离散可积系统^[1-2]是现代数学物理中重要的研究课题之一,是将经典可积系统中的微分算子$\partial_x$及变量$x$分别替换成差分算子$\Delta$和离散变量$n$.到目前为止,重要的离散可积系统,如离散KP系列^{[3-10, 18]},修正离散KP系列^[11-14]等都得到了研究.

与经典可积系统类似,流方程问题^[15-18]是离散可积系统中非常重要且基本的问题,如何将离散可积系统的流方程表示出来是离散可积系统研究的核心问题之一.本文主要研究修正离散KP系列的流方程问题.

首先,移动算子$\Lambda$和差分算子$\Delta$定义如下

(1) $\begin{eqnarray}\Lambda(v(n))&=&v(n+1), \end{eqnarray}$

(2) $\begin{eqnarray}\Delta(v(n))&=&v(n+1)-v(n)=(\Lambda-I)(v(n)).\end{eqnarray}$

对$\forall j\in Z$,算子$\Delta$满足Leibniz法则

(3) $\begin{eqnarray}\Delta^jv&=&\sum\limits_{i=0}^\infty C_j^i \Delta^i(v)(n+j-i)\Delta^{j-i}, \end{eqnarray}$

(4) $\begin{eqnarray}C_j^i&=&\frac{j(j-1)(j-2)\cdots(j-i+1)}{i!}.\end{eqnarray}$

修正离散KP系列^[11-14]的Lax方程为

(5) $\begin{eqnarray}L_{t_m}&=&[(L^m)_{\geqslant1}, L], \end{eqnarray}$

其中, $L$为Lax算子,形式如下

(6) $\begin{eqnarray}L=\sum\limits_{i=0}^\infty v_i\Delta^{-i+1}=v_0 \Delta+v_1+v_2\Delta^{-1}+\cdots, \end{eqnarray}$

这里, $v_i=v_i(n, x, t)=v_i(n, x, t_1, t_2, \cdots)$,并且记

(7) $\begin{eqnarray}(L^m)_{\geqslant1}&=&\sum\limits_{s=0}^{m-1}b_{sm}(n)\Delta^{m-s}.\end{eqnarray}$

从观察

(8) $\begin{equation} [(L^m)_{\geqslant1}, L]_{>1}=-[(L^m)_{<1}, L]_{>1}=0 \end{equation}$

出发,下面将给出$b_{sm}(n)$的递推公式.为此，先计算$[(L^m)_{\geq 1}, L]$,将(6)和(7)式代入即可.

(9) $\begin{eqnarray} [(L^m)_{\geq 1}, L]&=&\sum\limits_{i=0}^\infty\sum\limits_{s=0}^{m-1}[b_{sm}(n)\Delta^{m-s}, v_i\Delta^{-i+1}]\nonumber\\ &=&\sum\limits_{i=0}^\infty\sum\limits_{s=0}^{m-1}\sum\limits_{p=0}^\infty(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_i))\\&&-v_iC_{-i+1}^p\Delta^p (\Lambda^{-i+1-p}(b_{sm}(n))))\Delta^{m-s-p-i+1}\nonumber\\ &=&\sum\limits_{l=0}^\infty\sum\limits_{s=0}^{m-1}\sum\limits_{p=0}^l(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{l-p}))\\&&-v_{l-p}C_{p-l+1}^p\Delta^p (\Lambda^{1-l}(b_{sm}(n))))\Delta^{m-s-l+1}\nonumber\\ &=&\sum\limits_{k=0}^\infty\sum\limits_{s=0}^{m-1}\sum\limits_{p=0}^{k-s}(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\\&&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p (\Lambda^{1-k+s}(b_{sm}(n))))\Delta^{m-k+1}\nonumber\\ &=&\sum\limits_{k=0}^{m-1}\sum\limits_{s=0}^k\sum\limits_{p=0}^{k-s}(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\\&&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p (\Lambda^{1-k+s}(b_{sm}(n))))\Delta^{m-k+1}\nonumber\\ &&+\sum\limits_{k=m}^\infty\sum\limits_{s=0}^{m-1}\sum\limits_{p=0}^{k-s}(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\\&&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p (\Lambda^{1-k+s}(b_{sm}(n))))\Delta^{m-k+1}.\end{eqnarray} $

这里第三个等式中记$l=i+p$,在第四个等式中记$k=s+l$.因此

(10) $\begin{eqnarray} [(L^m)_{\geq 1}, L]_{>1}&=&\sum\limits_{k=0}^{m-1}\sum\limits_{s=0}^k\sum\limits_{p=0}^{k-s}(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\nonumber\\ &&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p(\Lambda^{1-k+s}(b_{sm}(n))))\Delta^{m-k+1}. \end{eqnarray}$

所以,由(8)和(10)式可知,对$k=0, 1, 2, \cdots, m-1$有

(11) $\begin{equation} \sum\limits_{s=0}^k\sum\limits_{p=0}^{k-s}(b_{sm}(n)C_{m-s}^p \Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))-v_{k-s-p}C_{p-k+s+1}^p\Delta^p (\Lambda^{1-k+s}(b_{sm}(n))))=0. \end{equation}$

特别地,当$k=0$时,有

(12) $\begin{equation} b_{0m}(n)\Lambda^m(v_0)-v_0\Lambda(b_{0m}(n))=0, \end{equation} $

对(12)式两边取自然对数,整理得

(13) $\begin{equation} ln(b_{0m}(n+1))-ln(b_{0m}(n))=ln(v_0(n+m))-ln(v_0(n)), \end{equation} $

即

(14) $\begin{equation} \Delta(ln(b_{0m}(n)))=\Delta((I+\Lambda+\Lambda^2+\cdots+\Lambda^{m-1})(ln(v_0(n)))), \end{equation} $

从而

(15) $\begin{equation} b_{0m}(n)=v_0(n)v_0(n+1)\cdots v_0(n+m-1), \ m=1, 2, 3, \cdots. \end{equation}$

对于$k>0$的情况,根据(11)式整理可得

(16) $\begin{eqnarray} -b_{km}(n)\Lambda^{m-k}(v_0)+v_0\Lambda(b_{km}(n))&=&\sum\limits_{s=0}^{k-1}\sum\limits_{p=0}^{k-s}(b_{sm}(n)C_{m-s}^p \Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\nonumber\\ &&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p (\Lambda^{1-k+s}(b_{sm}(n)))). \end{eqnarray} $

进一步可以将(16)的左边部分计算如下

(17) $\begin{eqnarray} v_0(n)\Lambda(b_{km}(n))-b_{km}(n)\Lambda^{m-k}(v_0(n))&=&(v_0(n)\Lambda-v_0(n+m-k))b_{km}(n)\nonumber\\ &=&(v_0(n)\Delta+v_0(n)-v_0(n+m-k))b_{km}(n), \end{eqnarray}$

为了求出$b_{km}(n)$,先看如下的引理.

引理 1  假设$v_0(n)\Delta+v_0(n)-v_0(n+m-k)=f(n)\Delta g(n)$,则

(18) $\begin{eqnarray} f(n)&=&v_0(n)v_0(n+1)\cdots v_0(n+m-k), \end{eqnarray}$

(19) $\begin{eqnarray}g(n)&=&\frac{1}{v_0(n)v_0(n+1)\cdots v_0(n+m-k-1)}. \end{eqnarray}$

证  由于$f(n)\Delta g(n)=f(n)g(n+1)\Delta+f(n)\Delta(g(n))$,与假设条件比较,得

$\begin{eqnarray}f(n)g(n+1)=v_0(n), \quadf(n)\Delta (g(n))=v_0(n)-v_0(n+m-k), \nonumber\end{eqnarray}$

两式相除得

(20) $\begin{equation}\frac{g(n)}{g(n+1)}=\frac{v_0(n+m-k)}{v_0(n)}.\nonumber\end{equation}$

对上式两边取自然对数,有

$\begin{eqnarray}-\Delta(ln(g(n)))&=&(\Lambda^{m-k}-I)(ln(v_0(n)))\nonumber\\&=&(\Lambda-I)(I+\Lambda+\Lambda^2+\cdots+\Lambda^{m-k-1})(ln(v_0(n)))\nonumber\\&=&\Delta\left((I+\Lambda+\cdots+\Lambda^{m-k-1})(ln(v_0(n)))\right), \nonumber\end{eqnarray}$

从而

(21) $\begin{equation}g(n)=\frac{1}{v_0(n)v_0(n+1)\cdots v_0(n+m-k-1)}, \nonumber\end{equation}$

将$g(n)$代入$f(n)g(n+1)=v_0(n)$中,得

(22) $\begin{equation}f(n)=\frac {v_0(n)}{g(n+1)}=v_0(n)v_0(n+1)\cdots v_0(n+m-k).\nonumber\end{equation}$

证毕.

由(16)式, (17)式及引理1,可得

(23) $\begin{eqnarray}(v_0(n)\Delta+v_0(n)-v_0(n+m-k))b_{km}(n)&=&(f(n)\Delta g(n))b_{km}(n)\nonumber\\ &=&\sum\limits_{s=0}^{k-1}\sum\limits_{p=0}^{k-s}\Big(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\\&&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p(\Lambda^{1-k+s}(b_{sm}(n)))\Big), \end{eqnarray}$

从而

(24) $\begin{eqnarray}b_{km}(n)&=&\frac{1}{g(n)}\Delta^{-1}\bigg(\frac{1}{f(n)}\sum\limits_{s=0}^{k-1}\sum\limits_{p=0}^{k-s}\Big(b_{sm}(n)C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\nonumber\\&&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p(\Lambda^{1-k+s}(b_{sm}(n)))\Big)\bigg).\end{eqnarray}$

综合上述,给出$b_{km}(n)$的表示形式.

定理 2

(25) $\begin{eqnarray}b_{0m}(n)&=&v_0(n)v_0(n+1)\cdots v_0(n+m-1), \nonumber\\b_{km}(n)&=&v_0(n)v_0(n+1)\cdots v_0(n+m-k-1)\Delta^{-1}\bigg(\frac{1}{v_0(n)\cdots v_0(n+m-k)}\nonumber\\&&\sum\limits_{s=0}^{k-1}\sum\limits_{p=0}^{k-s}\Big(b_{sm}C_{m-s}^p\Delta^p\big(\Lambda^{m-s-p}(v_{k-s-p})\big)-v_{k-s-p}C_{p-k+s+1}^p\Delta^p\big(\Lambda^{1-k+s}(b_{sm})\big)\Big)\bigg)\nonumber\\&&k=1, 2, \cdots, m-1.\end{eqnarray}$

由定理2罗列一些结果如下

(26) $\begin{eqnarray}(L)_{\geqslant1}&=&v_0(n)\Delta, \end{eqnarray}$

(27) $\begin{eqnarray}(L^2)_{\geqslant1}&=&v_0(n)v_0(n+1)\Delta^2+v_0(n)\Big(v_1(n+1)+v_1(n)+v_0(n+1)-v_0(n)\Big)\Delta, \end{eqnarray}$

(28) $\begin{eqnarray}(L^3)_{\geqslant1}&=&v_0(n)v_0(n+1)v_0(n+2)\Delta^3+v_0(n)v_0(n+1)\Big(v_1(n+2)\nonumber\\&&+v_1(n+1)+v_1(n)+2v_0(n+2)-v_0(n+1)-v_0(n)\Big)\Delta^2\nonumber\\&&+v_0(n)\Big(v_2(n+2)v_0(n+1)+v_2(n+1)v_0(n)+v_2(n)v_0(n-1)\nonumber\\&&+2v_1(n+2)v_0(n+1)+v_1(n+1)v_1(n+1)+v_1(n+1)v_1(n)\nonumber\\&&-v_1(n+1)v_0(n)+v_1(n)v_1(n)+v_1(n)v_0(n+1)-2v_1(n)v_0(n)\nonumber\\&&+v_0(n+2)v_0(n+1)-v_0(n+1)v_0(n+1)-v_0(n+1)v_0(n)\nonumber\\&&+v_0(n)v_0(n)\Big)\Delta, \end{eqnarray}$

(29) $\begin{eqnarray}(L^4)_{\geqslant1}&=&v_0(n)v_0(n+1)v_0(n+2)v_0(n+3)\Delta^4+v_0(n)v_0(n+1)v_0(n+2)\nonumber\\&&\Big(v_1(n+3)+v_1(n+2)+v_1(n+1)+v_1(n)+3v_0(n+3)-v_0(n+2)\nonumber\\&&-v_0(n+1)-v_0(n)\Big)\Delta^3+v_0(n)v_0(n+1)\Big(v_2(n+3)v_0(n+2)\nonumber\\&&+v_2(n+2)v_0(n+1)+v_2(n+1)v_0(n)+v_2(n)v_0(n-1)\nonumber\\&&+3v_1(n+3)v_0(n+2)+v_1(n+2)v_1(n+2)+v_1(n+2)v_1(n+1)\nonumber\\&&+v_1(n+2)v_1(n)+v_1(n+2)v_0(n+2)-v_1(n+2)v_0(n+1)\nonumber\\&&-v_1(n+2)v_0(n)+v_1(n+1)v_1(n+1)+v_1(n+1)v_1(n)\nonumber\\&&+2v_1(n+1)v_0(n+2)-2v_1(n+1)v_0(n+1)-v_1(n+1)v_0(n)\nonumber\\&&+v_1(n)v_1(n)+2v_1(n)v_0(n+2)-v_1(n)v_0(n+1)-2v_1(n)v_0(n)\nonumber\\&&+3v_0(n+3)v_0(n+2)-2v_0(n+2)v_0(n+2)-2v_0(n+2)v_0(n+1)\nonumber\\&&-2v_0(n+2)v_0(n)+v_0(n+1)v_0(n+1)+v_0(n+1)v_0(n)\nonumber\\&&+v_0(n)v_0(n)\Big)\Delta^2+v_0(n)\Big(v_3(n+3)v_0(n+2)v_0(n+1)\nonumber\\&&+v_3(n+2)v_0(n+1)v_0(n)+v_3(n+1)v_0(n)v_0(n-1)\nonumber\\&&+v_3(n)v_0(n-1)v_0(n-2)+3v_2(n+3)v_0(n+2)v_0(n+1)\nonumber\\&&+v_2(n+2)v_1(n+2)v_0(n+1)+2v_2(n+2)v_1(n+1)v_0(n+1)\nonumber\\&&+v_2(n+2)v_1(n)v_0(n+1)-v_2(n+2)v_0(n+2)v_0(n+1)\nonumber\\&&+2v_2(n+1)v_1(n+1)v_0(n)+2v_2(n+1)v_1(n)v_0(n)\nonumber\\&&-2v_2(n+1)v_0(n)v_0(n)+v_2(n+1)v_0(n)v_0(n-1)\nonumber\\&&+v_2(n)v_1(n+1)v_0(n-1)+2v_2(n)v_1(n)v_0(n-1)\nonumber\\&&+v_2(n)v_1(n-1)v_0(n-1)+v_2(n)v_0(n+1)v_0(n-1)\nonumber\\&&-2v_2(n)v_0(n)v_0(n-1)-v_2(n)v_0(n-1)v_0(n-1)\nonumber\\&&+v_2(n)v_0(n-1)v_0(n-2)+3v_1(n+3)v_0(n+2)v_0(n+1)\nonumber\\&&+2v_1(n+2)v_1(n+2)v_0(n+1)+2v_1(n+2)v_1(n+1)v_0(n+1)\nonumber\\&&+2v_1(n+2)v_1(n)v_0(n+1)-v_1(n+2)v_0(n+3)v_0(n+1)\nonumber\\&&-2v_1(n+2)v_0(n+1)v_0(n+1)-2v_1(n+2)v_0(n+1)v_0(n)\nonumber\\&&+v_1(n+1)v_1(n+1)v_1(n+1)+v_1(n+1)v_1(n+1)v_1(n)\nonumber\\&&-v_1(n+1)v_1(n+1)v_0(n+1)-v_1(n+1)v_1(n+1)v_0(n)\nonumber\\&&+v_1(n+1)v_1(n)v_1(n)-2v_1(n+1)v_1(n)v_0(n)\nonumber\\&&+v_1(n+1)v_0(n+2)v_0(n+1)-v_1(n+1)v_0(n+1)v_0(n+1)\nonumber\\&&+v_1(n+1)v_0(n)v_0(n)+v_1(n)v_1(n)v_1(n)+v_1(n)v_1(n)v_0(n+1)\nonumber\\&&-3v_1(n)v_1(n)v_0(n)+v_1(n)v_0(n+2)v_0(n+1)\nonumber\\&&-v_1(n)v_0(n+1)v_0(n+1)-2v_1(n)v_0(n+1)v_0(n)\nonumber\\&&+3v_1(n)v_0(n)v_0(n)+v_0(n+3)v_0(n+2)v_0(n+1)\nonumber\\&&-v_0(n+2)v_0(n+2)v_0(n+1)-v_0(n+2)v_0(n+1)v_0(n+1)\nonumber\\&&-v_0(n+2)v_0(n+1)v_0(n)+v_0(n+1)v_0(n+1)v_0(n+1)\nonumber\\&&+v_0(n+1)v_0(n+1)v_0(n)+v_0(n+1)v_0(n)v_0(n)\nonumber\\&&-v_0(n)v_0(n)v_0(n)\Big)\Delta.\end{eqnarray}$

有了以上的结论和结果,下面将给出修正离散KP系列的流方程的表示形式.根据(8)和(9)式,有

(30) $\begin{eqnarray}[(L^m)_{\geq 1}, L]&=&\sum\limits_{k=m}^\infty\sum\limits_{s=0}^{m-1}\sum\limits_{p=0}^{k-s}(b_{sm}C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\\&&-v_{k-s-p}C_{p-k+s+1}^p\Delta^p(\Lambda^{1-k+s}(b_{sm})))\Delta^{m-k+1}\nonumber\\&=&\sum\limits_{k=m}^\infty\sum\limits_{s=0}^{m-1}\sum\limits_{p=0}^{k-s}(b_{sm}C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\\&&-(-1)^pC_{k-s-2}^p\Delta^p(\Lambda^{1-k+s}(b_{sm})))\Delta^{m-k+1}\nonumber\\&=&\sum\limits_{k=m}^\infty\sum\limits_{s=0}^{m-1}(b_{sm}C_{m-s}^{k-s}\Delta^{k-s}(\Lambda^{m-s}(v_0))+b_{sm}C_{m-s}^{k-s-1}\Delta^{k-s-1}(\Lambda^{m-k-1}(v_1)))\nonumber\\&&+\sum\limits_{p=0}^{k-s-2}(b_{sm}C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s-p}))\\&&-(-1)^pC_{k-s-2}^p\Delta^p(\Lambda^{1-k+s}(b_{sm})))\Delta^{m-k+1}\nonumber\\&=&\sum\limits_{k\geqslant1}^\infty\sum\limits_{s=0}^{m-1}(b_{sm}C_{m-s}^{k-s}\Delta^{k-s}(\Lambda^{m-s}(v_0))+b_{sm}C_{m-s}^{k-s-1}\Delta^{k-s-1}(\Lambda^{m-k-1}(v_1)))\nonumber\\&&+\sum\limits_{p=0}^{k-s+m-3}(b_{sm}C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s+m-p-1}))\\&&+(-1)^pC_{k-s+m-3}^p\Delta^p(\Lambda^{2-k-m+s}(b_{sm})))\Delta^{-k+2}, \end{eqnarray}$

比较(30), (5)和(6)式,得到关于流方程的表示形式.

定理 3  对于修正离散KP系列,其流方程表示如下

(31) $\begin{eqnarray}v_{k, t_m}&=&\sum\limits_{s=0}^{m-1}(b_{sm}C_{m-s}^{k-s}\Delta^{k-s}(\Lambda^{m-s}(v_0))+b_{sm}C_{m-s}^{k-s-1}\Delta^{k-s-1}(\Lambda^{m-k-1}(v_1)))\nonumber\\&&+\sum\limits_{p=0}^{k-s+m-3}(b_{sm}C_{m-s}^p\Delta^p(\Lambda^{m-s-p}(v_{k-s+m-p-1}))\\&&+(-1)^pC_{k-s+m-3}^p\Delta^p(\Lambda^{2-k-m+s}(b_{sm}))).\end{eqnarray}$

依据(31)和(27)式,现给出$t_3$流如下

(32) $\begin{eqnarray}v_{0, t_3}&=&v_0(n)v_0(n+1)v_0(n+2)[v_0(n+3)+v_2(n+3)]+v_0(n-1)v_0(n)v_0(n+1), \quad \end{eqnarray}$

(33) $\begin{eqnarray}v_{1, t_3}&=&2v_0(n-2)v_0(n-1)v_0(n)-v_0(n-1)v_0(n)v_0(n+1)\nonumber\\&&+v_0(n)v_0(n+1)v_0(n+2)[-3v_0(n+3)+3v_0(n+4)+v_1(n+1)\nonumber\\&&-3v_2(n+2)+3v_2(n+3)+v_3(n+3)]\nonumber\\&&+v_0(n-1)v_0(n)[-v_0(n-1)-v_0(n)+2v_0(n+1)+v_1(n-1)\nonumber\\&&+v_1(n)+v_1(n+1)]+v_0(n)v_0(n+1)[v_0(n+2)+v_2(n+2)][-v_0(n)\nonumber\\&&-v_0(n+1)+2v_0(n+2)+v_1(n)+v_1(n+1)+v_1(n+2)].\end{eqnarray}$

本论文主要得到了修正离散KP系列的流方程递推公式,对于进一步研究修正离散KP系列有很大帮助.