本文研究如下柯西问题

(1.1) $ \begin{equation} \left\{ \begin{array}{lll} {\rm i}u_t = \Delta u+2\alpha u|u|^{2\alpha-2}\Delta (|u|^{2\alpha})+V(x)u+(W*|u|^2)u\\ \qquad +A_1|u|^{r_1-1}u+\cdots+A_m|u|^{r_m-1}u\\ \qquad -B_1|u|^{s_1-1}u-\cdots-B_n|u|^{s_n-1}u, \quad x\in {{\Bbb R}} ^N, \quad t>0\\ u(x, 0) = u_0(x), \quad x\in {{\Bbb R}} ^N, \end{array}\right. \end{equation} $

其中$ N\geq3 $,系数$ \alpha, r_1, \cdots , r_m, s_1, \cdots , s_n $,实值函数$ V(x) $, $ W(x) $满足如下条件

(CD1)  $ \alpha, r_1, \cdots , r_m, s_1, \cdots , s_n\in Z^+ $$ (Z^+ $表示正整数集), $ 1<r_1<\cdots<r_m $, $ 1<s_1<\cdots<s_n $, $ A_1, \cdots, $$ A_m\geq0, $$ B_1, \cdots, B_n\geq0 $;

(CD2)  $ V(x)\geq 0 $, $ V(x)\in {\mathfrak B}^{\infty}({{\Bbb R}} ^N) $, $ W(x) $是偶函数,并且$ \partial^KW(x)\in L^1({{\Bbb R}} ^N) $, $ K\in Z^+ $, $ W(x) = W_1(x)+W_2(x) $, $ W_1(x)\in L^{p}({{\Bbb R}} ^N) $, $ p>1 $, $ W_2(x)\in L^{\infty}({{\Bbb R}} ^N) $,其中

$ u_0 \in \Lambda: = \{v\in H^{1}({{\Bbb R}} ^N)| \int_{{{\Bbb R}} ^N}|\nabla (|v|^{2\alpha})|^2<\infty\}, $

$ {\mathfrak B}^{\infty}({{\Bbb R}} ^N) $定义为由所有属于$ C^{\infty}({{\Bbb R}} ^N) $且任意阶偏导数在$ {{\Bbb R}} ^N $中有界的函数构成的函数空间.

方程(1.1)通常出现在等离子体物理、流体力学以及凝聚态物理中.关于该问题解的适定性已有很多结果^[1-4].本文主要关注解的全局存在和有限时刻爆破现象.

首先给出方程(1.1)的解的全局存在和有限时刻爆破的定义.

定义1.1 假设$ u(x, t) $是方程(1.1)的解,如果$ u(x, t) $关于时间的最大存在区间是$ [0, +\infty) $,我们称$ u(x, t) $是全局存在的.反之,如果存在$ 0<T_0<+\infty $使得

(1.2) $ \begin{eqnarray} \lim\limits_{t\rightarrow T_0^-} \int_{{{\Bbb R}} ^N}\left[|\nabla u(x, t)|^2+|\nabla(|u(x, t)|^{2\alpha})|^2\right]{\rm d}x = +\infty, \end{eqnarray} $

我们称$ u(x, t) $会在有限时刻爆破.

其次,我们回顾下相关的文献. 1977年Glassey$ ^{[5]} $研究了非线性薛定谔方程$ {\rm i}u_t = \Delta u+F(|u|^2)u $的初值问题.给出了证明有限时刻爆破的方法和一些引理.此后越来越多的学者开始研究非线性薛定谔方程的全局存在和有限时刻爆破问题,特别是带有Hartree项的半线性薛定谔方程问题,参见文献[6-11].然而,研究拟线性薛定谔方程的全局存在和爆破现象的论文很少,参见文献[12-14].

定义方程(1.1)解的质量和能量如下.

定义1.2 $ \rm(i) $质量

$ m(u) = \left(\int_{{{\Bbb R}} ^N}|u(\cdot, t)|^2{\rm d}x\right)^{\frac{1}{2}}: = [M(u)]^{\frac{1}{2}}; $

$ \rm(ii) $能量

$ \begin{eqnarray*} E(u)& = &\frac{1}{2}\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x-\frac{1}{2}\int_{{{\Bbb R}} ^N}V|u|^2{\rm d}x-\frac{1}{4}\int_{{{\Bbb R}} ^N}(W*|u|^2)|u|^2{\rm d}x\nonumber\\ &&-\int_{{{\Bbb R}} ^N}\left(\frac{A_1}{r_1+1}|u|^{r_1+1}+\cdots+\frac{A_m}{r_m+1}|u|^{r_m+1}\right){\rm d}x\nonumber\\ &&+\int_{{{\Bbb R}} ^N}\left(\frac{B_1}{s_1+1}|u|^{s_1+1}+\cdots+\frac{B_n}{s_n+1}|u|^{s_n+1}\right){\rm d}x. \end{eqnarray*} $

我们的第一个结果建立了方程(1.1)解的全局存在的充分性条件.

定理1.1 假设$ u(x, t) $为方程(1.1)的解, $ u_0\in \Lambda $, $ E(u_0)\in (0, +\infty), M(u_0)\in (0, +\infty), $满足条件(CD1), (CD2),那么当下列任一条件成立时,解$ u(x, t) $全局存在.即对于任意$ t>0 $,

$ \int_{{{\Bbb R}} ^N}[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2]{\rm d}x\leq C. $

$ \rm(SA1) $  $ p>1 $, $ r_m<s_n $;

$ \rm(SA2) $  $ p>1 $, $ s_n\leq r_m $, $ r_m<\frac{4\alpha N-N+4}{N} $;

$ \rm(SA3) $ $ p>1 $, $ s_n\leq r_m $, $ r_m = \frac{4\alpha N-N+4}{N} $,对任意初值$ u_0 \in \Lambda $满足

$ C_2 = \frac{2A_m}{r_m+1}\|u_0\|_{L^2({{\Bbb R}} ^N)}^{\frac{4\alpha N-(r_m+1)(N-2)}{2\alpha N-N+2}}C_s^{\frac{(N-2)(r_m-1)}{4\alpha N-2N+4}}\leq\frac{1}{4}. $

这里$ C_s $表示为Sobolev不等式的最佳常数($ 2^* = \frac{2N}{N-2} $, $ N>2 $),

(1.3) $ \begin{eqnarray} \int_{{{\Bbb R}} ^N}w^{2^*}{\rm d}x\leq C_s\left(\int_{{{\Bbb R}} ^N}|\nabla w|^2{\rm d}x\right)^{\frac{2^*}{2}}, \quad w\in H^1(\mathbb{R^N}). \end{eqnarray} $

我们的第二个结果建立了方程(1.1)解的有限时刻爆破的充分性条件.

定理1.2 假设$ u(x, t) $为方程(1.1)的解, $ u_0\in \Lambda $, $ E(u_0)\in (-\infty, 0), M(u_0)\in (0, +\infty), $满足条件$ \rm(CD1), (CD2) $,并且

$ (x\cdot \nabla V)+(2\alpha N-N+2)V\leq 0, \quad (x\cdot \nabla W)+(2\alpha N-N+2)W\leq 0, $

$ |x|u_0\in L^2({{\Bbb R}} ^N), \quad \Im \int_{{{\Bbb R}} ^N}\bar{u_0}(x\cdot \nabla {u_0}){\rm d}x>0, $

存在正常数$ C_\varepsilon $满足

$ 2\left[(2\alpha-1)N+2\right]E(u_0)+C_\varepsilon M(u_0) \leq 0. $

那么当下列任一条件成立时,解$ u(x, t) $在有限时间$ T_0 $处爆破,也就是说

$ \lim\limits_{t\rightarrow T_0^-}\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x = +\infty. $

$ \rm(SA1) $ $ r_m>s_n $, $ r_m\geq4\alpha-1+\frac{4}{N} $;

$ \rm(SA2) $ $ r_m = s_n $, $ r_m\geq4\alpha-1+\frac{4}{N} $, $ A_m\geq B_n $;

$ \rm(SA3) $ $ r_m<s_n $, $ s_n\leq4\alpha-1+\frac{4}{N} $.

这里

$ T_0 = \frac{J(0)}{(2\alpha-1)N\cdot F(0)}, $

$ J(t) = \int_{{{\Bbb R}} ^N}|x|^2|u|^2{\rm d}x, \quad F(t) = \Im \int_{{{\Bbb R}} ^N}\bar{u}(x\cdot \nabla u){\rm d}x. $

本节将证明一个引理.以下用符号$ C $, $ C' $, $ \tilde{C} $, $ \hat{C} $, $ \cdots $, $ \varepsilon $定义与变量$ x $, $ t $无关的有限正常数.

引理2.1 假设$ u(x, t) $为方程(1.1)的解, $ u_0\in \Lambda $,则解$ u $在其存在的时间区间$ [0, t] $内满足

$ \rm(1) $质量守恒

$ m(u) = \left(\int_{{{\Bbb R}} ^N}|u(x, t)|^2{\rm d}x\right)^{\frac{1}{2}} = \left(\int_{{{\Bbb R}} ^N}|u_0(x)|^2{\rm d}x\right)^{\frac{1}{2}} = [M(u)]^{\frac{1}{2}} = [M(u_0)]^{\frac{1}{2}}; $

$ \rm(2) $能量守恒

$ \begin{eqnarray*} E(u)& = &\frac{1}{2}\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x-\frac{1}{2}\int_{{{\Bbb R}} ^N}V|u|^2{\rm d}x-\frac{1}{4}\int_{{{\Bbb R}} ^N}(W*|u|^2)|u|^2{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}\left(\frac{A_1}{r_1+1}|u|^{r_1+1}+\cdots+\frac{A_m}{r_m+1}|u|^{r_m+1}\right){\rm d}x\\ & &+\int_{{{\Bbb R}} ^N}\left(\frac{B_1}{s_1+1}|u|^{s_1+1}+\cdots+\frac{B_n}{s_n+1}|u|^{s_n+1}\right){\rm d}x = E(u_0); \end{eqnarray*} $

$ \rm(3) $

$ \frac{\partial}{\partial t} \int_{{{\Bbb R}} ^N}|x|^2|u|^2{\rm d}x = -4\Im \int_{{{\Bbb R}} ^N} \bar{u}(x\cdot \nabla u){\rm d}x; $

$ \rm(4) $

$ \begin{eqnarray*} \frac{\partial}{\partial t} \Im \int_{{{\Bbb R}} ^N} \bar{u}(x\cdot \nabla u){\rm d}x & = &-2\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x-\left[({2\alpha}-1)N+2\right]\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}(x\cdot {\nabla V})|u|^2{\rm d}x-{1\over2}\int_{{{\Bbb R}} ^N}\left[(x\cdot {\nabla W})*|u|^2\right]|u|^2{\rm d}x\\ &&+\int_{{{\Bbb R}} ^N}\left[\frac{NA_1(r_1-1)}{r_1+1}|u|^{r_1+1}+\cdots+\frac{NA_m(r_m-1)}{r_m+1}|u|^{r_m+1}\right]{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}\left[\frac{NB_1(s_1-1)}{s_1+1}|u|^{s_1+1}+\cdots+\frac{NB_n(s_n-1)}{s_n+1}|u|^{s_n+1}\right]{\rm d}x. \end{eqnarray*} $

证 (1) (1.1)式左右同乘以$ 2\bar{u} $,计算结果取虚部,得

(1.1) $ \begin{eqnarray} \frac{\partial}{\partial t}|u|^2 = \Im(2\bar{u}\Delta u) = \nabla \cdot(2\Im(\bar{u}\nabla u)), \end{eqnarray} $

然后在$ {{\Bbb R}} ^N\times [0, t] $上积分,得

$ M(u) = \int_{{{\Bbb R}} ^N}|u|^2{\rm d}x = \int_{{{\Bbb R}} ^N}|u_0|^2{\rm d}x = M(u_0). $

(2) (1.1)式左右同乘以$ 2\bar{u}_t $,计算结果取实部,然后在$ {{\Bbb R}} ^N\times [0, t] $上积分,得

$ \begin{eqnarray*} &&\frac{\partial}{\partial t}\left\{\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x-\int_{{{\Bbb R}} ^N}V|u|^2{\rm d}x-\frac{1}{2}\int_{{{\Bbb R}} ^N}(W*|u|^2)|u|^2{\rm d}x\right.\\ &&-\left.\int_{{{\Bbb R}} ^N}\left(\frac{2A_1}{r_1+1}|u|^{r_1+1}+\cdots+\frac{2A_m}{r_m+1}|u|^{r_m+1}\right){\rm d}x\right.\\ &&+\left.\int_{{{\Bbb R}} ^N}\left(\frac{2B_1}{s_1+1}|u|^{s_1+1}+\cdots+\frac{2B_n}{s_n+1}|u|^{s_n+1}\right){\rm d}x\right\} = 0, \end{eqnarray*} $

也就是说

$ \begin{eqnarray*} E(u)& = &\frac{1}{2}\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x-\frac{1}{2}\int_{{{\Bbb R}} ^N}V|u|^2{\rm d}x-\frac{1}{4}\int_{{{\Bbb R}} ^N}(W*|u|^2)|u|^2{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}\left(\frac{A_1}{r_1+1}|u|^{r_1+1}+\cdots+\frac{A_m}{r_m+1}|u|^{r_m+1}\right){\rm d}x\\ & &+\int_{{{\Bbb R}} ^N}\left(\frac{B_1}{s_1+1}|u|^{s_1+1}+\cdots+\frac{B_n}{s_n+1}|u|^{s_n+1}\right){\rm d}x = E(u_0). \end{eqnarray*} $

(3) (1.1)式左右同乘以$ |x|^2 $,然后在$ {{\Bbb R}} ^N $上积分,得

(2.2) $ \begin{eqnarray} \frac{\partial}{\partial t}\int_{{{\Bbb R}} ^N}|x|^2|u|^2{\rm d}x = \int_{{{\Bbb R}} ^N}|x|^2\nabla \cdot(2\Im (\bar{u}\nabla u)){\rm d}x = -4\Im \int_{{{\Bbb R}} ^N}\bar{u}(x\cdot \nabla u){\rm d}x. \end{eqnarray} $

(4)定义$ u(x, t) = a(x, t)+{\rm i}b(x, t) $,其中$ a(x, t) = \Re u(x, t) $, $ b(x, t) = \Im u(x, t) $.

(2.3) $ \begin{eqnarray} &&\frac{\partial}{\partial t}\Im \int_{{{\Bbb R}} ^N}\bar{u}(x\cdot \nabla u){\rm d}x\\ & = &\int_{{{\Bbb R}} ^N} \sum\limits_{k = 1}^N\left[x_k(b_t)_{x_k}a-x_k(a_t)_{x_k}b\right]{\rm d}x +\int_{{{\Bbb R}} ^N} \sum\limits_{k = 1}^N (x_ka_{x_k}\Delta a+x_kb_{x_k}\Delta b){\rm d}x\\ &&+\int_{{{\Bbb R}} ^N}\sum\limits_{k = 1}^N x_k(|u|^2)_{x_k}\alpha|u|^{2\alpha-2}\Delta(|u|^{2\alpha}){\rm d}x\\ &&+\frac{1}{2}\int_{{{\Bbb R}} ^N}\sum\limits_{k = 1}^N x_k(|u|^2)_{x_k}V{\rm d}x+\frac{1}{2}\int_{{{\Bbb R}} ^N}\sum\limits_{k = 1}^N x_k(|u|^2)_{x_k}(W*|u|^2){\rm d}x\\ &&+\frac{1}{2}\int_{{{\Bbb R}} ^N}\sum\limits_{k = 1}^N x_k(|u|^2)_{x_k}\left(A_1|u|^{r_1-1}+\cdots+A_m|u|^{r_m-1}\right){\rm d}x\\ &\quad&-\frac{1}{2}\int_{{{\Bbb R}} ^N}\sum\limits_{k = 1}^N x_k(|u|^2)_{x_k}\left(B_1|u|^{s_1-1}+\cdots+B_n|u|^{s_n-1}\right){\rm d}x\\ & = &N\int_{{{\Bbb R}} ^N}(a_tb-ab_t){\rm d}x+\int_{{{\Bbb R}} ^N}\sum\limits_{k = 1}^N(x_kb_{x_k}a_t-x_ka_{x_k}b_t){\rm d}x\\ &\quad&+\frac{N-2}{2}\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x+\frac{N-2}{2}\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\\ &\quad&-\frac{1}{2}\int_{{{\Bbb R}} ^N}(NV+x\cdot{\nabla V})|u|^2{\rm d}x-\frac{1}{2}\int_{{{\Bbb R}} ^N}\left[(NW+\frac{x\cdot{\nabla W}}{2})*|u|^2\right]|u|^2{\rm d}x\\ &\quad&-\int_{{{\Bbb R}} ^N}\left[\frac{N A_1}{r_1+1}|u|^{r_1+1}+\cdots+\frac{N A_m}{r_m+1}|u|^{r_m+1}\right]{\rm d}x\\ &&+\int_{{{\Bbb R}} ^N}\left[\frac{N B_1}{s_1+1}|u|^{s_1+1}+\cdots+\frac{N B_n}{s_n+1}|u|^{s_n+1}\right]{\rm d}x\\ & = &-N\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x+{2\alpha}N\int_{{{\Bbb R}} ^N}|u|^{2\alpha}\Delta(|u|^{2\alpha}){\rm d}x+N\int_{{{\Bbb R}} ^N}V|u|^2{\rm d}x\\ &\quad& +N\int_{{{\Bbb R}} ^N}(W*|u|^2)|u|^2{\rm d}x+N\int_{{{\Bbb R}} ^N}(A_1|u|^{r_1-1}+\cdots+A_m|u|^{r_m-1})|u|^2{\rm d}x\\ &\quad& -N\int_{{{\Bbb R}} ^N}(B_1|u|^{s_1-1}+\cdots+B_n|u|^{s_n-1})|u|^2{\rm d}x+(N-2)\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x\\ &\quad& +(N-2)\int_{{{\Bbb R}} ^N}|\nabla(|u|^{2\alpha})|^2{\rm d}x-\int_{{{\Bbb R}} ^N}(NV+x\cdot \nabla V)|u|^2{\rm d}x\\ &\quad& -\int_{{{\Bbb R}} ^N}\left[(NW+\frac{x\cdot \nabla W}{2})*|u|^2\right]|u|^2{\rm d}x\\ &\quad &-2\int_{{{\Bbb R}} ^N}\left(\frac{N A_1}{r_1+1}|u|^{r_1+1}+\cdots+\frac{N A_m}{r_m+1}|u|^{r_m+1}\right){\rm d}x\\ &\quad &+2\int_{{{\Bbb R}} ^N}\left(\frac{N B_1}{s_1+1}|u|^{s_1+1}+\cdots+\frac{N B_n}{s_n+1}|u|^{s_n+1}\right){\rm d}x\\ & = &-2\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x-\left[({2\alpha}-1)N+2\right]\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\\ &\quad& -\int_{{{\Bbb R}} ^N}(x\cdot {\nabla V})|u|^2{\rm d}x-{1\over2}\int_{{{\Bbb R}} ^N}\left[(x\cdot {\nabla W})*|u|^2\right]|u|^2{\rm d}x\\ &\quad &+\int_{{{\Bbb R}} ^N}\left[\frac{NA_1(r_1-1)}{r_1+1}|u|^{r_1+1}+\cdots+\frac{NA_m(r_m-1)}{r_m+1}|u|^{r_m+1}\right]{\rm d}x\\ &\quad& -\int_{{{\Bbb R}} ^N}\left[\frac{NB_1(s_1-1)}{s_1+1}|u|^{s_1+1}+\cdots+\frac{NB_n(s_n-1)}{s_n+1}|u|^{s_n+1}\right]{\rm d}x. \end{eqnarray} $

引理2.1证毕.

本节将给出定理1.1的证明,并建立方程(1.1)解的全局存在的充分性条件.

证 由引理2.1(2)能量守恒得

(3.1) $ \begin{eqnarray} &\quad &\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x\\ & = &\left(2E(u_0)+\int_{{{\Bbb R}} ^N}V|u|^2{\rm d}x+\frac{1}{2}\int_{{{\Bbb R}} ^N}(W*|u|^2)|u|^2{\rm d}x\right)\\ &\quad& +\left\{\int_{{{\Bbb R}} ^N}\left(\frac{2A_1}{r_1+1}|u|^{r_1+1}+\cdots+\frac{2A_m}{r_m+1}|u|^{r_m+1}\right){\rm d}x\right.\\ &\quad &\left.-\int_{{{\Bbb R}} ^N}\left(\frac{2B_1}{s_1+1}|u|^{s_1+1}+\cdots+\frac{2B_n}{s_n+1}|u|^{s_n+1}\right){\rm d}x\right\}: = (I)+{II}. \end{eqnarray} $

下面分别讨论(I), {II}式.

首先由质量守恒, H$ \ddot{\rm o} $lder不等式,插值不等式, Sobolev不等式计算得

(3.2) $ \begin{eqnarray} (I)& = &2E(u_0)+\int_{{{\Bbb R}} ^N}V|u|^2{\rm d}x+\frac{1}{2}\int_{{{\Bbb R}} ^N}(W*|u|^2)|u|^2{\rm d}x\\ &\leq &2E(u_0)+\|V\|_{L^{\infty}({{\Bbb R}} ^N)}\|u\|_{L^2({{\Bbb R}} ^N)}^2+\frac{1}{2}\|W_1\|_{L^{p}({{\Bbb R}} ^N)}\|u\|_{L^{\frac{4p}{2p-1}}({{\Bbb R}} ^N)}^4{}\\ &&+\frac{1}{2}\|W_2\|_{L^{\infty}({{\Bbb R}} ^N)}\|u\|_{L^2({{\Bbb R}} ^N)}^4\\ &\leq &C_0+\frac{1}{2}\|W_1\|_{L^{p}({{\Bbb R}} ^N)}\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{4p(2\alpha N-N+2)-4\alpha N}{p(2\alpha N-N+2)}}\|u\|_{L^{2\alpha \cdot 2^*}({{\Bbb R}} ^N)}^{\frac{4\alpha N}{p(2\alpha N-N+2)}}\\ &\leq &C_0+\frac{1}{2}\|W_1\|_{L^{p}({{\Bbb R}} ^N)}\|u_0\|_{L^2({{\Bbb R}} ^N)}^{\frac{4p(2\alpha N-N+2)-4\alpha N}{p(2\alpha N-N+2)}} C_s^{\frac{N-2}{p(2\alpha N-N+2)}}\left(\int_{{{\Bbb R}} ^N}|\nabla(|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N}{p(2\alpha N-N+2)}}\\ & = &C_0+C_1\left(\int_{{{\Bbb R}} ^N}|\nabla(|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N}{p(2\alpha N-N+2)}}, \end{eqnarray} $

其中

$ C_0 = 2E(u_0)+\|V\|_{L^{\infty}({{\Bbb R}} ^N)}\|u_0\|_{L^2({{\Bbb R}} ^N)}^2+\frac{1}{2}\|W_2\|_{L^{\infty}({{\Bbb R}} ^N)}\|u_0\|_{L^2({{\Bbb R}} ^N)}^4, $

$ C_1 = \frac{1}{2}\|W_1\|_{L^{p}({{\Bbb R}} ^N)}\|u_0\|_{L^2({{\Bbb R}} ^N)}^{\frac{4p(2\alpha N-N+2)-4\alpha N}{p(2\alpha N-N+2)}} C_s^{\frac{N-2}{p(2\alpha N-N+2)}}. $

将{II}式分为以下两种情况讨论.

(i) $ r_m<s_n $.取常数$ 0<\varepsilon_1, \cdots, \varepsilon_m\leq\frac{1}{m}<1 $,

(3.3) $ \begin{eqnarray} && \left(\frac{2A_1}{r_1+1}\|u\|_{L^{r_1+1}({{\Bbb R}} ^N)}^{r_1+1}+\cdots+\frac{2A_m}{r_m+1}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}\right)\\ &&-\left(\frac{2B_1}{s_1+1}\|u\|_{L^{s_1+1}({{\Bbb R}} ^N)}^{s_1+1}+\cdots+\frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}\right)\\ &\leq & \frac{2A_1}{r_1+1}\|u\|_{L^{r_1+1}({{\Bbb R}} ^N)}^{r_1+1}+\cdots+\frac{2A_m}{r_m+1}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}-\frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}\\ &\leq& \frac{2A_1}{r_1+1}\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{2(s_n-r_1)}{s_n-1}}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{\frac{(r_1+1)(s_n-1)-2(s_n-r_1)}{s_n-1}}\\ && +\cdots+\frac{2A_m}{r_m+1}\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{2(s_n-r_m)}{s_n-1}}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{\frac{(r_m+1)(s_n-1)-2(s_n-r_m)}{s_n-1}}-\frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}\\ &\leq& C_{\varepsilon_1}\|u\|_{L^2({{\Bbb R}} ^N)}^2+\varepsilon_1\cdot \frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}\\ && +\cdots+C_{\varepsilon_m}\|u\|_{L^2({{\Bbb R}} ^N)}^2+\varepsilon_m\cdot \frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}-\frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}\\ & = &\|u_0\|_{L^2({{\Bbb R}} ^N)}^2\cdot C_{\varepsilon}-\frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}\left(1-\varepsilon_1-\cdots-\varepsilon_m\right)\\ &\leq &M(u_0)\cdot C_{\varepsilon}, \end{eqnarray} $

其中$ C_{\varepsilon} = C_{\varepsilon_1}+\cdots+C_{\varepsilon_m} $.

(ii) $ r_m\geq s_n $.不失一般性,假设$ r_{m-1}<s_n $,取常数$ 0<\varepsilon_1, \cdots, \varepsilon_{m-1}\leq\frac{1}{m-1}<1 $,

(3.4) $ \begin{eqnarray} &&\left(\frac{2A_1}{r_1+1}\|u\|_{L^{r_1+1}({{\Bbb R}} ^N)}^{r_1+1}+\cdots+\frac{2A_m}{r_m+1}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}\right)\\ &&-\left(\frac{2B_1}{s_1+1}\|u\|_{L^{s_1+1}({{\Bbb R}} ^N)}^{s_1+1}+\cdots+\frac{2B_n}{s_n+1}\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}\right)\\ &\leq& M(u_0)\cdot C_{\varepsilon}+\frac{2A_m}{r_m+1}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}\\ &\leq &M(u_0)\cdot C_{\varepsilon}+\frac{2A_m}{r_m+1}\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{4\alpha N-(r_m+1)(N-2)}{2\alpha N-N+2}}\|u\|_{L^{2\alpha \cdot {2^*}}({{\Bbb R}} ^N)}^{\frac{2\alpha N(r_m+1)-4\alpha N}{2\alpha N-N+2}}\\ &\leq &M(u_0)\cdot C_{\varepsilon}+\frac{2A_m}{r_m+1}\|u_0\|_{L^2({{\Bbb R}} ^N)}^{\frac{4\alpha N-(r_m+1)(N-2)}{2\alpha N-N+2}}C_s^{\frac{(N-2)(r_m-1)}{4\alpha N-2N+4}}\left(\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N(r_m-1)}{4\alpha N-2N+4}}\\ & = & M(u_0)\cdot C_{\varepsilon}+C_2\left(\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N(r_m-1)}{4\alpha N-2N+4}}. \end{eqnarray} $

其中

$ C_2 = \frac{2A_m}{r_m+1}\|u_0\|_{L^2({{\Bbb R}} ^N)}^{\frac{4\alpha N-(r_m+1)(N-2)}{2\alpha N-N+2}}C_s^{\frac{(N-2)(r_m-1)}{4\alpha N-2N+4}}. $

下面依据定理1.1的三种情形依次证明之.

(SA1)  $ p>1 $, $ r_m<s_n $.在这种情况下

$ \frac{N}{p(2\alpha N-N+2)}<1, \quad \frac{4p}{2p-1}<2\alpha \cdot 2^*. $

然后用Young不等式,有

(3.5) $ \begin{eqnarray} && \int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x\\ &\leq& C_0+C_1\left(\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N}{p(2\alpha N-N+2)}}+M(u_0)\cdot{C}_{\varepsilon}\\ &\leq &C_0+M(u_0)\cdot{C}_{\varepsilon}+C_{\varepsilon_1}\cdot C_1^{\frac{p(2\alpha N-N+2)}{p(2\alpha N-N+2)-N}}+\frac{1}{2}\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x, \end{eqnarray} $

因此

$ \int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x\leq C. $

(SA2)  $ p>1 $, $ s_n\leq r_m< \frac{4\alpha N-N+4}{N} $.则有$ \frac{N}{p(2\alpha N-N+2)}<1 $, $ \frac{4p}{2p-1}<2\alpha \cdot 2^* $,

$ \frac{N(r_m-1)}{4\alpha N-2N+4}<1, \quad 1<r_m<2\alpha \cdot 2^*-1. $

从而得到

(3.6) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x\\ &\leq& C_0+C_1\left(\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N}{p(2\alpha N-N+2)}}+M(u_0)\cdot{C}_{\varepsilon} +C_2\left(\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N(r_m-1)}{4\alpha N-2N+4}}\\ &\leq& C'(u_0, p, \alpha)+\frac{1}{4}\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x+C_{\varepsilon_2}\cdot C_2^{\frac{4\alpha N-2N+4}{(4\alpha-r_m-1)N+4}}+\frac{1}{4}\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x, \end{eqnarray} $

这意味着

$ \int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x\leq C. $

(SA3)  $ p>1 $, $ s_n\leq r_m( = \frac{4\alpha N-N+4}{N}) $.因此$ \frac{N}{p(2\alpha N-N+2)}<1 $, $ \frac{4p}{2p-1}<2\alpha \cdot 2^* $, $ \frac{N(r_m-1)}{4\alpha N-2N+4} = 1, $$ 1<r_m<2\alpha \cdot 2^*-1 $.

若初值$ u_0 \in \Lambda $满足$ C_2\leq\frac{1}{4} $,则

(3.7) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x\\ &\leq& C_0+C_1\left(\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\right)^{\frac{N}{p(2\alpha N-N+2)}}+M(u_0)\cdot{C}_{\varepsilon}+C_2\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\\ &\leq &C'(u_0, p, \alpha)+\frac{1}{4}\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x+\frac{1}{4}\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x, \end{eqnarray} $

从而

$ \int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x\leq C. $

定理1.1证毕.

本节将给出定理1.2的证明,并建立方程(1.1)解的有限时刻爆破的充分性条件.

证 假设$ u(x, t) $为方程(1.1)的解, $ u_0\in \Lambda $,满足条件(CD1), (CD2),并且

(4.1) $ \begin{eqnarray} (x\cdot {\nabla V})+(2\alpha N-N+2)V\leq0, \quad (x\cdot {\nabla W})+(2\alpha N-N+2)W\leq0. \end{eqnarray} $

在$ u $存在的时间区间内,令

$ J(t) = \int_{{{\Bbb R}} ^N}|x|^2|u|^2{\rm d}x, \quad F(t) = \Im \int_{{{\Bbb R}} ^N}\bar{u}(x\cdot \nabla u){\rm d}x. $

由引理2.1(2)能量守恒得

(4.2) $ \begin{eqnarray} \frac{\partial}{\partial t}F(t)& = &-2\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x-\left[({2\alpha}-1)N+2\right]\int_{{{\Bbb R}} ^N}|\nabla (|u|^{2\alpha})|^2{\rm d}x\\ &\quad&-\int_{{{\Bbb R}} ^N}(x\cdot {\nabla V})|u|^2{\rm d}x-\frac{1}{2}\int_{{{\Bbb R}} ^N}\left[(x\cdot {\nabla W})*|u|^2\right]|u|^2{\rm d}x\\ &\quad&+\int_{{{\Bbb R}} ^N}\left[\frac{NA_1(r_1-1)}{r_1+1}|u|^{r_1+1}+\cdots+\frac{NA_m(r_m-1)}{r_m+1}|u|^{r_m+1}\right]{\rm d}x\\ &\quad &-\int_{{{\Bbb R}} ^N}\left[\frac{NB_1(s_1-1)}{s_1+1}|u|^{s_1+1}+\cdots+\frac{NB_n(s_n-1)}{s_n+1}|u|^{s_n+1}\right]{\rm d}x\\ & = &\left(-2\left[(2\alpha-1)N+2\right]E(u_0)+(2\alpha-1)N\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x\right.\\ &&\left.-\int_{{{\Bbb R}} ^N}\left[x\cdot {\nabla V}+(2\alpha N-N+2)V\right]|u|^2{\rm d}x\right.\\ &&\left.-\frac{1}{2}\int_{{{\Bbb R}} ^N}\left[(x\cdot {\nabla W}+(2\alpha N-N+2)W)*|u|^2\right]|u|^2{\rm d}x\right)\\ &&+\left(\int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}\right]{\rm d}x\right.\\ &&\left.-\int_{{{\Bbb R}} ^N}\left[NB_1(1-\frac{4\alpha+\frac{4}{N}}{s_1+1})|u|^{s_1+1}+\cdots+NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})|u|^{s_n+1}\right]{\rm d}x\right)\\ &: = &(I)+(II). \end{eqnarray} $

据假设(4.1)得

(4.3) $ \begin{eqnarray} (I)& = &-2\left[(2\alpha-1)N+2\right]E(u_0)+(2\alpha-1)N\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x\\ &\quad&-\int_{{{\Bbb R}} ^N}\left[x\cdot {\nabla V}+(2\alpha N-N+2)V\right]|u|^2{\rm d}x\\ &\quad &-\frac{1}{2}\int_{{{\Bbb R}} ^N}\left[(x\cdot {\nabla W}+(2\alpha N-N+2)W)*|u|^2\right]|u|^2{\rm d}x\\ &\geq &-2\left[(2\alpha-1)N+2\right]E(u_0)+(2\alpha-1)N\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x. \end{eqnarray} $

而(4.2)式中的(II)式需分为以下三种情况讨论.

(SA1)  $ r_m>s_n $, $ r_m\geq4\alpha-1+\frac{4}{N} $.取常数$ 0\leq k\leq m-1 $, $ 0\leq l\leq n $ ($ k, \ l\in N $)满足条件

$ 0<\tilde{\varepsilon}_1, \cdots, \tilde{\varepsilon}_{n+k-l}\leq\frac{1}{n+k-l}<1. $

假设

$ r_1<\cdots<r_k<4\alpha-1+\frac{4}{N}\leq r_{k+1}<\cdots<r_m, $

$ s_1<\cdots<s_l<4\alpha-1+\frac{4}{N}\leq s_{l+1}<\cdots<s_n. $

则由Young不等式得

(4.4) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}\right]{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}\left[NB_1(1-\frac{4\alpha+\frac{4}{N}}{s_1+1})|u|^{s_1+1}+\cdots+NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})|u|^{s_n+1}\right]{\rm d}x\\ &\geq& \int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+NA_k(1-\frac{4\alpha+\frac{4}{N}}{r_k+1})|u|^{r_k+1}\right]{\rm d}x\\ &&+\int_{{{\Bbb R}} ^N}NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}\left[NB_{l+1}(1-\frac{4\alpha+\frac{4}{N}}{s_{l+1}+1})|u|^{s_{l+1}+1}+\cdots+NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})|u|^{s_n+1}\right]{\rm d}x\\ &\geq& NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{2(r_m-r_1)}{r_m-1}}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{\frac{(r_1+1)(r_m-1)-2(r_m-r_1)}{r_m-1}}\\ && +\cdots+NA_k(1-\frac{4\alpha+\frac{4}{N}}{r_k+1})\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{2(r_m-r_k)}{r_m-1}}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{\frac{(r_k+1)(r_m-1)-2(r_m-r_k)}{r_m-1}}\\ &&+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}\\ &&-NB_{l+1}(1-\frac{4\alpha+\frac{4}{N}}{s_{l+1}+1})\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{2(r_m-s_{l+1})}{r_m-1}}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{\frac{(s_{l+1}+1)(r_m-1)-2(r_m-s_{l+1})}{r_m-1}}\\ && -\cdots-NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})\|u\|_{L^2({{\Bbb R}} ^N)}^{\frac{2(r_m-s_n)}{r_m-1}}\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{\frac{(s_n+1)(r_m-1)-2(r_m-s_n)}{r_m-1}}\\ & = &-\tilde{C}_{\varepsilon_1}\|u\|_{L^2({{\Bbb R}} ^N)}^2-\tilde{\varepsilon}_1\cdot NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}-\cdots-\tilde{C}_{\varepsilon_k}\|u\|_{L^2({{\Bbb R}} ^N)}^2\\ &&-\tilde{\varepsilon}_k\cdot NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}\\ && -\tilde{C}_{\varepsilon_{k+1}}\|u\|_{L^2({{\Bbb R}} ^N)}^2-\tilde{\varepsilon}_{k+1}\cdot NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}\\ && -\cdots-\tilde{C}_{\varepsilon_{n+k-l}}\|u\|_{L^2({{\Bbb R}} ^N)}^2-\tilde{\varepsilon}_{n+k-l}\cdot NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}\\ & = &-\|u_0\|_{L^2({{\Bbb R}} ^N)}^2\cdot\tilde{C}_{\varepsilon}+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}(1-\tilde{\varepsilon}_1-\cdots-\tilde{\varepsilon}_{n+k-l})\\ &\geq&-M(u_0)\cdot \tilde{C}_\varepsilon, \end{eqnarray} $

其中$ \tilde{C}_\varepsilon = \tilde{C}_{\varepsilon_1}+\cdots+\tilde{C}_{\varepsilon_{n+k-l}} $.

(SA2)  $ r_m = s_n $, $ r_m\geq4\alpha-1+\frac{4}{N} $, $ A_m\geq B_n $.取常数$ 0\leq k\leq m-1 $, $ 0\leq l\leq n-1 $ ($ k, \ l\in N $)满足条件

$ 0<\hat{\varepsilon}_1, \cdots, \hat{\varepsilon}_{n+k-l-1}\leq\frac{1}{n+k-l-1}<1. $

参考(SA1)的结果,有

(4.5) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}\right]{\rm d}x\\ && -\int_{{{\Bbb R}} ^N}\left[NB_1(1-\frac{4\alpha+\frac{4}{N}}{s_1+1})|u|^{s_1+1}+\cdots+NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})|u|^{s_n+1}\right]{\rm d}x\\ & = &\int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+N(A_m-B_n)(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}\right]{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}\left[NB_1(1-\frac{4\alpha+\frac{4}{N}}{s_1+1})|u|^{s_1+1}+\cdots+NB_{n-1}(1-\frac{4\alpha+\frac{4}{N}}{s_{n-1}+1})|u|^{s_{n-1}+1}\right]{\rm d}x\\ &\geq& -\|u_0\|_{L^2({{\Bbb R}} ^N)}^2\cdot\hat{C}_{\varepsilon}+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})\|u\|_{L^{r_m+1}({{\Bbb R}} ^N)}^{r_m+1}(1-\hat{\varepsilon}_1-\cdots-\hat{\varepsilon}_{n+k-l-1})\\ &\geq&-M(u_0)\cdot \hat{C}_\varepsilon, \end{eqnarray} $

其中$ \hat{C}_\varepsilon = \hat{C}_{\varepsilon_1}+\cdots+\hat{C}_{\varepsilon_{n+k-l-1}} $.

(SA3)  $ r_m<s_n $, $ s_n\leq4\alpha-1+\frac{4}{N} $.取常数$ 0<\bar{\varepsilon}_1, \cdots, \bar{\varepsilon}_m\leq\frac{1}{m}<1 $,则有

$ r_1<\cdots<r_m<4\alpha-1+\frac{4}{N}, $

$ s_1<\cdots<s_n\leq4\alpha-1+\frac{4}{N}. $

与(SA1)证明类似,有

(4.6) $ \begin{eqnarray} && \int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}\right]{\rm d}x\\ && -\int_{{{\Bbb R}} ^N}\left[NB_1(1-\frac{4\alpha+\frac{4}{N}}{s_1+1})|u|^{s_1+1}+\cdots+NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})|u|^{s_n+1}\right]{\rm d}x\\ &\geq &\int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}\right]{\rm d}x\\ &&-\int_{{{\Bbb R}} ^N}NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})|u|^{s_n+1}{\rm d}x\\ &\geq& -\|u_0\|_{L^2({{\Bbb R}} ^N)}^2\cdot\bar{C}_{\varepsilon}-NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})\|u\|_{L^{s_n+1}({{\Bbb R}} ^N)}^{s_n+1}(1-\bar{\varepsilon}_1-\cdots-\bar{\varepsilon}_m)\\ &\geq&-M(u_0)\cdot \bar{C}_\varepsilon, \end{eqnarray} $

其中$ \bar{C}_\varepsilon = \bar{C}_{\varepsilon_1}+\cdots+\bar{C}_{\varepsilon_m} $.

显然在上述三种情况下, (II)式都有

(4.7) $ \begin{eqnarray} && \int_{{{\Bbb R}} ^N}\left[NA_1(1-\frac{4\alpha+\frac{4}{N}}{r_1+1})|u|^{r_1+1}+\cdots+NA_m(1-\frac{4\alpha+\frac{4}{N}}{r_m+1})|u|^{r_m+1}\right]{\rm d}x\\ && -\int_{{{\Bbb R}} ^N}\left[NB_1(1-\frac{4\alpha+\frac{4}{N}}{s_1+1})|u|^{s_1+1}+\cdots+NB_n(1-\frac{4\alpha+\frac{4}{N}}{s_n+1})|u|^{s_n+1}\right]{\rm d}x\\ &\geq&-M(u_0)\cdot C_\varepsilon, \end{eqnarray} $

其中$ C_\varepsilon = {\rm min}(\tilde{C}_\varepsilon, \ \hat{C}_\varepsilon, \ \bar{C}_\varepsilon) $.

除了假设(4.1)和三种情况(SA1)–(SA3)外,如果下列条件满足

$ |x|u_0\in L^2({{\Bbb R}} ^N), \quad F(0) = \Im\int_{{{\Bbb R}} ^N}\bar{u_0}(x\cdot \nabla {u_0}){\rm d}x>0, $

(4.8) $ \begin{equation} 2\left[(2\alpha-1)N+2\right]E(u_0)+M(u_0)\cdot C_\varepsilon\leq 0, \end{equation} $

则根据(4.3), (4.7)和(4.8)式可得到

(4.9) $ \begin{eqnarray} \frac{\partial}{\partial t}F(t)&\geq& -2\left[(2\alpha-1)N+2\right]E(u_0)+(2\alpha-1)N\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x-M(u_0)\cdot C_\varepsilon\\ &\geq& (2\alpha-1)N\|\nabla u\|^2_{L^2({{{\Bbb R}} ^N})}\geq 0, \end{eqnarray} $

从而

$ F(t)\geq F(0)>0. $

由引理2.1(3)得

$ \frac{\partial}{\partial t}J(t) = -4F(t)<0, $

$ J(t)<J(0) = \int_{{{\Bbb R}} ^N}|x|^2|u_0|^2{\rm d}x. $

注意到条件$ |x|u_0\in L^2({{\Bbb R}} ^N) $,用Schwarz不等式得到

(4.10) $ \begin{eqnarray} \quad F(t)& = &\left|\Im \int_{{{\Bbb R}} ^N}\bar{u}(x\cdot \nabla u){\rm d}x\right|\leq \int_{{{\Bbb R}} ^N}\left|\bar{u}(x\cdot \nabla u)\right|{\rm d}x\\ &\leq &\left(\int_{{{\Bbb R}} ^N}x^2|u|^2{\rm d}x\right)^{\frac{1}{2}}\left(\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x\right)^{\frac{1}{2}}< J^{\frac{1}{2}}(0)\|\nabla u\|_{L^2({{\Bbb R}} ^N)}. \end{eqnarray} $

从而由(4.9)和(4.10)式得到

$ F^2(t)< J(0)\|\nabla u\|^2_{L^2({{\Bbb R}} ^N)}\leq \frac{J(0)}{(2\alpha-1)N}\cdot\frac{\partial}{\partial t}F(t), $

也就是说

(4.11) $ \begin{eqnarray} \frac{\partial}{\partial t}F(t)\geq \frac{(2\alpha-1)N}{J(0)}\cdot F^2(t). \end{eqnarray} $

将(4.11)式从$ 0 $到$ t $积分,得到

$ F(t)\geq \frac{F(0)\cdot J(0)}{J(0)-(2\alpha-1)N\cdot F(0)t}, $

从而

$ \|\nabla u\|_{L^2({{\Bbb R}} ^N)}> \frac{F(t)}{J^{\frac{1}{2}}(0)}\geq\frac{F(0)\cdot J^{\frac{1}{2}}(0)}{J(0)-(2\alpha-1)N\cdot F(0)t}. $

令

$ T_0 = \frac{J(0)}{(2\alpha-1)N\cdot F(0)}, $

得到

$ \lim\limits_{t\rightarrow T_0^-}\|\nabla u\|_{L^2({{\Bbb R}} ^N)} = +\infty. $

即

$ \lim\limits_{t\rightarrow T_0^-}\int_{{{\Bbb R}} ^N}\left[|\nabla u|^2+|\nabla (|u|^{2\alpha})|^2\right]{\rm d}x = +\infty. $

定理1.2证毕.