众所周知,许多经典的可积系统的初值问题可以由反散射方法来求解. 1997年, Fokas^[1]借鉴反散射变换的思想提出了统一变换方法,我们称之为Fokas方法,该方法很好的求解了可积方程的初边值问题^[2-5].就像全直线上的反散射方法一样, Fokas方法也是将初边值问题的解表示成相应的Riemann-Hilbert (RH)问题的解. 2012年, Lenells^[6]首次将此方法做了推广,并且研究了Degasperis-Procesi方程在半直线上的初边值问题^[7].在这之后,越来越多的学者开始关注可积系统的初边值问题^[8-23].此外,基于RH方法得到的可积方程的多孤子解在近几年也得到了广泛研究^[24-29].

2015年,张等^[30]从耦合高阶Chen-Lee-Liu方程

(1.1) $ \begin{eqnarray} \left\{\begin{array}{l} { } q_{t}+q_{xxx}+\frac{3}{2}{\rm i}qrq_{xx}-\frac{3}{4}q^2r^2q_{x}+\frac{3}{2}{\rm i}q_x^2r = 0, \\ { } r_{t}+r_{xxx}+\frac{3}{2}{\rm i}qrr_{xx}-\frac{3}{4}q^2r^2r_{x}-\frac{3}{2}{\rm i}r_x^2q = 0 \end{array}\right. \end{eqnarray} $

出发,在约化条件$ q = r^* $下,得到了高阶Chen-Lee-Liu(HOCLL)方程

(1.2) $ \begin{eqnarray} q_{t}+q_{xxx}+\frac{3}{2}{\rm i}|q|^2q_{xx}-\frac{3}{4}|q|^4q_{x}+\frac{3}{2}{\rm i}q_x^2q^* = 0, \end{eqnarray} $

其中$ q(x, t) $是复的光滑包络函数, $ t $和$ x $分别是时间和空间变量.并用达布变换方法得到了方程(1.2)的怪波解^[30].随后,在2018年,胡等^[29]基于RH方法得到了HOCLL方程的多孤子解.而在本文中,基于前人的工作,我们运用Fokas方法研究HOCLL方程在半直线$ \Omega = \{0<x<\infty, 0<t<T\} $上的初边值问题,其初值和边值条件定义如下

(1.3) $ \begin{eqnarray} \begin{array}{l} q_0(x) = q(x, t = 0);g_0(t) = q(x = 0, t), g_1(t) = q_{x}(x = 0, t), g_2(t) = q_{xx}(x = 0, t). \end{array} \end{eqnarray} $

本文结构如下:在第二部分,给出了Lax对的谱分析.在第三部分,证明了HOCLL方程初边值问题的解可用RH问题的形式解唯一表示.

HOCLL方程(1.2)有如下的Lax对^[30]

(2.1) $ \begin{eqnarray} \left\{ \begin{array}{l} { } \phi_{x} = M\phi = (-{\rm i}\lambda^2\sigma_3+\lambda\sigma_3Q+\frac{1}{4}{\rm i}\sigma_3Q^2)\phi, \\ { } \phi_{t} = N\phi = (-4{\rm i}\lambda^6\sigma_3+4\lambda^5\sigma_3Q+2{\rm i}\lambda^4\sigma_3Q^2+2\lambda^3 Z_0+\lambda^2Z_1+\lambda Z_2+\frac{1}{4}Z_3)\phi, \end{array}\right. \end{eqnarray} $

其中$ \lambda\in{\rm C} $是谱参数, $ \sigma_3, Q, Z_0, Z_1, Z_2 $和$ Z_3 $定义为

(2.2) $ \begin{eqnarray} \begin{array}{l} \sigma_3 = \left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right), {\qquad} Q = \left(\begin{array}{cc} 0 & q \\ q^* & 0 \end{array}\right), //{ } Z_0 = {\rm i}Q_x-\frac{1}{2}\sigma_3Q^3, Z_1 = [Q, Q_x]-\frac{1}{2}{\rm i}\sigma_3Q^4, // { } Z_2 = Q_{xx}\sigma_3+\frac{1}{2}{\rm i}QQ_xQ-\frac{3}{2}{\rm i}Q^2Q_x+\frac{1}{4}\sigma_3Q^5, // { } Z_3 = \frac{1}{4}{\rm i}\sigma_3Q^6+\frac{3}{2}{\rm i}Q^2[Q_x, Q]-{\rm i}\sigma_3(Q_{xx}Q+QQ_{xx}-Q_x^2), \end{array} \end{eqnarray} $

因此, Lax对方程(2.1)有下列形式

(2.3) $ \begin{eqnarray} \left\{ \begin{array}{l} \phi_{x}+{\rm i}\lambda^2\sigma_{3}\phi = U_1\phi, \\ \phi_{t}+4{\rm i}\lambda^6\sigma_{3}\phi = U_2\phi, \end{array}\right. \end{eqnarray} $

其中$ U_1 = \lambda\sigma_3Q+\frac{1}{4}{\rm i}\sigma_3Q^2 $和$ U_2 = 4\lambda^5\sigma_3Q+2{\rm i}\lambda^4\sigma_3Q^2+2\lambda^3 Z_0+\lambda^2Z_1+\lambda Z_2+\frac{1}{4}Z_3. $假设

(2.4) $ \begin{eqnarray} \phi(x, t;\lambda) = \psi(x, t;\lambda){\rm e}^{-{\rm i}(\lambda^2x+4\lambda^6t)\sigma_3}, 0<x<\infty, 0<t<T. \end{eqnarray} $

因此,方程(2.3)等价为

(2.5) $ \begin{eqnarray} \left\{ \begin{array}{l} \psi_x+{\rm i}\lambda^2[\sigma_3, \psi] = U_1\psi, \\ \psi_t+4{\rm i}\lambda^6[\sigma_3, \psi] = U_2\psi, \end{array}\right. \end{eqnarray} $

写成全微分形式有

(2.6) $ \begin{eqnarray} {\rm d}({\rm e}^{{\rm i}(\lambda^2x+4\lambda^6t )\hat{\sigma}_3}\psi(x, t;\lambda)) = W_1(x, t;\lambda), 0<x<\infty, 0<t<T, \end{eqnarray} $

其中

(2.7) $ \begin{eqnarray} W_1(x, t;\lambda) = {\rm e}^{{\rm i}(\lambda^2x+4\lambda^6t )\hat{\sigma}_3}(U_1{\rm d}x+U_2{\rm d}t)\psi(x, t;\lambda), \end{eqnarray} $

其中, $ \hat{\sigma}_{3} $是矩阵算子,作用结果为$ \hat{\sigma}_{3}A = [\sigma_{3}, A] $和$ {\rm e}^{\hat{\sigma}_{3}}A = {\rm e}^{\sigma_{3}}A{\rm e}^{-\sigma_{3}} $.

然后,我们寻找方程(2.1)如下形式的一个解

(2.8) $ \begin{equation} \psi(x, t;\lambda) = D_0+\sum\limits_{j = 1}^5\frac{D_j}{\lambda_j}+O(\frac{1}{\lambda^6}), \lambda\rightarrow\infty. \end{equation} $

将(2.8)式代入(2.5)式的$ x $部分,并比较$ \lambda $的同次幂系数得

(2.9) $ \begin{eqnarray} \begin{array}{l} \lambda^2: {\rm i}[\sigma_3, D_0] = 0, \\ \lambda^1: {\rm i}[\sigma_3, D_1] = \sigma_3QD_0, \\ { } \lambda^{0}:D_{0, x}+{\rm i}[\sigma_3, D_2] = \frac{{\rm i}}{4}\sigma_3 Q^2D_0+\sigma_3QD_1. \end{array} \end{eqnarray} $

从得$ O(\lambda^1) $得到$ D_0 $是一个对角矩阵,假设

$ D_0 = \left(\begin{array}{cc} D_0^{11} & 0 \\ 0 & D_0^{22} \end{array}\right). $

由$ O(\lambda^0) $得

(2.10) $ \begin{equation} D_1^{(o)} = -\frac{{\rm i}}{2}QD_0, \, D_{0, x} = -\frac{{\rm i}}{4}qr\sigma_3D_0, \end{equation} $

这里$ D_1^{(o)} $表示$ D_1 $的次对角线部分.对于方程(2.5)的$ t $部分,我们进行类似的分析,并且经过一段繁琐的计算可以得到

(2.11) $ \begin{eqnarray} D_{0, t} = (-\frac{{\rm i}}{16}q^3r^3-\frac{3}{8}qr^2q_x+\frac{3}{8}q^2rr_x+\frac{{\rm i}}{4}q_{xx}r+\frac{{\rm i}}{4}qr_{xx}-\frac{{\rm i}}{4}q_{x}r_{x})D_0\sigma_3. \end{eqnarray} $

注意到方程(2.1)的守恒律

$ {\rm i}(qr)_t = (\frac{{\rm i}}{4}q^3r^3+\frac{3}{2}qr^2q_x-\frac{3}{2}q^2rr_x-{\rm i}q_{xx}r-{\rm i}qr_{xx}+{\rm i}q_xr_x)_x. $

我们定义

(2.12) $ \begin{eqnarray} D_0(x, t) = \exp\bigg({\rm i}\int_{(x_0, t_0)}^{(x, t)}\Delta(x, t)\sigma_3\bigg), \quad\Delta(x, t) = \Delta_1(x, t){\rm d}x+\Delta_2(x, t){\rm d}t, \end{eqnarray} $

其中$ \Delta_1 = -\frac{1}{4}qr, \, \Delta_2 = -\frac{1}{16}q^3r^3+\frac{3{\rm i}}{8}qr^2q_x-\frac{3{\rm i}}{8}q^2rr_x+\frac{1}{4}q_{xx}r+\frac{1}{4}qr_{xx}-\frac{1}{4}q_xr_x $,不失一般性,在(2.12)式中,我们取$ (x_0, t_0) = (0, 0) $.

由于(2.12)式与积分路径无关,而且$ \Delta $是独立于$ \lambda $的函数,故引入特征函数$ \mu(x, t;\lambda) $,

(2.13) $ \begin{eqnarray} \psi(x, t;\lambda) = {\rm e}^{{\rm i}\int_{(0, 0)}^{(x, t)}\Delta\hat{\sigma}_3}\mu(x, t;\lambda)D_0(x, t), 0<x<\infty, 0<t<T. \end{eqnarray} $

因此, Lax对方程(2.5)可以写为

(2.14) $ \begin{eqnarray} {\rm d}({\rm e}^{{\rm i}(\lambda^2 x+4\lambda^6t)\hat{\sigma}_3}\mu(x, t;\lambda)) = W_2(x, t;\lambda), \end{eqnarray} $

其中

(2.15) $ \begin{equation} \begin{array}{cc} W_2(x, t;\lambda) = {\rm e}^{{\rm i}(\lambda^2 x+4\lambda^6t)\hat{\sigma}_3}V(x, t;\lambda)\mu(x, t;\lambda), \\ { } V(x, t;\lambda) = V_1(x, t;\lambda){\rm d}x+V_2(x, t;\lambda){\rm d}t = {\rm e}^{-{\rm i}\int_{(0, 0)}^{(x, t)} \Delta\hat{\sigma}_3}(U_1{\rm d}x+U_2{\rm d}t-{\rm i}\Delta\sigma_3). \end{array} \end{equation} $

由$ U(x, t;\lambda) $和$ \Delta $,我们得到

$ V_1(x, t;\lambda) = \left(\begin{array}{cc} { }\frac{{\rm i}}{2}qr\ & \lambda q{\rm e}^{-2{\rm i}\int_{(0, 0)}^{(x, t)}\Delta} \\ -\lambda r{\rm e}^{2{\rm i}\int_{(0, 0)}^{(x, t)}\Delta}\ &{ } -\frac{{\rm i}}{2}qr \end{array}\right), V_2(x, t;\lambda) = \left(\begin{array}{cc} V_2^{11} & V_2^{12}\\ V_2^{21} & -V_2^{11} \end{array}\right). $

$ V_2^{11} = 2{\rm i}\lambda^4|q|^2+\lambda^2(qr_x-q_xr-\frac{{\rm i}}{2}|q|^4)+\frac{{\rm i}}{16}|q|^6+\frac{3}{8}|q|^2qr_x -\frac{{\rm i}}{4}q_{xx}r-\frac{{\rm i}}{4}qr_{xx}+\frac{{\rm i}}{4}|q_x|^2, $

$ V_2^{12} = (4\lambda^5 q+\lambda^3(2{\rm i}q_x-|q|^2q) +\lambda(\frac{1}{4}|q|^4q-\frac{3{\rm i}}{2}|q|^2q_x+\frac{{\rm i}}{2}q^2r_x-q_{xx})){\rm e}^{-2{\rm i}\int_{(0, 0)}^{(x, t)}\Delta}, $

$ V_2^{21} = (-4\lambda^5 r+\lambda^3(2{\rm i}r_x+|q|^2r) +\lambda(-\frac{1}{4}|q|^4r-\frac{3{\rm i}}{2}|q|^2r_x+\frac{{\rm i}}{2}r^2q_x+r_{xx})){\rm e}^{2{\rm i}\int_{(0, 0)}^{(x, t)}\Delta}. $

因此,方程(2.14)可以写成

(2.16) $ \begin{eqnarray} \left\{ \begin{array}{l} \mu_x+{\rm i}\lambda^2[\sigma_3, \mu] = V_1\mu, \\ \mu_t+4{\rm i}\lambda^6[\sigma_3, \mu] = V_2\mu. \end{array}\right. \end{eqnarray} $

假设$ q(x, t) $在半直线区域$ \Omega = \{0<x<\infty, 0<t<T\} $内充分光滑,定义三个$ 2\times2 $矩阵值特征函数$ \{\mu_j(x, t, \lambda)\}_1^3 $

(2.17) $ \begin{eqnarray} \mu_j(x, t;\lambda) = {\rm I}+\int_{(x_j, t_j)}^{(x, t)}{\rm e}^{-{\rm i}(\lambda^2 x+4\lambda^6t)\hat{\sigma}_3}W_2(\xi, \tau, \lambda), 0<x<\infty, 0<t<T, \end{eqnarray} $

其中积分路径是从$ (x_j, t_j) $到$ (x, t) $,并且有$ (x_1, t_1) = (0, T ), (x_2 t_2) = (0, 0), (x_3, t_3) = (\infty, t) $ (如图 1所示).由于积分方程(2.19)与积分路径无关,我们选取如图 1所示特殊的积分路径,得到

(2.18) $ \begin{equation} \left\{ \begin{array}{l} { } \mu_1(x, t;\lambda) = {\rm I}+\int_{0}^{x}{\rm e}^{{\rm i}\lambda^2 (\xi-x)\hat{\sigma}_3}(V_1\mu_1)(\xi, t, \lambda){\rm d}\xi -{\rm e}^{-{\rm i}\lambda^2 x\hat{\sigma}_3}\int_{t}^{T}{\rm e}^{4{\rm i}\lambda^6(\tau-t)\hat{\sigma}_3}(V_2\mu_1)(0, \tau, \lambda){\rm d}\tau , \\ { } \mu_2(x, t;\lambda) = {\rm I}+\int_{0}^{x}{\rm e}^{{\rm i}\lambda^2 (\xi-x)\hat{\sigma}_3}(V_1\mu_2)(\xi, t, \lambda){\rm d}\xi +{\rm e}^{-{\rm i}\lambda^2 x\hat{\sigma}_3}\int_{0}^{t}{\rm e}^{4{\rm i}\lambda^6(\tau-t)\hat{\sigma}_3}(V_2\mu_2)(0, \tau, \lambda){\rm d}\tau , \\ { } \mu_3(x, t;\lambda) = {\rm I}-\int_{x}^{\infty}{\rm e}^{{\rm i}\lambda^2 (\xi-x)\hat{\sigma}_3}(V_1\mu_3)(\xi, t, \lambda){\rm d}\xi . \end{array}\right. \end{equation} $

假设$ [\mu_j]_k $是$ \mu_j $的第$ k $列向量,由方程(2.9)得到$ \mu_j(x, t, \lambda) $的第一、二列分别含有如下的指数项

(2.19) $ \begin{equation} {[\mu_j]}_{1}:\, {\rm e}^{-2{\rm i}[\lambda^2(\xi-x)+8\lambda^6(\tau-t)]};\quad {[\mu_j]}_{2}:\, {\rm e}^{2{\rm i}[\lambda^2(\xi-x)+8\lambda^6(\tau-t)]}. \end{equation} $

同时,在曲线上有如下的不等式

(2.20) $ \begin{equation} \begin{array}{l} \gamma_1 = (0, T)\rightarrow(x, t):\, x-\xi\geq 0, \, t-\tau\leq 0;\\ \gamma_2 = (0, 0)\rightarrow(x, t):\, x-\xi\geq 0, \, t-\tau\geq 0;\\ \gamma_3 = (\infty, t)\rightarrow(x, t):\, x-\xi\leq 0. \end{array} \end{equation} $

为了得到$ \{\mu_j(x, t, \lambda)\}_1^3 $在复$ \lambda $平面上的有界解析区域,我们用$ {\rm Im}{{\rm i}\lambda^2} = 0 $和$ {\rm Im}{{\rm i}\lambda^6} = 0 $把复$ \lambda $平面划分为六个区域(如图 2).不难发现特征函数$ \{\mu_j(x, t, \lambda)\}_1^3 $的有界解析区域分别为

(2.21) $ \begin{equation} \begin{array}{l} \mu_1:\, \lambda\in(D_2, D_5), \quad \mu_2:\, \lambda\in(D_1\cup D_3, D_4\cup D_6), \\ \mu_3:\, \lambda\in(D_4\cup D_5\cup D_6, D_1\cup D_2\cup D_3), \end{array} \end{equation} $

其中$ \{D_n\}_1^6 $表示六个开的、互不相交的复$ \lambda $平面内的子集,如图 2所示.

为了导出RH问题的跳跃矩阵,定义两个$ 2\times2 $矩阵值特征函数$ s(\lambda) $和$ S(\lambda) $如下

(2.22) $ \begin{equation} \left\{ \begin{array}{l} \mu_3(x, t;\lambda) = \mu_2(x, t;\lambda){\rm e}^{-{\rm i}(\lambda^2 x+4\lambda^6t)\hat{\sigma}_3}s(\lambda), \\ \mu_1(x, t;\lambda) = \mu_2(x, t;\lambda){\rm e}^{-{\rm i}(\lambda^2 x+4\lambda^6t)\hat{\sigma}_3}S(\lambda). \end{array}\right. \end{equation} $

在方程(2.22)第一、二个式中分别取$ (x, t) = (0, 0) $和$ (x, t) = (0, T) $,有

(2.23) $ \begin{equation} s(\lambda) = \mu_3(0, 0;\lambda), S(\lambda) = ({\rm e}^{4{\rm i}\lambda^6 T\hat{\sigma}_3}\mu_2(0, T;\lambda))^{-1}, \end{equation} $

由(2.22)和(2.23)式可以得到如下的全局关系

(2.24) $ \begin{equation} \mu_1(x, t;\lambda) = \mu_3(x, t;\lambda){\rm e}^{-{\rm i}(\lambda^2 x+4\lambda^6t)\hat{\sigma}_3}(s(\lambda))^{-1}S(\lambda), \lambda\in(D_2\cup D_4, D_1\cup D_3). \end{equation} $

特别地,由于$ \mu_1(0, t, \lambda) = {\rm I} $,我们有全局关系为

(2.25) $ \begin{equation} S^{-1}(\lambda)s(\lambda) = {\rm e}^{4{\rm i}\lambda^6T\hat\sigma_3}c(T, \lambda), \, \lambda\in(D_2\cup D_4, D_1\cup D_3), \end{equation} $

其中$ c(T, \lambda) = \mu_3(0, t, \lambda). $

命题2.1   (对称性)  特征函数

$ \mu_j(x, t;\lambda) = \left(\begin{array}{cc} \mu_j^{11}(x, t;\lambda) & \mu_j^{12}(x, t;\lambda) \\ \mu_j^{21}(x, t;\lambda) & \mu_j^{22}(x, t;\lambda) \end{array}\right)\ (j = 1, 2, 3.) $

有如下的对称性:

$ \bullet $$ \mu_j^{11}(x, t;\lambda) = \overline{\mu_j^{22}(x, t;\bar{\lambda})} $, $ \mu_j^{12}(x, t;\lambda) = \overline{\mu_j^{21}(x, t;\bar{\lambda})} $;

$ \bullet $$ \mu_j^{11}(x, t;-\lambda) = \mu_j^{11}(x, t;\lambda) $, $ \mu_j^{12}(x, t;-\lambda) = -\mu_j^{12}(x, t;\lambda) $,

$ \mu_j^{21}(x, t;-\lambda) = -\mu_j^{21}(x, t;\lambda) $, $ \mu_j^{22}(x, t;-\lambda) = \mu_j^{22}(x, t;\lambda) $.

命题2.2  由方程(2.22)和(2.23)定义的谱函数$ s(\lambda) $和$ S(\lambda) $有如下的积分形式

(2.26) $ \begin{eqnarray} \begin{array}{l} { } s(\lambda) = {\rm I}-\int_{0}^{\infty}{\rm e}^{{\rm i}\lambda^2 \xi \hat{\sigma}_3}(V_1\mu_3)(\xi, 0;\lambda){\rm d}\xi , \\ { } S^{-1}(\lambda) = {\rm I}+\int_{0}^{T}{\rm e}^{4{\rm i}\lambda^6\tau\hat{\sigma}_3}(V_2\mu_2)(0, \tau;\lambda){\rm d}\tau . \end{array} \end{eqnarray} $

我们定义矩阵函数$ s(\lambda) $和$ S(\lambda) $如下

(2.27) $ \begin{eqnarray} s(\lambda) = \left(\begin{array}{cc} \overline{a(\bar{\lambda})} & b(\lambda)\\ \overline{b(\bar{\lambda})} & a(\lambda) \end{array}\right), S(\lambda) = \left(\begin{array}{cc} \overline{A(\bar{\lambda})} & B(\lambda)\\ \overline{B(\bar{\lambda})}& A(\lambda)\\ \end{array}\right). \end{eqnarray} $

利用方程(2.23)和(2.26),我们得到

$ \left(\begin{array}{c} b(\lambda) \\ a(\lambda) \\ \end{array} \right) = \mu_3^{(2)}(0, 0;\lambda) = \left( \begin{array}{c} \mu_3^{12}(0, 0;\lambda) \\ \mu_3^{22}(0, 0;\lambda) \\ \end{array} \right), $

$ \left( \begin{array}{c} -{\rm e}^{-4{\rm i}\lambda^6T}B(\lambda) \\ \overline{A(\bar{\lambda})} \\ \end{array} \right) = \mu_2^{(2)}(0, T;\lambda) = \left( \begin{array}{c} \mu_2^{12}(0, T;\lambda) \\ \mu_2^{22}(0, T;\lambda) \\ \end{array} \right). $

$ a(-\lambda) = a(\lambda), \, b(-\lambda) = -b(\lambda), $

$ A(-\lambda) = A(\lambda), \, B(-\lambda) = -B(\lambda). $

$ {\rm det} s(\lambda) = {\rm det} S(\lambda) = 1. $

$ a(\lambda) = 1+O(\frac{1}{\lambda}), b(\lambda) = O(\frac{1}{\lambda}), \lambda\rightarrow\infty, \lambda\in D_1\cup D_2, $

$ A(\lambda) = 1+O(\frac{1}{\lambda}), B(\lambda) = O(\frac{1}{\lambda}), \lambda\rightarrow\infty, \lambda\in D_1\cup D_3. $

为了研究方便,我们假设

(2.28) $ \begin{eqnarray} \begin{array}{l} \theta(\lambda) = \lambda^2 x+4\lambda^6t, \quad \alpha(\lambda) = \overline{a(\bar{\lambda})}A(\lambda)-\overline{b(\bar{\lambda})}B(\lambda), \\ \beta(\lambda) = a(\lambda)B(\lambda)-b(\lambda)A(\lambda), \quad \delta(\lambda) = \overline{a(\bar{\lambda})}\beta(\lambda)+b(\lambda)\alpha(\lambda). \end{array} \end{eqnarray} $

矩阵函数$ M(x, t;\lambda) $定义为

(2.29) $ \begin{eqnarray} \begin{array}{l} { } M_{+}^{(1)}(x, t, \lambda) = (\frac{\mu_3^{D_4\cup D_5\cup D_6}(x, t, \lambda)}{\alpha(\lambda)}, \mu_1^{D_5}(x, t, \lambda)), \lambda\in D_5, \\ { } M_{-}^{(1)}(x, t, \lambda) = (\frac{\mu_3^{D_4\cup D_5\cup D_6}(x, t, \lambda)}{\overline{a(\bar{\lambda})}}, \mu_2^{D_4\cup D_6}(x, t, \lambda)), \lambda\in D_4\cup D_6, \\ { } M_{+}^{(2)}(x, t, \lambda) = (\mu_2^{D_1\cup D_3}(x, t, \lambda), \frac{\mu_3^{D_1\cup D_2\cup D_3}(x, t, \lambda)}{a(\lambda)}), \lambda\in D_1\cup D_3, \\ { } M_{-}^{(2)}(x, t, \lambda) = (\mu_1^{D_2}(x, t, \lambda), \frac{\mu_3^{D_1\cup D_2\cup D_3}(x, t, \lambda)}{\overline{\alpha(\bar{\lambda})}}), \lambda\in D_2. \end{array} \end{eqnarray} $

因此,我们有

(2.30) $ \begin{equation} {\rm det} M(x, t;\lambda) = 1, \, \, M(x, t;\lambda) = {\rm I}+O(\frac{1}{\lambda}), \lambda\rightarrow\infty. \end{equation} $

命题2.3  假设$ u(x, t) $充分光滑, $ \{\mu_j(x, t, \lambda)\}_1^3 $由方程(2.17)给出, $ M(x, t;\lambda) $由方程(2.29)给出,那么$ M(x, t;\lambda) $满足下面的跳跃条件

(2.31) $ \begin{eqnarray} M_{+}(x, t, \lambda) = M_{-}(x, t, \lambda)J(x, t, \lambda), \lambda^6 \in {{\Bbb R}}, \end{eqnarray} $

其中$ J(x, t, \lambda) $是跳跃矩阵,在不同的跳跃曲线上,跳跃矩阵可以分别表示为

(2.32) $ \begin{eqnarray} J(x, t, \lambda) = \left\{ \begin{array}{l} { } J_1(x, t, \lambda), {\rm Arg} \lambda = k\pi \cup k\pi+\frac{\pi}{2}, \\ { } J_2(x, t, \lambda), {\rm Arg} \lambda = k\pi+\frac{\pi}{6} \cup k\pi+\frac{\pi}{3}, \\ { } J_3(x, t, \lambda), {\rm Arg} \lambda = k\pi+\frac{2\pi}{3} \cup k\pi+\frac{5\pi}{6}, \end{array}\right. \end{eqnarray} $

其中

(2.33) $ \begin{eqnarray} &J_1(x, t, \lambda) = \left(\begin{array}{cc} 1 &{ } \frac{b(\lambda)}{\overline{a(\bar{\lambda})}}{\rm e}^{-2{\rm i}\theta(\lambda)} \\ { } -\frac{\overline{b(\bar{\lambda})}}{a(\lambda)}{\rm e}^{2{\rm i}\theta(\lambda)} &{ } \frac{1}{a(\lambda)\overline{a(\bar{\lambda})}} \end{array}\right), {}\\ &J_2(x, t, \lambda) = \left(\begin{array}{cc} { } \frac{a(\lambda)}{\overline{\alpha(\bar{\lambda})}} & 0 \\ { } -\overline{\delta(\bar{\lambda})} {\rm e}^{2{\rm i}\theta(\lambda)}& { } \frac{\overline{\alpha(\bar{\lambda})}}{a(\lambda)} \end{array}\right), J_3(x, t, \lambda) = \left(\begin{array}{cc} { } \frac{\overline{a(\bar{\lambda})}}{\alpha(\lambda)} & { } \delta(\lambda){\rm e}^{-2{\rm i}\theta(\lambda)} \\ 0 &{ } \frac{\alpha(\lambda)}{\overline{a(\bar{\lambda})}} \end{array} \right). \end{eqnarray} $

证  为了导出(2.31)式,我们把(2.22)和(2.27)式写成下面的形式

(2.34) $ \begin{eqnarray} && \left\{ \begin{array}{l} \overline{a(\bar{\lambda})}\mu_2^{D_1\cup D_3}+\overline{b(\bar{\lambda})}{\rm e}^{2{\rm i}\theta(\lambda)}\mu_2^{D_4\cup D_6} = \mu_3^{D_4\cup D_5\cup D_6}, \\ b(\lambda){\rm e}^{-2{\rm i}\theta(\lambda)}\mu_2^{D_1\cup D_3}+a(\lambda)\mu_2^{D_4\cup D_6} = \mu_3^{D_1\cup D_2\cup D_3}, \end{array}\right.{}\\ && \left\{ \begin{array}{l} \overline{A(\bar{\lambda})}\mu_2^{D_1\cup D_3}+\overline{B(\bar{\lambda})}{\rm e}^{2{\rm i}\theta(\lambda)}\mu_2^{D_4\cup D_6} = \mu_1^{D_2}, \\ B(\lambda){\rm e}^{-2{\rm i}\theta(\lambda)}\mu_2^{D_1\cup D_3}+A(\lambda)\mu_2^{D_4\cup D_6} = \mu_1^{D_5}, \end{array}\right.\\ && \left\{ \begin{array}{l} \overline{\alpha(\bar{\lambda})}\mu_3^{D_4\cup D_5\cup D_6}+\overline{\beta(\bar{\lambda})}{\rm e}^{2{\rm i}\theta(\lambda)}\mu_2^{D_1\cup D_2\cup D_3} = \mu_1^{D_2}, \\ \beta(\lambda){\rm e}^{-2{\rm i}\theta(\lambda)}\mu_3^{D_4\cup D_5\cup D_6}+\alpha(\lambda)\mu_2^{D_1\cup D_2\cup D_3} = \mu_1^{D_5}. \end{array}\right.{} \end{eqnarray} $

通过直接计算可以得到(2.33)式,并且$ J_i(x, t;\lambda)\ (i = 1, 2, 3, 4.) $满足跳跃条件(2.31),进一步地,我们可以得到$ M_{+}^{(1)}(x, t;\lambda) $和$ M_{-}^{(2)}(x, t;\lambda) $之间的跳跃关系为

$ M_{+}^{(1)}(x, t;\lambda) = M_{-}^{(2)}(x, t;\lambda)J_4(x, t;\lambda), $

其中$ J_4(x, t;\lambda) = J_2(x, t;\lambda)J_1^{-1}(x, t;\lambda)J_3(x, t;\lambda) $.

函数$ M(x, t;\lambda) $在复$ \lambda $平面上是解析函数,它可能的极点就是$ a(\lambda) $和$ \alpha(\lambda) $的零点.我们假设

$ \bullet $$ a(\lambda) $有$ 2n $个简单零点$ \{\varepsilon_j\}_{j = 1}^{2n} $, $ 2n = 2n_1+2n_2 $,且$ \{\varepsilon_j\}_{j = 1}^{2n_1} $在$ D_1\cup D_3 $中, $ \{\varepsilon_j\}_{j = 1}^{2n_2} $在$ D_4\cup D_6 $中.

$ \bullet $$ \alpha(\lambda) $有$ 2N $个简单零点$ \{\gamma_j\}_{j = 1}^{2N} $ ($ 2N = 2N_1+2N_2 $),且$ \{\varepsilon_j\}_{j = 1}^{2N_1} $在$ D_5 $中, $ \{\varepsilon_j\}_{j = 1}^{2N_2} $在$ D_2 $中.

$ \bullet $$ \alpha(\lambda) $和$ a(\lambda) $没有相同的零点.

命题2.4  记$ \dot{a}(\lambda) = \frac{{\rm d}a}{{\rm d}\lambda} $,我们有如下的留数条件:

(ⅰ) $ {\rm Res} \{[M(x, t;\lambda)]_{2} , \varepsilon_j\} = \frac{{\rm e}^{-2{\rm i}\theta(\varepsilon_j)} b(\varepsilon_j)}{\dot{a}(\varepsilon_j)}[M(x, t;\varepsilon_j)]_{1}, $$ j = 1, 2, \cdots, 2n_1 $.

(ⅱ) $ {\rm Res}\{[M(x, t;\lambda)]_{1} , \bar{\varepsilon}_j \} = \frac{{\rm e}^{2{\rm i}\theta(\bar{\varepsilon_j})}\overline{b(\bar{\varepsilon}_j)}}{\overline{\dot{a} (\bar{\varepsilon}_j)}}[M(x, t;\bar{\varepsilon}_j)]_{2} $, $ j = 1, 2, \cdots, 2n_2 $.

(ⅲ) $ {\rm Res}\{[M(x, t;\lambda)]_{1} , \gamma_j \} = \frac{{\rm e}^{2{\rm i}\theta(\gamma_j)}} {\dot{\alpha}(\gamma_j)\beta(\gamma_j)}[M(x, t;\gamma_j)]_{2} $, $ j = 1, 2, \cdots, 2N_1 $.

(ⅳ) $ {\rm Res}\{[M(x, t;\lambda)]_{2} , $$ \bar{\gamma}_j \} = \frac{{\rm e}^{-2{\rm i}\theta (\bar{\gamma}_j)}}{\overline{\dot{\alpha}(\bar{\gamma}_j)}\overline{\beta(\bar{\gamma_j})}} [M(x, t;\bar{\gamma_j})]_{1} $, $ j = 1, 2, \cdots, 2N_2 $.

证  这里我们只需证明(ⅰ),其余的留数计算与之类似.考虑

$ M_{+}^{(2)}(x, t;\lambda) = \bigg(\mu_2^{D_1\cup D_3}, \frac{\mu_3^{D_1\cup D_2\cup D_3}}{a(\lambda)}\bigg), $

$ a(\lambda) $的简单零点$ \{\varepsilon_j\}_{1}^{2n_1} $是$ \frac{\mu_3^{D_1\cup D_2\cup D_3}}{a(\lambda)} $的简单极点,那么有

$ {\rm Res}\bigg\{\frac{\mu_3^{D_1\cup D_2\cup D_3}(x, t;\lambda)}{a(\lambda)}, \varepsilon_j\bigg\} = \lim\limits_{\lambda\rightarrow\varepsilon_j}(\lambda-\varepsilon_j)\frac{\mu_3^{D_1\cup D_2\cup D_3}(x, t;\lambda)}{a(\lambda)} = \frac{\mu_3^{D_1\cup D_2\cup D_3}(x, t;\varepsilon_j)}{\dot{a}(\varepsilon_j)}. $

将$ \lambda = \varepsilon_j $代入(2.33)式的第二个方程得到

$ \mu_3^{D_1\cup D_2\cup D_3}(x, t;\varepsilon_j) = b(\varepsilon_j){\rm e}^{-2{\rm i}\theta(\varepsilon_j)}\mu_2^{D_1\cup D_2}(x, t;\varepsilon_j). $

进一步地,我们有

$ {\rm Res}\bigg\{\frac{\mu_3^{D_1\cup D_2\cup D_3}(x, t;\lambda)}{a(\lambda)}, \varepsilon_j\bigg\} = \frac{b(\varepsilon_j){\rm e}^{-2{\rm i}\theta(\varepsilon_j)}}{\dot{a}(\varepsilon_j)}\mu_2^{D_1\cup D_2}(x, t;\varepsilon_j). $

这就是(ⅰ)式,证毕.

在上一部分定义的$ M(x, t, \lambda) $满足一个与$ q(x, t) $的初边值数据有关的$ 2\times 2 $矩阵RH问题.通过解这个$ 2\times 2 $矩阵RH问题,对于所有的$ (x, t) $可以重建(1.2)的解,因此,我们有:

定理3.1  在方程(2.27)中,由$ a(\lambda), b(\lambda), A(\lambda) $和$ B(\lambda) $定义的谱函数$ S(\lambda) $和$ s(\lambda) $满足全局关系(2.25).即利用初边值数据$ q_0(x) = q(x, 0);g_0(t) = q(0, t), g_1(t) = q_{x}(0, t), g_2(t) = q_{xx}(0, t) $,我们由方程(2.22)定义了谱函数$ s(\lambda) $和$ S(\lambda) $,并且通过计算得到了跳跃矩阵$ J(x, t, \lambda) $,假设$ a(\lambda) $和$ \alpha(\lambda) $的简单极点分别为$ \{\varepsilon_j\}_{j = 1}^{2n} $和$ \{\gamma_j\}_{j = 1}^{2N} $,则矩阵函数$ M(x, t, \lambda) $是下面RH问题的解:

$ \bullet $$ M(x, t;\lambda) $是$ {\rm C}\setminus\{\lambda^6 \in {{\Bbb R}}\} $上的分片解析函数.

$ \bullet $在曲线$ \lambda^6 \in {{\Bbb R}} $上, $ M(x, t;\lambda) $满足如下的跳跃关系:

(3.1) $ \begin{equation} M_{+}(x, t;\lambda) = M_{-}(x, t;\lambda)J(x, t;\lambda). \end{equation} $

$ \bullet $$ M(x, t;\lambda) = {\rm I}+O(\frac{1}{\lambda}), \, \lambda\rightarrow\infty $.

$ \bullet $$ M(x, t;\lambda) $满足命题2.4中的留数条件.

因此, $ M(x, t;\lambda) $存在并且是唯一的,且势函数$ q(x, t) $可以由矩阵函数$ M(x, t, \lambda) $来给出:

(3.2) $ \begin{equation} { } q(x, t) = 2{\rm i}m(x, t){\rm e}^{2{\rm i}\int_{(0, 0)}^{(x, t)}\Delta}, {\qquad} { } m(x, t) = \lim\limits_{\lambda\rightarrow\infty}(\lambda M(x, t;\lambda))_{12}. \end{equation} $

此外, $ q(x, t) $是HOCLL方程(1.2)的解,并且满足方程(1.3)定义的初边值数据.

证  事实上,如果函数$ a(\lambda) $和$ \alpha(\lambda) $没有零点,那么$ 2\times2 $矩阵函数$ M(x, t;\lambda) $是一个非正则的RH问题.利用跳跃矩阵$ J(x, t;\lambda) $的对称条件,我们可以证明这个非正则的RH问题能够映射到一个耦合与代数方程系统的正则性的RH问题.

定理3.2   (消灭定理)定理3.1中的带有边界条件$ M(x, t;\lambda)\rightarrow 0, \, \, \, \lambda \rightarrow \infty $的RH问题有且仅有零解.

证  可以类似于文献[4-5]中的Dressing方法证明该消灭定理,这里省略.