自对偶Yang-Mills (SDYM)方程^[1]}$是最重要的偏微分方程之一, 无论是在数学物理领域还是在拓朴和几何领域^{[2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]}考虑到许多著名的可积方程如Korteweg-de Vries (KdV)方程、 非线性Schr\"{o}dinger方程、Kadomtsev-Petviashvili (KP) 方程和Davey-Stewartson (DS)方程能够从SDYM方程约化得到, 我们自然希望该类方程能包含“所有”的可积方程^[5] 近年来,SDYM方程已经被推广到时空非交换的情形^{[17, 18, 19, 20, 21]}

对于偏微分方程来说,特别是SDYM方程及各种推广形式, 达布变换提供了一种纯代数的方法用于构造精确解^{[6, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19]} 我们首先构造一类新的带负幂次谱参数的广义SDYM (gSDYM)方程. 对于如下的微分算子

\begin{equation}\label{b1} L_i(\lambda)=\frac{\partial}{\partial p_i}-a_i(\lambda) \frac{\partial}{\partial x_i}=\frac{\partial}{\partial p_i} -\left(\sum_{k=-M}^M a_{ik}\lambda^{k+1}\right)\frac{\partial} {\partial x_i},\qquad 1\leq i \leq n, \end{equation}

(1.1)

我们考虑谱问题

\begin{equation}\label{b2} L_i(\lambda)\Psi=-A_i(\lambda)\Psi= -\left(\sum_{l=-M}^M A_{il}\lambda^l\right)\Psi,\qquad 1\leq i \leq n, \end{equation}

(1.2)

这里$\lambda$是谱参数,$\Psi$是$N\times N$的特征函数矩阵, $a_{ik},-M\leq k \leq M,1\leq i \leq n$是常数, $p=(p_1,p_2,\cdots,p_n)$,$x=(x_1,x_2,\cdots,x_n)\in {\Bbb C}^n$或${\Bbb R}^n$是两个向量并且$A_{il}, -M\leq l \leq M,1\leq i \leq n$都是依赖于$p,x$的关于势函数的 $N\times N$矩阵. 注意到$L_i(\lambda),1\leq i \leq n$都是线性算子,我们计算 \begin{eqnarray*} L_j(\lambda)L_i(\lambda)\Psi &=&L_j(\lambda)\bigg(-\sum_{l=-M}^M A_{il}\lambda^l\Psi\bigg)\nonumber\\ &=&-\sum_{l=-M}^M A_{il}\lambda^l L_j(\lambda)\Psi-\sum_{l=-M}^M \lambda^l(L_j\left(\lambda)A_{il}\right)\Psi \nonumber\\ &=&\sum_{l=-M}^M \lambda^l A_{il}\sum_{k=-M}^M \lambda^k A_{jk}\Psi-\sum_{l=-M}^M \lambda^l\left(\frac{\partial A_{il}}{\partial p_j}-a_j(\lambda)\frac{\partial A_{il}}{\partial x_j}\right)\Psi \nonumber\\ &=&\sum_{m=-2M}^{2M}\lambda^m \bigg(\sum_{k+l=m \atop -M\leq k,l \leq M}A_{ik}A_{jl}\bigg) \Psi-\sum_{m=-M}^{M}\lambda^m \frac{\partial A_{im}}{\partial p_j}\Psi \nonumber\\ & & +\sum_{m=-2M}^{2M}\lambda^{m+1} \bigg(\sum_{k+l=m \atop -M\leq k,l \leq M}a_{jk} \frac{\partial A_{il}}{\partial x_j}\bigg)\Psi. \end{eqnarray*} 这样,相容性条件

\begin{equation}\label{b3} L_j(\lambda)L_i(\lambda)\Psi=L_i(\lambda)L_j(\lambda)\Psi,\qquad 1\leq i,j \leq n \end{equation}

(1.3)

能展开为 \begin{eqnarray*} & & \sum_{m=-2M}^{2M}\lambda^m \sum_{k+l=m \atop -M\leq k,l \leq M}A_{ik}A_{jl}-\sum_{m=-M}^{M}\lambda^m \frac{\partial A_{im}}{\partial p_j} +\sum_{m=-2M}^{2M} \lambda^{m+1}\sum_{k+l=m \atop -M\leq k,l \leq M}a_{jk} \frac{\partial A_{il}}{\partial x_j}\nonumber\\ & =&\sum_{m=-2M}^{2M}\lambda^m \sum_{k+l=m \atop -M\leq k,l \leq M}A_{jk}A_{il}-\sum_{m=-M}^{M}\lambda^m \frac{\partial A_{jm}}{\partial p_i} +\sum_{m=-2M}^{2M} \lambda^{m+1}\sum_{k+l=m \atop -M\leq k,l \leq M} a_{ik} \frac{\partial A_{jl}}{\partial x_i},\nonumber\\ & & \qquad \qquad \qquad \qquad \qquad \qquad 1\leq i,j \leq n. \end{eqnarray*} 令$\lambda^{-2M}$; $\lambda^m,-2M+1\leq m \leq -M-1, M+1\leq m \leq 2M$; $\lambda^m,-M\leq m \leq M$和 $\lambda^{2M+1}$的系数为零即可得如下的$2n$维gSDYM方程

\begin{equation}\label{b4} [A_{i,-M},A_{j,-M}]=0,\qquad 1\leq i,j\leq n, \end{equation}

(1.4)

\begin{eqnarray}\label{b5} & & \sum_{k+l=m \atop -M\leq k,l \leq M}[A_{ik},A_{jl}]+\sum_{k+l=m-1 \atop -M\leq k,l \leq M}\left(a_{jk}\frac{\partial A_{il}}{\partial x_j}-a_{ik}\frac{\partial A_{jl}}{\partial x_i}\right)=0,\nonumber\\ & & -2M+1\leq m \leq -M-1 \mbox{或} M+1\leq m \leq 2M,\qquad 1\leq i,j\leq n, \end{eqnarray}

(1.5)

\begin{eqnarray}\label{b6} & & \sum_{k+l=m \atop -M\leq k,l \leq M}[A_{ik},A_{jl}]-\frac{\partial A_{im}}{\partial p_j}+\frac{\partial A_{jm}}{\partial p_i}+\sum_{k+l=m-1 \atop -M\leq k,l \leq M}\left(a_{jk}\frac{\partial A_{il}}{\partial x_j}-a_{ik}\frac{\partial A_{jl}}{\partial x_i}\right)=0,\nonumber\\ & & -M\leq m \leq M,\qquad 1\leq i,j\leq n, \end{eqnarray}

(1.6)

\begin{eqnarray}\label{b7} a_{jM}\frac{\partial A_{iM}}{\partial x_j}-a_{iM}\frac{\partial A_{jM}}{\partial x_i}=0,\qquad 1\leq i,j\leq n, \end{eqnarray}

(1.7)

这里的$[\cdot,\cdot]$定义为矩阵Lie代数的Lie括号运算. 我们设$n,N\geq 2$,这是因为当$N=1$时矩阵$A_{ik},A_{jl}$为零矩阵, 而当$n=1$时上述的gSDYM自动成立. 这类新的gSDYM方程(1.4)--(1.7)包括著名的Takasaki情形、 Belavin-Zakharov情形、Ablowitz-Chakravarty-Takhtajan情形和Ma情形 (参见文献^[11]).

本文安排如下: 在第二节中,我们将构造gSDYM方程(1.4)--(1.7) 的达布变换. 第三节包括一些结论并且给出文献^[11]提到的一个公 开问题的解答.

方便起见,我们重新记谱问题(1.2)为 $$ L_i(\lambda)\Psi=\left(\frac{\partial}{\partial p_i}- a_i(\lambda)\frac{\partial}{\partial x_i}\right)\Psi= -A_i(\lambda)\Psi,\qquad 1\leq i \leq n. $$ 达布变换问题指的是我们需要从方程的精确解 $$ \Psi,\qquad A_i(\lambda)=\sum_{l=-M}^M A_{il}\lambda^l, 1\leq i\leq n $$ 出发构造新的精确解

\begin{equation}\label{c1} \widetilde{\Psi},\qquad \widetilde{A}_i(\lambda)= \sum_{l=-M}^M \widetilde{A}_{il}\lambda^l,1\leq i\leq n, \end{equation}

(2.1)

使得谱问题(1.2)恒成立. 本节的目的就是解决该问题. 我们将证明

\begin{equation}\label{c2} \widetilde{\Psi}=(\lambda I+\alpha)\Psi,\qquad I={\mbox{diag}}\Big(\underbrace{1,1,\cdots,1}_N\Big) \end{equation}

(2.2)

给出了可计算的新的特征函数矩阵,这里的矩阵$\alpha$需要满足若干 约束条件.

下面我们推导达布变换所需要满足的条件,首先有 \begin{eqnarray*} L_i(\lambda)\widetilde{\Psi} &=&L_i(\lambda)\left[(\lambda I+\alpha)\Psi\right] \\ &=&(L_i(\lambda)\alpha)\Psi+(\lambda I+\alpha)L_i(\lambda)\Psi \\ &=&(L_i(\lambda)\alpha)\Psi-(\lambda I+\alpha)A_i(\lambda)\Psi \\ &=&\left(\frac{\partial \alpha}{\partial p_i}-\sum_{k=-M}^Ma_{ik} \lambda^{k+1}\frac{\partial \alpha}{\partial x_i}\right)\Psi -(\lambda I+\alpha)\left(\sum_{l=-M}^MA_{il}\lambda^l\right)\Psi, 1\leq i \leq n \end{eqnarray*} 和 $$ -\widetilde{A}_i(\lambda)\widetilde{\Psi}= -\left(\sum_{l=-M}^M\widetilde{A}_{il}\lambda^l\right) (\lambda I+\alpha)\Psi,\qquad 1\leq i \leq n, $$ 平衡$L_i(\lambda)\widetilde{\Psi}=-\widetilde{A}_i(\lambda) \widetilde{\Psi},~ 1\leq i\leq n$关于$\lambda^{-M}$; $ \lambda^m,~ -M+1 \leq m \leq -1 $ 或$ 1\leq m\leq M$; $ \lambda^0$和$\lambda^{M+1}$的系数,得到对于$1\leq i\leq n$有 \begin{eqnarray*} & & -\alpha A_{i,-M}=-\widetilde{A}_{i,-M}\alpha,\\ & & -a_{i,m-1}\frac{\partial \alpha}{\partial x_i}-A_{i,m-1}-\alpha A_{i,m}=-\widetilde{A}_{i,m-1}-\widetilde{A}_{i,m}\alpha,\\ & & \qquad \qquad \qquad -M+1 \leq m \leq -1 {\mbox{或}} 1\leq m\leq M,\\ & & \frac{\partial \alpha}{\partial p_i}-a_{i,-1}\frac{\partial \alpha}{\partial x_i}-A_{i,-1}-\alpha A_{i,0}=-\widetilde{A}_{i,-1}-\widetilde{A}_{i,0}\alpha,\\ & & -a_{i,M}\frac{\partial \alpha}{\partial x_i}-A_{i,M}=-\widetilde{A}_{i,M}. \end{eqnarray*} 从这些等式中,我们可以得到矩阵$\alpha$的约束条件

\begin{equation}\label{c3} \frac{\partial \alpha}{\partial p_i}-a_{i,-1}\frac{\partial \alpha}{\partial x_i}-A_{i,-1}-\alpha A_{i,0}=-\widetilde{A}_{i,-1}-\widetilde{A}_{i,0}\alpha,\qquad 1\leq i\leq n \end{equation}

(2.3)

和决定$\widetilde{A}_{im}$的递归关系

\begin{equation}\label{c4} \widetilde{A}_{i,-M}=\alpha A_{i,-M} \alpha^{-1},\qquad 1\leq i\leq n, \end{equation}

(2.4)

\begin{equation}\label{c5} \widetilde{A}_{i,M}=A_{i,M}+a_{i,M}\frac{\partial \alpha}{\partial x_i},\qquad 1\leq i\leq n, \end{equation}

(2.5)

\begin{eqnarray}\label{c6} & & \widetilde{A}_{i,m-1}=A_{i,m-1}+a_{i,m-1}\frac{\partial \alpha}{\partial x_i}+\alpha A_{i,m}-\widetilde{A}_{i,m}\alpha,\nonumber\\ & & -M+1 \leq m \leq -1 {\mbox{or}} 1\leq m\leq M,\qquad 1\leq i\leq n. \end{eqnarray}

(2.6)

从式子(2.6),注意到当$-M+1\leq m\leq -1,\ 1\leq i\leq n$时, 我们能进一步计算 \begin{eqnarray*} \widetilde{A}_{im}&=&A_{i,m-1}\alpha^{-1}+\alpha A_{im}\alpha^{-1}+a_{i,m-1}\frac{\partial \alpha}{\partial x_i}\alpha^{-1}+\widetilde{A}_{i,m-1}\left(-\alpha^{-1}\right)\\ &=&A_{i,m-1}\alpha^{-1}+\alpha A_{im}\alpha^{-1}+a_{i,m-1}\frac{\partial \alpha}{\partial x_i}\alpha^{-1}\\ &&+\left[A_{i,m-2}\alpha^{-1}+\alpha A_{i,m-1}\alpha^{-1}+a_{i,m-2}\frac{\partial \alpha}{\partial x_i}\alpha^{-1}+\widetilde{A}_{i,m-2}\left(-\alpha^{-1}\right)\right]\left(-\alpha^{-1}\right)\\ &=&A_{i,m-1}\alpha^{-1}+\alpha A_{im}\alpha^{-1}+a_{i,m-1}\frac{\partial \alpha}{\partial x_i}\alpha^{-1}\\ &&+\left[A_{i,m-2}\alpha^{-1}+\alpha A_{i,m-1}\alpha^{-1}+a_{i,m-2}\frac{\partial \alpha}{\partial x_i}\alpha^{-1}\right]\left(-\alpha^{-1}\right)+\widetilde{A}_{i,m-2}\left(-\alpha^{-1}\right)^2\\ & &\qquad\vdots\\ &=&\sum_{k=0}^{M+m-1}\left(A_{i,m-k-1}\alpha^{-1}+\alpha A_{i,m-k}\alpha^{-1}+a_{i,m-k-1}\frac{\partial \alpha}{\partial x_i}\alpha^{-1}\right)\left(-\alpha^{-1}\right)^k\\ &&+\widetilde{A}_{i,-M}\left(-\alpha^{-1}\right)^{M+m}\\ &=&\sum_{k=0}^{M+m-1}\left(A_{i,m-k-1}\alpha^{-1}+a_{i,m-k-1}\frac{\partial \alpha}{\partial x_i}\alpha^{-1}\right)\left(-\alpha^{-1}\right)^k-\sum_{k=0}^{M+m}\alpha A_{i,m-k}(-\alpha^{-1})^{k+1}\\ & &-\alpha A_{i,m-(M+m)}\alpha^{-1}(-\alpha^{-1})^{M+m}+\widetilde{A}_{i,-M}(-\alpha^{-1})^{M+m}. \end{eqnarray*} 从而,把式子(2.4)代入上述方程,有

\begin{eqnarray}\label{c7} & & \widetilde{A}_{im}=-\sum_{k=0}^{M+m-1}\left(A_{i,m-k-1}+a_{i,m-k-1}\frac{\partial \alpha}{\partial x_i}\right)\left(-\alpha^{-1}\right)^{k+1}-\sum_{k=0}^{M+m}\alpha A_{i,m-k}(-\alpha^{-1})^{k+1},\nonumber\\ & &\qquad \qquad -M+1\leq m\leq -1,1\leq i\leq n. \end{eqnarray}

(2.7)

对于$1\leq m\leq M,\ 1\leq i\leq n$,通过类似的计算,有

\begin{eqnarray}\label{c8} \widetilde{A}_{i,m-1}&=&\sum_{k=0}^{M-m+1}\left(A_{i,m+k-1}+a_{i,m+k-1}\frac{\partial \alpha}{\partial x_i}\right)\left(-\alpha\right)^k+\sum_{k=0}^{M-m}\alpha A_{i,m+k}(-\alpha)^{k},\nonumber\\ & & 1\leq m\leq M,1\leq i\leq n. \end{eqnarray}

(2.8)

这样,利用式子(2.7)和(2.8),约束条件(2.3)在$1\leq i\leq n$下有

\begin{eqnarray}\label{c9} \frac{\partial \alpha}{\partial p_i} &=&a_{i,-1}\frac{\partial \alpha}{\partial x_i}+A_{i,-1}+\alpha A_{i,0}-\widetilde{A}_{i,-1}-\widetilde{A}_{i,0}\alpha\nonumber\\ & =&a_{i,-1}\frac{\partial \alpha}{\partial x_i}+A_{i,-1}+\alpha A_{i,0}+\left[\sum_{k=0}^{M-2}A_{i,-k-2}(-\alpha^{-1})^{k+1}+\sum_{k=0}^{M-2}a_{i,-k-2}\right.\nonumber\\ &&\left.\times\frac{\partial \alpha}{\partial x_i}(-\alpha^{-1})^{k+1}+\sum_{k=0}^{M-1}\alpha A_{i,-k-1}(-\alpha^{-1})^{k+1}\right]\nonumber\\ &&+\left[\sum_{k=0}^MA_{ik}(-\alpha)^{k+1}+\sum_{k=0}^Ma_{ik}\frac{\partial \alpha}{\partial x_i}(-\alpha)^{k+1}+\sum_{k=1}^{M}\alpha A_{ik}(-\alpha)^{k}\right]\nonumber\\ &=&\sum_{k=0}^MA_{ik}(-\alpha)^{k+1}+\sum_{k=0}^Ma_{ik}\frac{\partial \alpha}{\partial x_i}(-\alpha)^{k+1}+\sum_{k=1}^{M}\alpha A_{ik}(-\alpha)^{k}\nonumber\\ &&+\sum_{k=-M}^{-1}A_{ik}(-\alpha^{-1})^{-k-1}+\sum_{k=-M}^{-1}a_{ik}\frac{\partial \alpha}{\partial x_i}(-\alpha^{-1})^{-k-1}+\sum_{k=-M}^{0}\alpha A_{ik}(-\alpha^{-1})^{-k}\nonumber\\ &=&\sum_{k=-M}^MA_{ik}(-\alpha)^{k+1}+\sum_{k=-M}^Ma_{ik}\frac{\partial \alpha}{\partial x_i}(-\alpha)^{k+1}+\sum_{k=-M}^{M}\alpha A_{ik}(-\alpha)^{k}. \end{eqnarray}

(2.9)

这是最终的约束条件的表达式,其用于限制矩阵$\alpha$按照式子(2.2) 构造新的特征函数矩阵. 下面的定理提供了这样的一种矩阵.

定理2.1 令$h_s,1\leq s \leq N$是相应于谱参数$\lambda_1,\lambda_2, \cdots,\lambda_N$的$N$维列特征向量,也就是说,$N$维列特征向量 $h_s,1\leq s \leq N$满足

\begin{equation}\label{c10} \frac{\partial h_s}{\partial p_i}=a_i(\lambda_s) \frac{\partial h_s}{\partial x_i}-A_i(\lambda_s)h_s, \qquad 1\leq s\leq N,1\leq i \leq n. \end{equation}

(2.10)

假设矩阵 $$H=[h_1,h_2,\cdots,h_N] $$的行列式是非零的,则由

\begin{equation}\label{c11} \alpha=-H\Lambda H^{-1},\qquad \Lambda={\mbox{diag}} (\lambda_1,\lambda_2,\cdots,\lambda_N), \end{equation}

(2.11)

定义的矩阵满足约束条件(2.9). 因此,我们有达布变换 $$\widetilde{\Psi}=(\lambda I+\alpha)\Psi $$ 和由式子(2.4), (2.5),(2.7)和 (2.8)给出的精确解$\widetilde{A}_{il}, 1\leq i \leq n,-M\leq l \leq M$.

证 由逆矩阵的导数公式 $$ \frac{\partial H^{-1}}{\partial y_i}=-H^{-1}\frac{\partial H}{\partial y_i}H^{-1},\qquad y_i=p_i {\mbox{或}} x_i,1\leq i \leq n, $$ 对于任何$1\leq i \leq n$,有 $$ \frac{\partial \alpha}{\partial p_i}=-\frac{\partial H}{\partial p_i}\Lambda H^{-1}+H\Lambda H^{-1}\frac{\partial H}{\partial p_i}H^{-1}, $$ $$ \frac{\partial \alpha}{\partial x_i}=-\frac{\partial H}{\partial x_i}\Lambda H^{-1}+H\Lambda H^{-1}\frac{\partial H}{\partial x_i}H^{-1}. $$ 另一方面,从式子(2.10)有 \begin{eqnarray*} \frac{\partial H}{\partial p_i}&=&\sum_{k=-M}^M a_{ik}\frac{\partial H}{\partial x_i}\Lambda^{k+1}-\left[A_i(\lambda_1)h_1,A_i(\lambda_2)h_2,\cdots,A_i(\lambda_N)h_N\right]\\ &=&\sum_{k=-M}^M a_{ik}\frac{\partial H}{\partial x_i}\Lambda^{k+1}-\sum_{l=-M}^M A_{il}H\Lambda^{l},\qquad 1\leq i\leq n. \end{eqnarray*} 这样,我们可以计算得到 \begin{eqnarray*} \frac{\partial \alpha}{\partial p_i} &=&-\frac{\partial H}{\partial p_i}\Lambda H^{-1}+H\Lambda H^{-1}\frac{\partial H}{\partial p_i}H^{-1}\\ & =&-\sum_{k=-M}^M a_{ik}\frac{\partial H}{\partial x_i}\Lambda^{k+2}H^{-1}+\sum_{l=-M}^M A_{il}H\Lambda^{l+1}H^{-1}\\ & & +\sum_{k=-M}^M a_{ik}H\Lambda H^{-1}\frac{\partial H}{\partial x_i}\Lambda^{k+1}H^{-1}-H\Lambda H^{-1}\sum_{l=-M}^M A_{il}H\Lambda^{l}H^{-1}\\ & =&-\sum_{k=-M}^M a_{ik}\frac{\partial H}{\partial x_i}H^{-1}(-\alpha)^{k+2}+\sum_{l=-M}^M A_{il}(-\alpha)^{l+1}\\ & & +\sum_{k=-M}^M a_{ik}H\Lambda H^{-1}\frac{\partial H}{\partial x_i}H^{-1}(-\alpha)^{k+1}+\alpha\sum_{l=-M}^M A_{il}(-\alpha)^{l}\\ & =&\sum_{k=-M}^M a_{ik}\left(-\frac{\partial H}{\partial x_i}\Lambda H^{-1}+H\Lambda H^{-1}\frac{\partial H}{\partial x_i}H^{-1}\right)(-\alpha)^{k+1}\\ & & +\sum_{l=-M}^M A_{il}(-\alpha)^{l+1}+\alpha\sum_{l=-M}^M A_{il}(-\alpha)^{l}\\ & =&\sum_{k=-M}^M a_{ik}\frac{\partial \alpha}{\partial x_i}(-\alpha)^{k+1}+\sum_{l=-M}^M A_{il}(-\alpha)^{l+1}+\alpha\sum_{l=-M}^M A_{il}(-\alpha)^{l}. \end{eqnarray*} 这意味着由式子(2.11)定义的矩阵$\alpha$实际上是满足约束条件 (2.9)的.

这里需要指出的是我们要求特征值集合$\{\lambda_1, \lambda_2,\cdots,\lambda_N\}$中至少有两个不同的值. 否则只能得到原始解而不能构造出新解,因为矩阵$\alpha$变 为一个带常系数的单位矩阵. 方程(2.10)是线性的, 从而是精确可解的. 即给定一个精确解如平凡的零解, 我们也能依次利用达布变换构造一系列新的精确解. 事实上, 文献^[11]构造的Vandermonde行列式型和广义Cauchy行列式 型解也适用于本文的定理.

本文从谱问题(1.2)出发构造了带负幂次谱参数的gSDYM方程(1.4)--(1.7). 此类方程的达布变换的显式表达式由定理2.1给出. 利用达布变换, 文献^[11]给出的Vandermonde行列式和广义Cauchy行列式可以用来 构造精确解. 总之,本文直接推广了文献^[11]的结果. 另外,令 $$M=\max (M_1,M_2), $$ 则以下的谱问题 $$ \left[\frac{\partial}{\partial p_i}-\left(\sum_{k=-M_1}^{M_2} a_{ik}\lambda^{k+1}\right)\frac{\partial}{\partial x_i}\right]\Psi=-\left(\sum_{l=-M_1}^{M_2} A_{il}\lambda^l\right)\Psi,\qquad 1\leq i \leq n, $$ 可以看作问题(1.2)的一种简化.

文献^[11]提出了一个公开问题,即如下$2n$维SDYM型方程

\begin{eqnarray}\label{d1} & & \sum_{k+l=m \atop -M\leq k,l \leq M}[A_{ik},A_{jl}]+\sum_{k+l=m \atop -M\leq k,l \leq M}\left(a_{jk}\frac{\partial A_{il}}{\partial x_j}-a_{ik}\frac{\partial A_{jl}}{\partial x_i}\right)=0,\nonumber\\ & & -2M+1\leq m \leq -M-1 \mbox{或} M+1\leq m \leq 2M,\qquad 1\leq i,j\leq n, \end{eqnarray}

(3.1)

\begin{eqnarray}\label{d2} & & \sum_{k+l=m \atop -M\leq k,l \leq M}[A_{ik},A_{jl}]-\frac{\partial A_{im}}{\partial p_j}+\frac{\partial A_{jm}}{\partial p_i}+\sum_{k+l=m \atop -M\leq k,l \leq M}\left(a_{jk}\frac{\partial A_{il}}{\partial x_j}-a_{ik}\frac{\partial A_{jl}}{\partial x_i}\right)=0,\nonumber\\ & & -M\leq m \leq M,\qquad 1\leq i,j\leq n, \end{eqnarray}

(3.2)

的达布变换构造问题. 这里我们给出相应的达布变换如下.

方程(3.1)--(3.2)可由谱问题 $$ \left[\frac{\partial}{\partial p_i}-\left(\sum_{k=-M}^{M} a_{ik}\lambda^{k}\right)\frac{\partial}{\partial x_i}\right]\Psi=-\left(\sum_{l=-M}^{M} A_{il}\lambda^l\right)\Psi,\qquad 1\leq i \leq n $$ 的相容性条件得到. 达布变换有形式 $$ \widetilde{\Psi}=(\lambda I+\alpha)\Psi,\qquad I={\mbox{diag}}\Big(\underbrace{1,1,\cdots,1}_N\Big), $$ 这里 $$ \widetilde{A}_{i,M}= A_{i,M} ,\qquad 1\leq i\leq n, $$ $$ \widetilde{A}_{i,-M}=\alpha A_{i,-M}\alpha^{-1}+a_{i,-M}\frac{\partial \alpha}{\partial x_i}\alpha^{-1},\qquad 1\leq i\leq n, $$ \begin{eqnarray*} \widetilde{A}_{im}&=&-\sum_{k=0}^{M+m-1}A_{i,m-k-1}\left(-\alpha^{-1}\right)^{k+1}-\sum_{k=0}^{M+m}\left(a_{i,m-k}\frac{\partial \alpha}{\partial x_i}+\alpha A_{i,m-k}\right)(-\alpha^{-1})^{k+1},\nonumber\\ & & -M+1\leq m\leq -1,1\leq i\leq n. \end{eqnarray*} \begin{eqnarray*} \widetilde{A}_{i,m-1}&=&\sum_{k=0}^{M-m+1}A_{i,m+k-1}\left(-\alpha\right)^k+\sum_{k=0}^{M-m}\left(a_{i,m+k}\frac{\partial \alpha}{\partial x_i}+\alpha A_{i,m+k}\right)(-\alpha)^{k},\nonumber\\ & & 1\leq m\leq M,1\leq i\leq n. \end{eqnarray*} 矩阵$\alpha$的约束条件为 \begin{eqnarray*} \frac{\partial \alpha}{\partial p_i} &=&a_{i0}\frac{\partial \alpha}{\partial x_i}+A_{i,-1}+\alpha A_{i0}-\widetilde{A}_{i,-1}-\widetilde{A}_{i0}\alpha\nonumber\\ &=&\sum_{k=-M}^MA_{ik}(-\alpha)^{k+1}+\sum_{k=-M}^Ma_{ik}\frac{\partial \alpha}{\partial x_i}(-\alpha)^{k}+\sum_{k=-M}^{M}\alpha A_{ik}(-\alpha)^{k}, \end{eqnarray*} 并且对于$\alpha=-H\Lambda H^{-1}$,定理2.1同样有效.