从矩阵谱问题或Lax对出发, 零曲率公式提供了一个构造孤子梯队的有效方法.成功的例子包括Ablowitz-Kaup-Newell-Segur (AKNS)梯队, Kaup-Newell (KN)梯队, Wadati-Konno-Ichikawa (WKI)梯队, Korteweg-de Vries (KdV)梯队, Dirac梯队和Boiti-Pempinelli-Tu (BPT)梯队等^[1-38].基于矩阵谱问题, 孤子梯队通常具有双哈密顿结构从而是Liouville可积的.特别是当矩阵圈代数是半单的时候, 相应的双哈密顿结构能够利用迹恒等式得到, 参见文献[9, 11, 18-20, 22, 27-29].

经典的AKNS关于空间变量的谱问题是

$ \begin{eqnarray}\label{a1} \phi_x=(\lambda e_1+qe_2+pe_3)\phi =\left(\begin{array}{ccc}-\lambda~~&q\\ p~~&\lambda\end{array}\right)\phi, \end{eqnarray} $

(1.1)

这里 $e_1=\left(\begin{array}{ccc}-1~~&0\\ 0~~&1\end{array}\right), \ e_2=\left(\begin{array}{ccc}0~~&1\\ 0~~&0\end{array}\right), \ e_3=\left(\begin{array}{ccc}0~~&0\\ 1~~&0\end{array}\right) $ 是李代数sl $(2, {\Bbb R})$ 的基.文献[6, 13]的作者引入非线性项 $\alpha pq$ , 提出了如下的两分量扰动AKNS关于空间变量的谱问题

$ \begin{eqnarray}\label{yz} \phi_x=((\lambda+\alpha pq) e_1+qe_2+pe_3)\phi =\left(\begin{array}{ccc}-\lambda-\alpha pq~~&q\\ p~~&\lambda+\alpha pq\end{array}\right)\phi, \end{eqnarray} $

(1.2)

并构造了双哈密顿结构.我们将把上述谱问题(1.2)推广到多分量的情形, 即构造了相应的多分量孤子梯队和双哈密顿结构.第二节中, 我们考虑如下的基于李代数sl $(m+1, {\Bbb R})$ 的多分量扰动AKNS关于空间变量的谱问题

$ \begin{eqnarray}\label{GAKNS1} \phi_x=U(u, \lambda)\phi= \left(\begin{array}{ccc} -m(\lambda+\alpha r)&q\\ p&(\lambda+\alpha r)I_m \end{array}\right)\phi, \quad u=\left(\begin{array} {c} q^T \cr p \end{array}\right), \quad r=qp, \end{eqnarray} $

(1.3)

这里 $q=(q_1, q_2, \cdots, q_m)$ , $p=(p_1, p_2, \cdots, p_m)^{T}$ , $\lambda$ 是谱参数, $I_m$ 是 $m\times m$ 单位矩阵.当 $\alpha=0$ 时, 相应的谱问题已经在文献[12, 15, 20, 28, 34]中得到解决.特别地, $m=2$ 的情形约化为四分量的AKNS谱问题^[16].进而当 $m=1$ 时该谱问题约化为经典的AKNS情形.第三节将给出一些总结.

首先我们求解稳态零曲率方程

$ \begin{eqnarray} W_x=[U, W].\label{SZC1} \end{eqnarray} $

(2.1)

我们设 $W$ 为

$ \begin{eqnarray} W=\left(\begin{array}{ccc}a~~&b\\c~~&d\end{array}\right), \end{eqnarray} $

(2.2)

这里 $a$ 是标量, $b^T$ 和 $c$ 是 $m$ -维列向量, $d$ 是 $m\times m$ 矩阵.从而零曲率方程(2.1等价于

$ \begin{equation}\left\{\begin{array}{l} a_x=qc-bp, \\ d_x=pb-cq, \\ b_x=qd-aq-(m+1)(\lambda+\alpha r)b, \\ c_x=pa-dp+(m+1)(\lambda+\alpha r)c. \end{array}\right. \end{equation} $

(2.3)

所以我们可以得到

$ \mbox{tr}(d)_x=\mbox{tr}(d_x) =\mbox{tr}(pb-cq)=bp-qc=-a_x. $

接下来我们构造 $W$ 的形式解

$ \begin{eqnarray} W=\left(\begin{array}{ccc}a~~&b\\c~~&d\end{array}\right)= \sum\limits_{k=0}^{\infty}W_k\lambda^{-k}=\sum\limits_{k=0}^{\infty} \left(\begin{array}{ccc}a^{(k)}~~&b^{(k)}\\c^{(k)}~~&d^{(k)}\end{array}\right)\lambda^{-k}, \end{eqnarray} $

(2.4)

这里 $b^{(k)}, c^{(k)}$ 和 $d^{(k)}$ 设为

$ \begin{eqnarray} b^{(k)}=(b_1^{(k)}, b_2^{(k)}, \cdots, b_m^{(k)}), \quad c^{(k)}=(c_1^{(k)}, c_2^{(k)}, \cdots, c_m^{(k)})^T, \quad d^{(k)}=(d_{ij}^{(k)})_{m\times m}. \end{eqnarray} $

(2.5)

在上述假设下, 我们有如下递推关系式

$ \begin{equation}\label{RRS} \left\{\begin{array}{l} b^{(0)}=0, \quad c^{(0)}=0, \quad a^{(0)}_x=0, \quad d^{(0)}_x=0, \\ a^{(k)}_{x}=qc^{(k)}-b^{(k)}p, \ d^{(k)}_x=pb^{(k)}-c^{(k)}q, \\ b^{(k+1)}=\frac{1}{m+1}[qd^{(k)}-a^{(k)}q-b^{(k)}_x-(m+1)\alpha rb^{(k)}], \\ c^{(k+1)}=\frac{1}{m+1}[c^{(k)}_x-pa^{(k)}+d^{(k)}p-(m+1)\alpha rc^{(k)}], \quad k\ge 0.\\ \end{array}\right. \end{equation} $

(2.6)

从而对于 $b^{(k)}$ 和 $c^{(k)}$ 有

$ \begin{eqnarray} \left(\begin{array}{ccc} c^{(k+1)}\\{b^{(k+1)}}^T \end{array}\right)=L\left(\begin{array}{ccc}c^{(k)}\\{b^{(k)}}^T\end{array}\right), \qquad k\ge 0, \end{eqnarray} $

(2.7)

这里算子 $L$ 可写为

$ \begin{eqnarray} &L=\frac{1}{m+1}\\ &\left(\begin{array}{ccc} \begin{array}{l} \Big[\partial-\beta(m+1)r -\sum\limits_{i=1}^m p_i\partial^{-1}q_i\Big]I_m\\ -p\partial^{-1}q \end{array} & p\partial^{-1}p^T+(p\partial^{-1}p^T)^T\\ -q^T\partial^{-1}q-(q^T\partial^{-1}q)^T& \begin{array}{l} \Big[-\partial-\beta(m+1)r +\sum\limits_{i=1}^mq_i\partial^{-1}p_i\Big]I_m\\ +q^T\partial^{-1}p^T\end{array} \end{array}\right).\end{eqnarray} $

(2.8)

设初值为

$ \begin{eqnarray} a^{(0)}=-m, \qquad d^{(0)}=I_m, \label{initialvalues-1stsh} \end{eqnarray} $

(2.9)

并且令

$ \begin{eqnarray} W_k|_{u=0}=0, \qquad k\ge 1, \label{zeroconstantofintegration-1stsh} \end{eqnarray} $

(2.10)

则我们可以一致地确定序列 $\{a^{(k)}, b^{(k)}, c^{(k)}, d^{(k)}|\, \, k\ge 1\}$ 的表达式.例如前三组为

$ \begin{eqnarray*}\left\{\begin{array}{l} a^{(1)}=0, \quad b^{(1)}=q, \quad c^{(1)}=p, \quad d^{(1)}=(0)_{m\times m};\\ a^{(2)}=\frac{r}{m+1}, \ b^{(2)}=\frac{1}{m+1}[-q_x-(m+1)\alpha rq], \\ c^{(2)}=\frac{1}{m+1}[p_x-(m+1)\alpha rp], \ d^{(2)}=\frac{1}{m+1}(-pq);\\ a^{(3)}=\frac{1}{(m+1)^2}[qp_x-q_xp-2(m+1)\alpha r^2], \\ d^{(3)}=\frac{1}{(m+1)^2}[pq_x-p_xq+2(m+1)\alpha rpq], \\ b^{(3)}=\frac{1}{(m+1)^2}\left\{q_{xx}-2rq+(m+1)\beta[r_xq+2rq_x+(m+1)\alpha r^2q]\right\}, \\ c^{(3)}=\frac{1}{(m+1)^2}\left\{p_{xx}-2pr-(m+1)\beta[r_xp+2rp_x-(m+1)\alpha r^2p]\right\}.\end{array}\right. \end{eqnarray*} $

显然, (2.6), (2.9)和(2.10)式保证了 $W_k\in \textrm{sl}(m+1, {\Bbb R}), \quad k\ge 0$ .

更进一步, 我们记关于时间变量的矩阵谱问题为

$ \begin{eqnarray}\label{AP1}&& \phi_{t_n}=V^{(n)}\phi, \nonumber\\ && V^{(n)}=(\lambda^nW)_++\Delta_n=\sum\limits_{j=0}^nW_j\lambda^{n-j}+\left(\begin{array}{ccc}e_n~~&0\\ 0~~&h_n\end{array}\right), \qquad n\ge 0, \end{eqnarray} $

(2.11)

这里 $P_+$ 定义为 $\lambda$ 的多项式部分. (1.3)和(2.11)式的相容性条件, 即零曲率方程

$ \begin{eqnarray} U_{t_n}-V^{(n)}_x+[U, V^{(n)}]=0, \qquad n\ge 0, \end{eqnarray} $

(2.12)

给出了

$ \begin{equation}\left\{\begin{array}{l} e_{n_x}=-m\alpha r_{t_n}, \qquad h_{n_x}=\alpha r_{t_n}I_m, \\ q_{t_n}=-(m+1)b^{(n+1)}-qh_n+e_nq, \\ p_{t_n}=(m+1)c^{(n+1)}-pe_n+h_np, \end{array}\right.\qquad n\ge 0. \end{equation} $

(2.13)

取 $e_n=-mf_n$ 和 $h_n=f_nI_m$ , 我们有

$ \begin{equation}\left\{\begin{array}{l} {f_n}_x=\alpha r_{t_n}=\beta(m+1)(qc^{(n+1)}-b^{(n+1)}p), \\ q_{t_n}=-(m+1)(b^{(n+1)}+f_nq), \\ p_{t_n}=(m+1)(c^{(n+1)}+f_np), \qquad n\ge0.\end{array}\right. \end{equation} $

(2.14)

从而解第一个方程可以得到 $f_n=\alpha (m+1)a^{(n+1)}$ , 即有

$ \begin{equation} \Delta _n= \alpha (m+1) a^{(n+1)} \left(\begin{array}{ccc} -m~~ &0\\ 0~~& I_m \end{array}\right), \qquad n\ge 0. \end{equation} $

(2.15)

这样我们就得到了如下的多分量扰动AKNS孤子梯队

$ \begin{equation}\label{pAKNS} \left\{\begin{array}{l} q_{t_n}=-(m+1)[b^{(n+1)}+\alpha (m+1)\partial^{-1}(qc^{(n+1)}-b^{(n+1)}p)q], \\ p_{t_n}=(m+1)[c^{(n+1)}+\alpha (m+1)\partial^{-1}(qc^{(n+1)}-b^{(n+1)}p)p], \end{array}\right.\qquad n\ge 0. \end{equation} $

(2.16)

(2.16)式写成算子方程的形式有

$ \begin{eqnarray}\label{SH1} u_{t_n}=\left(\begin{array}{ccc}q^T\\p \end{array}\right)_{t_n}=K_n=R\left(\begin{array}{ccc}c^{(n+1)}\\{b^{(n+1)}}^T\end{array}\right), \qquad n\ge 0, \end{eqnarray} $

(2.17)

这里算子 $R$ 定义为

$ \begin{eqnarray} R=(m+1) \left(\begin{array}{ccc} -\alpha (m+1)q^T\partial^{-1}q~~&-I_m+\alpha (m+1)q^T\partial^{-1}p^T\\ I_m+\alpha (m+1)p\partial^{-1}q~~&-\alpha (m+1)p\partial^{-1}p^T \end{array}\right). \end{eqnarray} $

(2.18)

在该孤子梯队(2.17)中, 第一个非平凡的非线性系统为

$ \left\{\begin{array}{l} p_{t_2}=\frac{1}{m+1}(p_{xx}-2rp)-\alpha [2(pq)_xp+(m+1)\alpha r^2p], \\ q_{t_2}=-\frac{1}{m+1}(q_{xx}-2rq)-\alpha [2q(pq)_x-(m+1)\alpha r^2q].\end{array}\right. $

当 $m=2, ~ \alpha=0$ 时, 上述系统退化为著名的非线性Burgers系统.类似地, mKdV系统可以从(2.16)式的第二个非线性系统中得到.

最后, 我们利用如下的迹恒等式^{[9, 11, 27, 29]}

$ \frac{\delta}{\delta u}\int \mbox{tr}\Bigl(W\frac{\partial U}{\partial \lambda}\Bigr){\rm d}x=\lambda^{-\gamma}\frac{\partial}{\partial\lambda}\Bigl[\lambda^\gamma \mbox{tr}\Bigl(W\frac{\partial U}{\partial u}\Bigr)\Bigr] $

来构造孤子梯队(2.17)的双哈密顿结构.

直接计算有

$ \begin{equation}\label{TI1}\left\{\begin{array}{l} \mbox{tr}\Bigl(W\frac{\partial U}{\partial \lambda}\Bigr)=-ma+\mbox{tr}(d), \\ \mbox{tr}\Bigl(W\frac{\partial U}{\partial u}\Bigr) =-\alpha m a\left(\begin{array}{ccc}p\\ q^T\end{array}\right)+ \left(\begin{array}{ccc}c\\ b^T\end{array}\right)+\alpha \mbox{tr}(d)\left(\begin{array}{ccc}p\\q^T\end{array}\right). \end{array}\right. \end{equation} $

(2.19)

把(2.19)式代入上述迹恒等式, 平衡 $\lambda$ 的次数, 我们有

$ \begin{eqnarray*} &&\frac{\delta}{\delta u}\int\Bigl(-ma^{(k+1)} +\sum\limits_{i=1}^md_{ii}^{(k+1)}\Bigr){\rm d}x\\ &=&(\gamma-k) \Bigl[-\alpha m\left(\begin{array}{ccc}p\\ q^T\end{array}\right)a^{(k)}+\left(\begin{array}{ccc}c^{(k)}\\ {b^{(k)}}^T\end{array}\right)+\alpha \left(\begin{array}{ccc}p\\q^T\end{array}\right) \sum\limits_{i=1}^md_{ii}^{(k)} \Bigr]. \end{eqnarray*} $

为了确定常数 $\gamma$ 的值, 我们在上式中令 $k=1$ 即可得到 $\gamma=0$ .从而我们有

$ \begin{eqnarray} \frac{\delta}{\delta u}{\cal H}_k=\Bigl[-\alpha m\left(\begin{array}{ccc}p\\ q^T\end{array}\right)a^{(k)}+\left(\begin{array}{ccc}c^{(k)}\\ {b^{(k)}}^T\end{array}\right)+\alpha \left(\begin{array}{ccc}p\\q^T\end{array}\right)\sum\limits_{i=1}^md_{ii}^{(k)} \Bigr], \qquad k\ge 0. \end{eqnarray} $

(2.20)

这里的哈密顿泛函 ${\cal H}_k$ 定义为

$ \begin{eqnarray}&&{\cal H}_0=\alpha m(m+1)\int r {\rm d}x, \nonumber\\&& {\cal H}_k=\frac{\delta}{\delta u}\int\frac{\Bigl(ma^{(k+1)} -\sum\limits_{i=1}^md_{ii}^{(k+1)}\Bigr)}{k}{\rm d}x, \qquad k\ge 1. \end{eqnarray} $

(2.21)

为了给出哈密顿结构和哈密顿算子, 我们记

$ \begin{eqnarray} G^{(k)}=\Bigl[-\alpha m\left(\begin{array}{ccc}p\\ q^T\end{array}\right)a^{(k)}+\left(\begin{array}{ccc}c^{(k)}\\ {b^{(k)}}^T\end{array}\right)+\alpha \left(\begin{array}{ccc}p\\q^T\end{array}\right)\sum\limits_{i=1}^md_{ii}^{(k)} \Bigr], \qquad k\ge 0, \end{eqnarray} $

(2.22)

则有

$ \begin{eqnarray} G^{(k)}&=&-\alpha m\left(\begin{array}{ccc}p\\ q^T\end{array}\right)a^{(k)}+\left(\begin{array}{ccc}c^{(k)}\\ {b^{(k)}}^T\end{array}\right)+\alpha \left(\begin{array}{ccc}p\\ q^T\end{array}\right)\partial^{-1}(b^{(k)}p-qc^{(k)})\nonumber\\ &=&N^{-1}\left(\begin{array}{ccc} c^{(k)}\\{b^{(k)}}^T \end{array}\right), \qquad k\ge 0, \end{eqnarray} $

(2.23)

这里算子 $N^{-1}$ 是

$ \begin{eqnarray} N^{-1}=\left(\begin{array}{ccc} I_m-\alpha (m+1)p\partial^{-1}q~~&\alpha (m+1)p\partial^{-1}p^T\\ -\alpha (m+1)q^T\partial^{-1}q~~&I_m+\alpha (m+1)q^T\partial^{-1}p^T \end{array}\right). \end{eqnarray} $

(2.24)

从而我们有

$ \begin{eqnarray} &&\left(\begin{array}{ccc} c^{(k)}\\{b^{(k)}}^T \end{array}\right)=NG^{(k)}, \nonumber\\ &&N=\left(\begin{array}{ccc} I_m+\alpha (m+1)p\partial^{-1}q~~&-\alpha (m+1)p\partial^{-1}p^T\\ \alpha (m+1)q^T\partial^{-1}q~~&I_m-\alpha (m+1)q^T\partial^{-1}p^T \end{array}\right), \qquad k\ge 0. \end{eqnarray} $

(2.25)

这样我们已经可以得到孤子梯队(2.17)如下的哈密顿结构

$ \begin{eqnarray} u_{t_n}=K_n=J\frac{\delta {\cal H}_{n+1}}{\delta u}. \end{eqnarray} $

(2.26)

这里的哈密顿算子 $J$ 定义为

$ \begin{eqnarray} J=RN=(m+1)\left(\begin{array}{ccc}-2\alpha (m+1)q^T\partial^{-1}q~~&-I_m+2\alpha (m+1)q^T\partial^{-1}p^T\\ I_m+2\alpha (m+1)p\partial^{-1}q~~&-2\alpha (m+1)p\partial^{-1}p^T \end{array}\right). \end{eqnarray} $

(2.27)

从关系式 $G^{(n+1)}=\Psi G^{(n)}$ , $\Psi=N^{-1}LN$ , $K_{n+1}=\Phi K_n, \, \, n\ge 0$ 和 $J\Psi=\Phi J$ , 我们可以得到孤子梯队(2.17)的遗传递推算子

$ \begin{eqnarray} \Phi=\Psi^{\dagger}=N^\dagger L^\dagger N^{-\dagger}, \end{eqnarray} $

(2.28)

这里算子 $\Psi^\dagger$ 定义为 $\Psi$ 的共轭算子.从而 $\Phi=(\Phi_{ij})_{2\times 2}$ 的显式表达式可以算出

$ \begin{equation}\left\{\begin{array}{rl} \Phi_{11}=& \alpha ^2(m+1)q^T\partial^{-1}q\partial p\partial^{-1}p^T\\ & +\alpha q^T\partial^{-1}p^T\{[\partial+\alpha (m+1)r]I_m +\alpha (m+1)\partial q^T\partial^{-1}p^T\}\\ & +\frac{1}{m+1}\bigl\{q^T\partial^{-1}p^T-\bigl[\partial+\alpha (m+1)r-\sum\limits_{i=1}^{m}q_i\partial^{-1}p_i\bigr]I_m\\ & -\alpha (m+1)[\partial+\alpha (m+1)r]q^T\partial^{-1}p^T\bigr\}, \\ \Phi_{12}=& \frac{1}{m+1}\{q^T\partial^{-1}q+(q^T\partial^{-1}q)^T-\alpha (m+1)[\partial+\alpha (m+1)r]q^T\partial^{-1}q\}\\ & +\alpha ^2(m+1)q^T\partial^{-1}p^T\partial q^T\partial^{-1}q\\ & -\alpha q^T\partial^{-1}q\{[\partial-\alpha (m+1)r]I_m-\alpha (m+1)\partial p\partial^{-1}q\}, \\ \Phi_{21}=& -\frac{1}{m+1}\{p\partial^{-1}p^T+(p\partial^{-1}p^T)^T+\alpha (m+1)[\partial-\alpha (m+1)r]p\partial^{-1}p^T\}\\ & -\alpha ^2(m+1)p\partial^{-1}q\partial p\partial^{-1}p^T\\ & -\alpha p\partial^{-1}p^T\{[\partial+\alpha (m+1)r]I_m+\alpha (m+1)\partial q^T\partial^{-1}p^T\}, \\ \Phi_{22}=& -\alpha ^2(m+1)p\partial^{-1}p^T\partial q^T\partial^{-1}q\\ & +\alpha p\partial^{-1}q\{[\partial-\alpha (m+1)r]I_m-\alpha (m+1)\partial p\partial^{-1}q\}\\ & +\frac{1}{m+1}\bigl\{-p\partial^{-1}q+\bigl[\partial-\alpha (m+1)r-\sum\limits_{i=1}^mp_i\partial^{-1}q_i\bigr]I_m\\ & -\alpha (m+1)[\partial-\alpha (m+1)r]p\partial^{-1}q\bigr\}. \end{array}\right. \end{equation} $

(2.29)

所以我们得到了梯队(2.17)的双哈密顿结构

$ \begin{eqnarray}\label{BH1} u_{t_n}=K_n=J\frac{\delta {\cal H}_{n+1}}{\delta u}=M\frac{\delta {\cal H}_{n}}{\delta u}, \qquad n\ge 0. \end{eqnarray} $

(2.30)

这里第二个哈密顿算子 $M=(M_{ij})_{2\times 2}$ 由式子 $M=\Phi J$ 给出, 其显式表达式为

$ \begin{equation}\left\{\begin{array}{rl} M_{11}= & \alpha (m+1)\left[(\partial +\alpha (m+1)r)I_m-\alpha (m+1)q^T\partial^{-1}p^T\partial\right]q^T\partial^{-1}q\\ & +q^T\partial^{-1}q+(q^T\partial^{-1}q)^T\\ & -\alpha (m+1)q^T\partial^{-1}q[(\partial-\alpha (m+1)r)I_m+\alpha (m+1)\partial p\partial^{-1}q], \\ M_{12}=& -q^T\partial^{-1}p^T+\bigl[\partial+\alpha (m+1)r-\sum\limits_{i=1}^mq_i\partial^{-1}p_i\bigr]I_m\\ & -\alpha (m+1)[\partial+\alpha (m+1)r]q^T\partial^{-1}p^T +\alpha ^2(m+1)^2q^T\partial^{-1}q\partial p\partial^{-1}p^T\\ & -\alpha (m+1)q^T\partial^{-1}p^T[(\partial+\alpha (m+1)r)I_m+\alpha (m+1)\partial q^T\partial^{-1}p^T], \\ M_{21}=& -p\partial^{-1}q+\bigl[\partial-\alpha (m+1)r-\sum\limits_{i=1}^mp_i\partial^{-1}q_i\bigr]I_m\\ & +\alpha (m+1)[\partial-\alpha (m+1)r]p\partial^{-1}q +\alpha ^2(m+1)^2p\partial^{-1}p^T\partial q^T\partial^{-1}q\\ & +\alpha (m+1)p\partial^{-1}q[(\partial-\alpha (m+1)r)I_m+\alpha (m+1)\partial p\partial^{-1}q], \\ M_{22}=& -\alpha (m+1)\left[(\partial-\alpha (m+1)r)I_m+\alpha (m+1)p\partial^{-1}q\partial\right]p\partial^{-1}p^T\\ & +p\partial^{-1}p^T+(p\partial^{-1}p^T)^T\\ & +\alpha (m+1)p\partial^{-1}p^T[(\partial+\alpha (m+1)r)I_m-\alpha (m+1)\partial q^T\partial^{-1}p^T]. \end{array}\right. \end{equation} $

(2.31)

双哈密顿结构意味着该多分量扰动AKNS孤子梯队(2.17)是Liouville可积的^[36].另外守恒泛函 $\{{\cal H}_k\}_{k=0}^{\infty}$ 和对称 $\{K_k\}_{k=0}^{\infty}$ 有关系式

$ \begin{equation}\left\{\begin{array}{l} \{{\cal H}_k, {\cal H}_l\}_{J}=\int\Big ( \frac{\delta{\cal H}_k}{\delta u}\Big)^T J\frac{\delta {\cal H}_l}{\delta u}{\rm d}x=0, \\ \{{\cal H}_k, {\cal H}_l\}_{M}=\int \Big( \frac{\delta{\cal H}_k}{\delta u}\Big)^T M\frac{\delta {\cal H}_l}{\delta u}{\rm d}x=0, \end{array}\right. \qquad k, l\ge 0 \end{equation} $

(2.32)

和

$ \begin{eqnarray} [K_k, K_l]=K_k'(u)[K_l]-K_l'(u)[K_k]=0, \qquad k, l\ge 0. \end{eqnarray} $

(2.33)

通过引入一个非线性项 $r=qp=\sum\limits_{i=1}^mq_ip_i$ , 我们基于李代数sl $(m+1, {\Bbb R})$ 提出了一个新的矩阵谱问题.利用零曲率公式, 我们构造了相应的多分量扰动AKNS孤子梯队.最后, 我们利用迹恒等式建立了双哈密顿结构.在附录中我们列出了一些新的恒等式.

从如下的关于空间变量的矩阵谱问题

$ \phi_x=U(u, \lambda)\phi=\left(\begin{array}{ccc} 0&~-q~&-m(\lambda+\alpha r)\\ q^T&0&-p\\ m(\lambda+\alpha r)&p^T&0 \end{array}\right)\phi, \quad u=\left(\begin{array} {c} q^T \cr p \end{array}\right), \quad r=qq^T+p^Tp $

和 $W=\left(\begin{array}{ccc}0&~~-c~~&-a\\ c^T&d&-b\\ a&b^T&0\end{array}\right)$ 出发, 我们也可以构造基于李代数so $(m+2, {\Bbb R})$ 的多分量扰动AKNS孤子梯队.方便起见, 我们省略了类似的计算过程.

设 $b^{(k)}$ 和 $q=(q_1, \cdots, q_m)$ 是 $m$ -维行向量, $c^{(k)}$ 和 $p=(p_1, \cdots, p_m)^T$ 是 $m$ -维列向量, 容易证明下列恒等式成立.

$ \partial^{-1}(c^{(k)}q)p= \Big(\sum\limits_{j=1}^m p_j\partial^{-1}q_j\Big)(c^{(k)}), \quad p\partial^{-1}(b^{(k)}p)=p\partial^{-1}(p^T{b^{(k)}}^T), $

$ \partial^{-1}(pb^{(k)})p={(p\partial^{-1}p^T)}^T({b^{(k)}}^T), \quad \partial^{-1}({b^{(k)}}^Tp^T)q^T= \Big(\sum\limits_{j=1}^mq_j\partial^{-1}p_j\Big)({b^{(k)}}^T), $

$ q^T\partial^{-1}({c^{(k)}}^Tq^T)=q^T\partial^{-1}(qc^{(k)}), \quad\partial^{-1}(q^T{c^{(k)}}^T)q^T=(q^T\partial^{-1}q)^T(c^{(k)}), $

$ \Big(\sum\limits_{j=1}^mq_j\partial^{-1}p_j\Big)q^T\partial^{-1}p^T= (q^T\partial^{-1}q)^Tp\partial^{-1}p^T, $

$ \Big(\sum\limits_{j=1}^mq_j\partial^{-1}p_j\Big)q^T\partial^{-1}q= (q^T\partial^{-1}q)^Tp\partial^{-1}q, $

$ \Big(\sum\limits_{j=1}^mp_j\partial^{-1}q_j\Big)p\partial^{-1}p^T= (p\partial^{-1}p^T)^Tq^T\partial^{-1}p^T, $

$ \Big(\sum\limits_{j=1}^mp_j\partial^{-1}q_j\Big)p\partial^{-1}q=(p\partial^{-1}p^T)^Tq^T\partial^{-1}q, $

$ q^T\partial^{-1}p^T\Big(\sum\limits_{j=1}^mq_j\partial^{-1}p_j\Big)= q^T\partial^{-1}q(p\partial^{-1}p^T)^T, $

$ q^T\partial^{-1}q \Big(\sum\limits_{j=1}^mp_j\partial^{-1}q_j\Big)= q^T\partial^{-1}p^T(q^T\partial^{-1}q)^T, $

$ p\partial^{-1}p^T\Big(\sum\limits_{j=1}^mq_j\partial^{-1}p_j\Big)= p\partial^{-1}q(p\partial^{-1}p^T)^T, $

$ p\partial^{-1}q \Big(\sum\limits_{j=1}^mp_j\partial^{-1}q_j\Big)=p\partial^{-1}p^T(q^T\partial^{-1}q)^T. $