## 具有非零边界条件的混合 Chen-Lee-Liu 导数非线性薛定谔方程的单极解和双极解

1四川师范大学数学科学系 成都 610100

2电子科技大学数学科学系 成都611731

## Simple-Pole and Double-Pole Solutions for the Mixed Chen-Lee-Liu Derivative Nonlinear Schrödinger Equation with Nonzero Boundary Conditions

Wang Chunjiang,1,*, Zhang Jian,2

1School of Mathematical Sciences, Sichuan Normal University, Chengdu 610100

2School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu 611731

 基金资助: 国家自然科学基金.  11871138

 Fund supported: The NSFC.  11871138

Abstract

This paper is concerned with simple-pole and double-pole solutions for the mixed Chen-Lee-Liu derivative nonlinear Schrödinger equation with non-zero boundary conditions at infinity. By solving a direct scattering problem, the Jost eigenfunctions and scattering matrix are given, their symmetries and asymptotic behaviors are also presented. Then the inverse scattering problems are solved in terms of the matrix Riemann-Hilbert method. In addition, the trace formulae for analytic scattering coefficients and theta conditions are derived. Finally, the explicit formulae of double-pole solutions for the equation are obtained.

Keywords： Nonlinear Schrödinger equation ; Non-zero boundary conditions ; Inverse scattering ; Riemann-Hilbert problem ; Double-pole solution

Wang Chunjiang, Zhang Jian. Simple-Pole and Double-Pole Solutions for the Mixed Chen-Lee-Liu Derivative Nonlinear Schrödinger Equation with Nonzero Boundary Conditions. Acta Mathematica Scientia[J], 2023, 43(1): 101-122 doi:

## 1 引言

$\begin{matrix}\label{1.1} \left\{ \begin{array}{lll} {\rm i}r_t+r_{xx}+|r|^2r-{\rm i}|r|^2r_x=0,\ (t,x)\in\Bbb R\times\Bbb R,\\ \lim\limits_{x\rightarrow \pm\infty}|r(t,x)|=r_0=c>0,\ t\in\Bbb R. \end{array} \right. \end{matrix}$

Hu 等[5] 研究了半直线上(1.1)的初始边值问题. Zhang 等[6] 通过 Darboux 变换得到了(1.1)的高阶解. 然后, Fang 等 [7] 得到了具有零边界条件的(1.1)的孤子解. 最近, Zhao 和 Fan [8] 通过 Riemann-Hilbert 方法研究了具有非零边界条件的(1.1)的多孤子解.

$$$\label{1.2} \lim\limits_{x\rightarrow \pm\infty}r(t,x)=r_\pm {\rm e}^{\left(-\frac{\rm i}{2}r_0^2-{\rm i}\right)x -\left(\frac{3{\rm i}}{4}r_0^4+{\rm i}r_0^2+{\rm i}\right)t}.$$$

### 2.1 黎曼面与单值化坐标

$$$\phi_{x}=X\phi,\quad \phi_{t}=T\phi,$$$

$\begin{eqnarray*} X&=&{\rm i}k^2\sigma_{3}+kQ-\frac{\rm i}{2}\sigma_{3}+\frac{\rm i}{4}Q^2\sigma_{3},\\ T&=&-2{\rm i}k^4\sigma_{3}-2k^3Q+k^2(2{\rm i}\sigma_{3}-{\rm i}Q^2\sigma_{3})+k(Q+{\rm i}\sigma_{3}Q_x-\frac{1}{2}Q^3)\\ &&-\frac{\rm i}{2}\sigma_{3}-\frac{\rm i}{8}Q^4\sigma_{3}+\frac{1}{4}(QQ_x-Q_xQ), \end{eqnarray*}$
${Q = \left( {\begin{array}{*{20}{c}} 0~~&r\\ -r^*~~&0 \end{array}} \right),\quad \sigma_{3} = \left( {\begin{array}{*{20}{c}} 1~~&0\\ 0~~&{ - 1} \end{array}} \right)}.$

$\begin{matrix} \left\{ \begin{array}{lll} r \rightarrow r{\rm e}^{\left(-\frac{\rm i}{2}r_0^2-{\rm i}\right)x-\left(\frac{3{\rm i}}{4}r_0^4+{\rm i}r_0^2+{\rm i}\right)t},\\ \phi \rightarrow {\rm e}^{\left[\left(-\frac{\rm i}{4}r_0^2-\frac{\rm i}{2}\right)x-\left(\frac{3{\rm i}}{8}r_0^4 +\frac{\rm i}{2}r_0^2+\frac{\rm i}{2}\right)t\right]\sigma_{3}}\phi. \end{array} \right. \end{matrix}$

$\begin{matrix}\label{2.3} \left\{ \begin{array}{lll} {\rm i}r_t+r_{xx}+\frac{1}{2}r_0^4r-{\rm i}r_0^2r_x-2{\rm i}r_x-{\rm i}|r|^2r_x-\frac{1}{2}r|r|^2r_0^2=0,\\ \lim\limits_{x\to\pm\infty}r(x,t)=r_\pm,\ |r_\pm|=r_0. \end{array} \right. \end{matrix}$

$$$\label{2.5} \phi_{x}=X\phi,\quad \phi_{t}=T\phi,$$$

$\begin{eqnarray*} X&=&{\rm i}k^2\sigma_{3}-\frac{\rm i}{4}(|r|^2-r_0^2)\sigma_{3}+kQ,\\ T&=&-2{\rm i}k^4\sigma_{3}-2k^3Q+k^2(2{\rm i}\sigma_{3}-{\rm i}Q^2\sigma_{3})+k \left(\frac{1}{2}r_0^2Q+2Q+{\rm i}\sigma_{3} Q_x-\frac{1}{2}Q^3\right)\\ &&+\frac{1}{4}(QQ_x-Q_xQ) +\left(-\frac{\rm i}{4}|r|^2r_0^2-\frac{\rm i}{2}|r|^2+\frac{3{\rm i}}{8}r_0^4+\frac{\rm i}{2}r_0^2 -\frac{\rm i}{8}|r|^4\right)\sigma_3. \end{eqnarray*}$

$x\rightarrow\pm\infty$ 时, 考虑(2.4)的渐近散射问题

$\begin{matrix}\label{2.6} \psi_{x}=X_\pm\psi,\quad \psi_{t}=T_\pm\psi, \end{matrix}$

$X_{\pm}={\rm i}k^2\sigma_{3}+kQ_{\pm},\quad T_{\pm}=(-2k^2+2+r_0^2)X_{\pm},$
$Q_{\pm}= \left( \begin{array}{*{20}{c}} 0~~&r_{\pm}\\ -r_{\pm}^*~~&0 \end{array} \right).$

$k+{\rm i}r_0=r_1{\rm e}^{{\rm i} \theta_{1}},\quad k-{\rm i}r_0=r_2{\rm e}^{{\rm i} \theta_{2}},\quad -\frac{\pi}{2}<\theta_{j}<\frac{3\pi}{2},\quad j=1,2,$

$\begin{eqnarray*} \lambda(k)=\left\{ \begin{array}{lll} (r_1r_2)^{\frac{1}{2}}{\rm e}^{{\rm i}\left(\frac{\theta_1+\theta_2}{2}\right)},\quad & \ k\in S_1,\\ -(r_1r_2)^{\frac{1}{2}}{\rm e}^{{\rm i}\left(\frac{\theta_1+\theta_2}{2}\right)},\quad & \ k\in S_2. \end{array} \right. \end{eqnarray*}$

$z=k+\lambda,$

$k(z)=\frac{1}{2}(z-\frac{r_0^2}{z}),\quad \lambda(z)=\frac{1}{2}(z+\frac{r_0^2}{z}).$

$\begin{eqnarray*} D^+=\left \{z:{\rm RezImz}>0\right \},\quad D^-=\left \{z:{\rm RezImz}<0\right \},\quad \Sigma=\Bbb R \cup {\rm i}\Bbb R\setminus\{0\}. \end{eqnarray*}$

### 2.2 Jost 解的解析性

$\begin{matrix}\label{3.1} Y_{\pm}= \left( {\begin{array}{*{30}{c}} 1~~&\displaystyle{\frac{{\rm i}r_{\pm}}{z}}\\ \displaystyle{\frac{{\rm i}r_{\pm}^*}{z}}~~&1 \end{array}} \right)=I+\frac{\rm i}{z}\sigma_3Q_\pm. \end{matrix}$

$\bullet$$\phi_{-,1}, \phi_{+,2} 解析延伸到 D^- 且连续延伸到 D^-\cup \Sigma_0. \bullet$$\phi_{+,1}$, $\phi_{-,2}$ 解析延伸到 $D^+$ 且连续延伸到 $D^+\cup \Sigma_0$.

### 2.3 散射矩阵的解析性

$$$\label{4.3} \phi_+(x,t,z)=\phi_-(x,t,z)S(z),\quad z\in\Sigma_0,$$$

$s_{11}(z)=\frac{{\rm Wr} (\phi_{+,1},\phi_{-,2})}{\gamma},\quad s_{12}(z)=\frac{{\rm Wr}(\phi_{+,2},\phi_{-,2})}{\gamma},$
$s_{21}(z)=\frac{{\rm Wr}(\phi_{-,1},\phi_{+,1})}{\gamma},\quad s_{22}(z)=\frac{{\rm Wr}(\phi_{-,1},\phi_{+,2})}{\gamma},$

$\bullet$$s_{11}(z) 解析延拓到 D^+ 且连续延拓到 D^+\cup \Sigma_0. \bullet$$s_{22}(z)$ 解析延拓到 $D^-$ 且连续延拓到 $D^-\cup \Sigma_0$.

$\bullet$$s_{12}(z), s_{21}(z)$$z\in\Sigma_{0}$ 中连续.

$$$\rho(z)=\frac{s_{21}(z)}{s_{11}(z)},\quad \tilde{\rho}(z)=\frac{s_{12}(z)}{s_{22}(z)}.$$$

### 2.4 Jost 解和散射矩阵的对称性

$\bullet$ 第一对称性

$\begin{matrix}\label{5.1} \phi_\pm(x,t,z)=\sigma_2\phi_\pm^*(x,t,z^*)\sigma_2,\quad S(z)=\sigma_2S^*(z^*)\sigma_2,\quad \rho(z)=-\tilde{\rho}^*(z^*), \end{matrix}$

$\bullet$ 第二对称性

$\begin{matrix}\label{5.2} \phi_\pm(x,t,z)=\sigma_1\phi_\pm^*(x,t,-z^*)\sigma_1,\quad S(z)=\sigma_1S^*(-z^*)\sigma_1,\quad \rho(z)=\tilde{\rho}^*(-z^*), \end{matrix}$

$\bullet$ 第三对称性

$$$\label{5.3} \phi_\pm(x,t,z)=\frac{\rm i}{z}\phi_\pm(x,t,-\frac{r_0^2}{z})\sigma_3Q_\pm,\ \ S(z)=(\sigma_3Q_-)^{-1}S(-\frac{r_0^2}{z})\sigma_3Q_+, \ \ \rho(z)=\frac{r_-^*}{r_-}\tilde{\rho}(-\frac{r^2_0}{z}).$$$

$\phi(x,t,z)$ 是散射问题(2.4)的一个解, 即

$\phi_x=\left({\rm i}k^2\sigma_3-\frac{\rm i}{4}(|r|^2-r_0^2)\sigma_3+kQ\right)\phi,$

$[\sigma_2\phi^*(z^*)\sigma_2]_x=\sigma_2\left[\left(-{\rm i}(k^2(z^*))^*\sigma_3+\frac{\rm i}{4}(|r|^2-r_0^2)\sigma_3 +k^*(z^*)Q^*\right)\right]\phi^*(z^*)\sigma_2.$

$\sigma_2\sigma_3\sigma_2=-\sigma_3,\quad \sigma_2Q^*\sigma_2=Q,\quad k^*(z^*)=k(z),\quad \sigma_2^2=I,$

$[\sigma_2\phi^*(z^*)\sigma_2]_x=\left[{\rm i}k^2(z)\sigma_3-\frac{\rm i}{4}(|r|^2-r_0^2)\sigma_3+k(z)Q\right] [\sigma_2\phi^*(z^*)\sigma_2].$

$\sigma_2Y_\pm^*(z^*)\sigma_2=Y_\pm(z),\quad \sigma_2{\rm e}^{-{\rm i}\theta^*(z^*)\sigma_3}\sigma_2={\rm e}^{{\rm i}\theta(z)\sigma_3},$

$\sigma_2\phi^*(z^*)\sigma_2=Y_\pm(x,t,z){\rm e}^{{\rm i}\theta(z)\sigma_3}+O(1),\quad x \rightarrow \pm\infty.$

$\phi_\pm(x,t,z)=\sigma_2\phi_\pm^*(x,t,z^*)\sigma_2,$

$$$\label{5.9} \phi_{\pm,1}(z)={\rm i}\sigma_2\phi_{\pm,2}^*(z^*),\quad \phi_{\pm,2}(z)=-{\rm i}\sigma_2\phi_{\pm,1}^*(z^*).$$$

$\begin{eqnarray*} &&\sigma_1\sigma_3\sigma_1=-\sigma_3,\quad \sigma_1Q^*\sigma_1=-Q,\quad k^*(-z^*)=-k(z),\quad \\ &&\sigma_1Y_\pm^*(-z^*)\sigma_1=Y_\pm(z),\quad \sigma_1{\rm e}^{-{\rm i}\theta^*(-z^*)\sigma_3}\sigma_1={\rm e}^{{\rm i}\theta(z)\sigma_3}. \end{eqnarray*}$

$\phi_\pm(x,t,z)=\sigma_1\phi_\pm^*(x,t,-z^*)\sigma_1,$

$\phi_{\pm,1}(z)=\sigma_1\phi_{\pm,2}^*(-z^*),\quad \phi_{\pm,2}(z)=\sigma_1\phi_{\pm,1}^*(-z^*).$

$k(-\frac{r_0^2}{z})=k(z),\quad \lambda(-\frac{r_0^2}{z})=-\lambda(z),\quad \theta(-\frac{r_0^2}{z})=-\theta(z),$

$\begin{eqnarray*} A_+[z_n]&=&A_+[-z_n]=-\frac{r^*_-}{r_-}\frac{z^{2}_n}{r^2_0}A^*_+\left[-\frac{r^2_0}{z^*_n}\right]= -\frac{r^*_-}{r_-}\frac{z^{2}_n}{r^2_0}A^*_+\left[\frac{r^2_0}{z^*_n}\right]\\ &=&-A^*_-[z^*_n]=-A^*_-[-z^*_n]=\frac{r^*_-}{r_-}\frac{z^2_n}{r^2_0}A_- \left[-\frac{r^2_0}{z_n}\right]= \frac{r^*_-}{r_-}\frac{z^2_n}{r^2_0}A_-\left[\frac{r^2_0}{z_n}\right],\\ A_+[\omega_m]&=&A_+[-\omega_m]=- A^*_-[\omega^*_m]=-A^*_-[-\omega^*_m]. \end{eqnarray*}$

$z=-z_n$ 带入(2.26)式, 由 Jost 解的对称性可得

$\begin{eqnarray*} \mu_{+,1}(x,t,-z_n)=-b_+(z_n){\rm e}^{-2{\rm i}\theta(-z_n)}\mu_{-,2}(x,t,-z_n). \end{eqnarray*}$

$s_{11}(z)$ 是一个偶函数, 即 $s^{\prime}_{11}(-z_n)=-s_{11}^{\prime}(z_n)$, 则

$\begin{eqnarray*} A_+[z_n]=A_+[-z_n]. \end{eqnarray*}$

$\begin{eqnarray*} &&A_+\left[-\frac{r^2_0}{z^*_n}\right]=A_+\left[\frac{r^2_0}{z^*_n}\right], \ A_+\left[\omega_m\right]=A_+\left[-\omega_m\right], \\ && A_-\left[-\frac{r^2_0}{z_n}\right]=A_-\left[\frac{r^2_0}{z_n}\right],\ A_-\left[z^*_n\right]=A_-\left[-z^*_n\right],\ A_-\left[\omega^*_m\right]=A_-\left[-\omega^*_m\right]. \end{eqnarray*}$

$\sigma_2\mu^*_{+,2}(x,t,z^*_n)=b^*_-(z^*_n){\rm e}^{-2{\rm i}\theta^*(z^*_n)}\sigma_2\mu^*_{-,1}(x,t,z^*_n).$

$\mu_{+,1}(x,t,z_n)=-b^*_-(z^*_n){\rm e}^{-2{\rm i}\theta(z_n)}\mu_{-,2}(x,t,z_n).$

$z=-{\frac{r^2_0}{z^*_n}}$ 带入(2.27)式, 根据 Jost 解的第三对称性知

$\begin{eqnarray*} \mu_{+,1}\left(x,t,-\frac{r^2_0}{z^*_n}\right)=\frac{r^*_-}{r_+}b_-(z^*_n){\rm e}^{-2{\rm i}\theta \left(-\frac{r^2_0}{z^*_n}\right)} \mu_{-,2}\left(x,t,-\frac{r^2_0}{z^*_n}\right). \end{eqnarray*}$

$\mu_{+,2}\left(x,t,-\frac{r^2_0}{z_n}\right)=\frac{r_-}{r^*_+}b_+(z_n){\rm e}^{2{\rm i} \theta\left(-\frac{r^2_0}{z_n}\right)} \mu_{-,1}\left(x,t,-\frac{r^2_0}{z_n}\right).$

$\begin{matrix}\label{2.68} s_{11}\left(-\frac{r^2_0}{z^*_n}\right)=\frac{r_-}{r_+}s^*_{11}(z_n), \end{matrix}$

$s^{\prime}_{11}\left(-\frac{r^2_0}{z^*_n}\right)=\left(\frac{z^*_n}{r_0}\right)^2 \frac{r_-}{r_+}(s^*_{11}(z_n))^{\prime}.$

$s^{\prime}_{22}\left(-\frac{r^2_0}{z_n}\right)=\left(\frac{z_n}{r_0}\right)^2\frac{r_-}{r_+} (s^*_{22}(z^*_n))^{\prime}.$

$A_+[z_n]=-\frac{r^*_-}{r_-}\frac{z^2_n}{r^2_0}A^*_+\left[-\frac{r^2_n}{z^*_n}\right],\ A_+[z_n]=\frac{r^*_-}{r_-}\frac{z^2_n}{r^2_0}A_-\left[-\frac{r^2_n}{z_n}\right].$

### 2.6 Jost 解和散射矩阵的渐近性

$\begin{eqnarray*} \mu^{(0)}_{\pm,x}&=&-\frac{\rm i}{4}[\mu^{(2)}_\pm,\sigma_3]-\frac{\rm i}{4}[\sigma_3Q_\pm\mu^{(1)}_\pm,\sigma_3] -\frac{\rm i}{4}(|r|^2-r_0^2)\sigma_3\mu^{(0)}_\pm\\ &&+\frac{1}{2}\Delta Q_\pm\mu^{(1)}_\pm-\frac{\rm i}{2}\sigma_3Q_\pm\Delta Q_\pm\mu^{(0)}_\pm, \end{eqnarray*}$

$\begin{matrix}\label{6.11} \mu^{(0)}_{\pm}={\rm e}^{{\rm i}\nu_\pm(x,t)\sigma_3}, \end{matrix}$

$\nu_\pm(x,t)=\frac{1}{4}\int_{\pm\infty}^{x}(|r|^2-r_0^2){\rm d}y.$

$S(z)={\rm e}^{-{\rm i}\nu_0\sigma_3}+O(z^{-1}),\quad z\rightarrow\infty,$
$S(z)={\rm diag}\left(\frac{r_-}{r_+},\frac{r_+}{r_-}\right){\rm e}^{{\rm i}\nu_0\sigma_3}+O(z),\quad z\rightarrow0,$

$\begin{eqnarray*} \nu_0=\frac{1}{4}\int_{-\infty}^{+\infty}(|r|^2-r^2_0){\rm d}y. \end{eqnarray*}$

$z\rightarrow\infty$ 时, 有

$\begin{eqnarray*} s_{11}&=&\frac{{\rm Wr}(\phi_{+,1},\phi_{-,2})}{\gamma}\\ &=&\det \left( {\begin{array}{*{20}{c}} {\rm e}^{{\rm i}\nu_+}+O(z^{-1})~&O(z^{-1})\\ O(z^{-1})~&{\rm e}^{-{\rm i}\nu_-}+O(z^{-1}) \end{array}} \right)\left(1-\frac{r^2_0}{z^2}+\frac{r^4_0}{z^4}+\cdots\right)\\ &=&{\rm e}^{-{\rm i}\nu_0}+O(z^{-1}), \end{eqnarray*}$

$\nu_0=\frac{1}{4}\int_{-\infty}^{+\infty}(|r|^2-r^2_0){\rm d}y.$

$s_{22}=\frac{{\rm Wr}(\phi_{-,1},\phi_{+,2})}{\gamma}={\rm e}^{{\rm i}\nu_0}+O(z^{-1}),\quad s_{12}=O(z^{-1}),\quad s_{21}=O(z^{-1}).$

$z\rightarrow0$ 时, 有

$\begin{eqnarray*} s_{11}&=&\frac{{\rm Wr}(\phi_{+,1},\phi_{-,2})}{\gamma}\\ &=&\det \left( {\begin{array}{*{20}{c}} O(1)&\displaystyle{\frac{\rm i}{z}}{\rm e}^{{\rm i}\nu_-}r_-+O(1)\\ \displaystyle{\frac{\rm i}{z}}{\rm e}^{-{\rm i}\nu_+}r^*_++O(1)&O(1) \end{array}} \right)\left(\frac{z^2}{r^2_0}-\frac{z^4}{r^4_0}+\cdots\right)\\ &=&\frac{r_-}{r_+}{\rm e}^{{\rm i}\nu_0}+O(z), \end{eqnarray*}$

$s_{22}=\frac{{\rm Wr}(\phi_{-,1},\phi_{+,2})}{\gamma}=\frac{r_+}{r_-}{\rm e}^{-{\rm i}\nu_0}+O(z),\quad s_{12}=O(z),\quad s_{21}=O(z).$

### 3.1 广义矩阵 Riemann-Hilbert 问题

$\begin{matrix}\label{2.81} M(x,t,z)=\left\{ \begin{array}{lll} M^+(x,t,z)=\left(\frac{\mu_{+,1}(x,t,z)}{s_{11}(z)},\mu_{-,2}(x,t,z)\right),\ & \ z\in D^+, \\ M^-(x,t,z)=\left(\mu_{-,1}(x,t,z),\frac{\mu_{+,2}(x,t,z)}{s_{22}(z)}\right),\ & \ z\in D^-. \end{array} \right. \end{matrix}$

$\bullet$ 解析性: $M(x,t,z)$${\Bbb C}\setminus\Sigma 解析且有单极点. \bullet 跳跃条件 $$\label{8.2} M^{-}(x,t,z)=M^{+}(x,t,z)(I-G(x,t,z)),\quad z\in\Sigma,$$ 其中 $$\label{8.3} G(x,t,z)= \left( {\begin{array}{*{20}{c}} 0&-{\rm e}^{2{\rm i}\theta(z)}\tilde{\rho}(z)\\ {\rm e}^{-2{\rm i}\theta(z)}\rho(z)&\rho(z)\tilde{\rho}(z) \end{array}} \right).$$ \bullet 渐近性质 \begin{eqnarray*} &&M(x,t,z)\sim {\rm e}^{{\rm i}\nu_-\sigma_3}+O(z^{-1}), \quad z\rightarrow\infty,\\ &&M(x,t,z)\sim \frac{\rm i}{z}{\rm e}^{{\rm i}\nu_-\sigma_3}\sigma_3Q_-+O(1), \quad z\rightarrow0. \end{eqnarray*} \mu_\pm$$s_{ij}(z)$ 的解析性可得 $M(x,t,z)$ 的解析性. 此外, 由(2.14)式知

$\begin{eqnarray*} \mu_{-,1}(x,t,z)&=&\frac{\mu_{+,1}(x,t,z)}{s_{11}(z)}-\rho(z) {\rm e}^{-2{\rm i}\theta(z)}\mu_{-,2}(x,t,z),\\ \frac{\mu_{+,2}(x,t,z)}{s_{22}(z)}&=&\tilde{\rho}(z) {\rm e}^{2{\rm i}\theta(z)}\mu_{-,1}(x,t,z)+\mu_{-,2}(x,t,z)\\ &=&\tilde{\rho}(z) {\rm e}^{2{\rm i}\theta(z)}\frac{\mu_{+,1}(x,t,z)}{s_{11}(z)}+(1-\rho(z)\tilde{\rho}(z))\mu_{-,2}(x,t,z), \end{eqnarray*}$

$\begin{eqnarray*} M^+(x,t,z)&=&(\mu_{+,1},\mu_{-,2}) \left( {\begin{array}{*{20}{c}} \displaystyle{\frac{1}{s_{11}}}~&0\\ 0~&1 \end{array}} \right)\\ &=&\left( {\begin{array}{*{20}{c}} {\rm e}^{{\rm i}\nu_-}+O(z^{-1})&O(z^{-1})\\ O(z^{-1})&{\rm e}^{-{\rm i}\nu_-}+O(z^{-1}) \end{array}}\right)\\ &=&{\rm e}^{{\rm i}\nu_-\sigma_3}+O(z^{-1}). \end{eqnarray*}$

$z\rightarrow0$, 有

$\begin{eqnarray*} M^+(x,t,z)&=&(\mu_{+,1},\mu_{-,2}) \left( {\begin{array}{*{20}{c}} \displaystyle{\frac{1}{s_{11}}}~&0\\ 0~&1 \end{array}} \right)\\ &=&\left( {\begin{array}{*{20}{c}} O(1)&\displaystyle{\frac{\rm i}{z}}r_-{\rm e}^{{\rm i}\nu_-}+O(1)\\ \displaystyle{\frac{\rm i}{z}}r^*_-{\rm e}^{-{\rm i}\nu_-}+O(1)&O(1) \end{array}}\right)\\ &=&\frac{\rm i}{z}{\rm e}^{{\rm i}\nu_-\sigma_3}\sigma_3Q_-+O(1). \end{eqnarray*}$

$\begin{matrix}\label{8.8} \zeta_n= \left\{\begin{array}{lll} z_n, &n=1,\cdots,N_1, \\ -z_{n-N_1}, & n=N_{1}+1,\cdots,2N_1, \\ \displaystyle{\frac{r^2_0}{z^*_{n-2N_1}}}, & n=2N_{1}+1,\cdots,3N_1, \\ -\displaystyle{\frac{r^2_0}{z^*_{n-3N_1}}}, & n=3N_{1}+1,\cdots,4N_1, \\ w_{n-4N_1},& n=4N_{1}+1,\cdots,4N_1+N_2,\\ -w_{n-4N_1-N_2},& n=4N_{1}+N_2+1,\cdots,4N_{1}+2N_2. \end{array}\right. \end{matrix}$

$\hat{\zeta}_n=-{\frac{r^2_0}{\zeta_n}}$, $n=1,2,\cdots,4N_1+2N_2$.

$\begin{eqnarray*} &&M^{-}-{\rm e}^{{\rm i}\nu_-\sigma_3}-\frac{\rm i}{z}{\rm e}^{{\rm i}\nu_-\sigma_3}\sigma_3Q_- -\sum_{n=1}^{4N_1+2N_2}\frac{\mathop{{\rm Res}} \limits_{z=\hat{\zeta}_n}M^-}{z-\hat{\zeta}_n}-\sum_{n=1}^{4N_1+2N_2}\frac{\mathop{{\rm Res}} \limits_{z=\zeta_n}M^+}{z-\zeta_n}\\ &=&M^{+}-{\rm e}^{{\rm i}\nu_-\sigma_3}-\frac{\rm i}{z}{\rm e}^{{\rm i}\nu_-\sigma_3}\sigma_3Q_- -\sum_{n=1}^{4N_1+2N_2}\frac{\mathop{{\rm Res}} \limits_{z=\hat{\zeta}_n}M^-}{z-\hat{\zeta}_n}-\sum_{n=1}^{4N_1+2N_2}\frac{\mathop{{\rm Res}} \limits_{z=\zeta_n}M^+}{z-\zeta_n}-M^+G. \end{eqnarray*}$

$\begin{matrix}\label{2.90} M(x,t,z)&=&{\rm e}^{{\rm i}\nu_-\sigma_3}+\frac{\rm i}{z}{\rm e}^{{\rm i}\nu_-\sigma_3}\sigma_3Q_- +\sum_{n=1}^{4N_1+2N_2}\frac{\mathop{{\rm Res}} \limits_{z=\hat{\zeta}_n}M^-}{z-\hat{\zeta}_n}+\sum_{n=1}^{4N_1+2N_2}\frac{\mathop{{\rm Res}} \limits_{z=\zeta_n}M^+}{z-\zeta_n} \\ &&+\frac{1}{2{\rm i}\pi}\int_{\Sigma}\frac{M^+(x,t,\xi)}{\xi-z}G(x,t,\xi){\rm d}\xi. \end{matrix}$

### 3.2 重构公式

$\begin{eqnarray*} \frac{\mathop{{\rm Res}}\limits_{z=\zeta_n}M^+(x,t,z)}{z-\zeta_n} +\frac{\mathop{{\rm Res}}\limits_{z=\hat{\zeta}_n}M^-(x,t,z)}{z-\hat{\zeta}_n} =\left[C_n(z)\mu_{-,2}(x,t,\zeta_n),\hat{C}_n(z)\mu_{-,1}(x,t,\hat{\zeta}_n)\right], \end{eqnarray*}$

$C_n(z)=\frac{A_+[\zeta_n]{\rm e}^{-2{\rm i}\theta(x,t,\zeta_n)}}{z-\zeta_n},\quad \hat{C}_n(z)=\frac{A_-[\hat{\zeta}_n]{\rm e}^{2{\rm i}\theta(x,t,\hat{\zeta}_n)}}{z-\hat{\zeta}_n}.$

$$$\label{8.15} \mu_{-,2}(x,t,\zeta_n)=\left( {\begin{array}{*{20}{c}} \displaystyle{\frac{\rm i}{\zeta_n}}r_-{\rm e}^{{\rm i}\nu_-}\\ {\rm e}^{-{\rm i}\nu_-} \end{array}} \right)+\sum_{n=1}^{4N_1+2N_2}\hat{C}_n(\zeta_n)\mu_{-,1}(x,t,\hat{\zeta}_n) +\frac{1}{2{\rm i}\pi}\int_{\Sigma}\frac{(M^+G)_2(x,t,\xi)}{\xi-\zeta_n}{\rm d}\xi.$$$

$\begin{matrix}\label{8.17} &&M(x,t,z) \\ &=&{\rm e}^{{\rm i}\nu_-\sigma_3}+\frac{1}{z} \left( {\rm i}{\rm e}^{{\rm i}\nu_-\sigma_3}\sigma_3Q_- +\sum_{n=1}^{4N_1+2N_2}\left[A_+[\zeta_n]{\rm e}^{-2{\rm i}\theta(\zeta_n)}\mu_{-,2}(\zeta_n),A_-[\hat{\zeta}_n]{\rm e}^{2{\rm i}\theta(\hat{\zeta}_n)}\mu_{-,1}(\hat{\zeta}_n)\right]\right. \\ &&\left.-\frac{1}{2{\rm i}\pi}\int_{\Sigma}(M^+G)(x,t,\xi){\rm d}\xi\right)+O(z^{-2}). \end{matrix}$

$M=M^+$, 比较矩阵(3.7)式在$(1,2)$位置的元素, 则带有单极点势的重构公式为

$$$\label{8.18} r(x,t)=r_-{\rm e}^{2{\rm i}\nu_-}-{\rm e}^{{\rm i}\nu_-}\left({\rm i}\sum_{n=1}^{4N_1+2N_2} A_-[\hat{\zeta}_n]{\rm e}^{2{\rm i}\theta(x,t,\hat{\zeta}_n)} \mu_{-,11}(\hat{\zeta}_n) +\frac{1}{2\pi}\int_{\Sigma}(M^+G)_{12}(x,t,\xi){\rm d}\xi \right).$$$

### 3.3 迹公式和 $\theta$ 条件

$\begin{eqnarray*} &&\beta^+(z)=s_{11}(z)\prod_{n=1}^{4N_1+2N_2}\frac{z-\hat{\zeta}_n}{z-\zeta_n}{\rm e}^{{\rm i}\nu_0},\\ &&\beta^-(z)=s_{22}(z)\prod_{n=1}^{4N_1+2N_2}\frac{z-\zeta_n}{z-\hat{\zeta}_n}{\rm e}^{-{\rm i}\nu_0}. \end{eqnarray*}$

$z\rightarrow\infty$ 时, $\beta^\pm\rightarrow1$. 因此

$\beta^+(z)\beta^-(z)=\frac{1}{1+\rho(z)\rho^*(z^*)},\quad z\in\Sigma.$

$\log\beta^\pm(z)=\mp\frac{1}{2\pi {\rm i}}\int_\Sigma\frac{\log[1+\rho(\xi)\rho^*(\xi^*)]}{\xi-z}{\rm d}\xi,\quad z\in D^\pm.$

$s_{11}(z)=\exp\left[-\frac{1}{2{\rm i}\pi}\int_{\Sigma}\frac{\log[1+\rho(\xi)\rho^*(\xi^*)]}{\xi-z}{\rm d}\xi\right] \prod^{4N_1+2N_2}_{n=1}\frac{z-\zeta_n}{z-\hat{\zeta}_n}{\rm e}^{-{\rm i}\nu_0},\quad z\in D^+,$
$s_{22}(z)=\exp\left[\frac{1}{2{\rm i}\pi}\int_{\Sigma}\frac{\log[1+\rho(\xi)\rho^*(\xi^*)]}{\xi-z}{\rm d}\xi\right] \prod^{4N_1+2N_2}_{n=1}\frac{z-\hat{\zeta}_n}{z-\zeta_n}{\rm e}^{{\rm i}\nu_0},\qquad z\in D^-.$

$z\rightarrow0$ 时, 由(3.9)和(2.34)式知 $\theta$ 条件为

$\begin{matrix}\label{8.25} \arg(\frac{r_-}{r_+})+2\nu_0=\frac{1}{2\pi}\int_\Sigma\frac{\log[1+\rho(\xi)\rho^*(\xi^*)]} {\xi}{\rm d}\xi+8\sum_{n=1}^{N_1}\arg(z_n)+4\sum_{m=1}^{N_2} \arg(w_m). \end{matrix}$

### 3.4 单极解的无散射势

$\begin{matrix}\label{2.101} r(x,t)=r_-{\rm e}^{2{\rm i}\nu_-}+{\rm i}{\rm e}^{2{\rm i}\nu_-}\frac{\det\left[\begin{array}{*{20}{c}} M~& \beta\\ \alpha^T~&0 \end{array}\right]}{\det M}, \end{matrix}$

$\begin{eqnarray*} m_{kn}=\hat{C}_n(\zeta_k)-\frac{{\rm i}r_-}{\zeta_k}\delta_{k,n}, \ \beta_k=-\frac{{\rm i}r_-}{\zeta_k}, \ \alpha_n=A_-[\hat{\zeta}_n]{\rm e}^{2{\rm i}\theta(x,t,\hat{\zeta}_n)}. \end{eqnarray*}$

$\begin{matrix}\label{2.103} \mu_{-,2}(x,t,z)=\frac{{\rm i}r_-}{z}\mu_{-,1}(x,t,-\frac{r^2_0}{z}). \end{matrix}$

$\begin{eqnarray*} \left( {\begin{array}{*{20}{c}} \displaystyle{\frac{\rm i}{\zeta_n}}r_-{\rm e}^{{\rm i}\nu_-}\\ {\rm e}^{-{\rm i}\nu_-} \end{array}} \right)+\sum_{n=1}^{4N_1+2N_2}\left(\hat{C}_n(\zeta_n)-\frac{{\rm i}r_-}{\zeta_n}\delta_{k,n}\right) \mu_{-,1}(x,t,\hat{\zeta}_n)=0, \end{eqnarray*}$

$\begin{matrix}\label{2.104} \sum_{n=1}^{4N_1+2N_2}\left(\hat{C}_n(\zeta_n)-\frac{{\rm i}r_-}{\zeta_n}\delta_{kn}\right) \mu_{-,11}(x,t,\hat{\zeta}_n)=-\frac{\rm i}{\zeta_n}r_-{\rm e}^{{\rm i}\nu_-}. \end{matrix}$

## 4 双极解的直散射问题

$\begin{matrix}\label{10.2} \mathop{{\rm Res}}\limits_{z=z_0}\left[\frac{f}{g}\right]=\frac{2f^{\prime}(z_0)}{g^{\prime\prime}(z_0)} -\frac{2f(z_0)g^{\prime\prime\prime}(z_0)}{3(g^{\prime\prime}(z_0))^2}, \quad\mathop{P_{-2}}\limits_ {z=z_0}\left[\frac{f}{g}\right]=\frac{2f(z_0)}{g^{\prime\prime}(z_0)}. \end{matrix}$

$z_0$$s_{11} 的双极点, 可得 \begin{eqnarray*} \det\big(\mu_{+,1}^{\prime}(z_0)-b_+(z_0){\rm e}^{-2{\rm i}\theta(z_0)}\mu^{\prime}_{-,2}(z_0)+2{\rm i}\theta^{\prime}(z_0) b_+(z_0){\rm e}^{-2{\rm i}\theta(z_0)}\mu_{-,2}(z_0), {\rm e}^{-2{\rm i}\theta(z_0)}\mu_{-,2}(z_0)\big)=0, \end{eqnarray*} 存在一个常数 d_n 使得 \begin{matrix}\label{10.1} \mu_{+,1}^{\prime}(z_0)={\rm e}^{-2{\rm i}\theta(z_0)}\left[(d_+(z_0)-2{\rm i}b_+(z_0)\theta^{\prime} (z_0))\mu_{-,2}(z_0)+b_+(z_0)\mu^{\prime}_{-,2}(z_0)\right]. \end{matrix} 利用(2.26),(4.2)和(4.1)式, 有 \begin{matrix}\label{10.3} &&\mathop{P_{-2}}\limits_ {z=z_0}\left[\frac{\mu_{+,1}}{s_{11}}\right]=A_+[z_0]{\rm e}^{-2{\rm i}\theta(z_0)}\mu_{-,2}(z_0), \\ &&\mathop{{\rm Res}}\limits_{z=z_0}\left[\frac{\mu_{+,1}}{s_{11}}\right]=A_+[z_0]{\rm e}^{-2{\rm i}\theta(z_0)} \left[\mu^\prime_{-,2}(z_0)+\mu_{-,2}(z_0)(B_+[z_0]-2{\rm i}\theta^\prime(z_0))\right], \end{matrix} 其中 A_+[z_0]={\frac{2b_+[z_0]}{s^{\prime\prime}_{11}(z_0)}}, B_+[z_0]={\frac{d_+(z_0)}{b_+(z_0)}-\frac{s^{\prime\prime\prime}_{11}(z_0)} {3s^{\prime\prime}_{11}(z_0)}}, z_0 \in Z \cap D^+. z^*_0$$s_{22}$ 的双极点, 可得

$\begin{matrix}\label{10.4} \mu_{+,2}^{\prime}(z^*_0)={\rm e}^{2{\rm i}\theta(z^*_0)}\left[(d_-(z^*_0)+2{\rm i}b_-(z^*_0)\theta^{\prime} (z^*_0))\mu_{-,1}(z^*_0)+b_-(z^*_0)\mu^{\prime}_{-,1}(z^*_0)\right], \end{matrix}$

$\begin{matrix}\label{10.5} &&\mathop{P_{-2}}\limits_ {z=z^*_0}\left[\frac{\mu_{+,2}}{s_{22}}\right]=A_-[z^*_0]{\rm e}^{2{\rm i}\theta(z^*_0)}\mu_{-,1}(z^*_0), \\ &&\mathop{{\rm Res}}\limits_{z=z^*_0}\left[\frac{\mu_{+,2}}{s_{22}}\right]=A_-[z^*_0]{\rm e}^{2{\rm i}\theta(z^*_0)} \left[\mu^\prime_{-,1}(z^*_0)+\mu_{-,1}(z^*_0)(B_-[z^*_0]+2{\rm i}\theta^\prime(z^*_0))\right], \end{matrix}$

$\begin{eqnarray*} A_+[z_n]&=&-A_+[-z_n]=-\frac{r^*_-}{r_-}\frac{z^4_n}{r^4_0}A^*_+\left[-\frac{r^2_0}{z^*_n}\right]= \frac{r^*_-}{r_-}\frac{z^4_n}{r^4_0}A^*_+\left[\frac{r^2_0}{z^*_n}\right]\\ &=&-A^*_-[z^*_n]=A^*_-[-z^*_n]=\frac{r^*_-}{r_-}\frac{z^4_n}{r^4_0}A_- \left[-\frac{r^2_0}{z_n}\right]=-\frac{r^*_-}{r_-}\frac{z^4_n}{r^4_0}A_-\left[\frac{r^2_0}{z_n}\right],\\ A_+[\omega_m]&=&-A_+[-\omega_m]=-A^*_-[\omega^*_m]=A^*_-[-\omega^*_m], \\ B_+[z_n]&=&-B_+[-z_n]=\frac{r^2_0}{z^2_n}B^*_+\left[-\frac{r^2_0}{z^*_n}\right]+\frac{2}{z_n}= -\frac{r^2_0}{z^2_n}B^*_+\left[\frac{r^2_0}{z^*_n}\right]+\frac{2}{z_n}\\ &=&B^*_-[z^*_n]=-B^*_-[-z^*_n]=\frac{r^2_0}{z^2_n}B_-\left[-\frac{r^2_0}{z_n}\right]+\frac{2}{z_n}= -\frac{r^2_0}{z^2_n}B_-\left[\frac{r^2_0}{z_n}\right]+\frac{2}{z_n},\\ B_+[\omega_m]&=&-B_+[-\omega_m]= B^*_-[\omega^*_m]=-B^*_-[-\omega^*_m]. \end{eqnarray*}$

$\begin{eqnarray*} &&H=\left[ {\begin{array}{*{20}{c}} H^{(1,1)}~&H^{(1,2)}\\ H^{(2,1)}~&H^{(2,2)} \end{array}} \right],\quad H^{(m,j)}=(h^{(m,j)}_{s,n})_{(4N_1+2N_2)\times(4N_1+2N_2)}, \ m,j=1,2,\\ &&h^{(1,1)}_{s,n}=\hat{C}_n(\zeta_s)\left(\hat{D}_n+\frac{1}{\zeta_s-\hat{\zeta}_n}\right) -\frac{{\rm i}r_-}{\zeta_s}\delta_{s,n}, \quad h^{(1,2)}_{s,n}=\hat{C}_n(\zeta_s), \ s,n=1,2, \\ &&h^{(2,1)}_{s,n}=\frac{\hat{C}_n(\zeta_s)}{\zeta_s-\hat{\zeta}_n}\left(\hat{D}_n +\frac{2}{\zeta_s-\hat{\zeta}_n}\right)-\frac{{\rm i}r_-}{\zeta^2_s}\delta_{s,n}, \quad h^{(2,2)}_{s,n}=\frac{\hat{C}_n(\zeta_s)}{\zeta_s-\hat{\zeta}_n} +\frac{{\rm i}r_-r^2_0}{\zeta^3_s}\delta_{s,n}. \end{eqnarray*}$

$N=1$ 时,(5.9)式中双极解的表达式非常复杂, 因此没有明确给出. 然而, 在软件 Maple 的帮助下, 通过取不同参数, 得到的相应动力学结构.

## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Kundu A.

Landau-Lifshitz and higherorder nonlinear systems gauge generated from nonlinear Schrödinger type equations

J Math Phys, 1984, 25(12): 3433-3438

Kivshar Y S, Agrawal G P. Optical Solitons:From Fibers to Photonic Crystals. New York: Academic, 2003

Dysthe K B.

Note on the modification of the nonlinear Schödinger equation for application to deep water waves

Proc R Soc Lond A, 1979, 369(1736): 105-114

Chan H N, Chow K W, Kedziora D J.

Rogue wave modes for a derivative nonlinear Schrödinger model

Phys Rev E, 2014, 89(3): 032914

Hu B B, Zhang L, Zhang N.

On the Riemann-Hilbert problem for the mixed Chen-Lee-Liu derivative nonlinear Schrödinger equation

J Comput Appl Math, 2021, 390: 113393

Zhang Y S, Guo L J, Chabchoub A. Higher-order rogue wave dynamics for a derivative nonlinear Schrödinger equation. https://arxiv.org/pdf/1409.7923v2.pdf

Fang F, Hu B B, Zhang L. Riemann-Hilbert method and $N$-soliton solutions for the mixed Chen-Lee-Liu derivative nonlinear Schrödinger equation. https://arxiv.org/pdf/2004.03193.pdf

Zhao Y, Fan E G.

N-soliton solution for a higher-order Chen-Lee-Liu equation with nonzero boundary conditions

Modern Phys Lett B, 2020, 34(4): 2050054

Biondini G, Kovačič G.

Inverse scattering transform for the focusing nonlinear Schrödinger equation with nonzero boundary conditions

J Math Phys, 2014, 55(3): 031506

Pichler M, Biondini G.

On the focusing non-linear Schrödinger equation with non-zero boundary conditions and double poles

IMA J Appl Math, 2017, 82(1): 131-151

Wen L L, Zhang N, Fan E G.

$N$-soliton solution of the Kundu-Type equation via Riemann-Hilbert approach

Acta Mathematica Scientia, 2020, 40B(1): 113-126

Zhang B, Fan E G.

Riemann-Hilbert approach for a Schrödinger-type equation with nonzero boundary conditions

Modern Phys Lett B, 2021, 35(12): 2150208

Zhang G Q, Yan Z Z.

The derivative nonlinear Schrödinger equation with zero/nonzero boundary conditions: Inverse scattering transforms and N-double-pole solutions

J Nonlinear Sci, 2020, 30(2): 3089-3127

Zhang G Q, Yan Z Z.

Focusing and defocusing Hirota equations with non-zero boundary conditions: Inverse scattering transforms and soliton solutions

Commun Nonlinear Sci Numer Simulat, 2020, 80: 104927

Zhang G Q, Yan Z Z.

Focusing and defocusing mKdV equations with nonzero boundary conditions: Inverse scattering transforms and soliton interactions

Physica D, 2020, 410: 132521

Clarkson P A, Cosgrove C M.

Painlevé analysis of the non-linear Schrödinger family of equations

J Phys A: Math Gen, 1987, 20(8): 2003-2024

Faddeev L D, Takhtajan L A.

Hamiltonian Methods in the Theory of Solitons

Berlin: Springer, 1987

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