数学物理学报, 2023, 43(4): 1085-1122

聚焦 Kundu-Eckhaus 方程的反散射变换法: 阶跃振荡背景下的长时间渐进性

王贵贤,1, 王秀彬,2, 韩波,1,*

1哈尔滨工业大学 哈尔滨 150001

2中国矿业大学 江苏徐州 221116

Inverse Scattering Transform for the Focusing Kundu-Eckhaus Equation: Long-time Dynamics of the Steplike Oscillating Background

Wang Guixian,1, Wang XiuBin,2, Han Bo,1,*

1School of Mathematics, Harbin Institute of Technology, Harbin 150001

2School of Mathematics and Institute of Mathematical Physics, China University of Mining and Technology, Jiangsu Xuzhou 221116

通讯作者: *韩波, E-mail: bohan@hit.edu.cn

收稿日期: 2022-03-24   修回日期: 2023-01-10  

基金资助: 国家自然科学基金(12271129)
国家自然科学基金(12201622)

Received: 2022-03-24   Revised: 2023-01-10  

Fund supported: NSFC(12271129)
NSFC(12201622)

作者简介 About authors

王贵贤,E-mail:guixianwang@hit.edu.cn;

王秀彬,E-mail:xbwang@cumt.edu.cn

摘要

该文利用非线性速降法研究了阶跃振荡背景下聚焦 Kundu-Eckhaus 方程解的长时间渐进性问题. 在稀疏情况下, 当解趋于 $x$ 轴时, 其渐进性以平面波的形式呈现;当解趋于 $t$ 轴时, 其渐进性以缓慢衰减的形式呈现; 而在两个过渡扇区, 解的渐进性可表示为调制椭圆波函数. 此外, 在激波情况下, 解的渐进性可由依赖于亏格为3的黎曼曲面的超椭圆函数表示.该文所得结论有助于解释存在五次非线性项以及自频移效应的调制不稳定性下的非线性阶段.

关键词: 聚焦 Kundu-Eckhaus 方程; 反散射变换法; Riemann-Hilbert 问题; 非线性速降法

Abstract

In this paper, we study the long-time dynamics of the solution of the focusing Kundu-Eckhaus equation under steplike oscillating background via the nonlinear steepest descent method. In the rarefaction case, when the solution is near the $x$-axis, the form of the leading behavior is the plane waves, when the solution tends to the $t$-axis, the leading behavior decays slowly, and when the solution belongs to two transition sectors, the form of the leading behavior is the elliptic waves. Furthermore, in the shock case, the leading behavior is described by terms of hyperelliptic functions depended on a Riemann surface of genus 3. Our results may be useful to explain the nonlinear stage of modulation instability in presence of the the quintic nonlinear and the self-frequency shift effects.

Keywords: The focusing Kundu-Eckhaus equation; Inverse scattering transform; Riemann-Hilbert problem; The nonlinear steepest descent method

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本文引用格式

王贵贤, 王秀彬, 韩波. 聚焦 Kundu-Eckhaus 方程的反散射变换法: 阶跃振荡背景下的长时间渐进性[J]. 数学物理学报, 2023, 43(4): 1085-1122

Wang Guixian, Wang XiuBin, Han Bo. Inverse Scattering Transform for the Focusing Kundu-Eckhaus Equation: Long-time Dynamics of the Steplike Oscillating Background[J]. Acta Mathematica Scientia, 2023, 43(4): 1085-1122

1 引言

众所周知, 非线性薛定谔 (NLS) 方程

$\begin{matrix}{\rm i}u_{t}+\frac{1}{2}u_{xx}+|u|^{2}u=0\end{matrix}$

在数学物理领域有着重要的意义, 并且广泛应用于描述非线性波现象,如深海[1],非线性光学[2-3],大气[4],等离子体[5],玻色-爱因斯坦凝聚[6-7],甚至金融学[8]等. 然而, 为了描述高阶非线性效应在实际数学和物理系统中的贡献, 在 NLS 方程(1.1)中添加高阶非线性项是不可避免的. 高阶非线性项不仅可以扩展克尔型非线性的范畴,而且改变了输入场强对无线端脉冲宽度的 NLS 方程NLS的影响.因此, 对广义 NLS 方程进行研究是必要也是值得的. 本文主要研究了聚焦 Kundu-Eckhaus (KE) 方程

$\begin{matrix}{\rm i}u_{t}+\frac{1}{2}u_{xx}+|u|^{2}u+2\beta^{2}|u|^4u-2{\rm i}\beta(|u|^2)_{x}u=0,~~x \in \mathbb{R},~t\geq 0,\end{matrix}$

该方程是由 Kundu[9]提出, 其中 $\beta$ 为任意常数, $\beta^{2}$ 表示五次非线性系数, 方程(1.2)左端最后一项表示拉曼效应, 它是产生自频移的原因. 当 $\beta=0$ 时, 方程(1.2)可简化为 NLS 方程(1.1). 通过 Darboux 变换法和广义Darboux 变换法[10-12],聚焦 KE 方程是一个具有 Lax 对、孤子解和怪波解的完全可积系统[13-15].在适当的条件下对方程(1.2)应用反散射变换法,使得产生各种与物理应用[16]相关的解成为可能.

反散射变换法是二十世纪数学的一项重要成就[17-20],在无反射势情况下得到了许多非线性可积方程初值问题的精确解. 然而, 对于一般的初边值问题,并不能求解出显式解. 故针对非线性可积模型解的长时间渐近性的研究变得非常有意义. 近年来, 这一课题取得了许多进展,尤其是用于求解振荡 Riemann-Hilbert 问题的非线性速降法[21-22]. 基于该方法, 关于 NLS 方程(1.1)初边值问题解的长时间渐近性方面的研究成果[23-34]逐渐发展且丰富起来. 同时也涌现了对于具有多分量谱问题的非线性可积系统的相关研究[35-37].

虽然现已存在 KE 方程解的长时间渐进性的一些工作[38-41], 但大都关于初边值条件为零或简单非零边值条件的情况. 因此, 本文利用非线性速降法探索在阶跃振荡背景下聚焦 KE 方程(1.2)解的长时间渐进性. 当 $x\rightarrow\pm\infty$ 时, 初值趋于振荡波形式

$\begin{matrix}u(x,0)=u_{0}(x)\sim\left\{\begin{array}{c} A_{1}{\rm e}^{{\rm i}(\mu_{1}x+\theta_{1})},~~x\rightarrow -\infty, \\ A_{2}{\rm e}^{{\rm i}(\mu_{2}x+\theta_{2})},~~x\rightarrow +\infty, \end{array} \right.\end{matrix}$

其中 $\{A_{j},\mu_{j},\theta_{j}\}_{1}^{2}$ 为实常数且 $A_{j}>0$, $j=1, 2$.

为完善柯西问题(1.2)-(1.3), 边值条件需满足

$\begin{matrix}\int_{0}^{(-1)^{j}\infty}|u(x,t)-u_{0j}(x,t)|{\rm d}x<\infty,~~\forall~t\geq0,~j=1,2,\end{matrix}$

其中, $u_{0j}(x,t)$ 为聚焦 KE 方程(1.2)的平面波解, 形式为

$\begin{matrix}u_{0j}(x,t)=A_{j}{\rm e}^{{\rm i}(\theta_{j}+\mu_{j}x+\omega_{j}t)},~~\omega_{j}:=A_{j}^{2}-\frac{1}{2}\mu_{j}^{2}+2\beta^{2} A_{j}^{4},~~j=1,2.\end{matrix}$

接下来, 给出两条假设

(1) 本文所研究的柯西问题不存在孤子解;

(2) 假设存在某一紧集, 使得初值 $u_{0}(x)$ 在该集合之外仍具有相同的形式,

即存在 $B > 0$, 当 $x < -B$ 时, $u_{0}(x)=u_{01}(x,0)$; 当 $x > B$ 时, $u_{0}(x)=u_{02}(x,0)$.

本节将给出两条重要定理. 在第2节, 主要介绍了反散射变换法和基本的 Riemann-Hilbert 问题, 这是进行一系列推导的基础. 根据 $\tilde{\mu}_{1}\neq \tilde{\mu}_{2}$, 分为稀疏情况 ($\tilde{\mu}_{2}< \tilde{\mu}_{1}$) 和激波情况 ($\tilde{\mu}_{2}> \tilde{\mu}_{1}$). 这两种情况都值得研究. 在第3节, 针对稀疏情况, 将上半平面划分为平面波区域、椭圆波区域以及慢衰减区域, 并分别计算了这三个区域的渐进解.在第4节, 讨论了激波的情况, 利用依赖于亏格为3的 黎曼曲面的超椭圆函数描述了解的长时间渐进性. 最后, 针对所研究问题给出一些结论并引发讨论.

定理 1.1$\tilde{\mu}_{2}< \tilde{\mu}_{1}$ 时, 聚焦 KE 方程(1.2)的解可分为以下三种情况

$\bullet$$\xi<-2\tilde{\mu}_{1}-2\sqrt{2}A_{1}$$\xi>-2\tilde{\mu}_{2}+2\sqrt{2}A_{2}$ 时, 平面波区域的渐进解可表示为

$\begin{matrix}u(x,t)&=&{\rm e}^{-2{\rm i}\beta\int_{x}^{+\infty}\left(|u(y,t)|^{2}-A_{j}^{2}\right){\rm d}y}A_{j}{\rm e}^{{\rm i}[\mu_{j}x+(A_{j}^{2}-\frac{\mu_{j}^{2}}{2}+2\beta^{2}A_{j}^{4})t+\phi_{j}(\xi)]}\\&&+O\left(t^{-\frac{1}{2}}\right),~~(-1)^{j}\xi \gg 0,~j=1, 2,\end{matrix}$

其中, $\phi_{1}(-\infty)=\theta_{1}$, $\phi_{2}(+\infty)=\theta_{2}$.

$\bullet$$-2\tilde{\mu}_{1}-2\sqrt{2}A_{1}<\xi<-2\tilde{\mu}_{1}$$-2\tilde{\mu}_{2}<\xi<-2\tilde{\mu}_{2}+2\sqrt{2}A_{2}$ 时, 椭圆波区域的渐进解的形式为

$ \begin{matrix}u(x, t)&=&{\rm e}^{-2{\rm i}\beta\int_{x}^{+\infty}\left(|u(y,t)|^{2}-A_{j}^{2}\right){\rm d}y}\\&&\cdot(A_{j}+{\rm Im}\beta(\xi))\frac{\Theta\left(\frac{F_{j}t}{2\pi}+\frac{f_{j}}{2\pi}+\chi_{j}-U_{\infty}+b_{j}\right)\Theta\left(U_{\infty}+b_{j}\right)}{\Theta\left(\frac{F_{j}t}{2\pi}+\frac{f_{j}}{2\pi}+\chi_{j}+U_{\infty}+b_{j}\right)\Theta\left(-U_{\infty}+b_{j}\right)} \\&&\cdot{\rm e}^{{\rm i}[\mu_{j}x+(A_{j}^{2}-\frac{\mu_{j}^{2}}{2}+2\beta^{2}A_{j}^{4}+2g(\infty))t+2\hat{G}(\infty)+{\rm e}^{{\rm i}\phi_{j}}]}+O\left(t^{-\frac{1}{2}}\right),~~j=1,2,\end{matrix}$

其中, $b_{j}$, $f_{j}$, $\Theta(\lambda)$, $F_{j}$, $\chi_{j}$, $U_{\infty}$, $g(\infty)$ 以及 $\hat{G}(\infty)$ 在(3.11)和(3.13)式中有定义.

$\bullet$$-2\tilde{\mu}_{1}<\xi<-2\tilde{\mu}_{2}$ 时, 慢衰减区域的渐进解为

$ \begin{matrix}u(x,t)={\rm e}^{-2{\rm i}\beta\int_{x}^{+\infty}\left(|u(y,t)|^{2}-A_{j}^{2}\right){\rm d}y}b_{0}(\xi)t^{-1/2}{\rm e}^{{\rm i}\left[b_{1}(\xi)t+b_{2}(\xi)\log t+b_{3}(\xi)\right]}+o\left(t^{-\frac{1}{2}}\right),\end{matrix}$

其中 $b_{i}$, $i=0, 1, 2, 3$ 由(3.19)式定义.

定理 1.2$\tilde{\mu}_{2}> \tilde{\mu}_{1}$ 时, 设 ${\cal I}$ 为亏格为3的 黎曼曲面上的一个紧子集, 记为 ${\cal I}=(\xi_{1},\xi_{2})$. 对于任意 $\xi\in{\cal I}$, 当 $t\rightarrow\infty$ 时, 聚焦 KE 方程(1.2)在集合 ${\cal I}$ 上的渐进解可表示为

$\begin{matrix}u(x,t)={\rm e}^{-2{\rm i}\beta\int_{x}^{+\infty}\left(|u(y,t)|^{2}-A_{j}^{2}\right){\rm d}y}\left(E_{0}(\xi)+E_{1}(\xi)t^{-1/2}\right)+O\left(t^{-1}\ln t\right),\end{matrix}$

这里误差项关于 $\xi$ 是一致的, $E_{i}$, $i=0, 1$ 分别由(4.20)和(4.42)式定义.

注 1.1 众所周知, 聚焦 KE 方程(1.2)与 NLS 方程(1.1)的 Lax 对形式满足以下关系式

$\begin{matrix} \left\{\begin{array}{c} \Phi_{x}(x,t,\lambda)=\tilde{U}(x,t,\lambda)\Phi(x,t,\lambda), \\ \Phi_{t}(x,t,\lambda)=\tilde{V}(x,t,\lambda)\Phi(x,t,\lambda), \end{array} \right.\end{matrix}$

其中, $\tilde{U}$$\tilde{V}$ 分别表示为

$\begin{array}{l}\tilde{U}=\mathrm{e}^{-\mathrm{i} \beta \int|u|^{2} \mathrm{~d} x \sigma_{3}} U=-\mathrm{i} \lambda \sigma_{3}+\tilde{Q}, \\ \tilde{V}=\mathrm{e}^{-\mathrm{i} \beta \int|u|^{2} \mathrm{~d} x \sigma_{3}} V=-\mathrm{i} \lambda^{2} \sigma_{3}+\lambda \tilde{Q}+\frac{\mathrm{i}}{2}|u|^{2} \sigma_{3}-\frac{\mathrm{i}}{2} \tilde{Q}_{x} \sigma_{3}, \\ \tilde{Q}=\left(\begin{array}{cc}0 & u \mathrm{e}^{-2 \mathrm{i} \beta \int|u|^{2} \mathrm{~d} x} \\ -u^{*} \mathrm{e}^{2 \mathrm{i} \beta \int|u|^{2} \mathrm{~d} x} & 0\end{array}\right).\end{array}$

因此, 方程(1.2)和(1.1)势函数的关系为

$\begin{matrix}u=q{\rm e}^{2{\rm i}\beta\int|q|^2{\rm d}x}.\end{matrix}$

这似乎可以从 NLS 方程的 Lax 对出发, 求得势函数 $q(x,t)$, 然后根据(1.10)式, 从而得到聚焦 KE 方程的解 $u(x,t)$. 然而, 作为创新点之一, 本文直接基于聚焦 KE 方程的 Lax 对, 通过构造并求解相应的 Riemann-Hilbert 问题最终得到解的长时间渐进性. 更重要的是, 有效地避免了求解复杂的积分, 大大缩减了计算量.

2 反散射变换法和 Riemann-Hilbert问题

$\Gamma_{j}=[\bar{C}_{j},C_{j}]$, 其中 $C_{j}:= \tilde{\mu}_{j}+{\rm i}A_{j}$, $\tilde{\mu}_{j}:=-\frac{\mu_{j}}{2}+\beta A_{j}^{2}$, $j=1,2$. 进而 $\Gamma_{j}=\{\tilde{\mu}_{j}+{\rm i}a|~|a|\leq A_{j}\}$ 表示向上的垂直线段. $\Gamma= \mathbb{R} \cup\Gamma_{1}\cup\Gamma_{2}$.

$\tilde{\mu}_{1}\neq \tilde{\mu}_{2}$, 可分为稀疏情况 ($\tilde{\mu}_{2}< \tilde{\mu}_{1}$) 和激波情况 ($\tilde{\mu}_{2}> \tilde{\mu}_{1}$), 如图2.1所示.

图2.1

图2.1   基本的 Riemann-Hilbert 问题的轮廓线 $\Gamma$, 稀疏情况(左图)和激波情况(右图)


记上下复半平面分别为开集 ${\Bbb C}^{+}=\{{\rm Im} \lambda>0\}$${\Bbb C}^{-}=\{{\rm Im} \lambda<0\}$, 黎曼曲面为 $\overline{{\Bbb C}}={\Bbb C}\cup\{\infty\}$. 主分支的对数形式为 $\ln \lambda=\ln |\lambda|+{\rm i}\arg \lambda$, $\arg \lambda \in (-\pi,\pi]$.

除非另有说明, 本文将使用 $z^{\alpha}={\rm e}^{\alpha \ln z}$ 来定义所有复幂. 函数 $f(\lambda)$ 的 Schwarz 共轭形式为 $f^{\ast}(\lambda):=\overline{f(\bar{\lambda})}$. 此外, 边值的左右极限分别用下标 $+/-$ 表示. 泡利矩阵为

$\sigma_{1}:=\left(\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right), \quad \sigma_{2}:=\left(\begin{array}{cc}0 & -\mathrm{i} \\ \mathrm{i} & 0\end{array}\right), \quad \sigma_{3}:=\left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right)$

在以 $\Gamma$ 为界黎曼曲面 $\overline{{\Bbb C}}$ 上定义一个开集 $P$. 由所有函数 $f(\lambda)$ 组成的 Smirnoff 类 $\dot{E}^{2}(P)$ 具有性质

$\bullet$ 在开集 $P$ 上解析;

$\bullet$ 对于每个连通分量 $P_{j}$, 在 $P_{j}$ 中存在曲线 $\{B_{n}\}_{1}^{\infty}$, 使得 $B_{n}$ 覆盖 $P_{j}$ 的每个紧子集, 且满足 $\sup_{n\geq1}\|f\|_{L^{2}(B_{n})}<\infty$.

$E^{\infty}(P)$ 表示有界解析函数 $P\rightarrow{\Bbb C}$ 张成的空间.

$L^{2}$-范数[42,43]意义下, 定义 Riemann-Hilbert 问题

$\begin{matrix}\left\{\begin{array}{c} n\in {\Bbb I}+\dot{E}^{2}({\Bbb C}\setminus\Gamma),~~~~~~~~~~~~ \\ n_{+}(\lambda)=n_{-}(\lambda)J(\lambda), ~\lambda\in \Gamma. \end{array}\right.\end{matrix}$

在复共轭条件下, 由于所有的跳跃曲线均不变, 故跳跃矩阵满足

$\begin{matrix}J\equiv J(\lambda)=\left\{\begin{array}{c} \sigma_{2}J^{\ast}\sigma_{2},~~~~~~\lambda\in \Gamma\setminus\mathbb{R}, \\ \sigma_{2}(J^{\ast})^{-1}\sigma_{2},~~\lambda\in \Gamma\cap\mathbb{R}. \end{array}\right.\end{matrix}$

由 Riemann-Hilbert 问题(2.1)解的唯一性可知

$\begin{matrix}n=\sigma_{2}n^{\ast}\sigma_{2},~~\lambda\in {\Bbb C}\setminus\Gamma.\end{matrix}$

聚焦 KE 方程(1.2)的 Lax 对为

$\begin{matrix}\left\{\begin{array}{c} \Phi_{x}(x,t,\lambda)=X(x,t,\lambda)\Phi(x,t,\lambda), \\ \Phi_{t}(x,t,\lambda)=T(x,t,\lambda)\Phi(x,t,\lambda), \end{array} \right.\end{matrix}$

其中

$\begin{eqnarray*}&&X=-{\rm i}\lambda\sigma_{3}-{\rm i}\beta|u|^2\sigma_{3}+{\cal U},\\&&T=-{\rm i}\lambda^2\sigma_{3}+\lambda{\cal U}+\beta|u|^2{\cal U}+\frac{{\rm i}}{2}\left(|u|^2+4\beta^{2}|u|^4\right)\sigma_{3}-\frac{{\rm i}}{2}{\cal U}_{x}\sigma_{3}-\frac{\beta}{2}\left({\cal U}{\cal U}_{x}-{\cal U}_{x}{\cal U}\right),\\&&{\cal U}(x,t)=\left( \begin{array}{cc} 0 & u(x,t) \\ -\bar{u}(x,t) & 0 \\ \end{array}\right),\end{eqnarray*}$

这里 $u(x,t)=u_{0j}(x,t)$, $j=1, 2$ 由(1.5)式给出.

鉴于 Riemann-Hilbert 问题在反散射变换中发挥着重要的作用, 很自然通过构造合适的 Riemann-Hilbert 问题来求解原柯西问题.

${\Bbb C} \setminus \Gamma_{j}$, $j=1, 2$ 上定义以下解析函数

$\begin{matrix}\left\{\begin{array}{ll} U_{j}(\lambda)=\left[(\lambda+\frac{\mu_{j}}{2}-\beta A_{j}^{2})^{2}+A_{j}^{2}\right]^{\frac{1}{2}},\\[3mm] V_{j}(\lambda)=\left(\lambda-\frac{\mu_{j}}{2}+\beta A_{j}^{2}\right)U_{j}(\lambda),\\[3mm] \nu_{j}(\lambda)=\left(\frac{2\lambda+\mu_{j}-2\beta A_{j}^{2}-2{\rm i}A_{j}}{2\lambda +\mu_{j}-2\beta A_{j}^{2}+2{\rm i}A_{j}}\right)^{\frac{1}{4}},\\[5mm] \zeta_{j}(\lambda)=\left( \begin{array}{cc} \frac{1}{2}(\nu_{j}+\frac{1}{\nu_{j}}) & \frac{1}{2}{\rm e}^{{\rm i}\theta_{j}}(\nu_{j}-\frac{1}{\nu_{j}}) \\[3mm] \frac{1}{2}{\rm e}^{-{\rm i}\theta_{j}}(\frac{1}{\nu_{j}}-\nu_{j}) & \frac{1}{2}(\nu_{j}+\frac{1}{\nu_{j}}) \end{array} \right). \end{array} \right.\end{matrix}$

$\lambda\rightarrow\infty$ 时, 解析函数(2.5)可转化为

$\left\{\begin{array}{ll} U_{j}(\lambda)=\lambda+\frac{\mu_{j}}{2}-\beta A_{j}^{2}+O(\lambda^{-1}),\\[3mm] _{j}(\lambda)=\lambda^{2}+\frac{A_{j}^{2}}{2}-\frac{\mu_{j}^{2}}{4}+\beta^{2} A_{j}^{4}+O(\lambda^{-1}), \\[2mm] \nu_{j}(\lambda)=1+O(\lambda^{-1}),\\ \zeta_{j}(\lambda)={\Bbb I}+O(\lambda^{-1}), \end{array} \right.$

且有对称性

$U_{j}^{\ast}=U_{j}, ~V_{j}^{\ast}=V_{j},~\nu_{j}^{\ast}=\nu_{j}^{-1},~ \zeta_{j}^{\ast}=\sigma_{3}\sigma_{1}\zeta_{j}\sigma_{1}\sigma_{3}.$

进一步地, 构造满足 Lax 对(2.4)的背景解

$\begin{matrix}\Phi_{0j}(x,t,\lambda)&:=&{\rm e}^{\frac{{\rm i}}{2}(\mu_{j}x+\omega_{j}t)\sigma_{3}}M_{j}(\lambda){\rm e}^{-{\rm i}U_{j}\left[x+(\lambda-\frac{\mu_{j}}{2}+\beta A_{j}^{2})t\right]\sigma_{3}},\\M_{j}(\lambda)&:=&{\rm e}^{{\rm i}\frac{\theta_{j}}{2}\hat{\sigma}_{3}}\zeta_{j}(\lambda),~~j=1,2,\end{matrix}$

其中 ${\rm e}^{a\hat{\sigma}}{\cal A}={\rm e}^{a\sigma}{\cal A}{\rm e}^{-a\sigma}$. $\omega_{j}$$U_{j}$ 分别由(1.5)和(2.5)式给出.

对于 $\lambda\in {\Bbb C}\setminus\Gamma_{j}$, $M_{j}$$\Phi_{0j}$ 是解析的, 且满足对称条件(2.3), 故有

$\begin{matrix} \det\Phi_{0j}=\det M_{j}= \det\zeta_{j}\equiv 1.\end{matrix}$

$\Gamma_{j}$ 的传递方向可得, 当 $\lambda\in\Gamma_{j}$ 时, $\nu_{j+}(\lambda)={\rm i}\nu_{j-}(\lambda)$.

解析函数 $M_{j}$, $j = 1, 2$ 满足以下 Riemann-Hilbert 问题

$\left\{\begin{array}{l}M_{j} \in \mathbb{I}+\dot{E}^{2}\left(\mathbb{C} \backslash \Gamma_{j}\right), \\ M_{j+}(\lambda)=M_{j-}(\lambda)\left(\begin{array}{cc}0 & \mathrm{ie}^{\mathrm{i} \theta_{j}} \\ \mathrm{ie}^{-\mathrm{i} \theta_{j}} & 0\end{array}\right), \lambda \in \Gamma_{j}\end{array}\right.$

假设 $u(x,t)$ 是聚焦 KE 方程(1.2)在初边值条件(1.4)下的解, 则 Jost 解可定义为

$\begin{matrix}\Phi_{j}(x,t,\lambda):=T_{j}(x,t,\lambda){\rm e}^{-{\rm i}(U_{j}x+V_{j}t)\sigma_{3}},\end{matrix}$

其中 $U_{j}$, $V_{j}$ 由(2.5)式给出, 且 $T_{j}$, $j=1,2$ 的 Volterra 积分形式为

$\begin{matrix}T_{j}(x,t,\lambda)&=&{\rm e}^{\frac{{\rm i}}{2}(\mu_{j}x+\omega_{j}t)\sigma_{3}}{\rm e}^{{\rm i}\frac{\theta_{j}}{2}\hat{\sigma}_{3}}\zeta_{j}(\lambda)\\&&+\int_{(-1)^{j}\infty}^{x}\Phi_{0j}(x,t,\lambda)\Phi_{0j}^{-1}(y,t,\lambda)\Delta Q_{0j}(y,t)T_{j}(y,t,\lambda){\rm e}^{-{\rm i}U_{j}(y-x)\sigma_{3}}{\rm d}y,\end{matrix}$

这里 $\Delta Q_{0j}=Q-Q_{0j}$.

结合方程(2.8)和(2.9), Jost 解的 Volterra 积分形式为

$\begin{equation}\Phi_{j}(x,t,\lambda)=\Phi_{0j}(x,t,\lambda)+\int_{(-1)^{j}\infty}^{x}\Phi_{0j}(x,t,\lambda)\Phi_{0j}^{-1}(y,t,\lambda)\Delta Q_{0j}(y,t)\Phi_{j}(y,t,\lambda){\rm d}y,\end{equation}$

其中 $\Phi_{0j}(x,t,\lambda)$, $j=1,2$ 由(2.6)式给出.

接下来, 分析矩阵 $T_{j}$, $j=1,2$ 的解析性.

命题 2.1$T^{(i)}_{j}$ 为 矩阵 $T_{j}$ 的第 $i$ 列. $T_{1}^{(1)}$$T_{1}^{(2)}$ 分别在 ${\Bbb C}^{+}\setminus\Gamma_{1}$${\Bbb C}^{-}\setminus\Gamma_{1}$ 上解析并在 $\Gamma_{1}$ 上跳跃变形; $T_{2}^{(2)}$$T_{2}^{(1)}$ 分别在 ${\Bbb C}^{+}\setminus\Gamma_{2}$${\Bbb C}^{-}\setminus\Gamma_{2}$上解析并在 $\Gamma_{2}$ 上跳跃变形.

$\lambda\in \Gamma_{1}\cup\Gamma_{2}$ 时, 通过将(2.5)式中的 $M_{j}$, $U_{j}$, $V_{j}$ 和(2.6)式中的 $\Phi_{0j}$ 分别替换为 $M_{j\pm}$, $U_{j\pm}$, $V_{j\pm}$$\Phi_{0j\pm}$, 从而实现 Jost解 $\Phi_{j\pm}(x,t,\lambda)$ 以及矩阵 $T_{j\pm}$ 的构造, 且矩阵 $T_{j\pm}$ 仍为方程(2.9)的解.根据 $M_{j\pm}$, $U_{j\pm}$$\Phi_{0j\pm}$ 的对称性可得 $T_{j\pm}$$\Phi_{j\pm}$ 也满足(2.3)式.

由于 Jost 解 $\Phi_{j}(x, t, \lambda)$, $j=1,2$ 存在线性关系, 故存在一个不依赖于 $x$$t$ 的散射矩阵 $S(\lambda)$ 使得

$\begin{equation}\Phi_{2}(x, t,\lambda)=\Phi_{1}(x, t, \lambda)S(\lambda),~\lambda \in \mathbb{R},~\lambda\neq \tilde{\mu}_{1}, \tilde{\mu}_{2}.\end{equation}$

根据 $\Phi_{j}(x, t, \lambda)$ 的对称性, 有

$S(\lambda)=\left(\begin{array}{cc}a^{\ast}(\lambda) & b(\lambda) \\ -b^{\ast}(\lambda) & a(\lambda) \\\end{array}\right),~\det S(\lambda)=1.$

由命题 2.1 可知, $a(\lambda)$$a^{\ast}(\lambda)$ 分别在 ${\Bbb C}^{+} \setminus (\Gamma_{1}\cup\Gamma_{2})$${\Bbb C}^{-} \setminus (\Gamma_{1}\cup\Gamma_{2})$ 上解析, 且在 $\Gamma_{1}\cup\Gamma_{2}$ 上跳跃变形. 散射系数有如下渐进条件

$a(\lambda) = 1 + O(1/\lambda),~\lambda\in{\Bbb C}^{+},q b(\lambda) = O(1/\lambda),~\lambda\in \mathbb{R}.$

当(2.11)式中 $t=0$ 时, $a(\lambda)$$b(\lambda)$ 的定义只与初值 $u_{0}(x)$ 有关.

接下来, 介绍基本的 Riemann-Hilbert 问题.

首先定义一个 $2\times2$ 矩阵函数

$\begin{matrix}n(x, t, \lambda):=\left\{\begin{array}{ll} \left(a^{-1}\Phi_{1}^{(1)}~~\Phi_{2}^{(2)}\right){\rm e}^{{\rm i} \lambda(x+\lambda t)\sigma_{3}},&\lambda\in {\Bbb C}^{+}, \\[2mm] \left(\Phi_{2}^{(1)}~~\Phi_{1}^{(2)}(a^{\ast})^{-1}\right) {\rm e}^{{\rm i}\lambda(x+\lambda t)\sigma_{3}}, &\lambda\in {\Bbb C}^{-}, \end{array}\right.\end{matrix}$

这里, $\Phi_{j}(x,t,\lambda)$, $j=1,2$ 由(2.10)式定义. 函数 $n(x, t, \lambda)$ 表示基本的 Riemann-Hilbert 问题的解, 仅与初值 $u_{0}(x)$ 有关, 且满足对称关系(2.3).

矩阵函数 $n(x,t,\lambda)$ 的跳跃矩阵为

$\begin{matrix}J(x,t,\lambda)={\rm e}^{-{\rm i}\lambda(x+\lambda t)\hat{\sigma}_{3}}J_{0}(\lambda).\end{matrix}$

其中 $J_{0}(\lambda)$ 的具体表达式需进一步确定. 由 $n(x,t,\lambda)$ 的对称性可知跳跃矩阵 $J$$J_{0}$ 也满足对称关系(2.2).

根据假设(1), 对于 $\lambda\in \mathbb{R} $$\lambda\neq \tilde{\mu}_{1},~\tilde{\mu}_{2}$, 反射系数可定义为

$\begin{matrix}r(\lambda):=\frac{b^{\ast}(\lambda)}{a(\lambda)},~~\tilde{r}(\lambda):=\frac{b(\lambda)}{a(\lambda)}.\end{matrix}$

结合散射关系(2.11)和反射系数(2.14), 有

$J_{0}(\lambda)=\left(\begin{array}{cc}1+r r^{*} & r^{*} \\ r & 1\end{array}\right)=\left(\begin{array}{cc}1 & r^{*} \\ 0 & 1\end{array}\right)\left(\begin{array}{ll}1 & 0 \\ r & 1\end{array}\right), \lambda \in \mathbb{R}, \lambda \neq \tilde{\mu}_{1}, \tilde{\mu}_{2}$.

考虑到 $n(x, t, \lambda)$$\Gamma_{1}$$\Gamma_{2}$ 上的跳跃变形, 分为以下两种情况

情况(1) $\Gamma_{1}\cap\Gamma_{2}\neq\emptyset$$\Leftrightarrow$$\tilde{\mu}_{1}= \tilde{\mu}_{2}$.

情况(2) $\Gamma_{1}\cap\Gamma_{2}=\emptyset$$\Leftrightarrow$$\tilde{\mu}_{1}\neq\tilde{\mu}_{2}$.

现已存在情况(1)的相关研究[38-41], 故在这里重点探讨另一种情况.

引理 2.1$\tilde{\mu}_{1}\neq \tilde{\mu}_{2}$ 时, $J_{0}$ 可定义为

$J_{0}=\left\{\begin{array}{l}\left(\begin{array}{cc}1 & 0 \\ \frac{\mathrm{ie}^{-\mathrm{i} \theta_{1}}}{a_{+} a_{-}} & 1\end{array}\right), \lambda \in \Gamma_{1} \cap \mathbb{C}^{+}, \\ \left(\begin{array}{cc}\frac{a_{-}}{a_{+}} & \mathrm{ie}^{\mathrm{i} \theta_{2}} \\ 0 & \frac{a_{+}}{a_{-}}\end{array}\right), \lambda \in \Gamma_{2} \cap \mathbb{C}^{+},\end{array} \quad J_{0}=\left\{\begin{array}{c}\left(\begin{array}{cc}1 & \frac{\mathrm{ie} \mathrm{e}^{\mathrm{i} \theta_{1}}}{a_{+}^{*} a_{-}^{*}} \\ 0 & 1\end{array}\right), \lambda \in \Gamma_{1} \cap \mathbb{C}^{-}, \\ \left(\begin{array}{cc}\frac{a_{+}^{*}}{a_{-}^{*}} & 0 \\ \mathrm{ie}^{-\mathrm{i} \theta_{2}} & \frac{a_{-}^{*}}{a_{+}^{*}}\end{array}\right), \lambda \in \Gamma_{2} \cap \mathbb{C}^{-}.\end{array}\right.\right.$

$\lambda\in\Gamma_{1}\cup\Gamma_{2}$ 时,有

$\begin{eqnarray*}&&\Xi_{j}(x, t, \lambda)={\Bbb I}+\int_{(-1)^{j}\infty}^{x}\Upsilon_{0j}(x,y)\times\Delta Q_{0j}(y,t)\Xi_{j}(x, t, \lambda)\Upsilon_{0j}(y,x){\rm d}y,\\&&\Upsilon_{0j}(x,y):=\Phi_{0j}(x,t,\lambda)\Phi_{0j}^{-1}(y,t,\lambda),~j = 1, 2.\end{eqnarray*}$

对于任意给定的 $(y,t)$, 将(2.4)式中的 $u$$u_{0j}$ 代替, 使得函数 $\Upsilon_{0j}(x,y)$$x$ 部分的解. 特别地, 当 $x = y$ 时, 将退化为单位矩阵.由于(2.4)式中的矩阵 $X$ 是关于 $\lambda$ 的多项式, 故 $\Upsilon_{0j}(x,y)$ 也是关于 $\lambda\in\Gamma_{1}\cup\Gamma_{2}$ 的函数. 进而

$\begin{matrix}&&\Phi_{1\pm}=\Xi_{1}\Phi_{01\pm},~~\Phi_{2}=\Xi_{2}\Phi_{02},~~\lambda\in\Gamma_{1},\\&&\Phi_{2\pm}=\Xi_{2}\Phi_{02\pm},~~\Phi_{1}=\Xi_{1}\Phi_{01},~~\lambda\in\Gamma_{2}.\end{matrix}$

引入散射矩阵 $S_{\pm}(\lambda)$, 可得

$\begin{matrix}&&\Phi_{2}(x,t,\lambda)=\Phi_{1\pm}(x,t,\lambda)S_{\pm}(\lambda),~~\lambda\in\Gamma_{1},\\&&\Phi_{2\pm}=\Phi_{1}(x,t,\lambda)S_{\pm}(\lambda),~~\lambda\in\Gamma_{2}.\end{matrix}$

接下来分别讨论 $\lambda\in\Gamma_{1}$$\lambda\in\Gamma_{2}$ 两种情况.

对于 $\lambda\in\Gamma_{1}$, 由(2.17)和(2.18)式可知

$S_{\pm}(\lambda)=(\Phi_{01\pm})^{-1}(x, t,\lambda)\Xi_{1}^{-1}(x, t,\lambda)\Phi_{2}(x, t,\lambda).$

$x = t = 0$, 可得 $S_{\pm}(\lambda)=(M_{1\pm})^{-1}(\lambda)D_{1}(\lambda)$,其中, $M_{1\pm}(\lambda)$ 由(2.6)式 $j$$1\pm$ 时定义,

$\begin{matrix}D_{1}(\lambda):=\Xi_{1}^{-1}(0,0,\lambda)\Phi_{2}(0,0,\lambda).\end{matrix}$

结合(2.7), (2.8)和(2.11)式, 可计算出

$S_{-} S_{+}^{-1}=\left(M_{1-}\right)^{-1} M_{1+}=\left(\begin{array}{cc}0 & \mathrm{ie}^{\mathrm{i} \theta_{1}} \\ \mathrm{ie}^{-\mathrm{i} \theta_{1}} & 0\end{array}\right)$

等价于

$S_{-}(\lambda)=\left(\begin{array}{cc}0 & \mathrm{ie}^{\mathrm{i} \theta_{1}} \\ \mathrm{ie}^{-\mathrm{i} \theta_{1}} & 0\end{array}\right) S_{+}(\lambda), \quad \lambda \in \Gamma_{1}$

更具体地

$\begin{equation}S_{21-}={\rm i}{\rm e}^{-{\rm i}\theta_{1}}S_{11+},~~S_{22-}={\rm i}{\rm e}^{-{\rm i}\theta_{1}}S_{12+},~~\lambda\in\Gamma_{1}.\end{equation}$

根据(2.12)式, 跳跃关系为

$\left(\frac{\Phi_{1+}^{(1)}}{a_{+}} \Phi_{2}^{(2)}\right)=\left(\frac{\Phi_{1-}^{(1)}}{a_{-}} \Phi_{2}^{(2)}\right)\left(\begin{array}{ll}1 & 0 \\ c_{1} & 1\end{array}\right), \lambda \in \Gamma_{1} \cap \mathbb{C}^{+}$,

其中 $c_{1}\equiv c_{1}(\lambda)$.

基于散射关系(2.18)式,有

\begin{matrix} $\Phi_{1\pm}^{(2)}=S_{22\pm}\Phi_{2}^{(1)}-S_{21\pm}\Phi_{2}^{(2)}.\end{matrix}$

由于 $\det\Phi_{2}=1$, (2.23)式可转化为 $\det\left(\Phi_{1\pm}^{(1)}~\Phi_{2}^{(2)}\right)=S_{22\pm}$.

结合(2.11)式, 得到

$\begin{equation}S_{22\pm}=a_{\pm},~~\lambda\in\Gamma_{1}\cap{\Bbb C}^{+}.\end{equation}$

将(2.23)式带入到(2.24)式可得

$\frac{\Phi_{1+}^{(1)}}{a_{+}}-\frac{\Phi_{1-}^{(1)}}{a_{-}}=\left(\frac{S_{21-}}{S_{22-}}-\frac{S_{21+}}{S_{22+}}\right)\Phi_{2}^{(2)}.$

进一步, 考虑(2.21)和(2.22)式, 计算得到

$\begin{matrix}\frac{S_{21-}}{S_{22-}}-\frac{S_{21+}}{S_{22+}}=\frac{{\rm i}{\rm e}^{-{\rm i}\theta_{1}}}{a_{+}a_{-}}=c_{1}.\end{matrix}$

类似地, 对于 $\lambda\in\Gamma_{2}$, 首先定义

$S_{\pm}(\lambda)=\Phi_{1}^{-1}(x,t,\lambda)\Xi_{2}(x,t,\lambda)\Phi_{02\pm}(x,t,\lambda).$

其次, 当 $x = t = 0$ 时, $S_{\pm}(\lambda)=D_{2}(\lambda)M_{2\pm}(\lambda)$,其中

$\begin{matrix}D_{2}(\lambda):= \Phi_{1}^{-1}(0,0,\lambda)\Xi_{2}(0,0,\lambda).\end{matrix}$

基于上述推导过程, 最终求得 $c_{2}={\rm i}{\rm e}^{{\rm i}\theta_{2}}$.证毕.

根据假设(2), 当 $\tilde{\mu}_{1}\neq \tilde{\mu}_{2}$ 时, 存在 $B>0$, 使得初值 $u_{0}(x)$ 等价于

$\begin{matrix}u_{0}(x)\sim\left\{\begin{array}{ll} A_{1}{\rm e}^{{\rm i}(\mu_{1}x+\theta_{1})}, & x<-B, \\ A_{2}{\rm e}^{{\rm i}(\mu_{2}x+\theta_{2})}, & x>B.\end{array}\right.\end{matrix}$

此外, 散射系数 $a(\lambda)$$b(\lambda)$${\Bbb C} \setminus (\Gamma_{1}\cup\Gamma_{2})$ 上解析.

对于 $\lambda\in\Gamma_{1}\cup\Gamma_{2}$, 散射矩阵为

$S_{\pm} \equiv S_{\pm}(\lambda)=\left(\begin{array}{cc} a_{\pm}^{\ast}(\lambda) & b_{\pm}(\lambda) \\ -b_{\pm}^{\ast}(\lambda) & a_{\pm}(\lambda) \\ \end{array} \right).$

结合(2.11)和(2.20)式, 则有

$\begin{matrix}\left\{\begin{array}{ll} a_{+}=-{\rm i}{\rm e}^{-{\rm i}\theta_{1}}b_{-}, \\ b_{+}=-{\rm i}{\rm e}^{{\rm i}\theta_{1}}a_{-}, \end{array}\right.~\lambda\in\Gamma_{1};~~\left\{\begin{array}{ll} a_{+}=-{\rm i}{\rm e}^{-{\rm i}\theta_{2}}b_{-}^{\ast}, \\ b_{+}=-{\rm i}{\rm e}^{{\rm i}\theta_{2}}a_{-}^{\ast}, \end{array}\right.~\lambda\in\Gamma_{2}.\end{matrix}$

同样地, 由 $c_{2}$ 表达式以及(2.25)和(2.28)式可得

$\begin{matrix}r_{+}(\lambda)-r_{-}(\lambda)=\frac{{\rm i}{\rm e}^{-{\rm i}\theta_{1}}}{a_{+}a_{-}},~\lambda\in\Gamma_{1};~~\tilde{r}_{+}(\lambda)-\tilde{r}_{-}(\lambda)=\frac{{\rm i}{\rm e}^{{\rm i}\theta_{2}}}{a_{+}a_{-}},~\lambda\in \Gamma_{2}.\end{matrix}$

跳跃矩阵为

$J_{0}=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & 0 \\ r_{+}-r_{-} & 1\end{array}\right), & \lambda \in \Gamma_{1} \cap \mathbb{C}^{+}, \\ \left(\begin{array}{ll}\frac{a_{-}}{a_{+}} & \left(\tilde{r}_{+}-\tilde{r}_{-}\right) a_{+} a_{-} \\ 0 & \frac{a_{+}}{a_{-}}\end{array}\right), & \lambda \in \Gamma_{2} \cap \mathbb{C}^{+}, \\ \left(\begin{array}{ll}1 & r_{-}^{*}-r_{+}^{*} \\ 0 & 1\end{array}\right), & \lambda \in \Gamma_{1} \cap \mathbb{C}^{-}, \\ \left(\begin{array}{cc}\frac{a_{+}^{*}}{a_{-}^{*}} & 0 \\ \left(\tilde{r}_{-}^{*}-\tilde{r}_{+}^{*}\right) a_{+}^{*} a_{-}^{*} & \frac{a_{-}^{*}}{a_{+}^{*}}\end{array}\right), & \lambda \in \Gamma_{2} \cap \mathbb{C}^{-}.\end{array}\right.$

引理 2.2 散射系数和反射系数满足渐进性

$\begin{eqnarray*}&&a(\lambda)= 1 + O({\rm e}^{4B|{\rm Im}\lambda|}\lambda^{-1}),~~\lambda\rightarrow \infty,~~\lambda\in{\Bbb C},\\&&b(\lambda)= O({\rm e}^{4B|{\rm Im}\lambda|}\lambda^{-1}),~~\lambda\rightarrow \infty,~~\lambda \in {\Bbb C},\\&&r(\lambda)=O({\rm e}^{4B|{\rm Im}\lambda|}\lambda^{-1}),~~\lambda\rightarrow \infty,~~\lambda \in {\Bbb C}\cup\mathbb{R}.\end{eqnarray*}$

具体证明可参考文献[23].

参考引理 2.1 的证明过程, 有

$S_{\pm}(\lambda)=\left\{\begin{array}{ll} {\rm e}^{\frac{{\rm i}\theta_{2}}{2}}\hat{\sigma}_{3}\zeta_{1\pm}^{-1}(\lambda)D_{1}(\lambda), & \lambda\in \Gamma_{1}, \\[2mm] D_{2}(\lambda){\rm e}^{\frac{{\rm i}\theta_{2}}{2}}\hat{\sigma}_{3}\zeta_{2\pm}^{-1}(\lambda), & \lambda\in \Gamma_{2}, \end{array} \right.$

其中 $D_{1}(\lambda)$$D_{2}(\lambda)$ 分别由(2.19)和(2.26)式给出.对于 $\lambda\in{\Bbb C}\setminus\Gamma_{1}\cup\Gamma_{2}$, $D_{j}(\lambda)$ 是解析的, $\Xi_{j}$ 是可积的. 此外, $\theta_{1}$$\theta_{2}$ 分别在 $C_{2}$$C_{1}$ 的邻域内解析. 故 $D_{2}$$D_{1}$ 也分别在 $C_{2}$$C_{1}$ 的邻域内解析.

对于 $C_{2}$ 邻域内的任意 $\lambda$, 有

$a(\lambda)=\frac{1}{2\nu_{2}(\lambda)}\left(-((D_{2}(\lambda)))_{21}{\rm e}^{{\rm i}\theta_{2}}+((D_{2}(\lambda)))_{22}\right)+\frac{\nu_{2}(\lambda)}{2}\left((D_{2}(\lambda))_{21}{\rm e}^{{\rm i}\theta_{2}}+(D_{2}(\lambda))_{22}\right).$

而对于 $C_{1}$ 邻域内的情况通过以下命题给出.

命题 2.2 假设 $(1 + |x|)(u_{0}(x)-u_{01}(x)) \in L^{1}((-\infty, 0])$. 给定 $x\in\mathbb{R} $, $\epsilon \in(0, {\rm Im} C_{1})$, $B_{\epsilon}(C_{1})$$B_{\epsilon}(\bar{C}_{1})$ 分别表示以 $C_{1}$$\bar{C}_{1}$ 为圆心, 半径是 $\epsilon$ 的圆盘. 那么 Jost 解 $\Phi_{1}$ 满足

$\begin{eqnarray*}&&|\Phi_{1}^{(1)}(x,0,\lambda)|\leq B|\lambda-C_{1}|^{-\frac{1}{4}},~~\lambda\in B_{\epsilon}(C_{1})\setminus\Gamma_{1},\\&&|\Phi_{1}^{(1)}(x,0,\lambda)|\leq B|\lambda-\bar{C}_{1}|^{-\frac{1}{4}},~~\lambda\in B_{\epsilon}(\bar{C}_{1})\setminus\Gamma_{1}.\end{eqnarray*}$

对于纯阶跃振荡初值

$u_{0}(x)\sim\left\{\begin{array}{ll} A_{1}{\rm e}^{{\rm i}(\mu_{1}x+\theta_{1})}, & x<0, \\ A_{2}{\rm e}^{{\rm i}(\mu_{2}x+\theta_{2})}, & x>0. \end{array} \right.$

$x = t = 0$ 使得$S(\lambda)=M_{1}^{-1}(\lambda)M_{2}(\lambda)={\rm e}^{\frac{{\rm i}\theta_{1}}{2}\hat{\sigma}_{3}}\zeta_{1}^{-1}(\lambda){\rm e}^{\frac{{\rm i}\theta_{2}}{2}\hat{\sigma}_{3}}\zeta_{2}(\lambda)$.

进一步地

$\begin{eqnarray*}&&a=\frac{1}{4}\left[-{\rm e}^{-{\rm i}(\theta_{1}-\theta_{2})}(\nu_{1}-\nu_{1}^{-1})(\nu_{2}-\nu_{2}^{-1})+(\nu_{1}+\nu_{1}^{-1})(\nu_{2}+\nu_{2}^{-1})\right],\\&&b=\frac{1}{4}\left[{\rm e}^{{\rm i}\theta_{2}}(\nu_{1}+\nu_{1}^{-1})(\nu_{2}-\nu_{2}^{-1})-(\nu_{1}-\nu_{1}^{-1})(\nu_{2}+\nu_{2}^{-1})\right],\end{eqnarray*}$

其中, $\nu_{j}\equiv \nu_{j}(\lambda)$, $j = 1, 2$ 由(2.5)式给出.

$\lambda = C_{1}$ 时, 通过改变由(2.15)和(2.16)式定义的跳跃矩阵 $J_{0}$, 基本的 Riemann-Hilbert 问题可转变为 Riemann-Hilbert 问题(2.1), 并在 $\Gamma_{1}$$\Gamma_{2}$ 端点上所有奇异点的阶数至多是 $|\lambda-C_{j}|^{-\frac{1}{4}}$$|\lambda-\bar{C}_{j}|^{-\frac{1}{4}}$. 对于任意 $\lambda\in {\Bbb C}^{+}\cup\mathbb{R} \setminus\{C_{2}\}$, 假设 $a(\lambda)\neq 0$, 这保证解的唯一性.

接下来给出基本的 Riemann-Hilbert 问题的定义.

定义 2.1 给定 $r(\lambda)$, $\lambda\in \mathbb{R} $, 以及 $a_{\pm}(\lambda)$, $\lambda\in\Gamma_{1}\cup\Gamma_{2}\cap{\Bbb C}^{+}$. 矩阵函数 $n(x, t, \lambda)$$\lambda\in{\Bbb C}\setminus\Gamma$ 上解析且满足以下三个条件

(1) 跳跃条件(2.1), 其中跳跃矩阵 $J_{0}$ 由(2.15)和(2.16)式给出;

(2) 正则化条件: 当 $n(x, t, \lambda)\rightarrow {\Bbb I}$ 时, $\lambda \rightarrow\infty$;

(3) 在 $\Gamma_{1}$$\Gamma_{2}$ 端点上所有奇异点的阶数至多是 $|\lambda-C_{j}|^{-\frac{1}{4}}$$|\lambda-\bar{C}_{j}|^{-\frac{1}{4}}$.

$n(x, t, \lambda)$ 是基本的 Riemann-Hilbert 问题的解, 则柯西问题(1.2)-(1.3)的解 $u(x, t)$ 可表示为

$\begin{matrix}u(x, t){\rm e}^{2{\rm i}\beta\int_{x}^{+\infty}(|u(y,t)|^{2}-A^{2}){\rm d}y}=2{\rm i}\lim_{\lambda \rightarrow\infty}\lambda \left(n(x, t, \lambda)\right)_{12}.\end{matrix}$

3 稀疏情况: $\tilde{\mu}_{2}<\tilde{\mu}_{1}$}

非线性可积系统柯西问题的解可以通过构造并求解相关 Riemann-Hilbert 问题来表示, 这使得非线性速降法能够更好地分析解的长时间渐进性. 据我们所知, 该方法最初是针对初边值条件为零的情况[21]提出的. 为了解决非零初边值的问题, 出现了所谓的 $g$ 函数机制[22], 它与跳跃矩阵的某些项随着 $t$ 趋于无穷远时而呈指数增长或振荡有关.

利用函数 $g(\xi, \lambda)$ 代替最初的势函数

$\begin{matrix}\phi(\xi,\lambda):=\xi \lambda+\lambda^{2},~~\xi:=x/t,\end{matrix}$

从而实现在特定区域上的跳跃变形. 通过对跳跃矩阵进行适当的三角因式分解, 并对基本的Riemann-Hilbert 问题进行相应的变形, 使最初包含指数增长项的跳跃矩阵转变成具有特定结构的分段常数矩阵, 且这些矩阵仅与 $x$$t$ 有关. 而其他跳跃矩阵则指数衰减为单位矩阵.

考虑极限 Riemann-Hilbert 问题的结构, 利用与该问题相关的黎曼函数和黎曼曲面上的 Abel 积分计算得到该问题的显式解. 根据参数 $\xi$ 的取值, 可划分为不同的黎曼曲面.进一步, 基于参数 $A_{j}$, $\beta$ 以及 $\mu_{j}$$(j=1,2)$ 得到不同的情况. 对于每种情况, 选择一组适当的 $g$ 函数来分析其解的渐近行为. 另外, 所有 $g$ 函数都需要满足对称性 $g = g^{\ast}$, 及渐进性

$\begin{matrix}\frac{\partial}{\partial\lambda}g(\xi,\lambda)=\frac{\partial}{\partial \lambda}\phi(\xi,\lambda)+O\left(\lambda^{-2}\right)=2\lambda+\xi+O\left(\lambda^{-2}\right),~~\lambda\rightarrow\infty.\end{matrix}$

由此可推断出水平集 ${\rm Im} g(\xi,\lambda)$ 有两个无穷分支: 实轴以及渐近于垂线 ${\rm Re} \lambda =-\frac{\xi}{2}$ 的分支.实轴与水平集 ${\rm Im} g=0$ 的其他分支的交点称为 ${\rm Im} g$ 的实零点.

图3.1

图3.1   $\xi\gg0\mbox{ 时}$, ${\rm Im}\phi<0$ (粉色区域), ${\rm Im}\phi>0$ (白色区域). 稀疏情况(左图)和激波情况(右图)


基于 $\xi$ 的值以及在 $\Gamma_{1}$$\Gamma_{2}$ 上跳跃矩阵的性质, $g$ 函数被定义为

$\begin{matrix}g_{j}(\xi,\lambda):=\left[\left(\lambda-\frac{\mu_{j}}{2}+\beta A_{j}^{2}\right)+\xi\right]\sqrt{(\lambda+\frac{\mu_{j}}{2}-\beta A_{j}^{2})^{2}+A_{j}^{2}},\end{matrix}$

其中, 当 $\xi\ll 0$ 时, $j = 1$, 当 $\xi\ll 0$ 时, $j = 1$.

因此, 除了实轴外, 水平集 ${\rm Im} g=0$ 存在渐近于直线 ${\rm Re} \lambda =-\frac{\xi}{2}$ 的无穷分支, 以及连接 $C_{j}$$\bar{C}_{j}$ 的有界分支 $\Gamma_{j}$.

注3.1 接下来, 考虑包含相应 $g$ 函数定义的平方根分支切割的选择时, 需要将复平面 $\lambda$ 划分为两个区域. 特别地, 连接 $C_{j}$$\bar{C}_{j}$$g_{j}$ 被定义成线段 $(C_{j},\bar{C}_{j})$.

定义 $B_{1}=\frac{\mu_{1}}{2}-\beta A_{1}^{2}-\sqrt{2}A_{1}$$B_{2}=\frac{\mu_{2}}{2}-\beta A_{2}^{2}+\sqrt{2}A_{2}$. 上半复平面可被划分为五个区域, 如图3.2所示.

图3.2

图3.2   稀疏情况下的区域划分


$\xi<2B_{1}$$\xi>2B_{2}$ 时, 将探索平面波区域的渐进解.

假设 $\xi\gg 0$, 令 $g \equiv g_{2}(\xi, \lambda)$ 为该区域上的一个 $g$ 函数, 则关于 $\lambda$ 的导数表示为

$\begin{matrix}\frac{\partial}{\partial \lambda}g(\xi,\lambda)=2\frac{(\lambda-d_{1}(\xi))(\lambda-d_{2}(\xi))}{\sqrt{(\lambda-C_{2})(\lambda-\bar{C}_{2})}}.\end{matrix}$

${\rm Im}g_{2}(\xi,\lambda)=0$ 可计算出

$\begin{eqnarray*}&&d_{1}\equiv d_{1}(\xi)=-\frac{\mu_{2}-2\beta A_{2}^{2}}{4}-\frac{\xi}{4}-\frac{1}{4}\sqrt{(\xi-\mu_{2}+2\beta A_{2}^{2})^{2}-8A_{2}^{2}}\,\\&&d_{2}\equiv d_{2}(\xi)=-\frac{\mu_{2}-2\beta A_{2}^{2}}{2}-\frac{\xi}{4}+\frac{1}{4}\sqrt{(\xi-\mu_{2}+2\beta A_{2}^{2})^{2}-8A_{2}^{2}}\.\end{eqnarray*}$

此外 $\xi>-2d_{1}>-2d_{2}>\mu_{2}-2\beta A_{2}^{2}$.

图3.3

图3.3   $\xi>\xi_{m}$ (左图), $\xi=\xi_{m}$ (右图)


定义 $n^{(1)}$

$\begin{matrix}n^{(1)}(x,t,\lambda):={\rm e}^{-{\rm i}tg_{j}^{(0)}(\xi)\sigma_{3}}n(x,t,\lambda){\rm e}^{-{\rm i}t(g_{j}(\xi,\lambda)-\phi(\xi,\lambda))\sigma_{3}},~~j=1, 2,\end{matrix}$

其中

$\begin{matrix}&&g_{j}^{(0)}(\xi):=\omega_{j}+ \frac{\xi\mu_{j}}{2}=A_{j}^{2}-\frac{\mu_{j}^{2}}{2}+2\beta^{2}A_{j}^{4}+\frac{\xi\mu_{j}}{2},\\&&g_{j}(\xi,\lambda)=\lambda^{2}+\xi \lambda+g_{j}^{(0)}(\xi)+O\left(\lambda^{-1}\right),~~\lambda\rightarrow\infty.\end{matrix}$

对于 $\lambda\in\Gamma$, $n_{+}^{(1)}(x,t,\lambda)=n_{-}^{(1)}(x,t,\lambda)J^{(1)}(x,t,\lambda)$.

更具体地, 当 $\lambda\in\Gamma_{1}$ 时, 随着 $t \rightarrow\infty$, 跳跃矩阵 $J^{(1)}(x,t,\lambda)$ 可退化成单位矩阵.

$\lambda\in\Gamma_{2}\cap{\Bbb C^{+}}$ 时, 由(2.14)和(2.28)式可得

$\begin{matrix}J^{(1)}(x,t,\lambda)&=&\left( \begin{array}{cc} \frac{a_{-}(\lambda)}{a_{+}(\lambda)}{\rm e}^{{\rm i}t(g_{2+}(\xi,\lambda)-g_{2-}(\xi,\lambda))} & {\rm i}{\rm e}^{{\rm i}\theta_{2}} \\ 0 & \frac{a_{+}(\lambda)}{a_{-}(\lambda)}{\rm e}^{-{\rm i}t(g_{2+}(\xi,\lambda) -g_{2-}(\xi,\lambda))} \\ \end{array} \right)\\&=&\left( \begin{array}{cc} 1 & 0 \\ -r_{-}(\lambda){\rm e}^{2{\rm i}tg_{2-}(\xi,\lambda)} & 1 \\ \end{array} \right)\left( \begin{array}{cc} 0 & {\rm i}{\rm e}^{{\rm i}\theta_{2}} \\ {\rm i}{\rm e}^{-{\rm i}\theta_{2}} & 0 \\ \end{array} \right)\left( \begin{array}{cc} 1 & 0 \\ r_{+}(\lambda){\rm e}^{2{\rm i}tg_{2+}(\xi,\lambda)} & 1 \\ \end{array} \right).\end{matrix}$

类似地, 当 $\lambda\in\Gamma_{2}\cap{\Bbb C^{-}}$ 时, 有

$J^{(1)}(x, t, \lambda)=\left(\begin{array}{cc}1 & 0 \\ -r_{+}(\lambda) \mathrm{e}^{2 \mathrm{i} t g_{2+}(\xi, \lambda)} & 1\end{array}\right)\left(\begin{array}{cc}0 & \mathrm{ie}^{\mathrm{i} \theta_{2}} \\ \mathrm{ie}^{-\mathrm{i} \theta_{2}} & 0\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ r_{-}(\lambda) \mathrm{e}^{2 \mathrm{i} t g_{2-}(\xi, \lambda)} & 1\end{array}\right)$

根据文献[27,28], 可得两个修正的 Riemann-Hilbert 问题的解 $n^{mod_{-j}}$, $j=1,2$.

进一步地, 聚焦 KE 方程(1.2)在平面波区域的渐进解 $u(x,t)$ 可表示为

$\begin{matrix}u(x, t)&=&{\rm e}^{-2{\rm i}\beta\int_{x}^{+\infty}\left(|u(y,t)|^{2}-A_{j}^{2}\right){\rm d}y}A_{j}{\rm e}^{{\rm i}[\mu_{j}x+(A_{j}^{2}-\frac{\mu_{j}^{2}}{2}+2\beta^{2}A_{j}^{4})t+\phi_{j}(\xi)]}\\&&+O\big(t^{-\frac{1}{2}}\big),~~(-1)^{j}\xi \gg 0,~~j=1, 2,\end{matrix}$

其中 $\phi_{1}(-\infty)=\theta_{1}$, $\phi_{2}(+\infty)=\theta_{2}$.

$2B_{1}<\xi<-2\tilde{\mu}_{1}$$-2\tilde{\mu}_{2}<\xi<2B_{2}$ 时, 将探索椭圆波区域的渐进解.由于两个椭圆波区域推导过程类似, 接下来只分析 $\xi\in(-2\tilde{\mu}_{2},2B_{2})$ 的情况.

$\xi$ 逐渐逼近 $2B_{2}$ 时, 引入一个新的 $g$ 函数. $\frac{\partial g}{\partial \lambda}$ 表示从右平面波扇区到相邻椭圆波扇区的过渡, 形式为

$\begin{matrix}\frac{\partial}{\partial \lambda}g(\xi,\lambda)=2\frac{(\lambda-d(\xi))(\lambda-\varrho(\xi))(\lambda-\bar{\varrho}(\xi))}{\sqrt{(\lambda-C_{2})(\lambda-\bar{C}_{2})(\lambda-\varrho(\xi))(\lambda-\bar{\varrho}(\xi))}}.\end{matrix}$

此外参数 $d(\xi)$$\varrho(\xi)$ 有以下性质

(1) 当 $\lambda =\infty$ 时, 满足渐进性(3.2);

(2) 正则性 $\int_{C_{2}}^{\bar{C}_{2}}{\rm d}g=0$.

参考 Riemann-Hilbert 问题在文献[25]中进一步变形的过程.

$\lambda\in \Gamma_{j},~j=1,2$ 时, 有

$\begin{matrix}\left\{\begin{array}{ll} n^{mod}\in {\Bbb I}+\dot{E}^{2}({\Bbb C}\setminus(\Gamma_{1}\cup\Gamma_{2})),\\ n_{+}^{mod}(\lambda)={\rm i}n_{-}^{mod}(\lambda)\left(\begin{array}{cc} 0 & {\rm e}^{{\rm i}(P_{j}x+G_{j}t)}{\rm e}^{\theta_{j}} \\ {\rm e}^{-{\rm i}(P_{j}x+G_{j})}{\rm e}^{\theta_{j}} & 0 \\ \end{array} \right). \end{array}\right.\end{matrix}$

由(2.5)式, 可得 $\nu_{2}-\frac{1}{\nu_{2}}$ 的唯一零点 $\lambda_{2}$.

$\Lambda_{1}(\lambda_{2})$$\Lambda_{2}(\lambda_{2})$ 分别为 $\lambda_{2}$ 在第一和第二椭圆黎曼面上的原像, 则

$\begin{matrix}b_{2}=-\int_{\bar{C}_2}^{\Lambda_{2}(\lambda_{2})}\frac{w_{0}}{U_{2}(\lambda)}{\rm d}\lambda.\end{matrix}$

其中实常数 $w_{0}=\big(\oint_{\Gamma_{2}}\frac{{\rm d}\lambda}{U_{2}(\lambda)}\big)^{-1}$.

因此, 聚焦 KE 方程(1.2)在该区域渐进解可用亏格为 1 黎曼曲面

$\begin{matrix}f_{2}^{2} =(\lambda-C_{2})(\lambda-\bar{C}_{2})(\lambda-{\varrho}(\xi))(\lambda-\bar{\varrho}(\xi))\end{matrix}$

上调制椭圆波来表示, 其表达式为

$\begin{eqnarray*}u(x, t)&=&{\rm e}^{-2{\rm i}\beta\int_{x}^{+\infty}\left(|u(y,t)|^{2}-A_{2}^{2}\right){\rm d}y}\\&&\cdot(A_{2}+{\rm Im}\beta(\xi))\frac{\Theta\left(\frac{F_{2}t}{2\pi}+\frac{f_{2}}{2\pi}+\chi_{2}-U_{\infty}+b_{2}\right)\Theta\left(U_{\infty}+b_{2}\right)}{\Theta\left(\frac{F_{2}t}{2\pi}+\frac{f_{2}}{2\pi}+\chi_{2}+U_{\infty}+b_{2}\right)\Theta\left(-U_{\infty}+b_{2}\right)} \\&&\cdot{\rm e}^{{\rm i}[\mu_{2}x+(A_{2}^{2}-\frac{\mu_{2}^{2}}{2}+2\beta^{2}A_{2}^{4}+2g(\infty))t+2\hat{G}(\infty)+{\rm e}^{{\rm i}\phi_{2}}]}+O\left(t^{-\frac{1}{2}}\right),\end{eqnarray*}$

其中关于 $\tau\equiv\tau(\xi)$ 的 Jacobi theta 函数$\Theta(z)=\sum_{m \in \mathbb{Z}} \mathrm{e}^{2 \mathrm{i} \pi\left(\frac{1}{2} \tau m^{2}+m z\right)}, z \in \mathbb{C}$

$\begin{matrix}&&F_{2}=\left(\int_{C_2}^{\varrho}+\int_{\bar{C}_2}^{\bar{\varrho}}\right){\rm d}g,~U_{2}(\lambda)=\left[(\lambda+\frac{\mu_{2}}{2}-\beta A_{2}^{2})^{2}+A_{2}^{2}\right]^{\frac{1}{2}},\\&&\chi_{2}=-\ln({\rm i}{\rm e}^{-{\rm i}\phi_{2}})/(2\pi{\rm i}),~\hat{G}(\infty)=\frac{1}{2\pi {\rm i}}\int_{\Gamma_{2}}\frac{G(\lambda,f)}{U_{2}(\xi)}\xi{\rm d}\xi,\\&&U(\infty)=\int_{\bar{C}_2}^{\infty}\frac{w_{0}}{U_{2}(\lambda)}{\rm d}\lambda,~g(\infty)=\frac{1}{2\pi^{2}{\rm i}}\int_{\Gamma_{2}}\frac{1}{U_{2}(k)}\int_{-\infty}^{{\rm Re} \lambda}\frac{\ln(1+r^{2}(k))}{\xi-k}{\rm d}\xi {\rm d}k.\end{matrix}$

$-2\tilde{\mu}_{1}<\xi<-2\tilde{\mu}_{2}$ 时, 将探索慢衰减区域的渐进解.

首先定义一个 $g$ 函数使得在点 $\tilde{\mu}_{2}$ 处的导数满足渐进性(3.2). 当 $g(\xi, \lambda) =\phi(\xi, \lambda)$ 时, 所研究的柯西问题等价于求解零边值条件下聚焦 KE 方程的渐进解. 故有 $u(x, t) = O\big(t^{-\frac{1}{2}}\big)$. 为了得到在临界点 $\lambda = -\frac{\xi}{2} \in\mathbb{R} $ 上更精确的渐近解 $u(x,t)$, 需要进一步分析.

图3.4

图3.4   慢衰减情况下, ${\rm Im}\phi<0$ (粉色区域), ${\rm Im}\phi>0$ (白色区域)


根据文献[21], 通过对基本的 Riemann-Hilbert 问题变形可得到一个新的可解的 Riemann-Hilbert 问题.

定义 $\Gamma_{p} :=(-\infty, -\frac{\xi}{2})\cup \Gamma_{2}$. 设函数 $p(\lambda)\equiv p(\xi, \lambda)$ 是该 Riemann-Hilbert 问题在定义域 $\Gamma_{p}$ 的解, 则有跳越关系 $p_{+}(\lambda) = p_{-}(\lambda)J_{p}(\lambda)$,其中跳跃矩阵为

$\begin{matrix}J_{p}=\left\{\begin{array}{ll} 1+|r|^{2},~~&\lambda\in (-\infty, -\frac{\xi}{2}), \\[2mm] \frac{a_{-}}{a_{+}},~~~~&\lambda\in \Gamma_{2}\cap{\Bbb C}^{+}, \\[3mm] \frac{a_{+}^{\ast}}{a_{-}^{\ast}},~&\lambda\in \Gamma_{2}\cap{\Bbb C}^{-}, \end{array} \right.\end{matrix}$

此外, 当 $\lambda\rightarrow\infty$ 时, 有正则化条件 $p(\lambda)\rightarrow 1$.

根据柯西积分公式, 可得$p(\lambda)=\exp\left\{\frac{1}{2\pi{\rm i}}\int_{\Gamma_{p}}\frac{\log J_{p}(k)}{k-\lambda}{\rm d}k\right\}=p_{0}p_{1}p_{2}$,其中

$\begin{eqnarray*}&&p_{0}=\exp\left\{\frac{1}{2\pi{\rm i}}\int_{-\infty}^{-\frac{\xi}{2}}\frac{\log 1+|r(k)|^{2}}{k-\lambda}{\rm d}k\right\},\\&&p_{1}=\exp\left\{\frac{1}{2\pi{\rm i}}\int_{0}^{{\rm i}\infty}\frac{\log \frac{a_{-}(k)}{a_{+}(k)}}{k-\lambda}{\rm d}k\right\},\\&&p_{2}=\exp\left\{\frac{1}{2\pi{\rm i}}\int_{-{\rm i}\infty}^{0}\frac{\log \frac{a_{+}^{\ast}(k)}{a_{-}^{\ast}(k)}}{k-\lambda}{\rm d}k\right\}.\end{eqnarray*}$

$\lambda_{0}=-\frac{\xi}{2}$, 当 $\lambda\rightarrow \lambda_{0}$ 时, $p_{0}(\lambda)=\left(\lambda+\frac{\xi}{2}\right)^{{\rm i}\nu(-\frac{\xi}{2})}{\rm e}^{\chi(\lambda)}$,其中

$\begin{eqnarray*}&&\nu(-\frac{\xi}{2})=-\frac{1}{2\pi}\log(1+|r(-\frac{\xi}{2})|^{2})\in\mathbb{R},\\&&\chi(\lambda)=-\frac{1}{2\pi{\rm i}}\int_{-\infty}^{-\frac{\xi}{2}}\log(\lambda-k){\rm d}(1+|r(k)|^{2})\in {\rm i}\mathbb{R}.\end{eqnarray*}$

进一步地,$p_{1}(\lambda_{0})p_{2}(\lambda_{0})=\exp\bigg \{\frac{{\rm i}}{\pi}{\rm Im}\int_{0}^{\infty}\frac{\log \frac{a_{-}({\rm i}\tau)}{a_{+}({\rm i}\tau)}}{{\rm i}\tau+\frac{\xi}{2}}{\rm d}\tau\bigg\}$.

基于以上分析, 对基本的 Riemann-Hilbert 问题进行第一次变形.

$\lambda\in {\Bbb C}\setminus\Gamma$ 时, 定义

$n^{(1)}(x,t,\lambda):=n(x,t,\lambda)p(\xi,\lambda)^{-\sigma_{3}},$

其满足 $n_{+}^{(1)}(x,t,\lambda):=n_{-}^{(1)}(x,t,\lambda){\rm e}^{-{\rm i}t\phi(\xi,\lambda)\hat{\sigma}_{3}}J_{0}^{(1)}(\lambda)$,跳跃矩阵为 $J_{0}^{(1)} = p_{-}^{\sigma_{3}}J_{0}p_{+}^{-\sigma_{3}}$.

$\lambda\in \Gamma_{1}\cup\Gamma_{2}$ 时, $J_{0}^{(1)}$ 为三角矩阵, 且对角线上均为单位矩阵.

$\lambda\in\mathbb{R} $ 时, 则是由矩阵的乘积组成

$J_{0}^{(1)}=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & r^{*} p^{2} \\ 0 & 1\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ r p^{-2} & 1\end{array}\right), & \lambda \in\left(-\frac{\xi}{2},+\infty\right), \\ \left(\begin{array}{cc}1 & 0 \\ \frac{r}{p_{+} p_{-}} & 1\end{array}\right)\left(\begin{array}{cc}1 & r^{*} p_{+} p_{-} \\ 0 & 1\end{array}\right), & \lambda \in\left(-\infty,-\frac{\xi}{2}\right), \\ \left(\begin{array}{cc}1 & 0 \\ \frac{\mathrm{ie}^{-\mathrm{i} \theta_{1}}}{a_{+} a_{-}} & 1\end{array}\right), & \lambda \in \Gamma_{1} \cap \mathbb{C}^{+}, \\ \left(\begin{array}{ll}1 & \mathrm{ie}^{\mathrm{i} \theta_{2}} p_{+} p_{-} \\ 0 & 1\end{array}\right), & \lambda \in \Gamma_{2} \cap \mathbb{C}^{+}, \\ \sigma_{2} J_{0}^{(1) *} \sigma_{2}, & \lambda \in\left(\Gamma_{1} \cap \Gamma_{2}\right) \cap \mathbb{C}^{-}.\end{array}\right.$

第二次变形旨在减少在以 $-\frac{\xi}{2}$ 为中心的 $\Sigma_{cr}=\bigcup\limits_{j=1}^{4}L_{j}$ 上的跳跃矩阵. 首先定义

$\begin{matrix}n^{(2)}(x,t,\lambda):=n^{(1)}(x,t,\lambda){\rm e}^{-{\rm i}t\phi(\xi,\lambda)\hat{\sigma}_{3}}G(\lambda),\end{matrix}$

其中

$G \equiv G(\lambda)=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & 0 \\ -r p^{-2} & 1\end{array}\right), & \lambda \in P_{1}, \\ \mathbb{I}, & \lambda \in P_{2}, \\ \left(\begin{array}{cc}1 & -\tilde{r} a^{2} p^{2} \\ 0 & 1\end{array}\right), & \lambda \in P_{3}, \\ \left(\begin{array}{cc}1 & 0 \\ \tilde{r}^{*} a^{* 2} p^{-2} & 1\end{array}\right), & \lambda \in P_{4}, \\ \mathbb{I}, & \lambda \in P_{5}, \\ \left(\begin{array}{cc}1 & r^{*} p^{2} \\ 0 & 1\end{array}\right), & \lambda \in P_{6},\end{array}\right.$

反射系数 $r$$\tilde{r}$ 由(2.14)式给出.

然后, 满足跳越关系

$\begin{matrix}n_{+}^{(2)}(x,t,\lambda):=n_{-}^{(2)}(x,t,\lambda){\rm e}^{-{\rm i}t\phi(\xi,\lambda)\hat{\sigma}_{3}}J_{0}^{(2)}(\lambda),~~\lambda\in\Gamma\cup \Sigma_{cr},\end{matrix}$

跳跃矩阵为 $J_{0}^{(2)}=G_{-}^{-1}J_{0}^{(1)}G_{+}$,且有如下性质

$\bullet$ 对于 $\lambda\in\mathbb{R} $, $J_{0}^{(2)}= {\Bbb I}$;

$\bullet$ 对于 $\lambda \in \Sigma_{cr}$,

$J_{0}^{(2)}=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & 0 \\ r p^{-2} & 1\end{array}\right), & \lambda \in L_{1}, \\ \left(\begin{array}{cc}1 & \tilde{r} a^{2} p^{2} \\ 0 & 1\end{array}\right), & \lambda \in L_{2}, \\ \left(\begin{array}{cc}1 & 0 \\ -\tilde{r}^{*} a^{* 2} p^{-2} & 1\end{array}\right), \lambda \in L_{3}, \\ \left(\begin{array}{cc}1 & -r^{*} p^{2} \\ 0 & 1\end{array}\right), & \lambda \in L_{4}.\end{array}\right.$

通过将 $n^{(2)}$ 满足的 Riemann-Hilbert 问题与零边值条件的情况[21]进行对比, 发现唯一差异是近似中的一个只与 $\xi$ 有关的额外因素 $p(\lambda)$.

$\lambda \rightarrow -\frac{\xi}{2}$ 时, $p(\lambda)\sim\left(\lambda +\frac{\xi}{2}\right)^{{\rm i}\nu(-\frac{\xi}{2})}{\rm e}^{\tilde{\chi}(-\frac{\xi}{2})}$, 这里

$\tilde{\chi}(-\frac{\xi}{2})=\chi(-\frac{\xi}{2})+\frac{1}{\pi {\rm i}}\Im\int_{0}^{+\infty}\frac{\log\frac{a_{-}({\rm i}\tau)}{a_{+}({\rm i}\tau)}}{{\rm i}\tau+\frac{\xi}{2}}{\rm d}\tau.$

最后, 聚焦 KE 方程(1.2)在慢衰减区域的渐进解被定义为(1.8)式, 其中参数 $b_{j}$, $j=0, 1, 2, 3$ 表示为

$\begin{matrix}&&b_{0}=\sqrt{\frac{\log(1+|r(-\frac{\xi}{2})|^{2})}{4\pi}},~~b_{1}=\frac{\xi^{2}}{2},~~b_{2}=-\nu(-\frac{\xi}{2}),\\&&b_{3}=-3\log2\nu(-\frac{\xi}{2})+\arg\Gamma({\rm i}\nu(-\frac{\xi}{2}))-\arg r(-\frac{\xi}{2})-2{\rm i}\tilde{\chi}(-\frac{\xi}{2})+\frac{\pi}{4}.\end{matrix}$

4 激波情况: $\tilde{\mu}_{2}>\tilde{\mu}_{1}$

首先定义 $\hat{n}(x,t,\lambda):=n(x,t,\lambda)\nu_{1}^{\sigma_{3}}(\lambda),$其中 $\nu_{1}(\lambda)$ 由(2.5)式给出. 令 $\hat{a}:=a\nu_{1}$, $\hat{b}:=b\nu_{1}$, $\hat{r}:=\frac{\hat{b}^{\ast}}{\hat{a}}=\frac{b^{\ast}\nu_{1}^{\ast}}{a\nu_{1}}=r\nu_{1}^{-2}$.然后, 计算得到 $\hat{a}\hat{a}^{\ast}=aa^{\ast}$, $\hat{b}\hat{b}^{\ast}=bb^{\ast}$$|\hat{r}|^{2}=\hat{r}\hat{r}^{\ast}=rr^{\ast}=|r|^{2}$.

函数 $\hat{n}(x,t,\lambda)$ 的跳越关系通过跳跃矩阵

$\begin{matrix}\hat{J}_{0}=\nu_{1-}^{\sigma_{3}}J_{0}\nu_{1+}^{-\sigma_{3}}\end{matrix}$

表示, 其中 $J_{0}$ 由引理2.1给出. 此外,函数 $\hat{n}(x,t,\lambda)$ 和跳跃矩阵 $\hat{J}$ 满足对称性条件(2.2)和(2.3).

那么柯西问题(1.2)-(1.3)的解 $u(x,t)$ 可重述为

$\begin{matrix}u(x, t){\rm e}^{2{\rm i}\beta\int_{x}^{+\infty}(|u(y,t)|^{2}-A^{2}){\rm d}y}=2{\rm i}\lim_{\lambda \rightarrow\infty}\lambda\left(\hat{n}(x, t, \lambda)\right)_{12}.\end{matrix}$

针对激波情况, 需要分析解在亏格为 3的 黎曼曲面上的渐进性. 在该区域, $g$ 函数的导数存在5个零点, 分别为一个实数零点 $d\equiv d(\xi)$, 两对复共轭零点 $\{\alpha,\bar{\alpha}\}$$\{\varrho,\bar{\varrho}\}$, 其中 $\alpha\equiv \alpha(\xi)$$\varrho\equiv \varrho(\xi)$ 属于集合 ${\Bbb C}^{+}\setminus(\Gamma_{1}\cup\Gamma_{2})$. 接下来需要在每个跳跃曲线上定义合适的 $g$ 函数.

图4.1 中, 当 $A_{1} = A_{2} = 3/2$, $\xi = 1$ 时, 水平集 ${\rm Im} g = 0$.$\varrho$$\alpha$, $\bar{\alpha}$$\bar{\varrho}$, 以及 $\bar{\varrho}$$\varrho$ 的有向线段分别表示为 $\gamma_{(\varrho,\alpha)}$, $\gamma_{(\bar{\alpha},\bar{\varrho})}$$\gamma_{(\bar{\varrho},\varrho)}$.超椭圆曲面 $N\equiv N_{\alpha\varrho}$ 由所有点 $P := (\omega,\lambda)\in {\Bbb C}^{2}$ 构成, 且该曲面是紧的, 包含无穷远处的点以及

$\omega^{2}(\lambda) =(\lambda-C_{1})(\lambda-\bar{C}_{1})(\lambda-C_{2})(\lambda-\bar{C}_{2})(\lambda-\alpha(\xi))(\lambda-\bar{\alpha}(\xi))(\lambda-\varrho(\xi))(\lambda-\bar{\varrho}(\xi)).$

图4.1

图4.1   跳跃曲线(黑色实线), ${\rm Im} g <0$ (粉色区域), ${\rm Im} g >0$ (白色区域)


若分支曲线集 ${\cal C}:=\Gamma_{1}\cup\Gamma_{2}\cup\gamma_{(\varrho,\alpha)}\cup\gamma_{(\bar{\alpha},\bar{\varrho})}$ 包含 8 个分支点, 那么超椭圆曲面 $N$ 为复平面.当 $\lambda\in \overline{{\Bbb C}}\setminus {\cal C}$ 时, $N$ 的上下半平面分别记为 $\lambda^{+}$$\lambda^{-}$, 而且当 $\lambda\rightarrow\infty$ 时, 有$\omega = \lambda^{4} + O(\lambda^{3})$$\omega = -\lambda^{4} + O(\lambda^{3})$.$\varpi=(\varpi_{1}, \varpi_{2}, \varpi_{3})$${\cal H}^{1}(N)$ 对偶正则同调基 $\{a_{j},b_{j}\}_{1}^{3}$ 的正则化的基, 则有 $\int_{a_{i}}\varpi_{j}{\rm d}\lambda=\delta_{ij}$, 其中 $\varpi_{j}=\sum\limits_{l=1}^{3}A_{jl}\hat{\varpi}_{l}$,$\hat{\varpi}_{l}:=\frac{\lambda^{l-1}}{\omega}$,$\left(A^{-1}\right)_{jl}=\int_{a_{j}}\hat{\varpi}_{l}{\rm d}\lambda$,以及 $\tau_{jl}:=\int_{b_{j}}\varpi_{l}{\rm d}\lambda$$3\times3$ 周期矩阵.

对于 $z\in {\Bbb C}^{3}$, 关于 $\tau$ 的 Jacobi theta 函数 $\Theta(z):=\sum\limits_{M\in{\Bbb Z}^{3}}{\rm e}^{2\pi{\rm i}(\frac{1}{2}M^{T}\tau M+M^{T}z)}$, 满足

$\begin{eqnarray*}&&\Theta(z)=\Theta(-z),~~\Theta(z+{\rm e}^{(j)})=\Theta(z),\\&&\Theta(z+\tau^{(j)})={\rm e}^{2\pi{\rm i}(-\frac{\tau_{jj}}{2}-z_{j})}\Theta(z),~~j=1,2,3,~z\in {\Bbb C}^{3}.\end{eqnarray*}$

定义一个 Abel-型映射 $\varphi: N \rightarrow{\Bbb C}^{3}$. 对于 $D\in N$, $\varphi(D)=\int_{\bar{C}_{2}}^{D}\varpi {\rm d}\lambda$.

$\lambda\in {\Bbb C}\setminus{\cal C}$ 时, 引入一个新的 $g$ 函数, 关于 $\lambda$ 的导数为

$\begin{matrix}\frac{\partial}{\partial \lambda}g(\xi,\lambda)=2\frac{(\lambda-d(\xi))(\lambda-\alpha(\xi))(\lambda-\bar{\alpha}(\xi))(\lambda-\varrho(\xi))(\lambda-\bar{\varrho}(\xi))}{\omega(\lambda)},\end{matrix}$

其中, $d\equiv d(\xi)$ 为实数.

图4.2

图4.2   跳跃曲线(红色实线), ${\rm Im} g <0$ (粉色区域), ${\rm Im} g >0$ (白色区域)


定义 $\Gamma^{mod} := {\cal C}\cup\gamma_{(\bar{\varrho},\varrho)} = \Gamma_{1}\cup\Gamma_{2}\cup\gamma_{(\varrho,\alpha)}\cup \gamma_{(\bar{\alpha},\bar{\varrho})}\cup\gamma_{(\bar{\varrho},\varrho)}$.

对于 $\lambda\in{\Bbb C}\setminus\Gamma^{mod}$, $g$ 函数为

$\begin{matrix}g(\lambda)\equiv g(\xi, \lambda):=\int_{\bar{C}_{2}}^{\lambda}{\rm d}g,\end{matrix}$

且满足

$\bullet$ 对于任意 $\lambda\in\overline{{\Bbb C}}\setminus\Gamma^{mod}$, $g(\lambda)-\lambda^{2}-\xi \lambda$ 解析延拓到集合 $\Gamma^{mod}$;

$\bullet$ 对称性: $g=g^{\ast}$;

$\bullet$ 跳跃条件: 对于 $\lambda\in\gamma_{(\bar{\varrho},\varrho)}$, $g_{+}(\lambda)-g_{-}(\lambda)= 2H_{1}$, 以及

$\begin{eqnarray*}g_{+}(\lambda)+g_{-}(\lambda)=\left\{\begin{array}{ll} 2H_{2}, & \lambda\in\Gamma_{1}, \\ 2H_{3}, & \lambda\in\gamma_{(\varrho,\alpha)}\cup \gamma_{(\bar{\alpha},\bar{\varrho})}, \\ 0, & \lambda\in\Gamma_{2}, \end{array} \right.\end{eqnarray*}$

其中, $H_{j}\equiv H_{j}(\xi)$, $j=1, 2, 3$ 为实常数,

$H_{1}=\frac{g_{+}(\varrho)-g_{-}(\varrho)}{2}=\frac{g_{+}(\bar{\varrho})-g_{-}(\bar{\varrho})}{2}, H_{2}=g(C_{1})=g(\bar{C}_{1}), H_{3}=g(\alpha)=g(\bar{\alpha}).$

在某些情况下, 可以通过改变跳跃曲线定义新的 Riemann-Hilbert 问题.

图4.3

图4.3   跳跃曲线 $\Sigma$ (左图), 跳跃曲线 $\hat{\Sigma}= \Sigma\cup\gamma$ (右图)


根据图4.3, 引入矩阵 Riemann-Hilbert 问题

$\begin{equation}\left\{\begin{array}{ll} n\in {\Bbb I}+\dot{E}^{2}({\Bbb C}\setminus\Sigma), \\ n_{+}=n_{-}v, ~\lambda\in\Sigma, \end{array}\right.\end{equation}$

其中, $v : \Sigma\rightarrow GL(2,{\Bbb C})$ 为定义在 $\Sigma$ 上的跳跃矩阵.

引理 4.1$q$$q^{-1}\in E^{\infty}({\Bbb C}\setminus\hat{\Sigma})$$2\times2$ 的矩阵函数, 则 $n$ 满足 Riemann-Hilbert 问题(4.5)当且仅当对于 $\lambda\in{\Bbb C}\setminus\hat{\Sigma}$,$\hat{n}(\lambda):=n(\lambda)q(\lambda)$满足

$\begin{equation}\left\{\begin{array}{ll} \hat{n}\in {\Bbb I}+\dot{E}^{2}({\Bbb C}\setminus\hat{\Sigma}), \\ \hat{n}_{+}=\hat{n}_{-}\hat{v}, ~\lambda\in \hat{\Sigma}, \end{array}\right.\end{equation}$

其中跳跃矩阵为

$\begin{equation}\hat{v}=\left\{\begin{array}{ll} q_{-}^{-1}vq_{+},~~&\lambda\in\Sigma,\\ q_{-}^{-1}q_{+},~~~&\lambda\in\gamma. \end{array}\right.\end{equation}$

$q(\infty) \neq{\Bbb I}$ 时, 有跳跃曲线有变形.

引理 4.2$q$$q^{-1}\in E^{\infty}({\Bbb C}\setminus\hat{\Sigma})$$2\times2$ 的矩阵函数, 并在 $\lambda =\infty$ 处解析. 那么, $n$ 满足 Riemann-Hilbert 问题(4.5)当且仅当对于 $\lambda\in{\Bbb C}\setminus\hat{\Sigma}$,$\hat{n}(\lambda):=q(\infty)^{-1}n(\lambda)q(\lambda)$满足 Riemann-Hilbert 问题(4.6), 其中, 跳跃矩阵 $\hat{v}$ 由(4.7)式定义.

为得到 Riemann-Hilbert 问题(4.6)的解 $\hat{n}$ 的长时间渐进性, 必须对跳跃曲线进行一系列变换. 首先定义函数 $\hat{n}^{(j)}(x, t, \lambda)$, $j = 1, 2, \cdots, 5$, 其满足 Riemann-Hilbert 问题

$\begin{matrix}\left\{\begin{array}{ll} \hat{n}^{(j)}(x,t,\cdot)\in {\Bbb I}+\dot{E}^{2}({\Bbb C}\setminus\Gamma^{(j)}),~~~~~~~~~~~~~~~~~~~~~ \\ \hat{n}^{(j)}_{+}(x, t, \lambda)=\hat{n}^{(j)}_{-}(x, t, \lambda)\hat{v}^{(j)}(x, t,\lambda), ~\lambda\in \Gamma^{(j)}, \end{array}\right.\end{matrix}$

经过五次变形后, 可得到跳跃矩阵 $\hat{v}^{(5)}$.$t \rightarrow\infty$ 时, 可计算出聚焦 KE 方程(1.2)在激波情况下的渐进解.

类似于(2.2)和(2.3)式, 对于五次跳跃曲线变形, 跳跃矩阵均满足

$\begin{matrix}\hat{v}^{(j)}(x, t, \lambda)=\left\{\begin{array}{ll} \sigma_{2}\overline{\hat{v}^{(j)}(x, t, \bar{\lambda})}\sigma_{2}, & \lambda\in \Gamma^{(j)}\setminus\mathbb{R}, \\ \sigma_{2}\overline{\hat{v}^{(j)}(x, t, \bar{\lambda})}^{-1}\sigma_{2}, & \lambda\in \Gamma^{(j)}\cap\mathbb{R}, \end{array}\right.~~j=1, 2, \cdots, 5.\end{matrix}$

进而, 有

$\begin{matrix}\hat{n}^{(j)}=\sigma_{2}\overline{\hat{n}^{(j)}(x, t, \bar{\lambda})}\sigma_{2},~~\lambda\in {\Bbb C}\setminus\Gamma^{(j)},~~j=1, 2, \cdots, 5.\end{matrix}$

下面介绍跳跃曲线的第一次变形.

通过将(3.5)和(3.6)式中的 $n^{(1)}$ 替换成 $\hat{n}^{(1)}$ 的方式实现新的 Riemann-Hilbert 问题的定义. 相对应的跳跃曲线为$\Gamma^{(1)} := \mathbb{R} \cup\Gamma_{1}\cup\Gamma_{2}\cup\gamma_{(\varrho,\alpha)}\cup \gamma_{(\bar{\alpha},\bar{\varrho})}\cup\gamma_{(\bar{\varrho},\varrho)},$以及跳跃矩阵表达式为$\hat{v}^{(1)}(x,t,\lambda):={\rm e}^{-{\rm i}tg_{-}(\lambda)\sigma_{3}}\hat{J}_{0}(\lambda) {\rm e}^{{\rm i}tg_{+}(\lambda)\sigma_{3}}.$$\lambda\in\mathbb{R} \cup\Gamma_{1}\cup\Gamma_{2}$ 时, $\hat{J}_{0}(\lambda)$ 由(4.1)式给出, 当 $\lambda\in\gamma_{(\varrho,\alpha)}\cup \gamma_{(\bar{\alpha},\bar{\varrho})}\cup\gamma_{(\bar{\varrho},\varrho)}$ 时, $\hat{J}_{0}(\lambda)$ 则为单位矩阵.

下面介绍跳跃曲线的第二次变形.

通过对跳跃矩阵 $\hat{v}^{(1)}$ 进行因式分解发现, 对于 $\lambda\in(-\infty,d)$, 分解式是不正确的. 因此, 定义新的函数

$\begin{matrix}\hat{n}^{(2)}(x,t,\lambda):=\hat{n}^{(1)}(x,t,\lambda)\delta(\lambda)^{-\sigma_{3}},\end{matrix}$

其中, 对于 $\lambda\in{\Bbb C}\setminus(-\infty,d]$, 复函数

$\delta(\lambda)\equiv\delta(\xi,\lambda):={\rm e}^{\frac{1}{2\pi {\rm i}}\int_{-\infty}^{d}\frac{\ln(1+|\hat{r}(k)|^{2})}{k-\lambda}{\rm d}k}$

满足

$\bullet$$\delta(\lambda)$$\delta(\lambda)^{-1}$ 是解析有界的, 并且连续到 $(-\infty,-1)\cup(-1,d)$;

$\bullet$ 对称性: $\delta=(\delta^{\ast})^{-1}$;

$\bullet$ 跳跃条件: 对于 $\lambda\in(-\infty, -1)\cup(-1, d)$, $\delta_{+}=\delta_{-}(1+|r|^{2})$;

$\bullet$$\lambda\rightarrow\infty$ 时, $\delta(\lambda)=1+O\left(\lambda^{-1}\right)$关于 $\arg \lambda\in[2\pi]$ 是一致的.

由上述条件可知 $\delta(\xi,\cdot)^{\sigma_{3}}\in{\Bbb I}+(\dot{E}^{2}\cap E^{\infty})\left({\Bbb C}\setminus(\infty,d]\right)$. 跳跃矩阵有以下形式

$\begin{matrix} \hat{v}^{(2)}=\left(\begin{array}{cc} \delta_{-} & 0 \\ 0 & \delta_{-}^{-1} \\ \end{array} \right)\hat{v}^{(1)}\left( \begin{array}{cc} \delta_{+}^{-1} & 0 \\ 0 & \delta_{+} \\ \end{array} \right).\end{matrix}$

下面介绍跳跃曲线的第三次变形.

在集合 $\{U_{j}\}_{1}^{6}$ 上定义函数 $\hat{n}^{(3)}(x,t,\lambda)$ 使其在实轴上不存在跳跃关系. 该函数为

$\begin{equation}\hat{n}^{(3)}(x,t,\lambda):=\hat{n}^{(2)}(x,t,\lambda)\times \hat{J}_{j}^{(3)},~~j=1,2,\cdots,6,\end{equation}$

其中

$\begin{array}{l}\hat{J}_{1}^{(3)}=\left(\begin{array}{cc}1 & 0 \\ -\frac{\hat{r}^{2 \mathrm{i} t g}}{\delta^{2}} & 1\end{array}\right), \lambda \in U_{1}, \hat{J}_{2}^{(3)}=\mathbb{I}, \lambda \in U_{2}, \hat{J}_{3}^{(3)}=\left(\begin{array}{cc}1 & -\frac{\hat{r}^{*} \delta^{2}}{\left(1+\hat{r} \hat{r}^{*}\right) \mathrm{e}^{2 \mathrm{iitg}}} \\ 0 & 1\end{array}\right), \lambda \in U_{3}, \\ \hat{J}_{4}^{(3)}=\left(\begin{array}{cc}1 & 0 \\ \frac{\hat{r} \mathrm{e}^{2 \mathrm{i} t g}}{\left(1+\hat{r} \hat{r}^{*}\right) \delta^{2}} & 1\end{array}\right), \lambda \in U_{4}, \hat{J}_{5}^{(3)}=\mathbb{I}, \lambda \in U_{5}, \hat{J}_{6}^{(3)}=\left(\begin{array}{cc}1 & \frac{\hat{r}^{*} \delta^{2}}{\mathrm{e}^{2 \mathrm{i} t g}} \\ 0 & 1\end{array}\right), \lambda \in U_{6}. \\\end{array}$

结合(4.8), (4.11)和(4.12)式, 可计算出在 $\Gamma^{(3)}$ 上的跳跃矩阵为

$\begin{array}{l}\hat{v}_{1}^{(3)}=\left(\begin{array}{cc}1 & 0 \\ \hat{r} \delta^{-2} \mathrm{e}^{2 \mathrm{i} t g} & 1\end{array}\right), \quad \hat{v}_{2}^{(3)}=\left(\begin{array}{cc}1 & \hat{a} \hat{b} \delta^{2} \mathrm{e}^{-2 \mathrm{i} t g} \\ 0 & 1\end{array}\right), \\ \hat{v}_{3}^{(3)}=\left(\begin{array}{cc}0 & \mathrm{i} \mathrm{e}^{\mathrm{i} \theta_{1}} \hat{a}_{+} \hat{a}_{-} \delta^{2} \mathrm{e}^{-\mathrm{i} t\left(g_{+}+g_{-}\right)} \\ \frac{\mathrm{e}^{-\mathrm{i} \theta_{1}}}{\hat{a}_{+} \hat{a}_{-}} \delta^{-2} \mathrm{e}^{\mathrm{i} t\left(g_{+}+g_{-}\right)} & 0\end{array}\right), \\ \hat{v}_{4}^{(3)}=\left(\begin{array}{cc}0 & \mathrm{i} \nu_{1}^{2} \delta^{2} \mathrm{e}^{-\mathrm{i} t\left(g_{+}+g_{-}\right)} \\ \mathrm{i} \nu_{1}^{-2} \delta^{-2} \mathrm{e}^{\mathrm{i} t\left(g_{+}+g_{-}\right)} & 0\end{array}\right), \\ \hat{v}_{5}^{(3)}=\mathrm{e}^{-\mathrm{i} t g_{-} \sigma_{3}}\left(\begin{array}{cc}1 & -\hat{a} \hat{b} \delta^{2} \\ 0 & 1\end{array}\right) \mathrm{e}^{\mathrm{i} t g_{+} \sigma_{3}}, \hat{v}_{6}^{(3)}=\mathrm{e}^{-\mathrm{i} t g_{-} \sigma_{3}}\left(\begin{array}{cc}1 & -\hat{a} \hat{b} \delta^{2} \\ \hat{r} \delta^{-2} & \hat{a} \hat{a}^{*}\end{array}\right) \mathrm{e}^{\mathrm{i} t g_{+} \sigma_{3}},\end{array}$

由(4.9)式可知上述跳跃矩阵可延拓到下半平面.

下面介绍跳跃曲线第四次变形.

本次变形的目的是实现在 $\gamma_{(\varrho, \alpha)}\cup\gamma_{(\bar{\alpha},\bar{\varrho})}$ 非对角线上的跳跃, 以及在 $\gamma_{(\bar{\varrho}, \varrho)}$ 对角线上的跳跃. 因此,对于 $\lambda\in\{V_{j}\}_{1}^{4}$, 函数 $n^{(4)}$ 可表示为

$\hat{n}^{(4)}(x,t,\lambda):=\hat{n}^{(3)}(x,t,\lambda)\times \hat{J}_{j}^{(4)},$

其中

$\begin{array}{l}\hat{J}_{1}^{(4)}:=\left(\begin{array}{cc}1 & 0 \\ -\frac{\mathrm{e}^{2 \mathrm{i} t g}}{\hat{a} \hat{b} \delta^{2}} & 1\end{array}\right), \lambda \in V_{1}, \hat{J}_{2}^{(4)}:=\left(\begin{array}{cc}1 & 0 \\ \frac{\mathrm{e}^{2 \mathrm{i} t g}}{\hat{a} \hat{b} \delta^{2}} & 1\end{array}\right), \lambda \in V_{2}, \\ \hat{J}_{3}^{(4)}:=\left(\begin{array}{cc}1 & 0 \\ -\frac{\hat{b}^{*}}{\hat{a}^{2} \hat{a}^{*}} \delta^{-2} \mathrm{e}^{-2 \mathrm{i} t g} & 1\end{array}\right), \lambda \in V_{3}, \hat{J}_{4}^{(4)}:=\left(\begin{array}{cc}1 & -\frac{\hat{b}}{\hat{a}^{*}} \delta^{2} \mathrm{e}^{-2 \mathrm{i} t g} \\ 0 & 1\end{array}\right), \lambda \in V_{4} \cdot\end{array}$

由(4.10)式可知 $\hat{n}^{(4)}$ 可延伸到下半平面.

$\lambda\in{\Bbb C}^{+}\setminus\{V_{j}\}_{1}^{4}$ 时, 则有 $\hat{J}_{j}^{(4)}={\Bbb I}$.

根据(4.8), (4.11)和(4.12)式, 函数 $\hat{n}^{(4)}$$\Gamma^{(4)}$ 上的跳跃矩阵为

$\hat{v}_{1}^{(4)}=\left(\begin{array}{cc}1 & 0 \\ \frac{\hat{b}}{\hat{a}^{*}} \delta^{-2} \mathrm{e}^{2 \mathrm{i} t g} & 1\end{array}\right), \quad \hat{v}_{2}^{(4)}=\left(\begin{array}{cc}1 & \hat{a} \hat{b} \delta^{2} \mathrm{e}^{-2 \mathrm{i} t g} \\ 0 & 1\end{array}\right)$,
$\begin{array}{l}\hat{v}_{4}^{(4)}=\left(\begin{array}{cc}0 & \mathrm{i} \delta^{2} \nu_{1}^{2} \mathrm{e}^{-\mathrm{i} t\left(g_{+}+g_{-}\right)} \\ \mathrm{i} \delta^{-2} \nu_{1}^{-2} \mathrm{e}^{\mathrm{i} t\left(g_{+}+g_{-}\right)} & 0\end{array}\right), \quad \hat{v}_{5}^{(4)}=\hat{v}_{7}^{(4)}=\left(\begin{array}{cc}1 & 0 \\ -\frac{\mathrm{e}^{2 \mathrm{i} t g}}{\hat{a} \hat{b} \delta^{2}} & 1\end{array}\right), \\ \hat{v}_{6}^{(4)}=\left(\begin{array}{cc}0 & -\hat{a} \hat{b} \delta^{2} \mathrm{e}^{-\mathrm{i} t\left(g_{+}+g_{-}\right)} \\ \frac{\mathrm{e}^{\mathrm{i} t\left(g_{+}+g_{-}\right)}}{\hat{a} \hat{b} \delta^{2}} & 0\end{array}\right), \quad \hat{v}_{8}^{(4)}=\left(\begin{array}{cc}1 & 0 \\ \frac{\mathrm{e}^{2 \mathrm{i} t g}}{\hat{a}^{2} \hat{a}^{*} \hat{b} \delta^{2}} & 1\end{array}\right), \\ \hat{v}_{9}^{(4)}=\left(\begin{array}{cc}1 & 0 \\ -\frac{\hat{a}^{*} \mathrm{e}^{2 \mathrm{i} t g}}{\hat{b} \delta^{2}} & 1\end{array}\right), \quad \hat{v}_{10}^{(4)}=\left(\begin{array}{cc}1 & -\frac{b \delta^{2}}{\hat{a}^{*} \mathrm{e}^{2 \mathrm{i} t g}} \\ 0 & 1\end{array}\right) \\ \hat{v}_{11}^{(4)}=\left(\begin{array}{cc}\frac{\mathrm{e}^{\mathrm{i} t\left(g_{+}-g_{-}\right)}}{\hat{a} \hat{a}^{*}} & 0 \\ 0 & \hat{a} \hat{a}^{*} \mathrm{e}^{-\mathrm{i} t\left(g_{+}-g_{-}\right)}\end{array}\right), \quad \hat{v}_{12}^{(4)}=\left(\begin{array}{cc}1 & 0 \\ \frac{\mathrm{e}^{2 \mathrm{i} t g} \hat{b}^{*}}{\hat{a}^{2} \hat{a}^{*} \delta^{2}} & 1\end{array}\right), \\\end{array}$

由(4.9)式可知, 跳跃矩阵可延伸到下半平面.

下面介绍跳跃曲线的第五次变形.

本次变形旨在实现在四条分支曲线以及 $\gamma_{(\bar{\varrho}, \varrho)}$ 上的跳跃. 定义函数

$\begin{matrix}\hat{n}^{(5)}(x,t,\lambda):={\rm e}^{-{\rm i}h(\infty)\sigma_{3}}\hat{n}^{(4)}(x,t,\lambda){\rm e}^{{\rm i}h(\lambda)\sigma_{3}},\end{matrix}$

这里函数 $h(\lambda)\equiv h(\xi,\lambda)$ 在定义域 $ \overline{{\Bbb C}}\setminus\Gamma^{mod}$ 上解析, 同时满足对称性 $h=h^{\ast}$. ${\rm e}^{{\rm i}h(\lambda)\sigma_{3}}\in E^{\infty}\left(\overline{{\Bbb C}}\setminus \Gamma^{mod}\right)$.

进一步地, 跳跃矩阵可表示为

$\begin{equation}\hat{v}^{(5)}={\rm e}^{-{\rm i}h_{-}\sigma_{3}}\hat{v}^{(4)}{\rm e}^{{\rm i}h_{+}\sigma_{3}},\end{equation}$

且函数 $h_{\pm}$ 有以下性质

$\begin{matrix}&&h_{+}+h_{-}=\left\{\begin{array}{lll} 2\omega_{1}-{\rm i}\ln(\hat{a}_{+}\hat{a}_{-}\delta^{2}{\rm e}^{{\rm i}\theta_{1}}), & \lambda\in \Gamma_{1}\cap{\Bbb C}^{+}, \\ 2\omega_{1}+{\rm i}\ln(\hat{a}_{+}^{\ast}\hat{a}_{-}^{\ast}\delta^{-2}{\rm e}^{-{\rm i}\theta_{1}}), & \lambda\in \Gamma_{1}\cap{\Bbb C}^{-}, \\ 2\omega_{2}-{\rm i}\ln({\rm i}\hat{a}\hat{b}\delta^{2}), & \lambda\in \gamma_{(\varrho,\alpha)}, \\ 2\omega_{2}+{\rm i}\ln(-{\rm i}\hat{a}^{\ast}\hat{b}^{\ast}\delta^{-2}), & \lambda\in \gamma_{(\bar{\alpha},\bar{\varrho})}, \\ -{\rm i}\ln(\nu_{1}^{2}\delta^{2}), & \lambda\in \Gamma_{2}, \end{array}\right.\\&&h_{+}-h_{-}=\left\{\begin{array}{ll} 2\omega_{3}-{\rm i}\ln(aa^{\ast}), & \lambda\in \gamma_{(d,\varrho)}, \\ 2\omega_{3}+{\rm i}\ln(aa^{\ast}), & \lambda\in \gamma_{(\bar{\varrho},d)}, \end{array} \right.\end{matrix}$

其中 $\{\omega_{j}\}_{1}^{3}$ 为实数. 设 $h_{k}:= h_{+}+h_{-}$, 则

$\bullet$$\Gamma_{1}\cap{\Bbb C}^{+}$, $\Gamma_{1}\cap{\Bbb C}^{-}$, $\gamma_{(\varrho,\alpha)}$, $\gamma_{(\bar{\alpha},\bar{\varrho})}$$\Gamma_{2}$ 上连续;

$\bullet$ 对任意 $\lambda\in {\cal C}\cup\Gamma_{1}\cup\Gamma_{2} \cup\gamma_{(\varrho,\alpha)}\cup\gamma_{(\bar{\alpha},\bar{\varrho})}$, $h_{k}=h_{k}^{\ast}$.

$a(\lambda), b(\lambda)\neq 0$ 时, 上述条件很容易验证. 取 $\lambda =d$, 可计算得到 $\ln(aa^{\ast}) = \ln(|a(d)^{2}|)>0$, 则在构造(4.14)式中的对数值时除了要考虑上述条件外还要满足 $h_{+}-h_{-}$ 在跳跃曲线 $\gamma_{(d,\varrho)}$$\gamma_{(\bar{\varrho},d)}$ 上连续.

对于 $\lambda\in \Gamma^{mod}\setminus {\Bbb C}^{+}$, 引入

$\begin{matrix}{\cal H}(\lambda)\equiv {\cal H}(\xi,\lambda):=\left\{\begin{array}{ll} \frac{2\omega_{1}-{\rm i}\ln(\hat{a}_{+}\hat{a}_{-}\delta^{2}{\rm e}^{{\rm i}\theta_{1}})}{\omega_{+}}, & \lambda\in \Gamma_{1}\cap{\Bbb C}^{+}, \\[3mm] \frac{2\omega_{2}-{\rm i}\ln({\rm i}\hat{a}\hat{b}\delta^{2})}{\omega_{+}}, & \lambda\in \gamma_{(\varrho,\alpha)}, \\[3mm] \frac{2\omega_{3}-{\rm i}\ln(aa^{\ast})}{\omega}, & \lambda\in \gamma_{(d,\varrho)}, \\[3mm] -\frac{{\rm i}\ln(\delta^{2}\nu_{1}^{2})}{\omega_{+}}, & \lambda\in \Gamma_{2}\cap{\Bbb C}^{+}. \end{array}\right.\end{matrix}$

根据对称性, 函数 ${\cal H}$ 可延伸到 $\Gamma^{mod}\setminus {\Bbb C}^{-}$.

进而, 有

$\begin{equation}h(\lambda):=\frac{\omega(\lambda)}{2\pi {\rm i}}\int_{\Gamma^{mod}}\frac{{\cal H}(k)}{k-\lambda}{\rm d}k,~~\lambda\in\overline{{\Bbb C}}\setminus\Gamma^{mod}.\end{equation}$

对于使得函数 $h(\lambda)$ 满足以下条件的实数 $\omega_{j} \equiv \omega_{j}(\xi)$, $j = 1, 2, 3$ 是唯一的.

$\bullet$ 对称条件: $h=h^{\ast}$;

$\bullet$$\Gamma^{mod}$ 上, 跳跃矩阵满足(4.14)式;

$\bullet$ 对于 $\lambda\in\overline{{\Bbb C}}\setminus\Gamma^{mod}$, 函数 ${\rm e}^{{\rm i}h\sigma_{3}}$ 是解析有界的;

$\bullet$ 随着 $k\rightarrow\infty$, 则有$h(\lambda)=h(\infty)+O\left(\lambda^{-1}\right)$,且 $h(\infty)=-\frac{1}{2\pi{\rm i}}\int_{\Gamma^{mod}}k^{3}{\cal H}(k){\rm d}k$.

$t\rightarrow\infty$ 时, 跳跃矩阵的渐进性为

$\begin{matrix}\hat{v}^{(5)}\rightarrow{\Bbb I}, ~~\lambda\in\Gamma^{(5)}\setminus\Gamma^{mod},\end{matrix}$

此外, 解 $\hat{n}^{(5)}$ 无限接近满足 Riemann-Hilbert 问题(4.8)的解 $n^{mod}$.

相应地, 跳跃矩阵可表示为

$v^{\text {mod }}=\left\{\begin{array}{cc}\left(\begin{array}{cc}0 & \mathrm{ie}^{-2 \mathrm{i}\left(t H_{1}+\omega_{1}\right)} \\ \mathrm{ie}^{2 \mathrm{i}\left(t H_{1}+\omega_{1}\right)} & 0\end{array}\right), \lambda \in \Gamma_{1}, \\ \left(\begin{array}{cc}0 & \mathrm{ie}^{-2 \mathrm{i}\left(t H_{2}+\omega_{2}\right)} \\ \mathrm{ie}^{2 \mathrm{i}\left(t H_{2}+\omega_{2}\right)} & 0\end{array}\right), \lambda \in \gamma_{(\rho, \alpha)} \cup \gamma_{(\bar{\alpha}, \bar{e})}, \\ \left(\begin{array}{cc}\mathrm{e}^{2 \mathrm{i}\left(t H_{3}+\omega_{3}\right)} & 0 \\ 0 & \mathrm{e}^{-2 \mathrm{i}\left(t H_{3}+\omega_{3}\right)}\end{array}\right), & \lambda \in \gamma_{(\bar{\rho}, \underline{e})} \\ \mathrm{i} \sigma_{1}, & \lambda \in \Gamma_{2}.\end{array}\right.$

该矩阵与 $\lambda$ 无关, 对于 $\lambda\in\gamma_{(\bar{\varrho},\varrho)}$, 它是对角形式的, 而在四个分支曲线上则为非对角形式的.

对于 $\lambda\in{\Bbb C}\setminus\Gamma^{mod}$, 定义函数

$\nu(\lambda)\equiv\nu(\xi, \lambda):=\left(\frac{(\lambda-C_{1})(\lambda-C_{2})(\lambda-\alpha)(\lambda-\varrho)} {(\lambda-\bar{C}_{1})(\lambda-\bar{C}_{2})(\lambda-\bar{\alpha})(\lambda-\bar{\varrho})} \right)^{\frac{1}{4}}.$

随着 $\lambda\rightarrow\infty$, 则 $\nu(\lambda) = 1 + O\left(\lambda^{-1}\right)$.

定义 $N\rightarrow\overline{{\Bbb C}}$ 的一个亚纯函数 $\hat{\nu}\equiv\hat{\nu}(\xi,\cdot)$, 其中函数 $\nu^{2}$$-\nu^{2}$ 分别定义在上下黎曼曲面 $N$ 上. 换言之, 对于 $\lambda\in{\Bbb C}\setminus\Gamma^{mod}$, $\hat{\nu}(\lambda^{\pm})=\pm\nu^{2}(\lambda)$.

若函数 $\hat{\nu}$ 有四个简单零点 $C_{1}$, $\alpha$, $\varrho$, 和 $C_{2}$, 则 $\hat{\nu}$ 为一个四次函数. 进一步地, 通过计算得知 $\hat{\nu}-1$ 在黎曼曲面 $N$ 上有四个零点, 分别记为 $+\infty$, $D_{1}$, $D_{2}$, $D_{3}$.

$P:=D_{1}D_{2}D_{3}$. 向量 ${\cal K}$, 复向量 $l$ 和向量值函数 $v(t)$ 可定义为

$\begin{eqnarray*}&&{\cal K}\equiv {\cal K}(\xi):=\frac{1}{2}\left({\rm e}^{(1)}+{\rm e}^{(3)}+\tau^{(1)}+\tau^{(2)}+\tau^{(3)}\right)\in{\Bbb C}^{3},\\&&l\equiv l(\xi):=\sum\limits_{1}^{3}\varphi(D_{j})+{\cal K}\in{\Bbb C}^{3},~~\varphi(P):=\sum\limits_{1}^{3}\varphi(D_{j}),\\&&v(t)\equiv v(\xi,t)=-\frac{1}{\pi}\left(tH_{1}+\omega_{1}, tH_{2}+\omega_{2}, t(H_{2}-H_{3})+H_{2}-H_{3}+\frac{\pi}{2}\right).\end{eqnarray*}$

定理 4.1 对任意 $t\geq0$, $\{H_{i},\omega_{i}\}_{1}^{3}$ 为任意实数, Riemann-Hilbert 问题(4.8)存在唯一解 $n^{mod}(x,t,\lambda)$ 满足

$\begin{equation}2{\rm i}\lim_{\lambda\rightarrow\infty}\lambda(n^{mod}(x,t,\lambda))_{12}={\rm Im}(C_{1}+C_{2}+\alpha+\varrho)\frac{\Theta\left(\varphi(\infty^{+})+l\right)\Theta\left(\varphi(\infty^{+})-v(t)-l\right)}{\Theta\left(\varphi(\infty^{+})-l\right)\Theta\left(\varphi(\infty^{+})+v(t)+l\right)},\end{equation}$

且关于 $\arg\lambda\in[2\pi]$ 是一致的.

因此, $E_{0}$ 可表示为

$\begin{equation}E_{0}(\xi,t)=2{\rm i}{\rm e}^{2{\rm i}(tg(0)+h(\infty))}\lim_{\lambda\rightarrow\infty}\lambda(n^{mod}(x,t,\lambda))_{12}.\end{equation}$

通过对跳跃曲线进行五次变形最终得到 $\hat{n}^{(5)}(x,t,\lambda)$ 能较好的逼近 Riemann-Hilbert 问题(4.8)的解. 然而, 当 $\lambda$ 无限接近跳跃曲线 $\Gamma^{mod}$ 时, 这种近似解关于 $\lambda$ 不再是一致的. 故基于局部逼近原理, 当 $\lambda\in D_{\epsilon}(\alpha)$, $\lambda\in D_{\epsilon}(\varrho)$ 以及 $\lambda\in D_{\epsilon}(d)$ 时, 逼近解 $\hat{n}^{(5)}(x,t,\lambda)$ 的局部解 $n^{\alpha}(x,t,\lambda)$, $n^{\varrho}(x,t,\lambda)$$n^{d}(x,t,\lambda)$ 的具体表达式将会给出, 且对误差项进行改进.

$\lambda\in D_{\epsilon}(\alpha)\setminus\Gamma^{(5)}$ 时, 定义函数

$\begin{equation}n^{(\alpha_{0})}(x,t,\lambda):=\hat{n}^{(5)}(x,t,\lambda){\rm e}^{\left[\frac{{\rm 1}}{2}\ln\left(-\delta(\lambda)^{2}\hat{a}(\lambda)\hat{b}(\lambda)\right)-{\rm i}\left(tg(\alpha)+h(\lambda)\right)\right]\sigma_{3}},\end{equation}$

这里指数项是有界而且解析的.

$\lambda\in D_{\epsilon}(\alpha)\setminus\gamma_{(\varrho,\alpha)}$ 时, 函数 $g_{\alpha}(\lambda)\equiv g_{\alpha}(\xi,\lambda):=g(\lambda)-g(\alpha)$.

开集 $D_{\epsilon}(\alpha)$ 以及指向 $\alpha$ 的有向曲线 ${\cal S}_{i}$ 分别被定义为 $\left\{{\cal S}_{i}\right\}_{1}^{4}$$\left\{{\cal Y}_{i}\right\}_{1}^{4}$, 且${\cal Y}_{i} = \overline{{\cal S}}_{i-1}\cap\overline{{\cal S}}_{i}$, $i = 1, \cdots, 4$,其中 $\overline{{\cal S}}_{0}\equiv \overline{{\cal S}}_{4}$. 具体分布图如图4.4 所示.

图4.4

图4.4   开集 $D_{\epsilon}(\alpha)$ 以及集合 $\{{\cal S}_{i}\}_{1}^{4}$ 的分布图


鉴于(4.3)式中跳跃矩阵 $\hat{v}^{(5)}$, 可推断出 $n^{(\alpha_{0})}$ 在跳跃矩阵

$v^{\left(\alpha_{0}\right)}=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & -\mathrm{e}^{-2 \mathrm{i} t g_{\alpha}} \\ 0 & 1\end{array}\right), \lambda \in \mathcal{Y}_{1}, \\ \left(\begin{array}{cc}1 & 0 \\ \mathrm{e}^{2 \mathrm{i} t g_{\alpha}} & 1\end{array}\right), & \lambda \in \mathcal{Y}_{2} \cup \mathcal{Y}_{4} \\ \left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right), & \lambda \in \mathcal{Y}_{3}\end{array}\right.$

的作用下, 满足关系

$n_{+}^{(\alpha_{0})}(x,t,\lambda)=n_{-}^{(\alpha_{0})}(x,t,\lambda)v^{(\alpha0)}(x,t,\lambda),~~\lambda\in\Gamma^{(5)}\cap D_{\epsilon}(\alpha).$

对于 $\lambda\in D_{\epsilon}(\alpha)$, 利用局部变量变换, 定义一个新的变量

$\begin{matrix}&&\varpi\equiv\varpi(\lambda):=\left(\frac{3{\rm i}t}{2}g_{\alpha}(\lambda)\right)^{2/3},\\&&g_{\alpha}(\lambda) = \int_{\alpha}^{\lambda}\sqrt{k-\alpha}f(k){\rm d}k,\end{matrix}$

其中$f(\lambda)$ 是解析的, 且在 $\lambda= \alpha$ 处不为零.

方程(4.22)可重述为

$\begin{matrix}\varpi:=\left(\frac{3t}{2}\right)^{2/3}(\lambda-\alpha)\psi_{\alpha}(\lambda),\end{matrix}$

这里, 当 $\lambda\in D_{\epsilon}(\alpha)$ 时, $\psi_{\alpha}(\lambda)\equiv\psi_{\alpha}(\xi,\lambda)$ 为解析函数, 且 $\psi_{\alpha}(\alpha)\neq0$.

根据(4.23)式, 圆盘 $D_{\epsilon}(\alpha)$ 到原点邻域的映射可记为 $\lambda\mapsto\xi$.

由于在跳跃曲线 $\gamma(\varrho,\alpha)$${\rm Im} g_{\alpha}(\alpha)$ 为零, 那么 ${\cal Y}_{3}$ 上的跳跃曲线可映射到 $\mathbb{R} _{-}$ 上, 该映射表示为 $\lambda\mapsto\xi$, ${\cal Y}_{3}$ 上的跳跃曲线与 $\xi$ 有关.

方程(4.22)可进一步变形为 $\frac{4}{3}\varpi^{3/2}:=2{\rm i}tg_{\alpha}(\lambda)$.$n^{(\alpha0)}(\lambda)$$n^{Ai}(\varpi(\lambda))$$\alpha$ 附近的跳跃轨迹相同.

接下来定义在 $\alpha$ 附近 $\hat{n}^{(5)}$ 的逼近函数

$\begin{matrix}n^{\alpha}(x,t,\lambda)=Y_{\alpha}(x,t,\lambda)n^{Ai}(\varpi(\lambda)){\rm e}^{\left[-\frac{1}{2}\ln(-\delta^{2}\hat{a}\hat{b})+{\rm i}\left(tg(\alpha)+h(\lambda)\right)\right]\sigma_{3}},~~\lambda\in D_{\epsilon}(\alpha)\setminus\Gamma^{(5)},\end{matrix}$

其中, 函数

$\begin{matrix}Y_{\alpha}(x,t,\lambda):=n^{mod}{\rm e}^{\left[-\frac{1}{2}\ln(-\delta^{2}\hat{a}\hat{b})+{\rm i}\left(tg(\alpha)+h(\lambda)\right)\right]\sigma_{3}}n_{as,M}^{Ai}(\varpi(\lambda)),~~M\geq0\end{matrix}$

$D_{\epsilon}(\alpha)$ 上是解析的.

$\lambda\in {\cal Y}_{3}$ 时, 解析函数 $n_{as,M}^{Ai}(\varpi(\lambda))$ 满足关系式

$\begin{matrix}\left(n_{as,M}^{Ai}(\varpi(\lambda))\right)_{-}={\rm i}\sigma_{2}\left(n_{as,M}^{Ai}(\varpi(\lambda))\right)_{+}.\end{matrix}$

鉴于 $n^{mod}$ 为 Riemann-Hilbert 问题(4.8)的解, 故 $n^{mod}{\rm e}^{\left[-\frac{1}{2}\ln(-\delta^{2}\hat{a}\hat{b})+{\rm i}\left(tg(\alpha)+h(\lambda)\right)\right]\sigma_{3}}$ 满足 (4.26)式. 因此, 函数 ${\cal Y}_{\alpha}$$D_{\epsilon}(\alpha)$ 上解析.

由(4.24)和(4.25)式可知

$\begin{matrix}n^{\alpha}(\lambda)n^{mod}(\lambda)^{-1}={\Bbb I}+O(t^{-M-1}),~~t\rightarrow\infty,~\lambda\in\partial D_{\epsilon}(\alpha),\end{matrix}$

并且对于 $\lambda\in\partial D_{\epsilon}(\alpha)$ 是一致的.

$\lambda\in D_{\epsilon}(\varrho)\setminus\Gamma^{(5)}$ 时, 定义函数

$\begin{matrix}n^{(\varrho_{0})}(x,t,\lambda):=\hat{n}^{(5)}(x,t,\lambda){\rm e}^{\left[\frac{1}{2}\ln\left(\delta(\lambda)^{2}\hat{a}(\lambda)\hat{b}(\lambda)\right)-{\rm i}h(\lambda)\right]\sigma_{3}},\end{matrix}$

其中指数项是有界而且解析的.

开集 $D_{\epsilon}(\varrho)$ 以及指向 $\varrho$ 的有向曲线 ${\cal T}_{i}$ 分别被定义为 $\left\{{\cal T}_{i}\right\}_{1}^{5}$$\left\{{\cal Z}_{i}\right\}_{1}^{5}$, 且 ${\cal Z}_{i} = \overline{{\cal T}}_{i-1}\cap\overline{{\cal T}}_{i}$, $i = 1, \cdots, 5$,其中 $\overline{{\cal T}}_{0}\equiv \overline{{\cal T}}_{5}$. 具体分布图如图4.5 所示.

图4.5

图4.5   开集 $D_{\epsilon}(\varrho)$ 以及集合 $\{{\cal T}_{i}\}_{1}^{5}$ 的分布图


基于(4.13)式中跳跃矩阵 $\hat{v}^{(5)}$, 可推断出 $n^{(\varrho_{0})}$ 满足在跳跃矩阵

$v^{\left(\rho_{0}\right)}=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & -\frac{1}{a a^{*}} \mathrm{e}^{-2 \mathrm{i} t g} \\ 0 & 1\end{array}\right), & \lambda \in \mathcal{Z}_{1}, \\ \left(\begin{array}{cc}1 & 0 \\ a a^{*} \mathrm{e}^{2 \mathrm{i} t g} & 1\end{array}\right), & \lambda \in \mathcal{Z}_{2}, \\ \left(\begin{array}{cc}0 & \mathrm{e}^{-\mathrm{i} t\left(g_{+}+g_{-}\right)} \\ -\mathrm{e}^{\mathrm{i} t\left(g_{+}+g_{-}\right)} & 0\end{array}\right), & \lambda \in \mathcal{Z}_{3}, \\ \left(\begin{array}{cc}1 & 0 \\ \frac{1}{a a^{*}} \mathrm{e}^{2 \mathrm{i} t g} & 1\end{array}\right), & \lambda \in \mathcal{Z}_{4}\\ \left(\begin{array}{ll}\frac{\mathrm{e}^{\mathrm{i} t\left(g_{+}-g_{-}\right)}}{a a^{*}} & 0\\ 0 & \mathrm{e}^{-\mathrm{i} t\left(g_{+}-g_{-}\right)} a a^{*}\end{array}\right), & \lambda \in \mathcal{Z}_{5}, \end{array}\right.$

上的跳越关系

$n_{+}^{(\varrho_{0})}(x,t,\lambda)=n_{-}^{(\varrho_{0})}(x,t,\lambda)v^{(\varrho_{0})}(x,t,\lambda).$

对于 $\lambda\in D_{\epsilon}(\varrho)\setminus(\gamma_{\bar{\varrho},\varrho}\cup\gamma_{\varrho,\alpha})$, 函数 $g_{\varrho}(\lambda)\equiv g_{\varrho}(\xi,\lambda)$ 可表示为

$g_{\varrho}(\lambda):=\int_{\varrho}^{\lambda}{\rm d}g=\left\{\begin{array}{ll} g(\lambda)-g_{-}(\varrho), &\lambda\in {\cal T}_{1}\cup{\cal T}_{2}\cup{\cal T}_{5}, \\ g(\lambda)-g_{+}(\varrho), & \lambda\in {\cal T}_{3}\cup{\cal T}_{4}, \end{array} \right.$

进而, 函数 $n^{(\varrho_{1})}(x,t,\lambda)$$n^{(\varrho_{0})}(x,t,\lambda)$ 的关系为

$n^{(\varrho_{1})}(x,t,\lambda):=n^{(\varrho_{0})}(x,t,\lambda) A(\lambda),~~\lambda\in D_{\epsilon}(\varrho)\setminus\Gamma^{(5)}.$

其中, 分段亚纯函数

$A(\lambda)\equiv A(\xi,\lambda):=\left\{\begin{array}{ll} (aa^{\ast})^{-\sigma_{3}/2}{\rm e}^{-{\rm i}tg_{-}(\varrho)\sigma_{3}}, &\lambda\in {\cal T}_{1}\cup{\cal T}_{2}\cup{\cal T}_{5}, \\ (aa^{\ast})^{\sigma_{3}/2}{\rm e}^{-{\rm i}tg_{+}(\varrho)\sigma_{3}}, & \lambda\in {\cal T}_{3}\cup{\cal T}_{4}. \end{array} \right.$

根据

$\begin{eqnarray*}&&g_{+}+g_{-}=g_{+}(\varrho)+g_{-}(\varrho),~~\lambda\in \gamma_{(\varrho,\alpha)},\\&&g_{+}-g_{-}=g_{+}(\varrho)-g_{-}(\varrho),~~\lambda\in \gamma_{(d,\varrho)},\end{eqnarray*}$

可得出 $n^{(\varrho_{1})}$ 满足 $n_{+}^{(\varrho_{1})} = n_{-}^{(\varrho_{1})}v^{(\varrho_{1})}$, 其中

$v^{\left(\rho_{1}\right)}=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & -\mathrm{e}^{-2 \mathrm{i} t g_{e}} \\ 0 & 1\end{array}\right), \lambda \in \mathcal{Z}_{1}, \\ \left(\begin{array}{cc}1 & 0 \\ \mathrm{e}^{2 \mathrm{i} t g_{e}} & 1\end{array}\right), & \lambda \in \mathcal{Z}_{2} \cup \mathcal{Z}_{4}, \\ \left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right), & \lambda \in \mathcal{Z}_{3}, \\ \left(\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right), & \lambda \in \mathcal{Z}_{5}.\end{array}\right.$

$\lambda$ 逼近 $\alpha$ 的情况类似, 对于 $\lambda\in D_{\epsilon}(\varrho)\setminus\Gamma^{(5)}$, 有

$\begin{matrix}n^{\varrho}(x,t,\lambda)=Y_{\varrho}(x,t,\lambda)n^{Ai}(\varpi(\lambda))A(\lambda)^{-1}\times{\rm e}^{\left[\frac{1}{2}\ln\left(\delta(\lambda)^{2}\hat{a}(\lambda)\hat{b}(\lambda)\right)-{\rm i}h(\lambda)\right]\sigma_{3}},\end{matrix}$

而且函数

$\begin{matrix}Y_{\varrho}(x,t,\lambda):=n^{mod}{\rm e}^{\left[\frac{1}{2}\ln\left(\delta(\lambda)^{2}\hat{a}(\lambda)\hat{b}(\lambda)\right)-{\rm i}h(\lambda)\right]\sigma_{3}}A(\lambda)(n_{as,M}^{Ai}(\varpi(\lambda)))^{-1},~~M\geq0\end{matrix}$

$D_{\epsilon}(\varrho)$ 上解析.

对于 $\lambda\in {\cal Z}_{3}$, $\left(n_{as,M}^{Ai}(\varpi(\lambda))\right)_{+}=-{\rm i}\sigma_{2}\left(n_{as,M}^{Ai}(\varpi(\lambda))\right)_{-}$.

因此, 结合(4.29)及(4.30)式, 可推导出

$\begin{matrix}n^{\varrho}(\lambda)n^{mod}(\lambda)^{-1}={\Bbb I}+O\left(t^{-M-1}\right),~~t\rightarrow\infty,~\lambda\in\partial D_{\epsilon}(\varrho),\end{matrix}$

并且关于 $\lambda\in D_{\epsilon}(\varrho)$ 是一致的.

开集 $D_{\epsilon}(d)$ 以及指向 $d$ 的有向曲线 ${\cal R}_{i}$ 分别被定义为 $\left\{{\cal R}_{i}\right\}_{1}^{6}$$\left\{{\cal X}_{i}\right\}_{1}^{6}$, 并且 ${\cal X}_{i} = \overline{{\cal R}}_{i-1}\cap \overline{{\cal R}}_{i}$, $i = 1, \cdots, 6$,其中 $\overline{{\cal R}}_{0}\equiv\overline{{\cal R}}_{6}$. 具体分布如图4.6 所示.

图4.6

图4.6   开集 $D_{\epsilon}(d)$ 以及集合 $\{{\cal R}_{i}\}_{1}^{6}$ 的分布图


对于 $\lambda\in D_{\epsilon}(d)$, 定义

$g_{d}(\lambda):=\int_{d}^{\lambda}{\rm d}g=\left\{\begin{array}{cc} g(\lambda)-g_{-}(d), & \lambda\in {\cal R}_{1}\cup{\cal R}_{2}\cup{\cal R}_{6}, \\ g(\lambda)-g_{+}(d), & \lambda\in {\cal R}_{3}\cup{\cal R}_{4}\cup{\cal R}_{5}, \end{array} \right.$

及分段常数函数

$\begin{matrix}B(\lambda)\equiv B(\xi, \lambda):=\left\{\begin{array}{cc} {\rm e}^{-{\rm i}tg_{-}(d)\sigma_{3}}, & \lambda\in {\cal R}_{1}\cup{\cal R}_{2}\cup{\cal R}_{6}, \\ {\rm e}^{-{\rm i}tg_{+}(d)\sigma_{3}}, & \lambda\in {\cal R}_{3}\cup{\cal R}_{4}\cup{\cal R}_{5}. \end{array} \right.\end{matrix}$

进一步地, 对于 $\lambda\in D_{\epsilon}(d)\setminus\Gamma^{(5)}$,$n^{(d_{0})}(x,t,\lambda):=\hat{n}^{(5)}(x,t,\lambda){\rm e}^{-{\rm i}h\sigma_{3}}B(\lambda),$且满足 $n_{+}^{(d_{0})}=n_{-}^{(d_{0})}v^{(d_{0})}$, 其中跳跃矩阵 $v^{(d_{0})}$

为了去除跳跃曲线 $\gamma_{(\bar{\varrho},\varrho)}$, 引入复值函数

$\begin{equation}\tilde{\delta}(\xi,\lambda)\equiv\tilde{\delta}(\lambda):=\exp\left\{\frac{1}{2\pi {\rm i}}\left(\int_{\gamma_{(\bar{\varrho},d)}}\frac{\ln(a(k)\overline{a(\bar{k})})}{k-\lambda}{\rm d}k\!-\!\int_{\gamma_{(d,\varrho)}}\frac{\ln(a(k)\overline{a(\bar{k})})}{k-\lambda}{\rm d}k\right)\right\},\lambda\in {\Bbb C}\setminus\gamma_{(\bar{\varrho},\varrho)}.\end{equation}$

那么, 跳跃曲线需要满足

$\bullet$$\ln(a(k)\overline{a(\bar{k})})$ 是连续的;

$\bullet$ 对于 $k = d$, 是严格正的.

由于 $a(\lambda)$ 不存在零点或极点, 所以函数 $aa^{\ast}$ 在所有跳跃曲线上都是非零且有限的. 一般来说, 函数 $\tilde{\delta}$ 在点 $\varrho$$\bar{\varrho}$ 处是非奇异的. 接下来, 只需要考虑在 $d$ 点附近的情况.

首先介绍函数 $\tilde{\delta}$ 的性质.

$\bullet$$\lambda\in D_{\epsilon}(d)\setminus\gamma_{(\bar{\varrho},\varrho)}$ 时, $\tilde{\delta}(\lambda)$$\tilde{\delta}(\lambda)^{-1}$ 是解析有界的;

$\bullet$ 对称性: 当 $\lambda\in {\Bbb C}\setminus\gamma_{(\bar{\varrho},\varrho)}$ 时, $\tilde{\delta}=(\tilde{\delta}^{\ast})^{-1}$;

$\bullet$ 跳跃条件: $\tilde{\delta}_{+}=\tilde{\delta}_{-}\times\left\{\begin{array}{ll} (aa^{\ast})^{-1},~ & \lambda\in \gamma_{(d,\varrho)}, \\ aa^{\ast},~ & \lambda\in \gamma_{(\bar{\varrho},d)}. \end{array}\right.$

然后, 定义$n^{(d_{1})}(x,t,\lambda):=n^{(d_{0})}(x,t,\lambda)\tilde{\delta}(\lambda)^{-\sigma_{3}},~~\lambda\in D_{\epsilon}(d),$满足跳越关系 $n_{+}^{(d_{1})}=n_{-}^{(d_{1})}v^{(d_{1})}$, 其中跳跃矩阵为

在复平面 $z$ 上由开集 $D_{\epsilon}(d)$ 到原点邻域的投影表示为映射: $\lambda\mapsto z$, 并且该映射可产生最短的跳跃曲线.

首先, 引入一个新的变量 $z \equiv z(\lambda):=\sqrt{tg_{d}(\lambda)}$.

由于 $g_{d}(\lambda)$$\lambda = d$ 处存在重根, 故有 $z={\rm i}\sqrt{t}(\lambda-d)\psi_{d}(\lambda)$.

对于 $\lambda\in D_{\epsilon}(d)$, 函数$\psi_{d}(\lambda)\equiv \psi_{d}(\xi,\lambda)$ 是解析的. 根据 $\frac{\partial g}{\partial\lambda}$ 在(4.3)式中的定义, 可计算出 $\frac{\partial^{2} g}{\partial\lambda^{2}}< 0$.

进而, 跳跃曲线必须满足条件: $\psi_{d}(d) > 0$.

对于 $\lambda\in \overline{D_{\epsilon}(d)}$, 假定 ${\rm Re} d(\lambda) > 0$, 则 $2{\rm i}tg_{d}(\lambda)=2{\rm i}z^{2}$.

集合 $D_{\epsilon}(d)$ 到原点邻域的映射记为 $\lambda\mapsto z$. 通过一系列变形分别将跳跃曲线 ${\cal X}_{3}$${\cal X}_{6}$ 映射到 $\mathbb{R} _{-}$$\mathbb{R} _{+}$ 上; 跳跃曲线 ${\cal X}_{1}$, ${\cal X}_{2}$, ${\cal X}_{4}$, 以及 ${\cal X}_{5}$ 映射到复平面 $z$ 上, 且 $\arg z$ 分别为 $\frac{\pi}{4}$, $\frac{3\pi}{4}$, $-\frac{3\pi}{4}$$-\frac{\pi}{4}$.

接下来, 将分析当 $\lambda$ 趋于 $d$ 的邻域时, $\tilde{\delta}$ 的渐进性.

函数 $\ln(\lambda-d)$ 在跳跃曲线 $\gamma_{(d,\varrho)}$$\gamma_{(d,\bar{\varrho})}$ 上分别表示为

$\begin{eqnarray*}&&\ln_{\varrho}(\lambda-d)=\ln(\lambda-d),~~\lambda\in D_{\epsilon}(d)\setminus\gamma_{(d,\varrho)},\\&&\ln_{\bar{\varrho}}(\lambda-d)=\ln(\lambda-d),~~\lambda\in D_{\epsilon}(d)\setminus\gamma_{(\bar{\varrho},d)},\end{eqnarray*}$

其中, 对于 $\lambda> d$, $\ln_{\varrho}(\lambda-d)$$\ln_{\bar{\varrho}}(\lambda-d)$ 必须是严格正的.

进一步地

$\begin{eqnarray*}&&L_{\varrho}(k,\lambda):=\ln(\lambda-k),~~k\in\gamma_{(d,\varrho)},~~\lambda\in D_{\epsilon}\setminus\gamma_{(d,\varrho)}, \\&&L_{\bar{\varrho}}(k,\lambda):=\ln(\lambda-k),~~k\in\gamma_{(\bar{\varrho},d)},~~\lambda\in D_{\epsilon}\setminus\gamma_{(\bar{\varrho},d)}.\end{eqnarray*}$

跳跃曲线必须满足以下三条

$\bullet$ 对于任意 $\lambda\in D_{\epsilon}\setminus\gamma_{(d,\varrho)}$, $L_{\varrho}(k,\lambda)$ 关于 $k\in \gamma_{(d,\varrho)}$ 连续;

$\bullet$ 对于任意 $\lambda\in D_{\epsilon}\setminus\gamma_{(\bar{\varrho},d)}$, $L_{\bar{\varrho}}(k,\lambda)$ 关于 $k\in \gamma_{(\bar{\varrho},d)}$ 连续;

$\bullet$$k= d$ 时, $L_{\varrho}(d,\lambda) = \ln_{\varrho}(\lambda-d),~~L_{\bar{\varrho}}(d,\lambda) = \ln_{\bar{\varrho}}(\lambda-d)$.

基于 $\tilde{\delta}(\lambda)$$\delta(\lambda)$ 的性质, 由分部积分法可计算出

$\begin{eqnarray*}&&\tilde{\delta}(\lambda)={\rm e}^{{\rm i}\nu[\ln_{\varrho}(\lambda-d)+\ln_{\bar{\varrho}}(\lambda-d)]+\tilde{\chi}(\lambda)},~~\lambda\in {\Bbb C}\setminus\gamma_{(\bar{\varrho},\varrho)},\\&&\delta(\lambda)={\rm e}^{-{\rm i}\nu\ln(\lambda-d)+\chi(\lambda)},~~\lambda\in{\Bbb C}\setminus(-\infty,d],\end{eqnarray*}$

其中

$\begin{aligned} \nu \equiv \nu(\xi):= & \frac{1}{2 \pi} \ln \left(1+|u|^{2}\right)>0, \\ \tilde{\chi}(\lambda):= & \frac{1}{2 \pi \mathrm{i}} L_{\varrho}(\varrho, \lambda) \ln \left(1+r(\varrho) r^{*}(\varrho)\right)+\frac{1}{2 \pi \mathrm{i}} L_{\bar{\varrho}}(\bar{\varrho}, \lambda) \ln \left(1+r(\bar{\varrho}) r^{*}(\bar{\varrho})\right) \\ & -\frac{1}{2 \pi \mathrm{i}} \int_{\gamma_{(d, e)}} L_{\varrho}(k, \lambda) \mathrm{d} \ln \left(1+r(k) r^{*}(k)\right)+\frac{1}{2 \pi \mathrm{i}} \int_{\gamma_{(\bar{\rho}, d)}} L_{\bar{\varrho}}(k, \lambda) \mathrm{d} \ln \left(1+r(k) r^{*}(k)\right), \\ \chi(\lambda):= & \frac{1}{2 \pi \mathrm{i}}\left[\ln (\lambda+1) \ln \frac{1+\left|r_{+}(-1)\right|^{2}}{1+\left|r_{-}(-1)\right|^{2}}-\left(\int_{-\infty}^{-1}+\int_{-1}^{d}\right) \ln (\lambda-k) \mathrm{d} \ln \left(1+r(k) r^{*}(k)\right)\right].\end{aligned}$

$a=-\pi$ 时, $\ln_{-\pi}\lambda= \ln\lambda$.

定义函数$p(z)\equiv p(\xi, z):={\rm e}^{-{\rm i}\nu[\ln_{-\pi/2}(z)-\ln z-\ln_{0}(z)]},~ z\in{\Bbb C}\setminus(\mathbb{R} \cup {\rm i}\mathbb{R} _{-}).$$\delta_{0}(t)$, $\delta_{1}(\lambda)$$\delta_{2}(\lambda)$ 可表示为

$\begin{eqnarray*}&&\delta_{0}(t)\equiv\delta_{0}(\xi, t):={\rm e}^{\frac{\pi\nu}{2}}t^{-\frac{{\rm i}\nu}{2}}{\rm e}^{-{\rm i}\nu\ln\psi_{d}(d)}{\rm e}^{\chi(d)+\tilde{\chi}(d)},~~t>0,\\&&\delta_{1}(\lambda)\equiv\delta_{1}(\xi, \lambda):={\rm e}^{-{\rm i}\nu\ln\frac{\psi_{d}(\lambda)}{\psi_{d}(d)}}{\rm e}^{\chi(\lambda)-\chi(d)+\tilde{\chi}(d)-\tilde{\chi}(\lambda)},~~\lambda\in D_{\epsilon}(d),\\&&\delta_{2}(\lambda)\equiv\delta_{2}(\xi,\lambda)=p(z(\lambda))\delta_{0}(t)\delta_{1}(\lambda),~~\lambda\in D_{\epsilon}(d)\setminus\left((-\infty,d]\cup\gamma_{(\bar{\varrho},\varrho)}\right).\end{eqnarray*}$

定义跳跃曲线 $X := X_{1}\cup X_{2}\cup X_{3}\cup X_{4}\subset{\Bbb C}$, 其中, $X_{1}:=\{k{\rm e}^{\frac{{\rm i}\pi}{4}}| 0\leq k<\infty\}$, $X_{2}:=\{k{\rm e}^{\frac{3{\rm i}\pi}{4}}| 0\leq k<\infty\}$,$X_{3}:=\{k{\rm e}^{-\frac{3{\rm i}\pi}{4}}| 0\leq k<\infty\}$, $X_{4}:=\{k{\rm e}^{-\frac{{\rm i}\pi}{4}}| 0\leq k<\infty\}$, 方向如图4.7 所示.

图4.7

图4.7   跳跃曲线 $X$


函数 $\rho(u, z):= {\rm e}^{{\rm i}\nu(u)\ln_{-\pi/2}z}$ 在沿着负虚轴的跳跃曲线时等价于 $z^{i\nu(u)}$, 这里 $\nu(u):=\frac{1}{2\pi}\ln(1+|u|^{2})$.

定义一个新的 Riemann-Hilbert 问题

$\begin{equation}\left\{\begin{array}{ll} n^{X}(u,\cdot)\in {\Bbb I}+\dot{E}^{2}({\Bbb C}\setminus X),\\ n^{X}_{+}(u,z)=n^{X}_{-}(u,z)v^{X}(u,z), ~z\in X, \end{array}\right.\end{equation}$

跳跃矩阵 $v^{X}(u,z)$

$v^{X}(u, z):=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & 0 \\ u \rho(u, z)^{-2} \mathrm{e}^{2 \mathrm{i} z^{2}} & 1\end{array}\right), & \lambda \in X_{1}, \\ \left(\begin{array}{cc}1 & \bar{u} \rho(u, z)^{2} \mathrm{e}^{-2 \mathrm{i} z^{2}} \\ 0 & 1\end{array}\right), & \lambda \in X_{2}, \\ \left(\begin{array}{cc}u \\ -\frac{u}{1+|u|^{2}} \rho(u, z)^{-2} \mathrm{e}^{2 \mathrm{i} z^{2}} & 1\end{array}\right), & \lambda \in X_{3}, \\ \left(\begin{array}{cc}1 & -\frac{\bar{u}}{1+|u|^{2}} \rho(u, z)^{2} \mathrm{e}^{-2 \mathrm{i} z^{2}} \\ 0 & 1\end{array}\right), & \lambda \in X_{4}.\end{array}\right.$

值得注意的是, 矩阵 $v^{X}(u,z)$ 除了零点处均连续.

$z\rightarrow0$ 时, 矩阵有快速振荡的项并且满足$v^{X}(u,\cdot)-{\Bbb I}\in L^{2}(X)\setminus L^{\infty}(X)$.

根据抛物线柱面函数[44]}, 可知 Riemann-Hilbert 问题(4.35)有显式解.

对于 $\lambda\in D_{\epsilon}(d)\setminus\Gamma^{(5)}$, 在跳跃曲线 $X$ 定义函数

$n^{(d_{2})}(x,t,z(\lambda)):=n^{(d_{2})}(x,t,\lambda)\delta_{0}(t)^{\sigma_{3}},$

且关于 $z$ 是解析的, 满足跳跃条件$n_{+}^{(d_{2})}=n_{-}^{(d_{2})}v^{(d_{2})}$, 其中跳跃矩阵为

$v^{\left(d_{2}\right)}:=\left\{\begin{array}{ll}\left(\begin{array}{cc}1 & 0 \\ \frac{\hat{b}^{*}}{\hat{a}} \delta_{1}^{-2} p(z)^{-2} \mathrm{e}^{2 \mathrm{i} z^{2}} & 1\end{array}\right), & \arg z=\frac{\pi}{4}, \\ \left(\begin{array}{cc}1 & \frac{\hat{b}}{\hat{a}^{*}} \delta_{1}^{2} p(z)^{2} \mathrm{e}^{-2 \mathrm{i} z^{2}} \\ 0 & 1\end{array}\right), & \arg z=\frac{3 \pi}{4}, \\ \left(\begin{array}{cc}1 & 0 \\ -\frac{\hat{b}^{*}}{\hat{a}^{2} \hat{a}^{*}} \delta_{1}^{-2} p(z)^{-2} \mathrm{e}^{2 \mathrm{i} z^{2}} & 1\end{array}\right), & \arg z=\frac{5 \pi}{4}, \\ \left(\begin{array}{ll}1 & -\frac{b}{\left(\hat{a}^{*}\right)^{2} \hat{a}} \delta_{1}^{2} p(z)^{2} \mathrm{e}^{-2 \mathrm{i} z^{2}} \\ 0 & 1\end{array}\right), & \arg z=\frac{7 \pi}{4},\end{array}\right.$

这里

$p(z)=\rho(z)\times\left\{\begin{array}{ll} 1, & \arg z\in(0,\pi), \\ 1+|u|^{2}, & \arg z\in(-\pi,\pi/2), \\ (1+|u|^{2})^{-1}, & \arg z\in(-\pi/2,0), \end{array}\right.$

给定 $z$, 随着 $t\rightarrow\infty$, 有 $\hat{r}(\lambda(z))\rightarrow u$, 及 $\delta_{1}(\lambda(z))\rightarrow1$. 这表明当 $t$ 足够大时, 跳跃矩阵 $v^{(d_{2})}$ 无限逼近 $v^{X}$.

因此, 对于 $\lambda\in D_{\epsilon}(d)$, 随着 $t\rightarrow\infty$, $\hat{n}^{(5)}\rightarrow n^{X}\delta_{0}^{-\sigma_{3}}\tilde{\delta}^{\sigma_{3}}B(\lambda)^{-1}{\rm e}^{{\rm i}h\sigma_{3}}$.

为了更好地逼近 $\hat{n}^{(5)}$, 定义函数

$\begin{matrix}n^{d}(x,t,\lambda)=Y_{d}(x,t,\lambda)n^{X}(u,z(\lambda))\delta_{0}(t)^{-\sigma_{3}}\tilde{\delta}(\lambda)^{\sigma_{3}}B(\lambda)^{-1}{\rm e}^{{\rm i}h\sigma_{3}},\end{matrix}$

其中, 函数

$\begin{matrix}Y_{d}(x,t,\lambda):=n^{mod}(x,t,\lambda){\rm e}^{-{\rm i}h(\lambda)\sigma_{3}}B(\lambda)\tilde{\delta}(\lambda)^{-\sigma_{3}}\delta_{0}(t)^{\sigma_{3}}\end{matrix}$

$D_{\epsilon}(d)$ 上解析.

$\lambda\in D_{\epsilon}(d)$ 时, 函数$f(\lambda) :=n^{mod}(x,t,\lambda){\rm e}^{-{\rm i}h(\lambda)\sigma_{3}}$除了在跳跃曲线 ${\cal X}_{3}\cup {\cal X}_{6}$ 上都是解析的, 故有

$f_{+}=f_{-} \times\left\{\begin{array}{cc}\left(\begin{array}{cc}\frac{\mathrm{e}^{\mathrm{i} t\left(g_{+}-g_{-}\right)}}{a a^{*}} & 0 \\ 0 & a a^{*} \mathrm{e}^{-\mathrm{i} t\left(g_{+}-g_{-}\right)}\end{array}\right), \lambda \in \mathcal{X}_{3}, \\ \left(\begin{array}{cc}a a^{*} \mathrm{e}^{\mathrm{i} t\left(g_{+}-g_{-}\right)} & 0 \\ 0 & \frac{\mathrm{e}^{-\mathrm{i} t\left(g_{+}-g_{-}\right)}}{a a^{*}}\end{array}\right), \lambda \in \mathcal{X}_{6}.\end{array}\right.$

由(4.32)和(4.34)式可推断 $\tilde{\delta}(\lambda)^{\sigma_{3}}B(\lambda)^{-1}$ 满足在跳跃曲线 ${\cal X}_{3}\cup {\cal X}_{6}$ 上的关系式. 因此, 对于 $\lambda\in D_{\epsilon}(d)$, $Y_{d}$ 为解析函数.

集合 $\Gamma^{(5)}$$D_{\epsilon}(d)$ 上十字的并集, 记为 ${\cal X}:={\cal X}_{1}\cup{\cal X}_{2}\cup {\cal X}_{4}\cup {\cal X}_{5}$,跳跃曲线 $\gamma_{(\bar{\varrho},\varrho)}\cap D_{\epsilon}(d)={\cal X}_{3}\cup {\cal X}_{6}$.

对于 $\lambda\in D_{\epsilon}(d)\setminus\Gamma^{(5)}$, 由(4.37)式定义的函数 $n^{d}(x,t,\lambda)$ 是解析有界的.

对于 $\lambda\in D_{\epsilon}(d)$, 由(4.38)式定义的函数 $Y_{d}(x,t,\lambda)$ 也是解析有界的.

因此, 当$\lambda\in\gamma_{(\bar{\varrho},\varrho)}\cap D_{\epsilon}(d)$ 时, $n^{d}$ 满足关系式 $n_{+}^{d}=n_{-}^{d}v^{d}$, 其中跳跃矩阵 $v^{d}=\hat{v}^{(5)}$.

进一步地, 随着 $t\rightarrow\infty$, 则有

$\begin{matrix}\left\{\begin{array}{ll} \|\hat{v}^{(5)}-v^{d}\|_{L^{1}(\chi)}=O\left(t^{-1}\ln t\right), \\ \|\hat{v}^{(5)}-v^{d}\|_{L^{2}(\chi)}=O\left(t^{-3/4}\ln t\right), \\ \|\hat{v}^{(5)}-v^{d}\|_{L^{\infty}(\chi)}=O\left(t^{-1/2}\ln t\right). \end{array} \right.\end{matrix}$

此外, 当 $\lambda\in\partial D_{\epsilon}(d)$ 时, $n^{mod}(n^{d})^{-1}$ 满足

$\begin{matrix}\|n^{mod}(n^{d})^{-1}-{\Bbb I}\|_{L^{\infty}(\partial D_{\epsilon}(d))}=O\big(t^{-1/2}\big).\end{matrix}$

随着 $t\rightarrow\infty$,有

$\frac{1}{2\pi{\rm i}}\int_{\partial D_{\epsilon}(d)}(n^{mod}(n^{d})^{-1}-{\Bbb I}){\rm d}\lambda=\frac{Y_{d}(x,t,d)n_{1}^{X}Y_{d}(x,t,d)^{-1}}{\sqrt{t}\psi_{d}(d)}+O\left(t^{-1}\right),$

其中

$n_{1}^{X}(\xi) \equiv n_{1}^{X}:=\left(\begin{array}{cc}0 & -\mathrm{e}^{\pi \nu} \varrho^{X}(u) \\ \mathrm{e}^{\pi \nu} \overline{\varrho^{X}(u)} & 0\end{array}\right)$

因此, 系数 $E_{1}$ 可表示为

$\begin{matrix}E_{1}(\xi,t)=-2{\rm i}{\rm e}^{2{\rm i}(tg(0)+h(\infty))}\frac{\left(Y_{d}(x,t,d)n_{1}^{X}Y_{d}(x,t,d)^{-1}\right)_{12}}{\psi_{d}(d)}.\end{matrix}$

基于上述分析, 局部解 $n^{\bar{\alpha}}(x,t,\lambda)$$n^{\bar{\varrho}}(x,t,\lambda)$ 均满足对称性(4.10).

记五个开集的并为 ${\cal D}$. 更具体地,${\cal D}=D_{\epsilon}(\bar{\alpha})\cup D_{\epsilon}(\bar{\varrho})\cup D_{\epsilon}(\varrho)\cup D_{\epsilon}(\alpha)\cup D_{\epsilon}(d)$.

接下来引入近似解

$\begin{matrix}n^{app}:=\left\{\begin{array}{ll} n^{\alpha}, & \lambda\in D_{\epsilon}(\alpha), \\ n^{\varrho}, & \lambda\in D_{\epsilon}(\varrho), \\ n^{\bar{\alpha}}, & \lambda\in D_{\epsilon}(\bar{\alpha}), \\ n^{\bar{\varrho}}, & \lambda\in D_{\epsilon}(\bar{\varrho}), \\ n^{d}, & \lambda\in D_{\epsilon}(d), \\ n^{mod}, & \lambda\in {\Bbb C}\setminus{\cal D}, \end{array} \right.\end{matrix}$

其满足 Riemann-Hilbert 问题, 且跳跃曲线为

$\Gamma^{app}=\Gamma^{mod}\cup\partial {\cal D}\cup {\cal Y}\cup {\cal Y}^{\ast}\cup {\cal Z}\cup {\cal Z}^{\ast}\cup {\cal X},$

具体分布如图4.8所示.

图4.8

图4.8   跳跃曲线 $\Gamma^{app}$, ${\rm Im} g <0$ (粉色区域), ${\rm Im} g >0$ (白色区域)


设跳跃曲线

$\tilde{\Gamma} = (\Gamma^{(5)} \setminus(\Gamma^{mod}\cup\bar{{\cal D}}))\cup \partial {\cal D }\cup {\cal X}.$

具体分布如图4.9所示.函数 $\tilde{n}(x,t,\lambda)$ 可表示为 $\tilde{n}:=\hat{n}^{(5)}(n^{app})^{-1}$, 从而使得当 $t\rightarrow\infty$ 时, $\tilde{n}\rightarrow{\Bbb I}$, 同时对于 $\lambda\in \tilde{\Gamma}$, 其为 Riemann-Hilbert 问题(4.8)的解.

图4.9

图4.9   跳跃曲线 $\tilde{\Gamma}$, ${\rm Im} g <0$ (粉色区域), ${\rm Im} g >0$ (白色区域)


相应的跳跃矩阵可表示为

$\begin{matrix} \tilde{v}:=\left\{\begin{array}{ll} n^{mod}\hat{v}^{(5)}(n^{mod})^{-1}, & \lambda\in \tilde{\Gamma}\setminus\bar{{\cal D}},\\ n^{mod}(n^{\alpha})^{-1}, & \lambda\in \partial D_{\epsilon}(\alpha), \\ n^{mod}(n^{\varrho})^{-1}, & \lambda\in \partial D_{\epsilon}(\varrho), \\ n^{mod}(n^{\bar{\alpha}})^{-1}, & \lambda\in \partial D_{\epsilon}(\bar{\alpha}), \\ n^{mod}(n^{\bar{\varrho}})^{-1}, & \lambda\in \partial D_{\epsilon}(\bar{\varrho}), \\ n^{mod}(n^{d})^{-1}, & \lambda\in \partial D_{\epsilon}(d), \\ n_{-}^{d}\hat{v}^{(5)}(n_{+}^{d})^{-1}, & \lambda\in {\cal X}. \end{array} \right.\end{matrix}$

$\hat{\omega}=\tilde{v}-{\Bbb I}$.$t\rightarrow\infty$ 时, $\hat{v}^{(5)}-{\Bbb I}$$\Gamma^{(5)}\setminus\Gamma^{mod}$ 上以指数形式衰减. 更具体地

$\begin{matrix}\|\hat{\omega}\|_{(L^{1}\cap L^{2}\cap L^{\infty})(\tilde{\Gamma}\setminus\bar{{\cal D}})}=O\left({\rm e}^{-bt}\right),~~t\rightarrow\infty,\end{matrix}$

其中 $b \equiv b(\xi) > 0$.

由(4.27)和(4.31)式, 可计算出

$\begin{matrix}\|\hat{\omega}\|_{L^{\infty}(\partial D_{\epsilon}(\alpha)\cup\partial D_{\epsilon}(\varrho)\cup\partial D_{\epsilon}(\bar{\alpha})\cup\partial D_{\epsilon}(\bar{\varrho}))}=O\left(t^{-M}\right),~~t\rightarrow\infty,~~\forall M \geq 1.\end{matrix}$

同样地, 由(4.40)式, 可得

$\begin{equation}\|\hat{\omega}\|_{L^{\infty}(\partial D_{\epsilon}(d))}=O\big(t^{-1/2}\big),~~t\rightarrow\infty.\end{equation}$

对于 $\lambda\in{\cal X}$, $\hat{\omega}=n_{-}^{d}(\hat{v}^{(5)}-v^{d})(n_{+}^{d})^{-1}$.

根据(4.39)式, 随着 $t\rightarrow\infty$, 则有

$\begin{matrix}\left\{\begin{array}{ll} \|\hat{\omega}\|_{L^{1}(\chi)}=O\left(t^{-1}\ln t\right), \\ \|\hat{\omega}\|_{L^{2}(\chi)}=O\left(t^{-3/4}\ln t\right), \\ \|\hat{\omega}\|_{L^{\infty}(\chi)}=O\left(t^{-1/2}\ln t\right). \end{array} \right.\end{matrix}$

由(4.43), (4.44), (4.45)以及(4.46)式可知

$\left\{\begin{array}{ll} \|\hat{\omega}\|_{(L^{1}\cap L^{2})(\tilde{\Gamma})}=O\left(t^{-1/2}\right), \\ \|\hat{\omega}\|_{L^{\infty}(\tilde{\Gamma})}=O\left(t^{-1/2}\ln t\right). \end{array} \right.$

进而, 对于 $\lambda\in{\Bbb C}\setminus\tilde{\Gamma}$, 有柯西算子 $(\hat{{\cal C}}f)(\lambda)=\frac{1}{2\pi {\rm i}}\int_{\tilde{\Gamma}}\frac{f(k)}{k-\lambda}{\rm d}k$, 在 $\tilde{\Gamma}$ 左右两边的边值记为 $\hat{{\cal C}}_{\pm}f$. 根据文献[42]可知柯西算子的定义 $\hat{{\cal C}}_{\hat{\omega}}: L^{2}(\tilde{\Gamma})\rightarrow L^{2}(\tilde{\Gamma})$, 故有$\hat{{\cal C}}_{\hat{\omega}}f = \hat{{\cal C}}_{-}(f\hat{\omega})$.

$t\rightarrow\infty$ 时, $ \|\hat{{\cal C}}_{\hat{\omega}}\|_{{\cal B}(L^{2}(\tilde{\Gamma}))}\leq C\|\hat{\omega}\|_{L^{\infty}(\tilde{\Gamma})}=O\left(t^{-1/2}\ln t\right)$.特别地, ${\Bbb I}-\hat{{\cal C}}_{\hat{\omega}}(\xi,t,\cdot)\in B\big(L^{2}(\tilde{\Gamma})\big)$ 是可逆的.

对于 $\hat{d}(x, t,\lambda)\in{\Bbb I}+L^{2}(\tilde{\Gamma})$, 则 $\hat{d}:={\Bbb I}+({\Bbb I}-\hat{{\cal C}}_{\hat{\omega}})^{-1}\hat{{\cal C}}_{\hat{\omega}}$.

根据 Neumann 级数的标准估计, 可得

$\|\hat{d}-{\Bbb I}\|_{L^{2}(\tilde{\Gamma})}\leq B\frac{\|\hat{\omega}\|_{(L^{2})(\tilde{\Gamma})}}{1-\|\hat{{\cal C}}_{\hat{\omega}}\|_{{\cal B}(L^{2}(\tilde{\Gamma}))}}.$

$t\rightarrow\infty$ 时,有

$\begin{equation}\|\hat{d}(x,t,\cdot)-{\Bbb I}\|_{L^{2}(\tilde{\Gamma})}=O\big(t^{-1/2}\big).\end{equation}$

此外, 近似解 $\tilde{n}(x,t,\lambda)$ 是唯一的, 其形式为

$\tilde{n}(x,t,\lambda)={\Bbb I}+\hat{{\cal C}}(\hat{d}\hat{\omega})={\Bbb I}+\frac{1}{2\pi {\rm i}}\int_{\tilde{\Gamma}}\hat{d}(x,t,k)\hat{\omega}(x,t,k)\frac{{\rm d}k}{k-\lambda}.$

由于 $\hat{\omega}$ 随着 $\lambda\rightarrow\infty$ 指数衰减, 可知存在极限

$\begin{matrix}\lim_{\lambda\rightarrow\infty}\lambda\left(\tilde{n}(x,t,\lambda)-{\Bbb I}\right)=-\frac{1}{2\pi {\rm i}}\int_{\tilde{\Gamma}}\hat{d}(x,t,k)\hat{\omega}(x,t,k){\rm d}k,\end{matrix}$

关于 $\arg \lambda\in [2\pi]$ 是一致的.

根据(4.46)和(4.47)式, 可得

$O(\|\hat{\omega}\|_{L^{1}({\cal X})})+O(\|\hat{d}-{\Bbb I} \|_{L^{2}({\cal X})}\|\hat{\omega}\|_{L^{2}({\cal X})})=O\left(t^{-1}\ln t\right),~~t\rightarrow\infty.$

进一步地, (4.48)式可表示为

$\lim_{\lambda\rightarrow\infty}\lambda(\tilde{n}(x,t,\lambda)-{\Bbb I})=-\frac{Y_{d}(x,t,d)n_{1}^{X} Y_{d}(x,t,d)^{-1}}{\sqrt{t}\psi_{d}(d)}+O\left(t^{-1}\ln t\right),~~t\rightarrow\infty.$

综上所述, 近似解 $\hat{n}(x,t,\lambda)$ 可表示为

$\hat{n}(x,t,\lambda)={\rm e}^{{\rm i}\left(tg^{(0)}+h(\infty)\right)\sigma_{3}}\tilde{n}(x,t,\lambda)n^{mod}(x,t,\lambda){\rm e}^{-{\rm i}h\sigma_{3}}\delta^{\sigma_{3}}{\rm e}^{-{\rm i}t(g(\lambda)-\phi(\lambda))\sigma_{3}},~~\forall~\lambda\in U_{2}.$

基于(4.2)式, 聚焦 KE 方程(1.2)的解 $u(x,t)$ 有如下形式

$\begin{eqnarray*}u(x,t)&=&2{\rm i}{\rm e}^{-2{\rm i}\beta\int_{x}^{+\infty}(|u(y,t)|^{2}-A_{j}^{2}){\rm d}y}\lim_{\lambda\rightarrow\infty}\lambda(\hat{n}(x,t,\lambda))_{12}\\&=&2{\rm i}{\rm e}^{2{\rm i}\left(tg^{(0)}+h(\infty)-\beta\int_{x}^{+\infty}(|u(y,t)|^{2}-A_{j}^{2}){\rm d}y\right)}\left(\lim_{\lambda\rightarrow\infty}\lambda(n^{mod}(x,t,\lambda))_{12}\!+\!\lim_{\lambda\rightarrow\infty}\lambda(\tilde{n}(x,t,\lambda))_{12}\right).\end{eqnarray*}$

进而定理1.2成立.

5 结论

本文研究了聚焦 KE 方程(1.2)在无穷远处初值为两个不同的平面波形式(1.3)下解的长时间渐进性. 首先将柯西问题转化成基本的 Riemann-Hilbert 问题, 然后通过对该问题的跳跃曲线进行一系列的变形, 将其转化为可解的矩阵 Riemann-Hilbert 问题. 利用非线性速降法, 最终得到不同情况下的渐近解.所研究的非线性可积系统由于参数 $\beta$ 的存在而变得更加普遍. 鉴于五次非线性项和自频移效应, 可以描述更加复杂的物理现象, 进一步有助于理解非线性科学中调制不稳定性的阶段. 更重要的是, 本文所用到的非线性速降法为探索数学物理和工程中其他非线性可积模型的长时间渐近性提供了强有力的数学工具.

非局部可积系统是近年来数学物理领域研究的热点之一, 因此, 值得注意的是, 非局部 NLS 型方程解的长时间渐进性是否可以用非线性速降法进行分析? 这些问题将留待以后讨论.

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[本文引用: 1]

Wang X B, Han B.

A Riemann-Hilbert approach to a generalized nonlinear Schrödinger equation on the quarter plane

Math Phys Anal Geom, 2020, 23: 25

DOI:10.1007/s11040-020-09347-1      [本文引用: 2]

Guo B L, Nan L.

Long-time asymptotics for the Kundu-Eckhaus equation on the half-line

J Math Phys, 2018, 59: 061505

DOI:10.1063/1.5020996      URL     [本文引用: 2]

Wang D S, Wang X L.

Long-time asymptotics and the bright $N$-soliton solutions of the Kundu-Eckhaus equation via the Riemann-Hilbert approach

Nonlinear Anal Real World Appl, 2018, 41: 334-361

DOI:10.1016/j.nonrwa.2017.10.014      URL     [本文引用: 2]

Wang D S, Guo B L, Wang X L.

Long-time asymptotics of the focusing Kundu-Eckhaus equation with nonzero boundary conditions

J Differential Equations, 2019, 266: 5209-5253

DOI:10.1016/j.jde.2018.10.053      URL     [本文引用: 2]

Lenells J.

The nonlinear steepest descent method for Riemann-Hilbert problems of low regularity

Indiana Math J, 2017, 66(4): 1287-1332

DOI:10.1512/iumj.2017.66.6078      URL     [本文引用: 2]

Lenells J.

Matrix Riemann-Hilbert problems with jumps across Carleson contours

Monatsh Math, 2018, 186(1): 111-152

DOI:10.1007/s00605-017-1019-0      PMID:31258193      [本文引用: 1]

We develop a theory of -matrix Riemann-Hilbert problems for a class of jump contours and jump matrices of low regularity. Our basic assumption is that the contour is a finite union of simple closed Carleson curves in the Riemann sphere. In particular, unbounded contours with cusps, corners, and nontransversal intersections are allowed. We introduce a notion of -Riemann-Hilbert problem and establish basic uniqueness results and Fredholm properties. We also investigate the implications of Fredholmness for the unique solvability and prove a theorem on contour deformation.

Its A R.

Asymptotics of solutions of the nonlinear Schrödinger equation and isomonodromic deformations of systems of linear differential equations

Dokl akad nauk Sssr, 1981, 24: 452-456

[本文引用: 1]

Zakharov V E, Ostrovsky L A.

Modulation instability: the beginning

Physica D, 2009, 238: 540-548

DOI:10.1016/j.physd.2008.12.002      URL    

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