椭圆链式KP系统的可积性分析

doi:10.12677/pm.2024.145179

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椭圆链式KP系统的可积性分析
Integrability Analysis of Elliptical Chain KP System

DOI: 10.12677/pm.2024.145179, PDF, HTML, XML, 下载: 63 浏览: 122
作者: 王辛乙：上海理工大学理学院，上海
关键词: 椭圆链式KP系统；Lax组；柯西矩阵法；Elliptic Lattice KP System； Lax Triplet； Cauchy Matrix Method

摘要: 我们以前已经深入研究了求解椭圆链式KP方程的直接线性化方法，而求解该方程的柯西矩阵方法只是近年来的一个热门主题。本文将讨论基于柯西矩阵方法的椭圆链式KP方程的可积性，为下一步的求解做准备。本文首先从辅助向量u(2i)=(IXC)−1Pic出发得到椭圆链式KP系统的lax组，然后从Lax组中推导出椭圆链式KP系统来完成闭环。

Abstract: We have previously delved into direct linearization methods for solving the elliptic lattice KP system, and the Cauchy matrix method for solving this equation has only been a hot topic in recent years. This article will discuss the integrability of the elliptic lattice KP system based on the Cauchy matrix method, in preparation for the next step of solution. Firstly, starting from the auxiliary vectoru(2i)=(IXC)−1Pic, we obtain the lax system of the elliptic lattice KP system, and then derive the elliptic lattice KP system from the lax system to complete the closed-loop process.

文章引用：王辛乙. 椭圆链式KP系统的可积性分析[J]. 理论数学, 2024, 14(5): 219-225. https://doi.org/10.12677/pm.2024.145179

1. 引言

椭圆可积系统被认为是“顶部的可积系统” [1] ，通过周期性退化，它们可以简化为与有理函数相关的可积方程。现已知的例子是链式Landau-Lifshitz方程 [2] ，Adler的链式krichhever-novikov系统 [3] ，也称为Adler-bobenko-suris四边形方程 [4] 中的Q4方程，Adler-yamilov系统 [5] 和椭圆Korteweg-de Vries系统 [6] 。

本节要研究的椭圆KP方程是3 + 1维的，它是比2 + 1维椭圆Kdv方程更高阶的存在，当然由椭圆Kadomtsev-Petviashvili (KP)方程可以约化为椭圆Kdv方程，因此我们可以把椭圆KP方程看作是椭圆Kdv方程的推广。

椭圆链式KP系统的Lax表示与辅助向量在离散演化密切相关。在之前的研究中，我们已经通过直接线性化方法求出过椭圆KP的离散Lax对 [7] ，本文将介绍另一种方法——柯西矩阵法，通过Lax对的推导，能够更直观的感受椭圆KP方程的可积性。

2. 预备知识

本文考虑与椭圆曲线相关的链势Kadomtsev-Petviashvili (KP)方程的推广 [8] ：

$\begin{array}{l} (a - \tilde{u}) (b - c + \tilde{\dot{u}} - \tilde{\hat{u}}) + (b - \hat{u}) (c - a + \hat{\tilde{u}} - \hat{\dot{u}}) + (c - \dot{u}) (a - b + \dot{\hat{u}} - \dot{\tilde{u}}) \\ = g (\tilde{\hat{s^{'}}} (\tilde{s} - \hat{s}) + \hat{\dot{s^{'}}} (\hat{s} - \dot{s}) + \dot{\tilde{s^{'}}} (\dot{s} - \tilde{s})), \end{array}$ (2.1a)

$\begin{array}{l} \frac{(a - \dot{u}) \dot{\tilde{s}} - (b + \dot{u}) \hat{\dot{s}} + \hat{\dot{w}} - \tilde{\dot{w}}}{\dot{s}} + \frac{(b + \tilde{u}) \hat{\tilde{s}} - (c + \tilde{u}) \tilde{\dot{s}} + \tilde{\dot{w}} - \tilde{\hat{w}}}{\tilde{s}} \\ + \frac{(c + \hat{u}) \dot{\hat{s}} - (a + \hat{u}) \hat{\tilde{s}} + \hat{\tilde{w}} - \hat{\dot{w}}}{\hat{s}} = 0, \end{array}$ (2.1b)

$\begin{array}{l} \frac{(a - \dot{\tilde{u}}) \dot{s^{'}} - (c - \dot{\tilde{u}}) \tilde{s^{'}} + \dot{w^{'}} - \tilde{w^{'}}}{\dot{\tilde{s^{'}}}} + \frac{(c - \hat{\dot{u}}) \hat{s^{'}} - (b - \hat{\dot{u}}) \dot{s^{'}} + \hat{w^{'}} - \dot{w^{'}}}{\hat{\dot{s^{'}}}} \\ + \frac{(b - \tilde{\hat{u}}) \tilde{s^{'}} - (a - \hat{\tilde{u}}) \hat{s^{'}} + \tilde{w^{'}} - \hat{w^{'}}}{\hat{\tilde{s^{'}}}} = 0, \end{array}$ (2.1c)

$(a + u - \frac{\tilde{w}}{\tilde{s}}) (a - \tilde{u} + \frac{w}{s}) = a^{2} + (\underline{\tilde{y^{'}}} - \tilde{y}) - (y^{'} - y) + \frac{\tilde{w}}{\tilde{s}} (\tilde{u} - \tilde{\underline{u}}) - (\frac{1}{\tilde{s} \underline{s^{'}}} + 3 e_{1} + g \tilde{s^{'}} s),$ (2.1d)

$(b + u - \frac{\hat{w}}{\hat{s}}) (b - \hat{u} + \frac{w}{s}) = b^{2} + (\hat{\underline{y^{'}}} - \hat{y}) - (y^{'} - y) + \frac{\hat{w}}{\hat{s}} (\hat{u} - \hat{\underline{u}}) - (\frac{1}{\hat{s} \underline{s^{'}}} + 3 e_{1} + g \hat{s^{'}} s),$ (2.1e)

$(c + u - \frac{\dot{w}}{\dot{s}}) (c - \dot{u} + \frac{w}{s}) = c^{2} + (\dot{\underline{y^{'}}} - \dot{y}) - (y^{'} - y) + \frac{\dot{w}}{\dot{s}} (\dot{u} - \dot{\underline{u}}) - (\frac{1}{\dot{s} \underline{s^{'}}} + 3 e_{1} + g \dot{s^{'}} s),$ (2.1f)

$s^{'} \bar{w} = w^{'} \bar{s} .$ (2.1g)

其中 $a, b, c$ 是常数，符号~，^，˙，¯分别表示相对于链方向 $n, m, l, M$ 的位移 [9] 。(2.1)中的 $e_{1}, g \in ℂ$ 是椭圆曲线的模

$y^{2} = R (x) = \frac{1}{x} + 3 e_{1} + g x .$

我们将引入c和r上的色散关系：

$\tilde{c} = (a I + p), \hat{c} = (b I + p) c, \dot{c} = (b I + p) c, \bar{c} = - P c$ ， (2.2a)

$\tilde{r} = r {(a I - q)}^{- 1}, \hat{r} = r {(b I - q)}^{- 1}, \dot{r} = r {(c I - q)}^{- 1}, \bar{r} = - r Q^{- 1}$ ， (2.2b)

其中 $a, b, c$ 称为链参数，分别与链方向 $n, m, l$ 相关，M方向没有链参数。其中矩阵 $p, P \in ℂ_{N \times N}$ ， $q, Q \in ℂ_{N^{'} \times N^{'}}$ ， $X \in ℂ_{N \times N^{'}}$ ，列向量 $c \in ℂ_{N \times 1}$ ，行向量 $r \in ℂ_{1 \times N^{'}}$ 。

考虑一组标量，用 $S^{(i, j)}$ 表示，其中 $i, j \in ℤ$ ，写作：

$S^{(2 i, 2 j)} = r Q^{j} C {(I + X C)}^{- 1} P^{i} c$ ， (2.3a)

$S^{(2 i + 1, 2 j)} = r Q^{j} C {(I + X C)}^{- 1} P^{i} p c$ ， (2.3b)

其中， $C \in ℂ_{N^{'} \times N}$ 是一个辅助常数矩阵，使得 $I + X C$ 可逆。并且引入一些辅助向量 [10] ：

$u^{(2 i)} = {(I + X C)}^{- 1} P^{i} c, u^{(2 i + 1)} = {(I + X C)}^{- 1} P^{i} p c$ ， (2.4)

因此 $S^{(i, j)}$ 可以被写成

$S^{(2 i, 2 j)} = r Q^{j} C u^{(2 i)}$ , $S^{(2 i + 1, 2 j)} = r Q^{j} C u^{(2 i + 1)}$ ， (2.5)

$u^{(i)}$ 位移关系的显式计算可以利用式(2.3)实现。根据式(2.2)和式(2.4)的结构，结合 $u^{(i)}$ 的定义，可以推断出 $u^{(i)}$ 在 $n, m, l$ 方向上的位移关系具有相似性，可以通过交换位移和链参数从一个方向变换到另一个方向。

3. 椭圆KP方程的Lax组

通过对(2.3)两边对于矩阵的变换，我们可以得到可以得到 $u^{(i)}$ 的位移关系：

${\tilde{u}}^{(2 i)} = a u^{(2 i)} + u^{(2 i + 1)} - u^{(0)} {\tilde{S}}^{(2 i, 0)} + g u^{(- 2)} {\tilde{S}}^{(2 i, - 2)}$ ， (3.1a)

${\tilde{u}}^{(2 i + 1)} = a u^{(2 i + 1)} + u^{(2 i + 2)} + 3 e_{1} u^{(2 i)} + g u^{(2 i - 2)} - u^{(0)} {\tilde{S}}^{(2 i + 1, 0)} + g u^{(- 2)} {\tilde{S}}^{(2 i + 1, - 2)}$ ， (3.1b)

${\bar{u}}^{(2 i)} = - u^{(2 i + 2)} + u^{(1)} {\bar{S}}^{(2 i, 0)} - u^{(0)} {\bar{S}}^{(2 i, 1)}$ ， (3.1c)

${\bar{u}}^{(2 i + 1)} = - u^{(2 i + 3)} + u^{(1)} {\bar{S}}^{(2 i + 1, 0)} - u^{(0)} {\bar{S}}^{(2 i + 1, 1)}$ 。 (3.1d)

接着，让我们引入标量函数：

$\begin{array}{l} u = S^{(0, 0)}, s = S^{(- 2, 0)}, s^{'} = S^{(0, - 2)}, h = S^{(- 2, - 2)}, \\ v = 1 - S^{(- 1, 0)}, v^{'} = 1 - S^{(0, 1)}, w = 1 + S^{(- 2, 1)}, w^{'} = 1 + S^{(1, - 2)} . \end{array}$

结合式(3.1a)~式(3.1b)中 $i = 0$ 的具体情况，可以形成两个耦合线性方程：

${\tilde{u}}^{(0)} = (a - \tilde{u}) u^{(0)} + u^{(1)} - g \tilde{s^{'}} w {\underline{u}}^{(0)} + g \tilde{s^{'}} s {\underline{u}}^{(1)}$ ， (3.2a)

${\tilde{u}}^{(1)} = (3 e_{1} - \bar{y} - \tilde{y^{'}}) u^{(0)} + (a + \bar{u}) u^{(1)} - g \tilde{w^{'}} w {\underline{u}}^{(0)} + g \tilde{w^{'}} s {\underline{u}}^{(1)} - {\bar{u}}^{(0)}$ 。 (3.2b)

在一个2 × 1的块矩阵 $Φ = (\begin{matrix} u^{(0)} \\ u^{(1)} \end{matrix})$ ，每个式中，每个块都包含一个N阶的列向量，则可将上述两种关系结合为 $Φ$ 式，即：

$\tilde{Φ} = A_{1} Φ + B_{1} \underline{Φ} + J \bar{Φ}$ ， (3.3a)

其中：

$A_{1} = (\begin{matrix} a - \tilde{u} & 1 \\ 3 e_{1} - \bar{y} - \tilde{y^{'}} & a + \bar{u} \end{matrix})$ , $B_{1} = g (\begin{matrix} - \tilde{s^{'}} w & \tilde{s^{'}} s \\ - \tilde{w^{'}} w & \tilde{w^{'}} s \end{matrix})$ , $J = (\begin{matrix} 0 & 0 \\ - 1 & 0 \end{matrix})$ ， (3.3b)

类似地，我们有“−”和“˙”位移的类似方程

$\hat{Φ} = A_{2} Φ + B_{2} \underline{Φ} + J \bar{Φ}$ ， (3.3c)

$\dot{Φ} = A_{3} Φ + B_{3} \underline{Φ} + J \bar{Φ}$ ，(3.3d)

其中：

$A_{2} = (\begin{matrix} b - \hat{u} & 1 \\ 3 e_{1} - \bar{y} - \hat{y^{'}} & b + \bar{u} \end{matrix})$ , $B_{2} = g (\begin{matrix} - \hat{s^{'}} w & \hat{s^{'}} s \\ - \hat{w^{'}} w & \hat{w^{'}} s \end{matrix})$ ， (3.3e)

$A_{3} = (\begin{matrix} c - \dot{u} & 1 \\ 3 e_{1} - \bar{y} - \dot{y^{'}} & c + \bar{u} \end{matrix})$ , $B_{3} = g (\begin{matrix} - \dot{s^{'}} w & \dot{s^{'}} s \\ - \dot{w^{'}} w & \dot{w^{'}} s \end{matrix})$ 。(3.3f)

上述关系维度为2N × 2N的矩阵。当然它们也可以被理解为独立矩阵系统的集合，每个系统操作一个双分量向量，表示为N个解耦的2 × 2矩阵系统，其中：

${\tilde{Φ}}_{i} = A_{1} Φ_{i} + B_{1} {\underline{Φ}}_{i} + J {\bar{Φ}}_{i}$ ， (3.4a)

${\hat{Φ}}_{i} = A_{2} Φ_{i} + B_{2} {\underline{Φ}}_{i} + J {\bar{Φ}}_{i}$ ， (3.4b)

${\dot{Φ}}_{i} = A_{3} Φ_{i} + B_{3} {\underline{Φ}}_{i} + J {\bar{Φ}}_{i}$ ， (3.4c)

其中：

$Φ_{i} = (\begin{matrix} {(u^{(0)})}_{i} \\ {(u^{(1)})}_{i} \end{matrix}), i = 1, 2, \dots, N .$ (3.5)

${(u^{(j)})}_{i}$ 是N个分量向量 $u^{(j)}$ 的第i个分量，线性关系(3.55)的相容条件为 ${\hat{\tilde{Φ}}}_{i} = {\tilde{\hat{Φ}}}_{i}$ ， ${\dot{\tilde{Φ}}}_{i} = {\tilde{\dot{Φ}}}_{i}$ ，得到：

$\begin{array}{l} ({\hat{A}}_{1} J + J {\bar{A}}_{2} - {\tilde{A}}_{2} J - J {\bar{A}}_{1}) \bar{Φ} + ({\hat{A}}_{1} A_{2} + {\hat{B}}_{1} J + J {\bar{B}}_{2} - {\tilde{A}}_{2} A_{1} - {\tilde{B}}_{2} J - J {\bar{B}}_{1}) Φ \\ + ({\hat{A}}_{1} B_{2} + {\hat{B}}_{1} {\underline{A}}_{2} - {\tilde{A}}_{2} B_{1} - {\tilde{B}}_{2} {\underline{A}}_{1}) \underline{Φ} + ({\hat{B}}_{1} {\underline{B}}_{2} - {\tilde{B}}_{2} {\underline{B}}_{1}) \underline{\underline{Φ}} = 0 \end{array}$ ， (3.6a)

$\begin{array}{l} ({\dot{A}}_{1} J + J {\bar{A}}_{3} - {\tilde{A}}_{3} J - J {\bar{A}}_{1}) \bar{Φ} + ({\dot{A}}_{1} A_{3} + {\dot{B}}_{1} J + J {\bar{B}}_{3} - {\tilde{A}}_{3} A_{1} - {\tilde{B}}_{3} J - J {\bar{B}}_{1}) Φ \\ + ({\dot{A}}_{1} B_{3} + {\dot{B}}_{1} {\underline{A}}_{3} - {\tilde{A}}_{3} B_{1} - {\tilde{B}}_{3} {\underline{A}}_{1}) \underline{Φ} + ({\dot{B}}_{1} {\underline{B}}_{3} - {\tilde{B}}_{3} {\underline{B}}_{1}) \underline{\underline{Φ}} = 0 \end{array}$ ， (3.6b)

$\begin{array}{l} ({\hat{A}}_{3} J + J {\bar{A}}_{2} - {\dot{A}}_{2} J - J {\bar{A}}_{3}) \bar{Φ} + ({\hat{A}}_{3} A_{2} + {\hat{B}}_{3} J + J {\bar{B}}_{2} - {\dot{A}}_{2} A_{3} - {\dot{B}}_{2} J - J {\bar{B}}_{3}) Φ \\ + ({\hat{A}}_{3} B_{2} + {\hat{B}}_{3} {\underline{A}}_{2} - {\dot{A}}_{2} B_{3} - {\dot{B}}_{2} {\underline{A}}_{3}) \underline{Φ} + ({\hat{B}}_{3} {\underline{B}}_{2} - {\dot{B}}_{2} {\underline{B}}_{3}) \underline{\underline{Φ}} = 0 \end{array}$ 。 (3.6c)

由此得到了椭圆链式KP系统的Lax对，或者更准确地说是用Lax三重态来积分的证据 [11] 。

4. 由Lax组到KP方程

由相容条件(3.6a)可得：

$g \hat{s} \tilde{\hat{s^{'}}} + g \tilde{s} {\tilde{\hat{s}}}^{'} - (b - \hat{u}) (a - \tilde{\hat{u}}) + (a - \tilde{u}) (b - \tilde{\hat{u}}) + \hat{y^{'}} - \tilde{y^{'}} = 0$ ， (4.1a)

$\begin{array}{l} g \bar{\hat{s^{'}}} \bar{w} - g \bar{\tilde{s^{'}}} \bar{w} - g \hat{s} \tilde{\hat{w^{'}}} + g \tilde{s} \tilde{\hat{w^{'}}} + (a + \bar{\hat{u}}) (3 e_{1} - \bar{y} - \hat{y^{'}}) - (b + \bar{\tilde{u}}) (3 e_{1} - \bar{y} - \tilde{y^{'}}) \\ + (b - \hat{u}) (3 e_{1} - \bar{\hat{y}} - \tilde{\hat{y^{'}}}) - (a - \tilde{u}) (3 e_{1} - \bar{\tilde{y}} - \tilde{\hat{y^{'}}}) = 0 \end{array}$ ， (4.1b)

$- g s \hat{s^{'}} + g s \tilde{s^{'}} + (b + u) (a + \hat{u}) - (a + u) (b + \tilde{u}) - \hat{y} + \tilde{y} = 0$ ， (4.1c)

$\begin{array}{l} \frac{- \hat{s^{'}} (a - \hat{\tilde{u}}) + \tilde{s^{'}} (b - \hat{\tilde{u}}) - \hat{w^{'}} + \tilde{w^{'}}}{\hat{\tilde{s^{'}}}} \\ + \frac{- (b - \underline{\hat{u}}) \hat{w} + (a - \underline{\tilde{u}}) \tilde{w} + \hat{s} (3 e_{1} - y - \underline{\hat{y^{'}}}) - \tilde{s} (3 e_{1} - y - \underline{\tilde{y^{'}}})}{w} = 0 \end{array}$ ， (4.1d)

$\frac{\tilde{s} (a + u) - \hat{s} (b + u) + \hat{w} - \tilde{w}}{s} + \frac{\tilde{s^{'}} (b - \hat{\tilde{u}}) - \hat{s^{'}} (a - \hat{\tilde{u}}) - \hat{w^{'}} + \tilde{w^{'}}}{\hat{\tilde{s^{'}}}} = 0$ ， (4.1e)

$\begin{array}{l} - (a + \hat{\bar{u}}) w \hat{w^{'}} + (b + \tilde{\bar{u}}) w \tilde{w^{'}} - (b - \hat{\underline{u}}) \hat{w} \hat{\tilde{w^{'}}} + (a - \tilde{\underline{u}}) \tilde{w} \hat{\tilde{w^{'}}} + \hat{s} \hat{\tilde{w^{'}}} (3 e_{1} - y - \hat{\underline{y^{'}}}) \\ - \tilde{s} \hat{\tilde{w^{'}}} (3 e_{1} - y - \tilde{\underline{y^{'}}}) - \hat{s^{'}} w (3 e_{1} - \hat{\bar{y}} - \hat{\tilde{y^{'}}}) + \tilde{s^{'}} w (3 e_{1} - \tilde{\bar{y}} - \hat{\tilde{y^{'}}}) = 0 \end{array}$ ， (4.1f)

$\begin{array}{l} s (a \hat{w^{'}} + \hat{\bar{u}} \hat{w^{'}} - b \tilde{w^{'}} - \tilde{\bar{u}} \tilde{w^{'}} + \hat{s^{'}} (3 e_{1} - \hat{\bar{y}} - \hat{\tilde{y^{'}}}) + \tilde{s^{'}} (- 3 e_{1} + \tilde{\bar{y}} + \hat{\tilde{y^{'}}})) \\ + (- \tilde{s} (a + u) + \hat{s} (b + u) - \hat{w} + \tilde{w}) \hat{\tilde{w^{'}}} = 0 \end{array}$ ， (4.1g)

$s^{'} \bar{w} = w^{'} \bar{s}$ 。 (4.1h)

由相容条件(3.6b)可得：

$- g \dot{s} \tilde{\dot{s^{'}}} + g \tilde{s} \tilde{\dot{s^{'}}} - (c - \dot{u}) (a - \tilde{\dot{u}}) - (a - \tilde{u}) (b - \tilde{\dot{u}}) + \dot{y^{'}} - \tilde{y^{'}} = 0$ ， (4.2a)

$\begin{array}{l} g \bar{\dot{s^{'}}} \bar{w} - g \bar{\tilde{s^{'}}} \bar{w} - g \dot{s} \tilde{\dot{w^{'}}} + g \tilde{s} \tilde{\dot{w^{'}}} + (a + \bar{\dot{u}}) (3 e_{1} - \bar{y} - \dot{y^{'}}) - (c + \bar{\tilde{u}}) (3 e_{1} - \bar{y} - \tilde{y^{'}}) \\ + (c - \dot{u}) (3 e_{1} - \bar{\dot{y}} - \tilde{\dot{y^{'}}}) - (a - \tilde{u}) (3 e_{1} - \bar{\tilde{y}} - \tilde{\dot{y^{'}}}) = 0 \end{array}$ ， (4.2b)

$- g s \dot{s^{'}} + g s \tilde{s^{'}} + (c + u) (a + \dot{u}) - (a + u) (c + \tilde{u}) - \dot{y} + \tilde{y} = 0$ ， (4.2c)

$\begin{array}{l} \frac{- \tilde{s^{'}} (c - \dot{\tilde{u}}) + \dot{s^{'}} (a - \dot{\tilde{u}}) - \tilde{w^{'}} + \dot{w^{'}}}{\hat{\tilde{s^{'}}}} \\ + \frac{- (a - \tilde{\underline{u}}) \tilde{w} + (c - \dot{\underline{u}}) \dot{w} + \tilde{s} (3 e_{1} - y - \tilde{\underline{y^{'}}}) - \dot{s} (3 e_{1} - y - \dot{\underline{y^{'}}})}{w} = 0 \end{array}$ ， (4.2d)

$\frac{- \tilde{s} (a + u) + \dot{s} (c + u) - \dot{w} + \tilde{w}}{s} + \frac{\dot{s^{'}} (a - \dot{\tilde{u}}) + \tilde{s^{'}} (- c + \dot{\tilde{u}}) + \dot{w^{'}} - \tilde{w^{'}}}{\dot{\tilde{s^{'}}}} = 0$ ， (4.2e)

$\begin{array}{l} - (a + \dot{\bar{u}}) w \dot{w^{'}} + (c + \tilde{\bar{u}}) w \tilde{w^{'}} - (c - \dot{\underline{u}}) \dot{w} \dot{\tilde{w^{'}}} + (a - \tilde{\underline{u}}) \tilde{w} \dot{\tilde{w^{'}}} + \dot{s} \dot{\tilde{w^{'}}} (3 e_{1} - y - \dot{\underline{y^{'}}}) \\ - \tilde{s} \dot{\tilde{w^{'}}} (3 e_{1} - y - \tilde{\underline{y^{'}}}) - \dot{s^{'}} w (3 e_{1} - \dot{\bar{y}} - \dot{\tilde{y^{'}}}) + \tilde{s^{'}} w (3 e_{1} - \tilde{\bar{y}} - \dot{\tilde{y^{'}}}) = 0 \end{array}$ ， (4.2f)

$\begin{array}{l} s (a \dot{w^{'}} + \dot{\bar{u}} \dot{w^{'}} - c \tilde{w^{'}} - \tilde{\bar{u}} \tilde{w^{'}} + \dot{s^{'}} (3 e_{1} - \dot{\bar{y}} - \dot{\tilde{y^{'}}}) + \tilde{s^{'}} (- 3 e_{1} + \tilde{\bar{y}} + \dot{\tilde{y^{'}}})) \\ + (- \tilde{s} (a + u) + \dot{s} (c + u) - \dot{w} + \tilde{w}) \dot{\tilde{w^{'}}} = 0 \end{array}$ ， (4.2g)

$s^{'} \bar{w} = w^{'} \bar{s}$ 。 (4.2h)

由相容条件(3.6c)可得：

$- g \hat{s} \dot{\hat{s^{'}}} + g \dot{s} \dot{{\hat{s}}^{'}} + (b - \hat{u}) (c - \dot{\hat{u}}) - (c - \dot{u}) (b - \dot{\hat{u}}) - \hat{y^{'}} + \dot{y^{'}} = 0$ ， (4.3a)

$\begin{array}{l} g \bar{\hat{s^{'}}} \bar{w} - g \bar{\dot{s^{'}}} \bar{w} - g \hat{s} \dot{\hat{w^{'}}} + g \dot{s} \dot{\hat{w^{'}}} + (c + \bar{\hat{u}}) (3 e_{1} - \bar{y} - \hat{y^{'}}) - (b + \bar{\dot{u}}) (3 e_{1} - \bar{y} - \dot{y^{'}}) \\ + (b - \hat{u}) (3 e_{1} - \bar{\hat{y}} - \dot{\hat{y^{'}}}) - (c - \dot{u}) (3 e_{1} - \bar{\dot{y}} - \dot{\hat{y^{'}}}) = 0 \end{array}$ ， (4.3b)

$\begin{array}{l} - g s \hat{s^{'}} + g s \dot{s^{'}} + (b + u) (c + \hat{u}) - (c + u) (b + \dot{u}) - \hat{y} + \dot{y} = 0, \\ \frac{- \dot{s^{'}} (b - \dot{\hat{u}}) + \hat{s^{'}} (c - \dot{\hat{u}}) - \dot{w^{'}} + \hat{w^{'}}}{\dot{\hat{s^{'}}}}, \end{array}$ (4.3c)

$+ \frac{- (c - \dot{\underline{u}}) \dot{w} + (b - \hat{\underline{u}}) \hat{w} + \dot{s} (3 e_{1} - y - \dot{\underline{y^{'}}}) - \hat{s} (3 e_{1} - y - \hat{\underline{y^{'}}})}{w} = 0$ ， (4.3d)

$\frac{- \dot{s} (c + u) + \hat{s} (b + u) - \hat{w} + \dot{w}}{s} + \frac{\hat{s^{'}} (c - \hat{\dot{u}}) + \dot{s^{'}} (- b + \dot{\hat{u}}) + \hat{w^{'}} - \dot{w^{'}}}{\hat{\dot{s^{'}}}} = 0$ ，(4.3e)

$\begin{array}{l} - (c + \hat{\bar{u}}) w \hat{w^{'}} + (b + \dot{\bar{u}}) w \dot{w^{'}} - (b - \hat{\underline{u}}) \hat{w} \hat{\dot{w^{'}}} + (c - \dot{\underline{u}}) \dot{w} \hat{\dot{w^{'}}} + \hat{s} \hat{\dot{w^{'}}} (3 e_{1} - y - \hat{\underline{y^{'}}}) \\ - \tilde{s} \hat{\tilde{w^{'}}} (3 e_{1} - y - \tilde{\underline{y^{'}}}) - \dot{s^{'}} w (3 e_{1} - \hat{\bar{y}} - \hat{\dot{y^{'}}}) + \dot{s^{'}} w (3 e_{1} - \dot{\bar{y}} - \hat{\dot{y^{'}}}) = 0 \end{array}$ ， (4.3f)

$\begin{array}{l} s (c \hat{w^{'}} + \hat{\bar{u}} \hat{w^{'}} - b \dot{w^{'}} - \dot{\bar{u}} \dot{w^{'}} + \hat{s^{'}} (3 e_{1} - \hat{\bar{y}} - \hat{\dot{y^{'}}}) + \dot{s^{'}} (- 3 e_{1} + \dot{\bar{y}} + \hat{\dot{y^{'}}})) \\ + (- \dot{s} (c + u) + \hat{s} (b + u) - \hat{w} + \dot{w}) \hat{\dot{w^{'}}} = 0 \end{array}$ ， (4.3g)

$s^{'} \bar{w} = w^{'} \bar{s}$ 。 (4.3h)

通过方程式(4.1a)、(4.2a)和(4.3a)或(4.1c)^˙，(4.2c)^{^}以及(4.3c)^~我们可以得到(2.1a)。方程(1.1b)可以通过用方程(4.1e)^˙，(4.2e)^{^}和(4.3e)^~得到。将方程(4.1d)、(4.2d)和(4.3d)相加，或(4.1e)、(4.2e)和(4.3e)相加，得到(2.1c)。最后，计算(4.1c) + (4.1d) −(4.1e)、(4.2c) + (4.2d)−(4.2e)和(4.3c) + (4.3d)−(4.3e)将得到(1.1d)、(1.1e)和(1.1f)。

5. 结论

本文首先给出了椭圆KP方程的色散关系和辅助向量 $u^{(i)}$ ，从辅助向量 $u^{(i)}$ 出发通过矩阵变换得到了 $u^{(i)}$ 的位移关系式，当 $u^{(i)}$ 里 $i = 0$ 可求出两个耦合线性方程，进一步得到了椭圆KP方程的Lax组，接着我们从Lax组反推出了椭圆KP方程。

参考文献

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