{1}{2}} $, $ m\in N $且$ s\in[0, 1). $则$ q\in N, \; x\in\mathbb{R}^n $时, 以下估计成立

$ \left| \left( -\bigtriangleup \right) ^{m+s}\left< x \right> ^{-q} \right|\lesssim \left\{ \begin{array}{l} \left< x \right> ^{-n-2m}, \ \ \ s=0, \\ \left< x \right> ^{-n-2s}, \ \ \ \ s\in \left( 0, 1 \right).\\ \end{array} \right. $

引理4.2^[10] 假设$ \varphi _R\left( x \right) =\varphi \left( x/R \right) $, $ x \in\mathbb{R}^n $, 其中$ \varphi \left( x \right) =\left< x \right> ^{-q} $, $ \sigma \geqslant1 $, 则

$ \left( -\bigtriangleup \right) ^{\sigma}\left( \varphi _R \right) \left( x \right) =R^{-2\sigma}\left( \left( -\bigtriangleup \right) ^{\sigma}\varphi \right) \left( x/R \right). $

若$ \varphi \left( x \right)=\left< x \right> ^{-n-2s_{\sigma}}=\left( 1+|x|^2 \right) ^{-n/2-s_{\sigma}}, $其中当$ \sigma $是整数时, $ s_{\sigma} \in (0, 1) $; 当$ \sigma $是分数时, $ 0<s_{\sigma} \leqslant\sigma -[\sigma] $则

$ |\left( -\varDelta \right) ^{\sigma}\left< x \right> ^{-n-2s_{\sigma}}|\lesssim \left< x \right> ^{-n-2s_{\sigma}}. $

下证定理2.2

选择测试函数$ \eta \left( t \right) :=\left\{ \begin{array}{l} 1\; \; \; \; \; \; \; \; 0\leqslant t\leqslant \frac{1}{2}, \\ \text {递减}\; \; \; \frac{1}{2}\leqslant t\leqslant 1, \\ 0\; \; \; \; \; \; \; \; t\geqslant 1, \end{array} \right. $满足

$ \left( \eta \left( t \right) \right) ^{-\frac{p^{'}}{p}}\left( |\eta ^{'}\left( t \right) |^{p^{'}}+|\eta ^{''}\left( t \right) |^{p^{'}}+|\eta ^{'''}\left( t \right) |^{p^{'}} \right) \leqslant C, \; \; \; \; t\in \left[ \frac{1}{2}, 1 \right], $

其中$ \frac{1}{p}+\frac{1}{p^{'}}=1 $.

引入测试函数$ \psi _R\left( t, x \right) :=\eta _R\left( t \right)\varphi _R\left( x \right) :=\eta \left( t/R^{2\sigma} \right) \varphi \left( x/R \right) . $定义如下两个函数

$ \begin{align*} I_R:&=\int_0^{\infty}{\int_{\mathbb{R}^n}{|u\left( t, x \right) |^p\psi _R\left( t, x \right)}}dxdt, \ \\ \widetilde{I}_R:&=\int_0^{\infty}{\int_{\mathbb{R}^n}{|u\left( t, x \right) |^p\partial _t\psi _R\left( t, x \right)}}dxdt, \end{align*} $

两式相减并进行分部积分得

$ \begin{equation} \begin{aligned} &\; \; \; \; \; \; \; \; I_R-\widetilde{I}_R +\int_{\mathbb{R}^n}{u_1\left( x \right) \varphi _R\left( x \right)}dx+\left( I+\log \left( I+\left( -\bigtriangleup \right) ^{\sigma} \right) \right) ^{-1}\int_{\mathbb{R}^n}{u_0\left( x \right) \varphi _R\left( x \right)}dx \\ &=-\int_0^{\infty}{\int_{\mathbb{R}^n}{u\left( t, x \right) \left( \partial _{t}^{3}\psi _R\left( t, x \right) -\partial _{t}^{2}\psi _R\left( t, x \right) \right)}}dxdt\\ &\; \; \; \; +\int_0^{\infty}{\int_{\mathbb{R}^n}{\log}}\left( I+\left( -\bigtriangleup \right) ^{\sigma} \right) u\left( t, x \right) \left( \psi _R\left( t, x \right) -\psi _R^{'}\left( t, x \right) \right) dxdt\\ &\; \; \; \; +\int_0^{\infty}{\int_{\mathbb{R}^n}{\left( I+\log \left( I+\left( -\bigtriangleup \right) ^{\sigma} \right) \right) ^{-1}}} u\left( t, x \right)\left( \psi _R^{''}\left( t, x \right) -\psi _{R}^{'}\left( t, x \right) \right) dxdt\\ &=J_1+J_2+J_3, \end{aligned} \end{equation} $

(4.1)

由引理4.1, $ 4.2 $得

$ \begin{align*} | J_1 |&\lesssim \int_{\frac{R^{2\sigma}}{2}}^{R^{2\sigma}}{\int_{\mathbb{R}^n}{|u( t, x ) |\varphi _R}}( x) ( |\eta _{R}^{'''}( t ) |+|\eta _{R}^{''}( t ) ) |dxdt\\ &\lesssim I_{R}^{\frac{1}{p}}\left(\int_{\frac{R^{2\sigma}}{2}}^{R^{2\sigma}}\int_{\mathbb{R}^n}\eta _R( t, x) ^{-\frac{p^{'}}{p}}\varphi _R( x ) ( |\eta _{R}^{'''}( t ) |+|\eta _{R}^{''}( t ) )^{p^{'}} dxdt\right)^{\frac{1}{p^{'}}}\\ &\lesssim I_{R}^{\frac{1}{p}}\left( R^{-2\sigma p^{'}+2\sigma +n}\int_{\mathbb{R}^n}{\left< \widetilde{x} \right> ^{-n-2s_{\sigma}}d\widetilde{x}} \right) ^{\frac{1}{^{p^{'}}}}\\& \lesssim I_{R}^{\frac{1}{p}}R^{-2\sigma +\frac{2\sigma+n}{p^{'}}}, \\ \left| J_2 \right| &\lesssim \int_0^{\infty}{\int_{\mathbb{R}^n}{u\left( t, x \right) |\left( -\bigtriangleup \right) ^{\sigma}\psi _R}}\left( x \right) |dxdt +\int_0^{\infty}{\int_{\mathbb{R}^n}{u\left( t, x \right) |\left( -\bigtriangleup \right) ^{\sigma}\psi _{R}^{'}}}\left( t, x \right) |dxdt\\ &\lesssim I_{R}^{\frac{1}{p}}\left( \int_0^{R^{2\sigma}}{\int_{\mathbb{R}^n}{\eta _R\left( t \right) (\varphi _R\left( x \right)) ^{-\frac{p^{'}}{p}}|\left( -\bigtriangleup \right) ^{\sigma}\psi _R\left( t, x \right) |}}^{p^{'}}dxdt \right) ^{\frac{1}{^{p'}}}\\ &\; \; \; +I_{R}^{\frac{1}{p}}\left( \int_{\frac{R^{2\sigma}}{2}}^{R^{2\sigma}}{\int_{\mathbb{R}^n}{\eta _R\left( t \right) (\varphi _R\left( x \right)) ^{-\frac{p^{'}}{p}}|\left( -\bigtriangleup \right) ^{\sigma}\psi _{R}^{'}\left( t, x \right) |}}^{p'}dxdt \right) ^{\frac{1}{p'}}\\ &\lesssim I_{R}^{\frac{1}{p}}\left( R^{-2\sigma p^{'}+2\sigma +n}\int_{\mathbb{R}^n}{\left< \widetilde{x} \right> ^{-n-2s_{\sigma}}d\widetilde{x}} \right) ^{\frac{1}{^{p^{'}}}}\\& \lesssim I_{R}^{\frac{1}{p}}R^{-2\sigma +\frac{2\sigma +n}{p^{'}}}, \\ |J_3|&\lesssim \int_0^{\infty}\int_{\mathbb{R}^n}|u( t, x )| \varphi _R( |\eta ^{''}( t ) |+|\eta ^{'}( t)|)dxdt\\ &\lesssim I_{R}^{\frac{1}{p}}\left(\int_{\frac{R^{2\sigma}}{2}}^{R^{2\sigma}}\int_{\mathbb{R}^n}\varphi _R\eta _{R}^{-\frac{p^{'}}{p}}(|\eta ^{''}( t ) |+|\eta ^{'}( t )|)^{p^{'}}dxdt\right)^{\frac{1}{p^{'}}}\\ &\lesssim I_{R}^{\frac{1}{p}}\left( R^{-2\sigma p^{'}+2\sigma +n}\int_{\mathbb{R}^n}{\left< \widetilde{x} \right> ^{-n-2s_{\sigma}}d\widetilde{x}} \right) ^{\frac{1}{^{p^{'}}}}\\& \lesssim I_{R}^{\frac{1}{p}}R^{-2\sigma +\frac{2\sigma +n}{p^{'}}}. \end{align*} $

由Young不等式得

$ \begin{equation} \begin{aligned} &I_R-\widetilde{I}_R+\int_{\mathbb{R}^n}u_1(x)\varphi_R(x)dx+(I+\log(I+(-\bigtriangleup)^{\sigma}))^{-1}\int_{\mathbb{R}^n}u_0(x)\varphi_R(x)dx\\ &\lesssim I_R^{-\frac{1}{p}}R^{-2\sigma+\frac{2\sigma+n}{p^{'}}} \lesssim \frac{1}{p}I_R+\frac{1}{p^{'}}R^{-2\sigma p^{'}+2\sigma +n}, \end{aligned} \end{equation} $

(4.2)

即

$ \begin{equation} \frac{1}{p^{'}}I_R-\widetilde{I}_R+\displaystyle{\int}_{\mathbb{R}^n}u_1\left( x \right) \varphi _R\left( x \right) dx\lesssim \frac{1}{p^{'}}R^{-2\sigma p^{'}+2\sigma +n}, \end{equation} $

(4.3)

因为$ \eta(t) $单调递减, 即$ -\eta_{R}^{'}(t)\geqslant 0, -\widetilde{I}_R\geqslant0 $. 当$ m=1 $时, 若$ 1<p<1+\frac{2\sigma m}{n} $即

$ -2\sigma p^{'}+2\sigma+n<0, $

当$ R\rightarrow \infty $时,

$ \displaystyle{\int}_{\mathbb{R}^n}u_1( x ) \varphi _R( x ) dx\lesssim R^{-2\sigma p^{'}+2\sigma +n}\rightarrow0, $

与$ \int_{\mathbb{R}^n}u_1( x ) \varphi _R( x ) dx>0 $矛盾. 当$ m\in (1, 2) $时, 若$ 1<p<1+\frac{2\sigma m}{n} $即

$ n( 1-\tfrac{1}{m})>-2\sigma p^{'}+2\sigma +n, $

得

$ \begin{equation} \begin{aligned} \int_{\mathbb{R}^n}u_1( x) \varphi _R( x ) dx&\gtrsim \int_{|x|\leqslant R}u_1\left( x \right) \left< \dfrac{x}{R}\right> ^{-n-2s_{\sigma}}dx\\ &\gtrsim \int_{|x|\leqslant R}u_1\left( x \right) dx\\ &\gtrsim \int_{\mathbb{R}^n}|x|^{-\frac{n}{m}}\left( \log \left( |x| \right) \right) ^{-1}dx\\ &\gtrsim \left( \log \left( R \right) \right) ^{-1}R^{n\left( 1-\frac{1}{m} \right)}, \end{aligned} \end{equation} $

(4.4)

当$ R\rightarrow \infty $时,

$ ( \log ( R ) ) ^{-1}R^{n( 1-\frac{1}{m})}\lesssim \displaystyle{\int}_{\mathbb{R}^n}u_1( x) \varphi _R( x ) dx\lesssim R^{-2\sigma p^{'}+2\sigma +n}, $

即

$ \left( \log \left( R \right) \right) ^{-1}R^{n\left( 1-\frac{1}{m} \right)-(-2\sigma p^{'}+2\sigma +n)}\lesssim1, $

得$ n\left( 1-\frac{1}{m} \right) \leqslant -2\sigma p^{'}+2\sigma+n $与$ n\left( 1-\frac{1}{m} \right)>-2\sigma p^{'}+2\sigma +n $矛盾, 故定理$ 1.2 $得证.