由于超对称可积系统与超弦理论^[1-2]、超对称Yang-Mills理论^[3]以及超对称规范理论^[4]具有紧密联系,超对称可积系统在数学和理论物理领域引起了广泛关注.目前$ (1+1) $维超对称可积系统的结构和可积性质已经被广泛研究,参见文献[5-12].标准理论中的众多方法已推广到超对称可积系统中,如贝克隆变换^[14]、Painlevé测试^[15]、$ \tau $函数^[16]、达布变换^[17]、Hirota双线性方法^[18]和延拓结构理论等^[19].然而,对高维超对称可积系统的研究并不多,因此众多学者对高维超对称可积方程进行了研究. Saha和Chowhury^[20]提出了构造(2+1)维超对称可积系统的理论.有学者^[21]通过引入一般的辅助矩阵变量,构造了(2+1)维可积超对称海森堡铁磁链(HS)模型,进一步推出高维HS模型规范等价于超对称非线性薛定谔方程(NLSE).为了建立超对称可积系统,程等人^[22]提出了费米协变延拓结构理论,该理论被颜等人^[23]推广.费米协变延拓结构理论可以推导出(2+1)维超对称非线性演化方程的延拓结构, Lax表示和贝克隆变换.

超李代数在构造超对称可积系统中具有重要作用^[24-25].由于超代数在弦理论^[26]和对数共形场理论^[27]中具有重要应用,因此众多学者对超代数进行了广泛研究,例如关于超李代数$ osp(2n/2n) $的随机伊辛模型^[28].最近,通过利用齐性空间和增加维数的方法,有学者构造了基于超李代数$ osp(3/2) $的(2+1)维超对称可积方程^[29-30].

该文讨论了新的超对称可积系统的构造,并对其结构和可积性进行了研究.该文将利用超李代数的齐性空间和拓展系统维数的方法,构造基于超李代数$ osp(2/2) $的新的超对称可积方程.

经典的超李代数$ osp(2/2) $的矩阵生成元为^[9]

(2.1) $ \begin{eqnarray} \left( \begin{array}{cccc} a & 0 & x & x_1 \\ 0 & -a & y & y_1 \\ y_1 & x_1 & d & e \\ -y & -x & f & -d \\ \end{array} \right) , \end{eqnarray} $

其中$ (\tilde{a}, \tilde{d}, \tilde{e}, \tilde{f}) $和$ (\tilde{x}, \tilde{y}, \tilde{x}_1, \tilde{y}_1) $分别是超李代数$ osp(2/2) $的玻色生成元和费米生成元. $ \tilde{w} $表示$ w $对应的位置为$ 1 $,其他位置为$ 0 $的矩阵.

定义超李括号^[9]

(2.2) $ \begin{eqnarray} [X, Y] = XY-(-1)^{P(X)P(Y)}YX, \end{eqnarray} $

其中$ P(X) $和$ P(Y) $是生成元$ X $和$ Y $的宇称.

将超李代数$ osp(2/2) $分解为

(2.3) $ \begin{eqnarray} g = k\oplus m. \end{eqnarray} $

$ k, m $满足如下关系

(2.4) $ \begin{eqnarray} [k, k]\subset k, \ \ [k, m]\subset m, \ \ [m, m]\subset k, \end{eqnarray} $

其中

(2.5) $ \begin{eqnarray} k& = &\{\tilde{a}, \tilde{d}, \tilde{e}, \tilde{f}\}, {} \\ m& = &\{\tilde{x}, \tilde{y}, \tilde{x}_1, \tilde{y}_1\}. \end{eqnarray} $

将超李代数$ osp(2/2) $的费米、玻色生成元带入超李括号(2.2),可得其(反)交换关系.

考虑谱问题

(2.6) $ \begin{eqnarray} \phi_x& = &[\lambda A+Q(x, y, t)]\phi, {} \\ \phi_t& = &\lambda^{n-1}\phi_y+\sum\limits_{i = 0}^{n-2}B_{n-i}(x, y, t)\lambda^i\phi, \end{eqnarray} $

其中$ \lambda $是谱参数.常数矩阵$ A $属于$ g $的$ Cartan $子代数, $ Q(x, y, t) \in m $且$ B_{n-i}(x, y, t)\in g $.利用(2.6)式的可积性条件并比较$ \lambda^\mu (\mu = 0, \cdots, n) $的系数可得

(2.7) $ \begin{eqnarray} \partial_yQ& = &[A, B_2], {} \\ \partial_xB_{n-i}& = &[A, B_{n-i+1}]+[Q, B_{n-i}], {} \\ \partial_tQ& = &\partial_xB_n-[Q, B_n], \end{eqnarray} $

其中$ i = 1, \cdots, n-2 $.

考虑(2.7)式中$ n = 2 $的情形并假设$ A = \tilde{a}, B_2 = B_2^k+B_2^m $,其中$ B_2^k\in k, B_2^m\in m $,可得如下方程

(2.8) $ \begin{eqnarray} \partial_yQ& = &[A, B_2^m], \end{eqnarray} $

(2.9) $ \begin{eqnarray} \partial_xB_2^k& = &[Q, B_2^m], \end{eqnarray} $

(2.10) $ \begin{eqnarray} \partial_tQ& = &\partial_xB_2^m-[Q, B_2^k], \end{eqnarray} $

其中$ B_2^m \in m $和$ B_2^k \in k $.

根据方程(2.5),假设

(2.11) $ \begin{eqnarray} Q& = &q_1\tilde{x}+q_2\tilde{x_1}+q_3\tilde{y}+q_4\tilde{y_1}, \end{eqnarray} $

(2.12) $ \begin{eqnarray} B_2^m& = &s_1\tilde{x}+s_2\tilde{x_1}+s_3\tilde{y}+s_4\tilde{y_1}, \end{eqnarray} $

(2.13) $ \begin{eqnarray} B_2^k& = &M_1\tilde{a}+M_2\tilde{d}+M_3\tilde{e}+M_4\tilde{f}. \end{eqnarray} $

将(2.11)和(2.12)式代入(2.8)式,可得$ B_2^m $

(2.14) $ \begin{eqnarray} B_2^m& = &q_{1y}\tilde{x}+q_{2y}\tilde{x_1}-q_{3y}\tilde{y}-q_{4y}\tilde{y_1}. \end{eqnarray} $

将(2.12)和(2.13)式代入(2.9)式,可得

(2.15) $ \begin{eqnarray} M_1& = &\partial_x^{-1}(q_4q_{1y}-q_3q_{2y}+q_2q_{3y}-q_1q_{4y}), {} \\ M_2& = &\partial_x^{-1}(q_4q_{1y}+q_3q_{2y}-q_2q_{3y}-q_1q_{4y}), {} \\ M_3& = &\partial_x^{-1}(2q_4q_{2y}-2q_2q_{4y}), {} \\ M_4& = &\partial_x^{-1}(2q_1q_{3y}-2q_3q_{1y}). \end{eqnarray} $

将$ B_2^k $和$ B_2^m $代入(2.10)式,可得

(2.16) $ \begin{eqnarray} q_{1t}& = &q_{1xy}+q_1M_1-q_1M_2-q_2M_4, {} \\ q_{2t}& = &q_{2xy}+q_2M_1+q_2M_2-q_1M_3, {} \\ q_{3t}& = &-q_{3xy}-q_3M_1-q_3M_2-q_4M_4, {} \\ q_{4t}& = &-q_{4xy}-q_4M_1+q_4M_2-q_3M_3. \end{eqnarray} $

下面放松$ k $和$ m $的约束条件.假设$ k $和$ m $满足如下关系

(2.17) $ \begin{eqnarray} [k, k]\subset k, \ \ [k, m]\subset m, \ \ [m, m]\subset k \ \ \mbox{或} \ \ m, \end{eqnarray} $

其中

(2.18) $ \begin{eqnarray} k & = &\{\tilde{a}, \tilde{d}\}, {} \\ m& = &\{\tilde{e}, \tilde{f}, \tilde{x}, \tilde{y}, \tilde{x_1}, \tilde{y_1}\}. \end{eqnarray} $

考虑(2.7)式中$ n = 2 $的情形并假设$ A = \tilde{d}, B_n = B_n^k+B_n^m $,则由(2.7)式可得

(2.19) $ \begin{eqnarray} \partial_yQ& = &[A, B_2^m], \end{eqnarray} $

(2.20) $ \begin{eqnarray} \partial_xB_2^k& = &[Q, B_2^m]_k, \end{eqnarray} $

(2.21) $ \begin{eqnarray} \partial_tQ& = &\partial_xB_2^m-[Q, B_2^k]-[Q, B_2^m]_m, \end{eqnarray} $

其中$ B_2^m \in \bf{m} $, $ B_2^k \in \bf{k} $, $ [\cdot, \cdot]_k $表示生成元在子空间$ k $中的投影.利用性质(2.18),假设

(2.22) $ \begin{eqnarray} Q& = &q_1\tilde{e}+q_2\tilde{f}+q_3\tilde{x}+q_4\tilde{x_1}+q_5\tilde{y}+q_6\tilde{y_1}, \end{eqnarray} $

(2.23) $ \begin{eqnarray} B_2^m& = &s_1\tilde{e}+s_2\tilde{f}+s_3\tilde{x}+s_4\tilde{x_1}+s_5\tilde{y}+s_6\tilde{y_1}, \end{eqnarray} $

(2.24) $ \begin{eqnarray} B_2^k& = &M_1\tilde{a}+M_2\tilde{d}. \end{eqnarray} $

将(2.22)和(2.23)式代入(2.19)式,可得

(2.25) $ \begin{eqnarray} s_1& = &\frac{1}{2}q_{1y}, \ \ s_2 = -\frac{1}{2}q_{2y}, \ \ s_3 = -q_{3y}, {} \\ s_4& = &q_{4y}, \ \ \ s_5 = -q_{5y}, \ \ \ s_6 = q_{6y}, \end{eqnarray} $

其中$ q_1, q_2 $是费米变量, $ q_i \ (i = 3, \cdots, 6) $是玻色变量.将(2.23)和(2.24)式代入(2.20)式,可得

(2.26) $ \begin{eqnarray} M_1& = &\partial_x^{-1}(q_3q_{6y}-q_6q_{3y}+q_4q_{5y}-q_5q_{4y}), {} \\ M_2& = &\partial_x^{-1}[q_3q_{6y}-q_6q_{3y}+q_5q_{4y}-q_4q_{5y}-\frac{1}{2}(q_1q_{2y}+q_2q_{1y})]. \end{eqnarray} $

将$ B_2^k $和$ B_2^m $的表达式代入(2.21)式,可得(2+1)维超对称可积方程

(2.27) $ \begin{eqnarray} q_{1t}& = &\frac{1}{2}q_{1xy}+2q_1M_2-2q_4q_{6y}-2q_6q_{4y}, {} \\ q_{2t}& = &-\frac{1}{2}q_{2xy}-2q_2M_2-2q_3q_{5y}-2q_5q_{3y}, {} \\ q_{3t}& = &-q_{3xy}+q_3M_1-q_3M_2+q_2q_4y+\frac{1}{2}q_4q_{2y}, {} \\ q_{4t}& = &q_{4xy}+q_4M_1+q_4M_2-q_1q_{3y}-\frac{1}{2}q_3q_{1y}, {} \\ q_{5t}& = &-q_{5xy}-q_5M_1-q_5M_2+q_2q_{6y}+\frac{1}{2}q_6q_{2y}, {} \\ q_{6t}& = &q_{6xy}+q_6M_2-q_6M_1-q_1q_{5y}-\frac{1}{2}q_5q_{1y}. \end{eqnarray} $

接下来考虑用扩展系统维数的方法来构造(2+1)维超对称可积演化方程.该部分将利用与谱参数无关的二维Lax算子.

首先考虑Lax算子

(3.1) $ \begin{eqnarray} K_1(Q)& = &\partial_x+H_1\partial_y+Q, \end{eqnarray} $

(3.2) $ \begin{eqnarray} K_2(Q)& = &{\rm i}\partial_t+H_1\partial^2_y+A\partial_y+B. \end{eqnarray} $

根据可积性条件

(3.3) $ \begin{eqnarray} [K_1, K_2]\eta = 0, \end{eqnarray} $

利用$ \Psi, \partial_y\Psi, \partial_y^2\Psi $前的系数为$ 0 $,可得

(3.4) $ \begin{eqnarray} &&A = Q, \end{eqnarray} $

(3.5) $ \begin{eqnarray} &&\partial_xQ-H_1\partial_yQ+[H_1, B] = 0, \end{eqnarray} $

(3.6) $ \begin{eqnarray} &&\partial_xB+H_1\partial_yB+[Q, B]-{\rm i}\partial_tQ-H_1\partial_y^2Q-A\partial_yQ = 0. \end{eqnarray} $

假设$ B = (B_{ij}), (i, j = 1, \cdots, 4) $,将矩阵$ B $分解为两部分$ B = B^1+B^2 $,其中$ B^1 $由$ B_{ij} (i, j = 1, \cdots, 2, i = j = 3, i = j = 4) $为非零元素,其它元素为$ 0 $构成的矩阵.将$ H_1 = d $和(2.22)式代入(3.5)式,可得$ B^2 $有如下形式

(3.7) $ \begin{eqnarray} &&B_{13} = q_{3x}, \ \ \ \ B_{14} = -q_{4x}, \ \ \ B_{23} = q_{5x}, \ \ \ B_{24} = -q_{6x}, \ \ \ B_{31} = -q_{6x}+q_{6y}, {} \\ &&B_{32} = -q_{4x}+q_{4y}, \ \ B_{34} = \frac{1}{2}(q_{1y}-q_{1x}), \ \ B_{41} = -q_{5x}-q_{5y}, {} \\ &&B_{42} = -q_{3x}-q_{3y}, \ \ B_{43} = \frac{1}{2}(q_{2x}+q_{2y}). \end{eqnarray} $

根据方程(3.7),可从(3.6)式得到如下方程

(3.8) $ \begin{eqnarray} &&B_{11} = B_{22} = q_3q_6+q_4q_5, \ \ B_{12} = 2q_3q_4, \ \ B_{21} = 2q_5q_6, {} \\ &&B_{33} = (\partial_x+\partial_y)^{-1}(\partial_x-\partial_y)(q_3q_6+q_4q_5+\frac{1}{2}q_1q_2), {} \\ &&B_{44} = (\partial_x-\partial_y)^{-1}(\partial_x+\partial_y)(q_3q_6+q_4q_5-\frac{1}{2}q_1q_2). \end{eqnarray} $

将(3.7)和(3.8)式代入(3.6)式,可得如下新的超对称非线性演化方程

(3.9) $ \begin{eqnarray} {\rm i}q_{1t}& = &-q_{1yy}-(\partial_x+\partial_y)(q_4q_6)+q_1B_{44}, {} \\ {\rm i}q_{2t}& = &q_{2yy}+2\partial_y(q_3q_5)+q_2B_{33}-q_2B_{44}, {} \\ {\rm i}q_{3t}& = &q_{3yy}+(3q_4q_5+q_3q_6)q_3+q_{4x}q_2+\frac{1}{2}(q_{2x}+q_{2y})q_4-q_3B_{44}, {} \\ {\rm i}q_{4t}& = &-q_{4yy}+(3q_3q_6+q_4q_5)q_4-q_{3x}q_1+\frac{1}{2}(q_{1y}-q_{1x})q_3-q_4B_{33}, {} \\ {\rm i}q_{5t}& = &q_{5yy}+(3q_3q_6+q_4q_5)q_5+q_{6x}q_2+\frac{1}{2}(q_{2x}+q_{2y})q_6-q_5B_{44}, {} \\ {\rm i}q_{6t}& = &-q_{6yy}+(3q_4q_5+q_3q_6)q_6-q_{5x}q_1+\frac{1}{2}(q_{1y}-q_{1x})q_5-q_6B_{33}. \end{eqnarray} $

贝克隆变换是研究方程解的有效工具.贝克隆变换建立了偏微分方程的解之间的关联.

首先推导超对称可积方程(2.16)的贝克隆变换.假设$ Q $和$ Q' $是非线性系统的两组解

(4.1) $ \begin{eqnarray} \eta'_x& = &[\lambda A+Q']\eta', {} \\ \eta_x& = &[\lambda A+Q]\eta. \end{eqnarray} $

通过达布变换$ \eta' = D\eta $并假设$ D = D_0+\lambda D_1 $,可由(4.1)式推出

(4.2) $ \begin{eqnarray} D_1& = &A = \tilde{a}, \end{eqnarray} $

(4.3) $ \begin{eqnarray} D_{1x}& = &[A, D_0]+Q'A-AQ, \end{eqnarray} $

(4.4) $ \begin{eqnarray} D_{0x}& = &Q'D_0-D_0Q , \end{eqnarray} $

根据(4.2)和(4.3)式,可得$ D_0 $的非对角部分

(4.5) $ \begin{eqnarray} D^{off}_0 = \left( \begin{array}{cccc} 0 & 0 & q_1 & q_2 \\ 0 & 0 & q_3 & q_4 \\ q_4' & q_2' & 0 & 0 \\ -q_3' & -q_1' & 0 & 0 \\ \end{array} \right). \end{eqnarray} $

同时由方程(4.4)推出$ D_0 $的对角部分为

(4.6) $ \begin{eqnarray} D^{diag}_0 = \left( \begin{array}{cccc} \omega & 0 & 0 & 0 \\ 0 & \eta & 0 & 0 \\ 0 & 0 & \alpha & 0 \\ 0 & 0 & 0 & \beta \\ \end{array} \right), \end{eqnarray} $

其中$ \alpha, \beta $为常数,且

(4.7) $ \begin{eqnarray} &&\omega = \partial_x^{-1}(q_1'q_4'-q_1q_4+q_2q_3-q_2'q_3'), {} \\ &&\eta = \partial_x^{-1}(-q_1'q_4'+q_1q_4-q_2q_3+q_2'q_3'). \end{eqnarray} $

将$ D_0 = D^{diag}_0+D^{off}_0 $代入(4.4)式,可得(2.16)式的贝克隆变换为

(4.8) $ \begin{eqnarray} &&q'_{1x} = \eta q'_1-\beta q_1, q'_{2x} = \eta q'_2-\alpha q_2, {} \\ &&q'_{3x} = \omega q'_3-\beta q_3, q'_{4x} = \omega q'_4-\alpha q_4, {} \\ &&q_{1x} = \alpha q'_1-\omega q_1, q_{2x} = \beta q'_2-\omega q_2, {} \\ &&q_{3x} = \alpha q'_3-\eta q_3, q_{4x} = \beta q'_4-\eta q_4. \end{eqnarray} $

接下来该文将构造超对称可积系统(2.27)的贝克隆变换.根据达布变换$ \eta' = D\eta $并假设$ D = D_0+\lambda D_1 $,方程(4.1)可推出

(4.9) $ \begin{eqnarray} D_1& = &A = d, \end{eqnarray} $

(4.10) $ \begin{eqnarray} D_{1x}& = &[A, D_0]+Q'A-AQ, \end{eqnarray} $

(4.11) $ \begin{eqnarray} D_{0x}& = &Q'D_0-D_0Q , \end{eqnarray} $

其中$ Q $和$ Q' $为

(4.12) $ \begin{eqnarray} Q = \left( \begin{array}{ccccc} 0 & 0 & q_3 & q_4 \\ 0 & 0 & q_5 & q_6 \\ q_6 & q_4 & 0 & q_1 \\ -q_5 & -q_3 & q_2 & 0 \\ \end{array} \right) , \ \ \ \ Q' = \left( \begin{array}{ccccc} 0 & 0 & q'_3 &q'_4 \\ 0 & 0 & q'_4 & q'_6 \\ q'_6 & q'_4 & 0 & q'_1 \\ -q'_5 & -q'_3 & q'_2 & 0 \\ \end{array} \right). \end{eqnarray} $

为得到贝克隆变换,假设$ D_0 = (D_{ij}), (i, j = 1, \cdots, 4) $.通过与矩阵$ B $类似的分解,矩阵$ D_0 $分解为$ D_0 = D^1_0+D^2_0 $.将(4.9)和(4.12)式代入(4.10)式,可得$ D_0^2 $中的元素为

(4.13) $ \begin{eqnarray} &&D_{13} = q'_3, \ \ D_{14} = q'_4, \ \ D_{23} = q'_5, \ \ D_{24} = q'_6, \ {} \\ &&D_{31} = q_6, \ \ D_{32} = q_4, \ \ D_{34} = \frac{1}{2}(q_1+q'_1), \ {} \\ &&D_{41} = -q_5, \ \ D_{42} = -q_3, \ \ D_{43} = \frac{1}{2}(q_2+q'_2). \end{eqnarray} $

由(4.11)式可推出

(4.14) $ \begin{eqnarray} {(D_0^1)}_x& = &Q'(D_0^1+D_0^2)-(D_0^1+D_0^2)Q. \end{eqnarray} $

将(4.9)和(4.12)式代入(4.10)式,可得$ D^1_0 $中的元素为

(4.15) $ \begin{eqnarray} &&D_{11} = \alpha_{11}, \ \ D_{12} = \alpha_{12}, {} \\ &&D_{21} = \alpha_{21}, \ \ D_{22} = \alpha_{22}, {} \\ &&D_{33} = -\partial_x^{-1}(F_{36}+F_{45}+\frac{1}{2}F_{12}), {} \\ &&D_{44} = \partial_x^{-1}(F_{36}+F_{45}-\frac{1}{2}F_{12}), \end{eqnarray} $

其中

(4.16) $ \begin{eqnarray} F_{mn} = q_mq_n-q'_mq'_n. \end{eqnarray} $

根据(4.13)和(4.15)式,可推出$ D_0 = D_0^1+D_0^2 $的具体表达式.由(4.11)式和$ D_0 $,可得(2.27)式的贝克隆变换.

(4.17) $ \begin{eqnarray} &&q'_{3x} = \frac{1}{2}q'_4(q'_2-q_2)-D_{11}q_3-D_{12}q_5+q'_3D_{33}, {} \\ &&q'_{4x} = \frac{1}{2}q'_3(q'_1+q_1)-D_{11}q_4-D_{12}q_6-q_1q'_3+q'_4D_{44}, {} \\ &&q'_{5x} = \frac{1}{2}q'_6(q'_2+q_2)-D_{21}q_3-D_{22}q_5-q_2q'_6+q'_5D_{33}, {} \\ &&q'_{6x} = \frac{1}{2}q'_5(q'_1+q_1)-D_{21}q_4-D_{22}q_6-q_1q'_5+q'_6D_{44}, {} \\ &&q_{3x} = \frac{1}{2}(q_2-q'_2)q_4+q'_5D_{12}+q'_3D_{22}-q_3D_{44}, {} \\ &&q_{4x} = \frac{1}{2}(q'_1+q_1)q_3+q'_6D_{12}+q'_4D_{22}-q_4D_{33}-q_3q'_1, {} \\ &&q_{5x} = \frac{1}{2}(q_2-q'_2)q_6+q'_5D_{11}+q_3D_{21}+q_5D_{44}, {} \\ &&q_{6x} = \frac{1}{2}(q_1+q'_1)q_5+q'_6D_{11}+q'_4D_{21}-q_6D_{33}-q_5q'_1, {} \\ &&\frac{1}{2}(q_{1x}+q'_{1x}) = 2q'_4q'_6-2q_4q_6-q_1D_{33}+q'_1D_{44}, {} \\ &&\frac{1}{2}(q_{2x}+q'_{2x}) = -2q'_3q'_5+2q_3q_5+q'_2D_{33}-q_2D_{44}. \end{eqnarray} $

下面将推导方程(3.9)的贝克隆变换.假设Lax算子为

(4.18) $ \begin{equation} K_1(Q) = \partial_x+H_1\partial_y+Q, \quad K_1(Q') = \partial_x+H_1\partial_y+Q', \end{equation} $

其中$ Q $和$ Q' $为(4.12)式,且有

(4.19) $ \begin{equation} K_2(Q) = \partial_t+H_1\partial^2_y+A\partial_y+B, \; \; \; K_2(Q') = \partial_t+H_1\partial^2_y+A\partial_y+B', \end{equation} $

同时有如下关系

(4.20) $ \begin{eqnarray} [K_1, K_2]\eta = 0, \ \ [K'_1, K'_2]\eta' = 0, \end{eqnarray} $

其中$ \eta $和$ \eta' $分别为初始和终极Lax对的特征函数.

引入规范变换

(4.21) $ \begin{eqnarray} \eta' = D(Q, Q')\eta. \end{eqnarray} $

由此可得

(4.22) $ \begin{eqnarray} K_1(Q')D(Q, Q')-D(Q, Q')K_1(Q) = 0, {} \\ K_2(Q')D(Q, Q')-D(Q, Q')K_2(Q) = 0. \end{eqnarray} $

假设(4.22)式中$ D(Q, Q') = \rho\partial_y+D_0(Q, Q') $,且满足

(4.23) $ \begin{eqnarray} &&H_1 \rho = \rho H_1, \end{eqnarray} $

(4.24) $ \begin{eqnarray} &&[H_1, D_0]+Q'\rho-\rho Q = 0, \end{eqnarray} $

(4.25) $ \begin{eqnarray} &&\partial_xD_0+H_1\partial_yD_0+Q'D_0-\rho \partial_yQ-D_0Q = 0. \end{eqnarray} $

令$ H_1 = \tilde{d} $.由(4.23)式,设$ \rho = diag(\rho_1, \rho_2, \rho_3, \rho_4) $,其中$ \rho_i, (i = 1, \cdots, 4) $为常数.由方程(4.24)推出$ D^2_0 $中的元素为

(4.26) $ \begin{eqnarray} && D_{13} = \rho_3q'_3-\rho_1q_3, \ \ D_{14} = \rho_1q_4-\rho_4q'_4, \ \ D_{23} = \rho_3q'_5-\rho_2q_5, {} \\ && D_{24} = \rho_2q_6-\rho_4q'_6, \ \ D_{31} = \rho_3q_6-\rho_1q'_6, \ \ D_{32} = \rho_3q_4-\rho_2q'_4, {} \\ && D_{34} = -\frac{1}{2}(\rho_4q'_1-\rho_3q_1), \ \ D_{41} = \rho_4q_5-\rho_1q'_5, {} \\ && D_{42} = \rho_4q_3-\rho_2q'_3, \ \ D_{43} = \frac{1}{2}(\rho_3q'_2-\rho_4q_2). \end{eqnarray} $

由方程(4.25),可得$ D^1_0 $中的元素为

(4.27) $ \begin{eqnarray} &&D_{11} = \partial^{-1}_x[\rho_1(q'_3q'_6+q'_4q'_5+q_4q_5-q_3q_6)+2\rho_3q'_3q_6-2\rho_4q'_4q_5], {} \\ &&D_{12} = \partial^{-1}_x[2\rho_2q'_3q'_4-2\rho_1q_3q_4], {} \\ &&D_{21} = \partial^{-1}_x[2\rho_1q'_5q'_6-2\rho_4q_5q'_6], {} \\ &&D_{22} = \partial^{-1}_x[\rho_2(q'_3q'_6+q'_4q'_5-q_4q_5+q_3q_6)-2\rho_4q_3q'_6], {} \\ &&D_{33} = (\partial_x+\partial_y)^{-1}[\rho_3(q_3q_6+q_4q_5-q'_3q'_6-q'_4q'_5+\frac{1}{2}q_1q_2-\frac{1}{2}q'_1q'_2)], {} \\ &&D_{44} = (\partial_x-\partial_y)^{-1}[\rho_4(q_3q_6+q_4q_5-q'_3q'_6-q'_4q'_5+\frac{1}{2}q'_1q'_2-\frac{1}{2}q_1q_2)]. \end{eqnarray} $

根据$ D_0 = D_0^1+D_0^2 $,可得$ D_0 $具体表达式.将$ D_0 $代入(4.25)式即可得到方程(3.9)的贝克隆变换

(4.28) $ \begin{eqnarray} &&D_{13x}+q'_3D_{33}+q'_4D_{43}-D_{11}q_3-D_{12}q_5-D_{14}q_2-\rho_1q_{3y} = 0, {} \\ &&D_{14x}+q'_3D_{34}+q'_4D_{44}-D_{11}q_4-D_{12}q_6-D_{13}q_1-\rho_1q_{4y} = 0, {} \\ &&D_{23x}+q'_5D_{33}+q'_6D_{43}-D_{21}q_3-D_{22}q_5-D_{24}q_2-\rho_2q_{5y} = 0, {} \\ &&D_{24x}+q'_5D_{34}+q'_6D_{44}-D_{21}q_4-D_{22}q_6-D_{23}q_1-\rho_2q_{6y} = 0, {} \\ &&(\partial_x+\partial_y)D_{31}+q'_6D_{11}+q'_4D_{21}+q'_1D_{41}-q_6D_{33}+q_5D_{34}-\rho_3q_{6y} = 0, {} \\ &&(\partial_x+\partial_y)D_{32}+q'_6D_{12}+q'_4D_{22}+q'_1D_{42}-q_4D_{33}+q_3D_{34}-\rho_3q_{4y} = 0, {} \\ &&(\partial_x+\partial_y)D_{34}+q'_6D_{14}+q'_4D_{24}+q'_1D_{44}-q_4D_{31}-q_6D_{32}-q_1D_{33}-\rho_3q_{1y} = 0, {} \\ &&(\partial_x-\partial_y)D_{41}-q'_5D_{11}-q'_3D_{21}+q'_2D_{31}-q_6D_{43}+q_5D_{44}+\rho_4q_{5y} = 0, {} \\ &&(\partial_x-\partial_y)D_{42}-q'_5D_{12}-q'_3D_{22}+q'_2D_{32}-q_4D_{43}+q_3D_{44}+\rho_4q_{3y} = 0, \\ &&(\partial_x-\partial_y)D_{43}-q'_5D_{13}-q'_3D_{23}+q'_2D_{33}-q_3D_{41}-q_5D_{42}-q_2D_{44}-\rho_4q_{2y} = 0.{} \end{eqnarray} $

该文基于超李代数$ ops(2/2) $构造了新的(2+1)维超对称可积非线性演化方程.该文分别利用齐性空间以及扩展系统维数两种方法,构造了新的(2+1)维超对称系统.进一步给出了这些高维可积方程的贝克隆变换.超对称可积系统和超李代数在物理方面引起了广泛的关注.对于该文所提出的多个高维超对称可积方程在物理中的应用,值得在后续工作中进一步研究.