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.

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

(2.1) $ \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} $

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

(2.2) $ \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} $

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

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

其中

(2.4) $ \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} $

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

(2.5) $ \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} $

如果令

(2.6) $ 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}, $

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

(2.7) $ \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} $

从这些方程,可以得到下面的递推关系式

(2.8) $ \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} $

其中递推算子

(2.9) $ \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} $

$ 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

(2.10) $ \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} $

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

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

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

(2.12) $ \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} $

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). $

根据超迹恒等式^[32]

(3.1) $ \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} $

其中$ {\rm Str} $表示超迹.我们有

(3.2) $ {\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. $

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

(3.3) $ \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} $

取$ n = 0 $,可得$ \gamma = 0 $.于是

(3.4) $ \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} $

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

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

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

(3.6) $ \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} $

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

(4.1) $ \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} $

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

(4.2) $ \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} $

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

(4.3) $ \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} $

(4.4) $ \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} $

以及它的共轭谱问题

(4.5) $ \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} $

(4.6) $ \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} $

根据参考文献[34],有

(4.7) $ \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} $

于是我们就得到了下面的Bargmann对称约束

(4.8) $ \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} $

当$ n = 2 $时,有

(4.9) $ \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} $

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

(4.10) $ \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. $

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.