修正二分量Dullin-Gottwald-Holm (mDGH2)系统为

(1.1) $ \begin{equation} \left\{ \begin{array}{ll} u_{t}-\alpha^{2}u_{xxt}+c_{0}u_{x}+3uu_{x}+\beta u_{xxx} = \alpha^{2}(2u_{x}u_{xx}+uu_{xxx})+\gamma \rho\bar{\rho}_{x}, \; \; t>0, \; \; x\in{{\Bbb R}} , \\ \rho_{t}+(\rho u)_{x} = 0, \; \; t>0, \; \; x\in{{\Bbb R}} , \\ \rho = (1-\partial_{x}^{2})(\bar{\rho}-\bar{\rho}_{0}), \end{array} \right. \end{equation} $

其中常数$ \alpha^{2}(\alpha>0) $和$ \beta/c_{0} $是长尺度的平方, $ c_{0} = \sqrt{h\gamma}\geq0 $是静水在空间无限远处的线性波速,这里$ h>0 $为平均流体深度, $ \gamma>0 $表示浅水波应用中重力的向下恒定加速度, $ u(t, x) $表示流体速度. mDGH2系统(1.1)用平均或滤波后的密度$ \rho $表示$ \rho = (1-\partial_{x}^{2})(\bar{\rho}-\bar{\rho}_{0}) $,类似于动量和速度之间的关系,这里$ \bar{\rho}_{0} $被取为常数^{[14, 17-19]}.

值得注意的是, mDGH2系统(1.1)是以下DGH2系统的修正版本

(1.2) $ \begin{equation} \left\{ \begin{array}{ll} u_{t}-u_{xxt}+c_{0}u_{x}+3uu_{x}+\beta u_{xxx}-2u_{x}u_{xx}-uu_{xxx}+ \rho\rho_{x} = 0, \; \; t>0, \; \; x\in{{\Bbb R}} , \\ \rho_{t}+(\rho u)_{x} = 0, \; \; t>0, \; \; x\in{{\Bbb R}} , \end{array} \right. \end{equation} $

它是在文献[8]中通过运用Ivanov方法^[11]被推导出来.关于DGH2系统(1.2)的一些定性分析还有很多工作已被研究^[18-19].

一般情况下,在实验中很难避免能量耗散机制.为此, Ott和Sudan^[16]研究了KdV方程如何进行修正以用于包括耗散的影响和耗散对孤波解的影响. Ghidaglia^[7]研究了作为有限维动力系统的弱耗散KdV方程解的长时间行为.

同样,本文我们考虑以下的弱耗散mDGH2系统

(1.3) $ \begin{equation} \left\{ \begin{array}{ll} u_{t}-\alpha^{2}u_{xxt}+c_{0}u_{x}+3uu_{x}+\beta u_{xxx}+\lambda(u-\alpha^{2}u_{xx}) = \alpha^{2}(2u_{x}u_{xx}+uu_{xxx})+\gamma \rho\bar{\rho}_{x}, \\ \rho_{t}+(\rho u)_{x}+\lambda\rho = 0, \\ \rho = (1-\partial_{x}^{2})(\bar{\rho}-\bar{\rho}_{0}), \end{array} \right. \end{equation} $

其中$ t>0, \; x\in{{\Bbb R}} $,弱耗散项为$ \lambda(u-\alpha^{2}u_{xx}) $和$ \lambda\rho $,这里$ \lambda $是一个耗散参数.

若取$ \alpha = 1 $, $ \beta = 0 $, $ \gamma = -1 $, $ c_{0} = 0 $和$ \lambda = 0 $, mDGH2系统(1.3)可以约化为下面的mCH2系统

(1.4) $ \begin{equation} \left\{ \begin{array}{ll} u_{t}-u_{xxt}+3uu_{x}-2u_{x}u_{xx}-uu_{xxx}+\rho\bar{\rho}_{x} = 0, \; \; t>0, \; \; x\in{{\Bbb R}} , \\ \rho_{t}+(\rho u)_{x} = 0, \; \; t>0, \; \; x\in{{\Bbb R}} , \\ \rho = (1-\partial_{x}^{2})(\bar{\rho}-\bar{\rho}_{0}). \end{array} \right. \end{equation} $

mCH2系统(1.4)允许在速度和平均密度上有峰值的解^[10].在文献[10]中,作者通过分析确定了陡峭化的机制,该机制使得奇异解可以从光滑的,空间受限的初始数据中出现.他们发现流体速度中的爆破波并不意味着奇点的点态密度$ \rho $在垂直点的斜率.

若取$ \alpha = \beta = \gamma = 0 $, $ c_{0} = k $, $ \lambda = 0 $和$ \rho = 0 $, mDGH2系统(1.3)约化为CH方程^{[1, 2]}

(1.5) $ \begin{equation} u_{t}-u_{xxt}+ku_{x}+3uu_{x} = 2u_{x}u_{xx}+uu_{xxx}, \; \; t>0, \; \; x\in{{\Bbb R}} , \end{equation} $

该方程最早是由Fokas和Fuchssteiner在文献[6]中用它的双哈密顿结构正式推导出来的.在文献[4]中, Constantin和Lannes严格地用数学方法描述了CH方程(1.5)的流体动力学相关性. Constantin、Kolev^[5]和Ionescu-Kruse^[12]提供了CH方程在圆的微分同态群上作为测地线流方程的另一种推导.

当$ \rho = 0 $时, mDGH2系统(1.3)成为弱耗散DGH方程

(1.6) $ \begin{equation} u_{t}-\alpha^{2}u_{xxt}+c_{0}u_{x}+3uu_{x}+\beta u_{xxx}+\lambda(u-\alpha^{2}u_{xx}) = \alpha^{2}(2u_{x}u_{xx}+uu_{xxx}), \; \; t>0, \; \; x\in{{\Bbb R}} . \end{equation} $

在文献[15]中, Novruzov证明了初始数据在有限时间内导致弱耗散DGH方程(1.6)的解爆破的简单条件.

设$ y = u-\alpha^{2}u_{xx} $和$ v = \bar{\rho}-\bar{\rho}_{0} $, mDGH2系统(1.3)有如下形式

(1.7) $ \begin{equation} \left\{ \begin{array}{ll} &y_{t}+2d_{0}u_{x}+2u_{x}y+uy_{x}-\gamma \rho v_{x}+k y_{x}+\lambda y = 0, \; \; t>0, \; \; x\in{{\Bbb R}} , \\ &\rho_{t}+(\rho u)_{x}+\lambda\rho = 0, \; \; t>0, \; \; x\in{{\Bbb R}} , \\ &\rho = (1-\partial_{x}^{2})v, \end{array} \right. \end{equation} $

其中$ 2d_{0} = c_{0}+\frac{\beta}{\alpha^{2}} $和$ k = -\frac{\beta}{\alpha^{2}} $.

本文中,当$ \rho = 0 $时,我们的结果可以归结为弱耗散DGH方程(1.6)的相应结果,它的爆破在文献[15]中已经被研究.但是系统(1.3)和(1.2)之间有相当大的差异.系统(1.3)的爆破的发生受耗散参数的影响.值得注意的是,弱耗散项对mDCH2系统(1.3)的解没有影响.

论文的结构:第2节介绍了一些预备知识,包括局部的适定性和mDGH2系统(1.3)的精确爆破现象.第3节分析了全局存在性,证明了在条件$ \left(\|y_{0}\|_{L^{2}}^{2}+\|\rho_{0}\|^{2}_{L^{2}}\right)^{\frac{1}{2}}<\frac{4\lambda} {3} $下爆破现象是不可能发生的.爆破和爆破率分别在第4节和第5节中被研究.

引入$ G(x): = \frac{1}{2\alpha}e^{-|\frac{x}{\alpha}|} $, $ x\in{{\Bbb R}} $和算子$ \Lambda = (1-\alpha^{2}\partial_{x}^{2})^{-1} $.那么对所有$ f\in L^{2} $,有

(2.1) $ \begin{equation} G \ast f(x) = \Lambda f(x) = \int_{{{\Bbb R}} }G(x-\xi) f(\xi){\rm d}\xi, \end{equation} $

$ \partial_{x}^{2}G\ast f = \frac{1}{\alpha^{2}}(G\ast f-f), \; \; G\ast y = u, $

这里的符号‘$ \ast $’表示卷积.利用上述恒等式,将系统(1.3)写成双曲型准线性演化方程的等价形式

(2.2) $ \begin{equation} \left\{ \begin{array}{ll} { } u_{t}+\lambda u+uu_{x}+ku_{x} = -\partial_{x}G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2}v^{2} -\frac{\gamma}{2}v^{2}_{x}\right), \\ v_{t}+uv_{x}+\lambda v = -G\ast\left((u_{x}v_{x})_{x}+u_{x}v\right), \; \; t>0, \; \; x\in{{\Bbb R}} , \\ u(0, x) = u_{0}(x), \; \; x\in{{\Bbb R}} , \\ v(0, x) = v_{0}(x), \; \; x\in{{\Bbb R}} . \end{array} \right. \end{equation} $

利用Kato定理^[13],我们得到了具有初始数据$ (u_{0}, v_{0})\in H^{s}\times H^{s-1} $, $ s\geq \frac{5}{2} $的柯西问题(2.2)的局部适定性.更精确地说,我们得到了以下的局部适定性结果.

定理2.1  设$ (u_{0}, v_{0})\in H^{s}\times H^{s-1} $, $ s\geq\frac{5}{2} $,存在最大时间$ T = T(\|u_{0}\|_{H^{s}({{\Bbb R}} )}, $$ \|v_{0}\|_{H^{s-1}({{\Bbb R}} )})>0 $,和系统(2.2)的唯一解$ (u, v) $,

$ (u, v)\in C\left([0, T); H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} )\right) \cap C^{1}\left([0, T); H^{s-1}({{\Bbb R}} )\times H^{s-2}({{\Bbb R}} )\right). $

而且,该解连续依赖初始数据,即映射$ (u_{0}, v_{0})\mapsto(u, v) $

$ H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) \rightarrow C\left([0, T); H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} )\right) \cap C^{1}\left([0, T); H^{s-1}({{\Bbb R}} )\times H^{s-2}({{\Bbb R}} )\right) $

是连续的.

接下来,介绍标准粒子轨迹方法.我们考虑以下两个初值问题

(2.3) $ \left\{\begin{array}{l}q_{1 t}=u\left(t, q_{1}\right)+k, \quad 0 \leq t<T, x \in \mathbb{R}, \\q_{1}(0, x)=x, \quad x \in \mathbb{R}\end{array}\right. $

和

(2.4) $ \begin{equation} \left\{ \begin{array}{ll} &q_{2t} = u(t, q_{2}), \; \; 0\leq t<T, \; x\in {{\Bbb R}} , \\ &q_{2}(0, x) = x, \; \; x\in {{\Bbb R}} , \end{array} \right. \end{equation} $

其中$ u\in C^{1}\left([0, T), H^{s-1}\right) $表示具有初始数据$ (u_{0}, v_{0})\in H^{s}\times H^{s-1} $, $ s\geq\frac{5}{2} $的系统(2.2)的解$ (u, v) $的第一个分量, $ T>0 $是最大存在时间.将经典结果应用于常微分方程,我们能得到以下关于$ q_{i} $的结果.

引理2.1  设$ u\in C\left([0, T); H^{s}({{\Bbb R}} )\right) \cap C^{1}\left([0, T); H^{s-1}({{\Bbb R}} )\right) $, $ s\geq\frac{5}{2} $,则方程(2.3)和(2.4)分别有唯一解$ q\in C^{1}([0, T)\times {{\Bbb R}} ; {{\Bbb R}} ) $.此外, $ q_{i}(t, \cdot): {{\Bbb R}} \rightarrow{{\Bbb R}} $一个微分同胚映射对每个$ t\in [0, T) $

(2.5) $ \begin{equation} q_{itx}(t, x) = u_{x}(t, q_{i}(t, x))q_{i, x}(t, x), \end{equation} $

(2.6) $ \begin{equation} q_{ix}(t, x) = \exp\left(\int_{0}^{t}u_{x}(\tau, q_{i}(\tau, x)){\rm d}\tau\right)>0, \; \; i = 1, 2, \; \; (t, x)\in[0, T)\times {{\Bbb R}} . \end{equation} $

因此,任何函数$ v(t, \cdot) $的$ L^{\infty} $ -范数都保留在微分同态$ q_i(t, \cdot) $族下.

引理2.2  设$ (u_{0}, v_{0})\in H^{s}\times H^{s-1} $, $ s\geq\frac{5}{2} $, $ T>0 $是初始数据为$ (u_{0}, v_{0}) $时系统(2.2)的解$ (u, v) $的最大存在时间.那么

(2.7) $ \begin{equation} \rho(t, q_{2}(t, x))q_{2x} = e^{-\lambda t}\rho_{0}(x), \; \; (t, x)\in [0, T)\times {{\Bbb R}} . \end{equation} $

此外,如果存在一个$ x_{0}\in{{\Bbb R}} $有$ \rho_{0}(x_{0}) = 0 $,那么对任意$ t\in [0, T) $有$ \rho(t, q_{2}(t, x_{0})) = 0 $.

为了得到系统(2.2)解的精确爆破情况,我们给出下面的引理.

引理2.3  设$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $,其中$ s\geq\frac{5}{2} $,且$ T>0 $是初始数据为$ (u_{0}, v_{0}) $时系统(2.2)的解$ (u, v) $的最大时间.那么

(2.8) $ \begin{eqnarray} E(t) = \int_{{{\Bbb R}} }\left(u^{2}+\alpha^{2}u_{x}^{2}+v^{2}+v_{x}^{2}\right) {\rm d}x = e^{-2\lambda t}\int_{{{\Bbb R}} }\left(u_{0}^{2}+\alpha^{2}u_{0x}^{2}+v_{0}^{2}+v_{0x}^{2}\right) {\rm d}x. \end{eqnarray} $

而且

(2.9) $ \begin{eqnarray} \|u(t, \cdot)\|_{L^{\infty}({{\Bbb R}} )}^{2}+ \|v(t, \cdot)\|_{L^{\infty}({{\Bbb R}} )}^{2} \leq \frac{1}{2\alpha}\|u(t, \cdot)\|_{H^{1}_{\alpha}({{\Bbb R}} )}^{2} +\frac{1}{2}\|v(t, \cdot)\|_{H^{1}({{\Bbb R}} )}^{2}\leq C_{1}E(0), \end{eqnarray} $

其中$ C_{1} = \max\left\{\frac{1}{2\alpha}, \frac{1}{2}\right\} $.

证  系统(1.3)的第一个方程乘以$ u $,然后分部积分,我们有

(2.10) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }(u^{2}+\alpha^{2}u_{x}^{2}){\rm d}x & = &2\int_{{{\Bbb R}} }uu_{t}{\rm d}x+2\alpha^{2}\int_{{{\Bbb R}} }u_{x}u_{xt}{\rm d}x{} \\ & = &-2\gamma \int_{{{\Bbb R}} }uuv_{x}{\rm d}x+2\gamma \int_{{{\Bbb R}} }uv_{x}v_{xx}{\rm d}x -2\lambda \int_{{{\Bbb R}} }(u^{2}+\alpha^{2}u_{x}^{2}){\rm d}x. \end{eqnarray} $

同理,系统(1.3)的第二个等式能获得

(2.11) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }(v^{2}+v_{x}^{2}){\rm d}x & = &2\int_{{{\Bbb R}} }vv_{t}{\rm d}x+2\int_{{{\Bbb R}} }v_{x}v_{xt}{\rm d}x{} \\ & = &2\int_{{{\Bbb R}} }uv_{x}v{\rm d}x-2\int_{{{\Bbb R}} }uv_{x}v_{xx}{\rm d}x -2\lambda \int_{{{\Bbb R}} }(v^{2}+v_{x}^{2}){\rm d}x. \end{eqnarray} $

通过运用(2.10)和(2.11)式,有

(2.12) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }(u^{2}+\alpha^{2}u_{x}^{2}+v^{2}+v_{x}^{2}){\rm d}x = -2\lambda\int_{{{\Bbb R}} }(u^{2}+\alpha^{2}u_{x}^{2}+v^{2}+v_{x}^{2}){\rm d}x, \end{eqnarray} $

即

(2.13) $ \begin{equation} \int_{{{\Bbb R}} }(u^{2}+\alpha^{2}u_{x}^{2}+v^{2}+v_{x}^{2}){\rm d}x = e^{-2\lambda t}\int_{{{\Bbb R}} }(u_{0}^{2}+\alpha_{0}^{2}u_{0x}^{2}+v_{0}^{2}+v_{0x}^{2}){\rm d}x. \end{equation} $

因为

(2.14) $ \begin{eqnarray} u^{2}(t, x) = \int_{-\infty}^{x}uu_{x}{\rm d}x-\int_{x}^{\infty}uu_{x}{\rm d}x\leq \int_{{{\Bbb R}} }|uu_{x}|{\rm d}x\leq \frac{1}{2}\int_{{{\Bbb R}} }(u^{2}+u_{x}^{2}) {\rm d}x \end{eqnarray} $

和

(2.15) $ \begin{eqnarray} v^{2}(t, x) = \int_{-\infty}^{x}vv_{x}{\rm d}x-\int_{x}^{\infty}vv_{x}{\rm d}x\leq \int_{{{\Bbb R}} }|vv_{x}|{\rm d}x\leq \frac{1}{2}\int_{{{\Bbb R}} }(v^{2}+v_{x}^{2}) {\rm d}x, \end{eqnarray} $

那么,利用(2.13)式我们能得到(2.9)式.证毕.

引理2.4  设$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $,其中$ s\geq\frac{5}{2} $, $ T>0 $是初始数据为$ (u_{0}, v_{0}) $时系统(2.2)的解$ (u, v) $的最大时间.那么

(2.16) $ \begin{equation} \|v_{x}(t, \cdot)\|_{L^{\infty}}\leq e^{\lambda t}\|v_{0x}\|_{L^{\infty}} +\frac{C_{1}}{\lambda}(e^{\lambda t}-1)E(0), \; \; \mbox{对所有的}\; \; t\in [0, T). \end{equation} $

证  通过对系统(2.2)的第二个方程的$ x $求导,并利用引理2.3,可以看出这个引理是由直接计算得到的.

下面推导系统(2.2)强解的精确爆破情况.

定理2.2  设$ (u_{0}, v_{0})\in H^{s}\times H^{s-1} $, $ s\geq\frac{5}{2} $,且$ T>0 $是初始数据为$ (u_{0}, v_{0}) $时系统(2.2)的解$ (u, v) $的最大存在时间.那么,相应的解在有限时间内爆破,当且仅当

(2.17) $ \begin{equation} \lim\limits_{t\rightarrow T}\inf\limits_{x\in{{\Bbb R}} }\{u_{x}(t, x)\} = -\infty. \end{equation} $

证  将系统(1.3)的第一个方程乘以$ y = u-\alpha^{2}u_{xx} $,我们有

(2.18) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }y^{2}{\rm d}x& = & 2\int_{{{\Bbb R}} }yy_{t}{\rm d}x = 2\int_{{{\Bbb R}} }y\left(-2d_{0}u_{x}-ky_{x}-2u_{x}y-uy_{x}+\gamma \rho v_{x}\right){\rm d}x-2\lambda \int_{{{\Bbb R}} }y^{2}{\rm d}x{} \\ & = &-3\int_{{{\Bbb R}} }u_{x}y^{2}{\rm d}x+2\gamma \int_{{{\Bbb R}} }y\rho v_{x}{\rm d}x -2\lambda\int_{{{\Bbb R}} }y^{2}{\rm d}x. \end{eqnarray} $

同理,由系统(1.3)的第二个方程可知

(2.19) $ \begin{equation} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }\rho^{2}{\rm d}x = - \int_{{{\Bbb R}} }u_{x}\rho^{2}{\rm d}x-2\lambda\int_{{{\Bbb R}} }\rho^{2}{\rm d}x. \end{equation} $

通过使用(2.18)和(2.19)式,我们得到

(2.20) $ \begin{equation} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x = -3\int_{{{\Bbb R}} }u_{x}y^{2}{\rm d}x+2\gamma \int_{{{\Bbb R}} }y\rho v_{x}{\rm d}x - \int_{{{\Bbb R}} }u_{x}\rho^{2}{\rm d}x-2\lambda\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x. \end{equation} $

假设存在$ M>0 $使得

(2.21) $ \begin{equation} u_{x}(t, x)\geq -M, \; \; \forall (t, x)\in [0, T)\times {{\Bbb R}} . \end{equation} $

从(2.20)式和引理2.4可以得到

(2.22) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x&\leq& (3M-2\lambda)\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x{} \\ &&+\gamma\Big(e^{\lambda T}\|v_{0x}\|_{L^{2}}^{2} +\frac{C_{1}}{\lambda}E(0)(e^{\lambda T}-1)\Big) \int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x. \end{eqnarray} $

通过求解不等式(2.22),有

(2.23) $ \begin{eqnarray} &&\left(\|u(t, \cdot)\|_{H^{2}}^{2}+\|v(t, \cdot)\|_{H^{2}}^{2}\right){}\\ & = &\left(\|y(t, \cdot)\|_{L^{2}}^{2}+\|\rho(t, \cdot)\|_{L^{2}}^{2}\right){} \\ &\leq& \left(\|y_{0}\|_{L^{2}}^{2}+\|\rho_{0}\|_{L^{2}}^{2}\right) \exp\left(\Big(3M-2\lambda+\gamma e^{\lambda T} \|v_{0x}\|_{L^{\infty}}+\frac{C_{1}\gamma}{\lambda}(e^{\lambda T}-1)E(0)\Big)t\right) \end{eqnarray} $

对所有$ t\in[0, T) $成立.不等式(2.23)保证解$ (u, v) $在有限时间内不爆破.

因此,利用Sobolev的嵌入定理,我们可以证明如果

(2.24) $ \begin{equation} \lim\limits_{t\rightarrow T}\inf\limits_{x\in {{\Bbb R}} }\{u_{x}(t, x)\} = -\infty, \end{equation} $

则解在有限时间内单爆破.证毕.

在本节,我们将证明在条件$ \left(\|y_{0}\|_{L^{2}}^{2}+\|\rho_{0}\|^{2}_{L^{2}}\right)^{\frac{1}{2}}<\frac{4\lambda} {3} $下不可能发生“爆破”现象.

定理3.1  设$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $,其中$ s\geq\frac{5}{2} $.如果

(3.1) $ \begin{equation} \left(\|y_{0}\|_{L^{2}}^{2}+\|\rho_{0}\|^{2}_{L^{2}}\right)^{\frac{1}{2}}<\frac{4\lambda} {3}. \end{equation} $

则系统(1.3)的对应解$ u(t, x) $在时间上是全局存在的.

证  从(2.18)式,我们有

(3.2) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} }y^{2}{\rm d}x = -3\int_{{{\Bbb R}} }u_{x}y^{2}{\rm d}x+2\gamma \int_{{{\Bbb R}} }y\rho v_{x}{\rm d}x -2\lambda\int_{{{\Bbb R}} }y^{2}{\rm d}x. \end{eqnarray} $

将等式(3.2)乘以$ e^{2\lambda t} $,可得

(3.3) $ \begin{equation} \frac{\rm d}{{\rm d}t}\left(e^{2\lambda t}\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x\right) = -3e^{2\lambda t}\int_{{{\Bbb R}} }y^{2}u_{x}{\rm d}x+ 2\gamma e^{2\lambda t}\int_{{{\Bbb R}} }y\rho v_{x}{\rm d}x -e^{2\lambda t}\int_{{{\Bbb R}} }\rho^{2}u_{x}{\rm d}x. \end{equation} $

通过回顾函数$ G $的定义,再利用Young不等式,我们得到

(3.4) $ \begin{eqnarray} &&\|u_{x}\|_{L^{\infty}}\leq \|G_{x}\ast y\|_{L^{\infty}}\leq\|G_{x}\|_{L^{2}} \|y\|_{L^{2}}\leq \frac{1}{2}\|y\|_{L^{2}}, {} \\ &&\int_{{{\Bbb R}} }\rho^{2}u_{x}{\rm d}x\leq \|u_{x}\|_{L^{\infty}} \int_{{{\Bbb R}} }\rho^{2}{\rm d}x\leq \frac{1}{2}\|y\|_{L^{2}} \int_{{{\Bbb R}} }\rho^{2}{\rm d}x \end{eqnarray} $

和

(3.5) $ \begin{eqnarray} \int_{{{\Bbb R}} }y\rho v_{x}{\rm d}x\leq\|y\|_{L^{\infty}} \int_{{{\Bbb R}} }\rho v_{x}{\rm d}x = 0. \end{eqnarray} $

因此,有

(3.6) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\left(e^{2\lambda t}\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x\right) &\leq&\frac{3}{2}e^{2\lambda t}\left(\int_{{{\Bbb R}} } y^{2} {\rm d}x\right)^{\frac{1}{2}}\left(\int_{{{\Bbb R}} } y^{2} {\rm d}x\right)+ \frac{1}{2}e^{2\lambda t}\left(\int_{{{\Bbb R}} } y^{2} {\rm d}x\right)^{\frac{1}{2}}\left(\int_{{{\Bbb R}} } \rho^{2} {\rm d}x\right){} \\ &\leq&\frac{3}{2}e^{2\lambda t}\left(\int_{{{\Bbb R}} }( y^{2}+\rho^{2} ) {\rm d}x\right)^{\frac{3}{2}}{}\\ & = &\frac{3}{2}e^{-\lambda t}\left(e^{2\lambda t}\int_{{{\Bbb R}} }( y^{2}+\rho^{2} ) {\rm d}x\right)^{\frac{3}{2}}. \end{eqnarray} $

由上式可知

(3.7) $ \begin{equation} \frac{\rm d}{{\rm d}t}\left(e^{2\lambda t}\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x\right)^{-\frac{1}{2}} \geq-\frac{3}{4}e^{-\lambda t}. \end{equation} $

对不等式的$ t $从$ 0 $到$ t $积分得到

(3.8) $ \begin{eqnarray} \left(e^{2\lambda t}\int_{{{\Bbb R}} }(y^{2}+\rho^{2}){\rm d}x\right)^{-\frac{1}{2}} -\left(\int_{{{\Bbb R}} }(y_{0}^{2}+\rho_{0}^{2}){\rm d}x\right)^{-\frac{1}{2}} \geq\frac{3}{4\lambda}(e^{-\lambda t}-1)\geq-\frac{3}{4}. \end{eqnarray} $

然后,有

(3.9) $ \begin{equation} e^{\lambda t}\left(\|y\|_{L^{2}}^{2}+\|\rho\|_{L^{2}}^{2}\right)^{\frac{1}{2}} <\left(\left(\|y_{0}\|_{L^{2}}^{2}+\|\rho_{0}\|_{L^{2}}^{2}\right)^{-\frac{1}{2}} -\frac{3}{4\lambda}\right)^{-1}. \end{equation} $

基于

(3.10) $ \begin{equation} \left(\|y_{0}\|_{L^{2}}^{2}+\|\rho_{0}\|_{L^{2}}^{2}\right)^{\frac{1}{2}} -\frac{4\lambda}{3}<0, \end{equation} $

可得

(3.11) $ \begin{eqnarray} \|u_{x}\|_{L^{\infty}}\leq \frac{1}{2}\|y\|_{L^{2}} <\left(\|y\|_{L^{2}}^{2}+\|\rho\|_{L^{2}}^{2}\right)^{\frac{1}{2}} <e^{-\lambda t}\left(\left(\|y_{0}\|_{L^{2}}^{2}+\|\rho_{0}\|_{L^{2}}^{2}\right)^{-\frac{1}{2}} -\frac{3}{4\lambda}\right)^{-1}. \end{eqnarray} $

根据定理2.2,可以证明系统(1.3)的解在时间上是全局存在的.证毕

接下来,我们为系统(2.2)的强解建立奇异点的形式,并在初始数据上给出几个充分条件,以保证$ \alpha\leq 1 $发生爆破.

定理4.1  令$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $并且$ s\geq\frac{5}{2} $.如果$ u_{0} $满足下列不等式

(4.1) $ \begin{equation} \int_{{{\Bbb R}} }u_{0x}^{3}{\rm d}x<-3\lambda E(0)-\frac{E(0)}{2} \sqrt{36\lambda^{2}-\frac{6\gamma_{0}}{\alpha^{2}}E(0)-\frac{48d_{0}}{\alpha^{4}} \sqrt{C_{1}E(0)}}\ , \end{equation} $

其中$ \gamma_{0} = \max\{1, \gamma\} $,则系统(2.2)的相应解在有限时间内爆破.

证  通过对系统(2.2)的第一个方程关于$ x $进行微分,可以得到

(4.2) $ \begin{equation} u_{xt}+\lambda u_{x}+u_{x}^{2}+uu_{xx}+ku_{xx}+\partial_{x}^{2} G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right) = 0. \end{equation} $

应用$ G $的定义可以得到

(4.3) $ \begin{eqnarray} &&u_{xt}+\lambda u_{x}+\frac{1}{2}u_{x}^{2}+uu_{xx}+ku_{xx}+\frac{1}{\alpha^{2}} G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right){} \\ && -\frac{1}{\alpha^{2}} \left(2d_{0}u+u^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right) = 0. \end{eqnarray} $

引入

(4.4) $ \begin{equation} M(t) = \int_{{{\Bbb R}} }u_{x}^{3}{\rm d}x, \; \; t\geq 0. \end{equation} $

通过将(4.3)式与$ u^{2}_{x} $相乘,并使用以下关系

(4.5) $ \begin{eqnarray} &&\int_{{{\Bbb R}} }uu_{x}^{2}u_{xx}{\rm d}x = -\frac{1}{3} \int_{{{\Bbb R}} }u_{x}^{4}{\rm d}x, \; \; G\ast \left(u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}\right)\geq \frac{1}{2}u^{2}, {} \\ &&\|G\ast v_{x}^{2}\|_{L^{\infty}}\leq \|G\|_{L^{\infty}}\|v_{x}^{2}\|_{L^{1}} \leq\frac{1}{2\alpha}\|v_{x}^{2}\|_{L^{1}}, \end{eqnarray} $

我们推导出关于$ M(t) $的如下方程

(4.6) $ \begin{eqnarray} &&\frac{{\rm d}M(t)}{{\rm d}t}+3\lambda M(t){}\\ & = &-\frac{3}{2}\int_{{{\Bbb R}} }u_{x}^{4}{\rm d}x +\int_{{{\Bbb R}} }u_{x}^{4}{\rm d}x-\frac{3}{\alpha^{2}}\int_{{{\Bbb R}} }u_{x}^{2} G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2} v^{2}_{x}\right){\rm d}x{} \\ &&+\frac{3}{\alpha^{2}} \int_{{{\Bbb R}} }u_{x}^{2}\left(2d_{0}u+u^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2} v^{2}_{x}\right){\rm d}x{} \\ &\leq&-\frac{1}{2}\int_{{{\Bbb R}} }u_{x}^{4}{\rm d}x +\frac{3}{2\alpha^{2}}\int_{{{\Bbb R}} }u^{2}u_{x}^{2}{\rm d}x+ \frac{3\gamma}{4\alpha^{2}}\int_{{{\Bbb R}} }u_{x}^{2}v^{2}{\rm d}x +\frac{6d_{0}}{\alpha^{4}}\sqrt{C_{1}E^{3}(0)} +\frac{9\gamma}{4\alpha^{2}} \int_{{{\Bbb R}} }u_{x}^{2}G\ast v^{2}_{x}{\rm d}x.{}\\ \end{eqnarray} $

通过使用柯西-施瓦茨不等式,我们得到了

(4.7) $ \begin{equation} \left|\int_{{{\Bbb R}} }u_{x}^{3}{\rm d}x\right|\leq\frac{1}{\alpha} \left(\int_{{{\Bbb R}} }u_{x}^{4}{\rm d}x\right)^{\frac{1}{2}} \left(\alpha^{2}\int_{{{\Bbb R}} }u_{x}^{2}{\rm d}x\right)^{\frac{1}{2}}, \end{equation} $

因此

(4.8) $ \begin{equation} \int_{{{\Bbb R}} }u_{x}^{4}{\rm d}x\geq\frac{\alpha^{2}}{E(0)}\left(\int_{{{\Bbb R}} } u_{x}^{3}\right)^{2}. \end{equation} $

通过使用(2.14)和(2.15)式,进一步得到

(4.9) $ \begin{equation} \int_{{{\Bbb R}} }u^{2}u_{x}^{2}{\rm d}x\leq \|u\|_{L^{\infty}}^{2} \int_{{{\Bbb R}} }u_{x}^{2}{\rm d}x\leq \frac{1}{2} \|u\|_{H^{1}}^{2} \int_{{{\Bbb R}} }u_{x}^{2}{\rm d}x\leq \frac{1}{2} \|u\|_{H^{1}}^{4}, \end{equation} $

(4.10) $ \begin{equation} \int_{{{\Bbb R}} }u_{x}^{2}v^{2}{\rm d}x\leq \|v\|_{L^{\infty}}^{2} \int_{{{\Bbb R}} }u_{x}^{2}{\rm d}x\leq \frac{1}{2} \|v\|_{H^{1}}^{2} \int_{{{\Bbb R}} }u_{x}^{2}{\rm d}x\leq \frac{1}{2} \|u\|_{H^{1}}^{2}\|v\|_{H^{1}}^{2}. \end{equation} $

要注意的是

(4.11) $ \begin{equation} \|G\ast v_{x}^{2}\|_{L^{\infty}}\leq \|G\|_{L^{\infty}} \|v_{x}^{2}\|_{L^{1}}\leq \frac{1}{2} \|v\|_{H^{1}}^{2}. \end{equation} $

因此有

(4.12) $ \begin{eqnarray} \int_{{{\Bbb R}} }u_{x}^{2}G\ast v_{x}^{2}{\rm d}x\leq \frac{1}{2} \|u\|_{H^{1}}^{2}\|v\|_{H^{1}}^{2}. \end{eqnarray} $

因此,我们从上面的分析中得出

(4.13) $ \begin{eqnarray} \frac{{\rm d}M(t)}{{\rm d}t}+3\lambda M(t)&\leq& -\frac{1}{2E(0)} M^{2}(t)+\frac{3}{4\alpha^{2}}\|u\|_{H^{1}}^{4} +\frac{3\gamma}{\alpha^{2}}\|u\|_{H^{1}}^{2}\|v\|_{H^{1}}^{2}+ \frac{6d_{0}}{\alpha^{4}}\sqrt{C_{1}E^{3}(0)}{} \\ &\leq& -\frac{1}{2E(0)} M^{2}(t)+\frac{3\gamma_{0}}{4\alpha^{2}}E^{2}(0)+ \frac{6d_{0}}{\alpha^{4}}\sqrt{C_{1}E^{3}(0)}, \end{eqnarray} $

其中$ \gamma_{0} = \max\{1, \gamma\} $.我们可以进一步证明这一点

(4.14) $ \begin{eqnarray} \frac{{\rm d}M(t)}{{\rm d}t}&\leq& -\frac{1}{8E(0)}\left( 2M(t)+6\lambda E(0)-E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}\right){} \\ &&\times \left( 2M(t)+6\lambda E(0)+E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}\right). \end{eqnarray} $

根据假设,我们有$ \frac{{\rm d}M}{{\rm d}t}|_{t = 0}<0 $.通过考虑$ M(t) $的$ t $的连续性,可以进一步得到,对所有$ t\in[0, T) $有$ \frac{{\rm d}M}{{\rm d}t}<0 $.因此

(4.15) $ \begin{equation} \frac{{\rm d}M(t)}{{\rm d}t}<-3\lambda E(0)-\frac{1}{2}E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}\ , \; \mbox{所有} \; t\in[0, T). \end{equation} $

求解不等式(4.15),有

(4.16) $ \begin{eqnarray} &&\frac{2M(t)+6\lambda E(0)+E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}{2M(t)+6\lambda E(0)-E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}} {} \\ &&\times\exp\left(\frac{\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}{2}t\right)-1{} \\ &\leq& \frac{2E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}{2M(t)+6\lambda E(0)-E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}} \leq 0. \end{eqnarray} $

因为

(4.17) $ \begin{equation} 0<\frac{2M(t)+6\lambda E(0)+E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}{2M(t)+6\lambda E(0)-E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}<1, \end{equation} $

则存在一个$ T $

(4.18) $ \begin{eqnarray} &&\frac{2}{\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}{} \\ &&\times\ln\left( \frac{2M(t)+6\lambda E(0)+E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}{2M(t)+6\lambda E(0)-E(0)\sqrt{36\lambda^{2}+\frac{6\gamma_{0}}{\alpha^{2}}E(0)+\frac{48d_{0} }{\alpha^{4}}\sqrt{C_{1}E(0)}}}\right)\geq T, \end{eqnarray} $

满足

(4.19) $ \begin{equation} \lim\limits_{t\rightarrow T}M(t) = \lim\limits_{t\rightarrow T}\int_{{{\Bbb R}} } u_{x}^{3}{\rm d}x = -\infty. \end{equation} $

进一步考虑这一事实

(4.20) $ \begin{equation} \int_{{{\Bbb R}} }u_{x}^{3}{\rm d}x\geq \inf u_{x}(t, x) \int_{{{\Bbb R}} }u_{x}^{2}{\rm d}x\geq \inf u_{x}(t, x) E(0), \end{equation} $

我们可以证明

(4.21) $ \begin{equation} \lim\limits_{t\rightarrow T}\inf\limits_{x\in {{\Bbb R}} }u_{x}(t, x) = -\infty. \end{equation} $

证毕.

定理4.2  令$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $,且$ s\geq\frac{5}{2} $.如果存在$ x_{0}\in {{\Bbb R}} $满足$ y_{0}(x_{0}) = u_{0}(x_{0})-\alpha^{2}u_{0xx}(x_{0}) = 0 $, $ \rho_{0}(x_{0}) = \rho_{0}(x_{0})-\rho_{0xx}(x_{0}) = 0 $和下面的不等式

(4.22) $ \begin{eqnarray} &&\int_{-\infty}^{x_{0}}e^{\frac{\xi}{\alpha}}\left(y_{0}(\xi)+d_{0}\right){\rm d}\xi> \sqrt{\frac{\gamma(1+2\alpha^{2}C_{1})}{2}E(0)}\ , {} \\ && \int^{-\infty}_{x_{0}}e^{-\frac{\xi}{\alpha}}\left(y_{0}(\xi)+d_{0}\right){\rm d}\xi< -\sqrt{\frac{\gamma(1+2\alpha^{2}C_{1})}{2}E(0)}\ . \end{eqnarray} $

则对于$ \lambda<0 $和初始数据$ (u_{0}, v_{0}) $的系统(2.2)的解在有限时间爆破.

证  通过使用(2.3)式, (4.3)式和下列不等式

(4.23) $ \begin{equation} G\ast \left((u+d_{0})^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}\right)\geq \frac{1}{2} (u+d_{0})^{2}, \end{equation} $

可以得到

(4.24) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}u_{x}(t, q_{1}(t, x_{0}))+\lambda u_{x}(t, q_{1}(t, x_{0})){} \\ & = &(u_{xt}+uu_{xx}+ku_{xx})(t, q_{1}(t, x_{0}))+\lambda u_{x}(t, q_{1}(t, x_{0})){} \\ &\leq&-\frac{1}{2}u_{x}^{2}(t, q_{1}(t, x_{0}))+\frac{1}{2\alpha^{2}}(u+d_{0})^{2} (t, q_{1}(t, x_{0}))+\frac{\gamma}{2\alpha^{2}}v^{2}(t, q_{1}(t, x_{0})) -\frac{\gamma}{2\alpha^{2}}v_{x}^{2}(t, q_{1}(t, x_{0})) {} \\ &&+\frac{\gamma}{2\alpha^{2}}G\ast v_{x}^{2}(t, q_{1}(t, x_{0})) -\frac{\gamma}{2\alpha^{2}}G\ast v^{2}(t, q_{1}(t, x_{0})) +\frac{\gamma(1+2\alpha^{2} C_{1})}{4\alpha^{2}}. \end{eqnarray} $

在接下来的文章中,我们提出了三个论点来证明我们的结论.

论点1  对所有$ t $, $ y(t, q_{1}(t, x_{0}))+d_{0} = 0 $在它的预期使用期限内.

通过考虑(1.7)式中的第一个方程,并利用粒子轨迹法,我们得到了

(4.25) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\left(\left(y(t, q_{1}(t, x))+d_{0}\right)q_{1x}^{2}(t, x)\right) & = &\left(y_{t}+uy_{x}+2yu_{x}+ky_{x}+2d_{0}u_{x}\right)(t, q_{1}(t, x))q_{1x}^{2}(t, x){} \\ & = &(\gamma \rho v_{x}-\lambda y)(t, q_{1}(t, x))q_{1x}^{2}(t, x). \end{eqnarray} $

由于(2.4)式引入的$ q_{2}(\cdot, x) $是任意$ t\in[0, T) $的直线的微分同构,则有

(4.26) $ \begin{equation} q_{2}(t, \tilde{x}_{0}(t)) = q_{1}(t, x_{0}), \; \; t\in[0, T), \end{equation} $

其中$ \tilde{x}_{0}(t)\in {{\Bbb R}} $.对于$ t = 0 $,可以进一步得到

(4.27) $ \begin{equation} \tilde{x}_{0}(0) = q_{2}(0, \tilde{x}_{0}(0)) = q_{1}(0, x_{0}) = x_{0}. \end{equation} $

接下来,我们表明$ \rho(t, q_{1}(t, x_{0})) = 0 $.通过直接计算,则有

(4.28) $ \begin{equation} \frac{\rm d}{{\rm d}t}\rho(t, q_{2}(t, \tilde{x}_{0}(t))) = -(\rho u_{x})(t, q_{2}(t, \tilde{x}_{0}(t))). \end{equation} $

根据$ \rho_{0}(x_{0}) = 0 $,方程(4.28)积分得到

(4.29) $ \begin{eqnarray} \rho(t, q_{2}(t, \tilde{x}_{0}(t)))& = &\rho(0, q_{2}(0, \tilde{x}_{0}(0))) e^{-\int_{0}^{t}u_{x}(\tau, q_{2}(\tau, q_{2}(\tilde{x}_{0}(\tau)))){\rm d}\tau}{} \\ & = &\rho_{0}(x_{0})e^{-\int_{0}^{t}u_{x}(\tau, q_{2}(\tau, q_{2}(\tilde{x}_{0} (\tau)))){\rm d}\tau} = 0, \end{eqnarray} $

然后有

(4.30) $ \begin{equation} \rho(t, q_{1}(t, x_{0})) = \rho(t, q_{2}(t, \tilde{x}_{0}(t))) = 0. \end{equation} $

因此,可以获得

(4.31) $ \begin{equation} \frac{\rm d}{{\rm d}t}\left((y(t, q_{1}(t, x_{0}))+d_{0})q_{1x}^{2}(t, x_{0})\right) = -\rho(t, q_{1}(t, x_{0}))v_{x}(t, q_{1}(t, x_{0}))q_{1x}^{2}(t, x_{0}) = 0, \end{equation} $

这意味着

(4.32) $ \begin{equation} y(t, q_{1}(t, x_{0}))+d_{0} = y_{0}(x_{0})+d_{0} = 0. \end{equation} $

这就完成了论点1.

论点2  对所有$ x\in {{\Bbb R}} $和$ t>0 $, $ v_{x}^{2}(t, x)-v^{2}(t, x)\leq v_{x}^{2}(t, x_{0})-v^{2}(t, x_{0}) $.

由于引理2.2,则有

(4.33) $ \begin{equation} \left\{ \begin{array}{ll} \rho(t, q_{2}(t, x))\geq 0, \; \; \mbox{对}\; \; x\in(-\infty, x_{0}), \\ \rho(t, q_{2}(t, x))\leq 0, \; \; \mbox{对}\; \; x\in(x_{0}, \infty), \end{array} \right. \end{equation} $

对于所有$ t\geq0 $,由于$ v = G\ast \rho(t, x) $, $ x\in{{\Bbb R}} $, $ t\geq 0 $, $ v(t, x) $和$ v_{x}(t, x) $可以改写成

(4.34) $ \begin{equation} v(t, x) = \frac{1}{2\alpha}e^{-\frac{x}{\alpha}}\int_{-\infty}^{x} e^{\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi+\frac{1}{2\alpha}e^{\frac{x}{\alpha}} \int^{\infty}_{x} e^{-\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi, \end{equation} $

(4.35) $ \begin{equation} v_{x}(t, x) = -\frac{1}{2\alpha}e^{-\frac{x}{\alpha}}\int_{-\infty}^{x} e^{\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi+\frac{1}{2\alpha}e^{\frac{x}{\alpha}} \int^{\infty}_{x} e^{-\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi, \end{equation} $

对任何固定$ t $.对于$ x\leq q_{2}(t, x_{0}) $,有

(4.36) $ \begin{eqnarray} v_{x}^{2}(t, x)-v^{2}(t, x)& = &-\left(\int_{-\infty}^{q_{2}(t, x_{0})} e^{\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi- \int_{x}^{q_{2}(t, x_{0})} e^{\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi\right){} \\ &&\times \left(\int^{\infty}_{q_{2}(t, x_{0})} e^{-\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi+\int_{x}^{q_{2}(t, x_{0})} e^{-\frac{\xi}{\alpha}}\rho(t, \xi){\rm d}\xi\right){} \\ &\leq&v_{x}^{2}(t, q_{2}(t, x_{0}))-v^{2}(t, q_{2}(t, x_{0})). \end{eqnarray} $

循着同样的方法引出

(4.37) $ \begin{eqnarray} v_{x}^{2}(t, x)-v^{2}(t, x) \leq&v_{x}^{2}(t, q_{2}(t, x_{0}))-v^{2}(t, q_{2}(t, x_{0})), \end{eqnarray} $

对于$ x\geq q_{2}(t, x_{0}) $.

从(4.36)和(4.37)式,可以进一步得到

(4.38) $ \begin{eqnarray} v_{x}^{2}(t, x)-v^{2}(t, x) \leq v_{x}^{2}(t, q_{2}(t, x_{0}))-v^{2}(t, q_{2}(t, x_{0})), \end{eqnarray} $

对任何固定$ t $和$ x\in{{\Bbb R}} $.结合论点1和论点2,则有

(4.39) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}u_{x}(t, q_{1}(t, x_{0}))+\lambda u_{x}(t, q_{1}(t, x_{0})){} \\ &\leq&-\frac{1}{2}u_{x}^{2}(t, q_{1}(t, x_{0}))+\frac{1}{2\alpha^{2}}(u+d_{0})^{2} (t, q_{1}(t, x_{0}))+\frac{\gamma(1+2\alpha^{2} C_{1})}{4\alpha^{2}}. \end{eqnarray} $

论点3  在$ [0, T) $上, $ (u+d_{0})^{2}(t, q_{1}(t, x_{0})) -\alpha^{2}u_{x}^{2}(t, q_{1}(t, x_{0}))+\gamma(C_{1}+\frac{1}{2\alpha})E(0)<0 $,且$ u_{x}(t, q_{1}(t, x_{0}))<0 $是关于$ t $严格递减的.

假设存在这样一个$ t_{0} $

(4.40) $ \begin{equation} (u+d_{0})^{2}(t, q_{1}(t, x_{0})) -\alpha^{2}u_{x}(t, q_{1}(t, x_{0})) +\gamma(C_{1}+\frac{1}{2\alpha})E(0)<0, \; \; \mbox{在}\; \; [0, t_{0}), \end{equation} $

但是

(4.41) $ \begin{equation} (u+d_{0})^{2}(t_{0}, q_{1}(t_{0}, x_{0})) -\alpha^{2}u_{x}(t_{0}, q_{1}(t_{0}, x_{0})) +\gamma(C_{1}+\frac{1}{2\alpha})E(0)\geq0. \end{equation} $

由$ u(t, x) $和$ y(t, x) $的关系可知, $ u(t, x)+d_{0} $和$ u_{x}(t, x) $可以改写成

(4.42) $ \begin{equation} u(t, x)+d_{0} = \frac{1}{2\alpha}e^{-\frac{x}{\alpha}}\int_{-\infty}^{x} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi +\frac{1}{2\alpha}e^{\frac{x}{\alpha}} \int^{\infty}_{x} e^{-\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi, \end{equation} $

(4.43) $ \begin{equation} u_{x}(t, x) = -\frac{1}{2\alpha^{2}}e^{-\frac{x}{\alpha}}\int_{-\infty}^{x} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi+\frac{1}{2\alpha^{2}} e^{\frac{x}{\alpha}} \int^{\infty}_{x} e^{-\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi, \end{equation} $

引入

(4.44) $ \begin{equation} {\cal F}(t) = e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi, \end{equation} $

(4.45) $ \begin{equation} {\cal G}(t) = e^{\frac{q_{1}(t, x_{0})}{\alpha}}\int^{\infty}_{q_{1}(t, x_{0})} e^{-\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi. \end{equation} $

然后

(4.46) $ \begin{eqnarray} \frac{{\rm d}{\cal F}(t)}{{\rm d}t}& = &-\frac{1}{\alpha}(u(t, q_{1}(t, x_{0}))+d_{0}) e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi{} \\ &&+e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}y_{t}(t, \xi){\rm d}\xi. \end{eqnarray} $

通过分部积分,可以重写(4.46)式的第一项

(4.47) $ \begin{eqnarray} &&-\frac{1}{\alpha}(u(t, q_{1}(t, x_{0}))+k)e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi{} \\ & = &(\alpha uu_{x}-u^{2}-d_{0}u)(t, q_{1}(t, x_{0})) -\frac{1}{\alpha}e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi. \end{eqnarray} $

通过使用论点1,使用下面的等式

(4.48) $ \begin{eqnarray} &&-\alpha e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}(uu_{xx})(t, \xi){\rm d}\xi{} \\ & = &-\alpha(uu_{x})(t, q_{1}(t, x_{0})) +\frac{1}{2}u^{2}(t, q_{1}(t, x_{0})) +\alpha e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(u_{x}^{2}-\frac{1}{2\alpha^{2}}u^{2}\right)(t, \xi){\rm d}\xi.{}\\ \end{eqnarray} $

我们可以重写(4.46)式的第二项

(4.49) $ \begin{eqnarray} &&e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}y_{t}(t, \xi){\rm d}\xi{} \\ & = &-e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\Big(((y+d_{0})u)_{x}+\frac{1}{2}(u^{2}-\alpha^{2}u_{x}^{2})_{x}{} \\ &&+k(y+d_{0})_{x}+d_{0}u_{x}-\gamma\rho v_{x}+\lambda y\Big){\rm d}\xi{} \\ & = &\Big(\frac{\alpha^{2}}{2}u_{x}^{2}-\alpha uu_{x}-d_{0}u+\frac{\gamma}{2} (v^{2}-v_{x}^{2})\Big)(t, q_{1}(t, x_{0})){} \\ &&-\frac{\gamma}{2\alpha}e^{-\frac{q_{1}(t, x_{0})}{\alpha}} \int_{-\infty}^{q_{1}(t, x_{0})}e^{\frac{\xi}{\alpha}}(v^{2}-v_{x}^{2}){\rm d}\xi -\lambda e^{-\frac{q_{1}(t, x_{0})}{\alpha}}\int_{-\infty}^{q_{1}(t, x_{0})}e^{\frac{\xi}{\alpha}} y{\rm d}\xi{} \\ &&+\frac{1}{\alpha}e^{-\frac{q_{1}(t, x_{0})}{\alpha}} \int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2} +k(y+d_{0})+2d_{0}u\right){\rm d}\xi. \end{eqnarray} $

从以上分析,并根据事实

(4.50) $ \begin{equation} \int_{-\infty}^{x}e^{\frac{\xi}{\alpha}}\left( (u+d_{0})^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}\right) (t, \xi){\rm d}\xi\geq\frac{\alpha}{2}e^{\frac{x}{\alpha}}(u+d_{0})^{2}, \end{equation} $

则在$ [0, t_{0}) $上有

(4.51) $ \begin{eqnarray} \frac{{\rm d}{\cal F}(t)}{{\rm d}t}+\lambda {\cal F}(t) & = &\left(\frac{\alpha^{2}}{2}u_{x}^{2}-u^{2}-2d_{0}u\right)(t, q_{1}(t, x_{0})) -\frac{\gamma}{2\alpha}e^{-\frac{q_{1}(t, x_{0})}{\alpha}} \int_{-\infty}^{q_{1}(t, x_{0})}e^{\frac{\xi}{\alpha}}(v^{2}-v_{x}^{2}){\rm d}\xi{} \\ &&+\frac{1}{\alpha}e^{-\frac{q_{1}(t, x_{0})}{\alpha}} \int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2} +2d_{0}u\right){\rm d}\xi +\frac{\gamma}{2} (v^{2}-v_{x}^{2})(t, q_{1}(t, x_{0})){} \\ &\geq&\frac{1}{2}\left(\alpha^{2}u_{x}^{2}-(u+d_{0})^{2}\right)(t, q_{1}(t, x_{0})) -\frac{\gamma}{2}C_{1}E(0) -\frac{\gamma}{4\alpha^{2}}E(0)>0, . \end{eqnarray} $

根据同样的分析,则在$ [0, t_{0}) $上有

(4.52) $ \begin{eqnarray} \frac{{\rm d}{\cal G}(t)}{{\rm d}t}+\lambda{\cal G}(t) \leq\frac{1}{2}\left((u+d_{0})^{2}-\alpha^{2}u_{x}^{2}\right)(t, q_{1}(t, x_{0})) +\frac{\gamma}{2}C_{1}E(0) +\frac{\gamma}{4\alpha^{2}}E(0)<0. \end{eqnarray} $

值得注意的是,这个假设确保了$ {\cal F}(0)>0 $和$ {\cal G}(0)<0 $,由此可见(4.51)和(4.52)式在$ (0, t_{0}) $上分别为

(4.53) $ \begin{equation} {\cal F}(t)\geq e^{-\lambda t} {\cal F}(0)\; \; \mbox{和}\; \; {\cal G}(t)\leq e^{-\lambda t} {\cal G}(0). \end{equation} $

因此,则有

(4.54) $ \begin{eqnarray} &&(u+d_{0})^{2}(t, q_{1}(t, x_{0})) -\alpha^{2}u_{x}^{2}(t, q_{1}(t, x_{0})){} \\ & = &\frac{1}{\alpha^{2}}\int_{-\infty}^{q_{1}(t_{0}, x_{0})}e^{\frac{\xi}{\alpha}} (y(t_{0}, \xi)+d_{0}){\rm d}\xi\int^{\infty}_{q_{1}(t_{0}, x_{0})}e^{-\frac{\xi}{\alpha}} (y(t_{0}, \xi)+d_{0}){\rm d}\xi{} \\ &< &\left((u+d_{0})^{2}(x_{0}) -\alpha^{2}u_{x}(x_{0})\right)<-\gamma \left(C_{1}+\frac{1}{2\alpha^{2}}\right)E(0)<0, \end{eqnarray} $

对于$ \lambda<0 $.这是一个明显的矛盾.然后, $ u_{x}(t, q_{1}(t, x_{0})) $是严格递减.另一方面,有

(4.55) $ \begin{eqnarray} u_{x}(t, q_{1}(t, x_{0}))& = &-\frac{1}{2\alpha^{2}}e^{-\frac{q_{1}(t, x_{0})}{\alpha}} \int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi{} \\ &&+\frac{1}{2\alpha^{2}} e^{\frac{q_{1}(t, x_{0})}{\alpha}} \int^{\infty}_{q_{1}(t, x_{0})} e^{-\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi{} \\ &\leq&-\frac{1}{2\alpha^{2}}e^{-\frac{x_{0}}{\alpha}} \int_{-\infty}^{x_{0}} e^{\frac{\xi}{\alpha}}\left(y_{0}(\xi)+d_{0}\right){\rm d}\xi{} \\ &&+\frac{1}{2\alpha^{2}} e^{\frac{x_{0}}{\alpha}} \int^{\infty}_{x_{0}} e^{-\frac{\xi}{\alpha}}\left(y_{0}(\xi)+d_{0}\right){\rm d}\xi. \end{eqnarray} $

通过考虑初始假设,则有$ u_{x}(t, q_{1}(t, x_{0}))<0 $.这样,就完成了论点3的证明.

其次,通过引入

(4.56) $ \begin{eqnarray} h(t) = u_{x}(t, q_{1}(t, x_{0})), \; \; t>0, \end{eqnarray} $

则有

(4.57) $ \begin{eqnarray} \frac{{\rm d} h(t)}{{\rm d}t}+\lambda h(t)&\leq& \frac{1}{2\alpha^{2}}(u+d_{0})^{2}(t, q_{1}(t, x_{0})) -\frac{1}{2}u_{x}^{2}(t, q_{1}(t, x_{0}))+\frac{\gamma}{2\alpha^{2}} C_{1}E(0)+\frac{\gamma}{4\alpha^{4}}E(0){} \\ &\leq&\frac{1}{2\alpha^{2}}(u+d_{0})^{2}(x_{0}) -\frac{1}{2}u_{x}^{2}(x_{0})+\frac{\gamma}{2\alpha^{2}} C_{1}E(0)+\frac{\gamma}{4\alpha^{4}}E(0)<0. \end{eqnarray} $

假设相应的解在时间上是全局存在的.由于$ h(t) $在初始假设$ h(0)<0 $时是严格递减的,所以存在这样一个$ t_{1} $,对所有$ t>t_{1} $,有

(4.58) $ \begin{equation} h(t)<-\frac{1}{\alpha^{2}}\sqrt{2\alpha^{2}\left( \sqrt{C_{1}E(0)}+|d_{0}|\right)^{2}+\gamma(1+2\alpha^{2}C_{1})E(0)}: = -C<0. \end{equation} $

然后对$ t\in (t_{1}, \infty) $,有

(4.59) $ \begin{eqnarray} \frac{{\rm d} h(t)}{{\rm d}t}+\lambda h(t)&\leq& \frac{1}{2\alpha^{2}}(u+d_{0})^{2}(t, q_{1}(t, x_{0})) -\frac{1}{2}u_{x}^{2}(t, q_{1}(t, x_{0}))+\frac{\gamma}{2\alpha^{2}} C_{1}E(0)+\frac{\gamma}{4\alpha^{4}}E(0){} \\ &\leq&-\frac{1}{2}h^{2}(t)+\frac{2\alpha^{2}(\sqrt{C_{1}E(0)}+|d_{0}|)^{2}+ \gamma(1+2\alpha^{2}C_{1})E(0)}{4\alpha^{4}}{} \\ &\leq&-\frac{1}{4}h^{2}(t). \end{eqnarray} $

求解不等式(4.59)得到

(4.60) $ \begin{equation} h(t)\leq \frac{4\lambda h(t_{1})}{(h(t_{1})+4\lambda)e^{\lambda(t-t_{1})}-h(t_{1})}. \end{equation} $

我们可以证明当$ t\rightarrow t_{1}+\frac{1}{\lambda}\ln\left(\frac{h(t_{1})}{h(t_{1})+4\lambda}\right) $时$ h(t)\rightarrow\infty $.表明对应的解并不全局存在,即出现爆破波.

对于$ t_{1} $,我们回忆(4.57)式,有

(4.61) $ \begin{eqnarray} \frac{{\rm d} h(t)}{{\rm d}t}+\lambda h(t) \leq\frac{1}{2\alpha^{2}}(u+d_{0})^{2}(x_{0}) -\frac{1}{2}u_{x}^{2}(x_{0})+\frac{\gamma}{2\alpha^{2}} C_{1}E(0)+\frac{\gamma}{4\alpha^{4}}E(0). \end{eqnarray} $

如果$ h(0)<-C $,可以选择$ t_{1} = 0 $.另外,如果$ h(0)>-C $和$ h(t_{1}) = -e^{-\lambda t_{1}}C $,将(4.57)式从$ 0 $到$ t_{1} $积分得到

(4.62) $ \begin{eqnarray} -C-h(0)\leq \left(\frac{1}{2\alpha^{2}}(u_{0}(x_{0})+d_{0})^{2} -\frac{1}{2}u_{0x}^{2}(x_{0})+\frac{\gamma(1+2\alpha^{2}C_{1})}{4\alpha^{4}}E(0)\right)t_{1}. \end{eqnarray} $

因此,我们有

(4.63) $ \begin{equation} t_{1}\leq -\frac{C+h(0)}{\left(\frac{1}{2\alpha^{2}}(u_{0}(x_{0})+d_{0})^{2} -\frac{1}{2}u_{0x}^{2}(x_{0})+\frac{\gamma(1+2\alpha^{2}C_{1})}{4\alpha^{4}}E(0)\right)}. \end{equation} $

则时间$ t_{1} $可以被选择为

(4.64) $ \begin{equation} t_{1} = -\frac{C+h(0)}{\left(\frac{1}{2\alpha^{2}}(u_{0}(x_{0})+d_{0})^{2} -\frac{1}{2}u_{0x}^{2}(x_{0})+\frac{\gamma(1+2\alpha^{2}C_{1})}{4\alpha^{4}}E(0)\right)}. \end{equation} $

证毕.

基于定理4.2的证明,我们给出下面的结论.

定理4.3  设$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $,且$ s\geq\frac{5}{2} $.如果存在$ x_{0}\in {{\Bbb R}} $有$ y_{0}(x_{0}) = u_{0}(x_{0})-\alpha^{2}u_{0xx}(x_{0}) = 0 $, $ \rho_{0}(x_{0}) = \rho_{0}(x_{0})-\rho_{0xx}(x_{0}) = 0 $和不等式(4.22).设$ \lambda $满足不等式

(4.65) $ \begin{equation} \lambda<\frac{1}{8\alpha^{2}}e^{-\frac{q_{1}(t, x_{0})}{\alpha}} \int_{-\infty}^{q_{1}(t, x_{0})} e^{\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi-\frac{1}{8\alpha^{2}} e^{\frac{q_{1}(t, x_{0})}{\alpha}} \int^{\infty}_{q_{1}(t, x_{0})} e^{-\frac{\xi}{\alpha}}\left(y(t, \xi)+d_{0}\right){\rm d}\xi. \end{equation} $

那么具有初始数据$ (u_{0}, v_{0}) $的系统(2.2)的解在有限时间内爆破.

证  这个证明的前半部分类似于定理4.2.基于(4.59)式,则有

(4.66) $ \begin{eqnarray} \frac{{\rm d} h(t)}{{\rm d}t}+\lambda h(t) \leq -\frac{1}{4}h^{2}(t)\; \; \mbox{对}\; \; t\in (t_{1}, \infty). \end{eqnarray} $

当$ t\geq 0 $时$ h(t)<0 $,从(4.66)式中可以看出

(4.67) $ \begin{equation} \frac{\rm d}{{\rm d}t}\left(\frac{1}{h(t)}\right)-\frac{\lambda}{h(t)}\geq\frac{1}{4}, \; \; t\geq 0. \end{equation} $

从(4.67)式,则有

(4.68) $ \begin{equation} \left(\frac{1}{h(0)}+\frac{1}{4\lambda}\right)e^{\lambda t}-\frac{1}{4\lambda} \leq \frac{1}{h(t)}<0, \; \; t\geq 0. \end{equation} $

通过考虑定理的假设,我们可以推导出

(4.69) $ \begin{equation} \frac{1}{h(0)}+\frac{1}{4\lambda}>0, \end{equation} $

这表明

(4.70) $ \begin{equation} \left(\frac{1}{h(0)}+\frac{1}{4\lambda}\right)e^{\lambda t}\rightarrow\infty, \; \; \mbox{当}\; \; t\rightarrow\infty. \end{equation} $

结合(4.68)式,可以发现与定理4.2证明中所提供的假设相矛盾的地方.因此,有初始数据$ (u_{0}, v_{0}) $的系统(2.2)的解在有限时间内爆破.证毕

定理4.4  令$ (u_{0}, v_{0})\in H^{s}\times H^{s-1} $且$ s\geq\frac{5}{2} $为初始数据.如果存在一个$ x_{0}\in {{\Bbb R}} $满足

(4.71) $ \begin{equation} u'_{0}(x_{0})<-\lambda- \sqrt{\lambda^{2}+\frac{(\sqrt{C_{1}E(0)}+|d_{0}|)^{2}}{\alpha^{2}}+ \frac{\gamma(1+2\alpha^{2}C_{1})E(0)}{2\alpha^{4}}}. \end{equation} $

则系统(2.2)的相应解在有限时间内爆破.

证  根据定理4.2,则在$ [0, T) $上有

(4.72) $ \begin{eqnarray} \frac{{\rm d} h(t)}{{\rm d}t} &\leq&-\frac{1}{2}h^{2}(t)-\lambda h(t)+\frac{(\sqrt{C_{1}E(0)}+|d_{0}|)^{2}}{2\alpha^{2}}+ \frac{\gamma(1+2\alpha^{2}C_{1})E(0)}{4\alpha^{4}}{} \\ &\leq&-\frac{1}{2}\left(h(t)+\lambda+ \sqrt{\lambda^{2}+\frac{(\sqrt{C_{1}E(0)}+|d_{0}|)^{2}}{\alpha^{2}}+ \frac{\gamma(1+2\alpha^{2}C_{1})E(0)}{2\alpha^{4}}}\right){} \\ &&\times\left(h(t)+\lambda- \sqrt{\lambda^{2}+\frac{(\sqrt{C_{1}E(0)}+|d_{0}|)^{2}}{\alpha^{2}}+ \frac{\gamma(1+2\alpha^{2}C_{1})E(0)}{2\alpha^{4}}}\right), \end{eqnarray} $

其中$ h(t) = u_{x}(t, q_{1}(t, x_{0})) $.根据该定理的假设,可以得到$ \frac{{\rm d}h(t)}{{\rm d}t}|_{t = 0}< 0 $.然后,根据与定理4.1的证明相似的论证,可以很容易地得到这个定理的证明.证毕.

定理4.5  令$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $且$ s\geq\frac{5}{2} $.如果存在一个$ x_{0}\in {{\Bbb R}} $有

(4.73) $ \begin{eqnarray} u'_{0}(x_{0})<-\lambda+\sqrt{\lambda^{2}+\frac{1}{\alpha^{2}} (\sqrt{C_{1}E(0)}+|d_{0}|)^{2}+\frac{\gamma}{2\alpha}E(0)+\gamma C_{1}E(0)}. \end{eqnarray} $

那么,初始数据$ (u_{0}, v_{0}) $的系统(2.2)的解只在有限时间内存在,即发生爆破波.

证  假设$ (u, v) $是有初始数据$ (u_{0}, v_{0}) $的系统(2.2)的解.令$ T $为解的最大存在时间.引入

(4.74) $ \begin{equation} H(t) = \inf\limits_{x\in{{\Bbb R}} }\{u_{x}(t, x)\} = u_{x}(t, \xi(t)), \end{equation} $

这意味着$ H $几乎在$ [0, T) $上任何地方都是可微的,且有

(4.75) $ \begin{equation} \frac{{\rm d}H(t)}{{\rm d}t} = u_{tx}(t, \xi(t)). \end{equation} $

根据$ u_{xx}(t, \xi(t)) = 0 $和(4.3)式,则有

(4.76) $ \begin{eqnarray} \frac{{\rm d}H(t)}{{\rm d}t}& = &-H(t)^{2}-\lambda H(t)-\frac{1}{\alpha^{2}} G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right){} \\ & &+\frac{1}{\alpha^{2}} \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right){} \\ &\leq&-\frac{1}{2}H(t)^{2}-\lambda H(t)+\frac{1}{2\alpha^{2}} \left((\sqrt{C_{1}E(0)}+|d_{0}|)^{2}+\frac{\gamma}{2\alpha}E(0)+ \gamma C_{1}E(0)\right){} \\ & : = &-\frac{1}{2}H(t)^{2}-\lambda H(t)+H_{0} \end{eqnarray} $

通过使用(4.23),其中$ H_{0}>0 $可以被给出

(4.77) $ \begin{equation} H_{0} = \frac{1}{2\alpha^{2}} \left((\sqrt{C_{1}E(0)}+|d_{0}|)^{2}+\frac{\gamma}{2\alpha}E(0)+ \gamma C_{1}E(0)\right)>0. \end{equation} $

如果$ H(0)<-\lambda+\sqrt{\lambda^{2}+2H_{0}} $,则

(4.78) $ \begin{equation} \lim\limits_{t\rightarrow T}H(t) = -\infty, \end{equation} $

对一些$ T>0 $.因此,我们可以证明这个定理是成立的.证毕

在下面的定理中,我们证明了在周期情况下可能会发生爆破波.我们使用$ {\Bbb S}: = {{\Bbb R}} /{\Bbb Z} $作为单位圆.我们首先引入一个引理来证明这个定理.

引理4.1  令$ f(x)\in H^{2}({\Bbb S}) $.因此

(4.79) $ \begin{equation} f^{2}(x)\leq \frac{1}{\alpha}\|f(x)\|^{2}_{H_{\alpha}^{1}} +\left(\int_{{\Bbb S}}f(x){\rm d}x\right)^{2}. \end{equation} $

由于$ f(x) $在$ {\Bbb S} $上是连续的,因此存在一个点$ x_{0}\in {\Bbb S} $满足$ \int_{{\Bbb S}}f(x){\rm d}x = f(x_{0}) $.对于任何的$ x\in{\Bbb S} $,我们有

(4.80) $ \begin{eqnarray} \alpha\left(f^{2}(x)-\left(\int_{{\Bbb S}}f(x){\rm d}x\right)^{2}\right) = \alpha(f^{2}(x)-f^{2}(x_{0})) = \alpha\int_{x_{0}}^{x}2ff_{x}{\rm d}x\leq \|f(x)\|^{2}_{H^{1}_{\alpha}}. \end{eqnarray} $

不等式(4.80)产生引理.同样的方法则有

(4.81) $ \begin{equation} f^{2}(x)\leq \|f(x)\|^{2}_{H^{1}} +\left(\int_{{\Bbb S}}f(x){\rm d}x\right)^{2}. \end{equation} $

定理4.6  令$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $且$ s\geq\frac{5}{2} $和$ 2c_{0}+\frac{\beta}{\alpha^{2}} = 0 $为初始数据.如果存在$ x_{0}\in {\Bbb S} $,则

(4.82) $ \begin{eqnarray} u'_{0}(x_{0})<-\lambda-\sqrt{\lambda^{2}+\frac{2(1-C_{0})}{\alpha^{2}}D_{1} +\frac{\gamma}{2\alpha^{2}}E(0)+\frac{\gamma}{\alpha^{2}} D_{2}}\ , \end{eqnarray} $

其中$ C_{0} $, $ D_{1} $和$ D_{2} $被如下给出

(4.83) $ \begin{equation} C_{0} = \frac{1}{2}+\frac{\arctan\left(\sinh\left(\frac{1}{2\alpha}\right)\right)} {2\sinh\left(\frac{1}{\alpha}\right)+2\arctan\left(\sinh\left(\frac{1}{2\alpha}\right)\right) \sinh^{2}\left(\frac{1}{2\alpha}\right)}, \end{equation} $

(4.84) $ \begin{equation} D_{1} = \frac{1}{\alpha}E(0)+\left(\int_{{\Bbb S}}u_{0}(x){\rm d}x\right)^{2}, \; \; D_{2} = E(0)+\left(\int_{{\Bbb S}}v_{0}(x){\rm d}x\right)^{2}. \end{equation} $

则有初始数据$ (u_{0}, v_{0}) $的系统(2.2)的解在有限时间内爆破.

证  基于这个事实$ \int_{{\Bbb S}}u(t, x){\rm d}x $和$ \int_{{\Bbb S}}v(t, x){\rm d}x $关于时间$ t $的不变量,可以得到以下$ u $和$ v $的估计数

(4.85) $ \begin{equation} u^{2}(x)\leq\frac{1}{\alpha}\|u(x)\|_{H^{1}_{\alpha}}^{2} +\left(\int_{{\Bbb S}}u(x){\rm d}x\right)^{2} \leq\frac{1}{\alpha}E(0) +\left(\int_{{\Bbb S}}u_{0}(x){\rm d}x\right)^{2}: = D_{1}, \end{equation} $

(4.86) $ \begin{equation} v^{2}(x)\leq\frac{1}{\alpha}\|v(x)\|_{H^{1}_{\alpha}}^{2} +\left(\int_{{\Bbb S}}v(x){\rm d}x\right)^{2} \leq\frac{1}{\alpha}E(0) +\left(\int_{{\Bbb S}}v_{0}(x){\rm d}x\right)^{2}: = D_{2}, \end{equation} $

通过运用引理4.1.根据文献[20]的结果,进一步得到了Sobolev不等式

(4.87) $ \begin{equation} G\ast \left(u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}\right)(x)\geq C_{0} u^{2}(x). \end{equation} $

按照与定理4.5相同的方法,可以得到

(4.88) $ \begin{eqnarray} \frac{{\rm d}H(t)}{{\rm d}t}& = &-H(t)^{2}-\lambda H(t)-\frac{1}{\alpha^{2}} G\ast \left(u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right){} \\ & &+\frac{1}{\alpha^{2}} \left(u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right){} \\ &\leq&-\frac{1}{2}H(t)^{2}-\lambda H(t)+\frac{1-C_{0}}{\alpha^{2}}D_{1} +\frac{\gamma}{4\alpha^{3}}E(0)+\frac{\gamma}{2\alpha^{2}}D_{2}. \end{eqnarray} $

可以看出不等式(4.88)是一个Riccati类型的方程.根据与定理4.5相同的证明,可以看出这个定理是成立的.证毕.

注4.1  在最后两个定理的证明中,我们对$ u(t, x) $和$ v(t, x) $给出了不同的估计.按照文献$ [9] $中提供的相同方法,我们也可以找到一些有趣的例子来揭示它们之间的区别.

在本节,我们介绍了系统(2.2)爆破波解的爆破机制.

引理5.1^[3]  假设$ T>0 $和$ u\in C^{1}([0, T);H^{2}({{\Bbb R}} ) $,那么对于所有$ t\in[0, T) $,至少存在一个点$ \xi(t)\in {{\Bbb R}} $,使得

(5.1) $ \begin{equation} H(t): = \inf\limits_{x\in{{\Bbb R}} }u_{x}(t, x) = u_{x}(t, \xi(t)). \end{equation} $

函数$ H(t) $在$ (0, T) $上是可微的,且

(5.2) $ \begin{equation} \; \; \frac{{\rm d}H(t)}{{\rm d}t} = u_{tx}(t, \xi(t)). \end{equation} $

定理5.1  假设$ (u_{0}, v_{0})\in H^{s}({{\Bbb R}} )\times H^{s-1}({{\Bbb R}} ) $且$ s\geq\frac{5}{2} $是初始数据,其中$ T $为系统(2.2)对应解的最大存在时间.如果$ T $是有限的,则

(5.3) $ \begin{eqnarray} \lim\limits_{t\rightarrow T}\left(\inf\limits_{x\in{{\Bbb R}} } u_{x}(t, x) (T-t)\right) = -2. \end{eqnarray} $

证  根据定理2.2,有

(5.4) $ \begin{equation} \lim\limits_{t\rightarrow T}\inf\limits_{x\in{{\Bbb R}} }\{u_{x}(t, x)\} = -\infty. \end{equation} $

根据引理5.1,我们可以证明至少存在一个点$ \xi(t)\in{{\Bbb R}} $满足$ { } u_{x}(t, \xi(t)) = \inf_{x\in {{\Bbb R}} }\{u_{x}(t, x)\} $对所有$ t\in[0, T) $.引入

(5.5) $ \begin{equation} H(t) = u_{x}(t, \xi(t)) = \inf\limits_{x\in{{\Bbb R}} }\{u_{x}(t, x)\}, \; \; t\in[0, T), \end{equation} $

因此对于每一个$ t\in [0, T) $有$ u_{xx}(t, \xi(t)) = 0 $.从(4.3)式,则有

(5.6) $ \begin{eqnarray} \frac{{\rm d}H(t)}{{\rm d}t}& = &-H(t)^{2}-\lambda H(t)-\frac{1}{\alpha^{2}} G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right)(t, \xi(t)){} \\ & &+\frac{1}{\alpha^{2}} \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right)(t, \xi(t)). \end{eqnarray} $

通过运用引理2.3,进一步有

(5.7) $ \begin{eqnarray} &&\left| \left( G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right)\right)(t, \xi(t))\right|_{L^{\infty}}{} \\ &\leq&\|G\|_{L^{\infty}}\left\|2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right\|_{L^{1}}{} \\ &\leq&\tilde{\gamma}_{0} C_{1}E(0)+\sqrt{2d_{0}C_{1}E(0)}, \end{eqnarray} $

对所有$ t\in[0, T) $,其中$ \tilde{\gamma}_{0} = \max\{1, \gamma\} $.然后运用引理2.3,引理2.4和不等式(5.7),我们可以证明存在一个常数$ L>0 $满足

(5.8) $ \begin{eqnarray} &&\left|-\frac{1}{\alpha^{2}} G\ast \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right)(t, \xi(t))\right.{} \\ & &\left. +\frac{1}{\alpha^{2}} \left(2d_{0}u+u^{2}+\frac{\alpha^{2}}{2}u_{x}^{2}+\frac{\gamma}{2} v^{2}-\frac{\gamma}{2}v_{x}^{2}\right)(t, \xi(t))\right|\leq L, \; \; \mbox{即在}\; \; t\in[0, T). \end{eqnarray} $

根据(5.6)式,则有

(5.9) $ \begin{equation} \left|\frac{{\rm d}H(t)}{{\rm d}t}+\frac{1}{2}H(t)^{2}+\lambda H(t)\right| \leq L\; \; \mbox{即在}\; \; t\in[0, T). \end{equation} $

因此

(5.10) $ \begin{equation} \left|\frac{{\rm d}H(t)}{{\rm d}t}+\frac{1}{2}(H(t)+\lambda)^{2}\right| \leq \frac{1}{2}\lambda^{2}+L\; \; \mbox{即在}\; \; t\in[0, T). \end{equation} $

考虑(5.4)式生成$ { } \lim\limits_{t\uparrow T}\inf(H(t)+\lambda) = -\infty $.令$ \epsilon\in(0, \frac{1}{2}) $存在一个$ t_{1}\in (0, T) $满足$ H(t_{1})+\lambda<0 $和

(5.11) $ \begin{equation} (H(t_{1})+\lambda)^{2}>\frac{1}{\epsilon}\left(L+\frac{1}{2}\lambda^{2}\right). \end{equation} $

因此我们有

(5.12) $ \begin{equation} (H(t)+\lambda)^{2}>\frac{1}{\epsilon}\left(L+\frac{1}{2}\lambda^{2}\right), \; \; t\in[t_{1}, T). \end{equation} $

因为$ H(t)\in W_{\mbox{loc}}^{1, \infty} $是局部Lipschitz,其中一些$ \delta>0 $满足

(5.13) $ \begin{equation} (H(t)+\lambda)^{2}>\frac{1}{\epsilon}\left(L+\frac{1}{2}\lambda^{2}\right), \; \; t\in[t_{1}, t_{1}+\delta). \end{equation} $

考虑(5.10)和(5.12)式,获得

(5.14) $ \begin{equation} \frac{{\rm d}H(t)}{{\rm d}t} \leq \left(\epsilon-\frac{1}{2}\right)(H(t)+\lambda)^{2}\; \; \mbox{即在}\; \; t\in[t_{1}, t_{1}+\delta). \end{equation} $

由于$ H(t) $是局部Lipschitz,所以它也是连续的.因此通过在$ [t_{1}, t_{1}+\delta] $对不等式(5.14)积分,能够找到$ H(t_{1}+\delta)\leq H(t_{1}) $.因此有

(5.15) $ \begin{equation} H(t_{1}+\delta)+\lambda\leq H(t_{1})+\lambda<0. \end{equation} $

通过运用不等式(5.15),进一步有

(5.16) $ \begin{equation} (H(t_{1}+\delta)+\lambda)^{2}\geq (H(t_{1})+\lambda)^{2}>\frac{1}{\epsilon} (L+\frac{1}{2}\lambda^{2}). \end{equation} $

通过使用连续扩展,可以证明不等式(5.12)是新的.结合(5.10)和(5.12)式,则当$ t\in[t_{1}, T) $时有

(5.17) $ \begin{equation} \left|\frac{2\frac{{\rm d}H(t)}{{\rm d}t}+(H(t)+\lambda)^{2}}{2(H(t)+\lambda)^{2}} \right|<\epsilon. \end{equation} $

通过在$ (t, T) $上对(5.17)式积分,可以发现

(5.18) $ \begin{equation} \left|\frac{2+(H(t)+\lambda)(T-t)}{2(H(t)+\lambda)(T-t)}\right|<\epsilon, \; \; t\in[t_{1}, T). \end{equation} $

让$ \epsilon\rightarrow 0 $生成

(5.19) $ \begin{equation} \lim\limits_{t\rightarrow T}\left((H(t)+\lambda)(T-t)\right) = -2, \end{equation} $

这相当于

(5.20) $ \begin{equation} \lim\limits_{t\rightarrow T}\left(H(t)(T-t)\right) = -2. \end{equation} $

证毕.

注5.1  由定理4.1–4.6,可以看出,系统(1.3)强解的爆破现象受耗散参数的影响.然而,根据定理5.1,系统(1.3)强解的爆破率不受弱耗散项$ \lambda(u-\alpha^{2}u_{xx}) $和$ \lambda\rho $的影响.