数学物理学报, 2026, 46(1): 108-129

研究论文

非局域正流短脉冲方程解的长时间渐近

刘文豪,*, 张玉峰,

中国矿业大学数学学院 江苏徐州 221116

Long-Time Asymptotics of the Nonlocal Positive Flow Short-Pules Equation

Liu Wenhao,*, Zhang Yufeng,

School of Mathematics, China University of Mining and Technology, Jiangsu Xuzhou 221116

通讯作者: *刘文豪, Email: wenhao_1003@163.com

收稿日期: 2024-10-15   修回日期: 2025-02-17  

基金资助: 国家自然科学基金(12501336)
国家自然科学基金(12371256)

Received: 2024-10-15   Revised: 2025-02-17  

Fund supported: NSFC(12501336)
NSFC(12371256)

作者简介 About authors

张玉峰,Email:zhangyfcumt@163.com

摘要

基于弹性梁在张力作用下的非线性横向振荡现象, 非局域正流短脉冲方程首次被提出. 通过非线性最速下降方法, 该方程的 Cauchy 问题解的长时间渐近行为被讨论. 从其所满足的 WKI 型 Lax 对出发, 相应的初始 Riemann-Hilbert 问题以及解的重构公式被建立. 经过转向、延拓、截断以及缩放平移这一系列对跳跃围线的形变, 从而将初始 Riemann-Hilbert 问题转变为可用抛物柱面函数求解的模型 Riemann-Hilbert 问题. 最终, 非局域正流短脉冲方程解的长时间渐近行为被得到.

关键词： 可积系统; 非线性最速下降方法; 长时间渐近

Abstract

The nonlocal positive flow short-pules equation is first proposed based on the nonlinear transverse oscillation of elastic beam under tension in physics. By the nonlinear steepest descent method, the long-time asymptotics of the solution of the Cauchy problem for the equation is discussed. Starting from the WKI-type Lax pair it satisfies, the corresponding basic Riemann-Hilbert problem and reconstruction formula for the solution are established. Through a series of deformations such as reorientation, extending, cuting and rescaling, the basic Riemann-Hilbert problem is transformed into the model Riemann-Hilbert problem that can be solved using parabolic cylinder functions. Finally, the long-time asymptotics of the solution of the nonlocal positive flow short-pules equation is obtained.

Keywords： integrable system; nonlinear steepest descent method; long-time asymptotics

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本文引用格式

刘文豪, 张玉峰. 非局域正流短脉冲方程解的长时间渐近[J]. 数学物理学报, 2026, 46(1): 108-129

Liu Wenhao, Zhang Yufeng. Long-Time Asymptotics of the Nonlocal Positive Flow Short-Pules Equation[J]. Acta Mathematica Scientia, 2026, 46(1): 108-129

1 引言

Wadati 等人于 1979 年提出了下列耦合系统去描述弹性梁在张力作用下的非线性横向振荡现象[1]

$\begin{cases}q_{t}(x,t)+\left(\frac{q_{xx}(x,t)}{(1-r_{x}(x,t)q_{x}(x,t))^{\frac{3}{2}}}\right)_{x}=0,\\r_{t}(x,t)+\left(\frac{r_{xx}(x,t)}{(1-r_{x}(x,t)q_{x}(x,t))^{\frac{3}{2}}}\right)_{x}=0.\end{cases}$

当取 $r(x,t)=-q_{x}(-x,-t)$ 时, 上述系统即约化为所谓的非局域正流短脉冲方程

$q_{t}(x,t)+\left(\frac{q_{xx}(x,t)}{(1+q_{x}(-x,-t)q_{x}(x,t))^{\frac{3}{2}}}\right)_{x}=0,$

其初值 $q(x,0)=q^{0}(x)$ 属于 Schwartz 空间

$\mathscr{S}(\mathbb{R})=\left\{f(x)\in C^{\infty}(\mathbb{R}):\sup_{x\in\mathbb{R}}|x^{\alpha}\partial^{\beta}f(x)|<\infty,\forall\alpha,\beta\in\mathbb{N}\right\}.$

关于非局域系统最早可追溯到 2013 年由 Ablowitz 和 Musslimani 研究了具有 PT 对称性的非局域非线性薛定谔方程[2]. 从此掀起了研究非局域系统的热潮, 受到了非线性光学以及磁学等数学物理领域的广泛关注. Deift 和 Zhou 发展了反散射方法提出非线性最速下降方法研究了 mKdV 方程 Cauchy 问题解的长时间渐近行为[3], 对可积系统领域产生了深远的影响. 随后, 该方法迅速得到广泛的关注并被用来讨论一批与 $2\times2$ 矩阵谱问题甚至与高阶矩阵谱问题相联系的可积发展方程解的长时间渐近行为, 包括 Camassa-Holm 方程, short-pulse 方程, Degasperis-Procesi 方程等[4-13]. 近年来, Rybalko 和 Shepelsky 将非线性最速下降方法推广用于研究在零边界条件下非局域系统初值问题解的长时间渐近行为[14]. 更进一步地, 非局域系统在具有 step-like 型初值时的解的长时间渐近行为也被得到[15,16].

本文的主要目的是利用非线性最速下降方法研究 (1.2) 式的 Cauchy 问题解的长时间渐近行为, 从而比较其与局域正流短脉冲方程在相同类型初值情况下解的长时间渐近行为之间的不同. 主要结论如下

定理 1.1 对于 $\sqrt{-\frac{x}{12t}}<C$$(x<0)$, 初值 $q_{0}(x)$ 属于 Schwartz 空间的非局域正流短脉冲方程解的长时间渐近行为被表示如下

$\begin{aligned}q(x,t)=\frac{{\rm e}^{\frac{\pi}{2}\rm{Re}\nu+\rm{Im}\nu\log192tk_{0}^{3}}\rm{Re}\nu}{k_{0}^{2}\sqrt{24\pi tk_{0}}}\frac{|\Gamma({\rm i}\nu(k_{0}))|\cdot|\gamma_{1}(k_{0})|}{1+|\gamma_{1}(k_{0})\gamma_{2}(k_{0})|}\sin\{-16tk_{0}^{3}+\Upsilon\}+R(k_{0},t),\end{aligned}$

其中 $C$ 是一个固定的常数且

$\begin{aligned} &\Upsilon=\frac{3\pi}{4}+\frac{\pi\rm{Im}\nu}{2}+\rm{Re}\nu\log(192tk_{0}^{3})+\arg\Gamma^{\ast}(-i\nu)+\arg\frac{\gamma_{1}^{\ast}(k_{0})}{1+\gamma_{1}^{\ast}(k_{0})\gamma_{2}^{\ast}(k_{0})}+\log(-\rm{Im}\nu) \\ & +\frac{1}{\pi }\left(\int_{k_{0}}^{+\infty}+\int_{-\infty}^{-k_{0}}\right)\log\left(\frac{1+\gamma_{1}(\xi)\gamma_{2}(\xi)}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})}\right)\frac{\mathrm{d}\xi}{\xi-k}, \\ &\nu=-\frac{1}{2\pi}\log(1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})), \quad k_{0}=\sqrt{-\frac{y}{12t}}, \\ &x=y-\frac{1}{2\pi}\left(\int_{k_{0}}^{+\infty}+\int_{-\infty}^{-k_{0}}\right)\frac{\log(1+\gamma_{1}(s)\gamma_{2}(s))}{s^{2}}\mathrm{d}s+o(1), \end{aligned}$
$R(k_{0},t)=\left\{ \begin{aligned}&O\left(t^{-1}\log t\right), & \quad &\rm{Im}\nu(k_{0})=0,\\&O\left(t^{-1+|\rm{Im}\nu(k_{0})|}\log t\right), & \quad & \rm{Im}\nu(k_{0})\in\left(-\frac{1}{2},0\right)\cup\left(0,\frac{1}{2}\right),\end{aligned} \right.$

$\gamma_{1}(\cdot)$ 和 $\gamma_{2}(\cdot)$ 的定义见 (2.3) 和 (2.18) 式.

为了方便, 我们采用一些记号: (1) 对任意的矩阵 $G$, 我们定义 $|G|\triangleq\left(\sum_{{\rm i},j}|g_{{\rm i}j}|^{2}\right)^{\frac{1}{2}}=({\rm tr}G^{\dagger}G)^{\frac{1}{2}}$, 其中 $\dagger$ 表示共轭转置; (2) 对于任意两个量 $R$ 和 $S$, 如果存在一个常数 $C$ 使得 $|R|\leq CS$, 则称 $R\lesssim S$; (3) 对任意的矩阵函数 $H(\cdot)$, 定义 $\|H(\cdot)\|_{p}=\||H(\cdot)|\|_{p}$.

本文主要内容安排如下: 在第二章节, 通过 Riemann-Hilbert (RH) 方法我们构造出非局域正流短脉冲方程 (1.2) 的初始 RH 问题. 利用特征函数在 $k=0$ 以及 $k=\infty$ 附近的渐近行为, 解的重构公式也被表出. 第三小节遵循非线性最速下降方法不断的将初始 RH 问题进行形变, 最终得到可解的模型 RH 问题. 再利用抛物柱面函数, 方程 (1.2) 初值问题解的长时间渐近行为被得到.

2 谱分析

方程 (1.2) 所满足的 Lax 对为

$\phi_x=U\phi, \quad \phi_t=V\phi,$

这里

$\begin{aligned} U=-{\rm i}k\sigma_{3}+kP, \quad P=\begin{pmatrix} 0 & q_{x}(x,t) \\ -q_{x}(-x,-t) & 0 \\ \end{pmatrix},\quad V=\begin{pmatrix} V_{11} & V_{12} \\ V_{21} & -V_{11} \\ \end{pmatrix}, \end{aligned} $
$\begin{align*} V_{11}=&-\frac{4{\rm i}k^{3}}{m^{\frac{1}{2}}}-\frac{k^{2}(q_{x}(-x,-t)q_{xx}(x,t)+q_{x}(x,t)q_{xx}(-x,-t))}{m^{\frac{3}{2}}},\\ V_{12}=&\frac{4k^{3}q_{x}(x,t)}{m^{\frac{1}{2}}}+\frac{2{\rm i}k^{2}q_{xx}(x,t)}{m^{\frac{3}{2}}}-\left(\frac{kq_{xx}(x,t)}{m^{\frac{3}{2}}}\right)_{x},\\ V_{21}=&-\frac{4k^{3}q_{x}(-x,-t)}{m^{\frac{1}{2}}}-\frac{2{\rm i}k^{2}q_{xx}(-x,-t)}{m^{\frac{3}{2}}}-\left(\frac{kq_{xx}(-x,-t)}{m^{\frac{3}{2}}}\right)_{x}, \end{align*}$

其中 $m=1+q_{x}(x,t)q_{x}(-x,-t)$. 在零边界条件下, 根据 (2.1) 式引入规范变换

$\phi(x,t,k)=\psi^{0}(x,t,k){\rm e}^{-({\rm i}kx+4{\rm i}k^{3}t)\sigma_{3}}.$

则推导出关于 $\psi^{0}(x,t,k)$ 的 Lax 对为

$\psi^{0}_{x}=-{\rm i}k[\sigma_{3},\psi^{0}]+kP\psi^{0},$
$\psi^{0}_{t}=-4{\rm i}k^{3}[\sigma_{3},\psi^{0}]+(V+4{\rm i}k^{3}\sigma_{3})\psi^{0}.$

方程 (2.4a) 的 Jost 解可由下列 Volterra 积分方程表示

$\psi^{0}_{\pm}(x,t,k)=I+\int_{\pm\infty}^{x}{\rm e}^{-{\rm i}k(x-\xi)\sigma_{3}}kP\psi^{0}_{\pm}(x,t,k){\rm e}^{{\rm i}k(x-\xi)\sigma_{3}}\mathrm{d}\xi.$

借此可进一步求出 Jost 解 $\psi^{0}_{\pm}(x,t,k)$ 在 $k\to 0$ 时的渐近展开式是

$\psi^{0}_{\pm}(x,t,k)=I+\left(\begin{aligned}0 & q(x,t) q(-x,-t) & 0\end{aligned}\right)k+O(k^2).$

这里之所以给出该渐近展式是为后了续能够成功重构出方程 (1.2) 解的表达式. 接下来, 我们先来构造初始 RH 问题.

2.1 特征函数在 $k=\infty$ 附近的行为

首先考虑将矩阵 $U$ 对角化, 故引入变换

$\phi(x,t,k)=G(x,t)\Phi(x,t,k), \quad G=\sqrt{\frac{1+\sqrt{m}}{2\sqrt{m}}}\begin{pmatrix}1 & \frac{{\rm i}(1-\sqrt{m})}{q_{x}(-x,-t)} \\\frac{{\rm i}(1-\sqrt{m})}{q_{x}(x,t)} & 1 \\\end{pmatrix}.$

Lax 对 (2.1) 式变成等价形式

$\Phi_{x}=p_{x}\Phi+U^{p}\Phi,$
$\Phi_{t}=p_{t}\Phi+V^{p}\Phi,$

其中

$\begin{aligned}&p_{x}=-{\rm i}k\sqrt{m}\sigma_{3}, \quad p_{t}=-4{\rm i}k^{3}\sigma_{3}-{\rm i}k\frac{p_{t}^{1}-p_{t}^{2}+20q_{xx}(x,t)q_{xx}(-x,-t)}{8m^{3}},\\&p_{t}^{1}=5[q_{x}(x,t)q_{xx}(-x,-t)+q_{x}(-x,-t)q_{xx}(x,t)]^{2},\\&p_{t}^{2}=4m[q_{x}(-x,-t)q_{xxx}(x,t)+6q_{xx}(-x,-t)q_{xx}(x,t)+4q_{x}(x,t)q_{xxx}(-x,-t)],\end{aligned}$

以及

$\begin{aligned}&U^{p}=\left(\begin{aligned}\frac{q_{xx}(-x,-t)q_{x}(x,t)+q_{x}(-x,-t)q_{xx}(x,t)}{4\sqrt{m}(\sqrt{m}+1)}&\frac{{\rm i}(q_{xx}(-x,-t)q_{x}^{2}(x,t)+q_{xx}(x,t)(\sqrt{m}+1)^{2})}{4m(\sqrt{m}+1)}\cr-\frac{{\rm i}(q_{xx}(x,t)q^{2}_{x}(-x,-t)+q_{xx}(-x,-t)(\sqrt{m}+1)^{2})}{4m(\sqrt{m}+1)}&-\frac{q_{xx}(-x,-t)q_{x}(x,t)+q_{x}(-x,-t)q_{xx}(x,t)}{4\sqrt{m}(\sqrt{m}+1)}\end{aligned}\right),\\&V^{p}=(V^{p}_{{\rm i}j})_{2\times2},\end{aligned}$
$\begin{align*} V^{p}_{11}=&\frac{q_{x}(-x,-t)q_{xt}(x,t)+q_{xt}(-x,-t)q_{x}(x,t)}{4\sqrt{m}(\sqrt{m}+1)}\\ &-\frac{{\rm i}[(q_{xx}(-x,-t)q_{x}(x,t)-q_{xx}(x,t)q_{x}(-x,-t))^{2}+4mq_{xx}(-x,-t)q_{xx}(x,t)]}{8m^{3}}k,\\ V^{p}_{12}=&\frac{{\rm i}\left[q_{xt}(x,t)(\sqrt{m}+1)^{2}+q_{xt}(-x,-t)q_{x}^{2}(x,t)\right]}{4m(\sqrt{m}+1)}\\ &+\frac{{\rm i}\left[q_{xx}(x,t)(\sqrt{m}+1)^{2}+q_{xx}(-x,-t)q_{x}^{2}(x,t)\right]}{m^{\frac{3}{2}}(\sqrt{m}+1)}k^{2}\\ &+\frac{\left[mq_{xxx}(-x,-t)q_{x}^{2}(x,t)-V^{p+}_{12}-3(\sqrt{m}+1)q_{xx}(-x,-t)q_{xx}(x,t)q_{x}(x,t)\right]}{2m^{3}(\sqrt{m}+1)}k,\\ V^{p}_{21}=&\frac{{\rm i}\left[-q_{xt}(-x,-t)(\sqrt{m}+1)^{2}-q_{xt}(x,t)(q_{x}(-x,-t))^{2}\right]}{4m(\sqrt{m}+1)}\\ &+\frac{{\rm i}\left[-q_{xx}(-x,-t)(\sqrt{m}+1)^{2}-q_{xx}(x,t)(q_{x}(-x,-t))^{2}\right]}{m^{\frac{3}{2}}(\sqrt{m}+1)}k^{2}\\ &-\frac{\left[mq_{xxx}(x,t)(q_{x}(-x,-t))^{2}+V^{p+}_{21}+3(\sqrt{m}+1)q_{xx}(x,t)q_{xx}(-x,-t)q_{x}(-x,-t)\right]}{2m^{3}(\sqrt{m}+1)}k,\\ V^{p+}_{12}=&\left[\frac{3}{2}q_{xx}^{2}(x,t)q_{x}(-x,-t)-mq_{xxx}(x,t)\right](\sqrt{m}+1)^{2}-\frac{3}{2}(q_{xx}(-x,-t))^{2}q_{x}^{3}(x,t),\\ V^{p+}_{21}=&\left[\frac{3}{2}(q_{xx}(-x,-t))^{2}q_{x}(x,t)-mq_{xxx}(-x,-t)\right](\sqrt{m}+1)^{2}-\frac{3}{2}q_{xx}(x,t)(q_{x}(-x,-t))^{3}. \end{align*}$

方程 (1.2) 的第一个守恒律如下所示

$(\sqrt{m})_{t}=\left(\frac{p_{t}^{1}-p_{t}^{2}-20q_{xx}(x,t)q_{xx}^{\ast}(-x,-t)}{8m^{3}}\right)_{x},$

其保证了 $p_{xt}=p_{tx}$. 定义函数

$p(x,t,k)=-{\rm i}k\left(x+\int_{-\infty}^{x}(\sqrt{m}-1)\mathrm{d}\xi\right)-4{\rm i}k^{3}t,$

并引入新的变换

$\Psi(x,t,k)=\Phi(x,t,k){\rm e}^{-p\sigma_{3}}.$

结合 Lax 对(2.8)式得

$\Psi_{x}=[p_{x}\sigma_{3},\Psi]+U^{p}\Psi,$
$\Psi_{t}=[p_{t}\sigma_{3},\Psi]+V^{p}\Psi.$

类似地, 其 Jost 解 $\Psi_{\pm}(x,t,k)$ 可通过 Volterra 积分方程被表示为

$\Psi_{\pm}(x,t,k)=I+\int_{\pm\infty}^{x}{\rm e}^{p(x)\sigma_{3}-p(\xi)\sigma_{3}}U^{p}(x,t)\Psi_{\pm}(x,t,k){\rm e}^{p(\xi)\sigma_{3}-p(x)\sigma_{3}}\,\mathrm{d}\xi.$

取 $\Psi_{\pm}=(\Psi_{\pm1},\Psi_{\pm2})$.

注 2.1 不失一般性地, 初值 $q^{0}(x)$ 的选取保证在限制条件 $m(x,t)>0$ 情况下, 方程 (1.2) 与初始 RH 问题相联系的解仍存在且唯一.

命题 2.1 Jost 解 $\Psi_{\pm}(x,t,k)$ 满足下列一些性质

(Ⅰ) $\Psi_{+1}$ 和 $\Psi_{-2}$ 在复平面下半平面 $\mathbb{C}_{-}$ 上解析; $\Psi_{-1}$ 和 $\Psi_{+2}$ 在复平面上半平面 $\mathbb{C}_{+}$ 上解析;

(Ⅱ) $\Psi_{\pm}(x,t,k)=1$;

(Ⅲ) 它们满足下列对称关系

$\begin{aligned}\label{Psi pm symmtry} \Psi_{\pm}(x,t,k)=\begin{pmatrix} 1 & 0\\ 0 & -1 \\ \end{pmatrix}\Psi_{\pm}^{\ast}(x,t,-k^{\ast})\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}, \quad \Psi_{\pm}^{-1}(x,t,k)=\Psi_{\mp}^{\top}(-x,-t,-k); \end{aligned}$

(Ⅳ) 随着 $k\rightarrow\infty$, $\Psi_{\pm}(x,t,k)$ 的渐近展开为

$\Psi_{\pm}(x,t,k)=I+\sum_{i=1}^{\infty}\Psi_{\pm,i}^{(0)}(x,t)+O\left(\frac{1}{k}\right),$

其中 $\Psi_{\pm,{\rm i}}^{(0)}(x,t)=\int_{\pm\infty}^{x}U^{p}_{d}(\xi,t)\mathrm{d}\xi$, $\Psi_{\pm,{\rm i}+1}^{(0)}(x,t)=\int_{\pm\infty}^{x}U^{p}_{d}(\xi,t)\Psi_{\pm,{\rm i}}^{(0)}(\xi,t)\mathrm{d}\xi$, $U^{p}_{d}$ 表示 $U^{p}$ 的主对角元素.

2.2 散射矩阵

函数 $G\Psi_{\pm}{\rm e}^{p}$ 满足相同的微分方程 (2.1), 则存在一个散射矩阵 $S(k)$ 使得

$\Psi_{+}=\Psi_{-}{\rm e}^{p}S(k){\rm e}^{-p},\quad k\in\mathbb{R}, \quad \det S(k)=1.$

令 $S(k)=(S_{{\rm i}j}(k))_{2\times2}$, 那么从对称关系 (2.13) 式以及散射关系 (2.15) 式知

$\begin{aligned}S_{11}(k)&=S_{11}(-k)=S_{11}^{\ast}(-k^{\ast}), \quad S_{22}(k)=S_{22}(-k)=S_{22}^{\ast}(-k^{\ast}),\\S_{12}(k)&=S_{21}(-k)=-S_{12}^{\ast}(-k^{\ast}),\end{aligned}$

并且

$S_{11}(k)=\det(\Psi_{+1},\Psi_{-2}), \quad S_{22}(k)=\det(\Psi_{-1},\Psi_{+2}).$

此外, 取 $t=0$, 随着 $x\rightarrow-\infty$,

$\begin{aligned}S_{11}(k)&=1+\int_{-\infty}^{+\infty}(U^{p}_{11}\Psi_{+11}+U^{p}_{12}\Psi_{+21})(\xi,0,k)\mathrm{d}\xi,\\S_{22}(k)&=1+\int_{-\infty}^{+\infty}(U^{p}_{21}\Psi_{+12}+U^{p}_{22}\Psi_{+22})(\xi,0,k)\mathrm{d}\xi,\\S_{12}(k)&=\int_{-\infty}^{+\infty}{\rm e}^{-p(\xi)\hat{\sigma}_{3}}(U^{p}_{11}\Psi_{+12}+U^{p}_{12}\Psi_{+22})(\xi,0,k)\mathrm{d}\xi.\end{aligned}$

将 (2.14) 式代入到 (2.15) 式中, 随着 $k\rightarrow-\infty$ 有

$\begin{aligned}S_{12}(k)=&S_{21}(k)=O\left(\frac{1}{k}\right),\\S_{11}(k)=&1+\sum_{i=1}^{\infty}\left[\Psi_{-,{\rm i}}^{(0)}\right]_{22}+\sum_{i=1}^{\infty}\left[\Psi_{+,{\rm i}}^{(0)}\right]_{11}+\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\left[\Psi_{-,{\rm i}}^{(0)}\right]_{22}\left[\Psi_{+,j}^{(0)}\right]_{11}+O\left(\frac{1}{k}\right),\\S_{22}(k)=&1+\sum_{i=1}^{\infty}\left[\Psi_{-,{\rm i}}^{(0)}\right]_{11}+\sum_{i=1}^{\infty}\left[\Psi_{+,{\rm i}}^{(0)}\right]_{22}+\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\left[\Psi_{-,{\rm i}}^{(0)}\right]_{11}\left[\Psi_{+,j}^{(0)}\right]_{22}+O\left(\frac{1}{k}\right).\end{aligned}$

2.3 Riemann-Hilbert 问题

定义一个矩阵值函数 $M(x,t,k)$

$M(x,t,k)=\begin{cases}\left(\Psi_{-1}(x,t,k),\frac{\Psi_{+2}(x,t,k)}{S_{22}(k)}\right), & k\in \mathbb{C}_+,\\\left(\frac{\Psi_{+1}(x,t,k)}{S_{11}(k)},\Psi_{-2}(x,t,k)\right), & k\in \mathbb{C}_-.\end{cases}$

通过 (2.17) 式知道 $S_{22}(k)$ 在 $k\in\mathbb{C}_+$ 上解析, $S_{11}(k)$ 在 $k\in\mathbb{C}_-$ 上解析. 因此 $M(x,t,k)$ 在 $k\in\mathbb{C}\backslash\mathbb{R}$ 上是解析的, 并且其满足跳跃条件

$M_+(x,t,k)=M_-(x,t,k)J(x,t,k), \quad k\in\mathbb{R},$

其中

$\begin{gather*} J(x,t,k)=\left(\begin{array}{cc} 1 & {\rm e}^{2p}\gamma_{1}(k)\\ {\rm e}^{-2p}\gamma_{2}(k) & 1+\gamma_{1}(k)\gamma_{2}(k)\\ \end{array}\right), \quad \gamma_{1}(k)=\frac{S_{12}(k)}{S_{22}(k)}, \quad \gamma_{2}(k)=\frac{S_{12}^{\ast}(-k^{\ast})}{S_{11}(k)}.\label{gamma definition} \end{gather*}$

散射系数之间的对称关系 (2.16) 式暗示着

$\gamma_{1}(k)=-\gamma_{1}^{\ast}(-k^{\ast}), \quad \gamma_{2}(k)=-\gamma_{2}^{\ast}(-k^{\ast}).$

更多地, 从 $S(k)=1$ 推断出

$1+\gamma_{1}(k)\gamma_{2}(k)=\frac{1}{S_{11}(k)S_{22}(k)}.$

基于 (2.14) 式, $M(x,t,k)$ 随着 $k\rightarrow\infty$ 的渐近展开式被给出

$M_{\pm}(x,t,k)=M_{\infty\pm}(x,t)+O\left(\frac{1}{k}\right),$

这里 $M_{\infty\pm}(x,t)$ 可逆且与 $k$ 无关. 注意到 $\lim_{k\rightarrow\infty}J(k)=I$, 则

$M_{\infty+}(x,t)=M_{\infty-}(x,t)\triangleq M_{\infty}(x,t).$

定义

$\tilde{M}(x,t,k)=M_{\infty}^{-1}(x,t)M(x,t,k).$

函数 $\tilde{M}(x,t,k)$ 满足一个定义在有向围线 (方向向右) $\mathbb{R}$ 上的 RH 问题

$\begin{cases}\tilde{M}_{+}(x,t,k)=\tilde{M}_{-}(x,t,k)J(x,t,k), & k\in\mathbb{R},\\\tilde{M}(x,t,k)\rightarrow I, & k\rightarrow\infty.\end{cases}$

容易发现 Jost 解 $\Psi_{\pm}(x,t,k)$ 和 $\psi^{0}_{\pm}(x,t,k)$ 满足相同的微分方程, 进而可得

$\Psi_{+}(x,t,k)=G^{-1}(x,t)\psi^{0}_{+}(x,t,k){\rm e}^{{\rm i}k\int_{+\infty}^{x}(\sqrt{m}-1)\mathrm{d}\xi\sigma_{3}},\label{mu+}$
$\Psi_{-}(x,t,k)=G^{-1}(x,t)\psi^{0}_{-}(x,t,k){\rm e}^{-{\rm i}k\int_{x}^{-\infty}(\sqrt{m}-1)\mathrm{d}\xi\sigma_{3}}.\label{mu-}$

合并 (2.17), (2.24) 式和 (2.25) 式, 我们获得

$\begin{aligned}S_{11}(k)&=1-{\rm i}ck-\frac{c^{2}}{2}k^{2}+O(k^{3}), \quad k\to0,\\S_{22}(k)&=1+{\rm i}ck-\frac{c^{2}}{2}k^{2}+O(k^{3}), \quad k\to0,\end{aligned}$

其中，$c=\int_{-\infty}^{+\infty}(\sqrt{m}-1)\mathrm{d}\xi$ 是一个常数。自然而然地, $\tilde{M}(x,t,k)$ 随着 $k\to0$ 的渐近展开式可被表示为

$\begin{aligned} \begin{split}\label{tildeM asymptotic} \tilde{M}(x,t,k)=&M_{\infty}^{-1}(x,t)G^{-1}(x,t)\\ &\times\left(\begin{aligned}1-{\rm i}k\int_{x}^{-\infty}(\sqrt{m}(\xi,t)-1)\,\mathrm{d}\xi& kq(x,t)\cr kq(-x,-t)&1+{\rm i}k\int_{x}^{-\infty}(\sqrt{m}(\xi,t)-1)\,\mathrm{d}\xi\end{aligned}\right)+O(k^2). \end{split} \end{aligned}$

为了计算的方便, 引入尺度变换

$y=x-\int_{x}^{-\infty}(\sqrt{m}(\xi,t)-1)\,\mathrm{d}\xi\triangleq x-c_{-}(x,t).$

那么 RH 问题 (2.23) 在新的尺度下转换成

$\begin{cases}\hat{M}_{+}(y,t,k)=\hat{M}_{-}(y,t,k)\hat{J}(y,t,k), & k\in\mathbb{R},\\\hat{M}(y,t,k)\to I, & k\to\infty,\end{cases}$

其中

$\begin{aligned} \hat{J}(y,t,k)=\left(\begin{array}{cc} 1 & {\rm e}^{2{\rm i}t\theta(k)}\gamma_{1}(k)\\ {\rm e}^{-2{\rm i}t\theta(k)}\gamma_{2}(k) & 1+\gamma_{1}(k)\gamma_{2}(k)\\ \end{array}\right), \quad \theta(k)=-k\frac{y}{t}-4k^3.\label{gammahat definition} \end{aligned}$

因为跳跃矩阵 $\hat{J}(y,t,k)$ 是正定的, 由湮灭引理可知该 RH 问题的解存在且唯一[17]. 利用展式(2.27), 方程 (1.2) 解的重构公式可写为

$q(x,t)=q(y(x,t),t),$

其中

$x=y+\lim_{k\rightarrow0}\frac{\left(\hat{M}^{-1}(y,t,0)\hat{M}(y,t,k)\right)_{11}-1}{-{\rm i}k},$
$q(y,t)=\lim_{k\rightarrow0}\frac{\left(\hat{M}^{-1}(y,t,0)\hat{M}(y,t,k)\right)_{12}}{k}.$

3 长时间渐近

在本小节, 我们利用非线性最速下降方法来研究方程 (1.2) 初值问题解的长时间渐近. 经过对初始 RH 问题的一系列变换将其转化为等价的可用抛物柱面函数求解的模型 RH 问题, 最终给出位势 $q(x,t)$ 随着 $t\rightarrow\infty$ 的渐近主部. 根据(2.30)式中相函数 $\theta(k)$ 的表达式得两个驻相点分别为 $\pm k_{0}$, 其中 $k_{0}=\sqrt{-\frac{y}{12t}}$ ($y<0$). 相应地, 关于函数 $\text{Re}({\rm i}\theta(k))$ 的符号图表如图1 所示, 这对于我们后续分析函数的解析性有重要作用.

图1

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图1   $\text{Re}({\rm i}\theta(k))$ 的符号图表.


注意到跳跃矩阵 $\hat{J}(y,t,k)$ 有两种三角分解

$\hat{J}(y,t,k)\!=\!\left\{ \begin{aligned}\!\!\!&\left(\begin{array}{cc}1&0\\{\rm e}^{-2{\rm i}t\theta(k)}\gamma_{2}(k)&1\\\end{array}\right)\!\!\left(\begin{array}{cc}1&{\rm e}^{2{\rm i}t\theta(k)}\gamma_{1}(k)\\0&1\\\end{array}\right),\\\!\!\!&\left(\begin{array}{cc}1&{\rm e}^{2{\rm i}t\theta(k)}\frac{\gamma_{1}(k)}{1+\gamma_{1}(k)\gamma_{2}(k)}\\0&1\\\end{array}\right)\!\!\left(\begin{array}{cc}\frac{1}{1+\gamma_{1}(k)\gamma_{2}(k)}&0\\0&1+\gamma_{1}(k)\gamma_{2}(k)\\\end{array}\right)\!\!\left(\begin{array}{cc}1&0\\{\rm e}^{-2{\rm i}t\theta(k)}\frac{\gamma_{2}(k)}{1+\gamma_{1}(k)\gamma_{2}(k)}&1\\\end{array}\right)\!.\end{aligned} \right.$

接着引入一个标量函数 $\delta(k)$, 它满足的 RH 问题为

$\left\{ \begin{aligned}&\delta_{+}(k)=\delta_{-}(k)(1+\gamma_{1}(k)\gamma_{2}(k)),& \quad &|k|>k_{0},\\&\delta_{+}(k)=\delta_{-}(k),& \quad &|k|<k_{0},\\&\delta(k)\rightarrow 1,& \quad & k\rightarrow\infty.\end{aligned} \right.$

由 Plemelj 公式得 RH 问题 (3.2) 的解是

$\delta(k)=\left(\frac{k-k_{0}}{k+k_{0}}\right)^{-{\rm i}\nu}{\rm e}^{\chi(k)},$

其中

$\begin{align*} & \nu=-\frac{1}{2\pi}\log(1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})), \\ & \chi(k)=-\frac{1}{2\pi {\rm i}}\left(\int_{k_{0}}^{+\infty}+\int_{-\infty}^{-k_{0}}\right)\log\left(\frac{1+\gamma_{1}(\xi)\gamma_{2}(\xi)}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})}\right)\frac{\mathrm{d}\xi}{\xi-k}. \end{align*}$

需要强调的是这里 $\nu(k_{0})\in\mathbb{C}$, 且

$\text{Im}\nu(k_{0})=-\frac{1}{2\pi}\left(\int_{k_{0}}^{+\infty}+\int_{-\infty}^{-k_{0}}\right)\mathrm{d}\arg(1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})).$

对于 $k\in\mathbb{R}$, 假设

$|\text{Im}\nu(k)|<\frac{1}{2},$

那么 $\delta(k)$ 是单值的而且 $\delta(k)$ 在 $k=\pm k_{0}$ 处的奇性是平方可积的. 标量函数 $\delta(k)$ 的引入可使 $\hat{J}(y,t,k)$ 的两种三角分解写成统一的形式. 通过引入下列新的变换

$\hat{M}^\Delta(y,t,k)=\hat{M}(y,t,k)\left(\begin{array}{cc}\delta(k) & 0\\0 & \delta^{-1}(k)\\\end{array}\right).$

$\hat{M}^\Delta(y,t,k)$ 是下列定义在如图2 所示的有向围线上的 RH 问题的解

$\left\{ \begin{aligned}& \hat{M}^\Delta_{+}(y,t,k)=\hat{M}^\Delta_{-}(y,t,k)\hat{J}^{\Delta}(y,t,k),& \quad & k\in\mathbb{R},\\& \hat{M}^\Delta(y,t,k)\rightarrow I,& \quad & k\rightarrow\infty,\end{aligned} \right.$

图2

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图2   有向围线$\mathbb{R}$.


其中

$\hat{J}^{\Delta}(y,t,k)=(b_{-})^{-1}b_{+}=\left\{ \begin{aligned}&\left(\begin{array}{cc}1&{\rm e}^{2{\rm i}t\theta(k)}\delta_{-}^{2}(k)\rho_{2}(k)\\0&1\\\end{array}\right)\left(\begin{array}{cc}1&0\\{\rm e}^{-2{\rm i}t\theta(k)}\delta_{+}^{2}(k)\rho_{1}(k)&1\\\end{array}\right), \quad |k|<k_{0},\\&\left(\begin{array}{cc}1&{\rm e}^{2{\rm i}t\theta(k)}\delta_{-}^{2}(k)\rho_{4}(k)\\0&1\\\end{array}\right)\left(\begin{array}{cc}1&0\\{\rm e}^{-2{\rm i}t\theta(k)}\delta_{+}^{2}(k)\rho_{3}(k)&1\\\end{array}\right),\quad |k|>k_{0},\end{aligned} \right.$
$\rho_{1}(k)=-\gamma_{2}(k), \quad \rho_{2}(k)=-\gamma_{1}(k), \quad \rho_{3}(k)=\frac{\gamma_{2}(k)}{1+\gamma_{1}(k)\gamma_{2}(k)}, \quad \rho_{4}(k)=\frac{\gamma_{1}(k)}{1+\gamma_{1}(k)\gamma_{2}(k)}.$

根据 $\delta(k)$ 的定义, 随着 $k\rightarrow0$, 直接计算得

$\delta(k)=1+\delta_{1}k+O(k^{2}),$

其中

$\delta_{1}=\frac{\rm i}{2\pi}\left(\int_{k_{0}}^{+\infty}+\int_{-\infty}^{-k_{0}}\right)\frac{\log(1+\gamma_{1}(s)\gamma_{2}(s))}{s^{2}}\mathrm{d}s.$

再令 $\hat{M}(y,t,k)$ 和 $\hat{M}^\Delta(y,t,k)$ 随着 $k\rightarrow0$ 的展开式的形式为

$\begin{aligned}& \hat{M}(y,t,k)=\hat{M}_{0}(y,t)+\hat{M}_{1}(y,t)k+O(k^{2}),\\& \hat{M}^\Delta(y,t,k)=\hat{M}^\Delta_{0}(y,t)+\hat{M}^\Delta_{1}(y,t)k+O(k^{2}).\end{aligned}$

根据 (3.3) 式, 计算得到

$\begin{aligned}\label{relation between M tilde and M tilde dealta} \hat{M}_{0}(y,t)=\hat{M}^\Delta_{0}(y,t), \quad \hat{M}_{1}(y,t)=\hat{M}_{1}^{\Delta}(y,t)-\delta_{1}\hat{M}_{0}^{\Delta}(y,t)\sigma_{3}. \end{aligned}$

随之, (2.32) 和 (2.33) 式被转化为

$c_{-}={\rm i}\left(\left[(\hat{M}_{0}^{\Delta})^{-1}\hat{M}_{1}^{\Delta}\right]_{11}-\delta_{1}\right),$
$q(y,t)=\left[(\hat{M}_{0}^{\Delta})^{-1}\hat{M}_{1}^{\Delta}\right]_{12}.$

3.1 围线形变

本小节我们重新定义新的围线 $\Sigma=\mathbb{R}\cup L_{1}\cup L_{2}\cup L_{3}\cup L_{4}$ 如图3 所示, 并建立围线 $\Sigma$ 上的 RH 问题. 定义

$\begin{align*} L_{1\epsilon}&:\left\{k=k_{0}+k_{0}\alpha {\rm e}^{\frac{\pi {\rm i}}{4}}:-\sqrt{2}\leq\alpha<-\epsilon\right\}\cup\left\{k=-k_{0}+k_{0}\alpha {\rm e}^{\frac{3\pi {\rm i}}{4}}:-\sqrt{2}\leq\alpha<-\epsilon\right\},\\ L_{2\epsilon}&:\left\{k=k_{0}+k_{0}\alpha {\rm e}^{-\frac{\pi {\rm i}}{4}}:-\sqrt{2}\leq\alpha<-\epsilon\right\}\cup\left\{k=-k_{0}+k_{0}\alpha {\rm e}^{-\frac{3\pi {\rm i}}{4}}:-\sqrt{2}\leq\alpha<-\epsilon\right\}. \end{align*}$

图3

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图3   有向围线 $\Sigma$.


定理 3.1 谱函数 $\rho_{\rm i}(k)$$({\rm i}=1,2,3,4)$ 可被分解为

$\rho_{\rm i}(k)=R_{\rm i}(k)+h_{\rm i}^{1}(k)+h_{\rm i}^{2}(k), \quad k\in\mathbb{R},$

其中 $R_{i}(k)$ 是分段有理函数, $h_{\rm i}^{1}(k)$ 在 $\mathbb{R}$ 上解析, $h_{i}^{2}(k)$ 可解析延拓至 $L_{1}\cup L_{3}$ 上. 对任意的正整数 $l$ 以及 $j=1,2$, $R_{\rm i}$, $h_{\rm i}^{1}$ 和 $h_{\rm i}^{2}$ 有下列严格估计

$\begin{align*} & |{\rm e}^{-2{\rm i}t\theta(k)}h_{i}^{1}(k)|\lesssim\frac{1}{(1+|k|^{2})t^{l}}, \quad k\in\mathbb{R},\\ & |\frac{{\rm e}^{-2{\rm i}t\theta(k)}h_{i}^{1}(k)}{k^{j}}|\lesssim\frac{1}{(1+|k|^{2})t^{l}}, \quad k\in\mathbb{R}\setminus\{0\},\\ & |{\rm e}^{-2{\rm i}t\theta(k)}h_{i}^{2}(k)|\lesssim\frac{1}{(1+|k|^{2})t^{l}}, \quad\, k\in L_{1}\cup L_{3},\\ & |\frac{{\rm e}^{-2{\rm i}t\theta(k)}h_{i}^{2}(k)}{k^{j}}|\lesssim\frac{1}{(1+|k|^{2})t^{l}}, \quad k\in L_{1}\cup L_{3},\\ & |{\rm e}^{-2{\rm i}t\theta(k)}R_{i}(k)|\lesssim {\rm e}^{-16\epsilon^{2}k_{0}^{3}t}, \qquad k\in L_{1\epsilon}. \end{align*}$

对 (3.12) 式整体取 Schwartz 共轭

$\rho^{\ast}(k^{\ast})=R_{\rm i}^{\ast}(k^{\ast})+(h_{\rm i}^{1})^{\ast}(k^{\ast})+(h_{\rm i}^{2})^{\ast}(k^{\ast}).$

计算知道 ${\rm e}^{2{\rm i}t\theta(k)}(h_{\rm i}^{1})^{\ast}(k^{\ast})$, ${\rm e}^{2{\rm i}t\theta(k)}(h_{\rm i}^{2})^{\ast}(k^{\ast})$ 以及 ${\rm e}^{2{\rm i}t\theta(k)}R_{\rm i}^{\ast}(k^{\ast})$ 分别在 $\mathbb{R}\cup L_{2}\cup L_{4}$ 上有相同的估计.

证 类似的证明过程可在参考文献 [3,18] 中发现.

定理 3.1 表明定义在 (3.5) 式中的函数 $b_{+}$ 可被进一步分解

$b_{+}=(I+\omega_{+})=\begin{cases}b_{1+}^{o}b_{1+}^{a}=(I+\omega_{1+}^{o})(I+\omega_{1+}^{a}), \quad k<|k_{0}|,\\b_{3+}^{o}b_{3+}^{a}=(I+\omega_{3+}^{o})(I+\omega_{3+}^{a}), \quad k>|k_{0}|,\end{cases}$

其中

$\begin{align*} &b_{1+}^{o}b_{1+}^{a}=\begin{pmatrix} 1&0\\ {\rm e}^{-2{\rm i}t\theta(k)}\delta_{+}^{2}(k)h_{1}^{1}(k)&1 \end{pmatrix}\begin{pmatrix} 1&0\\ {\rm e}^{-2{\rm i}t\theta}\delta_{+}^{2}(k)[h_{1}^{2}(k)+R_{1}(k)]&1 \end{pmatrix},\\ &b_{3+}^{o}b_{3+}^{a}=\begin{pmatrix} 1&0\\ {\rm e}^{-2{\rm i}t\theta(k)}\delta_{+}^{2}(k)h_{3}^{1}(k)&1 \end{pmatrix}\begin{pmatrix} 1&0\\ {\rm e}^{-2{\rm i}t\theta}\delta_{+}^{2}(k)[h_{3}^{2}(k)+R_{3}(k)]&1 \end{pmatrix}. \end{align*}$

类似地,

$b_{-}=(I-\omega_{-})=\begin{cases}b_{2-}^{o}b_{2-}^{a}=(I-\omega_{2-}^{o})(I-\omega_{2-}^{a}), \quad k<|k_{0}|,\\b_{4-}^{o}b_{4-}^{a}=(I-\omega_{4-}^{o})(I-\omega_{4-}^{a}), \quad k>|k_{0}|,\end{cases}$

其中

$\begin{align*} & b_{2-}^{o}b_{2-}^{a}=\begin{pmatrix} 1 & -{\rm e}^{2{\rm i}t\theta}\delta_{-}^{-2}(k)h_{2}^{1}(k)\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -{\rm e}^{2{\rm i}t\theta}\delta_{-}^{-2}(k)[h_{2}^{2}(k)+R_{2}(k)]\\ 0 & 1 \end{pmatrix},\\ & b_{4-}^{o}b_{4-}^{a}=\begin{pmatrix} 1 & -{\rm e}^{2{\rm i}t\theta}\delta_{-}^{-2}(k)h_{4}^{1}(k)\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -{\rm e}^{2{\rm i}t\theta}\delta_{-}^{-2}(k)[h_{4}^{2}(k)+R_{4}(k)]\\ 0 & 1 \end{pmatrix}. \end{align*}$

令

$\begin{aligned}\label{Mhat sharp definition} \hat{M}^{\sharp}(y,t,k)=\left\{ \begin{aligned} & \hat{M}^{\Delta}(y,t,k),& \quad & k\in\Omega_{1}\cup\Omega_{2},\\ & \hat{M}^{\Delta}(y,t,k)(b_{3+}^{a})^{-1},& \quad & k\in\Omega_{3}\cup\Omega_{5},\\ & \hat{M}^{\Delta}(y,t,k)(b_{4-}^{a})^{-1},& \quad & k\in\Omega_{6}\cup\Omega_{8},\\ & \hat{M}^{\Delta}(y,t,k)(b_{1+}^{a})^{-1},& \quad & k\in\Omega_{4},\\ & \hat{M}^{\Delta}(y,t,k)(b_{2-}^{a})^{-1},& \quad & k\in\Omega_{7}. \end{aligned} \right. \end{aligned}$

则矩阵值函数 $\hat{M}^{\sharp}(y,t,k)$ 就满足一个定义在 $\Sigma$ 上的 RH 问题

$\begin{aligned}\label{Mhat sharp RHP} \left\{ \begin{aligned} & \hat{M}_{+}^{\sharp}(y,t,k)=\hat{M}_{-}^{\sharp}(y,t,k)\hat{J}^{\sharp}(y,t,k),& \quad & k\in\Sigma,\\ & \hat{M}^{\sharp}(y,t,k)\rightarrow I,& \quad & k\rightarrow\infty, \end{aligned} \right. \end{aligned}$

其中

$\hat{J}^{\sharp}(y,t,k)=(b_{-}^{\sharp})^{-1}b_{+}^{\sharp}=\left\{ \begin{aligned} & (b_{4-}^{o})^{-1}b_{3+}^{o}, & \quad & k\in(-\infty,-k_{0})\cup(k_{0},+\infty),\\ & (b_{2-}^{o})^{-1}b_{1+}^{o}, & \quad & k\in(-k_{0},k_{0}),\\ & (I)^{-1}b_{1+}^{a}, & \quad & k\in L_{1},\\ & (I)^{-1}b_{3+}^{a}, & \quad & k\in L_{3},\\ & (b_{2-}^{a})^{-1}I, & \quad & k\in L_{2},\\ & (b_{4-}^{a})^{-1}I, & \quad & k\in L_{4}.\end{aligned} \right.$

命题 3.1 散射系数 $\gamma_{1}(k)$ 和 $\gamma_{2}(k)$ 随着 $k\rightarrow0$ 有渐近展开为 $ \gamma_{1}(k)=\gamma_{2}(k)=O(k^{3}). $

命题3.1表明

$\begin{aligned}\hat{M}^{\sharp}(y,t,k)&=\hat{M}^{\sharp}_{0}(y,t)+\hat{M}^{\sharp}_{1}(y,t)k+O(k^{2}),\\&=\hat{M}^{\Delta}_{0}(y,t)+\hat{M}^{\Delta}_{1}(y,t)k+O(k^{2}), \quad k\rightarrow0.\end{aligned}$

故 (3.10) 和 (3.11) 式可写成

$c_{-}={\rm i}\left(\left[(\hat{M}_{0}^{\sharp})^{-1}\hat{M}_{1}^{\sharp}\right]_{11}-\delta_{1}\right),$
$q(y,t)=\left[(\hat{M}_{0}^{\sharp})^{-1}\hat{M}_{1}^{\sharp}\right]_{12}.$

定义函数

$\omega_{\pm}^{\sharp}=\pm(b_{\pm}^{\sharp}-I), \quad \omega^{\sharp}=\omega_{+}^{\sharp}+\omega_{-}^{\sharp},$

以及 Cauchy 算子 $C_{\pm}$

$(C_{\pm}f)(k)=\int_{\Sigma}\frac{f(\xi)}{\xi-k_{\pm}}\frac{\mathrm{d}\xi}{2\pi {\rm i}}, \quad k\in\Sigma, \quad f\in\mathscr{L}^{2}(\Sigma),$

这里 $k_{+}$ ($k_{-}$) 表示左 (右) 边界值. $C_{\pm}$ 是从 $\mathscr{L}^{2}(\Sigma)$ 到 $\mathscr{L}^{2}(\Sigma)$ 的有界算子, $C_{+}-C_{-}=1$. 另外再引入算子 $C_{\omega^{\sharp}}$: $\mathscr{L}^{2}(\Sigma)+\mathscr{L}^{\infty}(\Sigma)\rightarrow\mathscr{L}^{2}(\Sigma)$

$C_{\omega^{\sharp}}f=C_{+}(f\omega_{-}^{\sharp})+C_{-}(f\omega_{+}^{\sharp}).$

根据 Beals-Coifman 理论[19], 令 $\mu^{\sharp}(y,t,k)\in\mathscr{L}^{2}(\Sigma)+\mathscr{L}^{\infty}(\Sigma)$ 是下列奇异积分方程的解

$\mu^{\sharp}=I+C_{\omega^{\sharp}}\mu^{\sharp}.$

则 RH 问题(3.14)的解为

$\hat{M}^{\sharp}(y,t,k)=I+\int_{\Sigma}\frac{\mu^{\sharp}(y,t,\xi)\omega^{\sharp}(y,t,\xi)}{\xi-k}\frac{\mathrm{d}\xi}{2\pi {\rm i}}, \quad k\in\mathbb{C}\backslash\Sigma.$

将 (3.21) 式在 $k=0$ 处 Taylor 展开, 与 (3.15) 式比较 $k$ 的同次幂系数得

$\hat{M}_{0}^{\sharp}(y,t)=I+\int_{\Sigma}\frac{\mu^{\sharp}(y,t,\xi)\omega^{\sharp}(y,t,\xi)}{\xi}\frac{\mathrm{d}\xi}{2\pi {\rm i}},$
$\hat{M}_{1}^{\sharp}(y,t)=\int_{\Sigma}\frac{\mu^{\sharp}(y,t,\xi)\omega^{\sharp}(y,t,\xi)}{\xi^{2}}\frac{\mathrm{d}\xi}{2\pi {\rm i}}.$

3.2 围线的截断

定义截断后的围线为 $\Sigma^{\prime}=\Sigma\backslash(\mathbb{R}\cup L_{1\epsilon}\cup L_{2\epsilon})$ (如图4所示)

图4

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图4   有向围线 $\Sigma^{\prime}$.


为了实现将定义在 $\Sigma$ 上的 RH 问题等价的转换为定义在 $\Sigma^{\prime}$ 上的 RH 问题, 我们引入一个函数 $\omega^{\prime}$, 其在 $\Sigma\setminus\Sigma^{\prime}$ 上满足 $\omega^{\prime}=\omega^{\sharp}-\omega^{a}-\omega^{b}-\omega^{c}=\omega^{\sharp}-\omega^{e}=0$, 其中 $\omega^{a}=\omega^{\sharp}|_{\mathbb{R}}$ 仅在 $\mathbb{R}$ 有定义, 即在 $\Sigma$ 其它部分 $\omega^{a}=0$, 且由包含 $h_{\rm i}^{1}(k)$ 和 $(h_{\rm i}^{1})^{\ast}(k^{\ast})$ (${\rm i}=1,2,3,4$) 的项组成. 同理, $\omega^{b}=\omega^{\sharp}|_{L_{1}\cup L_{2}\cup L_{3}\cup L_{4}}$ 仅在 $L_{1}\cup L_{2}\cup L_{3}\cup L_{4}$ 上有定义且由包含 $h_{\rm i}^{2}(k)$ 和 $(h_{\rm i}^{2})^{\ast}(k^{\ast})$ 的项组成. $\omega^{c}=\omega^{\sharp}|_{L_{1\epsilon}\cup L_{2\epsilon}}$ 仅在 $L_{1\epsilon}\cup L_{2\epsilon}$ 上有定义且由包含 $R_{\rm i}(k)$ 和 $R_{\rm i}^{\ast}(k^{\ast})$ 的项组成. $\omega^{\prime}$ 仅在 $\Sigma^{\prime}$ 上有定义, 其主要贡献来自于有理项 $R_{\rm i}(k)$ 和 $R_{\rm i}^{\ast}(k^{\ast})$.

引理 3.1 随着 $t\rightarrow\infty$ 以及 $j=1,2$, 函数 $\omega^{a}$, $\omega^{b}$, $\omega^{c}$ 和 $\omega^{\prime}$ 的相关估计为

$\|\omega^{a}\|_{\mathscr{L}^{1}(\mathbb{R})\cap\mathscr{L}^{2}(\mathbb{R})\cap\mathscr{L}^{\infty}(\mathbb{R})}\lesssim t^{-l},$
$\left\|\frac{\omega^{a}}{k^{j}}\right\|_{\mathscr{L}^{1}(\mathbb{R})\cap\mathscr{L}^{2}(\mathbb{R})\cap\mathscr{L}^{\infty}(\mathbb{R})}\lesssim t^{-l},$
$\|\omega^{b}\|_{\mathscr{L}^{1}(L_{1}\cup L_{2}\cup L_{3}\cup L_{4})\cap\mathscr{L}^{2}(L_{1}\cup L_{2}\cup L_{3}\cup L_{4})\cap\mathscr{L}^{\infty}(L_{1}\cup L_{2}\cup L_{3}\cup L_{4})}\lesssim t^{-l},$
$\left\|\frac{\omega^{b}}{k^{j}}\right\|_{\mathscr{L}^{1}(L_{1}\cup L_{2}\cup L_{3}\cup L_{4})\cap\mathscr{L}^{2}(L_{1}\cup L_{2}\cup L_{3}\cup L_{4})\cap\mathscr{L}^{\infty}(L_{1}\cup L_{2}\cup L_{3}\cup L_{4})}\lesssim t^{-l},$
$\|\omega^{c}\|_{\mathscr{L}^{1}(L_{1\epsilon}\cup L_{2\epsilon})\cap\mathscr{L}^{2}(L_{1\epsilon}\cup L_{2\epsilon})\cap\mathscr{L}^{\infty}(L_{1\epsilon}\cup L_{2\epsilon})}\lesssim {\rm e}^{-16\epsilon^{2}k_{0}^{3}t},$
$\left\|\frac{\omega^{c}}{k^{j}}\right\|_{\mathscr{L}^{1}(L_{1\epsilon}\cup L_{2\epsilon})\cap\mathscr{L}^{2}(L_{1\epsilon}\cup L_{2\epsilon})\cap\mathscr{L}^{\infty}(L_{1\epsilon}\cup L_{2\epsilon})}\lesssim {\rm e}^{-16\epsilon^{2}k_{0}^{3}t},$
$\|\omega^{\prime}\|_{\mathscr{L}^{2}(\Sigma)}\lesssim t^{-\frac{1}{4}}, \quad \|\omega^{\prime}\|_{\mathscr{L}^{1}(\Sigma)}\lesssim t^{-\frac{1}{2}}.$

证 估计 (3.24)-(3.29) 式可由定理 3.1 中的结果给出. 又因为在 $k\in L_{3}$ 时

$|R_{3}(k)|=\frac{|\sum_{j=0}^{m}\mu_{j}(k-k_{0})^{j}|}{|(k-{\rm i})^{m+5}|}\lesssim\frac{1}{1+|k|^{5}}.$

而且 $\text{Re}(i\theta)\geq8\alpha^{2}k_{0}^{3}$, 那么立即可得

$|{\rm e}^{-2{\rm i}t\theta}\delta^{2}(k)R_{3}(k)|\lesssim {\rm e}^{-16\alpha^{2}k_{0}^{3}t}(1+|k|^{5})^{-1}.$

在 $k\in L_{1}$, $k\in L_{2}$ 以及 $k\in L_{4}$ 上也有相同的估计. 综上, 该引理得证.

定理 3.2 随着 $t\rightarrow\infty$, 我们得到下列结果

$\int_{\Sigma}\frac{\left((1-C_{\omega^{\sharp}})^{-1}I\right)(\xi)\omega^{\sharp}(y,t,\xi)}{\xi^{j}}\mathrm{d}\xi=\int_{\Sigma}\frac{\left((1-C_{\omega^{\prime}})^{-1}I\right)(\xi)\omega^{\prime}(y,t,\xi)}{\xi^{j}}\mathrm{d}\xi+O\left(t^{-l}\right).$

证 利用第二预解式

$\begin{aligned}&\int_{\Sigma}\frac{\left((1-C_{\omega^{\sharp}})^{-1}I\right)\omega^{\sharp}}{\xi^{j}}\mathrm{d}\xi=\int_{\Sigma}\frac{\left((1-C_{\omega^{\prime}})^{-1}I\right)\omega'}{\xi^{j}}\mathrm{d}\xi+\int_{\Sigma}\frac{\omega^{e}}{\xi^{j}}\mathrm{d}\xi+\int_{\Sigma}\frac{\left((1-C_{\omega^{\prime}})^{-1}(C_{\omega^{e}}I)\right)\omega^{\sharp}}{\xi^{j}}\mathrm{d}\xi\\&+\int_{\Sigma}\frac{\left((1-C_{\omega^{\prime}})^{-1}(C_{\omega^{e}}I)\right)\omega^{e}}{\xi^{j}}\mathrm{d}\xi+\int_{\Sigma}\frac{\left((1-C_{\omega^{\prime}})^{-1}C_{\omega^{e}}(1-C_{\omega^{\sharp}})^{-1}\right)(C_{\omega^{\sharp}}I)\omega^{\sharp}}{\xi^{j}}\mathrm{d}\xi.\end{aligned}$

结合引理 3.1 中的结果, (3.31) 式成立.

再引入奇异积分方程 $\mu^{\prime}=(1-C_{\omega^{\prime}})^{-1}I$, 那么

$\hat{M}^{\prime}(y,t,k)=I+\int_{\Sigma^{\prime}}\frac{\mu^{\prime}(y,t,\xi)\omega^{\prime}(y,t,\xi)}{\xi-k}\frac{\mathrm{d}\xi}{2\pi {\rm i}}$

是下列 RH 问题的解

$\left\{ \begin{aligned}& \hat{M}_{+}^{\prime}(y,t,k)=\hat{M}_{-}^{\prime}(y,t,k)\hat{J}^{\prime}(y,t,k),& \quad & k\in\Sigma^{\prime},\\& \hat{M}^{\prime}(y,t,k)\rightarrow I,& \quad & k\rightarrow\infty,\end{aligned} \right.$

其中

$\hat{J}^{\prime}=(b_{-}^{\prime})^{-1}b_{+}^{\prime}, \quad b_{\pm}^{\prime}=I\pm\omega_{\pm}^{\prime}, \quad \omega^{\prime}=\omega_{+}^{\prime}+\omega_{-}^{\prime},$
$b_{+}^{\prime}=\left\{ \begin{aligned}&\left(\begin{array}{cc}1&0\\{\rm e}^{-2{\rm i}t\theta}\delta^{2}(k)R_{1}(k)&1\\\end{array}\right), & \quad & k\in L_{1},\\&\left(\begin{array}{cc}1&0\\{\rm e}^{-2{\rm i}t\theta}\delta^{2}(k)R_{3}(k)& 1 \\ \end{array}\right), & \quad & k\in L_{3},\\&I, & \quad & k\in L_{2}\cup L_{4},\end{aligned} \right.$
$b_{-}^{\prime}=\left\{ \begin{aligned}&I, & \quad & k\in L_{1}\cup L_{3},\\&\left(\begin{array}{cc}1&-{\rm e}^{2{\rm i}t\theta}R_{2}(k)\delta^{-2}(k)\\0&1\\\end{array}\right), & \quad & k\in L_{2},\\&\left(\begin{array}{cc}1&-{\rm e}^{2{\rm i}t\theta}R_{4}(k)\delta^{-2}(k)\\0&1\\\end{array}\right), & \quad & k\in L_{4}.\end{aligned} \right.$

令 $\hat{M}^{\prime}(y,t,k)$ 在 $k=0$ 处也有展开式为

$\hat{M}^{\prime}(y,t,k)=\hat{M}_{0}^{\prime}(y,t)+\hat{M}_{1}^{\prime}(y,t)k+O(k^{2}),$

其中

$\hat{M}_{0}^{\prime}(y,t)=I+\int_{\Sigma^{\prime}}\frac{((1-C_{\omega^{\prime}})^{-1}I)(\xi)\omega^{\prime}(\xi)}{\xi}\frac{\mathrm{d}\xi}{2\pi {\rm i}},$
$\hat{M}_{1}^{\prime}(y,t)=\int_{\Sigma^{\prime}}\frac{((1-C_{\omega^{\prime}})^{-1}I)(\xi)\omega^{\prime}(\xi)}{\xi^{2}}\frac{\mathrm{d}\xi}{2\pi {\rm i}}.$

所以 (3.16) 和 (3.17) 式有与 $\hat{M}^{\prime}(y,t,k)$ 有关的形式

$\begin{aligned}-ic_{-}=&\left(I+\hat{M}_{0}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{22}\left(\hat{M}_{1}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{11}\\&-\left(\hat{M}_{0}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{12}\left(\hat{M}_{1}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{21}-\delta_{1},\end{aligned}$

以及

$\begin{aligned}q(y,t)=&\left(I+\hat{M}_{0}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{22}\left(\hat{M}_{1}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{12}\\&-\left(\hat{M}_{0}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{12}\left(\hat{M}_{1}^{\prime}(y,t)+O\left(t^{-l}\right)\right)_{22}.\end{aligned}$

3.3 分离两端不相交的围线

在上一小节, 通过相应的估计, 围线 $\Sigma$ 被截断为由 $\Sigma_{A}^{\prime}$ 和 $\Sigma_{B}^{\prime}$ 不相交的两部分组成的围线 $\Sigma^{\prime}$. 定义 $\omega_{\pm}^{\prime}=\omega_{A\pm}^{\prime}+\omega_{B\pm}^{\prime}$, 其中

$\omega_{A\pm}^{\prime}(k)=\left\{ \begin{aligned}&\omega_{\pm}^{\prime}(k),& \quad & k\in\Sigma_{A}^{\prime},\\&0,& \quad & k\in\Sigma_{B}^{\prime},\end{aligned} \right.\quad \omega_{B\pm}^{\prime}(k)=\left\{ \begin{aligned}&0,& \quad & k\in\Sigma_{A}^{\prime},\\&\omega_{\pm}^{\prime}(k),& \quad & k\in\Sigma_{B}^{\prime}.\end{aligned} \right.$

类似的引入算子 $C_{\omega_{A}^{\prime}}$ 和 $C_{\omega_{B}^{\prime}}$: $\mathscr{L}^{2}(\Sigma)+\mathscr{L}^{\infty}(\Sigma)\rightarrow\mathscr{L}^{2}(\Sigma)$. 对于 $\beta\neq\gamma\in{A,B}$, 我们有

$\|C_{\omega_{\gamma}^{\prime}}C_{\omega_{\beta}^{\prime}}\|_{\mathscr{L}^{2}(\Sigma^{\prime})}\lesssim t^{-\frac{1}{2}},$
$\|C_{\omega_{\gamma}^{\prime}}C_{\omega_{\beta}^{\prime}}\|_{\mathscr{L}^{\infty}(\Sigma^{\prime})\rightarrow\mathscr{L}^{2}(\Sigma^{\prime})}\lesssim t^{-\frac{3}{4}}.$

引理 3.2 As $t\rightarrow\infty$,

$\begin{aligned}\int_{\Sigma^{\prime}}\frac{\left((1-C_{\omega^{\prime}})^{-1}I\right)(\xi)\omega^{\prime}(\xi)}{\xi^{j}}\mathrm{d}\xi=&\int_{\Sigma_{A}^{\prime}}\frac{\left((1-C_{\omega_{A}^{\prime}})^{-1}I\right)(\xi)\omega_{A}^{\prime}(\xi)}{\xi^{j}}\mathrm{d}\xi\\&+\int_{\Sigma_{B}^{\prime}}\frac{\left((1-C_{\omega_{B}^{\prime}})^{-1}I\right)(\xi)\omega_{B}^{\prime}(\xi)}{\xi^{j}}\mathrm{d}\xi+O\left(t^{-1}\right).\end{aligned}$

证

利用(3.40)式, 该引理可被证明. 具体过程参见参考文献[20,引理 3.6]. 为了方便, 取

$\begin{aligned}M_{0}^{AB}&=\frac{1}{2\pi {\rm i}}\int_{\Sigma_{A}^{\prime}}\frac{\left((1-C_{\omega_{A}^{\prime}})^{-1}I\right)\omega_{A}^{\prime}}{\xi}\mathrm{d}\xi+\frac{1}{2\pi {\rm i}}\int_{\Sigma_{B}^{\prime}}\frac{\left((1-C_{\omega_{B}^{\prime}})^{-1}I\right)\omega_{B}^{\prime}}{\xi}\mathrm{d}\xi,\\M_{1}^{AB}&=\frac{1}{2\pi {\rm i}}\int_{\Sigma_{A}^{\prime}}\frac{\left((1-C_{\omega_{A}^{\prime}})^{-1}I\right)\omega_{A}^{\prime}}{\xi^{2}}\mathrm{d}\xi+\frac{1}{2\pi {\rm i}}\int_{\Sigma_{B}^{\prime}}\frac{\left((1-C_{\omega_{B}^{\prime}})^{-1}I\right)\omega_{B}^{\prime}}{\xi^{2}}\mathrm{d}\xi.\end{aligned}$

故, (3.38) 和 (3.39) 式被转换为

$\begin{aligned}-ic_{-}=&\left(I+M_{0}^{AB}+O\left(t^{-1}\right)\right)_{22}\left(M_{1}^{AB}+O\left(t^{-1}\right)\right)_{11}\\&-\left(M_{0}^{AB}+O\left(t^{-1}\right)\right)_{12}\left(M_{1}^{AB}+O\left(t^{-1}\right)\right)_{21}-\delta_{1}\end{aligned}$

和

$\begin{aligned}q(y,t)=&\left(I+M_{0}^{AB}+O\left(t^{-1}\right)\right)_{22}\left(M_{1}^{AB}+O\left(t^{-1}\right)\right)_{12}\\&-\left(M_{0}^{AB}+O\left(t^{-1}\right)\right)_{12}\left(M_{1}^{AB}+O\left(t^{-1}\right)\right)_{22}.\end{aligned}$

3.4 缩放平移算子及 RH 问题的进一步约化

本小节目的是建立过原点十字交叉线上的模型 RH 问题. 首先, 令 $\Sigma_{A}$ 和 $\Sigma_{B}$ 如图5所示

图5

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图5   有向围线 $\Sigma_{A}$ 以及 $\Sigma_{B}$.


再定义缩放平移算子

$\begin{aligned}N_{A}:&\mathscr{L}^{2}(\hat{\Sigma}_{A}^{\prime})\rightarrow\mathscr{L}^{2}(\Sigma_{A}),\\&f(k)\rightarrow(N_{A}f)(k)=f\left(\frac{k}{\sqrt{48tk_{0}}}-k_{0}\right),\end{aligned}$
$\begin{aligned}N_{B}:&\mathscr{L}^{2}(\hat{\Sigma}_{B}^{\prime})\rightarrow\mathscr{L}^{2}(\Sigma_{B}),\\&f(k)\rightarrow(N_{B}f)(k)=f\left(\frac{k}{\sqrt{48tk_{0}}}+k_{0}\right).\end{aligned}$

利用算子作用得

$N_{A}\left({\rm e}^{-{\rm i}t\theta}\delta(k)\right)=\delta_{A}^{0}\delta_{A}^{1}(k), \quad N_{B}\left({\rm e}^{-{\rm i}t\theta}\delta(k)\right)=\delta_{B}^{0}\delta_{B}^{1}(k),$

其中

$\begin{aligned}\delta_{A}^{0}=&(192tk_{0}^{3})^{-{\rm i}\nu/2}{\rm e}^{8{\rm i}tk_{0}^{3}}{\rm e}^{\chi(-k_{0})},\\\delta_{B}^{0}=&(192tk_{0}^{3})^{{\rm i}\nu/2}{\rm e}^{-8{\rm i}tk_{0}^{3}}{\rm e}^{\chi(k_{0})}.\end{aligned}$

设

$\begin{aligned}\hat{L}_{A1}=\{k=\alpha k_{0}\sqrt{48tk_{0}}{\rm e}^{^{\frac{3\pi {\rm i}}{4}}}:-\epsilon<\alpha<0\},\quad \hat{L}_{A3}=\{k=\alpha k_{0}\sqrt{48tk_{0}}{\rm e}^{^{\frac{3\pi {\rm i}}{4}}}:0<\alpha<+\infty\},\\\hat{L}_{B1}=\{k=\alpha k_{0}\sqrt{48tk_{0}}{\rm e}^{^{\frac{3\pi {\rm i}}{4}}}:-\epsilon<\alpha<0\},\quad \hat{L}_{B3}=\{k=\alpha k_{0}\sqrt{48tk_{0}}{\rm e}^{^{\frac{3\pi {\rm i}}{4}}}:0<\alpha<+\infty\}.\end{aligned}$

引理 3.3 随着 $t\rightarrow\infty$,

$\begin{gather*} |N_{A}(R_{1}(k))(\delta_{A}^{1})^{2}-R_{1}(-k_{0}+)(-k)^{2{\rm i}\nu}{\rm e}^{-{\rm i}k^{2}/2}|\lesssim t^{-\frac{1}{2}+|\rm{Im}\nu(k_{0})|}\log t, \quad k\in \hat{L}_{A1},\\ |N_{A}(R_{3}(k))(\delta_{A}^{1})^{2}-R_{3}(-k_{0}-)(-k)^{2{\rm i}\nu}{\rm e}^{-{\rm i}k^{2}/2}|\lesssim t^{-\frac{1}{2}+|\rm{Im}\nu(k_{0})|}\log t, \quad k\in \hat{L}_{A3},\\ \hspace{-.8cm}|N_{B}(R_{1}(k))(\delta_{B}^{1})^{2}-R_{1}(k_{0}-)k^{-2{\rm i}\nu}{\rm e}^{{\rm i}k^{2}/2}|\lesssim t^{-\frac{1}{2}+|\rm{Im}\nu(k_{0})|}\log t, \quad k\in \hat{L}_{B1},\\ \hspace{-.8cm}|N_{B}(R_{3}(k))(\delta_{B}^{1})^{2}-R_{3}(k_{0}+)k^{-2{\rm i}\nu}{\rm e}^{{\rm i}k^{2}/2}|\lesssim t^{-\frac{1}{2}+|\rm{Im}\nu(k_{0})|}\log t, \quad k\in \hat{L}_{B3}, \end{gather*}$

其中

$\begin{aligned}R_{1}(-k_{0}+)&=-\gamma_{2}(-k_{0}), \quad R_{1}(k_{0}-)=-\gamma_{2}(k_{0}),\\R_{3}(-k_{0}-)&=\frac{\gamma_{2}(-k_{0})}{1+\gamma_{1}(-k_{0})\gamma_{2}(-k_{0})}, \quad R_{3}(k_{0}+)=\frac{\gamma_{2}(k_{0})}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})}.\end{aligned}$

证 详见文献 [3,引理 3.35].

引入新的函数 $\omega_{A}$ 和 $\omega_{B}$,

$\omega_{A}=N_{A}\hat{\omega}_{A}^{\prime}, \quad \omega_{B}=N_{B}\hat{\omega}_{B}^{\prime},$

直接计算发现下列关系成立

$C_{\hat{\omega}_{A}^{\prime}}=N_{A}^{-1}C_{\omega_{A}}N_{A}, \quad C_{\hat{\omega}_{B}^{\prime}}=N_{B}^{-1}C_{\omega_{B}}N_{B}.$

而且

$\hat{\omega}=\hat{\omega}_{+}=\left\{ \begin{aligned}&\left(\begin{array}{cc}0&0\\(N_{A}s_{1})(k)&0\\\end{array}\right), & \quad & k\in\hat{L}_{A1},\\&\left(\begin{array}{cc}0&0\\(N_{A}s_{3})(k)&0\\\end{array}\right), & \quad & k\in \hat{L}_{A3},\\&\left(\begin{array}{cc}0&0\\(N_{B}s_{1})(k)&0\\\end{array}\right), & \quad & k\in \hat{L}_{B1},\\&\left(\begin{array}{cc}0&0\\(N_{B}s_{3})(k)&0\\\end{array}\right), & \quad & k\in \hat{L}_{B3},\end{aligned} \right. \quad \hat{\omega}=\hat{\omega}_{-}=\left\{ \begin{aligned}&\left(\begin{array}{cc}0&(N_{A}s_{2})(k)\\0&0\\\end{array}\right), & \quad & k\in\hat{L}_{A1}^{\ast},\\&\left(\begin{array}{cc}0&(N_{A}s_{4})(k)\\0&0\\\end{array}\right), & \quad & k\in \hat{L}_{A3}^{\ast},\\&\left(\begin{array}{cc}0&(N_{B}s_{2})(k)\\0&0\\\end{array}\right), & \quad & k\in \hat{L}_{B1}^{\ast},\\&\left(\begin{array}{cc}0&(N_{B}s_{4})(k)\\0&0\\\end{array}\right), & \quad & k\in \hat{L}_{B3}^{\ast},\end{aligned} \right.$

其中

$\begin{aligned}s_{1}(k)={\rm e}^{-2{\rm i}t\theta}\delta^{2}(k)R_{1}(k), \quad s_{2}(k)=-{\rm e}^{2{\rm i}t\theta}\delta^{-2}(k)R_{2}(k),\\s_{3}(k)={\rm e}^{-2{\rm i}t\theta}\delta^{2}(k)R_{3}(k), \quad s_{4}(k)=-{\rm e}^{2{\rm i}t\theta}\delta^{-2}(k)R_{4}(k).\end{aligned}$

所以围线 $\Sigma_{A}$ 上的跳跃矩阵 $J^{A^{0}}=(I-\omega_{A^{0}-})^{-1}(I+\omega_{A^{0}+})$ 可被构造如下

$\omega_{A^{0}}=\omega_{A^{0}+}=\left\{ \begin{aligned}&\begin{pmatrix}0 & 0 \\-(\delta_{A}^{0})^{2}(-k)^{2{\rm i}\nu}{\rm e}^{-\frac{1}{2}{\rm i}k^{2}}\gamma_{2}(-k_{0}) & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{A}^{1},\\&\begin{pmatrix}0 & 0 \\(\delta_{A}^{0})^{2}(-k)^{2{\rm i}\nu}{\rm e}^{-\frac{1}{2}{\rm i}k^{2}}\frac{\gamma_{2}(-k_{0})}{1+\gamma_{1}(-k_{0})\gamma_{2}(-k_{0})} & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{A}^{3},\end{aligned} \right.$
$\omega_{A^{0}}=\omega_{A^{0}-}=\left\{ \begin{aligned}&\begin{pmatrix}0 & -(\delta_{A}^{0})^{-2}(-k)^{-2{\rm i}\nu}{\rm e}^{\frac{1}{2}{\rm i}k^{2}}\gamma_{1}(-k_{0}) \\0 & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{A}^{2},\\&\begin{pmatrix}0 & (\delta_{A}^{0})^{-2}(-k)^{-2{\rm i}\nu}{\rm e}^{\frac{1}{2}{\rm i}k^{2}}\frac{\gamma_{1}(-k_{0})}{1+\gamma_{1}(-k_{0})\gamma_{2}(-k_{0})} \\0 & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{A}^{4}.\end{aligned} \right.$

引理 3.3 中的结果暗示了

$\left\|\frac{\omega_{A}-\omega_{A^{0}}}{k^{j}}\right\|_{\mathscr{L}^{1}(\Sigma_{A})\cup\mathscr{L}^{2}(\Sigma_{A})\cup\mathscr{L}^{\infty}(\Sigma_{A})}\lesssim t^{-\frac{1}{2}+|\text{Im}\nu(k_{0})|}\log t, \quad t\rightarrow\infty.$

类似可得 $\Sigma_{B}$ 上的跳跃矩阵 $J^{B^{0}}=(I-\omega_{B^{0}-})^{-1}(I+\omega_{B^{0}+})$ 为

$\omega_{B^{0}}=\omega_{B^{0}+}=\left\{ \begin{aligned}&\begin{pmatrix}0 & 0 \\(\delta_{B}^{0})^{2}k^{-2{\rm i}\nu}{\rm e}^{\frac{1}{2}{\rm i}k^{2}}\frac{\gamma_{2}(k_{0})}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})} & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{B}^{2},\\&\begin{pmatrix}0 & 0 \\-(\delta_{B}^{0})^{2}k^{-2{\rm i}\nu}{\rm e}^{\frac{1}{2}{\rm i}k^{2}}\gamma_{2}(k_{0}) & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{B}^{4},\end{aligned} \right.$
$\omega_{B^{0}}=\omega_{B^{0}-}=\left\{ \begin{aligned}&\begin{pmatrix}0 & (\delta_{B}^{0})^{-2}k^{2{\rm i}\nu}{\rm e}^{-\frac{1}{2}{\rm i}k^{2}}\frac{\gamma_{1}(k_{0})}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})} \\0 & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{B}^{1},\\&\begin{pmatrix}0 & -(\delta_{B}^{0})^{-2}k^{2{\rm i}\nu}{\rm e}^{-\frac{1}{2}{\rm i}k^{2}}\gamma_{1}(k_{0}) \\0 & 0 \\\end{pmatrix},& \quad & k\in\Sigma_{B}^{3}.\end{aligned} \right.$

引理 3.4 随着 $t\rightarrow\infty$, 则

$\begin{aligned}\int_{\Sigma_{A}^{\prime}}\frac{\left((1-C_{\omega_{A}^{\prime}})^{-1}I\right)(\xi)\omega_{A}^{\prime}(\xi)}{\xi^{j}}\mathrm{d}\xi=&\frac{1}{\sqrt{48tk_{0}}}\int_{\Sigma_{A}}\frac{\left((1-C_{\omega_{A^{0}}})^{-1}I\right)(\xi)\omega_{A^{0}}(\xi)}{(N_{A}\xi)^{j}}\mathrm{d}\xi\\&+O\left(t^{-1+|\rm{Im}\nu(k_{0})|}\log t\right).\end{aligned}$

证 合并 (3.45) 和 (3.47) 式推算出

$\begin{aligned}&\frac{1}{2\pi {\rm i}}\int_{\Sigma_{A}^{\prime}}\frac{\left((1-C_{\omega_{A}^{\prime}})^{-1}I\right)(\xi)\omega_{A}^{\prime}(\xi)}{\xi^{j}}\mathrm{d}\xi\\=&\frac{1}{2\pi {\rm i}}\int_{\hat{\Sigma}_{A}^{\prime}}\frac{\left(N_{A}^{-1}(1-C_{\omega_{A}})^{-1}N_{A}I\right)(\xi)\hat{\omega}_{A}^{\prime}(\xi)}{\xi^{j}}\mathrm{d}\xi\\=&\frac{1}{2\pi {\rm i}}\int_{\hat{\Sigma}_{A}^{\prime}}\frac{\left((1-C_{\omega_{A}})^{-1}I\right)\left((\xi+k_{0})\sqrt{48tk_{0}}\right)N_{A}\hat{\omega}_{A}^{\prime}\left((\xi+k_{0})\sqrt{48tk_{0}}\right)}{\xi^{j}}\mathrm{d}\xi\\=&\frac{1}{2\pi {\rm i}}\frac{1}{\sqrt{48tk_{0}}}\int_{\Sigma_{A}}\frac{\left((1-C_{\omega_{A}})^{-1}I\right)(\xi)\omega_{A}(\xi)}{(N_{A}\xi)^{j}}\mathrm{d}\xi.\end{aligned}$

除此之外

$\int_{\Sigma_{A}}\frac{\left((1-C_{\omega_{A}})^{-1}I\right)(\xi)\omega_{A}(\xi)}{(N_{A}\xi)^{j}}\mathrm{d}\xi=\int_{\Sigma_{A}}\frac{\left((1-C_{\omega_{A^{0}}})^{-1}I\right)(\xi)\omega_{A^{0}}(\xi)}{(N_{A}\xi)^{j}}\mathrm{d}\xi+O\left(t^{-\frac{1}{2}+|\text{Im}\nu(k_{0})|}\log t\right).$

最终, (3.53) 式显然成立.

另一方面, 随着 $t\rightarrow\infty$ 也有

$\begin{aligned}\int_{\Sigma_{B}^{\prime}}\frac{\left((1-C_{\omega_{B}^{\prime}})^{-1}I\right)(\xi)\omega_{B}^{\prime}(\xi)}{\xi^{j}}\mathrm{d}\xi=&\frac{1}{\sqrt{48tk_{0}}}\int_{\Sigma_{B}}\frac{\left((1-C_{\omega_{B^{0}}})^{-1}I\right)(\xi)\omega_{B^{0}}(\xi)}{(N_{B}\xi)^{j}}\mathrm{d}\xi\\&+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right).\end{aligned}$

若假设

$M^{B^{0}}(y,t,k)=I+\int_{\Sigma_{B}}\frac{\left((1-C_{\omega_{B^{0}}})^{-1}I\right)(\xi)\omega_{B^{0}}(\xi)}{\xi-k}\frac{\mathrm{d}\xi}{2\pi i}, \quad k\in\mathbb{C}\backslash\Sigma_{B}.$

我们有下列定理成立.

定理 3.3$M^{B^{0}}(y,t,k)$ 所满足的 RH 问题为

$\begin{aligned}\label{MB0 RHP} \left\{ \begin{aligned} &M_{+}^{B^{0}}(y,t,k)=M_{-}^{B^{0}}(y,t,k)J^{B^{0}}(y,t,k),& \quad & k\in\Sigma_{B},\\ &M^{B^{0}}(y,t,k)\rightarrow I,& \quad & k\rightarrow\infty. \end{aligned} \right. \end{aligned}$

将 $M^{B^{0}}(y,t,k)$ 的展开式设为

$M^{B^{0}}(k)=I-\frac{M^{B^{0}}_{1}}{k}+O(k^{-2}), \quad k\rightarrow\infty.$

则

$\label{(M^{B^{0}}_{1})_{12}}(M^{B^{0}}_{1})_{12}={\rm i}(\delta_{B}^{0})^{-2}\beta_{B}^{12}, \quad \beta_{B}^{12}=\frac{{\rm e}^{\frac{\pi v}{2}-\frac{3\pi {\rm i}}{4}}\nu\Gamma({\rm i}\nu)}{\sqrt{2\pi}}\frac{\gamma_{1}(k_{0})}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})}.$

证 借助于(3.55)式引入变换

$\tilde{\Phi}(k)=T_{B}(k)k^{{\rm i}\nu\sigma_{3}}{\rm e}^{-\frac{1}{4}{\rm i}k^{2}\sigma_{3}}, \quad T_{B}(k)=(\delta_{B}^{0})^{\sigma_{3}}M^{B^{0}}(k)(\delta_{B}^{0})^{-\sigma_{3}}.$

通过 Liouville 定理,

$\frac{\mathrm{d}\tilde{\Phi}(k)}{\mathrm{d}k}+\frac{1}{2}{\rm i}k\sigma_{3}\tilde{\Phi}(k)=\beta_{B}\tilde{\Phi}(k),$

其中

$\beta_{B}=-\frac{1}{2}{\rm i}(\delta_{B}^{0})^{\sigma_{3}}\left[\sigma_{3},M^{B^{0}}_{1}\right](\delta_{B}^{0})^{-\sigma_{3}}=\left(\begin{array}{cc}0&\beta_{B}^{12}\\\beta_{B}^{21}&0\\\end{array}\right).$

从上式易推出

$(M^{B^{0}}_{1})_{12}(M^{B^{0}}_{1})_{12}={\rm i}(\delta_{B}^{0})^{-2}\beta_{B}^{12}.$

众所周知, Weber 方程

$\frac{\mathrm{d}^{2}g(\zeta)}{\mathrm{d}\zeta^{2}}+\left(a+\frac{1}{2}-\frac{\zeta^{2}}{4}\right)g(\zeta)=0$

的解有下列形式

$g(\zeta)=c_{1}D_{a}(\zeta)+c_{2}D_{a}(-\zeta),$

其中 $D_{a}(\cdot)$ 是抛物柱面函数且

$\frac{\mathrm{d}D_{a}(\zeta)}{\mathrm{d}\zeta}+\frac{\zeta}{2}D_{a}(\zeta)-aD_{a-1}(\zeta)=0,$
$D_{a}(\pm\zeta)=\frac{\Gamma(a+1){\rm e}^{\frac{{\rm i}\pi a}{2}}}{\sqrt{2\pi}}D_{-a-1}(\pm {\rm i}\zeta)+\frac{\Gamma(a+1){\rm e}^{-\frac{{\rm i}\pi a}{2}}}{\sqrt{2\pi}}D_{-a-1}(\mp {\rm i}\zeta).$

取 $a={\rm i}\beta_{B}^{12}\beta_{B}^{21}$. 利用 Weber 方程及其解的形式可得

$\begin{aligned}&\tilde{\Phi}_{11}(k)=c_{3}D_{a}\left({\rm e}^{-\frac{3\pi {\rm i}}{4}}k\right)+c_{4}D_{a}\left({\rm e}^{\frac{\pi {\rm i}}{4}}k\right),\\&\beta_{B}^{12}\tilde{\Phi}_{22}(k)=c_{5}D_{-a}\left({\rm e}^{\frac{3\pi {\rm i}}{4}}k\right)+c_{6}D_{-a}\left({\rm e}^{-\frac{\pi {\rm i}}{4}}k\right),\end{aligned}$

其中 $c_{3}$, $c_{4}$, $c_{5}$ 和 $c_{6}$ 都是常数. 当 $\arg k\in(-\frac{\pi}{4},\frac{\pi}{4})$, 随着 $k\rightarrow\infty$, (3.58) 式暗示我们

$\tilde{\Phi}_{11}(k)k^{-{\rm i}\nu}{\rm e}^{\frac{{\rm i}k^{2}}{4}}\rightarrow1, \quad \tilde{\Phi}_{22}(k)k^{{\rm i}\nu}{\rm e}^{-\frac{{\rm i}k^{2}}{4}}\rightarrow1.$

根据抛物柱面函数随着 $k\rightarrow\infty$ 的展开式 (见文献 [21]) 计算出

$\begin{aligned}&\tilde{\Phi}_{11}(k)={\rm e}^{\frac{\pi \nu}{4}}D_{a}\left({\rm e}^{\frac{\pi {\rm i}}{4}}k\right), \quad \nu=\beta_{12}\beta_{21},\\&\beta_{12}\tilde{\Phi}_{22}(k)=\beta_{12}{\rm e}^{\frac{\pi v}{4}}D_{-a}\left({\rm e}^{-\frac{\pi {\rm i}}{4}}k\right),\\&\tilde{\Phi}_{12}(k)=\beta_{12}{\rm e}^{\frac{\pi(v-3{\rm i})}{4}}D_{-a-1}\left({\rm e}^{-\frac{\pi {\rm i}}{4}}k\right).\end{aligned}$

另外, 对于 $\arg k\in(-\frac{3\pi}{4},-\frac{\pi}{4})$,

$\begin{aligned}&\tilde{\Phi}_{11}(k)={\rm e}^{\frac{\pi \nu}{4}}D_{a}\left({\rm e}^{\frac{\pi {\rm i}}{4}}k\right),\\&\beta_{12}\tilde{\Phi}_{22}(k)=\beta_{12}{\rm e}^{-\frac{3\pi v}{4}}D_{-a}\left({\rm e}^{\frac{3\pi {\rm i}}{4}}k\right),\\&\tilde{\Phi}_{12}(k)=\beta_{12}{\rm e}^{\frac{\pi({\rm i}-3\nu)}{4}}D_{-a-1}\left({\rm e}^{\frac{3\pi {\rm i}}{4}}k\right).\end{aligned}$

沿着 $\arg k=-\frac{\pi}{4}$,

$\tilde{\Phi}_{+}(k)=\tilde{\Phi}_{-}(k)\left(\begin{array}{cc}1&\frac{\gamma_{1}(k_{0})}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})}\\0&1\\\end{array}\right),$

从上式得

$(\tilde{\Phi}_{+})_{12}(k)=(\tilde{\Phi}_{-})_{11}(k)\frac{\gamma_{1}(k_{0})}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})}+(\tilde{\Phi}_{-})_{12}(k).$

将 (3.63), (3.64) 式和 (3.65) 式代入到 (3.67) 式, 我们最终获得

$\beta_{B}^{12}=\frac{{\rm e}^{\frac{\pi v}{2}-\frac{3\pi {\rm i}}{4}}\nu\Gamma({\rm i}\nu)}{\sqrt{2\pi}}\frac{\gamma_{1}(k_{0})}{1+\gamma_{1}(k_{0})\gamma_{2}(k_{0})}.$

另一方面, 设

$M^{A^{0}}(y,t,k)=I+\int_{\Sigma_{A}}\frac{\left((1-C_{\omega_{A^{0}}})^{-1}I\right)(\xi)\omega_{A^{0}}(\xi)}{\xi-k}\frac{\mathrm{d}\xi}{2\pi {\rm i}}, \quad k\in\mathbb{C}\backslash\Sigma_{A}.$

并假设 $M^{A^{0}}(k)$ 的展式为

$M^{A^{0}}(k)=I-\frac{M^{A^{0}}_{1}}{k}+O(k^{-2}), \quad k\rightarrow\infty.$

因为 (2.21), (3.49), (3.50), (3.51) 式和 (3.52) 式表明跳跃矩阵 $J^{A^{0}}$ 和 $J^{B^{0}}$ 之间存在对称关系为

$J^{A^{0}}(k)=\sigma_{3}\left(J^{B^{0}}(-k^{\ast})\right)^{\ast}\sigma_{3}.$

基于 RH 问题的解是存在且唯一的, 那么

$M^{A^{0}}(k)=\sigma_{3}\left(M^{B^{0}}(-k^{\ast})\right)^{\ast}\sigma_{3}.$

比较 (3.56) 和 (3.69) 式中 $k$ 的系数并结合上式就有

$\label{(M^{A^{0}}_{1})_{12}}\left(M^{A^{0}}\right)_{12}=-\left(M^{B^{0}}\right)_{12}.$

考虑到随着 $t\rightarrow\infty$

$\begin{aligned}&\frac{1}{2\pi {\rm i}}\frac{1}{\sqrt{48tk_{0}}}\int_{\Sigma_{A}}\frac{\left((1-C_{\omega_{A^{0}}})^{-1}I\right)(\xi)\omega_{A^{0}}(\xi)}{N_{A}\xi}\mathrm{d}\xi+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&M^{A^{0}}\left(k_{0}\sqrt{48tk_{0}}\right)-I+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&-\frac{1}{k_{0}\sqrt{48tk_{0}}}M^{A^{0}}_{1}+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right),\end{aligned}$
$\begin{aligned}&\frac{1}{2\pi {\rm i}}\frac{1}{\sqrt{48tk_{0}}}\int_{\Sigma_{A}}\frac{\left((1-C_{\omega_{A^{0}}})^{-1}I\right)(\xi)\omega_{A^{0}}(\xi)}{(N_{A}\xi)^{2}}\mathrm{d}\xi+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&\sqrt{48tk_{0}}\frac{\mathrm{d}M^{A^{0}}}{\mathrm{d}k}\bigg|_{k=k_{0}\sqrt{48tk_{0}}}+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&\frac{1}{k_{0}^{2}\sqrt{48tk_{0}}}M^{A^{0}}_{1}+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right),\end{aligned}$

以及

$\begin{aligned}&\frac{1}{2\pi {\rm i}}\frac{1}{\sqrt{48tk_{0}}}\int_{\Sigma_{B}}\frac{\left((1-C_{\omega_{B^{0}}})^{-1}I\right)(\xi)\omega_{B^{0}}(\xi)}{N_{B}\xi}\mathrm{d}\xi+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&M^{B^{0}}\left(-k_{0}\sqrt{48tk_{0}}\right)-I+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&\frac{1}{k_{0}\sqrt{48tk_{0}}}M^{B^{0}}_{1}+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right),\end{aligned}$
$\begin{aligned}&\frac{1}{2\pi {\rm i}}\frac{1}{\sqrt{48tk_{0}}}\int_{\Sigma_{B}}\frac{\left((1-C_{\omega_{B^{0}}})^{-1}I\right)(\xi)\omega_{B^{0}}(\xi)}{(N_{B}\xi)^{2}}\mathrm{d}\xi+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&\sqrt{48tk_{0}}\frac{\mathrm{d}M^{B^{0}}}{\mathrm{d}k}\bigg|_{k=-k_{0}\sqrt{48tk_{0}}}+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right)\\=&\frac{1}{k_{0}^{2}\sqrt{48tk_{0}}}M^{B^{0}}_{1}+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right).\end{aligned}$

方程 (3.43) 和 (3.44) 可被整理为下列形式

$x=y-\frac{1}{2\pi}\left(\int_{k_{0}}^{+\infty}+\int_{-\infty}^{-k_{0}}\right)\frac{\log(1+\gamma_{1}(s)\gamma_{2}(s))}{s^{2}}\mathrm{d}s+o(1),$
$\begin{aligned}q(y,t)=\left(\frac{1}{k_{0}^{2}\sqrt{48tk_{0}}}\sigma_{3}\left(M^{B^{0}}_{1}\right)^{\ast}\sigma_{3}+\frac{1}{k_{0}^{2}\sqrt{48tk_{0}}}M^{B^{0}}_{1}\right)_{12}+O\left(t^{-1+|\text{Im}\nu(k_{0})|}\log t\right).\end{aligned}$

代入 (3.60), (3.72) 式和 (3.73)式到 (3.74) 式中, 我们最终得到定理1.1.

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