考虑如下高阶Camassa-Holm型方程

(1.1) $ \begin{eqnarray} (1-\partial_x^2)u_t=u^ku_{xxx}+bu^{k-1}u_xu_{xx}-(b+1)u^ku_x, \end{eqnarray} $

其中整数$ k\geq 2 $, 常数$ b \in {{\Bbb R}} . $方程(1.1) 的物理背景可见Grayshan和Himonas^[12]. 如果$ k=1 $, $ b=2 $, 方程(1.1) 变为Camassa-Holm方程^[2]. 当$ k=1 $, $ b=3 $, 方程(1.1) 转化为Degasperis-Procesi方程^[9]. 当$ k=2 $, $ b=3 $, 方程(1.1) 变为Novikov方程^[23].

近年来, 许多学者研究了方程(1.1) 的动力学性质. Himonas和Thompson^[18]研究了方程(1.1) 的两种性质: 持续性和唯一延拓性. Grayshan和Himonas ^[12]研究了方程(1.1) 的孤立波解. 应用Galerkin近似方法, Himonas和Holliman ^[16]讨论了方程(1.1) 在Sobolev空间中的局部适定性. 假设初值具有紧支集, 文献[17]研究了方程(1.1) 的解在无穷远处的渐近行为. Anco等^[1]在一定假设下讨论了方程(1.1) 的一类孤立子解. 利用Lie对称性分析, Wei等^[26]给出了方程(1.1) 的一些守恒律. 文献[15]讨论了方程(1.1) 的位势紧性和长时间行为. 方程(1.1)解的详细渐近描述在文献[13]中给出. 与方程(1.1) 有关的其他结果, 参见文献[3-8, 10, 11, 14, 19-22, 24, 25, 27, 29].

对于任意整数$ k\geq 2 $和常数$ b \in {{\Bbb R}} $, 考察方程(1.1) 整体弱解的存在性是比较困难的. 我们考虑如下方程

(1.2) $ \begin{eqnarray} (1-\partial_x^2)u_t=u^ku_{xxx}+(k+1)u^{k-1}u_xu_{xx}-(k+2)u^ku_x, \end{eqnarray} $

这里常数$ k\geq2 $, 方程(1.2) 是方程(1.1) 的一种特殊形式.

受文献[4, 28] 启发, 我们将研究方程(1.2) 的整体弱解. 如果初值不满足符号条件, 文献[4, 28]在空间$ H^1({{\Bbb R}} ) $中研究了经典Camasa-Holm方程整体弱解的存在性. 假设初值不满足符号条件, 我们将证明方程(1.2) 整体弱解的存在性. 本文的主要贡献是建立了方程(1.2) 粘性问题解的高阶可积性估计. 这个高阶估计不同于文献[4, 28] 中的高阶估计. 本文主要的研究方法来自文献[4, 28].

文章的第二节给出相关引理. 第三节给出本文的主要结果及其证明.

考虑方程(1.2) 的初值问题

(2.1) $ \begin{eqnarray} \left\{\begin{array}{l} { } u_{t}+u^ku_x+\Lambda^{-2}\Big((k+1)u^ku_x+\frac{k-1}{k}(u^k)_xu_{xx}+(u^k)_{xx}u_x \Big)=0, \\ u(0, x)=u_0(x), \; \; x\in{{\Bbb R}} . \end{array}\right. \end{eqnarray} $

问题(2.1) 等价于如下问题

(2.2) $ \begin{eqnarray} \left\{\begin{array}{l} { } u_{t}+u^ku_x+\frac{\partial B}{\partial x}=0 , \\ { } \frac{\partial B}{\partial x}=\Lambda^{-2}\Big((u^{k+1})_x+k(u^{k-1}u^2_{x})_x+u^{k-1}u_xu_{xx} \Big) , \\ u(0, x)=u_0(x), \end{array}\right. \end{eqnarray} $

这里$ \Lambda^{-2}=(1-\partial_x^2)^{-1} $. 由方程(1.2) 得

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }\left(u^{2}+u_{x}^{2}\right) {\rm d}x & =&2\int_{{{\Bbb R}} }u(u_{t}-u_{txx}) {\rm d}x\\ &=&2\int_{{{\Bbb R}} }u\left(u^ku_{xxx}+(k+1)u^{k-1}u_xu_{xx}-(k+2)u^ku_x\right){\rm d}x \\ & =&0. \end{eqnarray*} $

即$ \int_{{{\Bbb R}} }\left(u^{2}+u_{x}^{2}\right) {\rm d}x=\int_{{{\Bbb R}} }\left(u_{0}^{2}+u_{0 x}^{2}\right) {\rm d}x. $

参考文献[4, 28] 对整体弱解的定义, 我们给出如下定义.

定义2.1  一个连续函数$ u:[0, \infty) \times {{\Bbb R}} \rightarrow {{\Bbb R}} $称为问题(2.1) 或(2.2) 的整体弱解, 如果以下条件满足:

$ \rm (i) $$ u \in C([0, \infty) \times{{\Bbb R}} ) \cap L^{\infty}\left([0, \infty) ; H^{1}({{\Bbb R}} )\right); $

$ \rm (ii) $$ \|u(t, .)\|_{H^{1}({{\Bbb R}} )} \leq\left\|u_{0}\right\|_{H^{1}({{\Bbb R}} )}; $

$ \rm (iii) $$ u=u(t, x) $在分布意义下满足方程(2.1).

定义算子

$ \begin{eqnarray} \phi(x)=\left\{\begin{array}{ll} e^{\frac{1}{x^{2}-1}}, & |x|<1, \\ 0, & |x| \geq 1. \end{array}\right.{} \end{eqnarray} $

引入光滑算子$ \phi_{\varepsilon}(x)=\varepsilon^{-\frac{1}{4}} \phi\big(\varepsilon^{-\frac{1}{4}} x\big); $$ 0<\varepsilon<1, u_{\varepsilon, 0}=\phi_{\varepsilon} * u_{0} $, $ u_{\varepsilon, 0} \in C^{\infty}, $$ u_{0} \in H^{s}, s>0 $ ($ * $代表卷积). 为了研究方程(1.2) 整体弱解的存在性, 我们讨论如下粘性近似问题

(2.3) $ \begin{eqnarray} \left\{\begin{array}{l} { } \frac{\partial u_{\varepsilon}}{\partial t}+u^k \frac{\partial u_{\varepsilon}}{\partial x}+\frac{\partial B_{\varepsilon}}{\partial x}=\varepsilon \frac{\partial^{2} u_{\varepsilon}}{\partial x^{2}} , \\ u(0, x)=u_{\varepsilon, 0}(x), \end{array}\right. \end{eqnarray} $

这里

$ \frac{\partial B_{\varepsilon}}{\partial x}=\Lambda^{-2}\Big((u_{\varepsilon}^{k+1})_x+k(u_{\varepsilon}^{k-1}u_{\varepsilon}^2u_{x})_x+u_{\varepsilon}^{k-1} u_{\varepsilon x}u_{\varepsilon xx}\Big). $

(2.4) $ \begin{equation} \|u_\varepsilon\|_{L^\infty({{\Bbb R}} )}\leq\|u_\varepsilon\|_{H^1({{\Bbb R}} )}\leq \|u_{\varepsilon, 0} \|_{H^{1}({{\Bbb R}} )} \leq\|u_{0}\|_{H^{1}({{\Bbb R}} )}, \quad u_{\varepsilon, 0} \rightarrow u_{0}\quad (H^{1}({{\Bbb R}} )). \end{equation} $

对(2.3)式关于变量$ x $求导数, 令$ p_{\varepsilon}(t, x)= \frac{\partial u_{\varepsilon}(t, x)}{\partial x} $, 我们得到

(2.5) $ \begin{eqnarray} &&\frac{\partial p_{\varepsilon}}{\partial t}+u^k_{\varepsilon}p_{\varepsilon x}+\frac{1}{2}u^{k-1}_{\varepsilon}p^2_{\varepsilon}-\varepsilon p_{\varepsilon xx}{}\\ &=&u^{k+1}_{\varepsilon}-\Lambda^{-2}\Big(u^{k+1}_{\varepsilon}+(k-\frac{1}{2})u^{k-1}_{\varepsilon}p^2_{\varepsilon}+ (\frac{k-1}{2}u^{k-2}_{\varepsilon}p^3_{\varepsilon})_x \Big){}\\ &=&Q_{\varepsilon}(t, x). \end{eqnarray} $

在下面的讨论中, 我们用$ c $表示与参数$ \varepsilon $无关的任何正常数.

引理2.1  假设$ u_{0}(x)\in H^{s}({{\Bbb R}} ), $常数$ \sigma \geq 2, $则柯西问题(2.3) 有唯一解$ u_{\varepsilon} \in $$ C\big([0, \infty) ; $$ H^{\sigma}({{\Bbb R}} )\big) $且

(2.6) $ \begin{eqnarray} \int_{{{\Bbb R}} } \Big[u_{\varepsilon}^{2}+(\frac{\partial u_{\varepsilon}}{\partial x})^{2}\Big] {\rm d}x+2 \varepsilon \int_{0}^{t} \int_{{{\Bbb R}} } \Big[(\frac{\partial u_{\varepsilon}}{\partial x})^{2} +(\frac{\partial^{2} u_{\varepsilon}}{\partial x^{2}})^{2}\Big](s, x) {\rm d}x{\rm d}s =\left\|u_{\varepsilon, 0}\right\|_{H^{1}({{\Bbb R}} )}^{2} \end{eqnarray} $

或

$ \left\|u_{\varepsilon}(t, .)\right\|_{H^{1}({{\Bbb R}} )}^{2}+2 \varepsilon \int_{0}^{t}\left\| \frac{\partial u_{\varepsilon}}{\partial x}(s, .)\right\|_{H^{1}({{\Bbb R}} )}^{2}{\rm d}s=\left\|u_{\varepsilon, 0}\right\|_{H^{1}({{\Bbb R}} )}^{2}. \label{k10}{\nonumber} $

证   根据文献[5]的结论, 我们知道问题(2.3) 存在唯一解$ u_\varepsilon(t, x)\in C([0, \infty); H^\sigma({{\Bbb R}} )), $使用(2.3)式, 我们有

$ \begin{eqnarray*} &&\frac{\rm d}{{\rm d}t} \int_{{{\Bbb R}} }\Big(u^2_{\varepsilon}+u_x^2\Big){\rm d}x =2\int_{{{\Bbb R}} } u_{\varepsilon}(u_{\varepsilon t}-u_{\varepsilon txx}) {\rm d}x\\ & =&2\int_{{{\Bbb R}} } \Big( u^k_{\varepsilon}u_{\varepsilon xxx}+(k+1)u^{k-1}_{\varepsilon}u_{\varepsilon x}u_{\varepsilon xx}-(k+2)u_{\varepsilon}^ku_{\varepsilon x}+\varepsilon(u_{\varepsilon xx}-u_{\varepsilon xxxx})\Big) {\rm d}x \\ & =&-2\varepsilon \int_{{{\Bbb R}} }(u_{\varepsilon x}^2+u_{\varepsilon xx}^2){\rm d}x. \label{K11}{\nonumber} \end{eqnarray*} $

上式对$ t $积分便得到(2.6) 式成立.

引理2.2  假设$ u_{0}(x)\in H^{1}({{\Bbb R}} ), $则存在正常数$ c=c(\left\|u_{0}\right\|_{H^{1}({{\Bbb R}} )}) $使得

(2.7) $ \begin{equation} \|B_{\varepsilon}(t, \cdot)\|_{L^{\infty}({{\Bbb R}} )}\leq c, \quad \|B_{\varepsilon}(t, \cdot)\|_{L^{1}({{\Bbb R}} )}\leq c, \quad \|B_{\varepsilon}(t, \cdot)\|_{L^{2}({{\Bbb R}} )}\leq c, \end{equation} $

(2.8) $ \begin{equation} \Big\|\frac{\partial B_{\varepsilon}}{\partial x}(t, \cdot) \Big\|_{L^{\infty}({{\Bbb R}} )}\leq c, \quad \Big\|\frac{\partial B_{\varepsilon}}{\partial x}(t, \cdot)\Big\|_{L^{1}({{\Bbb R}} )} \leq c, \quad \Big\|\frac{\partial B_{\varepsilon}}{\partial x}(t, \cdot)\Big\|_{L^{2}({{\Bbb R}} )} \leq c, \end{equation} $

(2.9) $ \begin{equation} \|Q_{\varepsilon}(t, \cdot)\|_{L^{\infty}({{\Bbb R}} )}\leq c, \quad \|Q_{\varepsilon}(t, \cdot)\|_{L^{1}({{\Bbb R}} )}\leq c, \end{equation} $

(2.10) $ \begin{equation} \|Q_{\varepsilon}(t, \cdot)\|_{L^{2}({{\Bbb R}} )}\leq c, \quad \Big \|\frac{\partial Q_{\varepsilon}}{\partial x}(t, \cdot)\Big\|_{L^{2}({{\Bbb R}} )} \leq c. \end{equation} $

证   我们有

(2.11) $ \begin{eqnarray} \Lambda^{-2}U(x)=\frac{1}{2} \int_{{{\Bbb R}} } e^{-|x-y|}U(y) {\rm d}y, \quad U(x) \in L^{r}({{\Bbb R}} ), \quad 1\leq r <\infty \end{eqnarray} $

以及

(2.12) $ \begin{eqnarray} \Big|\Lambda^{-2}U_x(x)\Big|&=&\Big|\frac{1}{2} \int_{{{\Bbb R}} } e^{-|x-y|}\frac{\partial U(y)}{\partial y} {\rm d}y\Big|\\ &=&\Big|-\frac{1}{2}e^{-x}\int_{-\infty}^xU(y){\rm d}y+\frac{1}{2}e^x\int_x^\infty e^{-y}U(y){\rm d}y\Big|\\ &\leq &\frac{1}{2}\int_{-\infty}^{\infty}e^{-|x-y|}|U(y)|{\rm d}y. \end{eqnarray} $

使用(2.4), (2.11)和(2.12)式, 函数$ B_\varepsilon(t, x) $和Tonelli定理, 我们有

$ \| B_{\varepsilon}(t, \cdot)\|_{L^{\infty}({{\Bbb R}} )}= \bigg\| \Lambda^{-2}\Big(u^{k+1}_{\varepsilon}+ku^{k-1}_{\varepsilon}u^2_{\varepsilon x}-\frac{1}{2}\int^{x}_{-\infty} u^{k-1}(u^2_{\xi})_\xi {\rm d}\xi \Big)\bigg\|_{L^{\infty}({{\Bbb R}} )}\leq c, {\nonumber} $

$ \Big\|\frac{\partial B_{\varepsilon}}{\partial x}(t, \cdot)\Big\|_{L^{\infty}({{\Bbb R}} )}= \bigg\|\Lambda^{-2}\Big(u^{k+1}_{\varepsilon}+ku^{k-1}_{\varepsilon}u^2_{\varepsilon x}- \frac{1}{2}\int^{x}_{-\infty} u^{k-1}(u^2_{\xi})_\xi {\rm d}\xi \Big)_{x}\bigg\|_{L^{\infty}({{\Bbb R}} )} \leq c{\nonumber} $

和

(2.13) $ \begin{eqnarray} \| B_{\varepsilon}(t, \cdot)\|_{L^{1}({{\Bbb R}} )} \leq c, \quad \Big\|\frac{\partial B_{\varepsilon}} {\partial x}(t, \cdot)\Big\|_{L^{1}({{\Bbb R}} )} \leq c. \end{eqnarray} $

事实上, 我们得到

$ \| B_{\varepsilon}(t, \cdot)\|_{L^{2}({{\Bbb R}} )}^{2} \leq\| B_{\varepsilon}(t, \cdot)\|_{L^\infty({{\Bbb R}} )} \leq \|B_{\varepsilon}(t, \cdot)\|_{L^{1}({{\Bbb R}} )} \leq c $

和

$ \Big \|\frac{\partial B_{\varepsilon}(t, \cdot)}{\partial x}\Big\|_{L^{2}({{\Bbb R}} )}^{2} \leq \Big\|\frac{\partial B_{\varepsilon}(t, \cdot)}{\partial x}\Big\|_{L^\infty({{\Bbb R}} )} \Big\|\frac{\partial B_{\varepsilon}(t, \cdot)}{\partial x}\Big\|_{L^{1}({{\Bbb R}} )} \leq c.{\nonumber} $

这样我们推导出了(2.7)和(2.8) 式. 类似地, 使用(2.4), (2.5), (2.11) 和(2.12) 式有

$ \| Q_\varepsilon\|_{L^\infty({{\Bbb R}} )}\leq c, \quad \| Q_\varepsilon\|_{L^1({{\Bbb R}} )}\leq c, \quad \| Q_\varepsilon\|_{L^2({{\Bbb R}} )}\leq c, \label{K26}{\nonumber} $

于是我们证明了(2.9)和(2.10) 式成立.

引理2.3   若$ 0<t<T $, 则存在正常数$ c_0=c_0 (\left\| u_0 \right\|_{H^1({{\Bbb R}} )}, \left\| \frac{\partial u_{\varepsilon}}{\partial x} \right\|_{L^\infty}) $使得

(2.14) $ \begin{eqnarray} \int_{0}^{T} \int_{{{\Bbb R}} } \Big(\frac{\partial u_{\varepsilon}}{\partial x}\Big)^{2k}{\rm d}x{\rm d}t\leq c_0(1+T) e^{c_0T}. \end{eqnarray} $

证   为了便于书写, 我们记$ u=u_{\varepsilon}, $$ Q=Q_{\varepsilon}(t, x) $. 由方程(2.5), 我们有

$ \begin{eqnarray} u_{tx}+\frac{1}{2}u^{k-1}u^2_{x}+u^{k}u_{xx}-\varepsilon u_{xxx}=Q. \label{K28}{} \end{eqnarray} $

因为

$ \begin{eqnarray} \int_{\mathbb R} u^ku_{x}^{2k-1}u_{xx}{\rm d}x=-\frac{1}{2}\int_{\mathbb R}u^{k-1}u_{x}^{2k+1}{\rm d}x, \label{K29}{} \end{eqnarray} $

则

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\int_{\mathbb R}u^{2k}_x{\rm d}x &=&2k\int_{\mathbb R}u^{2k-1}u_{xt}{\rm d}x\\ &=&2k\int_{\mathbb R}u^{2k-1}_x(Q-u^ku_{xx}-\frac{1}{2}u^{k-1}u^2_x+\varepsilon u_{xxx}){\rm d}x\\ & =&2k\int_{\mathbb R}u^{2k-1}_x(Q+\varepsilon u_{xxx}){\rm d}x. \label{K30}{\nonumber} \end{eqnarray*} $

由此可得

$ \frac{\rm d}{{\rm d}t}\int_{\mathbb R} u_{x}^{2k}{\rm d}x-2k\varepsilon\int_{\mathbb R}u_{x}^{2k-1}u_{xxxx}{\rm d}x=2k\int_{\mathbb R}u_{x}^{2k-1}Q{\rm d}x. \label{K31}{\nonumber} $

又因为

$ -2k\varepsilon\int_{\mathbb R}u_{x}^{2k-1}u_{xxxx}{\rm d}x=(2k-1)\int_{\mathbb R}u_x^{2k-2}u_{xx}^2{\rm d}x > 0, \label{K32}{\nonumber} $

我们有

(2.15) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t} \int_{\mathbb R}u^{2k}_x{\rm d}x \leq 2k\int_{\mathbb R} u^{2k-1}_{x}Q{\rm d}x\leq2k\int_{\mathbb R} |u^{2k-1}_{x}||Q|{\rm d}x. \end{eqnarray} $

使用H$ \ddot{\rm o} $lder不等式和(2.15) 式得

(2.16) $ \begin{eqnarray} \int_{\mathbb R}u^{2k}_x{\rm d}x-\int_{\mathbb R}u^{2k}_x(0, x){\rm d}x&\leq&2k\int_{0}^{T}\int_{\mathbb R} |u^{2k-1}_{x}||Q|{\rm d}x{\rm d}t {} \\ &\leq&2k \int^T_0 \Big(\int_{\mathbb R}|Q|^{2k}{\rm d}x\Big)^{\frac{1}{2k}}\Big( \int_{\mathbb R}|u_x|^{2k}{\rm d}x \Big)^{\frac{2k-1}{2k}}{\rm d}t. \end{eqnarray} $

使用(2.16) 式和Gronwall不等式, 我们推知(2.14) 式成立.

引理2.4   假设$ k\geq 2 $为奇数, 设$ u_{\varepsilon}=u_{\varepsilon}(t, x) $是方程(2.3) 的唯一解. 对任意$ t>0 $, 则存在正常数$ c=c(\|u_{0}\|_{H^{1}({{\Bbb R}} )}) $使得

$ \begin{eqnarray} \frac{\partial u_{\varepsilon}(t, x)}{\partial x} \leq \frac{2}{t}+c. \label{K35}{} \end{eqnarray} $

证   因$ \|u_{\varepsilon}(t, x)\| \leq c $, $ \|Q_{\varepsilon}(t, x) \|\leq c, $$ k-1 $是偶数, 则

(2.17) $ \begin{eqnarray} \frac{\partial p_{\varepsilon}}{\partial t}+u^k_\varepsilon \frac{\partial p_{\varepsilon}}{\partial x}- \varepsilon \frac{\partial^{2} p_{\varepsilon}}{\partial x^{2}}+\frac{1}{2}u^{k-1}_\varepsilon p_{\varepsilon}^{2} =Q_{\varepsilon}(t, x) \leq c. \end{eqnarray} $

假设$ H=H(t) $满足

(2.18) $ \begin{eqnarray} \frac{{\rm d} H}{{\rm d} t}+u^{k-1}_\varepsilon H^{2}=c, \quad t>0, \quad H(0)=\left\|\frac{\partial u_{\varepsilon, 0}}{\partial x}\right\|_{L^{\infty}}. \end{eqnarray} $

于是知$ H=H(t) $是方程(2.17) 的一个上解. 使用文献[28] 中相同的方法分析, 我们得

$ \begin{eqnarray} H(t) \leq F(t)=\frac{2}{t}+c, \quad t>0. \end{eqnarray} $

引理2.4得证.

引理2.5  存在序列$ \left\{\varepsilon_{j}\right\}_{j \in {\Bbb N}}(\varepsilon_{j}\rightarrow0) $, 函数$ u \in L^{\infty}\left([0, \infty) ; H^{1}({{\Bbb R}} )\right) \cap H^{1}([0, T] \times {{\Bbb R}} ) $使得

(2.19) $ \begin{eqnarray} &&u_{\varepsilon_{j}} \rightharpoonup u \quad ( H^{1}([0, T] \times {{\Bbb R}} )), \quad T>0, \end{eqnarray} $

(2.20) $ \begin{eqnarray} &&u_{\varepsilon_{j}} \rightarrow u\quad ( L_{\rm loc}^{\infty}([0, \infty) \times {{\Bbb R}} )), \quad T>0. \end{eqnarray} $

证   对任意$ T>0 $, 使用引理2.1和(2.3) 式, 有

(2.21) $ \begin{eqnarray} \left\|\frac{\partial u_{\varepsilon}}{\partial t}\right\|_{L^{2}({{\Bbb R}} )}, \quad \left\|\frac{\partial u_{\varepsilon}}{\partial t}\right\|_{L^{2}([0, T] \times{{\Bbb R}} )} \leq c. \end{eqnarray} $

因为$ \left\{u_{\varepsilon}\right\} $在空间$ L^{\infty}\left([0, \infty) ; H^{1}({{\Bbb R}} )\right) \cap H^{1}([0, T] \times{{\Bbb R}} ) $一致有界, 则(2.19) 式成立. 对于任意$ 0 \leq s $, $ t \leq T $, 我们有

$ \left\|u_{\varepsilon}(t, .)-u_{\varepsilon}(s, .)\right\|_{L^{2}({{\Bbb R}} )}^{2} =\int_{{{\Bbb R}} }\left(\int_{s}^{t} \frac{\partial u_{\varepsilon}}{\partial t}(\tau, x) {\rm d}\tau\right)^{2} {\rm d}x \leq|t-s| \int_{{{\Bbb R}} } \int_{0}^{T}\left(\frac{\partial u_{\varepsilon}}{\partial t}(\tau, x)\right)^{2} {\rm d}\tau {\rm d}x. \label{K42}{\nonumber} $

因$ \left\{u_{\varepsilon}\right\} $在空间$ L^{\infty}\left([0, T] ; H^{1}({{\Bbb R}} )\right) $一致有界且$ H^{1}({{\Bbb R}} ) \subset L_{\rm loc}^{\infty} \subset L_{\rm loc}^{2}({{\Bbb R}} ) $. 由文献[4]中的引理5得到(2.20) 式成立.

引理2.6   假设$ u_{0}(x)\in H^{1}({{\Bbb R}} ) $, 则序列$ \{Q_{\varepsilon}(t, x)\}_\varepsilon $在空间$ W_{\rm loc}^{1, 1}([0, \infty) \times{{\Bbb R}} ) $一致有界. 特别地, 存在序列$ \left\{\varepsilon_{j}\right\}_{j \in N} \rightarrow0 $, 函数$ Q\in L^{\infty}([0, T);W^{1, \infty}({{\Bbb R}} )) $和$ 1<r<\infty $使得

(2.22) $ \begin{eqnarray} Q_{\varepsilon_{j}} \rightarrow Q \quad (L_{{\rm loc}}^{r}([0, T) \times{{\Bbb R}} )). \end{eqnarray} $

证   为了书写方便, 记$ u=u_{\varepsilon}(t, x) $, $ p=p_{\varepsilon}(t, x) $. 使用(2.5)式有

(2.23) $ \begin{eqnarray} \frac{\partial Q_\varepsilon}{\partial t} &=&\Big(u^{k+1}-\Lambda^{-2}\Big[u^{k+1}+(k-\frac{1}{2})u^{k-1}p^2+(\frac{k-1}{2}u^{k-2}p^2)_x \Big]\Big)_t {} \\ &=&(k+1)u^ku_t-\Lambda^{-2}\Big(\Big[(k+1)u^k+(k-\frac{1}{2})(k-1)u^{k-2}p^2\Big]u_t\Big){} \\ &&-\Lambda^{-2}\Big( 2(k-\frac{1}{2})u^{k-1}p(Q-\frac{1}{2}u^{k-1}p^2-u^kp_x+\varepsilon p_{xx})\Big){} \\ &&-\Lambda^{-2}\Big(\frac{k-1}{2}\partial_{x}[(k-2)u^{k-3}u_tp^3]\Big){} \\ &&-\Lambda^{-2}\Big(\frac{k-1}{2}\partial_{x}[3u^{k-2}p^2(Q-\frac{1}{2}u^{k-1}p^2-u^kp_x+\varepsilon p_{xx} )]\Big){} \\ &=&(k+1)u^ku_t-I_{1}-I_{2}-I_{3}-I_{4}. \end{eqnarray} $

使用引理2.2, (2.4)和(2.21) 式有

(2.24) $ \begin{eqnarray} \int_{{{\Bbb R}} }|u^ku_t|{\rm d}x \leq \Big(\int_{{{\Bbb R}} }u^{2k}{\rm d}x\Big)^{\frac{1}{2}} \Big(\int_{{{\Bbb R}} }u^2_t{\rm d}x\Big)^{\frac{1}{2}}\leq c \end{eqnarray} $

以及

(2.25) $ \begin{eqnarray} \int_{{{\Bbb R}} } |I_1| {\rm d}x &=&\Big|\frac{1}{2}\int_{{{\Bbb R}} }e^{|x-y|}\Big((k+1)u^ku_t+(k-\frac{1}{2})(k-1)u^{k-2}u_tp^2\Big){\rm d}y\Big|{} \\ &\leq &c\Big[(k+1)\Big(\int_{{{\Bbb R}} }u^{2k}{\rm d}y\Big)^{\frac{1}{2}} \Big(\int_{{{\Bbb R}} }u_t^2{\rm d}y\Big)^{\frac{1}{2}}{} \\ &&+(k-\frac{1}{2})(k-1)\Big(\int_{{{\Bbb R}} }u_t^2{\rm d}y\Big)^{\frac{1}{2}}\Big(\int_{{{\Bbb R}} }p^4{\rm d}y\Big)^{\frac{1}{2}} \Big] . \end{eqnarray} $

从而有

(2.26) $ \begin{eqnarray} \int^t_0\int_{{{\Bbb R}} }|I_{1}|{\rm d}x{\rm d}t\leq c. \end{eqnarray} $

注意到

$ \frac{\partial}{\partial x}(u^{2k-1}p^2)=(2k-1)u^{2k-2}p^3+2u^{2k-1}pp_x \label{K48}{\nonumber} $

和

$ \frac{\partial}{\partial x}(u^{k-1}pp_x)=(k-1)u^{k-2}p^2p_x+u^{k-1}p_{x}^2+u^{k-1}pp_{xx}. \label{K49}{\nonumber} $

我们得到

$ \begin{eqnarray*} |I_2|&=&\Big|(k-\frac{1}{2})\int_{{{\Bbb R}} }e^{|x-y|}\Big(u^{k-1}pQ-\frac{1}{2}u^{2k-2}p^3-u^{2k-1}pp_y+\varepsilon u^{k-1}pp_{yy}\big){\rm d}y\Big|{\nonumber} \\ & \leq &c \Big|\int_{{{\Bbb R}} } e^{|x-y|}\Big(u^{k-1}pQ-\frac{1}{2}u^{2k-2}p^3+u^{2k-1}p^2+(2k-1)u^{2k-1}p^3\Big){\rm d}y\Big|{\nonumber} \\ && +c\varepsilon\Big|\int_{{{\Bbb R}} } e^{|x-y|}\Big(u^{k-1}pp_y+(k-1)u^{k-2}p^2p_y+u^{k-1}p^2_y\Big){\rm d}y\Big|. \label{K50}{\nonumber} \end{eqnarray*} $

于是有

(2.27) $ \begin{equation} \int^t_0\int_{{{\Bbb R}} }|I_{2}|{\rm d}x{\rm d}t\leq c. \end{equation} $

使用(2.4)和(2.12) 式得

$ \begin{eqnarray} |I_3|=\Big|\frac{(k-1)(k-2)}{4}\int_{{{\Bbb R}} }e^{|x-y|}\partial_{y}(u^{k-3}u_tp^3){\rm d}y\Big|{} \leq c \Big|\int_{{{\Bbb R}} }e^{|x-y|}u^tp^3 {\rm d}y \Big|. \label{K52} \end{eqnarray} $

进一步得

(2.28) $ \begin{equation} \int^t_0\int_{{{\Bbb R}} }|I_{3}|{\rm d}x{\rm d}t\leq c. \end{equation} $

因为

$ \frac{\partial}{\partial x}(u^{k-2}p^2p_x)=(k-2)u^{k-3}p^3p_x+u^{k-2}pp_{x}^2+u^{k-2}p^2p_{xx}, \label{K54}{\nonumber} $

于是有

$ \begin{eqnarray*} |I_4|&=&\Big|\frac{3(k-1)}{4}\int_{{{\Bbb R}} }e^{|x-y|}\Big(u^{k-2}Qp-\frac{1}{2}u^{2k-3}p^4-u^{2k-2}p^2p_y+\varepsilon u^{k-2}p^2p_{yy}\Big){\rm d}y \Big|{\nonumber} \\ &\leq& c\Big|\int_{{{\Bbb R}} }e^{|x-y|}\Big(|u^{k-2}Qp|+|\frac{1}{2}u^{2k-3}p^4|+|u^{2k-2}p^2p_y|\Big){\rm d}y\Big|{\nonumber} \\ &&+c\varepsilon \Big| \int_{{{\Bbb R}} }e^{|x-y|}\Big(|u^{k-2}p^2p_{y}|+|u^{k-2}p^3p_y|+|u^{k-2}pp_y^2|{\rm d}y\Big)\Big|. \label{K55}{\nonumber} \end{eqnarray*} $

使用引理2.1–2.3和(2.4) 式得

(2.29) $ \begin{eqnarray} \int^t_0\int_{{{\Bbb R}} }|I_{4}|{\rm d}x{\rm d}t\leq c. \end{eqnarray} $

使用(2.23), (2.24), (2.26)–(2.29) 式和引理2.2, 我们推导出$ \{{Q_{\varepsilon}}\}_{\varepsilon} $在空间$ W^{1, 1}_{{\rm loc}}([0, \infty)\times {{\Bbb R}} ) $有界. 结合文献[24]中引理3.3得(2.22)式.

后续的讨论中, 我们使用上横线来表示空间$ L^{r}[(0, \infty) \times{{\Bbb R}} ) $$ (1<r<2k) $中的上极限.

引理2.7   存在序列$ \left\{\varepsilon_{j}\right\}_{j \in N}, \varepsilon_{j} \rightarrow 0, $函数$ p \in L_{{\rm loc}}^{r}([0, \infty) \times{{\Bbb R}} )) $和$ \overline{p^{2}}\in L_{{\rm loc}}^{r_1}([0, \infty) \times{{\Bbb R}} )), $如果$ 1<r<2k, 1<r_1<k $, 则

(2.30) $ \begin{eqnarray} & &p_{\varepsilon_{j}} \rightharpoonup p \quad L_{{\rm loc}}^{r}([0, \infty) \times{{\Bbb R}} ), \quad p_{\varepsilon_{j}} \stackrel{*}{\rightharpoonup} p \quad L_{{\rm loc}}^{\infty}\left([0, \infty) ; L^{2}({{\Bbb R}} )\right), \end{eqnarray} $

(2.31) $ \begin{eqnarray} & &p_{\varepsilon_{j}}^{2} \rightharpoonup \overline{p^{2}} \quad L_{{\rm loc}}^{r_1}([0, \infty) \times{{\Bbb R}} ). \end{eqnarray} $

此外, 对任意$ (t, x) \in[0, \infty) \times{{\Bbb R}} $有

(2.32) $ \begin{equation} p^{2}(t, x) \leq \overline{p^{2}}(t, x) \end{equation} $

和

(2.33) $ \begin{equation} \frac{\partial u}{\partial x}=p \quad \mbox{(在分布意义下).} \end{equation} $

证   由引理2.5知(2.30)和(2.31) 式成立. 结合引理2.5和(2.30) 式, 我们得到(2.33) 式成立.

我们用$ \left\{u_{\varepsilon_{j}}\right\}_{j \in N}, \left\{p_{\varepsilon_{j}}\right\}_{j \in N}, $$ \left\{Q_{\varepsilon_{j}}\right\}_{j \in N} $表示$ \left\{u_{\varepsilon}\right\}_{\varepsilon>0}, \left\{p_{\varepsilon}\right\}_{\varepsilon>0}, $$ \left\{Q_{\varepsilon}\right\}_{\varepsilon>0} $. 假设$ G \in C^{1}({{\Bbb R}} ) $为任意凸函数, $ G^{\prime} $有界且在$ {{\Bbb R}} $上Lipschitz连续, 利用(2.30) 式, 我们有

$ G(p_\varepsilon) \rightharpoonup \overline{G(p)}{\quad} L_{{\rm loc}}^{r}([0, \infty) \times{{\Bbb R}} ) \label{K61}, {\nonumber} $

$ G(p_\varepsilon) \stackrel{*}{\rightharpoonup} \overline{G(p)}{\quad} L_{{\rm loc}}^{\infty}\left([0, \infty) ; L^{2}({{\Bbb R}} )\right). \label{K62}{\nonumber} $

由(2.5) 式可得

(2.34) $ \begin{eqnarray} && \frac{\partial}{\partial t} G(p_\varepsilon)+\frac{1}{k}\frac{\partial}{\partial x}\left(u^k_{\varepsilon} G(p_\varepsilon)\right)-\varepsilon \frac{\partial^{2}}{\partial x^{2}} G(p_\varepsilon)+\varepsilon G^{\prime \prime}\left(p_{\varepsilon}\right)\left(\frac{\partial p_{\varepsilon}}{\partial x}\right)^{2} {} \\ & =&u^{k-1}_{\varepsilon x}p_{\varepsilon} G(p_\varepsilon)-\frac{1}{2}u^{k-1}_{\varepsilon}p_{\varepsilon}^{2}G^{\prime}(p_\varepsilon)-\frac{k-1}{k}u^k_{\varepsilon}\frac{\partial G(p_\varepsilon)}{\partial x}+Q_\varepsilon G^{\prime}(p_\varepsilon). \end{eqnarray} $

引理2.8   假设凸函数$ G \in C^{1}({{\Bbb R}} ) $, $ G^{\prime} $有界且Lipschitz连续. 在分布意义下, 则

(2.35) $ \begin{equation} \frac{\overline{\partial G(p)}}{\partial t}+\frac{1}{k}\frac{\partial}{\partial x}\left(u^k \overline{G(p)}\right) \leq u^{k-1}\overline{pG(p)}-\frac{1}{2}u^{k-1}\overline{p^{2} G^{\prime}(p) }-\frac{k-1}{k}u^k\overline{\frac{\partial G(p)}{\partial x}}+Q\overline{G^{\prime}(p)}, \end{equation} $

这里$ \overline{pG(p)} $和$ \overline{G^{\prime}(p) p^{2}} $表示$ p_{\varepsilon} G(p_\varepsilon) $和$ G^{\prime}(p_\varepsilon) p_{\varepsilon}^{2} $在空间$ L_{\mbox{loc}}^{r_1}([0, \infty) \times {{\Bbb R}} ) $($ 1<r_1<k $) 中的上极限.

证   应用引理2.5–2.7, 在(2.34) 式中令$ \varepsilon \rightarrow 0 $, 即得(2.35) 式.

注2.1   我们有

(2.36) $ \begin{eqnarray} p=p_{+}+p_{-}=\overline{p_{+}}+\overline{p_{-}}, \quad p^{2}=\left(p_{+}\right)^{2}+\left(p_{-}\right)^{2}, \quad \overline{p^{2}}=\overline{\left(p_{+}\right)^{2}}+\overline{\left(p_{-}\right)^{2}}. \end{eqnarray} $

令$ \eta_{+}:=\eta_{\chi[0, +\infty)}(\eta), \eta_{-}:=\eta_{\chi(-\infty, 0]}(\eta) $, ($ \eta \in{{\Bbb R}} $). 由引理2.4得

(2.37) $ \begin{eqnarray} p_{\varepsilon}(t, x), \quad p(t, x) \leq \frac{2}{t}+c, \quad 0<t<T, \quad x \in{{\Bbb R}} . \end{eqnarray} $

引理2.9   对任意$ t \geq 0, x \in {{\Bbb R}} $, 在分布意义下, 则

(2.38) $ \begin{eqnarray} \frac{\partial p}{\partial t}+\frac{1}{k}\frac{\partial}{\partial x}\left(u^k_{\varepsilon}p\right)=\frac{1}{2}u^{k-1} \overline{p^{2}}-(\frac{k-1}{k})u^kp_x+Q(t, x). \end{eqnarray} $

证   使用引理2.5–2.7, 在(2.5) 式中令$ \varepsilon \rightarrow 0 $, 即得(2.38)式.

引理2.10  假设凸函数$ G \in C^{1}({{\Bbb R}} ) $, $ G^{\prime}\in L^{\infty}({{\Bbb R}} ) $. 对任意$ T>0, $在分布意义下, 有

(2.39) $ \begin{eqnarray} &&\frac{\partial G(p)}{\partial t}+\frac{1}{k}\frac{\partial}{\partial x}\left(u^kG(p)\right) {}\\ &=& u^{k-1}p G(p)-u^{k-1}p^2G^{\prime}(p)+\frac{1}{2}u^{k-1}\overline{p^2}G^{\prime}(p)-\frac{k-1}{k}u^{k}\frac{\partial G(p) }{\partial x }+Q G^{\prime}(p). \end{eqnarray} $

证   假设$ \left\{w_{\delta}\right\}_{\delta} $是定义在$ (-\infty, \infty) $上的光滑函数。令$ p_{\delta}(t, x):=\left(p(t, .) * w_{\delta}\right)(x) $($ * $表示变量$ x $的卷积), 那么

(2.40) $ \begin{eqnarray} \frac{\partial G\left(p_{\delta}\right)}{\partial t} &=& G^{\prime}\left(p_{\delta}\right) \frac{\partial p_{\delta}}{\partial t}{} \\ & =& G^{\prime}(p_{\delta})\Big(-\frac{1}{k}\frac{\partial}{\partial x}(u^k p_{\delta}) * w_{\delta}+\frac{1}{2} u^{k-1}\bar{p}^{2}_{\delta} * w_{\delta}-\frac{k-1}{k}u^{k}\frac{\partial p_{\delta}}{\partial x } * w_{\delta}+Q* w_{\delta}\Big){} \\ &=&G^{\prime}(p_{\delta})\Big(-u^{k-1} p_{\delta}^2 * w_{\delta}-\frac{1}{k}u^{k}p_{\delta x} * w_{\delta}\Big) {} \\ & &+G^{\prime}(p_{\delta})\Big(\frac{1}{2} u^{k-1}\bar{p}^{2}_{\delta} * w_{\delta} -\frac{k-1}{k}u^{k}\frac{p_{\delta}}{x}* w_{\delta} +Q* w_{\delta}\Big) \end{eqnarray} $

且

(2.41) $ \begin{eqnarray} \frac{1}{k}\frac{\partial}{\partial x}\left(u^kG(p_{\delta})\right)=u^{k-1}p_{\delta}G(p_{\delta})+\frac{1}{k}u^k G^{\prime}(p_{\delta})(P_{\delta x}* w_{\delta}). \end{eqnarray} $

因凸函数$ G \in C^{1}({{\Bbb R}} ) $, $ G^{\prime}\in L^{\infty}({{\Bbb R}} ), $在(2.40)和(2.41) 式中令$ \delta \rightarrow 0 $, 即证.

按照文献[4, 28] 的思路, 我们将证明$ p_{\varepsilon}^2 $是强收敛的.

引理2.11^[28]  若$ u_{0} \in H^{1}({{\Bbb R}} ), $则

(2.42) $ \begin{eqnarray} \lim _{t \rightarrow 0} \int_{{{\Bbb R}} } p^{2}(t, x) {\rm d}x=\lim _{t \rightarrow 0} \int_{{{\Bbb R}} } \overline{p^{2}}(t, x) {\rm d}x=\int_{{{\Bbb R}} }\left(\frac{\partial u_{0}}{\partial x}\right)^{2}{\rm d}x. \end{eqnarray} $

引理2.12^[28]  若$ u_{0} \in H^{1}({{\Bbb R}} ) $, $ M>0 $, 则

(2.43) $ \begin{eqnarray} \lim _{t \rightarrow 0} \int_{{{\Bbb R}} }\left(\overline{G_{M}^{\pm}(p)}(t, x)-G_{M}^{\pm}(p)(t, x)\right) {\rm d}x=0, \end{eqnarray} $

这里

(2.44) $ \begin{eqnarray} G_{M}(\rho):=\left\{\begin{array}{ll} { } \frac{1}{2} \rho^2, & \mbox{ 如果 }\ |\rho| \leq M, \\ { } M|\rho|-\frac{1}{2} M^{2}, & \mbox{ 如果 }\ |\rho|>M, \end{array}\right. \end{eqnarray} $

且$ G_M^+(\rho)=G_M(\rho)\chi_{[0, +\infty)}(\rho) $, $ G_M^{-}(\rho)=G_M(\rho)\chi_{(-\infty, 0]}(\rho) $, $ \rho\in(-\infty, \infty). $

引理2.13^[28]  假设$ M>0, $$ G_M(\rho) $满足(2.44) 式, 则

(2.45) $ \begin{eqnarray} \left\{\begin{array}{l} { } G_{M}(\rho)=\frac{1}{2} \rho^2-\frac{1}{2}(M-|\rho|)^{2} \chi_{(-\infty, -M) \cap(M, \infty)}(\rho), \\ G_{M}^{\prime}(\rho)=\rho+(M-|\rho|) {\rm sign}(\rho) \chi_{(-\infty, -M) \cap(M, \infty)}(\rho) , \\ { } G_{M}^{+}(\rho)=\frac{1}{2}\left(\rho_{+}\right)^{2}-\frac{1}{2}(M-\rho)^{2} \chi_{(M, \infty)}(\rho) , \\ \left(G_{M}^{+}\right)^{\prime}(\rho)=\rho_{+}+(M-\rho) \chi_{(M, \infty)}(\rho) , \\ { } G_{M}^{-}(\rho)=\frac{1}{2}\left(\rho_{-}\right)^{2}-\frac{1}{2}(M+\rho)^{2} \chi_{(-\infty, -M)}(\rho) , \\ \left(G_{M}^{-}\right)^{\prime}(\rho)=\rho_{-}-(M+\rho) \chi_{(-\infty, -M)}(\rho). \end{array}\right. \end{eqnarray} $

引理2.14  若$ u_{0} \in H^{1}({{\Bbb R}} ), $奇数$ k\geq 2 $, 对任意$ t>0 $, 则有

$ \begin{eqnarray*} \frac{1}{2} \int_{{{\Bbb R}} }\left(\overline{\left(p_{+}\right)^{2}}-p_{+}^{2}\right)(t, x) {\rm d}x &\leq& -\frac{k-1}{k}\int_{0}^{t} \int_{{{\Bbb R}} }u^{k}\left(\overline{\frac{\partial G_{M}^{+}(p)}{x}} -\frac{\partial G_{M}^{+}(p)}{x}\right){\rm d}x{\rm d}s{\nonumber} \\ && +\int_{0}^{t} \int_{{{\Bbb R}} } Q(s, x)\left[\overline{p_{+}}(s, x)-p_{+}(s, x)\right] {\rm d}x{\rm d}s. \label{K75}{\nonumber} \end{eqnarray*} $

证   对任意$ T>0, $假设$ M>0 $充分大, 由(2.35) 式减去(2.39) 式并结合$ G_{M}^{+} $, 可得

(2.46) $ \begin{eqnarray} &&\frac{\partial}{\partial t}\left(\overline{G_{M}^{+}(p)}-G_{M}^{+}(p)\right)+\frac{1}{k}\frac{\partial}{\partial x}\left(u^k\left[\overline{G_{M}^{+}(p)}-G_{M}^{+}(p)\right]\right) {}\\ &\leq &u^{k-1} \left(\overline{pG_{M}^{+}(p)}-p G_{M}^{+}(p)\right)-\frac{1}{2}u^{k-1}\left(\overline{p^{2}(G_{M}^{+})^{\prime}(p)}-p^{2}\left(G_{M}^{+}\right)^{\prime}(p)\right) {}\\ &&-\frac{1}{2}u^{k-1}\left(\overline{p^{2}}-p^{2}\right)(G_{M}^{+})^{\prime}(p)-\frac{k-1}{k}u^{k}\left(\overline{\frac{\partial G_{M}^{+}(p)}{x}}-\frac{\partial G_{M}^{+}(p)}{x}\right){}\\ &&+Q(t, x)\left(\overline{\left(G_{M}^{+}\right)^{\prime}(p)}-\left(G_{M}^{+}\right)^{\prime}(p)\right). \end{eqnarray} $

注意到$ G_{M}^{+} $是增函数, 则有

(2.47) $ \begin{eqnarray} -\frac{u^{k-1}}{2}\left(\overline{p^{2}}-p^{2}\right)\left(G_{M}^{+}\right)^{\prime}(p) \leq 0. \end{eqnarray} $

使用引理2.13知

(2.48) $ \begin{eqnarray} \begin{array}{l} { } p G_{M}^{+}(p)-\frac{1}{2} p^{2}\left(G_{M}^{+}\right)^{\prime}(p)=-\frac{M}{2} p(M-p) \chi_{(M, \infty)(p)}, \\ { } \overline{pG_{M}^{+}(p)}-\frac{1}{2} \overline{p^{2}\left(G_{M}^{+}\right)^{\prime}(p)}=-\frac{M}{2} \overline{p(M-p) \chi_{(M, \infty)}(p)}. \end{array} \end{eqnarray} $

结合(2.36)和(2.37)式, 在空间$ \Omega_{M}=(\frac{2}{M-c}, \infty)\times{{\Bbb R}} $中, 我们能找到一个足够大的正常数$ M $使得

(2.49) $ \begin{eqnarray} pG_{M}^{+}(p)-\frac{1}{2} p^{2}\left(G_{M}^{+}\right)^{\prime}(p)=\overline{q G_{M}^{+}(p)}-\frac{1}{2} \overline{p^{2}\left(G_{M}^{+}\right)^{\prime}(p)}=0 \end{eqnarray} $

且

(2.50) $ \begin{eqnarray} G_{M}^{+}=\frac{1}{2}\left(p_{+}\right)^{2}, \quad\left(G_{M}^{+}\right)^{\prime}(p)=p_{+}, \quad \overline{G_{M}^{+}(p)}=\frac{1}{2} \overline{\left(p_{+}\right)^{2}}, \quad \overline{\left(G_{M}^{+}\right)^{\prime}(p)}=\overline{p_{+}}. \end{eqnarray} $

由(2.46)–(2.50) 式得

(2.51) $ \begin{eqnarray} &&\frac{\partial}{\partial t}\left(\overline{G_{M}^{+}(p)}-G_{M}^{+}(p)\right) +\frac{1}{k}\frac{\partial}{\partial x}\left(u^k\left[\overline{G_{M}^{+}(p)}-G_{M}^{+}(p)\right]\right) {}\\ & \leq& -\frac{k-1}{k}u^{k}\left(\overline{\frac{\partial G_{M}^{+}(p)}{x}}-\frac{\partial G_{M}^{+}(p)}{x}\right)+Q(t, x)\left(\overline{\left(G_{M}^{+}\right)^{\prime}(p)}-\left(G_{M}^{+}\right)^{\prime}(p)\right). \end{eqnarray} $

对(2.51) 式在$ (\frac{2}{M-c}, t) \times{{\Bbb R}} $积分得

(2.52) $ \begin{eqnarray} \frac{1}{2} \int_{{{\Bbb R}} }\left(\overline{\left(p_{+}\right)^{2}}-p_{+}^{2}\right)(t, x) {\rm d}x &\leq &\int_{{{\Bbb R}} }\left[\overline{G_{M}^{+}(p)}(\frac{2}{M-c}, x)-G_{M}^{+}(p)(\frac{2}{M-c}, x)\right] {\rm d}x{} \\ & &-\frac{k-1}{k}\int^{t}_{\frac{2}{M-c}} \int_{{\Bbb R}} u^{k}\left(\overline{\frac{\partial G_{M}^{+}(p)}{x}}-\frac{\partial G_{M}^{+}(p)}{x}\right){\rm d}x{\rm d}s{} \\ & &+\int_{\frac{2}{M-c}}^{t} \int_{{{\Bbb R}} } Q(s, x)\left[\overline{p_{+}}(s, x)-p_{+}(s, x)\right] {\rm d}x{\rm d}s. \end{eqnarray} $

令$ M \rightarrow \infty $, 应用引理2.12, 即得引理2.14的证明.

引理2.15   假设$ u_{0} \in H^{1}({{\Bbb R}} ), $则

$ \begin{eqnarray*} \int\left(\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right)(t, x) {\rm d}x &\leq &\frac{u^{k-1}M^{2}}{2} \int_{0}^{t} \int_{{{\Bbb R}} } \overline{(M+p) \chi_{(-\infty, -M)}(p)} {\rm d}x{\rm d}s\\ && -\frac{u^{k-1}M^{2}}{2} \int_{0}^{t} \int_{{{\Bbb R}} } (M+p) \chi_{(-\infty, -M)}(p) {\rm d}x{\rm d}s\\ &&+u^{k-1}M \int_{0}^{t}\int_{{{\Bbb R}} } \left[\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right] {\rm d}x{\rm d}s \\ && +\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } \left(\overline{p_{+}^{2}}-p_{+}^{2}\right){\rm d}x{\rm d}s\\ &&-\frac{k-1}{k}\int_{0}^{t} \int_{{{\Bbb R}} }u^{k}\left(\overline{\frac{\partial G_{M}^{-}(p)}{x}} -\frac{\partial G_{M}^{-}(p)}{x}\right){\rm d}x{\rm d}s\\ && +\int_{0}^{t} \int_{{{\Bbb R}} } Q(t, x)\left(\overline{\left(G_{M}^{-}\right)^{\prime}(p)} -\left(G_{M}^{-}\right)^{\prime}(p)\right) {\rm d}x{\rm d}s. \label{K83}{\nonumber} \end{eqnarray*} $

证   假设常数$ M $充分大, 使用引理2.8、2.10和$ G_{M}^{-}, $我们推导出

$ \begin{eqnarray*} && \frac{\partial}{\partial t}\left(\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right)+\frac{1}{k}\frac{\partial}{\partial x}\left(u^k\left[\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right]\right) \\ &\leq& u^{k-1} \left(\overline{pG_{M}^{-}(p)}-p G_{M}^{-}(p)\right)-\frac{1}{2}u^{k-1}\left(\overline{p^{2}(G_{M}^{-})^{\prime}(p)}-p^{2}\left(G_{M}^{-}\right)^{\prime}(p)\right) \\ &&-\frac{1}{2}u^{k-1}\left(\overline{p^{2}}-p^{2}\right)(G_{M}^{-})^{\prime}(p)-\frac{k-1}{k}u^{k}\left(\overline{\frac{\partial G_{M}^{-}(p)}{x}}-\frac{\partial G_{M}^{-}(p)}{x}\right)\\ &&+Q(t, x)\left(\overline{\left(G_{M}^{-}\right)^{\prime}(p)}-\left(G_{M}^{-}\right)^{\prime}(p)\right). \label{K84}{\nonumber} \end{eqnarray*} $

因为$ -M \leq\left(G_{M}^{-}\right)^{\prime} \leq 0, $则

(2.53) $ \begin{eqnarray} -\frac{u^{k-1}}{2}\left(\overline{p^{2}}-p^{2}\right)\left(G_{M}^{-}\right)^{\prime}(p) \leq \frac{u^{k-1}M}{2}\left(\overline{p^{2}}-p^{2}\right). \end{eqnarray} $

使用注2.1和引理2.13有

(2.54) $ \begin{eqnarray} \begin{array}{l} { } pG_{M}^{-}(p)-\frac{1}{2} p^{2}\left(G_{M}^{-}\right)^{\prime}(p)=-\frac{M}{2} p(M-p) \chi_{(-\infty, -M)}(p), \\ { } \overline{p G_{M}^{-}(p)}-\frac{u^{k-1}}{2} \overline{p^{2}\left(G_{M}^{-}\right)^{\prime}(p)}=-\frac{M}{2} \overline{p(M-p) \chi_{(-\infty, -M)}(p)}. \end{array} \end{eqnarray} $

使用(2.53)和(2.54) 式得

(2.55) $ \begin{eqnarray} && \frac{\partial}{\partial t}\Big(\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\Big) +\frac{1}{k}\frac{\partial}{\partial x}\Big(u^k\Big[\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\Big]\Big){} \\ &\leq&-\frac{u^{k-1}M}{2}\overline{p(M+p) \chi_{(-\infty, -M)}(p)}+\frac{u^{k-1}M}{2}p(M+p) \chi_{(-\infty, -M)}(p) +\frac{u^{k-1}M}{2} (\overline{p^{2}}-p^{2}){}\\ &&-\frac{k-1}{k}u^{k}\left(\overline{\frac{\partial G_{M}^{-}(p)}{x}} -\frac{\partial G_{M}^{-}(p)}{x}\right) +Q(t, x)\Big(\overline{(G_{M}^{-})^{\prime}(p)}-(G_{M}^{-})^{\prime}(p)\Big). \end{eqnarray} $

对(2.55) 式在$ (0, t) \times{{\Bbb R}} $积分, 我们有

(2.56) $ \begin{eqnarray} && \int_{{{\Bbb R}} }(\overline{G_{M}^{-}(p)}-G_{M}^{-}(p))(t, x) {\rm d}x {}\\ &\leq&-\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } \overline{p(M+p) \chi_{(-\infty, -M)}(p)} {\rm d}x{\rm d}s {}\\ &&+\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } p(M+p) \chi_{(-\infty, -M)}(p) {\rm d}x{\rm d}s+\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } \left(\overline{p^{2}}-p^{2}\right) {\rm d}x{\rm d}s {}\\ && -\frac{k-1}{k}\int_{0}^{t} \int_{{{\Bbb R}} }u^{k}\left(\overline{\frac{\partial G_{M}^{-} (p)}{x}}-\frac{\partial G_{M}^{-}(p)}{x}\right){\rm d}x{\rm d}s{}\\ &&+\int_{0}^{t} \int_{{{\Bbb R}} } Q(t, x)\left(\overline{\left(G_{M}^{-}\right)^{\prime}(p)} -\left(G_{M}^{-}\right)^{\prime}(p)\right) {\rm d}x{\rm d}s. \end{eqnarray} $

使用引理2.14有

(2.57) $ \begin{equation} \overline{G_{M}^{-}(p)}-G_{M}^{-}(p)=\frac{1}{2}\left(\overline{\left(p_{-}\right)^{2}}-\left(p_{-}\right)^{2}\right)+\frac{1}{2}(M+p)^{2} \chi_{(-\infty, -M)}(p) -\frac{1}{2} \overline{(M+p)^{2} \chi_{(-\infty, -M)}(p)}. \end{equation} $

使用(2.56)、(2.57) 式和注2.1有

$ \begin{eqnarray*} && \int_{{{\Bbb R}} }\left(\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right)(t, x) {\rm d}x {\nonumber}\\ &\leq&-\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} }\overline{q(M+p) \chi_{(-\infty, -M)}(p)}{\rm d}x{\rm d}s{\nonumber} \\ && +\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } q(M+p) \chi_{(-\infty, -M)}(p) {\rm d}x{\rm d}s{\nonumber} \\ && +u^{k-1}M \int_{0}^{t} \int \left[\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right] {\rm d}x {\rm d}s -\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } (M+p)^{2} \chi_{(-\infty, -M)}(p) {\rm d}x{\rm d}s{\nonumber} \\ && +\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } \overline {(M+p)^{2} \chi_{(-\infty, -M)}(p) }{\rm d}x{\rm d}s +\frac{u^{k-1}M}{2} \int_{0}^{t} \int \left(\overline{p_{+}^{2}}-p_{+}^{2}\right) {\rm d}x{\rm d}s{\nonumber} \\ && -\frac{k-1}{k}\int_{0}^{t} \int_{{{\Bbb R}} }u^{k}\left(\overline{\frac{\partial G_{M}^{-}( p)}{x}}-\frac{\partial G_{M}^{-}(p)}{x}\right){\rm d}x{\rm d}s\\ &&+\int_{0}^{t} \int_{{{\Bbb R}} } Q(t, x)\left(\overline{\left(G_{M}^{-}\right)^{\prime}(p)} -\left(G_{M}^{-}\right)^{\prime}(p)\right). \end{eqnarray*} $

使用等式$ M(M+p)^{2}-Mp(M+p)=M^{2}(M+p), $便完成引理2.15的证明.

引理2.16   假设$ u_0\in H^1({{\Bbb R}} ) $, 则下式在$ [0, \infty) \times(-\infty, \infty) $几乎处处成立

$ \begin{eqnarray} \overline{p^{2}}=p^{2}. \label{K90}{} \end{eqnarray} $

证   使用引理2.14和2.15, 我们有

(2.58) $ \begin{eqnarray} && \int_{{{\Bbb R}} }\left(\frac{1}{2}\left[\overline{\left(p_{+}\right)^{2}}-\left(p_{+}\right)^{2}\right]+\left[\overline{G_{M}^{-}}-G_{M}^{-}\right]\right)(t, x) {\rm d}x {}\\ &\leq& \frac{u^{k-1}M^{2}}{2} \int_{0}^{t} \int_{{{\Bbb R}} } \overline{(M+p) \chi_{(-\infty, -M)} (p)} {\rm d}x{\rm d}s{}\\ && -\frac{u^{k-1}M^{2}}{2}\int_{0}^{t} \int_{{{\Bbb R}} } (M+p) \chi_{(-\infty, -M)}(p) {\rm d}x{\rm d}s +u^{k-1}M \int_{0}^{t} \int_{{{\Bbb R}} } \left[\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right] {\rm d}x{\rm d}s{}\\ && +\frac{u^{k-1}M}{2} \int_{0}^{t} \int_{{{\Bbb R}} } \left(\overline{p_{+}^{2}}-p_{+}^{2}\right) {\rm d}x{\rm d}s -\frac{k-1}{k}\int_{0}^{t} \int_{{{\Bbb R}} }u^{k}\left(\overline{\frac{\partial G_{M}^{-}(p)}{x}} -\frac{\partial G_{M}^{-}(p)}{x}\right){\rm d}x{\rm d}s{}\\ && +\int_{0}^{t} \int_{{{\Bbb R}} } Q(s, x)\left(\left[\overline{p_{+}}-p_{+}\right] +\left[\overline{\left(G_{M}^{-}\right)^{\prime}(p)}-\left(G_{M}^{-}\right)^{\prime}(p)\right]\right) {\rm d}x{\rm d}s. \end{eqnarray} $

事实上, 从引理2.6可知, 存在一个正常数$ W $使得

(2.59) $ \begin{eqnarray} \|Q(t, x)\|_{L^{\infty}([0, T) \times{{\Bbb R}} )} \leq W. \end{eqnarray} $

应用注2.1和引理2.13有

$ p_{+}+\left(G_{M}^{-}\right)^{\prime}(p)=p-(M+p) \chi_{(-\infty, -M)}, $

$ \overline{p_{+}}+\overline{\left(G_{M}^{-}\right)^{\prime}(p)}=p-\overline{(M+p) \chi_{(-\infty, -M)}(p)}. $

因为函数$ G \rightarrow G_{+}+\left(G_{M}^{\prime}\right)^{\prime}(\rho) $是凸的, 则

$ 0 \leq[\overline{p_{+}}-p_{+}]+[\overline{(G_{M}^{-})^{\prime}(p)}-(G_{M}^{-})^{\prime}(p)] =(M+p) \chi_{(-\infty, -M)}-\overline{(M+p) \chi_{(-\infty, -M)}(p)}. \label{K94}{\nonumber} $

由(2.59) 式得

$ \begin{eqnarray*} &&Q(s, x)\left(\left[\overline{p_{+}}(s, x)-p_{+}(s, x)\right]+\left[\overline{\left(G_{M}^{-}\right)^{\prime}(p)}-\left(G_{M}^{-}\right)^{\prime}(p)\right]\right)\\ &\leq&-W\left(\overline{(M+p) \chi_{(-\infty, -M)}(p)}-(M+p) \chi_{(-\infty, -M)}(p)\right). \label{K95}{\nonumber} \end{eqnarray*} $

让$ M $充分大, 我们有

(2.60) $ \begin{eqnarray} &&\frac{u^{k-1} M^2}{2}\overline{(M+p) \chi_{(-\infty, -M)}(p)}-\frac{u^{k-1}M^{2}}{2}(M+p) \chi_{(-\infty, -M)}(p){}\\ &&+Q(s, x)\left(\Big[\overline{p_{+}}(s, x)-p_{+}(s, x)\Big]+\left[(\overline{G_{M}^{-}})^{\prime}(p)-\left(G_{M}^{-}\right)^{\prime}(p)\right]\right){}\\ &\leq&\left(\frac{ u^{k-1}M^{2}}{2}-W\right)\left(\overline{(M+p) \chi_{(-\infty, -M)}(p)}-(M+p) \chi_{(-\infty, -M)}(p)\right) \leq 0. \end{eqnarray} $

由(2.58)和(2.60) 式得

$ \begin{eqnarray*} 0&\leq& \int_{{{\Bbb R}} }\left(\frac{1}{2}\left[\overline{\left(p_{+}\right)^{2}}-\left(p_{+}\right)^{2}\right]+\left[\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right]\right)(t, x) {\rm d}x\\ &\leq &c M \int_{0}^{t} \int_{{{\Bbb R}} }\left(\frac{1}{2}\left[\overline{\left(p_{+}\right)^{2}}-p_{+}^{2}\right]+\left[\overline{G_{M}^{-}(p)}-G_{M}^{-}(p)\right]\right) {\rm d}s{\rm d}x. \label{K97}{\nonumber} \end{eqnarray*} $

使用Gronwall不等式, 我们有

$ 0 \leq \int_{{{\Bbb R}} }\left(\frac{1}{2}\left[\overline{\left(p_{+}\right)^{2}}-\left(p_{+}\right)^{2}\right]+\left[\overline{G_{R}^{-}(p)}-G_{R}^{-}(p)\right]\right)(t, x) {\rm d}x \leq 0. \label{K98}{\nonumber} $

使用Fatou引理, 注2.1和(2.32)式, 令$ M \rightarrow \infty $, 则

$ 0 \leq \int_{{{\Bbb R}} }\left(\overline{p^{2}}-p^{2}\right)(t, x) {\rm d}x \leq 0, \quad t>0, \label{K99}{\nonumber} $

于是便知引理2.16成立.

定理3.1   假设$ u_{0}(x) \in H^{1}({{\Bbb R}} ), $那么柯西问题(2.1) 或(2.2) 至少存在一个整体弱解$ u(t, x) $且满足如下性质:

$ \rm (a) $存在正常数$ c=c(\|u_{0}\|_{H^{1}({{\Bbb R}} )} $使得

(3.1) $ \begin{eqnarray} \frac{\partial u(t, x)}{\partial x} \leq \frac{2}{t}+c, \quad (t, x) \in[0, \infty) \times{{\Bbb R}} . \end{eqnarray} $

$ \rm (b) $存在依赖$ \left\|u_{0}\right\|_{H^{1}({{\Bbb R}} )} $, $ T $和$ k $的正常数$ c_0 $, 使得

(3.2) $ \begin{eqnarray} \int_{0}^{T} \int_{{{\Bbb R}} }\left|\frac{\partial u(t, x)}{\partial x}\right|^{2k} {\rm d}x{\rm d}t \leq c_0. \end{eqnarray} $

证   (a) 和(b) 可由引理2.3和2.4直接得到, 由引理2.16, 在空间$ L^2_{{\rm loc}}([0, \infty)\times{{\Bbb R}} ) $中, 我们有

$ p_\varepsilon\rightarrow p \quad (\varepsilon \rightarrow 0). \label{K102}{\nonumber} $

于是, 我们知道$ u $是问题(2.2) 的一个整体弱解(在定义2.1意义下).