近年来,随着可积系统的发展,超可积系统越来被重视.自从1986年, Chowdhury和Roy开始考虑超迹恒等式^[1], 1997年胡星标给出了超迹恒等式,但他没有给出其严格的证明^[2].后来马文秀给出了超迹恒等式的系统证明和常数$ \gamma $的计算公式^[3],并作为应用给出了超AKNS族和超Dirac族的超Hamilton结构.之后,很多超可积族及其超Hamilton结构被给出^[4-9].在文献[10]中给出了超AKNS族在隐式对称约束下的双非线性化,它在超可积系统发展中具有重要意义.

带自相容源的孤子方程越来越被重视,它的研究在物理方面具有重要应用.例如,带自相容源非线性Schrödinger方程描述了等离子物理和固体物理的相关性质.现已有多种方法给出带自相容源孤子方程的精确解,例如利用反散射变换方法^[11]、对称约束导出方法^[12].带自相容源方程族已有一些研究成果^[13-15],近年来,超JM族和广义超AKNS族的自相容源也被给出^[16-17].

守恒律在考虑孤子族的可积性方面起了重要作用. 1968年, Miura等人首次发现了KdV方程的无穷守恒律^[18].之后,人们寻找了多种方法求方程的守恒律,例如:直接通过Lax对方法^[19]、通过特征函数形式解和Bäcklund变换方法^[19]、Lagrangian方法等^[20-21].

本文安排如下:第2节,给出一个新的广义超孤子族;第3节,利用超迹恒等式给出所得广义超孤子族的超双- Hamilton结构;第4节,给出广义超孤子族的自相容源;第5节,给出广义超孤子族的无穷守恒律;最后,给出一些结论和探讨.

基于Lie超代数$ B(0, 1) $一组基

(2.1) $ \begin{eqnarray} && e_1 = \frac{1}{2}\left( \begin{array}{ccc} 1&{\quad} 0{\quad} &0 \\ 0&-1&0\\ 0&0&0\\ \end{array} \right), {\quad} e_2 = \frac{1}{2}\left( \begin{array}{ccc} 0 &{\quad} 1{\quad}&0 \\ 1&0&0 \\ 0&0&0\\ \end{array} \right), {\quad} e_3 = \frac{1}{2}\left( \begin{array}{ccc} 0&{\quad} 1{\quad}&0 \\ -1&0&0 \\ 0&0&0\\ \end{array} \right), {}\\ && e_4 = \frac{1}{2}\left( \begin{array}{ccc} 0 &{\quad} 0{\quad} &1\\ 0&0&0 \\ 0&-1&0\\ \end{array} \right), {\quad} e_5 = \frac{1}{2}\left( \begin{array}{ccc} 0 &{\quad} 0{\quad}&0\\ 0&0&1 \\ 1&0&0\\ \end{array} \right), \end{eqnarray} $

其中$ e_1, e_2, e_3 $是偶元, $ e_4, e_5 $是奇元,它们满足如下的交换关系$ [e_1, e_2] = e_3, [e_1, e_3] = e_2, $$ [e_1, e_4] = [e_2, e_5] = [e_3, e_5] = \frac{1}{2}e_4, [e_2, e_4] = [e_5, e_1] = [e_4, e_3] = \frac{1}{2}e_5, $$ [e_2, e_3] = -e_1, $$ [e_4, e_4]_+ = -\frac{1}{2}(e_2+e_3), $$ [e_4, e_5]_+ = \frac{1}{2}e_1, [e_5, e_5]_+ = \frac{1}{2}(e_2-e_3). $

受以前工作启发^{[7, 22]},考虑如下的谱问题

(2.2) $ \begin{eqnarray} \left\{ \begin{array}{ll} \varphi_x = U\varphi, \\ \varphi_t = V\varphi, \\ \end{array}\right. \end{eqnarray} $

其中

$ U = \frac{1}{2}\left( \begin{array}{ccc} \lambda^{-1}+w&{\quad} u_1+u_2{\quad} &u_3\\ u_1-u_2&-\lambda^{-1}-w &u_4\\ u_4&-u_3&0\\ \end{array} \right), {\quad} V = \frac{1}{2}\left( \begin{array}{ccc} A& {\quad} B+C{\quad} &\delta\\ B-C&-A&\rho \\ \rho&-\delta&0\\ \end{array} \right), $

这里$ w = \mu(u_{1}^{2}-u_{2}^{2}+2u_3u_4) $, $ \mu $是常数, $ \lambda $是谱参数, $ u_1 $和$ u_2 $是偶变量, $ u_3 $和$ u_4 $是奇变量.很显然,在谱问题(2.2)中令$ \mu = 0 $,则可约化成标准的超孤子族情形^[23].

设

(2.3) $ \begin{equation} 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}, \delta = \sum\limits_{m\geq 0} \delta_m\lambda^{m}, \rho = \sum\limits_{m\geq 0} \rho_m\lambda^{m}. \end{equation} $

利用驻定零曲率方程$ V_{x} = [U, V], $并比较两端$ \lambda $同次幂系数,得

(2.4) $ \begin{eqnarray} &&a_{mx} = u_2b_m-u_1c_m+\frac{1}{2}u_3\rho_m+\frac{1}{2}u_4\delta_m, {}\\ &&b_{mx} = c_{m+1}-u_2a_m-\frac{1}{2}u_3\delta_m+\frac{1}{2}u_4\rho_m+wc_m, {}\\ &&c_{mx} = b_{m+1}-u_1a_m-\frac{1}{2}u_3\delta_m-\frac{1}{2}u_4\rho_m+wb_m, \\ &&\delta_{mx} = \frac{1}{2}\delta_{m+1}+\frac{1}{2}(u_1+u_2)\rho_m-\frac{1}{2}u_3a_m-\frac{1}{2}u_4(b_m+c_m)+\frac{1}{2}w\delta_m, {}\\ &&\rho_{mx} = -\frac{1}{2}\rho_{m+1}+\frac{1}{2}(u_1-u_2)\delta_m-\frac{1}{2}u_3(b_m-c_m)+\frac{1}{2}u_4a_m-\frac{1}{2}w\rho_m, {} \end{eqnarray} $

于是得到递归关系

(2.5) $ \begin{equation} \left(\begin{array}{c} b_{m+1}\\ c_{m+1}\\ \rho_{m+1}\\ \delta_{m+1} \end{array}\right) = L\left(\begin{array}{c} b_{m}\\ -c_{m}\\ -\rho_{m}\\ -\delta_{m} \end{array}\right), \quad m\geqslant 0, \end{equation} $

这里递归算子$ L $如下

$ L = \left(\begin{array}{cccc} u_1\partial^{-1}u_2-w&\partial-u_1\partial^{-1}u_1& { } \frac{1}{2}u_1\partial^{-1}u_3+\frac{1}{2}u_4&{ }\ \frac{1}{2} u_1\partial^{-1}u_4+\frac{1}{2}u_3\\ \partial+u_2\partial^{-1}u_2&- u_2\partial^{-1}u_1-w& { } \frac{1}{2}u_2\partial^{-1}u_3-\frac{1}{2}u_4&{ }\ \frac{1}{2} u_2\partial^{-1}u_4+\frac{1}{2}u_3\\ -u_3 +u_4\partial^{-1}u_2&u_3- u_4\partial^{-1}u_1& { } -2\partial+\frac{1}{2}\partial^{-1}u_3-w&\ { } u_1-u_2+\frac{1}{2} u_4\partial^{-1}u_4\\ u_4+u_3\partial^{-1}u_2&\ u_4- u_3\partial^{-1}u_1\ &{ } -u_1-u_2+\frac{1}{2}u_3\partial^{-1}u_3&\ { } 2\partial+\frac{1}{2} u_3\partial^{-1}u_4-w \end{array}\right). $

取初值

(2.6) $ \begin{equation} a_0 = 1, \quad b_0 = c_0 = \delta_0 = \rho_0 = 0. \end{equation} $

由递归关系(2.4),可得前几组如下

(2.7) $ \begin{eqnarray} & &a_1 = 0, b_1 = u_1, c_1 = u_2, \delta_1 = u_3, \rho_1 = u_4, a_2 = \frac{1}{2}u_2^{2}-\frac{1}{2}u_1^{2}-u_3u_4, {}\\ &&b_2 = u_{2x}-wu_1, c_2 = u_{1x}-wu_2, \delta_{2} = 2u_{3x}-wu_3, \rho_2 = -2u_{4x}-wu_4, {}\\ &&a_3 = u_2u_{1x}-u_1u_{2x}+2(u_4u_{3x}-u_{4x}u_3)+w(u_1^{2}-u_2^{2}+2u_3u_4), {}\\ &&b_3 = u_{1xx}+\frac{1}{2}u_1u_{2}^{2}-\frac{1}{2}u_1^{3}-u_1u_3u_4+u_3u_{3x}-u_4u_{4x}-w_xu_2-2wu_{2x}+w^{2}u_1, \\ &&c_3 = u_{2xx}+\frac{1}{2}u_{2}^{3}-\frac{1}{2}u_2u_1^{2}-u_2u_3u_4+u_3u_{3x}+u_4u_{4x}-w_xu_1-2wu_{1x}+w^{2}u_2, {}\\ &&\delta_3 = 4u_{3xx}+\frac{1}{2}u_3u_{2}^{2}-\frac{1}{2}u_3u_1^{2}+u_4u_{1x}+u_4u_{2x}+2u_1u_{4x}+2u_2u_{4x}-2w_xu_3-4wu_{3x}+w^{2}u_3, {}\\ &&\rho_3 = 4u_{4xx}+\frac{1}{2}u_4u_{2}^{2}-\frac{1}{2}u_4u_1^{2}+u_3u_{1x}-u_3u_{2x}+2u_1u_{3x}-2u_2u_{3x}+2w_xu_4+4wu_{4x}+w^{2}u_4.{} \end{eqnarray} $

考虑辅助谱问题

(2.8) $ \begin{equation} \varphi_{t_{n}} = V^{(n)}\varphi, \end{equation} $

其中

$ V^{(n)} = \sum\limits_{m = 0}^{n}\frac{1}{2}\left( \begin{array}{ccc} a_m&{\quad} b_m+c_m{\quad} &\delta_m\\ b_m-c_m&-a_m&\rho_m \\ \rho_m& -\delta_m&0\\ \end{array} \right)\lambda^{-n+m}+\Delta_n, \quad n\geq0, $

这里$ \Delta_n $是修正项,但在标准的超孤子族^[23]中不会出现.令

(2.9) $ \begin{equation} \Delta_n = \frac{1}{2}\left( \begin{array}{ccc} a& {\quad} b+c{\quad} &e\\ b-c&-a&f \\ f& -e&0\\ \end{array} \right), \end{equation} $

谱问题(2.2)和(2.8)的相容性导出了下面的零曲率方程

(2.10) $ \begin{equation} U_{t_{n}}-V^{(n)}_{x}+[U, V^{(n)}] = 0, \end{equation} $

其中$ n\geq0 $.利用关系(2.4)给出

(2.11) $ \begin{eqnarray} \left\{ \begin{array}{ll} w_{t_{n}} = a_x, b = c = e = f = 0, \\ { } u_{1t_{n}} = b_{nx}-wc_n+u_2a_n+\frac{1}{2}u_3\delta_n-\frac{1}{2}u_4\rho_n+u_2a = c_{n+1}+u_2a, \\ { } u_{2t_{n}} = c_{nx}-wb_n+u_1a+\frac{1}{2}u_3\delta_n+\frac{1}{2}u_4\rho_n = b_{n+1}+u_1a, \\ { } u_{3t_{n}} = \delta_{nx}-\frac{1}{2}w\delta_n-\frac{1}{2}u_1\rho_n-\frac{1}{2}u_2\rho_n+\frac{1}{2}u_3a_n +\frac{1}{2}u_4b_n+\frac{1}{2}u_4c_n+\frac{1}{2}u_3a\\ { } \, {\qquad} = \frac{1}{2}\delta_{n+1}+\frac{1}{2}u_3a, \\ { } u_{4t_{n}} = \rho_{nx}+\frac{1}{2}w\rho_n-\frac{1}{2}u_1\delta_n+\frac{1}{2}u_2\delta_n+\frac{1}{2}u_3b_n -\frac{1}{2}u_3c_n-\frac{1}{2}u_4a_n-\frac{1}{2}u_4a\\ {\qquad}{ }\, = -\frac{1}{2}\rho_{n+1}-\frac{1}{2}u_4a, \end{array}\right. \end{eqnarray} $

易知

(2.12) $ \begin{equation} (u_2^{2}-u_1^{2}+2u_4u_3)_{tn} = 2u_2b_{n+1}-2u_1c_{n+1}+u_3\rho_{n+1}+u_4\delta_{n+1} = 2a_{n+1x}. \end{equation} $

取$ a = -2\mu a_{n+1} $,可得下面的广义超孤子族

(2.13) $ \begin{equation} u_{t_n} = \left(\begin{array}{c} u_1\\ u_2\\ u_3\\ u_4\\ \end{array}\right)_{t_n} = \left(\begin{array}{c} c_{n+1}-2\mu u_2a_{n+1}\\ b_{n+1}-2\mu u_1a_{n+1}\\ { } \frac{1}{2}\delta_{n+1}-\mu u_3a_{n+1}\\ { } -\frac{1}{2}\rho_{n+1}+\mu u_4a_{n+1} \end{array}\right), \end{equation} $

其中$ n\geq0 $.在(2.13)式中令$ \mu = 0 $,则可约化为标准的超孤子族^[23].

若在(2.13)式中取$ n = 1 $,可得一族平凡流.若取$ n = 2 $,超孤子族(2.13)约化为2 -阶超非线性可积耦合方程

(2.14) $ \begin{eqnarray} \left\{ \begin{array}{rl} u_{1t_2} = &{ } u_{2xx}+\frac{1}{2}u_2^{3}-\frac{1}{2}u_2u_1^{2}-u_2u_3u_4+u_3u_{3x}+u_4u_{4x} +2\mu(u_{4x}u_3u_1-u_{3x}u_4u_1-2u_1^{2}u_{1x}\\ &{ } +2u_1u_2u_{2x} -2u_3u_4u_{1x}-2u_2u_4u_{3x}+2u_2u_{4x}u_3)-\mu^{2}u_2(u_1^{2}-u_2^{2}+2u_3u_4), \\ u_{2t_{2}} = &{ } u_{1xx}-\frac{1}{2}u_1^{3}+\frac{1}{2}u_1u_2^{2}-u_1u_3u_4+u_3u_{3x}-u_4u_{4x}\\ &{ } +2\mu(u_{4x}u_3u_2-u_{3x}u_4u_2-2u_1u_2u_{1x} -2u_1u_4u_{3x}+2u_1u_{4x}u_3-2u_3u_4u_{2x}+2u_2^{2}u_{2x})\\ &{ } -\mu^{2}u_1(u_1^{2}-u_2^{2}+2u_3u_4)^{2}, \\ u_{3t_2} = &{ } 2u_{3xx}+\frac{1}{4}u_3u_2^{2}-\frac{1}{4}u_3u_1^{2}+\frac{1}{2}u_4u_{1x} +\frac{1}{2}u_4u_{2x}+u_1u_{4x}+u_2u_{4x}\\ &{ } +\mu(u_{3}u_1u_{2x}-u_3u_2u_{1x} +2u_2u_{2x}u_{3}-2u_1u_{1x}u_3-2u_1^{2}u_{3x}+2u_2^{2}u_{3x}-4u_3u_4u_{3x})\\ &{ } -\frac{1}{2}\mu^{2}u_3(u_1^{2}-u_2^{2}+2u_3u_4), \\ u_{4t_2} = &{ } -2u_{4xx}-\frac{1}{4}u_4u_2^{2}+\frac{1}{4}u_4u_1^{2}-u_1u_{3x}+u_2u_{3x} +\frac{1}{2}u_3u_{2x}-\frac{1}{2}u_3u_{1x}\\ &{ } +\mu(u_{4}u_2u_{1x}-u_4u_1u_{2x} -2u_1u_{1x}u_{4}+2u_2u_{2x}u_4-2u_1^{2}u_{4x}+2u_2^{2}u_{4x}-4u_3u_4u_{4x})\\ &{ } +\frac{1}{2}\mu^{2}u_4(u_1^{2}-u_2^{2}+2u_3u_4)^{2}. \end{array}\right. \end{eqnarray} $

它的Lax对由(2.2)中的$ U $和$ V^{(2)} $确定,这里

(2.15) $ \begin{equation} V^{(2)} = \left( \begin{array}{ccc} V_{11}^{(2)}&{\quad} V_{12}^{(2)}{\quad} &V_{13}^{(2)}\\ V_{21}^{(2)}&-V_{11}^{(2)}&V_{23}^{(2)} \\ V_{23}^{(2)}& -V_{13}^{(2)}&0\\ \end{array} \right), \end{equation} $

其中

(2.16) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } V_{11}^{(2)} = \frac{1}{2}\lambda^{-2}+\frac{1}{4}u_2^{2}-\frac{1}{4}u_1^{2}-\frac{1}{4}u_3u_4-\mu(u_2u_{1x}-u_1u_{2x}+2u_4u_{3x}-2u_{4x}u_3)\\ \quad\quad\quad-\mu^{2}(u_1^{2}-u_2^{2}+2u_3u_4), \\ { } V_{12}^{(2)} = \frac{1}{2}(u_1+u_2)\lambda^{-1}+\frac{1}{2}(u_1+u_2)_x-\frac{1}{2}\mu(u_1^{2}-u_2^{2}+2u_3u_4)(u_1+u_2), \\ { } V_{13}^{(2)} = \frac{1}{2}u_3\lambda^{-1}+u_{3x}-\frac{1}{2}\mu(u_1^{2}-u_2^{2}+2u_3u_4)u_3, \\ { } V_{21}^{(2)} = \frac{1}{2}(u_1-u_2)\lambda^{-1}+\frac{1}{2}(u_2-u_1)_x+\frac{1}{2}\mu(u_1^{2}-u_2^{2}+2u_3u_4)(u_2-u_1), \\ { } V_{23}^{(2)} = \frac{1}{2}u_4\lambda^{-1}-u_{4x}-\frac{1}{2}\mu(u_1^{2}-u_2^{2}+2u_3u_4)u_4. \end{array}\right. \end{eqnarray} $

本节将利用如下的超迹恒等式^[3-4]建立广义超孤子族(2.13)的超Hamilton结构

(3.1) $ \begin{equation} \frac{\delta}{\delta u}\int Str(V\frac{\partial U}{\partial\lambda}){\rm d}x = \lambda^{-\gamma}\frac{\partial}{\partial\lambda}\lambda^{\gamma}Str(\frac{\partial U}{\partial u}V), \end{equation} $

常数$ \gamma $为

(3.2) $ \begin{equation} \gamma = -\frac{\lambda}{2}\frac{\rm d}{{\rm d}\lambda}\ln|Str(VV)|. \end{equation} $

通过直接的计算得

(3.3) $ \begin{eqnarray} &&{ } Str(V\frac{\partial U}{\partial\lambda}) = -\frac{1}{2}\lambda^{-2}A, \quad Str(\frac{\partial U}{\partial u_1}V) = \frac{1}{2}(B+2\mu u_1A), {}\\ && Str(\frac{\partial U}{\partial u_2}V) = -\frac{1}{2}(C+2\mu u_2A), {\quad} { } Str(\frac{\partial U}{\partial u_3}V) = \frac{1}{2}(-\rho+2\mu u_4A), {}\\ &&Str(\frac{\partial U}{\partial u_4}V) = \frac{1}{2}(-\delta+2\mu u_3A). \end{eqnarray} $

把上面结果代入超迹恒等式(3.1),并比较两端$ \lambda^{n+2} $同次幂系数

(3.4) $ \begin{equation} \frac{\delta}{\delta u}\int( -a_{n+2}){\rm d}x = (\gamma+n+1)\left(\begin{array}{c} b_{n+1}+2\mu u_1a_{n+1}\\ -c_{n+1}-2\mu u_2a_{n+1}\\ -\rho_{n+1}+2\mu u_4a_{n+1}\\ -\delta_{n+1}+2\mu u_3a_{n+1} \end{array}\right), \quad n\geq0. \end{equation} $

在(3.4)式中令$ n = 0 $,得$ \gamma = 0 $.于是有

(3.5) $ \begin{equation} H_{n+1} = \int-\frac{a_{n+2}}{n+1}{\rm d}x, \quad \frac{\delta H_{n+1}}{\delta u} = \left(\begin{array}{c} b_{n+1}+2\mu u_1a_{n+1}\\ -c_{n+1}-2\mu u_2a_{n+1}\\ -\rho_{n+1}+2\mu u_4a_{n+1}\\ -\delta_{n+1}+2\mu u_3a_{n+1} \end{array}\right), {\quad} n\geqslant 0. \end{equation} $

另外,易得

(3.6) $ \begin{equation} \left(\begin{array}{c} b_{n+1}\\ c_{n+1}\\ \rho_{n+1}\\ \delta_{n+1} \end{array}\right) = R_{1}\left(\begin{array}{c} b_{n+1}+2\mu u_1a_{n+1}\\ -c_{n+1}-2\mu u_2a_{n+1}\\ -\rho_{n+1}+2\mu u_4a_{n+1}\\ -\delta_{n+1}+2\mu u_3a_{n+1} \end{array}\right), {\quad} n\geqslant 0, \end{equation} $

其中$ R_{1} $为

(3.7) $ \begin{equation} R_1 = \left(\begin{array}{cccc} 1-2\mu u_1\partial^{-1}u_2&-2\mu u_1\partial^{-1}u_1&\mu u_1\partial^{-1}u_3&\mu u_1\partial^{-1}u_4\\ -2\mu u_2\partial^{-1}u_2&\ -1-2\mu u_2\partial^{-1}u_1\ &\mu u_2\partial^{-1}u_3&\mu u_2\partial^{-1}u_4\\ 2\mu u_4\partial^{-1}u_2&2\mu u_4\partial^{-1}u_1&-1-\mu u_4\partial^{-1}u_3&-\mu u_4\partial^{-1}u_4\\ 2\mu u_3\partial^{-1}u_2&2\mu u_3\partial^{-1}u_1&-\mu u_3\partial^{-1}u_3&-1-\mu u_3\partial^{-1}u_4 \end{array}\right). \end{equation} $

因此,广义超孤子族(2.13)具有如下的Hamilton结构

(3.8) $ \begin{equation} u_{t_n} = R_2\left(\begin{array}{c} b_{n+1}\\ c_{n+1}\\ \rho_{n+1}\\ \delta_{n+1}\\ \end{array}\right) = R_2R_1\left(\begin{array}{c} b_{n+1}+2\mu u_1a_{n+1}\\ -c_{n+1}-2\mu u_2a_{n+1}\\ -\rho_{n+1}+2\mu u_4a_{n+1}\\ -\delta_{n+1}+2\mu u_3a_{n+1} \end{array}\right) = J\frac{\delta H_{n+1}}{\delta u}, \quad n\geq0, \end{equation} $

其中

(3.9) $ \begin{equation} R_2 = \left(\begin{array}{cccc} -2\mu u_2\partial^{-1}u_2&1+2\mu u_2\partial^{-1}u_1&-\mu u_2\partial^{-1}u_3&-\mu u_2\partial^{-1}u_4\\ 1-2\mu u_1\partial^{-1}u_2&2\mu u_1\partial^{-1}u_1&-\mu u_1\partial^{-1}u_3&-\mu u_1\partial^{-1}u_4\\ -\mu u_3\partial^{-1}u_2&\mu u_3\partial^{-1}u_1& { } -\frac{1}{2}\mu u_3\partial^{-1}u_3&{ } -\frac{1}{2}-\frac{1}{2}\mu u_3\partial^{-1}u_4\\ \mu u_4\partial^{-1}u_2&-\mu u_4\partial^{-1}u_1& { } \frac{1}{2}+\frac{1}{2}\mu u_4\partial^{-1}u_3&{ } \frac{1}{2}\mu u_4\partial^{-1}u_4 \end{array}\right), \end{equation} $

超Hamiltonian算子

(3.10) $ \begin{equation} J = R_2R_1 = \left(\begin{array}{cccc} -4\mu u_2\partial^{-1}u_2&-1-4\mu u_2\partial^{-1}u_1&2\mu u_2\partial^{-1}u_3&2\mu u_2\partial^{-1}u_4\\ 1-4\mu u_1\partial^{-1}u_2&-4\mu u_1\partial^{-1}u_1&2\mu u_1\partial^{-1}u_3&2\mu u_1\partial^{-1}u_4\\ -2\mu u_3\partial^{-1}u_2&-2\mu u_3\partial^{-1}u_1&\mu u_3\partial^{-1}u_3& { } \frac{1}{2}+\mu u_3\partial^{-1}u_4\\ 2\mu u_4\partial^{-1}u_2&2\mu u_4\partial^{-1}u_1& { } -\frac{1}{2}-\mu u_4\partial^{-1}u_3&\mu u_4\partial^{-1}u_4 \end{array}\right). \end{equation} $

另外,利用递归关系(2.4),广义超孤子族(2.13)可写为如下形式

(3.11) $ \begin{equation} u_{t_n} = R_2L\left(\begin{array}{c} b_{n}\\ c_{n}\\ \rho_{n}\\ \delta_{n}\\ \end{array}\right) = R_2LR_1\left(\begin{array}{c} b_{n}+2\mu u_1a_{n}\\ -c_{n}-2\mu u_2a_{n}\\ -\rho_{n}+2\mu u_4a_{n}\\ -\delta_{n}+2\mu u_3a_{n} \end{array}\right) = M\frac{\delta H_{n}}{\delta u}, \quad n\geq0, \end{equation} $

其中$ M = R_2LR_1 = (M_{ij})_{4\times4} $是第二个超Hamilton算子.

考虑线性系统

(4.1) $ \begin{equation} \left(\begin{array}{c} \varphi_{1j}\\ \varphi_{2j} \\ \varphi_{3j}\\ \end{array}\right)_{x} = U\left(\begin{array}{c} \varphi_{1j}\\ \varphi_{2j} \\ \varphi_{3j}\\ \end{array}\right), \quad \left(\begin{array}{c} \varphi_{1j}\\ \varphi_{2j} \\ \varphi_{3j}\\ \end{array}\right)_{t} = V\left(\begin{array}{c} \varphi_{1j}\\ \varphi_{2j} \\ \varphi_{3j}\\ \end{array}\right). \end{equation} $

基于文献[24],可得

(4.2) $ \begin{equation} \frac{\delta \lambda_{j}}{\delta u_i} = \frac{1}{3}Str(\Psi_j\frac{\partial U(u, \lambda_j)}{\delta u_i}), \ \ i = 1, 2, \cdots4, \end{equation} $

这里$ Str $表示矩阵的超迹和

(4.3) $ \begin{equation} \Psi_j = \left( \begin{array}{ccc} \varphi_{1j} \varphi_{2j}& -\varphi_{1j} ^{2} &\varphi_{1j}\varphi_{3j} \\ \varphi_{2j} ^{2}&{\quad} - \varphi_{1j} \varphi_{2j}{\quad} & \varphi_{2j} \varphi_{3j}\\ \varphi_{2j}\varphi_{3j}&- \varphi_{1j} \varphi_{3j}&0\\ \end{array} \right), \quad j = 1, 2, \cdots N. \end{equation} $

利用(4.1)式,可得$ \frac{\delta \lambda_{j}}{\delta u} $为

(4.4) $ \begin{equation} \sum\limits_{j = 1}^N \frac{\delta\lambda_j}{\delta u_i} = \sum\limits_{j = 1}^N \left( \begin{array}{cccccc} { } Str(\Psi_j\frac{\delta U}{\delta u_1})\\ { } Str(\Psi_j\frac{\delta U}{\delta u_2})\\ { } Str(\Psi_j\frac{\delta U}{\delta u_3})\\ { } Str(\Psi_j\frac{\delta U}{\delta u_4})\\ \end{array} \right) = \left( \begin{array}{cccccc} { } 2\mu u_1\langle\Phi_1, \Phi_2\rangle-\frac{1}{2}\langle\Phi_1, \Phi_1\rangle+\frac{1}{2}\langle\Phi_2, \Phi_2\rangle\\ { } -2\mu u_2\langle\Phi_1, \Phi_2\rangle+\frac{1}{2}\langle\Phi_1, \Phi_1\rangle+\frac{1}{2}\langle\Phi_2, \Phi_2\rangle\\ 2\mu u_4\langle\Phi_1, \Phi_2\rangle-\langle\Phi_2, \Phi_3\rangle\\ 2\mu u_3\langle\Phi_1, \Phi_2\rangle+\langle\Phi_1, \Phi_3\rangle\\ \end{array} \right), \end{equation} $

其中$ \Phi_i = (\varphi_{i1}, \cdots\varphi_{iN})^{T}, i = 1, 2, 3. $

因此,带自相容源的广义超孤子族(2.13)如下

(4.5) $ \begin{eqnarray} u_{t_n}& = &\left(\begin{array}{cccc} u_1 \\ u_2 \\ u_3 \\ u_4 \\ \end{array}\right)_{t_n} = J\frac{\delta H_{n+1}}{\delta u_i}+J\sum\limits_{j = 1}^N \frac{\delta\lambda_j}{\delta u_i}{}\\ & = &J\left(\begin{array}{c} b_{n+1}+2\mu u_1a_{n+1}\\ -c_{n+1}-2\mu u_2a_{n+1}\\ -\rho_{n+1}+2\mu u_4a_{n+1}\\ -\delta_{n+1}+2\mu u_3a_{n+1} \end{array}\right)+J\left( \begin{array}{cccccc} { } 2\mu u_1\langle\Phi_1, \Phi_2\rangle-\frac{1}{2}\langle\Phi_1, \Phi_1\rangle+\frac{1}{2}\langle\Phi_2, \Phi_2\rangle\\ { } -2\mu u_2\langle\Phi_1, \Phi_2\rangle+\frac{1}{2}\langle\Phi_1, \Phi_1\rangle+\frac{1}{2}\langle\Phi_2, \Phi_2\rangle\\ 2\mu u_4\langle\Phi_1, \Phi_2\rangle-\langle\Phi_2, \Phi_3\rangle\\ 2\mu u_3\langle\Phi_1, \Phi_2\rangle+\langle\Phi_1, \Phi_3\rangle\\ \end{array} \right). \end{eqnarray} $

取$ n = 2 $,可得一组带自相容源的超孤子方程

(4.6) $ \begin{eqnarray} \left\{ \begin{array}{rl} u_{1t_2} = &{ } u_{2xx}+\frac{1}{2}u_2^{3}-\frac{1}{2}u_2u_1^{2}-u_2u_3u_4+u_3u_{3x}+u_4u_{4x} +2\mu(u_{4x}u_3u_1-u_{3x}u_4u_1\\ &{ } -2u_1^{2}u_{1x} +2u_1u_2u_{2x}-2u_3u_4u_{1x}-2u_2u_4u_{3x}+2u_2u_{4x}u_3)\\ &{ } -\mu^{2}u_2(u_1^{2}-u_2^{2}+2u_3u_4) +2\mu u_1\sum\limits_{j = 1}^N \varphi_{1j}\varphi_{2j}-\frac{1}{2}\sum\limits_{j = 1}^N (\varphi_{1j}^{2}-\varphi_{2j}^2), \\ u_{2t_{2}} = &{ } u_{1xx}-\frac{1}{2}u_1^{3}+\frac{1}{2}u_1u_2^{2}-u_1u_3u_4+u_3u_{3x}-u_4u_{4x} +2\mu(u_{4x}u_3u_2-u_{3x}u_4u_2\\ &{ } -2u_1u_2u_{1x} -2u_1u_4u_{3x}+2u_1u_{4x}u_3-2u_3u_4u_{2x}+2u_2^{2}u_{2x})\\ &{ } -\mu^{2}u_1(u_1^{2}-u_2^{2}+2u_3u_4)^{2} -2\mu u_1\sum\limits_{j = 1}^N \varphi_{1j}\varphi_{2j}+\frac{1}{2}\sum\limits_{j = 1}^N (\varphi_{1j}^{2}+\varphi_{2j}^2), \\ u_{3t_2} = &{ } 2u_{3xx}+\frac{1}{4}u_3u_2^{2}-\frac{1}{4}u_3u_1^{2}+\frac{1}{2}u_4u_{1x}+\frac{1}{2}u_4u_{2x}+u_1u_{4x} +u_2u_{4x} +\mu(u_{3}u_1u_{2x}\\ &-u_3u_2u_{1x} +2u_2u_{2x}u_{3}-2u_1u_{1x}u_3-2u_1^{2}u_{3x}+2u_2^{2}u_{3x}-4u_3u_4u_{3x})\\ &{ } -\frac{1}{2}\mu^{2}u_3(u_1^{2}-u_2^{2}+2u_3u_4) +2\mu u_4\sum\limits_{j = 1}^N \varphi_{1j}\varphi_{2j}-\sum\limits_{j = 1}^N \varphi_{2j}\varphi_{3j}, \\ u_{4t_2} = &{ } -2u_{4xx}-\frac{1}{4}u_4u_2^{2}+\frac{1}{4}u_4u_1^{2}-u_1u_{3x}+u_2u_{3x}+\frac{1}{2}u_3u_{2x} -\frac{1}{2}u_3u_{1x}\\ &+\mu(u_{4}u_2u_{1x}-u_4u_1u_{2x} -2u_1u_{1x}u_{4}+2u_2u_{2x}u_4-2u_1^{2}u_{4x}+2u_2^{2}u_{4x} \\ &{ }-4u_3u_4u_{4x}) +\frac{1}{2}\mu^{2}u_4(u_1^{2}-u_2^{2}+2u_3u_4)^{2} +2\mu u_3\sum\limits_{j = 1}^N \varphi_{1j}\varphi_{2j}+\sum\limits_{j = 1}^N \varphi_{1j}\varphi_{3j}, \\ \varphi_{1jx} = &{ } \frac{1}{2}(\lambda^{-1}+w)\varphi_{1j}+\frac{1}{2}(u_1+u_2)\varphi_{2j}+\frac{1}{2}u_3\varphi_{3j}, \\ \varphi_{2jx} = &{ } \frac{1}{2}(u_1-u_2)\varphi_{1j}-\frac{1}{2}(\lambda^{-1}+w)\varphi_{2j}+\frac{1}{2}u_4\varphi_{3j}, \quad j = 1, \cdots N.\\ \varphi_{3jx} = &{ } \frac{1}{2}u_4\varphi_{1j}-\frac{1}{2}u_3\varphi_{2j}, \end{array} \right. \end{eqnarray} $

本节给出广义超孤子族的守恒律.引进变量

(5.1) $ \begin{equation} K = \frac{\varphi_{2}}{\varphi_{1}}, \quad G = \frac{\varphi_{3}}{\varphi_{1}}, \end{equation} $

可得

(5.2) $ \begin{equation} \begin{array}{l} { } K_{x} = \frac{1}{2}(u_1-u_2)-(\lambda^{-1}+w) K+\frac{1}{2}u_4G-\frac{1}{2}(u_1+u_2)K^{2}-\frac{1}{2}u_3 KG, \\ { } G_{x} = \frac{1}{2}u_4-\frac{1}{2}u_3K-\frac{1}{2}(\lambda ^{-1}+w)G-\frac{1}{2}(u_1+u_2)GK-\frac{1}{2}u_3G^{2}. \end{array} \end{equation} $

将$ K $和$ G $按$ \lambda $级数展开如下

(5.3) $ \begin{equation} K = \sum\limits_{j = 1}^\infty k_j\lambda^{j}, \ \ G = \sum\limits_{j = 1}^\infty g_j\lambda^{j}. \end{equation} $

把(5.3)式代入(5.2)式,并比较$ \lambda $的同次幂系数,得$ k_j $和$ g_j $的递归关系式如下

(5.4) $ \begin{equation} \begin{array}{l} { } k_{j+1} = -k_{jx}-wk_j+\frac{1}{2}u_4g_j-\frac{1}{2}(u_1+u_2)\sum\limits_{l = 1}^{j-1}k_lk_{j-l}-\frac{1}{2}u_3\sum\limits_{l = 1}^{j-1}k_lg_{j-l}, \\ { } g_{j+1} = -2g_{jx}+u_3k_j-wg_j-(u_1+u_2)\sum\limits_{l = 1}^{j-1}g_lk_{j-l}-u_3\sum\limits_{l = 1}^{j-1}g_lg_{j-l}, \quad j\geq2. \end{array} \end{equation} $

其$ k_j $和$ g_j $的前几项为

(5.5) $ \begin{eqnarray} k_1& = &\frac{1}{2}(u_1-u_2), g_1 = u_4, k_2 = \frac{1}{2}(u_2-u_1)x-\frac{1}{2}\mu(u_1-u_2)(u_1^{2}-u_2^{2}+2u_3u_4), {}\\ g_2& = &-2u_{4x}-\frac{1}{2}(u_1-u_2)u_3-\mu(u_1^{2}-u_2^{2})u_4, {}\\ k_3& = &\frac{1}{2}(u_1-u_2)_{xx}+\frac{1}{2}w_{x}(u_1-u_2)+\frac{1}{2}w^{2}(u_1-u_2)-\frac{1}{8}(u_1+u_2)(u_1-u_2)^{2} {}\\ &&+w(u_1-u_2)_x-u_4u_{4x}, {}\\ g_3& = &4u_{4xx}+\frac{3}{2}(u_2-u_1)_xu_3-(u_1-u_2)u_{3x}+(u_1-u_2w u_3)+\frac{1}{2}(u_1^{2}-u_2^{2})u_4 {}\\ &&+2wu_{4x}+w^{2}u_4+4w(u_1u_{1x}-u_2u_{2x}), \cdots. \end{eqnarray} $

由于

(5.6) $ \begin{equation} \frac{\partial}{\partial t} \bigg[\frac{1}{2}(\lambda^{-1}+w)+\frac{1}{2}(u_1+u_2)K+\frac{1}{2}u_3G\bigg] = \frac{\partial}{\partial x}\bigg[\frac{1}{2}A+\frac{1}{2}(B+C)K+\frac{1}{2}\delta G\bigg], \end{equation} $

设$ \sigma = \frac{1}{2}(\lambda^{-1}+w)+\frac{1}{2}(u_1+u_2)K+\frac{1}{2}u_3G $, $ \theta = \frac{1}{2}A+\frac{1}{2}(B+C)K+\frac{1}{2}\delta G $,则(5.6)式可写成$ \sigma_t = \theta_x $,这正是守恒律的标准形式.利用(2.14)式得

(5.7) $ \begin{equation} \begin{array}{l} { } A = \lambda^{-2}+\frac{1}{2}u_2^{2}-\frac{1}{2}u_1^{2}-u_3u_4, \\ B = u_1\lambda^{-1}+u_{2x}-\mu u_1(u_1^{2}-u_2^{2}+2u_3u_4), \\ C = u_2\lambda^{-1}+u_{1x}-\mu u_2(u_1^{2}-u_2^{2}+2u_3u_4), \\ \delta = u_3\lambda^{-1}+2u_{3x}-\mu u_3(u_1^{2}-u_2^{2}). \end{array} \end{equation} $

将$ \sigma $和$ \theta $按$ \lambda $的级数展开,相应的系数分别被称为守恒密度和流

(5.8) $ \begin{equation} \sigma = \frac{1}{2}\lambda^{-1}+\frac{1}{2}w+\sum\limits_{j = 1}^\infty \sigma_j\lambda^{j}, \ \ \theta = \frac{1}{2}(\lambda^{-2}+\frac{1}{2}u_2^{2}-\frac{1}{2}u_1^{2}-u_3u_4)+\sum\limits_{j = 1}^\infty \theta_j\lambda^{j}, \end{equation} $

其前2个守恒密度和流为

(5.9) $ \begin{eqnarray} \sigma_1 & = & \frac{1}{4}(u_1^{2}-u_2^{2})+\frac{1}{2}u_3u_4 , {}\\ \theta _1& = &\frac{1}{4}(u_1+u_2)[(u_2-u_1)x-\mu(u_1-u_2)(u_1^{2}-u_2^{2}+2u_3u_4)]{}\\ &&+\frac{1}{4}(u_1-u_2)[u_{1x}+\mu(u_2^{3} -u_1^{3}+u_1u_2^{2}-u_2u_1^{2}-2u_1u_3u_4-2u_2u_3u_4)]{}\\ &&-\frac{1}{2}u_3[2u_{4x}+\frac{1}{2}(u_1-u_2)u_3+\mu(u_1^{2}-u_2^{2})u_4] +\frac{1}{2}(2u_{3x}-\mu u_3u_1^{2}+\mu u_3u_2^{2})u_4, {} \\ \sigma _2 & = & \frac{1}{4}(u_1+u_2)[(u_2-u_1)_x-\mu(u_1-u_2)(u_1^{2}-u_2^{2}+2u_3u_4)]-u_3[u_{4x}+\frac{1}{2}\mu(u_1^{2}-u_2^{2})u_4] , {}\\ \theta_2& = &\frac{1}{2}(u_1+u_2)k_{3}+\frac{1}{4}[u_{1x}+\mu(u_2^{3}-u_1^{3}+u_1u_2^{2} -u_2u_1^{2}-2u_1u_3u_4-2u_2u_3u_4)]{}\\ &&\times[(u_2-u_1)x\mu(u_1-u_2)(u_1^{2}-u_2^{2}+2u_3u_4)]+\frac{1}{2}u_3g_{3}- \frac{1}{2}(2u_{3x}-\mu u_3u_1^{2}+\mu u_3u_2^{2}){}\\ &&\times [2u_{4x}+\frac{1}{2}(u_1-u_2)u_3+\mu(u_1^{2}-u_2^{2})u_4], \end{eqnarray} $

其中$ k_3 $和$ g_3 $由(5.5)式给出.而$ \sigma_j $和$ \theta_j $的递归关系如下

(5.10) $ \begin{eqnarray} \sigma_j& = &\frac{1}{2}(u_1+u_2)k_{j}+\frac{1}{2}u_3g_j, {}\\ \theta_j& = &\frac{1}{2}(u_1+u_2)k_{j+1}+\frac{1}{2}[u_{1x}+\mu(u_2^{3}-u_1^{3}+u_1u_2^{2}-u_2u_1^{2}-2u_1u_3u_4-2u_2u_3u_4)]k_j{}\\ &&+\frac{1}{2}u_3g_{j+1}+\frac{1}{2}(2u_{3x}-\mu u_3u_1^{2}+\mu u_3u_2^{2})g_j, \end{eqnarray} $

$ k_j $和$ g_j $可分别由(5.4)式推出.我们可以给出(2.14)式的前2个守恒律如下

(5.11) $ \begin{equation} \sigma_{1t} = \theta_{1x}, \quad \sigma_{2t} = \theta_{2x}, \end{equation} $

其中$ \sigma_1, \theta_1, \sigma_2 $和$ \theta_2 $由(5.9)式定义.则(2.13)式的无穷守恒律可由方程(5.2)-(5.11)容易地给出.

利用Lie超代数$ B(0, 1) $,我们建立了新的广义超孤子族,借助超迹恒等式将它写成双-Hamilton结构形式.同时,也给出了广义超孤子族的自相容源和守恒律.需要特别指出,广义超孤子族涉及到费米变量,扩展的费米变量运算构成一Grassmann代数.有时称(2.13)式为费米扩展,而对于其它的超可积系统,是否可以用类似的方法给出它的费米扩展?另外,在文献[25-26]中给出了超AKNS族和超经典Boussinesq族的双非线性化,本文是否也可以给出其双非线性化?近来,怪波解等^[27-32]给出可积系统精确解一类特殊的可积性.作为约化,我们得到了非线性可积方程,如何获得约化方程的解是项重要和困难的工作,下一步我们将考虑这个问题.