近年来, 离散可积系统研究已经成为人们关注的课题, 这是因为离散可积系统与元胞自动机、辛算法以及DNA研究等有着密切的联系^[1-7], 从而有着十分广阔的应用前景, 许多离散可积系已经被许多作者系统地研究了, 如Ablowitz-Ladik晶格方程^[2]、Toda晶格方程^[3]和微分-差分KdV方程^[4]等.此外, 作为经典可积系统离散形式, 可积辛映射也成为研究热点^[5-7], 非线性化方法建立了孤立子方程与有限维完全可积系的联系^[8-16], 现在非线性化方法也可以应用到离散可积系统.通过离散可积系统Lax对的非线性化, 可以将离散可积系统求解问题分解成一个可积辛映射与有限维可积系统的求解问题^{[3-14, 17-18]}.

本文引入一个离散矩阵谱问题, 通过离散零曲率方程, 导出一族离散可积系统, 利用迹恒等式建立了其Hamilton结构, 证明了其Liouville可积性.通过双Bargmann约束, 离散可积系Lax对与共轭Lax对空间部分与时间部分, 分别被非线性化为一个辛映射与一族有限维常微分方程, 通过寻找2N个函数独立且两两对合的运动积分, 证明它在Liouville意义下是完全可积的, 进一步给出了离散可积系解的表示.

首先引入以下离散谱问题

$E\varphi _n =U_n\varphi _n,{\begin{array}{*{20}c}&{U_n=U(u_n,\lambda )=\left( {{\begin{array}{*{20}c} 0 ~~& {u_n^2} \\ {\frac{1}{u_n}} ~~& {\frac{\lambda}{u_n}} \\ \end{array} }} \right)} \\ \end{array} },{\begin{array}{*{20}c} \\ \end{array} }\varphi =\left( {{\begin{array}{*{20}c} {\varphi _n^1 } \\ {\varphi _n^2 } \\ \end{array} }} \right),$

(2.1)

其中位移算子$E$, 算子$E$的逆$E^{-1}$及其差分算子${D}$定义如下

$ (Ef)(n)=f(n+1)\,,\,\,\,\,\,\,\,\,(E^{-1}f)(n)=f(n-1),\,\,\,\,\,\,n\in Z,$

(2.2)

$ (Df)(n)=f(n+1)-f(n)=(E-1)f(n)\,\,\,\,,\,\,n\in Z,$

(2.3)

记$f_n=f(n)$, 并假设$u_n=u(n,t)$是定义在$Z\times {\Bbb R}$上的实函数, $\lambda $是谱参数并且$\lambda _t =0$.

为了得到离散可积系统, 我们首先来解离散谱问题(2.1)的驻定离散零曲率方程

$ (E\Gamma_n )U_n-U_n\Gamma_n =0,$

(2.4)

其中$\Gamma_n =\left( {{\begin{array}{cc} A_n&B_n \\ C_n&{-A_n} \\ \end{array} }} \right)$.方程(2.4)容易写成下面的方程

$ \left\{{\begin{array}{ll} {\frac{1}{u_n}B_{n+1}-u_n^2C_n=0,} \\[3mm] {u_n^2(A_{n+1}+A_n)+\frac{\lambda}{u_n} B_{n+1}=0,} \\[3mm] {-\frac{1}{u_n}(A_{n+1}+A_n)-\frac{\lambda}{u_n} C_n=0,} \\[3mm] {u_n^2C_{n+1}-\frac{1}{u_n}B_n-\frac{\lambda}{u_n} (A_{n+1}-A_n)=0.} \\[3mm] \end{array} }\right.$

(2.5)

令$A_n=\sum\limits_{m=0}^\infty {A_n^{(m)} \lambda ^{-2m}},B_n=\sum\limits_{m=0}^\infty {B_n^{(m)} \lambda ^{-2m+1}},C_n=\sum\limits_{m=0}^\infty {C_n^{(m)} \lambda ^{-2m+1}} $, 将这些表达式代入方程(2.5)可得

$ B_{n+1}^{(0)} =0,~~C_{n+1}^{(0)} =0,~~A_{n+1}^{(0)} -A_n^{(0)} =0,$

(2.6)

$ \left\{\begin{array}{ll} B_{n+1}^{(m)} =u_n^3C_n^{(m)},&m\ge 0,\\ B_{n+1}^{(m+1)} =-u_n^3(A_{n+1}^{(m+1)} +A_n^{(m)} ),&m\ge 0,\\ C_n^{(m+1)} =-(A_{n+1}^{(m+1)} +A_n^{(m)}),&m\ge 0,\\ (A_{n+1}^{(m+1)} -A_n^{(m)} )+B_n^{(m)} -u_n^3C_{n+1}^{(m)} =0,~~&m\ge 0. \end{array}\right.$

(2.7)

如果选择初值$A_n^{(0)} =-\frac{1}{2}$, 并且在利用(2.7)式计算$A_m (m\ge 1)$的过程中, 差分算子$D$的逆运算选择零常数, 即要求$A_n^{(m)} \left| {_{\left[u_n \right]=0} =0} \right.(m\ge 1)$, 其中$\left[u_n \right]=(u_n,Eu_n,E^{-1}u_n,\cdots )$.这样由(2.7)式可以唯一确定出$A_n^{(m)},B_n^{(m)},C_n^{(m)} (m\ge 1)$, 前几个量计算如下

$\begin{array}{l} A_n^{(1)} =(u_n^{(-1)})^3,~~B_{n+1}^{(1)} =(u_n^{(-1)})^3,~~C_n^{(1)} =1,\\ A_n^{(2)} =-(u_n^{(-1)}u_n)^3-(u_n^{(-1)}u_n^{(-2)})^3-(u_n^{(-1)})^6,\\ B_{n+1}^{(2)} =-(u_n^{(-1)}u_n^{(-2)})^3-(u_n^{(-1)})^6,~~C_n^{(2)} =-u_n^3-(u_n^{(-1)})^3,\cdots . \end{array} $

令$f=\sum\limits_{m\in Z} {f_m \lambda ^m} $, $f_+ =\sum\limits_{m\ge 0} {f_m \lambda ^m} $, 于是有

$\left( {\Gamma_n \lambda ^{2m}} \right)_+ =\left( {{\begin{array}{ccc} {\sum\limits_{i=0}^m {A_n^{(ⅰ)} \lambda ^{2m-2i}} } ~ {\sum\limits_{i=0}^m {B_n^{(ⅰ)} \lambda ^{2m-2i+1}} } \\[4mm] {\sum\limits_{i=0}^m {C_n^{(ⅰ)} \lambda ^{2m-2i+1}} } ~ {-\sum\limits_{i=0}^m {A_n^{(ⅰ)} \lambda ^{2m-2i}} } \\ \end{array} }} \right),$

容易验证

$(E(\Gamma_n \lambda ^{2m})_+ )U_n-U_n(\Gamma_n \lambda ^{2m})_+ =\left( {{\begin{array}{*{20}c} 0&~~{u_n^2(A_{n+1}^{(m)} +A_n^{(m)} )} \\[2mm] {-\frac{1}{u_n}(A_{n+1}^{(m)} +A_n^{(m)} )} &~ 0 \end{array} }} \right),$

为了得到离散可积系统, 选择修正项

$\Delta_n^{(m)} =\left( {{\begin{array}{*{20}c} {\frac{2}{3}C_n^{(m+1)}} ~~& 0 \\[3mm] 0~&{-\frac{1}{3}C_n^{(m+1)}} \end{array} }} \right),$

并且令$V_n^{(m)} =(\Gamma_n \lambda ^{2m})_+ +\Delta _n^{(m)} $, 则离散零曲率方程

$ U_{nt_m } =(EV_n^{(m)} )U_n-U_nV_n^{(m)},~~m\ge 0$

(2.8)

等价于以下的离散非线性可积系统

$u_{nt_m } =-\frac{1}{3}u_n(A_{n+2}^{(m)} -A_n^{(m)} ),~~m\ge 0.$

(2.9)

当$m=1$时, 得到系统(2.9)中第一个非线性可积系统

$ u_{nt_1 } =-\frac{1}{3}u_n(u_{n+1}^3-u_{n-1}^3),$

(2.10)

当$m=2$时, 得到系统(2.9)中第二个非线性可积系统

$ u_{nt_2 } =\frac{1}{3}u_nu_{n+1}^3(u_{n+2}^3+u_{n+1}^3+u_{n}^3)-\frac{1}{3}u_nu_{n-1}^3(u_{n}^3+u_{n-1}^3+u_{n-2}^3).$

(2.11)

下面利用迹恒等式来建立离散可积系统(2.9)的Hamilton结构.

令

$R_n=\Gamma_n U_n^{-1}=\left( {{\begin{array}{cc} {\frac{A_n\lambda+B_n}{u_n^2}} ~&{u_nA_n} \\[3mm] {\frac{-C_n\lambda-A_n}{u_n^2}} ~~& {u_nC_n} \end{array} }} \right),$

并且利用Killing-Cartan形式

$\left\langle {M,N} \right\rangle =tr(MN),$

其中$M,N$是同阶矩阵, 于是有

$\left\langle {R_n,\frac{\partial U_n}{\partial \lambda }} \right\rangle =C,{\begin{array}{*{20}c} \\ \end{array} }\left\langle {R_n,\frac{\partial U_n}{\partial u_n}} \right\rangle =-3\frac{A_n+C_n\lambda }{u_n}=3\frac{A_{n+1}}{u_n} . $

利用迹恒等式^[2]和变分恒等式^[19], 可以得到

$ \frac{\delta }{\delta u_n}\sum\limits_{n\in Z} {\left\langle {R_n,\frac{\partial U_n}{\partial \lambda }} \right\rangle } =\left( {\lambda ^{-\varepsilon }(\frac{\partial }{\partial \lambda })\lambda ^\varepsilon } \right)\left\langle {R_n,\frac{\partial U_n}{\partial u_n}} \right\rangle,$

(3.1)

$ \frac{\delta }{\delta u_n}\sum\limits_{n\in Z} C_n =\left( {\lambda ^{-\varepsilon }(\frac{\partial }{\partial \lambda })\lambda ^\varepsilon } \right)\frac{3A_{n+1}}{u_n}.$

(3.2)

将表达式$A_n=\sum\limits_{m=0}^\infty {A_n^{(m)} \lambda ^{-2m}},B_n=\sum\limits_{m=0}^\infty {B_n^{(m)} \lambda ^{-2m+1}},C_n=\sum\limits_{m=0}^\infty {C_n^{(m)} \lambda ^{-2m+1}} $代入(3.2)式, 并比较(3.2)式两端$\lambda ^{-2m-1}$的系数可得

$ \frac{\delta }{\delta u_n}\sum\limits_{n\in Z} {C_n^{(m+1)}} =\frac{3(\varepsilon -2m)}{u_n}A_{n+1}^{(m)},$

(3.3)

在上式中令$m=0$, 可得$\varepsilon =0$, 于是有

$ \frac{\delta }{\delta u_n}\sum\limits_{n\in Z} {\frac{C_n^{(m+1)}}{-6m}} =\frac{A_{n+1}^{(m+1)} }{u_n},~~m\ge 1.$

(3.4)

设

$ \tilde {H}_n^{(m)}=\sum\limits_{n\in Z} {\frac{C_n^{(m+1)}}{-6m}},~~m\ge 1,$

(3.5)

这样就有

$ \frac{\delta \tilde {H}_n^{(m)} }{\delta u_n}=\frac{A_{n+1}^{(m+1)} }{u_n},~~m\ge 1.$

(3.6)

从而(2.9)式可以改写为以下Hamilton形式

$ u_{nt_m } =J\frac{\delta \tilde {H}_n^{(m)} }{\delta u_n},~~m\ge 1,$

(3.7)

其中Hamilton算子$J=-\frac{1}{3}u_n(E-E^{-1})u_n.$由(2.7)式得

$\frac{A_{n+1}^{(m)} }{u_n}=L\frac{A_{n+1}^{(m-1)} }{u_n},$

其中递推算子$L=-\frac{1}{u_n}(1-E^{(-1)})^{-1}u_n^3(1+E)u_n-\frac{1}{u_n}(1-E)^{-1}u_n^3(1+E^{-1})u_n.$

利用Hamilton算子$J$可以定义以下Poisson括号^[2]

$\left\{ {f,g} \right\}_J =\sum\limits_{n\in Z} {\left(J\frac{\delta f_n}{\delta u_n},~\frac{\delta g_n}{\delta u_n}\right)},$

(3.8)

并且容易验证

$(JL)^\ast =-JL,~~\left\{ {\tilde {H}_m,\tilde {H}_l } \right\}_J =0,~~m,l\ge 1. $

在上式中上标*表示算子的共轭运算, 即$R^\ast $为算子$R$的共轭算子, 并且有

$ (\tilde {H}_n^{(m)} )_{t_l } = {\left\langle {\frac{\delta \tilde {H}_n^{(m)} }{\delta u_n},u_{nt_l } }\right \rangle} = \left\langle \frac{\delta \tilde {H}_n^{(m)} }{\delta u_n},J\frac{\delta \tilde {H}_n^{(l)} }{\delta u_n} \right\rangle=\left\{ {\tilde {H}_n^{(m)},\tilde {H}_n^{(l)} } \right\}_J =0,~~m,l\ge 1. $

因此有守恒量$\left\{ {\tilde {H}_n^{(m)} } \right\}_{m=1}^\infty $关于Poisson括号(3.8)两两对合, 由此推出离散可积系族(2.9)中每一个非线性微分-差分方程都是Liouville可积的.

下面讨论离散可积系统(2.9)的对称约束.

考虑谱问题(2.1)的共轭谱问题

$E^{-1}\psi =(E^{-1}U_n^T(u_n,\lambda ))\psi,~~\psi =\left( {{\begin{array}{*{20}c} {\psi_n^1 } \\ {\psi_n^2 } \\ \end{array} }} \right)$

(4.1)

与相应时间部分

$\psi _{t_m } =-(V_n^{(m)} (u_n,\lambda ))^T\psi,$

(4.2)

由(4.1)与(4.2)式的相容性条件$(E^{-1}\psi )_{t_m } =E^{-1}(\psi _{t_m } )$可以得到

$ E^{-1}U_{nt_m }^T =(E^{-1}U_n^T)(V_n^{(m)} )^T-(E^{-1}(V_n^{(m)} )^T)(E^{-1}U_n^T).$

(4.3)

通过验证容易发现(4.3)式等价于(2.8)式, 这样(4.3)式为离散可积系统(2.9)的另一个离散零曲率表示, 而(4.1)与(4.2)式称为离散可积系统(2.9)的共轭Lax对, 利用共轭Lax对(4.1)与(4.2)式, 可以计算特征值$\lambda $相对于位势$u_n$的变分导数.令$\lambda _1 \,,\,\lambda _2 \,,\,\cdots \,\lambda _N $是$N$个不同的特征值, 则有

$\left( {{\begin{array}{*{20}c} {E\varphi _n^{1j} } \\ {E\varphi _n^{2j} } \\ \end{array} }} \right)=U(u_n,\lambda _j )\left( {{\begin{array}{*{20}c} {\varphi _n^{1j} } \\ {\varphi _n^{2j} } \\ \end{array} }} \right)\,,\left( \begin{array}{*{20}c} E^{-1}\psi _n^{1j} \\ E^{-1}\psi _n^{2j} \\ \end{array} \right)=(E^{-1}U^T(u_n,\lambda_j ))\left( {{\begin{array}{*{20}c} {\psi _n^{1j} } \\ {\psi _n^{2j} } \\ \end{array} }} \right)\,,\,\,\,$

(4.4)

$ \left( \begin{array}{*{20}c} \varphi _n^{1j} \\ \varphi _n^{2j} \\ \end{array} \right)_{t_m } =V_n^{(m)} (u_n,\lambda _j )\left( \begin{array}{*{20}c} \varphi _n^{1j} \\ \varphi _n^{2j} \\ \end{array} \right)\,,\left( \begin{array}{*{20}c} \psi _n^{1j} \\ \psi _n^{2j} \\ \end{array} \right)_{t_m} =-(V_n^{(m)})^ T(u_n,\lambda _j )\left( \begin{array}{*{20}c} \psi _n^{1j} \\ \psi _n^{2j} \\ \end{array} \right)\,,$

(4.5)

其中$1\le j\le N$.于是有

$(E\varphi ^{1j}_n,\,E\varphi ^{2j} _n )=(\varphi ^ {1j}_n,\varphi ^{2j}_n)U(u_n,\lambda _j )^T,1\le j\le N,$

(4.6)

$(E\psi _n^{1j},E\psi _n^{2j} )=(\psi _n^{1j},\psi ^{2j} _n )U(u_n,\lambda _j )^{-1},1\le j\le N.$

(4.7)

根据文献[11], 可求得特征值$\lambda _j(1\le j\le N) $对位势$u_n$的变分导数

$\frac{\delta \lambda _j }{\delta u_n}=\alpha _j \frac{-\varphi _n^{1j} \psi _n^{1j} +2\varphi _n^{2j} \psi _n^{2j} -3\lambda _j \varphi _n^{2j} \psi _n^{1j} }{u_n},~~~1\le j\le N.$

(4.8)

记

$\Phi _i =(\varphi _n^{i1},\varphi _n^{i2},\cdots,\,\varphi _n^{iN} )^T,~~\Psi _i =(\psi _n^{i1},\psi _n^{i2},\cdots,\,\psi _n^{iN} )^T,~ i=1,2,$

以及

$\Lambda =diag(\lambda _1,\lambda _2,\cdots,\lambda _N ),$

考虑以下Bargmann约束

$J\frac{\delta \tilde {H}_n^{(1)} }{\delta u_n}=J\sum\limits_{j=0}^N \frac{\delta \lambda _j }{\delta u_n},$

(4.9)

其中$\alpha _j =1(1\le j\le N)$.即有

$\frac{\delta \tilde {H}_n^{(1)} }{\delta u_n}=\frac{-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle }{u_n},$

(4.10)

从(4.10)式中可以解出位势

$u_n=(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{1/3},$

(4.11)

其中$\left\langle {.\,,\,.} \right\rangle $为${\Bbb R}^N$中标准内积.将(4.10)式代入(4.6)和(4.8)式, 得非线性化离散Bargmann系统

$ \left\{\begin{array}{rl} E\Phi _1 =&(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{2/3}\Phi _2,\\ E\Phi _2 =&(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-1/3}\Phi _1 \\ &+(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-1/3} \Lambda \Phi _2,\\ E\Psi _1 =&-(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-2/3}\Lambda \Psi _1 \\ &+(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-2/3} \Psi _2,\\ E\Psi _2 =&(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{1/3}\Psi _1. \end{array}\right.$

(4.12)

下面证明:由(4.4)式可以给出一个辛映射.为此, 定义$f_i =f_i (\Phi _1,\Phi _2,\Psi _1,\Psi _2 )\,,g_i =g_i (\Phi _1,\Phi _2,\Psi _1,\Psi _2 )$如下

$ \left\{\begin{array}{l}\begin{array}{l} f_i =-(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-2/3}\lambda _i \psi _n^{1i} \\ {\begin{array}{*{20}c} ~~~~+{(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-2/3}} \\ \end{array} }\psi _n^{2i},{\begin{array}{*{20}c}&{1\le i\le N,} \\ \end{array} } \end{array}\\ f_{N+i} =(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{1/3}\psi _n^{1i},{\begin{array}{*{20}c}&{1\le i\le N,} \\ \end{array} } \\ g_i =(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{2/3}\varphi _n^{2i},{\begin{array}{*{20}c} {{\begin{array}{*{20}c}&\\ \end{array} }}&{1\le i\le N,} \\ \end{array} } \\ \begin{array}{l} g_{N+i} =(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-1/3}\varphi _n^{1i} \\ {\begin{array}{*{20}c} {{\begin{array}{*{20}c}&\\ \end{array} }} &~~~~~~+ {(-\left\langle {\Phi _1,\Psi _1 } \right\rangle +2\left\langle {\Phi _2,\Psi _2 } \right\rangle -3\left\langle {\Lambda \Phi _2,\Psi _1 } \right\rangle )^{-1/3}} \\ \end{array} }\lambda _i \varphi _n^{2i},{\begin{array}{*{20}c}&{1\le i\le N.} \\ \end{array} } \\ \end{array}\end{array}\right.$

(4.13)

对于给定的函数$f=f(\Phi _1,\Phi _2,\Psi _1,\Psi _2 )$与$g=g(\Phi _1,\Phi _2,\Psi _1,\Psi _2 )$, 定义Poisson括号

$ \left\{ {f\,,\,g} \right\}=\sum\limits_{i=1}^2 {\sum\limits_{j=1}^N {\bigg(\frac{\partial f}{\partial \psi _n^{ij} }\frac{\partial g}{\partial \varphi _n^{ij} }-\frac{\partial f}{\partial \varphi _n^{ij} }\frac{\partial g}{\partial \psi _n^{ij} }\bigg)} }=\sum\limits_{i=1}^2 { \bigg(\left\langle {\frac{\partial f}{\partial \Psi _i },\frac{\partial g}{\partial \Phi _i }} \right\rangle -\left\langle {\frac{\partial f}{\partial \Phi _i },\frac{\partial g}{\partial \Psi _i }} \right\rangle } \bigg).$

(4.14)

命题1 如果令

$H(\Psi _1,\Psi _2,\Phi _1,\Phi _2 )=(E\Psi _1,E\Psi _2,E\Phi _1,E\Phi _2 ),$

(4.15)

则(4.15)式定义了一个辛映射.

证 通过计算^{[5-6, 13-14]}可得

$\left\{ {f_i,f_j } \right\}=\left\{ {g_i,g_j } \right\}=0\,,\,\,\left\{ {f_i,g_j } \right\}=\delta _{ij},~~1\le i,j\le 2N.$

(4.16)

下面考虑Lax对与共轭Lax对时间部分(4.5)的非线性化, 我们将证明它是Liouville可积有限维Hamilton系统, 将(4.3)式代入(4.11)式, 可得

$\begin{array}{l}\left( {{\begin{array}{*{20}c} {\varphi _n^{1j} } \\ {\varphi _n^{2j} } \\ \end{array} }} \right)_{t_m } =\left. {V_n^{(m)}(u_n,\lambda _j )} \right|_B \left( {{\begin{array}{*{20}c} {\varphi _n^{1j} } \\ {\varphi _n^{2j} } \\ \end{array} }} \right),~~j=1,2,\cdots N,\\[5mm] \left( {{\begin{array}{*{20}c} {\psi _n^{1j} } \\ {\psi _n^{2j} } \\ \end{array} }} \right)_{t_m } =-\left. {V_n^{(m)T}(u_n,\lambda _j )} \right|_B \left( {{\begin{array}{*{20}c} {\psi _n^{1j} } \\ {\psi _n^{2j} } \\ \end{array} }} \right),~~j=1,2,\cdots N,\end{array}$

(4.17)

这里下标B表示将(4.3)式代入相应的表达式$V_n^{(m)}(u_n,\lambda _j )$和$V_n^{(m)T}(u_n,\lambda _j )$.令

$\begin{array}{l} \tilde {A}_0 =-\frac{1}{2},\,\,\,\tilde {B}_0 =0,\,\,\,\tilde {C}_0 =0,~~\tilde {B}_1 =2\left\langle {\Phi _1,\Psi _1 } \right\rangle -\left\langle {\Phi _2,\Psi _2 } \right\rangle,~~\tilde {C}_1 =1,\\ [3mm] \tilde {A}_m =\frac{3}{2}(\left\langle {\Lambda ^{2m-2}\Phi _1,\Psi _1 } \right\rangle -\left\langle {\Lambda ^{2m-2}\Phi _2,\Psi _2 } \right\rangle ),\\[2mm] \tilde {B}_m =3\left\langle {\Lambda ^{2m-3}\Phi _{1,} \Psi _2 } \right\rangle,m\ge 2,\\ \tilde {C}_m =3\left\langle {\Lambda ^{2m-3}\Phi _2,\Psi _1 } \right\rangle,m\ge 2,\end{array}$

(4.18)

并且

$\begin{array}{l} \tilde {A}=\sum\limits_{m=0}^\infty {\tilde {A}_n^{(m)} \lambda ^{-2m}},\tilde {B}=\sum\limits_{m=0}^\infty {\tilde {B}_n^{(m)} \lambda ^{-2m+1}},\tilde {C}=\sum\limits_{m=0}^\infty {\tilde {C}_n^{(m)} \lambda ^{-2m+1}},\\[4mm] \tilde {A}^2+\tilde {B}\tilde {C}=\sum\limits_{j=0}^\infty {\tilde {F}_n^{(j)} \lambda ^{-2j}} \,,\,\,\,\,\\[4mm] \tilde {F}_n^{(0)} =\frac{1}{4},\tilde {F}_n^{(m)} =\sum\limits_{i=0}^m {\tilde {A}_n^{(ⅰ)} \tilde {A}_n^{(m-i)} +\sum\limits_{i=1}^m \tilde {B}_n^{(ⅰ)} \tilde {C}_n^{(m+1-i)},~~m\ge 1} .\end{array}$

(4.19)

证毕.

命题2  $D\tilde {F}_n^{(m)} =0,m\ge 0$.

证 容易验证$\left\{ {\tilde {A}_n^{(m)} } \right\}_{m=0}^\infty $, $\left\{ {\tilde {B}_n^{(m)} } \right\}_{m=0}^\infty $, $\left\{ {\tilde {C}_n^{(m)} } \right\}_{m=0}^\infty $满足(2.6)与(2.7)式, 从而

$\tilde {\Gamma }=\left( {{\begin{array}{*{20}c} {\tilde {A}_n} ~& {\tilde {B}_n} \\ {\tilde {C}_n} ~& {-\tilde {A}_n} \\ \end{array} }} \right)= \left( {{\begin{array}{*{20}c} {\sum\limits_{m=0}^\infty {\tilde {A}_n^{(m)} \lambda ^{-2m}} }&{\sum\limits_{m=0}^\infty {\tilde {B}_n^{(m)} \lambda ^{-2m+1}} } \\[4mm] {\sum\limits_{m=0}^\infty {\tilde {C}_n^{(m)} \lambda ^{-2m+1}} }&{-\sum\limits_{m=0}^\infty {\tilde {A}_n^{(m)} \lambda ^{-2m}} } \\ \end{array} }} \right) $

满足(2.4)式, 于是有

$D(\tilde {A}_n^2+\tilde {B}_n\tilde {C}_n)=0.$

由(4.19)式得

$ \tilde {F}_n^{(0)} =\frac{1}{4}\,,\,\,\,\tilde {F}_n^{(1)} =-\frac{3(\left\langle {\Phi _1,\Psi _1 } \right\rangle -\left\langle {\Phi _2 \Psi _2 } \right\rangle )}{2}+2\left\langle {\Phi _1,\Psi _1 } \right\rangle -\left\langle {\Phi _2,\Psi _2 } \right\rangle,$

$\begin{eqnarray} \tilde {F}_n^{(m)} &=&-\frac{3}{2}(\left\langle {\Lambda ^{2m-2}\Phi _1 ,\Psi _1 } \right\rangle -\left\langle {\Lambda ^{2m-2}\Phi _2 ,\Psi _2 } \right\rangle )\\ &&+ \frac{9}{4}\sum\limits_{i=1}^{m-1} {(\left\langle {\Lambda ^{2i-2}\Phi _1 ,\Psi _1 } \right\rangle -\left\langle {\Lambda ^{2i-2}\Phi _2 ,\Psi _2 } \right\rangle )(} \left\langle {\Lambda ^{2m-2i-2}\Phi _1 ,\Psi _1 } \right\rangle -\left\langle {\Lambda ^{2m-2i-2}\Phi _2 ,\Psi _2 } \right\rangle ) \\ &&+9\sum\limits_{i=2}^{m-1} {\left\langle {\Lambda ^{2i-3}\Phi _1 ,\Psi _2 } \right\rangle \left\langle {\Lambda ^{2m-2i-1}\Phi _2 ,\Psi _1 } \right\rangle } +6(\left\langle {\Phi _1 ,\Psi _1 } \right\rangle -\left\langle {\Phi _2 ,\Psi _2 } \right\rangle )\left\langle {\Lambda ^{2m-3}\Phi _2 ,\Psi _1 } \right\rangle .\\ \end{eqnarray}$

(4.20)

这时容易验证(4.17)式可以改写成以下Hamilton系统

$D(\tilde {A}_n^2+\tilde {B}_n\tilde {C}_n)=0,\Psi _{it_m } =-\frac{\partial ({1 \over 3}\tilde {F}_n^{(m+1)} )}{\partial \Phi _i },\Phi _{it_m } =\frac{\partial ({1 \over 3}\tilde {F}_n^{(m+1)} )}{\partial \Psi _i },i=1,2.$

(4.21)

根据文献[5, 10]可以得到

$2\frac{\rm d}{{\rm d}t_m }(\tilde {A}_n^2+\tilde {B}_n\tilde {C}_n)=\frac{\rm d}{{\rm d}t_m }tr\tilde {\Gamma }_n^2=\frac{\rm d}{{\rm d}t_m }tr[\tilde {V}_n^{(m)},\tilde {\Gamma }_n^2]=0,$

即

$\frac{\rm d}{{\rm d}t_m }\tilde {F}_n^{(l)} =0,m,l=1,2,\cdots. $

并且有

$\{\tilde {F}_n^{(l)},\tilde {F}_n^{(m+l)} \}=-\frac{\rm d}{{\rm d}t_m }\tilde {F}_n^{(l)} =0,~~l,m=1,2,\cdots. $

这说明$\tilde {F}_n^{(m)} (m\ge 0)$是离散Hamilton可积系统(2.9)的运动积分并且关于Poisson括号(4.16)两两对合.此外, 令

$\overline {F }_n^{(j)} =\varphi _n^{1j} \psi _n^{1j} +\varphi _n^{2j} \psi _n^{2j},~~1\le j\le N,$

通过直接计算可得

$ \{ \tilde {F}_n^{(m+1)},\bar {F}_n^{(j)} \}=\frac{{\rm d}\bar {F}_n^{(j)} }{{\rm d}t_m }=0,~1\le j\le N,m\ge 0,\{ \bar {F}_n^{(ⅰ)},\bar {F}_n^{(j)} \}=0,1\le i,j\le N.$

证毕.

命题3  $\tilde {F}_n^{(m+1)} (1\le m\le N),\bar {F}_n^{(j)} (1\le j\le N)$, 在${\Bbb R}^{4N}$的某一区域内是函数无关的.

证 由(4.21)式得

$\left\{\begin{array}{rl} \left. {\frac{\partial \tilde {F}_n^{(m+1)} }{\partial \Phi _1 }} \right|_{\Phi _2 =0,\Psi _2 =\frac{1}{3}\Lambda \Psi _1 } =& -\frac{1}{2}\Lambda ^{2m}\Psi _1 +\frac{9}{2}\sum\limits_{i=1}^m {\left\langle {\Lambda ^{2i-2}\Phi _1,\Psi _1 } \right\rangle } \Lambda ^{2m-2i}\Psi _1,\\[5mm] \left. {\frac{\partial \tilde {F}_n^{(m+1)} }{\partial \Phi _2 }} \right|_{\Phi _2 =0,\Psi _2 =\frac{1}{3}\Lambda \Psi _1 } =& \frac{1}{2}\Lambda ^{2m+1}\Psi _1 +6\left\langle {\Phi _1,\Psi _1 } \right\rangle \Lambda ^{2m}\Psi _1 \\[4mm] & +\frac{3}{2}\sum\limits_{i=1}^m {\left\langle {\Lambda ^{2i-2}\Phi _1,\Psi _1 } \right\rangle } \Lambda ^{2m-2i+1}\Psi _1 .\end{array}\right.$

取$\sum\limits_{j=1}^N {(\psi_n^{1j})^2 \ne 0} $, 并记$\Phi _1 =(\varphi _n^{11,} \varphi _n^{12},\cdots \varphi _n^{1N} )^T$为以下线性齐次方程组的解

$ \begin{array}{l} \varphi _n^{11} \psi _n^{11} +\varphi _n^{12} \psi _n^{12} +\cdots \varphi _n^{1N} \psi _n^{1N} =0,\\ \lambda _1^2 \varphi _n^{11} \psi _n^{11} +\lambda _2^2 \varphi _n^{12} \psi _n^{12} +\cdots +\lambda _N^2 \varphi _n^{1N} \psi _n^{1N} =0,\\ \cdots \cdots \cdots \\ \lambda _1^{2N-2} \varphi _n^{11} \psi _n^{11} +\lambda _2^{2N-2} \varphi _n^{12} \psi _n^{12} +\cdots +\lambda _N^{2N-2} \varphi _n^{1N} \psi _n^{1N} =0. \\ \end{array}$

(4.22)

由于Vandermode行列式

$\left| {{\begin{array}{*{20}c} 1&1&\cdots &~ 1 \\ {\lambda _1^2 }&{\lambda _2^2 }&\cdots &~ {\lambda _N^2 } \\&\cdots&\cdots&\\ {\lambda _1^{2N-2} }&~~{\lambda _2^{2N-2} } ~&\cdots &~ {\lambda _N^{2N-2} } \\ \end{array} }} \right|\ne 0,$

所以$\Phi _1 =(\varphi _n^{11},\varphi _n^{12,} \cdots \varphi _n^{1N} )^T$能由(4.22)式唯一地确定.这时有

$\frac{\partial \tilde {F}_n^{(m+1)} }{\partial \Phi _1 }=-\frac{1}{2}\Lambda ^{2m}\Psi _1,\frac{\partial \tilde {F}_n^{(m+1)} }{\partial \Phi _2 }=\frac{1}{2}\Lambda ^{2m+1}\Psi _1,m\ge 0.$

另外, 通过直接计算可得

$\frac{\partial \bar {F}_n^{(j)} }{\partial \varphi _n^{il} }=\psi _n^{ij} \delta _n^{jl},i=1,2;j,l=1,2\cdots N. $

于是

$ \det \left( {{\begin{array}{*{30}c} {\frac{\partial \bar {F}_1 }{\partial \Phi _1 }}&\cdots&{\frac{\partial \bar {F}_N }{\partial \Phi _1 }}& {\frac{\partial \tilde {F}_1 }{\partial \Phi _1 }}&\cdots& {\frac{\partial \tilde {F}_N }{\partial \Phi _1 }} \\[3mm] {\frac{\partial \bar {F}_1 }{\partial \Phi _2 }}&\cdots&{\frac{\partial \bar {F}_N }{\partial \Phi _2 }}& {\frac{\partial \tilde {F}_1 }{\partial \Phi _2 }}&\cdots& {\frac{\partial \tilde {F}_N }{\partial \Phi _2 }} \\ \end{array} }} \right)\\ =(\frac{-1}{2})^N\det \left( {{\begin{array}{*{20}c} {\psi _{11} }&\cdots&0&{\psi _{11} }&\cdots&{\lambda _1^{2N-2} \psi _{11} } \\ \vdots&\ddots&\vdots&\vdots&\ddots&\vdots \\ 0&\cdots&{\psi _{1N} }&{\psi _{1N} }&\cdots&{\lambda _N^{2N-2} \psi _{1N} } \\ {\frac{\psi _{11} }{3}}&\cdots&0&{-\lambda _1 \psi _{11} }&\cdots&{-\lambda _1^{2N-1} \psi _{11} } \\ \vdots&\ddots&\vdots&\vdots&\ddots&\vdots \\ 0&\cdots & {\frac{\psi _{1N} }{3}}&{-\lambda _N \psi _{1N} }&\cdots&{-\lambda _N^{2N-1} \psi _{1N} } \\ \end{array} }} \right) \\ =(-\frac{2}{3})^N(\prod\limits_{j=1}^N {\lambda _j \psi _{1j}^2 )\left| {{\begin{array}{*{20}c} 1&{\lambda _1^2 }&\cdots&{\lambda _1^{2N-2} } \\ 1&{\lambda _2^2 }&\cdots&{\lambda _2^{2N-2} } \\&\cdots&\cdots&\\ 1&{\lambda _N^2 }&\cdots&{\lambda _N^{2N-2} } \\ \end{array} }} \right|}.$

因此函数$\bar {F}_n^{(j)},\tilde {F}_n^{(j+1)} (j=1,2\cdots N)$在${\Bbb R}^{4N}$的某一区域上是函数独立的.

综上所述, 有以下定理.

定理1 辛映射(4.15)与有限维的Hamilton系统族(2.9)在Liouville意义下都是完全可积的.

定理2 令$(\Phi _i (n,t_m ),\Psi _i (n,t_m ))\ (i=1,2)$为以下系统

$\frac{\partial \Psi _i }{\partial t_m }=-\frac{\partial ({1 \over 3}\tilde {F}_n^{(m+1)} )}{\partial \Phi _i },~ {\frac{\partial \Phi _i }{\partial t_m }=\frac{\partial ({1 \over 3}\tilde {F}_n^{(m+1)} )}{\partial \Psi _i }},~~i=1,2,$

(4.23)

满足初值$(\Phi _i,\Psi _i )\left| {_{t_m =0} =(\Phi _i (0),\Psi _i (0))\ (i=1,2)} \right.$的解, 其中$\Phi _i (0),\Psi _i (0)\ (i=1,2)$为任意常数, 并且令

$(\Phi _i (n,t_m ),\Psi _i (n,t_m ))=H^n(\Phi _i (t_m ),\Psi _i (t_m )),i=1,2.$

(4.24)

则

$u(n,t_m )=(-\left\langle {\Phi _1 (n,t_m ),\Psi _1 (n,t_m )} \right\rangle +2\left\langle {\Phi _2 (n,t_m ),\Psi _2 (n,t_m )} \right\rangle \\ +3\left\langle {\Lambda \Phi _2 (n,t_m ),\Psi _1 (n,t_m )} \right\rangle )^{1/3}$

(4.25)

为离散可积系(2.9)的解.

证 根据(4.12)式, (4.15)式, (4.23)与(4.24)式, 有

$\begin{eqnarray*} \frac{\partial a(n,t_m )}{\partial t_m }&=&\frac{1}{3(a(n,t_m ))^2} \bigg(-\left\langle {\frac{\partial \Phi _1 (n,t_m )}{\partial t_m },\Psi _1 (n,t_m )} \right\rangle -\left\langle {\frac{\partial \Psi _1 (n,t_m )}{\partial t_m },\Phi _1 (n,t_m )} \right\rangle\bigg) \\&& +2\left\langle {\frac{\partial \Phi _2 (n,t_m )}{\partial t_m },\Psi _2 (n,t_m )} \right\rangle +2\left\langle {\frac{\partial \Psi _2 (n,t_m )}{\partial t_m },\Phi _2 (n,t_m )} \right\rangle \\ &&-3\left\langle {\frac{\partial \Phi _2 (n,t_m )}{\partial t_m },\Lambda \Psi _1 (n,t_m )} \right\rangle -3\left\langle {\frac{\partial \Psi _1 (n,t_m )}{\partial t_m },\Lambda \Phi _2 (n,t_m )} \right\rangle \\ &=&-u(n,t_m )\left\langle {\Lambda ^{2m-1}\Phi _2 (n,t_m ),\Psi _1 (n,t_m )} \right\rangle \\ &&+ \frac{1}{(u(n,t_m ))^2}(\left\langle {\Lambda ^{2m-1}\Phi _2 (n,t_m ),\Psi _1 (n,t_m )} \right\rangle \\ &&-\left\langle {\Lambda ^{2m+1}\Phi _2 (n,t_m ),\Psi _1 (n,t_m )} \right\rangle -\left\langle {\Lambda ^{2m}\Phi _1 (n,t_m ),\Psi _1 (n,t_m )} \right\rangle\\ && +\left\langle {\Lambda ^{2m}\Phi _2 (n,t_m ),\Psi _2 (n,t_m )} \right\rangle )\\ &=&\frac{u(n,t_m )}{3}(\tilde {A}_{n+2}^{(m)} -\tilde {A}_n^{(m)} ). \end{eqnarray*}$

证毕.

由定理2可以得出, 离散可积系统(2.9)的解, 可以由系统(4.23)解加上可积辛映射的迭代(4.24)得出.