上海理工大学学报  2020, Vol. 42 Issue (5): 424-429, 447   PDF    
七阶Kaup-Kupershmidt方程的经典李群分析和精确解
卢毅辉, 胡恒春     
上海理工大学 理学院,上海 200093
摘要: 为丰富七阶Kaup-Kupershmidt(KK)方程的解,利用经典李群分析得到了七阶Kaup-Kupershmidt(KK)方程对应的无穷小,进而得到了两种不同形式的约化方程,最后,通过对约化方程进行求解,得到了有理函数解、雅可比椭圆函数解、双曲函数解、三角函数解和幂级数解,同时,给出了幂级数解的收敛性的证明。
关键词: 七阶KK方程     经典李群分析     精确解    
Classical Lie group analysis and exact solutions of the seventh-order Kaup-Kupershmidt equation
LU Yihui, HU Hengchun     
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: The classical Lie group analysis was used to obtain the corresponding infinitesimals of the seventh-order Kaup-Kupershmidt (KK) equation, and two different reduction equations were obtained. Many different forms of exact solutions of the seventh-order KK equation were achieved by solving the reduction equations, such as the rational function solution, the Jacobi elliptic function solution, the hyperbolic function solutions, the trigonometric function solutions and the power series solution, meanwhile, the convergence of the power series in the solution was proved.
Key words: seventh-order Kaup-Kupershmidt equation     classical Lie group analysis     exact solutions    
1 问题的提出

非线性发展方程可以描述自然界中的各种非线性现象,并且在物理、化学和生物等学科中都有广泛的应用。因此,研究这些非线性发展方程的解及解的性质具有重要的理论和现实意义。到目前为止,学者们已经提出许多有效的方法来研究非线性模型的可积性质和精确解,例如:反散射方法、Hirota双线性方法、扩展的tanh方法、指数展开法、Painlevé分析法等[1-10]

KdV型方程是最经典的可积模型之一,它可以描述浅水波、等离子体中的离子声波和大气或海洋中的内波。同时,KdV方程族中描述表面波和内部波、重力–毛细管波以及水波与浮冰间相互作用的高阶可积元等问题也得到了广泛的研究[11]。这里,研究高阶KdV型方程之一的七阶Kaup-Kupershmidt(KK)方程[12-13],KK方程形式如下:

$ \begin{split}{u_t} + &2\;016{u^3}{u_x} + 630u_x^3 + 2\;268u{u_x}{u_{2x}} + 504{u^2}{u_{3x}} +\\ &252{u_{2x}}{u_{3x}} + 147{u_x}{u_{4x}} + 42u{u_{5x}} + {u_{7x}} = 0\\[-10pt]\end{split} $ (1)

这是七阶非线性方程

$\begin{split}{u_t} +& a{u^3}{u_x} + bu_x^3 + cu{u_x}{u_{2x}} + d{u^2}{u_{3x}} + e{u_{2x}}{u_{3x}} + \\ &f{u_x}{u_{4x}} + gu{u_{5x}} + {u_{7x}} = 0\end{split} $ (2)

通过固定常数

$\begin{split}&a = 2\;016,\;b = 630,\;c = 2\;268,\;d = 504,\;e = 252,\\ &f = 147,\;g = 42\end{split}$

得到的一个特殊情形。方程(2)被Pomeau[14]用来讨论标准KdV方程在奇异摄动作用下的结构稳定性,并且是在小波幅下,通过对欧拉方程在三阶以下的简单波运动的渐近展开而得到的[15]。此外,高阶色散项和非线性的相互作用在物理学中有重要应用[16]。不同于KdV方程,方程(2)还较好地描述了较短波长陡波的演化过程。在文献[17-18]中已给出了方程(1)的Cole-Hopf变换、Lax对、Hirota双线性形式和孤子解。目前为止,未见通过经典李群分析来研究方程(1)的文献。

本文通过群变换求得了方程(1)对应的无穷小及相应的约化方程,并采用不同的求解方法对约化方程进行求解,得到了有理函数解、雅可比椭圆函数解、双曲函数解、三角函数解和幂级数解,最后,对后期工作进行了展望。

2 七阶KK方程的对称和对称约化

首先,考虑单参数李群的无穷小变换

$\begin{split}&x \to x + \varepsilon \xi (x,t,u),t \to t + \varepsilon \tau (x,t,u),\\ & u \to u + \varepsilon \eta (x,t,u)\end{split}$ (3)

式中, $\varepsilon $ 为无穷小参数。

该变换群的向量场可以表示为

${{V}} = \xi (x,t,u)\partial x + \tau (x,t,u)\partial t + \eta (x,t,u)\partial u$ (4)

式中, $\xi ,\;\tau ,\;\eta$ $x,\;t,\;u$ 的待定函数。

方程(1)的对称方程为

$\Pr {{{V}}^{(7)}}(\varDelta )\left| {_{\varDelta = 0} = 0} \right.$

式中, $\Pr {{{V}}^{(7)}}$ 表示 ${{V}}$ 的七阶延拓,并且

$\begin{split}\varDelta = &{u_t} + 2\;016{u^3}{u_x} + 630u_x^3 + 2\;268u{u_x}{u_{2x}} + 504{u^2}{u_{3x}} +\\ &252{u_{2x}}{u_{3x}} + 147{u_x}{u_{4x}} + 42u{u_{5x}} + {u_{7x}}\end{split}$

这个延拓 $\Pr {{V}}$ 依赖于方程

$\Pr {{{V}}^{(7)}} = \eta \partial u + {\eta ^x}\partial {u_x} + {\eta ^{2x}}\partial {u_{2x}} + \cdots + {\eta ^{7x}}\partial {u_{7x}}$

系数函数 ${\eta ^{kx}}(k = 1,2, \cdots ,7)$

${\eta ^{kx}} = {\rm D}_x^k(\eta - \tau {u_t} - \xi {u_x}) + \tau {u_{kxt}} + \xi {u_{(k + 1)x}},\;k = 1,2, \cdots ,7$

在这里 ${{\rm D}_x}$ 表示全微分算子,定义为

${{\rm D}_x} = \partial x + {u_x}\partial u + {u_{tx}}\partial {u_t} + {u_{2x}}\partial {u_x} + \cdots $

将上述延拓公式代入方程(1)的李对称条件,然后令 $u$ 及其偏导数的系数为0,得到以下超定方程组:

$ \begin{array}{l} {\xi _t} = {\xi _u} = {\xi _{xx}} = {\tau _x} = {\tau _u} = {\eta _x} = {\eta _t} = {\eta _{uu}} = 0,\\ \eta + u{\eta _u} + u{\tau _t} - 3u{\xi _x} = 0,{\rm{ }}{\eta _u} - 5{\xi _x} + {\tau _t} = 0,\\ u{\tau _t} - u{\xi _x} + 3\eta = 0,{\rm{ }}u{\tau _t} + 2\eta - 3u{\xi _x} = 0,\\ {\tau _t} - 7{\xi _x} = 0,{\rm{ }}{\tau _t} + 2{\eta _u} - 3{\xi _x} = 0 \end{array} $

求解得到

${{V}} = ({c_1}x + {c_3})\frac{\partial }{{\partial x}} + (7{c_1}t + {c_2})\frac{\partial }{{\partial t}} + ( - 2{c_1}u)\frac{\partial }{{\partial u}}$ (5)

那么,方程(1)的无穷小生成子构成由以下线性无关算子张成的三维李代数:

$ {{{V}}_1} = \frac{\partial }{{\partial x}},{{{V}}_2} = \frac{\partial }{{\partial t}},{{{V}}_3} = x\frac{\partial }{{\partial x}} + 7t\frac{\partial }{{\partial t}} - 2u\frac{\partial }{{\partial u}} \\ $

经过直接计算,3个算子的李括号运算是封闭的,具体结果如下:

$\begin{split} &\left[ {{{{V}}_1},{{{V}}_1}} \right] = \left[ {{{{V}}_2},{{{V}}_2}} \right] = \left[ {{{{V}}_3},{{{V}}_3}} \right] = 0,\;\\ &\left[ {{{{V}}_1},{{{V}}_2}} \right] =- \left[ {{{{V}}_2},{{{V}}_1}} \right] = 0, \\ &\left[ {{{{V}}_1},{{{V}}_3}} \right] = - \left[ {{{{V}}_3},{{{V}}_1}} \right] ={{{V}}_1},\;\\ &\left[ {{{{V}}_2},{{{V}}_3}} \right] = - \left[ {{{{V}}_3},{{{V}}_2}} \right] = 7{{{V}}_2} \end{split} $

方程(1)的对应于无穷小生成子的单参数群 ${g_i}$

$\begin{split}&{g_1}:(x + \varepsilon ,t,u),\;{g_2}:(x,t + \varepsilon ,u),\;\\ &{g_3}:(x{{\rm{e}}^\varepsilon },t{{\rm{e}}^{7\varepsilon }},u{{\rm{e}}^{ - 2\varepsilon }})\end{split}$ (6)

式中: ${g_1},\;{g_2}$ 表示方程(1)解的时空平移不变性; ${g_3}$ 表示方程(1)解的伽利略伸缩不变性。从式(6)容易看出方程(1)有解。

${u_1} = f(x - \varepsilon ,t),\;{u_2} = f(x,t - \varepsilon ),\;{u_3} = {{\rm{e}}^{ - 2\varepsilon }}f(x{{\rm{e}}^{ - \varepsilon }},t{{\rm{e}}^{ - 7\varepsilon }})$

函数 $f$ 是方程(1)的任意已知解。接下来,从特征方程

$\dfrac{{{\rm{d}}x}}{{{c_1}x + {c_3}}} = \dfrac{{{\rm{d}}t}}{{7{c_1}t + {c_2}}} = \dfrac{{{\rm{d}}u}}{{ - 2{c_1}u}}$

入手可以求得相似变换和相似变量,进而得到方程(1)的约化方程。下面通过选取不同的任意常数分别讨论:

a. 当 ${c_1} = 0$ 时,得方程(1)对应的相似变量和相似变换

$z = x - kt,\;u(x,t) = f(z),\;(k = {c_3}/{c_2},{c_2} \ne 0)$

$f(z)$ 满足常微分方程

$\begin{split} &- kf' + 2\;016{f^3}f' + 630{f'^3} + 2\;268ff'f'' + \\ &\quad 504{f^2}{f^{(3)}} +252f''{f^{(3)}} + 147f'{f^{(4)}} +\\ & \quad42f{f^{(5)}} + {f^{(7)}} = 0 \end{split}$ (7)

b. 当 ${c_1} \ne 0$ 时,得到对应的相似变量和相似变换

$z = \dfrac{{x + \dfrac{c_3}{c_1}}}{{{{\left(t + \dfrac{c_2}{7{c_1}}\right)}^{\frac{1}{7}}}}},\;u(x,t) = \dfrac{{f(z)}}{{{{\left(t +\dfrac {c_2}{7{c_1}}\right)}^{\frac{2}{7}}}}}$

$f(z)$ 满足常微分方程

$\begin{split}& -\frac{2}{7}f - \dfrac{1}{7}zf' + 2\;016{f^3}f' + 630{f'^3} + \\ &\quad2\;268ff'f'' +504{f^2}{f^{(3)}}+252f''{f^{(3)}}+\\ &\quad147f'{f^{(4)}}+42f{f^{(5)}} + {f^{(7)}}=0 \end{split}$ (8)
3 约化方程的求解

上文通过群变换得到了方程(1)对应的无穷小和约化方程,本节将用扩展tanh方法、微分方程展开法、幂级数法来寻求约化方程的精确解。

3.1 利用扩展的tanh方法来求解约化方程(7)

该算法由Wazwa在文献[4]中提出。假设方程(7)的解可以写成如下形式:

$f(z) = \sum\limits_{i = - n}^n {{A_i}H{{(z)}^i}} $

这里 $H(z)$ 应满足黎卡提方程

$H'(z) = 1 - {H^2}(z)$

且有解

$H(z) = \tanh z$

平衡方程(7)中最高阶导数项与最高阶非线性项,可以确定正整数 $n$ 的值, ${A_i}$ 是待定的系数。主项平衡得到 $n = 2$ ,因此,方程(7)的解具有形式

$\begin{split}f(z) =& {A_{ - 2}}{\tanh ^{ - 2}}\ z + {A_{ - 1}}{\tanh ^{ - 1}}\ z+\\ &{A_0} + {A_1}\tanh \ z +{A_2}{\tanh ^2}\ z\end{split}$ (9)

将方程(9)代入方程(7),并且令 ${\tanh ^i}(z)$ 的系数等于0,得到3种类型的精确解:

$ \begin{array}{l} {\bf{a}} .\; u(x,t) = \dfrac{1}{3} - \dfrac{1}{2}{\tanh ^2}\left(x + \dfrac{4}{3}t\right)\\ {\bf{b}}.\; u(x,t) = - \dfrac{1}{2}{\tanh ^{ - 2}}\left(x + \dfrac{{256}}{3}t\right) + \dfrac{1}{3} - \dfrac{1}{2}{\tanh ^2}\left(x + \dfrac{{256}}{3}t\right)\\ {\bf{c}}.\; u(x,t) = \dfrac{1}{3} - \dfrac{1}{2}{\tanh ^{ - 2}}\left(x + \dfrac{4}{3}t\right) \end{array} $
3.2 利用微分方程展开法来求解方程(7)的精确解

假设方程(7)的解有以下形式

$f(z) = \sum\limits_{i = 1}^n {{P_i}{\phi ^i}(z)} + {P_0}$

$\phi '$ 是下面一阶常微分方程的解

$\phi '(z) = \varepsilon \sqrt {\sum\limits_{j = 0}^r {{h_j}{\phi ^j}} } $ (10)

式中: $\varepsilon = \pm 1$ $r$ 是一个正整数; ${P_i},{h_0}, \cdots ,{h_r}$ 是待定常数。

通过平衡方程(7)中的 ${f^{(7)}}$ $f{f^{(5)}}$ ,可以得到 $n$ $r$ 的关系

$r = n + 2$

这里,仅考虑 $r = 4$ ,那么 $n = 2$ ,因此方程(7)的解可以表示为

$f = {P_2}{\phi ^2} + {P_1}\phi + {P_0}$ (11)

$\varepsilon = 1$ ,将方程(11)和(10)代入方程(7)并令 ${\phi ^i}{\phi ^{'}}$ 的系数为0,可以得到相应的代数方程,解之得:

$\begin{split} {\bf{a}} .\; k =& - \frac{1}{{48}}h_2^3 - \dfrac{9}{{32}}h_3^2{h_0} + \frac{3}{{32}}{h_1}{h_2}{h_3},{P_0} = \frac{1}{{24}}{h_2},\\{P_1} =& - \frac{1}{8}{h_3},{P_2} = {h_4} = 0 \end{split}$

式中, ${h_0},{h_1},{h_2},{h_3}$ 为任意常数。

$ \begin{split} {\bf{b}}. \;k =& - \dfrac{{72{h_0}{h_2}P_2^4 + 63P_2^2P_1^2h_2^2 + 216P_2^3P_1^2{h_0} + 189{P_2}P_1^4{h_2} + 18P_2^3 + {P_1}{h_1}{h_2} + 8P_2^3h_2^3 - 27P_2^4 + h_1^2 + 189h_1^6}}{{{\rm{6}}P_2^3}}\\ {h_3} =& - 4{P_1},{h_4} = - 2{P_2},{P_0} = - \dfrac{1}{{12}}\dfrac{{3P_1^2 + 2{P_2}{h_2}}}{{{P_2}}} \end{split} $

式中, ${h_0},{h_1},{h_2},{P_1},{P_2}$ 为任意常数。任意常数的不同选取对应于不同的精确解,得出如下结果。

情形1

a. $\;{h_0} = {h_1} = {h_4} = 0,{h_2} > 0$

$ f = - \frac{1}{{24}}{h_2} + \frac{1}{8}{h_2}\sec {{\rm{h}}^2}\left(\frac{{\sqrt {{h_2}} }}{2}z\right)$

b. $\;{h_0} = {h_1} = {h_4} = 0,{h_2} < 0 $

$f = - \frac{1}{{24}}{h_2} + \frac{1}{8}{h_2}{\rm{sec}}{{\rm{h}}^2}\left(\frac{{\sqrt { - {h_2}} }}{2}z\right)$

c. $\;{h_0} = {h_1} = {h_4} = {h_2} = 0$

$f = - \frac{1}{{8{z^2}}}$

d. $\;{h_4} = 0,{h_3} > 0$

$f = - \dfrac{1}{{24}}{h_2} + \frac{1}{8}{h_3}\wp \left(\dfrac{{\sqrt {{h_3}} }}{2}z,{g_2},{g_3}\right)$

式中, $z = x - kt,{g_2} = - 4{h_1}/{h_3},{g_3} = - 4{h_0}/{h_3},$ $\wp $ 表示魏尔斯特拉斯雅可比椭圆函数。

情形2 ${P_1} = 0$ ,那么 ${h_3} = 0$ ,当 ${h_1} = {h_3} = 0$ 时,有以下7组雅可比椭圆函数解:

a. $\;{h_2} = 2{m^2} - 1,{h_0} = 1 - {m^2},{h_4} = - {m^2}$

$f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{cn}}(z))^2}$

b. $\;{h_2} = 2{m^2} - 1,{h_0} = - {m^2},{h_4} = 1 - {m^2}$

$f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{nc}}(z))^2}$

c. $\;{h_2} = 2{m^2} - 1,{h_0} = 1,{h_4} = ({m^2} - 1){m^2}$

$f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{sd}}(z))^2}$

d. $\;{h_2} = 2{m^2} - 1,{h_0} = ({m^2} - 1){m^2},{h_4} = 1$

$f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{ds}}(z))^2}$

e. $\;{h_2} = \dfrac{{1 - 2{m^2}}}{2},{h_0} = {h_4} = \dfrac{1}{4}$

$f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{ns}}(z) \pm {\rm{cs}}(z))^2}$

f. $\;{h_2} = \dfrac{{1 - 2{m^2}}}{2},{h_0} = {h_4} = \dfrac{{1 - {m^2}}}{4}$

$f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{nc}}(z) \pm {\rm{sc}}(z))^2}$

g. $\;{h_2} = \dfrac{{1 - 2{m^2}}}{2},{h_0} = \dfrac{{{m^2}}}{4},{h_4} = \dfrac{1}{4}$

$f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{ns}}(z) \pm {\rm{ds}}(z))^2}$

对于情形2,也有钟状孤波解、三角函数周期解、扭状孤波解,均可从情形1中通过适当的变换推导出来。此外,当 ${h_1} = {h_3} = 0$ ${h_0} = {h_4} = 1$ ${h_2} = - 2$ 时,该方法就退化为扩展的tanh法。

3.3 方程(8)的幂级数解

对于约化方程(8),考虑到其复杂性,这里用幂级数法来研究它的精确解,即寻找以下形式的幂级数解

$f(z) = \sum\limits_{n = 0}^\infty {{C_n}{z^n}} $ (12)

将式(12)代入方程(8),比较 ${z^n}$ 的系数,可以得到以下递推公式

$\begin{split} {C_{n + 7}} =& - \frac{1}{{(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)(n + 7)}}\left[ { - \frac{2}{7}{C_n} - \frac{1}{7}n{C_n}}+ \right. \\ & 2\;016\sum\limits_{j = 0}^i {\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_j}{C_{i - j}}{C_{k - i}}{C_{n - k + 1}}} } } (n - k + 1)+ \\ & 630\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 1}}(i + 1)(k - i + 1)(n - k + 1)} } +\\ & 2\;268\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 2}}(k - i + 1)(n - k + 2)(n - k + 1)} } +\\ & 504\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_i}{C_{k - i}}{C_{n - k + 3}}(n - k + 3)(n - k + 2)(n - k + 1)} } +\\ & 252\sum\limits_{k = 0}^n {{C_{k + 2}}{C_{n - k + 3}}(k + 2)(k + 1)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ & 147\sum\limits_{k = 0}^n {{C_{k + 1}}{C_{n - k + 4}}(k + 1)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ &\left. { 42\sum\limits_{k = 0}^n {{C_k}{C_{n - k + 5}}(n - k + 5)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} } \right] \\ \end{split} $

因此,对于任意选定的常数 ${C_i}(i = 0,1, \cdots ,6)$ ,序列 $\left\{ {{C_n}} \right\}_{n = 0}^\infty $ 的每一项都可以由递推公式唯一确定。为了确保幂级数(12)的收敛性,下面给出幂级数(12)的收敛性证明。

$\begin{array}{l} | {{C_{n + 7}}} | \leqslant 2\;268\left( {\displaystyle\sum\limits_{j = 0}^i {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {| {{C_j}} || {{C_{i - j}}} || {{C_{k - i}}} || {{C_{n - k + 1}}} |} } } + \displaystyle\sum\limits_{k = 0}^n {\left| {{C_k}} \right|\left| {{C_{n - k + 5}}} \right|} }+\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {\left| {{C_{i + 1}}} \right|\left| {{C_{k - i + 1}}} \right|\left| {{C_{n - k + 1}}} \right| + } } \right. \\ \;\;\;\;\;\;\;\; \displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {\left| {{C_{i + 1}}} \right|\left| {{C_{k - i + 1}}} \right|\left| {{C_{n - k + 2}}} \right|} } +\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {\left| {{C_i}} \right|\left| {{C_{k - i}}} \right|\left| {{C_{n - k + 3}}} \right|} } + \displaystyle\sum\limits_{k = 0}^n {\left| {{C_{k + 2}}} \right|\left| {{C_{n - k + 3}}} \right|}+\\ \;\;\;\;\;\;\;\; \left. { \displaystyle\sum\limits_{k = 0}^n {\left| {{C_{k + 1}}} \right|} \left| {{C_{n - k + 4}}} \right| + \left| {{C_n}} \right|} \right){\rm{, }}\;n = 0,1,2, \cdots \\ \end{array} $

现在,重新定义一个幂级数

$\delta = R(z) = \sum\limits_{n = 0}^\infty {{r^n}{z^n}} $

$ \left| {{r_0}} \right| = \left| {{C_0}} \right|,...,\left| {{r_6}} \right| = \left| {{C_6}} \right|$ ,且

$\begin{split} {r_{n + 7}} = &2\;268\left( \displaystyle\sum\limits_{j = 0}^i \displaystyle\sum\limits_{i = 0}^k \displaystyle\sum\limits_{k = 0}^n {r_j}{r_{i - j}}{r_{k - i}}{r_{n - k + 1}} +\right. \\ &\displaystyle\sum\limits_{k = 0}^n {{r_k}{r_{n - k + 5}}} +\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 1}} +\\ &\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 2}} +\\ &\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_i}{r_{k - i}}} } {r_{n - k + 3}} + \displaystyle\sum\limits_{k = 0}^n {{r_{k + 2}}{r_{n - k + 3}}}+ \\ &\left.{\displaystyle\sum\limits_{k = 0}^n {{r_{k + 1}}{r_{n - k + 4}}} + {r_n}} \right),\;n = 0,1,2, \cdots \end{split}$

容易看出 $\left| {{C_n}} \right| \leqslant \left| {{r_n}} \right|$ 。换句话说,幂级数 $\delta = R(z) = \displaystyle\sum\limits_{n = 0}^\infty {{r^n}{z^n}} $ 是幂级数(12)的强级数。利用数学分析的相关知识可知幂级数 $\delta = R(z)$ 有正的收敛半径。事实上,通过计算,有

$\begin{split} R(z) &= {r_0} + {r_1}z + {r_2}{z^2} + {r_3}{z^3} + {r_4}{z^4} + {r_5}{z^5} + {r_6}{z^6} + \\ &\displaystyle\sum\limits_{n = 0}^\infty {{r_{n + 7}}{z^{n + 7}}}{\rm{ }} = {r_0} + {r_1}z + {r_2}{z^2} + {r_3}{z^3} + {r_4}{z^4} + {r_5}{z^5} +\\ &{r_6}{z^6}+2\;268\left( {\displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{j = 0}^i {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_j}{r_{i - j}}{r_{k - i}}{r_{n - k + 1}} + } } } } } \right.\\ &\displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{k = 0}^n {{r_k}{r_{n - k + 5}}} }+ \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 1}} }+\\ & \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 2}}} + \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_i}{r_{k - i}}} } {r_{n - k + 3}} + }\\ &\left.\displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{k = 0}^n {{r_{k + 2}}{r_{n - k + 3}}} }{ + \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{k = 0}^n {{r_{k + 1}}{r_{n - k + 4}}} + {r_n}} } \right){z^{n + 7}} \end{split} $

进一步考虑隐函数方程

$\begin{split} F(z,&\delta ) = \delta - {r_0} - {r_1}z - {r_2}{z^2} - {r_3}{z^3} - {r_4}{z^4} - {r_5}{z^5} - {r_6}{z^6} -\\ &2\;268[{z^6}{R^3}(R - {r_0}) + {z^4}{(R - {r_0})^3} + {z^3}{(R - {r_0})^2}. \\ &(R - {r_0} - {r_1}z)+ {z^4}{R^2}(R - {r_0} - {r_1}z - {r_2}{z^2}) + \\ &{z^2}(R - {r_0} - {r_1}z)(R - {r_0} - {r_1}z - {r_2}{z^2})+\\ &{z^2}(R - {r_0}) (R - {r_0} - {r_1}z - {r_2}{z^2} - {r_3}{z^3})+\\ &{z^2}R(R - {r_0} - {r_1}z - {r_2}{z^2} - {r_3}{z^3} - {r_4}{z^4}) + {z^7}R] = 0 \end{split} $

由于 $F(z,\delta )$ $(0,{r_0})$ 的邻域内是解析函数且 $F(0,{r_0}) = 0$ ${F'_\delta }(0,{r_0}) = 1 \ne 0$ ,通过隐函数定理[19]可知, $\delta = R(z)$ 在点 $(0,{r_0})$ 邻域内是解析的且有正的收敛半径,也就是说在点 $(0,{r_0})$ 邻域内幂级数(12)是收敛的。再由判别级数收敛的比较判别法可知,幂级数解(12)是方程(8)的精确解,具体表达形式如下所示:

$\begin{array}{l} u = \dfrac{{{C_0}}}{A} + \dfrac{{{C_1}z}}{A} + \dfrac{{{C_2}{z^2}}}{A} + \dfrac{{{C_3}{z^3}}}{A} + \dfrac{{{C_4}{z^4}}}{A} + \dfrac{{{C_5}{z^5}}}{A} + \dfrac{{{C_6}{z^6}}}{A} -\\ \;\;\;\;\;\;\;\; \displaystyle\sum\limits_{n = 0}^\infty {\frac{1}{{(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)(n + 7)}}\left[ { - \frac{2}{7}{C_n} - \frac{1}{7}n{C_n}} \right.} +\\ \;\;\;\;\;\;\;\; 630\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 1}}(i + 1)(k - i + 1)(n - k + 1)} } +\\ \;\;\;\;\;\;\;\; 2\;268\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 2}}(k - i + 1)(n - k + 2)(n - k + 1)} } {\rm{ + }}\\ \;\;\;\;\;\;\;\;504\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{C_i}{C_{k - i}}{C_{n - k + 3}}(n - k + 3)(n - k + 2)(n - k + 1)} } +\\ \;\;\;\;\;\;\;\; 252\displaystyle\sum\limits_{k = 0}^n {{C_{k + 2}}{C_{n - k + 3}}(k + 2)(k + 1)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ \;\;\;\;\;\;\;\; 147\displaystyle\sum\limits_{k = 0}^n {{C_{k + 1}}{C_{n - k + 4}}(k + 1)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ \left. {\;\;\;\;\;\;\;\; 42\displaystyle\sum\limits_{k = 0}^n {{C_k}{C_{n - k + 5}}(n - k + 5)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} } \right]\dfrac{{{z^{n + 7}}}}{A} \\ \end{array} $

式中, $A = {\left(t + \dfrac{c_2}{7{c_1}}\right)^{\frac{2}{7}}}$ $z = \dfrac{{x + \dfrac{c_3}{c_1}}}{{{{\left(t + \dfrac{c_2}{7{c_1}}\right)}^{\frac{1}{7}}}}}$

4 结 论

通过经典李群方法对七阶KK方程进行了研究,得到了该方程对应的无穷小,进而得到了不同形式的约化方程。最后,通过求解约化方程得到了多种形式的精确解,包括有理解、椭圆函数解、三角函数解、双曲函数解、幂级数解,且给出了幂级数解收敛性的证明。

通过本文的分析可以看出,在解决非线性发展方程时,可以通过李群变换法巧妙地对原偏微分方程进行约化,进而通过对约化方程的求解来获得原方程的解。但是随着方程维数的增加,其约化难度将会变得困难许多。另外,如何对得到的约化方程进行有效处理使其转化为我们熟知的方程,亦即探讨约化方程与已知方程的联系是一个难点问题。目前对约化方程的处理大多借助于函数展开,例如,扩展的tanh展开法、Riccati展开法、微分方程展开法、幂级数法等。但是,能否通过某些适当的变换将约化方程巧妙地转化为我们容易解决的方程或者是否有更为有效的方法来处理约化方程值得今后深入研究。

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