Jacobi椭圆函数有理展开法和应用

doi:10.12677/pm.2024.1412409

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Jacobi椭圆函数有理展开法和应用
A Jacobi Elliptic Function Rational Expansion Method and Its Applications

DOI: 10.12677/pm.2024.1412409, PDF, HTML, XML,
作者: 吕大昭：北京建筑大学理学院，北京；崔艳英：北京工业大学耿丹学院信息工程学院，北京
关键词: Jacobi椭圆函数；有理解；双周期解；非线性波动方程；Jacobi Elliptic Function； Rational Solutions； Doubly Periodic Solutions； Nonlinear Wave Equation

摘要: 一个新的广义的Jacobi椭圆函数有理展开法被提出来构造非线性波动方程的有理解。利用这个直接有效的方法，获得了许多关于Jacobi椭圆函数的有理解。当模数m→0或1时，这些解退化为相应的关于孤立波或三角函数的有理解。

Abstract: A new general Jacobi elliptic function rational expansion procedure is presented for constructing rational solutions of nonlinear wave equations in terms of the Jacobi elliptic function. As a consequence, many new rational form Jacobi elliptic function solutions are obtained by this powerful and direct method. Moreover, the corresponding rational form solitary wave solutions and rational form trigonometric function solutions are also obtained when the modulus m→0 or 1.

文章引用：吕大昭, 崔艳英. Jacobi椭圆函数有理展开法和应用[J]. 理论数学, 2024, 14(12): 84-89. https://doi.org/10.12677/pm.2024.1412409

1. 引言

因为许多数学物理中的非线性波动方程拥有有理解，因此直接寻找非线性波动方程的有理解具有理论和现实的双重意义。特别是调查具有双周期的Jacobi椭圆函数的有理解是极为重要的，因为它的极限情形可以退化为相应的孤立波或三角函数的有理解。所以很多方法[1]被发展出来求有理解。最近，王[2]和陈[3]提出了Jacobi椭圆函数有理展开法用来构造非线性波动方程的有理解。稍后，他们[4]又推广了Jacobi椭圆函数有理展开法用来获得更多的有理解。但是，这些方法[1]-[4]仅仅求出特殊类型的有理解。在本文，我们提出了一个新的广义的Jacobi椭圆函数有理展开法用来统一构造非线性波动方程的更多有理解。

2. 广义的Jacobi椭圆函数有理展开法

下面我们简单说明我们的广义的Jacobi椭圆函数有理展开法。

步骤1：约化PDE到ODE

利用行波解约化 $u = u (ξ), ξ = k (x - λ t)$ ，其中 $k$ 和 $λ$ 分别是波数和波速，我们把偏微分方程

$P (u, u_{x}, u_{t}, u_{x x}, u_{x t}, u_{t t}, \dots) = 0$ (1)

约化到常微分方程

$G (u, \frac{d u}{d ξ}, \frac{d^{2} u}{d ξ^{2}}, \dots) = 0$ (2)

步骤2：引入有限幂级数形式解

为了寻找双周期有理解，我们假设常微分方程(2)有下面的有理形式解

$\begin{array}{l} u = \frac{a_{0} + a_{1} s n ξ + b_{1} c n ξ + c_{1} d n ξ + \sum_{i = 2}^{N} s n^{i - 2} ξ (a_{i} s n^{2} ξ + b_{i} s n ξ c n ξ + c_{i} s n ξ d n ξ + d_{i} c n ξ d n ξ)}{A_{0} + A_{1} s n ξ + B_{1} c n ξ + C_{1} d n ξ + \sum_{i = 2}^{M} s n^{i - 2} ξ (A_{i} s n^{2} ξ + B_{i} s n ξ c n ξ + C_{i} s n ξ d n ξ + D_{i} c n ξ d n ξ)} \\ = \frac{P_{N} (s n ξ, c n ξ, d n ξ)}{P_{M} (s n ξ, c n ξ, d n ξ)} \end{array}$ (3)

注意(3)式中的M和N是任意的，不是通过平衡常微分方程(2)中的最高阶导数项和非线性项来确定的。

步骤3：导出代数方程组

把(3)代入(2)，我们获得一个关于Jacobi椭圆函数的方程，然后通分，化简，合并同类项，得到关于未知量 $k, λ, a_{0}, a_{i}, b_{i}, c_{i}, d_{i + 1}, A_{0}, A_{i}, B_{i}, C_{i}, D_{i + 1} (i = 1, 2, \dots) .$ 的非线性代数方程组，利用著名的吴文俊消元法[5]求解，我们省略求解过程。

步骤4：获得Jacobi椭圆函数有理解

把求得的 $k, λ, a_{0}, a_{i}, b_{i}, c_{i}, d_{i + 1}, A_{0}, A_{i}, B_{i}, C_{i}, D_{i + 1} (i = 1, 2, \dots) .$ 的值代入(3)式，最后我们获得了非线性波动方程(1)的广义的Jacobi椭圆函数有理解。

3. 对KdV方程的应用

考虑KdV方程

$u_{t} + 6 u u_{x} + u_{x x x} = 0$ (4)

利用行波解约化 $u = u (ξ), ξ = k (x - λ t)$ ，我们得到

$- λ \frac{d u}{d ξ} + 6 u \frac{d u}{d ξ} + k^{2} \frac{d^{3} u}{d ξ^{3}} = 0$ (5)

积分一次，我们有

$- λ u + 3 u^{2} + k^{2} \frac{d^{2} u}{d ξ^{2}} + c = 0$

其中c是积分常数。

因此，利用上面步骤2~4，我们能够获得KdV方程丰富的有理形式的Jacobi椭圆函数解如下。其中 $I = \sqrt{- 1}$ 是虚数单位，m是模数， $ξ = k (x - λ t)$ 。

当N = M = 1时，我们得到

$u_{1} = \frac{a_{0} + (a_{0} - k^{2} m^{2} A_{1} + k^{2} A_{1}) s n ξ}{A_{1} + A_{1} s n ξ}, λ = - \frac{k^{2} m^{2} A_{1} - 5 k^{2} A_{1} - 6 a_{0}}{A_{1}}$

$u_{2} = \frac{- C_{1} m^{2} k^{2} + c_{1} + c_{1} d n ξ}{C_{1} + C_{1} d n ξ}, λ = - \frac{C_{1} m^{2} k^{2} + 4 C_{1} k^{2} - 6 c_{1}}{C_{1}}$

$u_{3} = \frac{b_{1} - \frac{1}{2} k^{2} B_{1} + b_{1} c n ξ + \frac{1}{2} k^{2} B_{1} d n ξ}{B_{1} + B_{1} c n ξ}, λ = - \frac{k^{2} m^{2} B_{1} - 6 b_{1} + k^{2} B_{1}}{B_{1}}$

$u_{4} = \frac{b_{1} c n ξ + (C_{1} m^{2} k^{2} - C_{1} k^{2} - b_{1}) d n ξ}{- C_{1} c n ξ + C_{1} d n ξ}, λ = \frac{5 C_{1} m^{2} k^{2} - 6 b_{1} - C_{1} k^{2}}{C_{1}}$

$u_{5} = \frac{(- \frac{1}{2} m I C_{1} k^{2} + m c_{1} I) s n ξ + \frac{1}{2} m C_{1} k^{2} c n ξ + c_{1} d n ξ}{m C_{1} I s n ξ + C_{1} d n ξ}, λ = - \frac{C_{1} m^{2} k^{2} - 6 c_{1} + C_{1} k^{2}}{C_{1}}$

$u_{6} = \frac{C_{1} m^{2} k^{2} + c_{1} m - k^{2} C_{1} m - c_{1} + m c_{1} c n ξ + c_{1} d n ξ}{m C_{1} - C_{1} + m C_{1} c n ξ + C_{1} d n ξ}, λ = - \frac{C_{1} m^{2} k^{2} - 3 k^{2} C_{1} m + C_{1} k^{2} - 6 c_{1}}{C_{1}}$

当N = M = 2时，我们得到

$u_{7} = \frac{- k^{2} A_{2} + a_{2} s n^{2} ξ + k^{2} A_{2} c n ξ d n ξ}{A_{2} s n^{2} ξ}, λ = - \frac{k^{2} A_{2} + k^{2} m^{2} A_{2} - 6 a_{2}}{A_{2}}$

$u_{8} = \frac{a_{0} + (- a_{0} + A_{0} k^{2} m^{2} - A_{0} k^{2}) s n^{2} ξ - \sqrt{1 - m^{2}} k^{2} A_{0} s n ξ d n ξ}{A_{0} - A_{0} s n^{2} ξ}, λ = - \frac{4 A_{0} k^{2} m^{2} - 5 A_{0} k^{2} - 6 a_{0}}{A_{0}}$

$u_{9} = \frac{(k^{2} m^{2} C_{2} + c_{2} - k^{2} C_{2}) \sqrt{1 - m^{2}} + k^{2} m^{2} \sqrt{1 - m^{2}} C_{2} s n^{2} ξ + c_{2} s n ξ d n ξ}{\sqrt{1 - m^{2}} C_{2} + C_{2} s n ξ d n ξ}, λ = \frac{- k^{2} C_{2} + 2 k^{2} m^{2} C_{2} + 6 c_{2}}{C_{2}}$

$u_{10} = \frac{k^{2} D_{2} m + (d_{2} m - k^{2} D_{2} m^{3} - k^{2} D_{2} m) s n^{2} ξ + d_{2} c n ξ d n ξ}{D_{2} m s n^{2} ξ + D_{2} c n ξ d n ξ}, λ = - \frac{k^{2} D_{2} - 6 d_{2} + D_{2} k^{2} m^{2}}{D_{2}}$

$u_{11} = \frac{(- D_{2} k^{2} - d_{2}) \sqrt{1 - m^{2}} s n ξ c n ξ + d_{2} c n ξ d n ξ}{- \sqrt{1 - m^{2}} D_{2} s n ξ c n ξ + D_{2} c n ξ d n ξ}, λ = - \frac{- 5 k^{2} D_{2} + 4 D_{2} k^{2} m^{2} - 6 d_{2}}{D_{2}}$

$u_{12} = \frac{- k^{2} D_{2} - 2 k^{2} D_{2} m + d_{2} - D_{2} k^{2} m^{2} + m d_{2} s n^{2} ξ + d_{2} c n ξ d n ξ}{D_{2} + m D_{2} s n^{2} ξ + D_{2} c n ξ d n ξ}, λ = - \frac{D_{2} k^{2} m^{2} + 6 k^{2} D_{2} m + k^{2} D_{2} - 6 d_{2}}{D_{2}}$

$u_{13} = \frac{k^{2} (- 1 + m^{2}) c n ξ d n ξ}{- 1 + s n^{2} ξ + c n ξ d n ξ}, λ = (5 m^{2} - 1) k^{2}$

$u_{14} = \frac{k^{2} m (m + 1) I s n ξ d n ξ}{- 1 + m^{2} s n^{2} ξ - m I s n ξ d n ξ - I s n ξ d n ξ - c n ξ d n ξ}, λ = - k^{2} m^{2} - 3 k^{2} m - k^{2}$

当N = M = 3时，我们得到

$u_{15} = \frac{\frac{4}{3} k^{2} \sqrt{- 6 - 6 m^{6} + 9 m^{2} + 9 m^{4} - 6 α} (- 1 - m^{6} + 2 α + 3 m^{2} (m^{4} - m^{2} + 1) s n^{2} ξ)}{(m^{4} - m^{2} + 1) (\sqrt{- 6 - 6 m^{6} + 9 m^{2} + 9 m^{4} - 6 α} + 9 m^{2} s n ξ c n ξ d n ξ)}, λ = \frac{4 k^{2} α}{m^{4} - m^{2} + 1}, α = \sqrt{{(m^{4} - m^{2} + 1)}^{3}}$

当N = M = 4时，我们得到

$u_{16} = - \frac{1}{6} \frac{(8 k^{2} - λ + 8 k^{2} m^{2}) m^{2} s n^{4} ξ + (λ m^{2} - 40 k^{2} m^{2} + 4 k^{2} m^{4} + λ + 4 k^{2}) s n^{2} ξ - λ + 8 k^{2} + 8 k^{2} m^{2}}{1 - m^{2} s n^{2} ξ - s n^{2} ξ + m^{2} s n^{4} ξ}, λ = λ$

$u_{17} = - \frac{1}{6} \frac{12 k^{2} m^{4} s n^{4} ξ + (- 4 k^{2} m^{2} - λ - 16 k^{2}) m^{2} s n^{2} ξ + λ - 8 k^{2} m^{2} + 16 k^{2}}{- 1 + m^{2} s n^{2} ξ}, λ = λ$

$u_{18} = - \frac{1}{6} \frac{(- 16 k^{2} - λ + 8 k^{2} m^{2}) s n^{4} ξ + (16 k^{2} + λ + 4 k^{2} m^{2}) s n^{2} ξ - 12 k^{2}}{s n^{2} ξ (- 1 + s n^{2} ξ)}, λ = λ$

注1：当 $N = M \geq 5$ 时，KdV方程也具有丰富的有理形式的Jacobi椭圆函数解，例如

$\begin{array}{l} u_{19} = \\ - \frac{1}{6} \frac{12 k^{2} m^{4} s n^{8} ξ - (16 k^{2} m^{2} + 16 k^{2} + λ) m^{2} s n^{6} ξ + (8 k^{2} m^{2} + 16 k^{2} + 16 k^{2} m^{4} + λ + λ m^{2}) s n^{4} ξ - (16 k^{2} m^{2} + 16 k^{2} + λ) s n^{2} ξ + 12 k^{2}}{s n^{2} ξ (1 - m^{2} s n^{2} ξ - s n^{2} ξ + m^{2} s n^{4} ξ)} \end{array}$

但是，N和M越大，求解代数方程组[5]的计算量越大。为了节省篇幅，我们省略它们。

注2：考虑到下面的关系

$\begin{array}{l} (1) s n ξ, c n ξ, 和 d n ξ \\ (2) n s ξ = \frac{1}{s n ξ}, c s ξ = \frac{c n ξ}{s n ξ}, 和 d s ξ = \frac{d n ξ}{s n ξ} \\ (3) s c ξ = \frac{s n ξ}{c n ξ}, n c ξ = \frac{1}{c n ξ}, 和 d c ξ = \frac{d n ξ}{c n ξ} \\ (4) s d ξ = \frac{s n ξ}{d n ξ}, c d ξ = \frac{c n ξ}{d n ξ}, 和 n d ξ = \frac{1}{d n ξ} \end{array}$

如果 $u = \frac{P_{N} (n s ξ, c s ξ, d s ξ)}{P_{M} (n s ξ, c s ξ, d s ξ)}$ ，那么可以利用上面的关系，通过通分转化成(3)的形式。其它情形类似。

注3：在合适的代数变换 $c n^{2} ξ = 1 - s n^{2} ξ, d n^{2} ξ = 1 - m^{2} s n^{2} ξ, m^{2} c n^{2} ξ + 1 - m^{2} = d n^{2} ξ$ 和因式分解之下，一些有理解可以变成多个单项式的和。比如 $u_{17}$ 和 $u_{19}$ 也可以表示为

$\begin{array}{l} u_{17} = \frac{1}{6} λ + \frac{8}{3} k^{2} - \frac{4}{3} k^{2} m^{2} - 2 k^{2} m^{2} s n^{2} ξ + 2 k^{2} m^{2} (1 - m^{2}) s d^{2} ξ, λ = λ \\ u_{19} = \frac{1}{6} λ + \frac{8}{3} k^{2} + \frac{8}{3} k^{2} m^{2} - 2 k^{2} m^{2} s n^{2} ξ - 2 k^{2} n s^{2} ξ - 2 k^{2} m^{2} c d^{2} ξ - 2 k^{2} d c^{2} ξ, λ = λ \end{array}$

这一点和注2互相呼应，互相印证。

注4：显然，由于我们的方法的形式解(3)更加广泛，所以比以前那些求Jacobi椭圆函数的有理解的方法[1]-[4]更加有效。这一点从我们已经获得的丰富的Jacobi椭圆函数的有理解可以证实。

注5：当 $P_{M} (s n ξ, c n ξ, d n ξ) = 1$ 时， $u (ξ) = \frac{P_{N} (s n ξ, c n ξ, d n ξ)}{P_{M} (s n ξ, c n ξ, d n ξ)} = P_{N} (s n ξ, c n ξ, d n ξ)$ ，本文方法变成了已知方法[6]-[8]，所以我们的方法比[6]-[8]更加有效，因为使用上面的方法[6]-[8]都不能求得这样丰富的有理解。

注6：当 $m = 0$ 时，我们得到

$\begin{array}{l} (1) s n (ξ, 0) = \sin ξ, c n (ξ, 0) = \cos ξ, d n (ξ, 0) = 1; \\ (2) n s (ξ, 0) = \csc ξ, c s (ξ, 0) = \cot ξ, d s (ξ, 0) = \csc ξ; \\ (3) s c (ξ, 0) = \tan ξ, n c (ξ, 0) = \sec ξ, d c (ξ, 0) = \sec ξ; \\ (4) s d (ξ, 0) = \sin ξ, c d (ξ, 0) = \cos ξ, n d (ξ, 0) = 1. \end{array}$

当 $m = 1$ 时，我们有

$\begin{array}{l} (1) s n (ξ, 1) = \tanh ξ, c n (ξ, 1) = sech ξ, d n (ξ, 1) = sech ξ; \\ (2) n s (ξ, 1) = \coth ξ, c s (ξ, 1) = csch ξ, d s (ξ, 1) = csch ξ; \\ (3) s c (ξ, 1) = \sinh ξ, n c (ξ, 1) = \cosh ξ, d c (ξ, 1) = 1; \\ (4) s d (ξ, 1) = \sinh ξ, c d (ξ, 1) = 1, n d (ξ, 1) = \cosh ξ . \end{array}$

容易看得出来，当模数m→0或1时，这些解退化为相应的关于孤立波或三角函数的有理解。为了节省篇幅，我们省略它们。

4. 解的结构

为了更好地了解有理解的结构，我们画出了解 $u_{17}$ 和 $u_{19}$ 的立体图(左图)和 $t = 0$ 的截面图(右图)，其中参数均取值为 $k = 1, λ = 2, m = \frac{1}{2}$ ，网格取值150 × 150。从中我们可以直观地看到有理解具有两种宏观结构：光滑型(分母永不为零)和爆破型(分母在某一时刻为零)。

解 $u_{17}$ 立体图解 $u_{17}$ 在 $t = 0$ 截面图

解 $(- u_{19})$ 立体图解 $(- u_{19})$ 在 $t = 0$ 截面图

5. 总结

本文，我们提出了一个新的广义的Jacobi椭圆函数有理展开法，获得了KdV方程的丰富的关于Jacobi椭圆函数的有理解。为了节省篇幅，我们省略其中的许多解。这种方法也可以应用于其他非线性波动方程。

参考文献

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