超广义Burgers方程族的非线性可积耦合及其Bargmann对称约束

超广义Burgers方程族的非线性可积耦合及其Bargmann对称约束

方芳^,, 胡贝贝, 张玲

Nonlinear Integrable Couplings and Bargmann Symmetry Constraint of Super Generalized-Burgers Hierarchy

Fang Fang^,, Hu Beibei, Zhang Ling

通讯作者: 方芳, E-mail: fangfang7679@163.com

收稿日期: 2019-03-19

基金资助:

国家自然科学基金.  11601055
安徽省自然科学研究项目基金.  KJ2015B02
安徽省教育厅自然科学研究项目基金.  KJ2017B10

Received: 2019-03-19

Fund supported:

the NSFC.  11601055
the Natural Science Research Project of Anhui Province.  KJ2015B02
the Natural Science Research Project of Anhui Provincial Education Department.  KJ2017B10

摘要

基于李超代数，构造了超广义Burgers方程族的非线性可积耦合，并且利用超级恒等式得到了它的超Hamilton结构.此外，该文计算出超广义Burgers方程族的非线性可积耦合的Bargmann对称约束.

关键词： 李超代数 ; 超广义Burgers方程族的非线性可积耦合 ; 超Hamilton结构 ; 可积耦合 ; Bargmann对称约束

Abstract

With the help of the enlarging Lie super algebra, we construct nonlinear integrable couplings for coupled generalized-Burgers hierarchy in this paper. Then, we establish its super-Hamiltonian structures by utilizing super trace identity. Furthermore, we obtain the Bargmann Symmetry Constraint of super generalized-Burgers hierarchy.

Keywords： Enlarging Lie super algebra ; Super generalized-Burgers hierarchy ; Integrable couplings ; Super-Hamiltonian structures ; Bargmann symmetry constraint

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本文引用格式

方芳, 胡贝贝, 张玲. 超广义Burgers方程族的非线性可积耦合及其Bargmann对称约束. 数学物理学报[J], 2020, 40(3): 694-704 doi:

Fang Fang, Hu Beibei, Zhang Ling. Nonlinear Integrable Couplings and Bargmann Symmetry Constraint of Super Generalized-Burgers Hierarchy. Acta Mathematica Scientia[J], 2020, 40(3): 694-704 doi:

1 引言

随着孤立子理论的发展,超可积方程引起了大量学者的研究兴趣,各个专业领域已有很多成果^[1-10].近几年来,超可积方程族的非线性可积耦合成为当今数学领域和物理邻域的热点问题.可积耦合这一概念首先是由Fuchssteiner^[11]在研究可积系统的无中心Virasoro对称代数和孤立子时提出来的.学者们研究出了一些经典孤立子方程的可积耦合的方法,例如: AKNS方程族, Levi方程族, DLW方程族等^[12-13].马教授和朱教授在文章^[16-17]中首次提出非线性可积耦合这一概念,并且提供了建立孤立子方程族的非线性可积耦合的一般方法.利用这个方法几个超可积方程族的非线性可积耦合就被研究出来了,其中包括超BKK方程族,超经典Boussinesq方程族和Levi方程族^[18-21].

在对可积系统研究的过程中,人们发现通过Bargmann对称约束的计算可以揭示许多可积系统之间存在的内在联系. 1988年,曹策问教授通过谱问题中位势函数与特征函数之间的一种约束条件,将原有谱问题约束成为有限维Hamilton系统^[22].例如CKdV方程族相关的谱问题在Bargmann对称约束下分别被非线性化为完全可积的Bargmann系统^[23].另外,还有一些超可积方程族的非线性化也被研究过^[24-29].

本文在第2节里利用李超代数,构建超广义Burgers方程族的非线性可积耦合.第3节中,在相应李超代数的基础上,通过超迹恒等式得到了超广义Burgers方程族的非线性可积耦合的超Hamilton结构,并且推出了它的几个特殊情况.最后在第4节中,计算出超广义Burgers方程族的非线性可积耦合的Bargmann对称约束.

2 超广义Burgers方程族的非线性可积耦合

2010年杨洪翔等^[30]在李超代数的基础上,利用超迹恒等式^[31],构建了超广义Burgers方程族及其Hamilton结构.其等谱问题如下

$\begin{equation} \phi_x = M\phi = \left(\begin{array}{ccccc} -\lambda+w & \lambda v &\lambda\alpha\\ -1& \lambda-w &\beta\\ \beta&-\lambda\alpha&0 \end{array}\right)\phi, \end{equation}$

(2.1)

其中 $\phi = (\phi_1, \phi_2, \phi_3)^{{\rm T}}, u = (w, v, \alpha, \beta)^{{\rm T}}, \lambda$ 是谱参数, $w, v$ 是偶变量, $\alpha, \beta$ 是奇变量.

现在让我们来将超广义Burgers方程族进行非线性可积耦合.首先引入扩大等谱问题

$\begin{equation} \phi_x = \tilde{M}\phi = \left(\begin{array}{ccccc} -\lambda+w &\ \lambda v\ & q & \lambda s& \lambda\alpha\\ -1&\ \lambda-w\ &0&-q&\beta\\ 0&0&-\lambda+w+q&\ \lambda(v+s)\ &0\\ 0&0&-1&\ \lambda-w-q\ &0\\ \beta &-\lambda\alpha &-\beta &\lambda\alpha &0 \end{array}\right)\phi, \end{equation}$

(2.2)

其中 $\phi = (\phi_1, \phi_2, \phi_3, \phi_4, \phi_5)^{{\rm T}}, \tilde{u} = (w, v, \alpha, \beta, q, s)^{{\rm T}}, \lambda$ 谱参数, $w, v, q, s, \phi_1, \phi_2, \phi_3, \phi_4$ 是偶变量 $\alpha, \beta, \phi_5$ 是奇变量.

由于扩大的零曲率方程

$\begin{equation} \tilde{N}_x = [\tilde{M}, \tilde{N}], \end{equation}$

(2.3)

其中

$\begin{equation} \tilde{N} = \left(\begin{array}{ccccc} A&\lambda B&E&\lambda F&\lambda\rho\\ C&-A&G&-E&\delta\\ 0&0&A+E& \lambda(B+F) & 0\\ 0&0&C+G&-A-E&0\\ \delta& -\lambda\rho & -\delta&\lambda\rho&0\end{array}\right), \end{equation}$

(2.4)

其中 $A, B, C, E, F, G$ 是偶变量, $\rho, \delta$ 是奇变量.

将 $\tilde{M}$ 和 $\tilde{N}$ 代入方程(2.3),我们可以得到如下的结果

$\begin{equation} \left\{\begin{array}{l} A_{x} = \lambda vC+\lambda B+\lambda\alpha\delta+\lambda\beta\rho, ~ B_{x} = -2\lambda B+2wB-2vA-2\lambda\alpha\rho, \\ C_{x} = 2\lambda C-2wC-2A+2\beta\delta, \\ E_{x} = \lambda(v+s)G+\lambda F+\lambda sC-\lambda\alpha\delta-\lambda\beta\rho, \\ F_{x} = -2\lambda F+2(w+q)F-2(v+s)E+2qB-2sA+2\lambda\alpha\rho, \\ G_{x} = 2\lambda G-2(w+q)G-2E-2qC-2\beta\delta, \\ \rho_{x} = -\lambda\rho+w\rho+v\delta-\alpha A-\beta B, ~ \delta_{x} = \lambda\delta-w\delta+\beta A-\lambda\rho-\lambda\alpha C. \end{array}\right. \end{equation}$

(2.5)

如果令

$A = \sum\limits_{m \geq 0}a_m\lambda^{-m}, B = \sum\limits_{m \geq 0}b_m\lambda^{-m}, C = \sum\limits_{m \geq 0}c_m\lambda^{-m}, E = \sum\limits_{m \geq 0}e_m\lambda^{-m}, \\ F = \sum\limits_{m \geq 0}f_m\lambda^{-m}, G = \sum\limits_{m \geq 0}g_m\lambda^{-m}, \rho = \sum\limits_{m \geq 0}\rho_m\lambda^{-m}, \delta = \sum\limits_{m \geq 0}\delta_m\lambda^{-m},$

(2.6)

则通过比较方程(2.5)两边同次数的 $\lambda$ 的系数,我们立即可得

$\begin{equation} \left\{\begin{array}{l} a_{m, x} = vc_{m+1}+b_{m+1}+\alpha\delta_{m+1}+\beta\rho_{m+1}, \\ e_{m, x} = (v+s)g_{m+1}+f_{m+1}+sc_{m+1}-\alpha\delta{m+1}-\beta\rho_{m+1}, \\ b_{m+1} = -\frac{1}{2}b_{m, x}+wb_{m}-va_{m}-\alpha\rho_{m+1}, \\ c_{m+1} = \frac{1}{2}c_{m, x}+wc_{m}+a_{m}-\beta\delta_{m}, \\ f_{m+1} = -\frac{1}{2}f_{m, x}+(w+q)f_{m}-(v+s)e_{m}+qb_{m}-sa_{m}+\alpha\rho_{m+1}, \\ g_{m+1} = \frac{1}{2}g_{m, x}+(w+q)g_{m}+e_{m}+qc_{m}+\beta\delta_{m}, \\ \rho_{m+1} = -\rho_{m, x}+w\rho_{m}+v\delta_{m}-\alpha a{m}-\beta b_{m}, \\ \delta_{m+1} = \delta_{m, x}+w\delta_{m}-\beta a{m}+\rho_{m+1}+\alpha c_{m+1}, \\ \end{array}\right. \end{equation}$

(2.7)

从这些方程,可以得到下面的递推关系式

$\begin{equation} \left\{\begin{array}{l} (4a_n+2e_n, 2c_{n+1}+g_{n+1}, -2\delta_{n+1}, 2\rho_{n+1}, 2a_n+2e_n, c_{n+1}+g_{n+1})^{\rm T}\\ = L( 4a_{n-1}+2e_{n-1}, 2c_n+g_n, -2\delta_n, 2\rho_n, 2a_{n-1}+2e_{n-1}, c_n+g_n)^{\rm T}, \\ a_n = \partial ^{-1}(vc_{n+1}+b_{n+1}+\alpha\delta_{n+1}+\beta\rho_{n+1}), \\ e_n = \partial ^{-1}((v+s)g_{n+1}+f_{n+1}+sc_{n+1}-\alpha\delta{n+1}-\beta\rho_{n+1}), \end{array}\right. \end{equation}$

(2.8)

其中递推算子

$\begin{equation} Ł = \left(\begin{array}{ccc} Ł_1 & L_2 & L_3 \\ L_4& L_5 & -L_4 \\ 0 & 0 & L_1+L_3 \end{array} \right). \end{equation}$

(2.9)

$L_1 = \left(\begin{array}{ccccc} \partial ^{-1}w\partial -\frac{1}{2}\partial & 2\partial ^{-1}v\partial +\partial ^{-1}v_{x} \\ \frac{1}{2}\partial ^{-1}w\partial -\frac{1}{4}\partial & \partial ^{-1}v\partial +\frac{1}{2}\partial ^{-1}v_{x}+w+\frac{1}{2}\partial \end{array}\right), \\ L_2 = \left(\begin{array}{ccccc} -\frac{3}{2}\partial ^{-1}\alpha\partial -\frac{1}{2}\partial ^{-1}\alpha_{x} & -\frac{1}{2}\partial ^{-1}\beta\partial +\frac{1}{2}\partial ^{-1}\beta_{x}\\ -\frac{3}{4}\partial ^{-1}\alpha\partial -\frac{1}{4}\partial ^{-1}\alpha_{x}+\frac{1}{2}\beta & -\frac{1}{4}\partial ^{-1}\beta+\frac{1}{4}\partial ^{-1}\beta_{x} \end{array}\right) , \\ L_3 = \left(\begin{array}{ccccc} \partial ^{-1}q\partial & 2\partial ^{-1}s\partial +\partial ^{-1}s_{x} \\ \frac{1}{2}\partial ^{-1}q\partial & \partial ^{-1}s\partial +\frac{1}{2}\partial ^{-1}s_{x}+q \end{array}\right),$

$L_4 = \left(\begin{array}{ccccc} \beta\partial ^{-1}w\partial +\frac{1}{2}\beta\partial & 2\beta\partial ^{-1}v\partial +\beta\partial ^{-1}v_{x}-2v\beta-\alpha\partial -2w\alpha \\ -\alpha\partial ^{-1}w\partial +\frac{1}{2}\alpha\partial -\beta\partial & -2\alpha\partial ^{-1}v-\alpha\partial ^{-1}v_{x}+2v\beta \end{array}\right), \\ L_5 = \left(\begin{array}{ccccc} \partial +w-\frac{3}{2}\beta\partial ^{-1}\alpha\partial -\frac{1}{2}\beta\partial ^{-1}\alpha_{x}+v-2\alpha\beta\ & \frac{1}{2}\beta\partial ^{-1}\beta_{x}-\frac{1}{2}\beta\partial ^{-1}\beta\partial +\partial -w \\ -v+\frac{1}{2}\alpha\partial ^{-1}\alpha_{x}+\frac{3}{2}\alpha\partial ^{-1}\alpha\partial +\alpha\beta\ & -\partial +w-\frac{1}{2}\alpha\partial ^{-1}\beta_{x}+\frac{1}{2}\alpha\partial ^{-1}\beta\partial \end{array}\right).$

令初始值 $a_0 = 1, e_0 = \epsilon, b_0 = c_0 = e_0 = f_0 = g_0 = \rho_0 = \delta_0 = 0$ ,则所有其它 $a_j, b_j, c_j, \rho_j, \delta_j$ $(j \geq 1)$ 都可以通过递推关系式(2.8)唯一给出.其中前几项分别为

$a_1 = \frac{1}{2}v-\alpha\beta, b_1 = -v, c_1 = 1, f_1 = -(v+s)\epsilon-s, g_1 = \epsilon, \rho_1 = -\alpha, \delta_1 = -\beta, \\ e_1 = \frac{1}{2}(v+s)\epsilon+\frac{1}{2}s+\alpha\beta, \\ a_2 = -\frac{1}{4}v_x-\alpha\beta_x+\alpha_x\beta+\frac{3}{2}\alpha\alpha_x+wv+\frac{3}{8}v^2-\Big(\frac{3}{2}v+2w\Big)\alpha\beta, \\ b_2 = \frac{1}{2}v_x-\alpha\alpha_x-\frac{1}{2}v^2-wv+v\alpha\beta, ~ c_2 = w+\frac{1}{2}v-\alpha\beta, \\ e_2 = \Big[\frac{3}{8}s^2+\frac{3}{4}sv+(q+w)s+qv-\frac{1}{4}s_x\Big](1+\epsilon)+ \Big(wv-\frac{1}{4}v_x+\frac{3}{8}v^2\Big)\epsilon \\ \quad\quad\ +\frac{3}{2}v\alpha\beta+\alpha_x\beta+\alpha\beta_x+2w\alpha\beta-\frac{3}{2}\alpha\alpha_x, \\ f_2 = \frac{1}{2}(v_x+s_x)\epsilon+\frac{1}{2}s_x-(w+q)(v+s)\epsilon-(w+q)s-\frac{1}{2}(v+s)^2\epsilon \\ \quad\quad\ -\frac{1}{2}(v+s)s-v\alpha\beta-qv-\frac{1}{2}sv+\alpha\alpha_x, \\ g_2 = \Big(\frac{1}{2}v+\frac{1}{2}s+w+q\Big)\epsilon+\frac{1}{2}s+q+\alpha\beta, \\ \rho_2 = \alpha_x-w\alpha-\frac{1}{2}v\alpha, ~ \delta_2 = -\beta_x+\alpha_x-w\beta-\frac{1}{2}v\beta, \\ b_3 = -w^2v-\frac{3}{8}v^3+\frac{3}{2}v^2\alpha\beta-\frac{1}{4}v_{xx}+\frac{1}{2}w_xv-\frac{3}{2}wv^2 +\frac{3}{2}v\alpha\beta_x-3v\alpha\alpha_x \\ \quad\quad\ +3wv\alpha\beta-3w\alpha\alpha_x+\frac{3}{2}\alpha\alpha_{xx}+\frac{3}{4}vv_x+wv_x-\frac{3}{2}v\alpha_x\beta, \\ c_3 = \frac{1}{2}w_x+\frac{3}{2}\alpha_x\beta-\frac{3}{2}\alpha\beta_x+w^2+\frac{3}{2}wv-3w\alpha\beta+\frac{3}{2}\alpha\alpha_x+\frac{3}{8}v^2-\frac{3}{2}v\alpha\beta+\beta\beta_x, \\ \rho_3 = -\alpha_{xx}+w_x\alpha+\frac{3}{4}v_x\alpha+2w\alpha_x-w^2\alpha-v\beta_x+\frac{3}{2}v\alpha_x-\frac{3}{2}wv\alpha-\frac{1}{2}v_x\beta-\frac{3}{8}v^2\alpha, \\ \delta_3 = -\beta_{xx}-w_x\beta-2w\beta_x+3w\alpha_x-w^2\beta-\frac{3}{2}wv\beta-\frac{3}{4}v_x\beta -\frac{3}{8}v^2\beta+\frac{3}{4}v_x\alpha \\ \quad\quad\ -\frac{3}{2}v\beta_x+\frac{3}{2}v\alpha_x+\frac{3}{2}w_x\alpha.$

现在,考虑与谱问题(2.2)相关的辅助谱问题 $\phi_{t_n} = \tilde{N}^{(n)}\phi,$ 其中

$\begin{equation} \tilde{N}^{(n)} = \tilde{N}_{+}^{(n)}+\Delta_n = \sum^n_{m = 0}\left(\begin{array}{ccccc} a_m&\lambda b_m&e_m&\lambda f_m&\lambda\rho_m\\ c_m&-a_m&g_m&-e_m&\delta_m\\ 0&0&a_m+e_m&\ \lambda(b_m+f_m)\ &0\\ 0&0&c_m+g_m&-a_m-e_m&0\\ \delta_m&\ -\lambda\rho_m&\ -\delta_m&\lambda\rho_m&0\end{array}\right) \lambda^{n-m} +\Delta_n, \end{equation}$

(2.10)

其中修正项

$\Delta_n = \left(\begin{array}{ccccc} -c_{n+1}&0&-g_{n+1}&0&0\\ 0& c_{n+1} & 0&g_{n+1}&0\\ 0&0&-(c_{n+1}+g_{n+1})&0&0\\ 0&0&0& c_{n+1}+g_{n+1} & 0\\ 0&0&0&0&0\end{array}\right),$

将方程(2.2)和方程(2.10)代入零曲率方程

$\begin{equation} \tilde{M}_{t_n}-\tilde{N}_x^{(n)}+[\tilde{M}, \tilde{N}^{(n)}] = 0, \end{equation}$

(2.11)

其中 $n \geq 0$ .利用方程(2.5),就得到了超广义Burgers方程族的非线性可积耦合

$\begin{equation} {\tilde{u}_{t_n} = \left(\begin{array}{ccccc} w\\ v\\ \alpha\\ \beta\\ q\\ s\end{array}\right)_{t_n} = \left(\begin{array}{ccccc} a_{n, x}-c_{n+1, x}, \\ -2(b_{n+1}+vc_{n+1}+\alpha\rho_{n+1}), \\ -\alpha c_{n+1}-\rho_{n+1}, \\ \delta_{n+1}-\alpha c_{n+1}-\rho_{n+1}+\beta c_{n+1}, \\ e_{n, x}-2g_{n+1, x}, \\ 2(\alpha-\beta)\rho_{n+1}-2e_{n, x}-2\alpha\delta_{n+1} \end{array}\right) = J\left(\begin{array}{ccccc} 4a_n+2e_n\\ 2c_{n+1}+g_{n+1}\\ -2\delta_{n+1}\\ 2\rho_{n+1}\\ 2a_n+2e_n\\ c_{n+1}+g_{n+1}\end{array}\right), n \geq 0}, \end{equation}$

(2.12)

其中

$J = \left(\begin{array}{ccccccccc} \frac{1}{2}\partial &-\partial &0 & 0& -\frac{1}{2}\partial &\partial \\ -\partial &0& -\alpha& \beta-\alpha & \partial &0 \\ 0&-\alpha& 0& -\frac{1}{2} &0&\alpha\\ 0& \beta-\alpha & -\frac{1}{2}& -\frac{1}{2}&0&\alpha-\beta\\ -\frac{1}{2}\partial &\partial &0&0&\partial &-2\partial \\ \partial &0&\alpha& \alpha-\beta &-2\partial &0 \end{array}\right).$

3 超Hamilton结构

根据超迹恒等式^[32]

$\begin{equation} \frac{\delta}{\delta \tilde{u}}\int{\rm Str}\left(\tilde{N}\frac{\partial \tilde{M}}{\partial \lambda}\right){\rm d}x = \left(\lambda^{-\gamma}\frac{\partial}{\partial\lambda}\lambda^{\gamma}\right){\rm Str}\left(\frac{\partial \tilde{M}}{\partial \tilde{u}}\tilde{N}\right), \end{equation}$

(3.1)

其中 ${\rm Str}$ 表示超迹.我们有

${\rm Str}\left(\tilde{N}\frac{\partial \tilde{M}}{\partial \lambda}\right) = -4A+2vC-2E+sC+(v+s)G+2\alpha\delta, \\ {\rm Str}\left(\frac{\partial \tilde{M}}{\partial w}\tilde{N}\right) = 4A+2E, ~ {\rm Str}\left(\frac{\partial \tilde{M}}{\partial v}\tilde{N}\right) = 2\lambda C+\lambda G, ~ {\rm Str}\left(\frac{\partial \tilde{M}}{\partial \alpha}\tilde{N}\right) = -2\lambda\delta, \\ {\rm Str}\left(\frac{\partial \tilde{M}}{\partial \beta}\tilde{N}\right) = 2\lambda\rho, ~ {\rm Str}\left(\frac{\partial \tilde{M}}{\partial q}\tilde{N}\right) = 2A+2E, ~ {\rm Str}\left(\frac{\partial \tilde{M}}{\partial s}\tilde{N}\right) = \lambda C+\lambda G.$

(3.2)

把方程(3.2)代入方程(3.1),比较方程(3.1)两端 $\lambda^{-n-1}$ 的系数,可以得到

$\begin{equation} \frac{\delta}{\delta \tilde{u}}\int(-4a_{n+1}+2vc_{n+1}-2e{n+1}+sc_{n+1}+(v+s)g_{n+1}+2\alpha\delta_{n+1}){\rm d}x\\ = (\gamma-n)\left(\begin{array}{ccccc} 4a_{n+1}+2e_{n+1}\\ 2c_{n+1}+g_{n+1}\\ -2\delta_{n+1}\\ 2\rho_{n+1}\\ 2a_{n+1}+2e_{n+1}\\ c_{n+1}+g_{n+1} \end{array}\right). \end{equation}$

(3.3)

取 $n = 0$ ,可得 $\gamma = 0$ .于是

$\begin{equation} \frac{\delta \tilde{H}_{n+1}}{\delta \tilde{u}} = \left(\begin{array}{ccccc} 4a_n+2e_n\\ 2c_{n+1}+g_{n+1}\\ -2\delta_{n+1}\\ 2\rho_{n+1}\\ 2a_n+2e_n\\ c_{n+1}+g_{n+1} \end{array}\right), \end{equation}$

(3.4)

其中 $\tilde{H}_{n+1} = -\frac{2}{n}\int(-4a_{n+1}+2vc_{n+1}-2e_{n+1}+sc_{n+1}+(v+s)g_{n+1}+2\alpha\delta_{n+1}){\rm d}x$ .则超广义Burgers方程族的非线性可积耦合(2.15)具有下面的超Hamilton形式

$\begin{equation} {\tilde{u}_{t_n} = K_n = J\frac{\delta \tilde{H}_n}{\delta \tilde{u}}}. \end{equation}$

(3.5)

当 $n = 2$ 时,超广义Burgers方程族的非线性可积耦合(2.15)的第1个非平凡的非线性方程组是其第2个流,即

$\begin{equation} \left\{ \begin{array}{l} w_{t_2} = -\frac{1}{2}w_{xx}-\frac{1}{4}v_{xx}-2ww_x-\frac{1}{2}(wv)_x+\frac{1}{2}(\alpha\beta)_{xx}+(w\alpha\beta)_x-[(\alpha_x-\beta_x)\beta]_x, \\ v_{t_2} = \frac{1}{2}v_{xx}-2(wv)_x-\frac{3}{2}vv_x+(v\alpha\beta)_x-\alpha\alpha_{xx}+2w(\alpha\alpha_x+\beta\beta_x+\alpha_x\beta), \\ \alpha_{t_2} = \alpha_{xx}-\frac{1}{2}(w\alpha)_x-\frac{3}{2}w_x\alpha-\frac{3}{2}(v\alpha)_x+\frac{1}{2}(v\beta)_x+\frac{1}{2}v\beta_x+\frac{3}{4}v_x\alpha+\frac{1}{2}\alpha_x\beta\\ \ \quad\quad+\alpha\beta(2\alpha_x-\beta_x), \\ \beta_{t_2} = \alpha_{xx}-\beta_{xx}-(w\beta)_x-\frac{1}{2}(v\beta)_x+\frac{1}{2}w_x\beta+\frac{1}{4}v_x\beta-\frac{1}{2}\beta(\alpha\beta)_x+w(\alpha-\beta)_x, \\ q_{t_2} = (1+\epsilon)\Big[\frac{1}{4}s_x+\frac{1}{2}q_x+(\frac{1}{2}v+w)q+(\frac{1}{2}s+q)(w+q)\Big]_x \\ \ \quad\quad+(\frac{1}{4}v_x+\frac{1}{2}w_x+\frac{1}{2}wv+w^2)_x\epsilon+(\beta\beta_x-w\alpha\beta-\frac{1}{2}\alpha_x\beta+\frac{1}{2}\alpha\beta_x)_x, \\ s_{t_2} = (1+\epsilon)\Big[\frac{1}{2}s_x-\frac{3}{4}s^2-\frac{3}{2}sv-2(w+q)s-2qv\Big]_x+(\frac{1}{2}v_x-2wv-\frac{3}{4}v^2)_x\epsilon\\ \ \quad\quad+2v\beta\beta_x+\alpha\alpha_{xx}-2w\alpha\alpha_x-v_x\alpha\beta-2v\alpha\beta_x. \end{array} \right. \end{equation}$

(3.6)

4 Bargmann对称约束

为了计算超广义Burgers方程族的非线性耦合的Bargmann对称约束,我们考虑引进谱问题(2.2)的共轭谱问题

$\begin{equation} \psi_x = -\tilde{M}^{{\rm St}}\psi\\ = \left(\begin{array}{ccccc} \lambda-w&1&0&0&\beta\\ -\lambda v& -\lambda+w & 0&0&-\lambda\alpha\\ -q&0&\lambda-w-q&1&-\beta\\ -\lambda s&q&-\lambda v-\lambda s& -\lambda+w+q & \lambda\alpha\\ -\lambda\alpha&-\beta&0&0&0 \end{array}\right)\psi, \end{equation}$

(4.1)

其中 ${\rm St}$ 表示对矩阵取超转置, $\psi$ 表示共轭特征函数.

由参考文献[33]里的引理,不难得到谱参数 $\lambda$ 关于位势 $\tilde{u}$ 的变分导数

$\begin{equation} \frac{\delta\lambda}{\delta \tilde{u}} = \frac{1}{E}\left(\begin{array}{ccccc} \psi_1\phi_1-\psi_2\phi_2+\psi_3\phi_3-\psi_4\phi_4\\ \lambda(\psi_1\phi_2+\psi_3\phi_4)\\ \lambda(\psi_5\phi_2-\psi_5\phi_4+\psi_1\phi_5)\\ -\psi_5\phi_1+\psi_2\phi_3+\psi_2\phi_5\\ (\psi_1+\psi_3)\phi_3-(\psi_2+\psi_4)\phi_4\\ \lambda(\psi_1+\psi_3)\phi_4\end{array}\right), \end{equation}$

(4.2)

其中 $E = \int(\psi_1\phi_1-v\psi_1\phi_2-\psi_2\phi_2-\alpha\psi_5\phi_2+\psi_3\phi_3-s\psi_1\phi_4-(v+s)\psi_3\phi_4-\psi_4\phi_4+\psi_5\phi_4-\alpha\psi_1\phi_5){\rm d}x.$ 取 $\tilde{N}$ 个不同的谱参数 $\lambda_j, j = 1, 2, \cdots, N$ ,我们考虑下面的谱问题

$\begin{equation} \left(\begin{array}{ccccc} \phi_{1j}\\ \phi_{2j}\\ \phi_{3j}\\ \phi_{4j}\\ \phi_{5j}\end{array}\right)_x = \left(\begin{array}{ccccc} -\lambda_j+w&\lambda v_j&q&\lambda_js&\lambda_j\alpha\\ -1&\lambda_j-w&0&-q&\beta\\ 0&0&-\lambda_j+w+q&\lambda_j(v+s)&0\\ 0&0&-1&\lambda_j-w-q&0\\ \beta&-\lambda_j\alpha&-\beta&\lambda_j\alpha&0 \end{array}\right)\left(\begin{array}{ccccc} \phi_{1j}\\ \phi_{2j}\\ \phi_{3j}\\ \phi_{4j}\\ \phi_{5j}\end{array}\right), \end{equation}$

(4.3)

$\begin{equation} \left(\begin{array}{ccccc} \phi_{1j}\\ \phi_{2j}\\ \phi_{3j}\\ \phi_{4j}\\ \phi_{5j}\end{array}\right)_t = \left(\begin{array}{ccccc} A&\lambda B&E&\lambda F&\lambda\rho\\ C&-A&G&-E&\delta\\ 0&0&A+E& \lambda(B+F) &0\\ 0&0&C+G&-A-E&0\\ \delta& -\lambda\rho & -\delta&\lambda\rho&0 \end{array}\right)\left(\begin{array}{ccccc} \phi_{1j}\\ \phi_{2j}\\ \phi_{3j}\\ \phi_{4j}\\ \phi_{5j}\end{array}\right), \end{equation}$

(4.4)

以及它的共轭谱问题

$\begin{equation} \left(\begin{array}{ccccc} \psi_{1j}\\ \psi_{2j}\\ \psi_{3j}\\ \psi_{4j}\\ \psi_{5j}\end{array}\right)_x = \left(\begin{array}{ccccc} \lambda_j-w&1&0&0&\beta\\ -\lambda_j v&-\lambda_j+w&0&0&-\lambda_j\alpha\\ -q&0&\lambda_j-w-q&1&-\beta\\ -\lambda_j s&q&-\lambda_j v-\lambda_j s&-\lambda_j+w+q&\lambda_j\alpha\\ -\lambda_j\alpha&-\beta&0&0&0 \end{array}\right)\left(\begin{array}{ccccc} \phi_{1j}\\ \phi_{2j}\\ \phi_{3j}\\ \phi_{4j}\\ \phi_{5j}\end{array}\right), \end{equation}$

(4.5)

$\begin{equation} \left(\begin{array}{ccccc} \psi_{1j}\\ \psi_{2j}\\ \psi_{3j}\\ \psi_{4j}\\ \psi_{5j}\end{array}\right)_t = \left(\begin{array}{ccccc} -A&-C&0&0&\delta\\ -\lambda B&A&0&0&-\lambda\rho\\ -E&-G&-A-E& -C-G &-\delta\\ -\lambda F&E&-\lambda(B+F)&A+E&\lambda\rho\\ -\lambda\rho& -\delta &0&0&0 \end{array}\right)\left(\begin{array}{ccccc} \psi_{1j}\\ \psi_{2j}\\ \psi_{3j}\\ \psi_{4j}\\ \psi_{5j}\end{array}\right). \end{equation}$

(4.6)

根据参考文献[34],有

$\begin{equation} \frac{\delta \tilde{H_k}}{\delta \tilde{u}}+\sum\limits^N_{j = 1}E_j\frac{\delta\lambda_j}{\delta \tilde{u}} = 0. \end{equation}$

(4.7)

于是我们就得到了下面的Bargmann对称约束

$\begin{equation} \tilde{u}_{t_n} = J\frac{\delta \tilde{H_n}}{\delta \tilde{u}}+J\sum\limits^N_{j = 1}\frac{\delta\lambda_j}{\delta \tilde{u}}. \end{equation}$

(4.8)

当 $n = 2$ 时,有

$\begin{equation} \left\{\begin{array}{l} w_{t_{2}} = -\frac{1}{2}w_{xx}-\frac{1}{4}v_{xx}-2ww_x-\frac{1}{2}(wv)_x+\frac{1}{2}(\alpha\beta)_{xx} +(w\alpha\beta)_x-[(\alpha_x-\beta_x)\beta]_x\\ +\frac{1}{2}\partial (\langle\Psi_1, \Phi_1\rangle-\langle\Psi_2, \Phi_2\rangle-2\Lambda\langle\Psi_1, \Phi_2\rangle-\langle\Psi_1, \Phi_3\rangle+\langle\Psi_2, \Phi_4\rangle+2\Lambda\langle\Psi_1, \Phi_4\rangle), \\ v_{t_{2}} = \frac{1}{2}v_{xx}-2(wv)_x-\frac{3}{2}vv_x+(v\alpha\beta)_x-\alpha\alpha_{xx}+2w(\alpha\alpha_x+\beta\beta_x+\alpha_x\alpha)\\ ~+\partial (\langle\Psi_1, \Phi_3\rangle-\langle\Psi_2, \Phi_4\rangle-\langle\Psi_4, \Phi_4\rangle-\langle\Psi_1, \Phi_1\rangle+\langle\Psi_2, \Phi_2\rangle+\langle\Psi_4, \Phi_3\rangle)\\ ~-\alpha\Lambda(\langle\Psi_5, \Phi_2\rangle-\langle\Psi_2, \Phi_5\rangle+\langle\Psi_1, \Phi_5\rangle)+(\beta-\alpha)(\langle\Psi_2, \Phi_3\rangle+\langle\Psi_2, \Phi_5\rangle -\langle\Psi_5, \Phi_1\rangle), \\ \alpha_{t_2} = \alpha_{xx}-\frac{1}{2}(w\alpha)_x-\frac{3}{2}w_x\alpha-\frac{3}{2}(v\alpha)_x+\frac{1}{2}(v\beta)_x+\frac{1}{2}v\beta_x+\frac{3}{4}v_x\alpha+\frac{1}{2}\alpha_x\beta\\ ~+\alpha\beta(2\alpha_x-\beta_x)+\alpha\Lambda(\langle\Psi_1, \Phi_4\rangle-\langle\Psi_1, \Phi_2\rangle)-\frac{1}{2}(\langle\Psi_2, \Phi_3\rangle+\langle\Psi_2, \Phi_5\rangle -\langle\Psi_5, \Phi_1\rangle), \\ \beta_{t_2} = \alpha_{xx}-\beta_{xx}-(w\beta)_x-\frac{1}{2}(v\beta)_x+\frac{1}{2}w_x\beta+\frac{1}{4}v_x\beta-\frac{1}{2}\beta(\alpha\beta)_x+w(\alpha-\beta)_x\\ ~+(\alpha-\beta)\Lambda(\langle\Psi_1, \Phi_4\rangle-\langle\Psi_1, \Phi_2\rangle)-\frac{1}{2}\Lambda(\langle\Psi_5, \Phi_2\rangle-\langle\Psi_5, \Phi_4\rangle+\langle\Psi_1, \Phi_5\rangle)\\ ~-\frac{1}{2}(\langle\Psi_2, \Phi_3\rangle-\langle\Psi_5, \Phi_1\rangle+\langle\Psi_2, \Phi_5\rangle), \\ q_{t_2} = (1+\epsilon)\Big[\frac{1}{4}s_x+\frac{1}{2}q_x+(\frac{1}{2}v+w)q+(\frac{1}{2}s+q)(w+q)\Big]_x +(\frac{1}{4}v_x+\frac{1}{2}w_x+\frac{1}{2}wv+w^2)_x\epsilon\\ ~+(\beta\beta_x-w\alpha\beta-\frac{1}{2}\alpha_x\beta+\frac{1}{2}\alpha\beta_x)_x +\partial \Lambda(\langle\Psi_1, \Phi_2\rangle-\langle\Psi_3, \Phi_4\rangle-2\langle\Psi_1, \Phi_4\rangle)\\ ~ +\partial (\langle\Psi_1, \Phi_3\rangle+\frac{1}{2}\langle\Psi_3, \Phi_3\rangle-\langle\Psi_2, \Phi_4\rangle -\frac{1}{2}\langle\Psi_4, \Phi_4\rangle-\frac{1}{2}\langle\Psi_1, \Phi_1\rangle +\frac{1}{2}\langle\Psi_2, \Phi_2\rangle), \\ s_{t_2} = (1+\epsilon) \Big[\frac{1}{2}s_x-\frac{3}{4}s^2-\frac{3}{2}sv-2(w+q)s-2qv\Big]_x+(\frac{1}{2}v_x-2wv-\frac{3}{4}v^2)_x\epsilon\\ ~+2v\beta\beta_x+\alpha\alpha_{xx}-2w\alpha\alpha_x-v_x\alpha\beta-2v\alpha\beta_x+\alpha \Lambda(\langle\Psi_2, \Phi_2\rangle-\langle\Psi_5, \Phi_4\rangle\\ ~+\langle\Psi_1, \Phi_5\rangle)+(\alpha-\beta)(\langle\Psi_2, \Phi_3\rangle+\langle\Psi_2, \Phi_5\rangle-\langle\Psi_5, \Phi_1\rangle)-\partial (\langle\Psi_1, \Phi_1\rangle\\ ~-\langle\Psi_2, \Phi_2\rangle-2\langle\Psi_1, \Phi_3\rangle-\langle\Psi_3, \Phi_3\rangle+2\langle\Psi_2, \Phi_4\rangle+\langle\Psi_4, \Phi_4\rangle), \end{array}\right. \end{equation}$

(4.9)

其中 $\Phi_i = (\phi_{i1}, \phi_{i2}, \cdots, \phi_{iN})^{\rm T}, \Psi_i = (\psi_{i1}, \psi_{i2}, \cdots, \psi_{iN}))^{\rm T}, i = 1, 2, 3, 4, 5$ , $\langle, \rangle$ 表示欧式空间 $\mathbb{R}^N$ 中的标准内积.

在(4.9)式中,令 $\epsilon = 0, w = -q, v = -s, \Phi_1 = -\Phi_3, \Phi_2 = -\Phi_4$ ,可以得到

$\left\{\begin{array}{l} w_{t_{2}} = -\frac{1}{2}w_{xx}-\frac{1}{4}v_{xx}-2ww_x-\frac{1}{2}(wv)_x+\frac{1}{2}(\alpha\beta)_{xx}+(w\alpha\beta)_x-[(\alpha_x-\beta_x)\beta]_x\\ +\frac{1}{2}\partial (\langle\Psi_1, \Phi_1\rangle-\langle\Psi_2, \Phi_2\rangle-2\Lambda\langle\Psi_1, \Phi_2\rangle-\langle\Psi_1, \Phi_3\rangle+\langle\Psi_2, \Phi_4\rangle+2\Lambda\langle\Psi_1, \Phi_4\rangle), \\ v_{t_{2}} = \frac{1}{2}v_{xx}-2(wv)_x-\frac{3}{2}vv_x+(v\alpha\beta)_x-\alpha\alpha_{xx}+2w(\alpha\alpha_x+\beta\beta_x+\alpha_x\alpha)\\ ~+\partial (\langle\Psi_1, \Phi_3\rangle-\langle\Psi_2, \Phi_4\rangle-\langle\Psi_4, \Phi_4\rangle-\langle\Psi_1, \Phi_1\rangle+\langle\Psi_2, \Phi_2\rangle+\langle\Psi_4, \Phi_3\rangle)\\ ~-\alpha\Lambda(\langle\Psi_5, \Phi_2\rangle-\langle\Psi_2, \Phi_5\rangle+\langle\Psi_1, \Phi_5\rangle)+(\beta-\alpha)(\langle\Psi_2, \Phi_3\rangle+\langle\Psi_2, \Phi_5\rangle -\langle\Psi_5, \Phi_1\rangle), \\ \alpha_{t_2} = \alpha_{xx}-\frac{1}{2}(w\alpha)_x-\frac{3}{2}w_x\alpha-\frac{3}{2}(v\alpha)_x+\frac{1}{2}(v\beta)_x+\frac{1}{2}v\beta_x+\frac{3}{4}v_x\alpha+\frac{1}{2}\alpha_x\beta\\ ~+\alpha\beta(2\alpha_x-\beta_x)+\alpha\Lambda(\langle\Psi_1, \Phi_4\rangle-\langle\Psi_1, \Phi_2\rangle)-\frac{1}{2}(\langle\Psi_2, \Phi_3\rangle+\langle\Psi_2, \Phi_5\rangle -\langle\Psi_5, \Phi_1\rangle), \\ \beta_{t_2} = \alpha_{xx}-\beta_{xx}-(w\beta)_x-\frac{1}{2}(v\beta)_x+\frac{1}{2}w_x\beta+\frac{1}{4}v_x\beta-\frac{1}{2}\beta(\alpha\beta)_x+w(\alpha-\beta)_x\\ ~+(\alpha-\beta)\Lambda(\langle\Psi_1, \Phi_4\rangle-\langle\Psi_1, \Phi_2\rangle)-\frac{1}{2}\Lambda(\langle\Psi_5, \Phi_2\rangle-\langle\Psi_5, \Phi_4\rangle+\langle\Psi_1, \Phi_5\rangle)\\ ~-\frac{1}{2}(\langle\Psi_2, \Phi_3\rangle-\langle\Psi_5, \Phi_1\rangle+\langle\Psi_2, \Phi_5\rangle). \end{array}\right.$

(4.10)

5 小结

在这篇文章中,我们根据李超代数不仅建立了超广义Burgers方程族的非线性耦合,而且还根据超迹恒等式计算出了它的超Hamilton结构.另外,通过相对应的共轭谱问题计算出超广义Burgers方程族的非线性耦合的Bargmann对称约束.

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A super AKNS scheme

1985

... 随着孤立子理论的发展,超可积方程引起了大量学者的研究兴趣,各个专业领域已有很多成果^[1-10].近几年来,超可积方程族的非线性可积耦合成为当今数学领域和物理邻域的热点问题.可积耦合这一概念首先是由Fuchssteiner^[11]在研究可积系统的无中心Virasoro对称代数和孤立子时提出来的.学者们研究出了一些经典孤立子方程的可积耦合的方法,例如: AKNS方程族, Levi方程族, DLW方程族等^[12-13].马教授和朱教授在文章^[16-17]中首次提出非线性可积耦合这一概念,并且提供了建立孤立子方程族的非线性可积耦合的一般方法.利用这个方法几个超可积方程族的非线性可积耦合就被研究出来了,其中包括超BKK方程族,超经典Boussinesq方程族和Levi方程族^[18-21]. ...

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1986

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1984

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2008

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2011

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2014

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2017

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2017

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2014

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2007

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2008

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2010

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2010

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2011

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2010

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2013

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2015

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2014

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2018

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1988

... 在对可积系统研究的过程中,人们发现通过Bargmann对称约束的计算可以揭示许多可积系统之间存在的内在联系. 1988年,曹策问教授通过谱问题中位势函数与特征函数之间的一种约束条件,将原有谱问题约束成为有限维Hamilton系统^[22].例如CKdV方程族相关的谱问题在Bargmann对称约束下分别被非线性化为完全可积的Bargmann系统^[23].另外,还有一些超可积方程族的非线性化也被研究过^[24-29]. ...

AKN族的Lax组的非线性化

1989

AKN族的Lax组的非线性化

1989

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1989

Nolinearization of the Lax system for AKNS hierarchy

1990

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1990

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1991

A classical integrable system and the involutive representation of solutions of the KdV equation

1991

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2010

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1997

... 2010年杨洪翔等^[30]在李超代数的基础上,利用超迹恒等式^[31],构建了超广义Burgers方程族及其Hamilton结构.其等谱问题如下 ...

Hamiltonian and Super-Hamiltonian systems of a hierarchy of soliton equations

2010

... 2010年杨洪翔等^[30]在李超代数的基础上,利用超迹恒等式^[31],构建了超广义Burgers方程族及其Hamilton结构.其等谱问题如下 ...

A supertrance identity and its applications to superintegrable systems

2008

... 根据超迹恒等式^[32] ...

Binary nonlinearization of the super AKNS system under an implicit symmetry constraint

2009

... 由参考文献[33]里的引理,不难得到谱参数

$\lambda$

关于位势

$\tilde{u}$

的变分导数 ...

Integration of the soliton hierarchy with self-consistent sources

2000

... 根据参考文献[34],有 ...

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