孤子方程是一种描述自然界中发生的非线性现象重要模型,在物理学的各个领域都有广泛应用,比如: 等离子物理、磁流体等. 研究孤子方程的精确解不仅有利于了解方程的本质属性和代数结构,而且有助于对它们所反映的现实自然现象进行分析和研究. 目前在孤子理论中已经有一批行之有效求精确解的方法,如反散射方法^{[1, 2]}、代数几何方法^{[3, 4, 5]}、Hirota双线性方法^{[6, 7, 8]}、Lax对非线性化方法^{[9, 10, 11, 12, 13]},以及其它一些方法^{[14, 15, 16, 17]}.其中Lax对非线性化方法是研究$(1+1)$ -维和$(2+1)$ -维孤子方程的有效工具,它将$(1+1)$-维孤子方程分离成相容的常微分方程,可以证明是Liouville意义下的有限维完全可积系. $(2+1)$-维孤子方程可以采用相似的方法分离成$(1+1)$ -维孤子方程,然后再分离成相容的常微分方程.

本文我们将构造$(2+1)$ -维广义Broer-Kaup-Kupershmidt孤子方程的拟周期解. 主要内容安排如下: 在第二节,$(2+1)$-维广义Broer-Kaup-Kupershmidt孤子方程分离成两个相容的$(1+1)$ -维孤子方程; 在第三节,在特征函数与位势之间的Bargmann约束下,谱问题被非线性化,并证明了原谱问题对应的Hamilton守恒积分的对合性; 在第四节,借助于椭圆坐标和拟Abel-Jacobi坐标,证明了此有限维Hamilton系统的完全可积性; 在第五节,通过Abel-Jacobi坐标反演,利用Riemann $\theta$函数得到了孤子方程的拟周期解.

考虑新谱问题如下

\begin{eqnarray}\phi_{x}=U\phi,U= \left(\begin{array}{cc}\lambda+u ~~&v+\beta u\\ 1~~&- \lambda-u\end{array}\right),\phi=\left(\begin{array}{l} p\\ q \end{array} \right),\end{eqnarray}

(2.1)

其中$\lambda$是谱参数,$u,v$是位势,$\beta$ 是常数. 定义线性映射 $\sigma_{\lambda}:{\Bbb C}^{3}\mapsto sl(2,{\Bbb C})$

\begin{equation} \sigma_{\lambda}(\alpha)=\left( \begin{array}{cc} \alpha_1+\lambda\alpha_3 ~~&\alpha_2+\beta\alpha_1\\ \alpha_3~~&-\alpha_1-\lambda\alpha_3\end{array} \right),~~\alpha\in {\Bbb C}^{3}. \end{equation}

(2.2)

设$V=\sigma_{\lambda}\ (G),G\in {\Bbb C}^{3}$,直接计算给出

\begin{eqnarray} V_{x}-[U,V]=\sigma_{\lambda}\{(K-\lambda J)G\} ,\end{eqnarray}

(2.3)

其中$K,J$是Lenard算子

\begin{equation} K=\left( \begin{array}{ccc} \partial+\beta&1&-(v+\beta u)\\ 2v-\beta^{2}&~~\partial-\beta-2u~~&\beta(v+\beta u)\\ -2&0&\partial+2u\end{array}\right),~~J=\left(\begin{array}{ccc}-2&0&2u\\4\beta&~~2~~&-2(v+2\beta u)\\ 0&0&0\end{array} \right),\end{equation}

(2.4)

这里$\partial=\partial/\partial x$. 令$G=\sum\limits_{m=0}^{\infty}\lambda^{-m}g_{m-1}$,$g_m=(g_m^{1},g_m^{2},g_m^{3})^{T}\in {\Bbb C}^{3}$,可得下面的命题.

命题2.1 令 $V=\sigma_{\lambda}\ (G)$,则$V_x-[U,V]=0$ 的充分必要条件是 $Kg_m=Jg_{m+1},Jg_{-1}=0,$ $ m=-1,0,1,\cdots .$

Lenard序列$g_m$和广义Broer-Kaup-Kupershmidt向量场$X_m$的递归关系定义如下

\begin{eqnarray} \begin{array}{l} Kg_m=Jg_{m+1},Jg_{-1}=0,\\ X_m=PJg_m=PKg_{m-1},m=0,1,2,\cdots,\end{array}\end{eqnarray}

(2.5)

其中投射 $P$: $\gamma=(\gamma_1,\gamma_2,\gamma_3)^{T}\mapsto(\gamma_1,\gamma_2)^{T}$. 序列$g_m(m\geq-1)$的分量式为

\begin{eqnarray}\left\{\begin{array}{l} g_{m+1}^{1}=\frac{1}{2}(\partial+2u)g_{m+1}^{3},\\ g_{m+1}^{2}=-2\beta g_{m+1}^{1}+(v+2\beta u)g_{m+1}^{3}+\frac{1}{2}(2v-\beta^{2})g_{m}^{1} \\\qquad\qquad +\frac{1}{2}(\partial-\beta-2u)g_{m}^{2} +\frac{1}{2}\beta(v+\beta u)g_{m}^{3},\\ g_{m+1 x}^{3}=-g_{m x}^{1}-\beta g_{m}^{1}-g_{m}^{2}+(v+\beta u)g_{m }^{3},\end{array}\right.\end{eqnarray}

(2.6)

其中

\begin{eqnarray} g_{-1}&=&(u,v,1),\\ g_{0}&=&(-\frac{1}{2}u_x-u^{2},\frac{1}{2}v_x-uv+\beta u_x,-u),\\ g_{1}&=&(\frac{1}{4}u_{xx}-\frac{1}{4}v_{x}-\frac{1}{4}\beta u_{x}-\frac{1}{2}uv-\frac{1}{2}\beta u^{2}+\frac{3}{2}uu_x+u^{3},\frac{1}{4}v_{xx}+\frac{1}{4}\beta v_{x}+\frac{1}{4}\beta ^{2}u_{x}\\ && -\frac{1}{2}vu_x-\frac{1}{2}v^{2}-\frac{1}{2}\beta uv-uv_x+u^{2}v-3\beta uu_x,\frac{1}{2}u_x-\frac{1}{2}v-\frac{1}{2}\beta u+u^{2}),\\ g_{2}&=&(-\frac{1}{8}u_{xxx}-\frac{3}{4}u_{x}^{2}+\frac{3}{4}u_xv+\frac{3}{4}uv_x+\frac{3}{2}u^{2}v+\frac{3}{2}\beta u^{3}-uu_{xx}-u^{4}-3u^{2}u_x,\\ &&\frac{1}{8}v_{xxx}+\frac{1}{4}\beta u_{xxx}-\frac{1}{4}vu_{xx}-\frac{3}{4}uv_{xx}-\frac{3}{4}u_xv_x-\frac{3}{4}vv_x+\frac{3}{4}\beta^{2}uu_x\\ && -\frac{3}{2}\beta vu_x+\frac{3}{2}uv^{2}+\frac{3}{2}uvu_x+\frac{3}{2}u^{2}v_x+\frac{5}{2}\beta u^{2}v-\beta uv_x-\beta u^{2}v-u^{3}v\\ && +6\beta u^{2}u_x,-\frac{1}{4}u_{xx}-\frac{3}{2}uu_x+\frac{3}{2}uv+\frac{3}{2}\beta u^{2}-u^{3}),\cdots . \end{eqnarray}

(2.7)

令

\begin{equation} G_N=(\lambda^{N}G)_+=\sum_{m=1}^{N}\lambda^{N-m}g_{m-1},V_N=\sigma_{\lambda}(G_N). \end{equation}

(2.8)

从零曲率方程 $U_{t_{N}}- V_{N,x}+ [U,V_N]=0$,可得广义Broer-Kaup-Kupershmidt孤子族

\begin{equation} \left(\begin{array}{c} u\\ v\end{array}\right)_{t_{N}}=X_N. \end{equation}

(2.9)

设 $t_1=y,t_2=t$,则有

\begin{eqnarray}\left\{\begin{array}{l} u_y=(-\frac{1}{2}u_x+\frac{1}{2}v+\frac{1}{2}\beta u-u^{2})_x,\\ v_y=(\frac{1}{2}v_x-\frac{1}{2}\beta v-\frac{1}{2}\beta^{2}u+\beta u_x-\beta u^{2}-2uv)_x,\end{array}\right.\end{eqnarray}

(2.10)

\begin{eqnarray}\left\{\begin{array}{l} u_t=(\frac{1}{4}u_{xx}+\frac{3}{2}uu_x-\frac{3}{2}uv+u^{3})_x,\\ v_t=(\frac{1}{4}v_{xx}-\frac{3}{4} v^{2}+\frac{3}{4}\beta^{2}u^{2}-\frac{3}{2}uv_x+2\beta u^{3}-3\beta uu_x+3u^{2}v)_x+\beta uv_x. \end{array}\right.\end{eqnarray}

(2.11)

若 $(u(x,y,t),v(x,y,t))$ 是方程(2.10)和(2.11)式的相容解,则 $(u(x,y,t),v(x,y,t))$ 是下面$(2+1)$ -维广义Broer-Kaup-Kupershmidt孤子方程的解

\begin{eqnarray}\left\{\begin{array}{l} u_t=(\frac{1}{4}u_{xx}+\frac{3}{2}\beta u^{2}-2u^{3}-3u\partial^{-1}u_y)_x,\\ v_t=(\frac{1}{4}v_{xx}-\frac{3}{4} v^{2}-\frac{9}{4}\beta^{2}u^{2}-\frac{3}{2}uv_x-3\beta uv+3u^{2}v+6\beta u\partial^{-1}u_y+8\beta u^{3})_x+\beta uv_x. \end{array}\right.\end{eqnarray}

(2.12)

若令 $\beta=0$,则方程(2.10)约化成经典的Broer-Kaup-Kupershmidt孤子方程^[18]

\begin{eqnarray}\left\{\begin{array}{l} u_t=-\frac{1}{2}u_{xx}+\frac{1}{2}v_x-2uu_x,\\ v_t=\frac{1}{2}v_{xx}-2uv_x-2vu_x. \end{array}\right.\end{eqnarray}

(2.13)

若取 $v=\beta=0$,则方程(2.10)的第一个方程约化成Burgers方程

\begin{eqnarray} u_t=-\frac{1}{2}u_{xx}-2uu_x. \end{eqnarray}

(2.14)

若在方程(2.12)的第二个方程中取 $\beta=0$ ,可得 $(2+1)$ -维mKdV方程

\begin{eqnarray} u_t=(\frac{1}{4}u_{xx}-2u^{3}-3u\partial^{-1}u_y)_x. \end{eqnarray}

(2.15)

设 $\alpha_{j},1\leq j\leq N$是谱问题(2.1)的$N$个互异特征值,$\phi=(p_{j},q_{j})^{T}$是相应的特征函数,考虑谱问题的分量形式

\begin{eqnarray} \left( \begin{array}{l} p_j\\ q_j \end{array} \right)_x= \left( \begin{array}{cc} \alpha_j+u &v+\beta u\\ 1&- \alpha_j -u\end{array} \right)\left( \begin{array}{l} p_j\\ q_j \end{array} \right),1\leq j \leq N. \end{eqnarray}

(3.1)

直接的计算给出 $KS_j=\alpha_jJS_j$,其中 $S_j=(p_jq_j-\alpha_jq_j^{2},-p_j^{2}+\beta\alpha_jq_j^{2}-\beta p_jq_j,q_j^{2})$. 考虑Bargmann约束

\begin{eqnarray} g_0=\sum_{j=1}^{N}S_j,\end{eqnarray}

(3.2)

利用(3.1)式可得其精确表示为

\begin{eqnarray}\left\{\begin{array}{l} u=-\langle q,q\rangle,\\ v=-2\langle p,q\rangle+\beta\langle q,q\rangle,\end{array}\right. \end{eqnarray}

(3.3)

其中 $p=(p_1,\cdots,p_N)^{T},q=(q_1,\cdots,q_N)^{T}$,$\langle\cdot,\cdot\rangle$ 是${\Bbb R}^{N}$上的内积. 把(3.3)式代入(3.1)式,则谱问题(2.1)被非线性化为

\begin{eqnarray}\left\{\begin{array}{l} p_x=(\Lambda-\langle q,q\rangle)p-2\langle p,q\rangle q=-\frac{\partial H}{\partial q},\\ q_x=p-(\Lambda-\langle q,q\rangle)q=\frac{\partial H}{\partial p},\end{array}\right.\end{eqnarray}

(3.4)

这里 $ \Lambda= diag(\alpha_1,\cdots,\alpha_N)$ 和

\begin{eqnarray} H=\frac{1}{2}\langle p,p\rangle-\langle\Lambda p,q\rangle+\langle p,q\rangle\langle q,q\rangle. \end{eqnarray}

(3.5)

构造Lenard特征值问题的解如下

\begin{eqnarray} G_\lambda=g_{-1}+\sum_{j=1}^{N}\frac{S_j}{\lambda-\alpha_j},\end{eqnarray}

(3.6)

易证 $(K-\lambda J)G_\lambda = 0$. 于是,驻定Lax方程 $V_x-[U,V]=0$ 沿 $x$ -流有解

\begin{eqnarray} V_\lambda=\sigma_\lambda(G_\lambda)=\left( \begin{array}{cc} \lambda+Q_\lambda(p,q)~~&-2\langle p,q\rangle-Q_{\lambda}(p,p)\\ 1+Q_{\lambda}(q,q)~~&- \lambda-Q_{\lambda}(p,q)\end{array}\right)=\left(\begin{array}{cc}V_{11}~~&V_{12}\\ V_{21}~~&-V_{11}\end{array} \right),\end{eqnarray}

(3.7)

其中 $Q_{\lambda}(\xi,\eta)=\sum\limits_{j=1}^{N}\frac{\xi_j\eta_j}{\lambda-\alpha_j} =\sum\limits_{n=0}^{\infty}\frac{\langle\Lambda^{n}\xi,\eta\rangle}{\lambda^{n+1}}$. 由驻定方程 $V_x-[U,V]=0$ 可知,$F_\lambda=\frac{1}{2}\det V_\lambda$ 沿 $x$ -流是不变的,因此可得(3.4)式的积分母函数为

\begin{eqnarray} F_{\lambda}=\frac{1}{2}\det V_{\lambda}=-\frac{\lambda^{2}}{2}+\sum\limits_{n=0}^{\infty}\lambda^{-n-1}F_{n},\end{eqnarray}

(3.8)

其中

\begin{eqnarray}\begin{array}{rl} F_0=&-\langle\Lambda p,q\rangle+\frac{1}{2}\langle p,p\rangle+\langle p,q\rangle\langle q,q\rangle=H,\\ F_n=&-\langle\Lambda^{n+1} p,q\rangle+\frac{1}{2}\langle\Lambda^{n} p,p\rangle+\langle p,q\rangle\langle\Lambda^{n} q,q\rangle\\ & +\frac{1}{2}\sum_{i+j=n-1}[\langle\Lambda^{i} q,q\rangle\langle\Lambda^{j}p,p\rangle-\langle\Lambda^{i}p,q\rangle\langle\Lambda^{j}p,q\rangle]. \end{array} \end{eqnarray}

(3.9)

视母函数$F_{\lambda}$为辛空间 $({\Bbb R}^{2N}$,d$p\wedge {\rm d}q)$ 的一个Hamilton函数,由光滑函数 $F$和$G$的Poisson括号定义^[19]

\begin{eqnarray}\{F,G\}=\sum\limits_{j=1}^{N}\bigg(\frac{\partial F}{\partial q_{j}} \frac{\partial G}{\partial p_{j}}-\frac{\partial F}{\partial p_{j}}\frac{\partial G}{\partial q_{j}}\bigg),\end{eqnarray}

(3.10)

设$F_{\lambda}$的流变量为 $t_{\lambda}$,直接给出正则方程

\begin{eqnarray} \frac{\rm d}{{\rm d}t_\lambda}\left( \begin{array}{l} p_k\\ q_k \end{array} \right)= \left( \begin{array}{l} -\frac{\partial F_\lambda}{\partial q_k}\\ \frac{\partial F_\lambda}{\partial p_k} \end{array} \right)=W(\lambda,\alpha_k)\left( \begin{array}{l} p_k\\ q_k \end{array} \right),1\leq k \leq N,\end{eqnarray}

(3.11)

其中

\begin{equation} W(\lambda,\alpha_k)=\frac{V_\lambda}{\lambda-\alpha_k}+V_0(\lambda),V_0(\lambda)=\left( \begin{array}{cc} -V_{21}~~&0~~\\ 0&V_{21} \end{array} \right). \end{equation}

(3.12)

命题3.1 沿 $t_{\lambda}$ -流,Lax矩阵$V_{\mu}$满足Lax方程

\[\frac{{\text{d}}}{{{\text{d}}{t_\lambda }}}{V_\mu } = [\;W(\lambda ,\mu ),{V_\mu }],\forall \lambda ,\mu \in \mathbb{C}\]

(3.13)

和

\begin{equation} \{F_{\lambda},F_{\mu}\}=0,\forall \lambda,\mu \in{\Bbb C},\end{equation}

(3.14)

\begin{equation} \{F_{m},F_{n}\}=0,\forall m,n=0,1,2,\cdots.\end{equation}

(3.15)

因为$F_\lambda,V_{\lambda}^{12}$,$V_{\lambda}^{21}$都是$\lambda$的有理函数,且$V_{\lambda}^{12},V_{\lambda}^{21}$在$\alpha_{j}$处有单一极点,故可设

\begin{eqnarray} V_{\lambda}^{12}=-2\langle p,q\rangle-Q_\lambda(p,p)=-2\langle p,q\rangle\frac{m(\lambda)}{a(\lambda)},~~V_{\lambda}^{21}=1+Q_\lambda(q,q)=\frac{n(\lambda)}{a(\lambda)} , \end{eqnarray}

(4.1)

其中 $$m(\lambda)=\prod_{j=1}^{N}(\lambda-\mu_j),~~n(\lambda)=\prod_{j=1}^{N}(\lambda-\nu_j),~~a(\lambda)=\prod_{j=1}^{N}(\lambda-\alpha_j), $$ 则$\{\mu_j\}$和$\{\nu_j\}$被称为Hamilton系统的两族椭圆坐标^[20]. 若将$V_{\lambda}^{12},V_{\lambda}^{21}$按$\lambda^{-1}$的幂进行展开,可得

\begin{eqnarray} \langle q,q\rangle=\sum\limits_{j=1}^{N}(\alpha_{j}-\nu_{j}),~~ \langle p,p\rangle=2\langle p,q\rangle\sum\limits_{j=1}^{N}(\alpha_{j}-\mu_{j}),\end{eqnarray}

(4.2)

于是有

\begin{eqnarray}\begin{array}{l} u=-\langle q,q\rangle=\sum\limits_{j=1}^{N}(\nu_{j}-\alpha_j),\\ v=-2\langle p,q\rangle+\beta\langle q,q\rangle=\exp\bigg\{2\partial^{-1}\sum\limits_{j=1}^{N}(\nu_{j}-\mu_j)\bigg\}-\beta\sum\limits_{j=1}^{N}(\nu_{j}-\alpha_j). \end{array}\end{eqnarray}

(4..3)

命题4.1 椭圆坐标沿 $t_{\lambda}$ -流满足演化方程

\begin{eqnarray}\left\{\begin{array}{l}laystyle{\frac{1}{2\sqrt{R(\mu_{k})}}\frac{{\rm d}\mu_{k}}{{\rm d}t_{\lambda}}=\frac{m(\lambda)}{a(\lambda)}\frac{1}{(\lambda-\mu_{k})m'(\mu_{k})}},\\laystyle{\frac{1}{2\sqrt{R(\nu_{k})}}\frac{{\rm d}\nu_{k}}{{\rm d}t_{\lambda}}=-\frac{n(\lambda)}{a(\lambda)}\frac{1}{(\lambda-\nu_{k})n'(\nu_{k})}},\end{array}\right.\end{eqnarray}

(4.4)

其中 $$R(\lambda)=a(\lambda)b(\lambda),~~b(\lambda)=\prod_{j=1}^{N+2}(\lambda-\beta_j),~~{\rm deg}R(\lambda)=2N+2. $$

证 令 $\lambda=\mu_j,\nu_j$ 分别代入(4.1)式中,得

$$V_{\mu_j}^{11}=\frac{\sqrt{R(\mu_j)}}{a(\mu_j)},V_{\nu_j}^{11}=\frac{\sqrt{R(\nu_j)}}{a(\nu_j)}. $$ 借助Lax方程 $\frac{\rm d}{{\rm d}t_{\lambda}}V_{\mu}=[W(\lambda,\mu),V_{\mu}]$,有

\begin{eqnarray} \begin{array}{l} \frac{\rm d}{{\rm d}t_{\lambda}}V_{\mu}^{12}=2\bigg(\frac{V_{\lambda}^{11}}{\lambda-\mu}-V_{\lambda}^{21}\bigg)V_{\mu}^{12} -2\frac{V_{\lambda}^{12}}{\lambda-\mu}V_{\mu}^{11} ,\\ \frac{\rm d}{{\rm d}t_{\lambda}}V_{\mu}^{21}=2\frac{V_{\lambda}^{21}}{\lambda-\mu}V_{\mu}^{11} -2\bigg(\frac{V_{\lambda}^{11}}{\lambda-\mu}-V_{\lambda}^{21}\bigg)V_{\mu}^{21},\end{array}\end{eqnarray}

(4.5)

分别令 $\mu=\mu_j$,$\nu=\nu_j$,直接计算可得(4.4)式.

利用多项式的插值公式,可得

\begin{eqnarray} \sum_{k=1}^{N}\frac{\mu_{k}^{N-j}}{2\sqrt{R(\mu_k)}}\frac{{\rm d}\mu_k}{{\rm d}t_\lambda}=-\frac{\lambda^{N-j}}{a(\lambda)},\sum_{k=1}^{N}\frac{\nu_{k}^{N-j}}{2\sqrt{R(\nu_k)}}\frac{{\rm d}\nu_k}{{\rm d}t_\lambda}=\frac{\lambda^{N-j}}{a(\lambda)}. \end{eqnarray}

(4.6)

对某一固定的$\lambda_0$,引入拟Abel-Jacobi坐标

\begin{eqnarray} \tilde{\phi}_j=\sum_{k=1}^{N}\int_{\lambda_0}^{\mu_k}\frac{\lambda^{N-j}{\rm d}\lambda}{2\sqrt{R(\mu)}},\tilde{\psi}_j=\sum_{k=1}^{N}\int_{\lambda_0}^{\nu_k}\frac{\lambda^{N-j}{\rm d}\lambda}{2\sqrt{R(\nu)}},j=1,2,\cdots,N,\end{eqnarray}

(4.7)

于是有

\begin{eqnarray} \frac{{\rm d}\tilde{\phi}_j}{{\rm d}t_\lambda}=-\frac{\lambda^{N-j}}{a(\lambda)},\frac{{\rm d}\tilde{\psi}_j} {{\rm d}t_\lambda}=-\frac{\lambda^{N-j}}{a(\lambda)}. \end{eqnarray}

(4.8)

命题4.2 由(3.9)式给出的 $F_{0},F_{1},\cdots ,F_{N-1}$ 是函数独立的.

证

$$\frac{{\rm d}\tilde{\phi}_j}{{\rm d}t_\lambda}=\{\tilde{\phi}_j,F_\lambda\}=\sum_{k=0}^{\infty}\frac{1}{\lambda^{k+1}}\{\tilde{\phi}_j,F_k\}=\sum_{k=0}^{\infty}\frac{1}{\lambda^{k+1}}\frac{{\rm d}\tilde{\phi_j}}{{\rm d}t_k}. $$ 另一方面 $$\frac{\lambda^{N-j}}{a(\lambda)}=\frac{1}{\lambda^{j}}\frac{\lambda^{N}}{(\lambda-\alpha_1)\cdots(\lambda-\alpha_N)} =\frac{1}{\lambda^{j}}\bigg(1+\frac{A_1}{\lambda}+\frac{A_2}{\lambda^{2}}+\cdots\bigg), $$ 其中 $$A_1=\sigma_1,~~A_k=\frac{1}{k}\bigg(\sigma_k+\sum_{i+j=k,i,j\geq1}\sigma_iA_j\bigg),~~\sigma_k=\sum_{j=1}^{N}\alpha^{k}_j,~~k\geq2. $$ 根据(4.8)式的第一个表达式,可得 $$\frac{{\rm d}\tilde{\phi}_j}{{\rm d}t_k}=A_{k-j+1,} k\geq1. $$ 则有

\begin{eqnarray} (\frac{{\rm d}\tilde{\phi}}{{\rm d}t_0},\cdots,\frac{{\rm d}\tilde{\phi}}{{\rm d}t_{N-1}})=\left( \begin{array}{ccccc} 1 &~~A_1~~&A_2&~~\cdots~~&A_{N-1}\\ &1&A_1&\cdots&A_{N-2}\\ &&\ddots&\ddots&\vdots\\&&&\ddots&A_1\\ &&&&1\\ \end{array} \right),\end{eqnarray}

(4.9)

其中 $\tilde{\phi}=(\tilde{\phi}_1,\cdots,\tilde{\phi}_N)^{T}$,设 $\sum\limits_{k=0}^{N-1}\zeta_k\omega^{2}{\rm d}F_k=0$,于是 $$0=\sum_{k=0}^{N-1}\zeta_k\omega^{2}(I{\rm d}F_k,I{\rm d}\tilde{\phi}_j)=\sum_{k=0}^{N-1}\zeta_k\frac{{\rm d}\tilde{\phi}_j}{{\rm d}t_k},~~1\leq j\leq N. $$ 由(4.9)式可知行列式为1,则 $\zeta_0=\cdots=\zeta_{N-1}=0$,因此 $F_{0},F_{1},\cdots ,F_{N-1}$ 是函数独立的.

定理4.1 有限维Hamilton系统(3.4)在Liouville意义下是完全可积的.

对于本文中所对应的超椭圆函数曲线 $\Gamma:\zeta^{2}-4R(\lambda)=0$,其亏格为$N$. 取椭圆曲线$\Gamma$上正则闭链 $a_{1},a_{2},\cdots ,a_{N}$; $b_{1},b_{2},\cdots ,b_{N}$,使其满足

$$a_{i}\circ a_{j}=0,\ \ b_{i}\circ b_{j}=0,\ \ a_{i}\circ b_{j}=\delta_{ij},\ i,j=1,2,\cdots ,N. $$则$\Gamma$上的$N$个全纯微分基底 $$\tilde{\omega}_{l}=laystyle{\frac{\lambda^{N-l}{\rm d}\lambda}{2\sqrt{R(\lambda)}} l=1,2,\cdots ,N} $$是线性无关的. 令 $C=(C_{il})_{N\times N}$ 为周期矩阵 $(A_{lj})_{N\times N}$ 的逆

\begin{eqnarray} C=(A_{lj})^{-1}_{N\times N},A_{lj}=\int_{a_{j}}\tilde{\omega}_{l}. \end{eqnarray}

(5.1)

则$\tilde{\omega}$经线性规范变换为

\begin{eqnarray} \omega_j=\sum_{l=1}^{N}C_{jl}\tilde{\omega_l},\omega=(\omega_1,\cdots,\omega_N)^{T}=C\tilde{\omega},\end{eqnarray}

(5.2)

使其满足

\begin{eqnarray} \int_{a_j}\omega_i=\delta_{ij},\int_{b_j}\omega_i=B_{ij},1\leq i,j\leq N,\end{eqnarray}

(5.3)

其中 $B=(B_{ij})_{N\times N}$ 对称且虚部正定,因而可定义$\Gamma$上的$\theta$函数^{[21, 22]}

\begin{eqnarray} \theta(\zeta)=\sum\limits_{Z\in {\Bbb Z}^{N}}\exp (\pi {\rm i}\langle BZ,Z\rangle +2\pi {\rm i}\langle\zeta,Z\rangle),\ \zeta\in {\Bbb C}^{N}. \end{eqnarray}

(5.4)

此时,对于固定的 $P_{0}\in\Gamma$,引进Abel映射${\cal A}(P)$和Abel-Jacobi坐标如下

\begin{eqnarray*} {\cal A}(P)=\int_{P_{0}}^{P}\omega,{\cal A}(\sum n_{k}P_{k})=\sum n_{k}{\cal A}(P_{k}),\end{eqnarray*} \begin{eqnarray} \phi={\cal A}\bigg(\sum\limits_{k=1}^{N}P(\mu_{k})\bigg)= \sum\limits_{k=1}^{N}\int_{P_{0}}^{P(\mu_{k})}\omega,\psi={\cal A}\bigg(\sum\limits_{k=1}^{N}P(\nu_{k})\bigg)= \sum\limits_{k=1}^{N}\int_{P_{0}}^{P(\nu_{k})}\omega. \end{eqnarray}

(5.5)

令 $S_k=\lambda^{k}_1+\cdots+\lambda^{k}_{2N+2}$,那么 $\frac{1}{\sqrt{R_{*}(z)}}=\sum\limits_{k=0}^{N}R_kz^{k}$ 系数满足递推公式

\begin{eqnarray} R_0=1,~~R_1=\frac{1}{2}S_1,~~R_k=\frac{1}{2k}\bigg(S_k+\sum_{i+j=k,i,j\geq1}S_jR_k\bigg). \end{eqnarray}

(5.6)

设 $C_1,\cdots,C_N$ 是矩阵$C$的列向量,通过直接的计算可得

\begin{eqnarray} \frac{1}{2\sqrt{R_{*}(z)}}(C_1+C_2z+\cdots+C_Nz^{N-1})=\sum_{k=0}^{\infty}\Omega_kz^{k},\end{eqnarray}

(5.7)

其系数为

\begin{eqnarray} \Omega_k=\frac{1}{2}(R_kC_1+\cdots+R_1C_k+C_{k+1}),~~1\leq k \leq N. \end{eqnarray}

(5.8)

命题5.1 利用Abel-Jacobi坐标,$H_k$ -流得以直化

\begin{eqnarray} \frac{{\rm d}\phi}{{\rm d}\tau_k}=\Omega_k,\frac{{\rm d}\psi}{{\rm d}\tau_k}=-\Omega_k. \end{eqnarray}

(5.9)

证 从(4.8)式可得

$$\frac{{\rm d}\tilde{\phi}}{{\rm d}t_\lambda}=\frac{\lambda^{N}}{2\sqrt{R(\lambda)}}(\lambda^{-1},\cdots,\lambda^{-N}), $$ $$\frac{{\rm d}\phi}{{\rm d}t_\lambda}=C\frac{{\rm d}\tilde{\phi}}{{\rm d}t_\lambda}=\frac{\lambda^{N}}{2\sqrt{R(\lambda)}}(C_1\lambda^{-1},\cdots,C_N\lambda^{-N})=\sum_{k=1}^{\infty}\Omega_k\lambda^{-k-1}, $$通过比较$\lambda^{-k-1}$的系数可得(5.9)式的第一个等式,同理,(5.9)式的第二个等式也可证.

经过直化的方程(5.9)很容易被积出 $\phi=\phi_0+\sum\Omega_kt_k$. 从"Abel-Jacobi窗口"中观察,$H_k$ -流和$X_k$ -流二者的解都是线性函数

\begin{eqnarray}\begin{array}{l} \hbox{ } H_{k}: \phi=\phi_{0}+\Omega_{k}t_{k},\psi=\psi_{0}-\Omega_{k}t_{k},\\\hbox{ } X_{k}:\phi=\phi_{0}+\Omega_{0}x+\Omega_{k}t_{k},\psi=\psi_{0}-\Omega_{0}x-\Omega_{k}t_{k}. \end{array} \end{eqnarray}

(5.10)

根据Riemann定理知,存在常向量 $M_{1},M_{2}\in{\Bbb C}^{N}$,使得 $\theta({\cal A}(P(\lambda))-\phi-M_{1})$ 有$N$个零点,分别为 $\lambda=\mu_{1},\cdots ,\mu_{N}$; $\theta({\cal A}(P(\lambda))-\psi-M_{2})$ 也有$N$个零点,分别为 $\lambda=\nu_{1},\cdots ,\nu_{N}$. 对于同一个$\lambda$,在$\Gamma$的不同页上有两个点$(\lambda,\sqrt{R}(\lambda))$和$(\lambda,-\sqrt{R}(\lambda))$. 因此,在$\infty_{s}(s=1,2)$处的局部坐标下,可得

$${\cal A}(P(z^{-1}))=-\eta_{s}+\frac{(-1)^{s-1}}{2}\sum\limits_{k=1}^{\infty}\frac{1}{k}\Omega_{k}z^{k}, $$ 其中 $ {\eta_{s}=-\int^{P_{0}}_{\infty_{s}}\omega}.$ 由文献[10, 22]的结论可得

\begin{equation} \sum_{j=1}^{N}\mu_j=I_1(\Gamma)+\frac{1}{2}\partial \ln\frac{\theta_1}{\theta_2},\end{equation}

(5.11)

\begin{equation} \sum_{j=1}^{N}\nu_j=I_1(\Gamma)-\frac{1}{2}\partial \ln\frac{\theta^{*}_1}{\theta^{*}_2},\end{equation}

(5.12)

其中 $I_1(\Gamma)=\sum\limits_{j=1}^{N}\int_{a_j}\lambda\omega_j$,$\theta_s=(\phi+K+\eta_s)$,$\theta^{*}_s=(-\psi-K-\eta_s),s=1,2$,$K$是一个常数.

命题5.2 广义Broer-Kaup-Kupershmidt方程(2.10)的拟周期解为

\begin{eqnarray}\left\{\begin{array}{rl} u(x,y)=& N_1-\frac{1}{2}\partial \ln\frac{\theta(\Omega_0x+\Omega_1y-\psi_0-K-\eta_1)} {\theta(\Omega_0x+\Omega_1y-\psi_0-K-\eta_2)},\\ v(x,y)=& -\beta N_1+\frac{1}{2}\beta\partial \ln\frac{\theta(\Omega_0x+\Omega_1y-\psi_0-K-\eta_1)} {\theta(\Omega_0x+\Omega_1y-\psi_0-K-\eta_2)}\\& +N_2\frac{\theta(\Omega_0x+\Omega_1y+\phi_0+K+\eta_2) \theta(\Omega_0x+\Omega_1y-\psi_0-K-\eta_2)} { \theta(\Omega_0x+\Omega_1y+\phi_0+K+\eta_1) \theta(\Omega_0x+\Omega_1y-\psi_0-K-\eta_1)},\end{array}\right.\end{eqnarray}

(5.13)

其中$N_1,N_2$为常数.

命题5.3 广义Broer-Kaup-Kupershmidt方程(2.11)的拟周期解为

\begin{eqnarray}\left\{\begin{array}{rl} u(x,t)=& N_1-\frac{1}{2}\partial \ln\frac{\theta(\Omega_0x+\Omega_2t-\psi_0-K-\eta_1)} {\theta(\Omega_0x+\Omega_2t-\psi_0-K-\eta_2)},\\ v(x,t)=& -\beta N_1+\frac{1}{2}\beta\partial \ln \frac{\theta(\Omega_0x+\Omega_2t-\psi_0-K-\eta_1)} {\theta(\Omega_0x+\Omega_2t-\psi_0-K-\eta_2)}\\& +N_2\frac{\theta(\Omega_0x+\Omega_2t+\phi_0+K+\eta_2) \theta(\Omega_0x+\Omega_2t-\psi_0-K-\eta_2)} { \theta(\Omega_0x+\Omega_2t+\phi_0+K+\eta_1) \theta(\Omega_0x+\Omega_2t-\psi_0-K-\eta_1)},\end{array}\right.\end{eqnarray}

(5.14)

其中$N_1,N_2$为常数.

命题5.4 $(2+1)$ -维广义Broer-Kaup-Kupershmidt方程(2.12)的拟周期解为

\begin{eqnarray}\left\{\begin{array}{rl}u(x,y,t)=& N_1-\frac{1}{2}\partial\ln\frac{\theta(\Omega_0x+\Omega_1y+\Omega_2t-\psi_0-K-\eta_1)}{\theta(\Omega_0x+\Omega_1y+\Omega_2t-\psi_0-K-\eta_2)},\\v(x,y,t)=& -\beta N_1+\frac{1}{2}\beta\partial\ln\frac{\theta(\Omega_0x+\Omega_1y+\Omega_2t-\psi_0-K-\eta_1)}{\theta(\Omega_0x+\Omega_1y+\Omega_2t-\psi_0-K-\eta_2)}\\& +N_2\frac{\theta(\Omega_0x+\Omega_1y+\Omega_2t+\phi_0+K+\eta_2)\theta(\Omega_0x+\Omega_1y+\Omega_2t-\psi_0-K-\eta_2)}{\theta(\Omega_0x+\Omega_1y+\Omega_2t+\phi_0+K+\eta_1)\theta(\Omega_0x+\Omega_1y+\Omega_2t-\psi_0-K-\eta_1)},\end{array}\right.\end{eqnarray}

(5.15)

其中$N_1,N_2$为常数.

在本文中,一个$(2+1)$ -维广义Broer-Kaup-Kupershmidt 孤子方程被分解为相容的常微分方程,借助于Riemann $\theta$函数给出了$(2+1)$ -维广义Broer-Kaup-Kupershmidt孤子方程的拟周期解. 在求解过程中,利用生成函数证明了对合性和函数独立性,通过引入Abel-Jacobi坐标把流进行了拉直.