2. 滁州学院 数学与金融学院, 安徽 滁州 239000
2. School of Mathematics and Finance, Chuzhou University, Chuzhou Anhui 239000, China
自Gardner, Green, Kruskal, Miüra发现了反散射变换以来, 一直到20世纪90年代, 反散射变换几乎只是用来分析纯初值问题, 但是在现实自然界中, 越来越多的自然现象需要考虑边值条件, 这样就自然地需要考虑初边值问题来取代初值问题. 1997年, Fokas[1]基于反散射变换的思想首次提出了统一变换方法, 很好地求解了可积方程的初边值问题.在过去的20年里, 该方法已经用来分析了一些具有
众所周知, 非线性薛定谔方程
$ \begin{align} {\rm i}q_{t}+q_{xx}+2|q|^2q = 0 \end{align} $ | (1) |
是描述非线性光纤、水波、等离子体中孤子传输的一个重要可积系统.在文献[17]中, 高阶非线性薛定谔方程, 也就是Sasa-Satsuma方程
$ \begin{align} u_{t}+u_{xxx}+6|u|^2u_{x}+3u(|u|^2)_x = 0 \end{align} $ | (2) |
也描述了超短光脉冲的传播特性.近来, Geng和Wu[18]提出了新的广义Sasa-Satsuma方程
$ \begin{align} u_{t}+u_{xxx}-6a|u|^2u_{x}-6bu^2u_{x}-3au(|u|^2)_x-3b^*u^*(|u|^2)_x = 0, \end{align} $ | (3) |
其中
$ \begin{align} u_0(x) = u(x, t = 0);\, q_0(t) = u(x = 0, t);\, q_1(t) = u_{x}(x = 0, t), q_2(t) = u_{xx}(x = 0, t). \end{align} $ | (4) |
本文结构如下:在第1节中, 定义了Lax对的两类特征函数
考虑广义Sasa-Satsuma方程(3)的
$ \begin{align} \left\{\!\!\begin{array}{l} \Psi_x = U\Psi = ({\rm i}\lambda\sigma+Q)\Psi, \\ \Psi_t = V\Psi = (4{\rm i}\lambda^3\sigma+4\lambda^2 Q+2{\rm i}\lambda(Q^2+Q_x)\sigma \\ +Q_xQ-QQ_x-Q_{xx}+2Q^3)\Psi. \end{array}\right. \end{align} $ | (5) |
其中:
$ \begin{align} \begin{array}{l} \sigma = \left(\begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&-1\end{array} \right), Q = \left(\begin{array}{ccc} 0&0&u\\ 0&0&u^*\\ au^*+bu&au+b^*u^*&0\end{array} \right). \end{array} \end{align} $ | (6) |
事实上, Lax对方程(5)可以写成
$ \begin{align} \left\{\!\!\begin{array}{l} \Psi_x-{\rm i}\lambda\sigma = F\Psi, \\ \Psi_t-4{\rm i}\lambda^3\sigma = G\Psi. \end{array}\right. \end{align} $ | (7) |
其中:
假设
$ \begin{align} \Psi(x, t, \lambda) = \Phi(x, t, \lambda){\rm e}^{{\rm i}(\lambda\sigma x+4\lambda^3\sigma t)} \end{align} $ | (8) |
引入一个新的特征函数
$ \begin{align} \left\{\!\!\begin{array}{l} \Phi_x-{\rm i}\lambda[\sigma, \Phi] = F\Phi, \\ \Phi_t-{\rm i}\lambda^3[\sigma, \Phi] = G\Phi.\\ \end{array}\right. \end{align} $ | (9) |
因此, 方程(9)可以写成如下的全微分形式
$ \begin{align} {\mathrm{d}}({{\rm e}}^{-({{\rm i}}\lambda\hat\sigma x+ 4{{\rm i}}\lambda^3\hat\sigma t)}\Phi) = W(x, t, \lambda), \end{align} $ | (10) |
其中
$ \begin{align} W(x, t, \lambda) = {\rm e}^{-({\rm i}\lambda x+2{\rm i}\lambda^2t)\hat\sigma}(F\mathrm{d}x+G\mathrm{d}t)\Phi. \end{align} $ | (11) |
这里的
由Volterra积分方程定义(9)的3个特征函数
$ \begin{align} \Phi_j(x, t, \lambda) = {\rm I}+\int_{\gamma_j}{\rm e}^{({\rm i}\lambda x+2{\rm i}\lambda^2t)\hat\sigma}W_j(x, t, \lambda), \quad j = 1, 2, 3. \end{align} $ | (12) |
其中: I是一个
假设
$ \left\{ {\begin{array}{*{20}{c}} {{[\Phi_j]}_{1}:\, {\rm e}^{-2{\rm i}\lambda(x-\xi)-8{\rm i}\lambda^3(t-\tau)};\, {[\Phi_j]}_{2}:\, {\rm e}^{-2{\rm i}\lambda(x-\xi)-8{\rm i}\lambda^3(t-\tau)};} \hfill \\ {{[\Phi_j]}_{3}:\, {\rm e}^{2{\rm i}\lambda(x-\xi)+8{\rm i}\lambda^3(t-\tau)}, \, {\rm e}^{2{\rm i}\lambda(x-\xi)+8{\rm i}\lambda^3(t-\tau)}.} \hfill \\ \end{array}} \right. $ | (13) |
同时, 在曲线上有如下的不等式
$ \begin{align} \gamma_1:\, x-\xi\geqslant 0, \, t-\tau\leqslant 0;\quad \gamma_2:\, x-\xi\geqslant 0, \, t-\tau\geqslant 0;\quad \gamma_3:\, x-\xi\leqslant 0. \end{align} $ | (14) |
为了得到
$ \begin{align*} & D_1 = \Big\{\lambda\in {\mathbb{C}}\Big|{\rm arg} \lambda\in\Big(0, \frac{\pi}{3}\Big)\cup\Big(\frac{2\pi}{3}, \pi\Big)\Big\}, \quad D_2 = \Big\{\lambda\in {\mathbb{C}}\Big|{\rm arg} \lambda\in\Big(\frac{\pi}{3}, \frac{2\pi}{3}\Big)\Big\}, \\ &D_3 = \Big\{\lambda\in {\mathbb{C}}\Big|{\rm arg} \lambda\in\Big(\frac{4\pi}{3}, \frac{5\pi}{3}\Big)\Big\}, \quad D_4 = \Big\{\lambda\in {\mathbb{C}}\Big|{\rm arg} \lambda\in\Big(\pi, \frac{4\pi}{3}\Big)\cup\Big(\frac{5\pi}{3}, 2\pi\Big)\Big\}. \end{align*} $ |
不难发现特征函数
$ \begin{align} \Phi_1:\, \lambda\in(D_3, D_3, D_2), \, \Phi_2:\, \lambda\in(D_4, D_4, D_1), \, \Phi_3:\, \lambda\in(C_{+}, C_{+}, C_{-}), \end{align} $ | (15) |
其中
$\left\{ {\begin{array}{*{20}{c}} {{D_1} = \{ \lambda \in {\mathbb{C}}|{\rm{Re}}\;{a_1} = {\rm{Re}}\;{a_2} < {\rm{Re}}\;{a_3},{\mkern 1mu} {\rm{Re}}\;{b_1} = {\rm{Re}}\;{b_2} < {\rm{Re}}\;{b_3}\} ,}\\ {{D_2} = \{ \lambda \in {\mathbb{C}}|{\rm{Re}}\;{a_1} = {\rm{Re}}\;{a_2} < {\rm{Re}}\;{a_3},{\mkern 1mu} {\rm{Re}}\;{b_1} = {\rm{Re}}\;{b_2} > {\rm{Re}}\;{b_3}\} ,}\\ {{D_3} = \{ \lambda \in {\mathbb{C}}|{\rm{Re}}\;{a_1} = {\rm{Re}}\;{a_2}{\rm{ > Re}}\;{a_3},{\mkern 1mu} {\rm{Re}}\;{b_1} = {\rm{Re}}\;{b_2} < Re\;{b_3}\} ,}\\ {{D_4} = \{ \lambda \in {\mathbb{C}}|{\rm{Re}}\;{a_1} = {\rm{Re}}\;{a_2}{\rm{ > Re}}\;{a_3},{\mkern 1mu} {\rm{Re}}\;{b_1} = {\rm{Re}}\;{b_2}{\rm{ > Re}}\;{b_3}\} ,} \end{array}} \right.$ | (16) |
其中
实际上, 对于
对任一
$ \begin{align} (H_n(x, t, \lambda))_{ij} = \delta_{ij}+\int_{\gamma_{ij}^n}({\rm e}^{({\rm i}\lambda x+4{\rm i}\lambda^3t)\hat\sigma}W_n(\xi, \tau, \lambda))_{ij}, \quad i, j = 1, 2, 3, \end{align} $ | (17) |
其中
$ \begin{align} \gamma_{ij}^n = \left\{\!\!\begin{array}{l} \gamma_1, \, \mbox{如果} \, {\rm Re}\; a_i(\lambda)<{\rm Re}\; a_j(\lambda) \, \mbox{且} \, {\rm Re}\; b_i(\lambda)\geqslant {\rm Re}\; b_j(\lambda), \\ \gamma_2, \, \mbox{如果} \, {\rm Re}\; a_i(\lambda)<{\rm Re}\; a_j(\lambda) \, \mbox{且} \, {\rm Re}\; b_i(\lambda)<{\rm Re}\; b_j(\lambda), \lambda\in D_n, \\ \gamma_3, \, \mbox{如果} \, {\rm Re}\; a_i(\lambda) \geqslant {\rm Re}\; a_j(\lambda). \end{array}\right. \end{align} $ | (18) |
按照
$\left\{ {\begin{array}{*{20}{c}} {{\gamma ^1} = \left( {\begin{array}{*{20}{c}} {{\gamma _3}}&{{\gamma _3}}&{{\gamma _2}}\\ {{\gamma _3}}&{{\gamma _3}}&{{\gamma _2}}\\ {{\gamma _3}}&{{\gamma _3}}&{{\gamma _3}} \end{array}} \right),\quad {\gamma ^2} = \left( {\begin{array}{*{20}{c}} {{\gamma _3}}&{{\gamma _3}}&{{\gamma _1}}\\ {{\gamma _3}}&{{\gamma _3}}&{{\gamma _1}}\\ {{\gamma _3}}&{{\gamma _3}}&{{\gamma _3}} \end{array}} \right),}\\ {{\gamma ^3} = \left( {\begin{array}{*{20}{c}} {{\gamma _3}}&{{\gamma _3}}&{{\gamma _3}}\\ {{\gamma _3}}&{{\gamma _3}}&{{\gamma _3}}\\ {{\gamma _1}}&{{\gamma _3}}&{{\gamma _3}} \end{array}} \right),\quad {\gamma ^4} = \left( {\begin{array}{*{20}{c}} {{\gamma _3}}&{{\gamma _3}}&{{\gamma _3}}\\ {{\gamma _3}}&{{\gamma _3}}&{{\gamma _3}}\\ {{\gamma _2}}&{{\gamma _2}}&{{\gamma _3}} \end{array}} \right).} \end{array}} \right.$ | (19) |
下面的命题保证了上述定义的
命题 1.1 对于每一个
$ \begin{align} H_n(x, t, \lambda) = {\rm I}+O\Big(\frac{1}{\lambda}\Big), \quad \lambda\in D_n, \quad \lambda\rightarrow\infty, \quad n = 1, 2, 3, 4. \end{align} $ | (20) |
证 明 相关的有界解析性质在文献[6]的附录B中已给出.将下面的展开式
$ \begin{align*} H = H_0+\frac{H^{(1)}}{\lambda}+\frac{H^{(2)}}{\lambda^2}+\cdots, \quad \lambda\rightarrow\infty, \end{align*} $ |
代入到Lax对方程(9)中并比较
我们还需要分析矩阵值函数
$ Z^A = \left(\begin{array}{ccc} m_{11}(Z)&-m_{12}(Z)&m_{13}(Z)\\ -m_{21}(Z)&m_{22}(Z)&-m_{23}(Z)\\ m_{31}(Z)&-m_{32}(Z)&m_{33}(Z) \end{array} \right), $ |
其中
$ \begin{align} \left\{\!\!\begin{array}{l} \Phi_x^A+{\rm i}\lambda[\sigma, \Phi^A] = -F^{\rm T}\Phi^A, \\ \Phi_t^A+4{\rm i}\lambda^3[\sigma, \Phi^A] = -G^{\rm T}\Phi^A, \\ \end{array}\right. \end{align} $ | (21) |
其中
$ \begin{align} \Phi_j^A(x, t, \lambda) = {\rm I}-\int_{\gamma_j}{\rm e}^{-({\rm i} \lambda(x-\xi)-4{\rm i}\lambda^3(t-\tau))\hat\sigma} (F^{\rm T}{\rm d}\xi+G^{\rm T}{\rm d}\tau), \quad j = 1, 2, 3. \end{align} $ | (22) |
然后可得到共轭特征函数如下的有界解析性质
$ \begin{align} \Phi_1^A : \lambda\in(D_2, D_2, D_3), \, \Phi_2^A :\, \, \lambda\in(D_1, D_1, D_4), \, \Phi_3^A :\, \, \lambda\in(C_{-}, C_{-}, C_{+}). \end{align} $ | (23) |
实际上, 对于
新的谱函数
$ \begin{align} S_n(\lambda) = H_n(0, 0, \lambda), \quad \lambda\in D_n, n = 1, 2, 3, 4. \end{align} $ | (24) |
假设
$ \begin{align} H_n(x, t, \lambda) = H_m(x, t, \lambda)J_{m, n}(x, t, \lambda), \lambda\in \bar D_n\cup \bar D_m, \, n, m = 1, 2, 3, 4, \, n\neq m, \end{align} $ | (25) |
其中
$ \begin{align} J_{m, n}(x, t, \lambda) = {\rm e}^{({\rm i}\lambda x+4{\rm i}\lambda^3t)\hat\sigma}(S_m^{-1}(\lambda)S_n(\lambda)). \end{align} $ | (26) |
此外, 由
$ \left\{ {\begin{array}{*{20}{c}} {\Phi_3(x, t, \lambda) = \Phi_2(x, t, \lambda){\rm e}^{({\rm i}\lambda x+4{\rm i}\lambda^3t)\hat\sigma} s(\lambda), } \hfill \\ {\Phi_1(x, t, \lambda) = \Phi_2(x, t, \lambda){\rm e}^{({\rm i}\lambda x+4{\rm i}\lambda^3t)\hat\sigma} S(\lambda).} \end{array}} \right. $ | (27) |
由于
$ \begin{align} s(\lambda) = \Phi_3(0, 0, \lambda), \quad S(\lambda) = \Phi_1(0, 0, \lambda) = {\rm e}^{-2{\rm i}\lambda^2T\hat\sigma}\Phi_2^{-1}(0, T, \lambda). \end{align} $ | (28) |
按照式(12)中
$ \begin{align*} &s(\lambda) = {\rm I}-\int_{0}^{\infty}{\rm e}^{-{\rm i}\lambda \xi\hat\sigma}(F\Phi_3)(\xi, 0, \lambda){\rm d}\xi, \\ &S(\lambda) = {\rm I}-\int_{0}^{T}{\rm e}^{-4{\rm i}\lambda^3\tau\hat\sigma}(G\Phi_1)(0, \tau, \lambda){\rm d}\tau = \Big[{\rm I}+\int_{0}^{T}{\rm e}^{-4{\rm i}\lambda^3\tau\hat\sigma}(G\Phi_2)(0, \tau, \lambda){\rm d}\tau\Big]^{-1}, \end{align*} $ |
其中
$ \begin{align*} &\Phi_1(0, t, \lambda) = {\rm I}-\int_{t}^{T}{\rm e}^{4{\rm i}\lambda^3(t-\tau)\hat\sigma}(G\Phi_1)(0, \tau, \lambda){\rm d}\tau, \lambda\in(D_1\cup D_3, D_1\cup D_3, D_2\cup D_4), \\ &\Phi_2(0, t, \lambda) = {\rm I}+\int_{0}^{T}{\rm e}^{4{\rm i}\lambda^3(t-\tau)\hat\sigma}(G\Phi_2)(0, \tau, \lambda){\rm d}\tau, \lambda\in(D_2\cup D_4, D_2\cup D_4, D_1\cup D_3), \\ &\Phi_3(x, 0, \lambda) = {\rm I}-\int_{x}^{\infty}{\rm e}^{{\rm i}\lambda(x-\xi)\hat\sigma}(F\Phi_3)(\xi, 0, \lambda){\rm d}\xi, \lambda\in(C_{+}, C_{+}, C_{-}). \end{align*} $ |
由
$ \left\{ {\begin{array}{*{20}{c}} {s(\lambda): \lambda\in(C_{+}, C_{+}, C_{-}), \quad S(\lambda): \lambda\in(D_1\cup D_3, D_1\cup D_3, D_2\cup D_4), } \hfill \\ {s^A(\lambda): \lambda\in(C_{-}, C_{-}, C_{+}), \quad S^A(\lambda): \lambda\in(D_2\cup D_4, D_2\cup D_4, D_1\cup D_3).} \end{array}} \right. $ | (29) |
并且
$ \begin{align} H_n(x, t, \lambda) = \Phi_2(x, t, \lambda){\rm e}^{({\rm i}\lambda x +4{\rm i}\lambda^3t)\hat\sigma} S_n(\lambda), \quad \lambda\in D_n. \end{align} $ | (30) |
命题 1.2[6, 13-16] 由方程(30)定义的
$ \begin{align} \begin{array}{l} S_1(\lambda) = \left(\begin{array}{ccc} s_{11} & s_{12}&0\\ s_{21} & s_{22}&0\\ s_{31} & s_{32}&\frac{1}{m_{33}(s)} \end{array} \right), \quad S_2(\lambda) = \left(\begin{array}{ccc} s_{11} & s_{12}&\frac{S_{13}}{(S^{\rm T}s^A)_{33}}\\ s_{21} & s_{22}&\frac{S_{23}}{(S^{\rm T}s^A)_{33}}\\ s_{31} & s_{32}&\frac{S_{33}}{(S^{\rm T}s^A)_{33}} \end{array} \right), \hfill \\ S_3(\lambda) = \left(\begin{array}{ccc} S_{11}^{(3)} & S_{12}^{(3)} & s_{13}\\ S_{21}^{(3)}& S_{22}^{(3)} & s_{23}\\ S_{31}^{(3)} & S_{32}^{(3)}& s_{33} \end{array} \right), \quad S_4(\lambda) = \left(\begin{array}{ccc} \frac{m_{22}(s)}{s_{33}}&\frac{m_{21}(s)}{s_{33}} & s_{13}\\ \frac{m_{12}(s)}{s_{33}}&\frac{m_{11}(s)}{s_{33}} & s_{23}\\ 0&0 & s_{33} \end{array}\right). \end{array} \end{align} $ | (31) |
其中
$ \begin{align*} &S_{11}^{(3)} = \frac{m_{22}(s)m_{33}(S)-m_{32}(s)m_{23}(S)}{(s^{\rm T}S^A)_{33}}, \quad S_{12}^{(3)} = \frac{m_{21}(s)m_{33}(S)-m_{31}(s)m_{23}(S)}{(s^{\rm T}S^A)_{33}}, \\ &S_{21}^{(3)} = \frac{m_{12}(s)m_{33}(S)-m_{32}(s)m_{13}(S)}{(s^{\rm T}S^A)_{33}}, \quad S_{22}^{(3)} = \frac{m_{11}(s)m_{33}(S)-m_{31}(s)m_{13}(S)}{(s^{\rm T}S^A)_{33}}, \\ &S_{31}^{(3)} = \frac{m_{12}(s)m_{23}(S)-m_{22}(s)m_{13}(S)}{(s^{\rm T}S^A)_{33}}, \quad S_{32}^{(3)} = \frac{m_{11}(s)m_{23}(S)-m_{21}(s)m_{13}(S)}{(s^{\rm T}S^A)_{33}}. \end{align*} $ |
这里
$ \begin{align*} (S^{\rm T}s^A)_{33} = S_{13}m_{13}(s)-S_{23}m_{23}(s)+S_{33}m_{33}(s), \\ (s^{\rm T}S^A)_{33} = s_{13}m_{13}(S)-s_{23}m_{23}(S)+s_{33}m_{33}(S). \end{align*} $ |
由于
假设 1.3 假设
且上面的零点互不相同, 同时假设
命题 1.4 假设
$ \begin{align} {\rm Res}_{\lambda = \lambda_j}[H_1]_3 = \frac{s_{12}(\lambda_j)[H(\lambda_j)]_1-s_{11}(\lambda_j)[H(\lambda_j)]_2}{\dot{m_{33}(s)(\lambda_j)}m_{23}(s)(\lambda_j)} {\rm e}^{\theta_{31}(\lambda_j)} , \quad 1\leqslant j\leqslant n_0, \quad\lambda_j\in D_1. \end{align} $ | (32) |
$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {\rm{Re}}{{\rm{s}}_{\lambda = {\lambda _j}}}{[{H_2}]_3} = \frac{{{s_{32}}({\lambda _j}){S_{13}}({\lambda _j}) - {s_{32}}({\lambda _j}){S_{33}}({\lambda _j})}}{{\mathop {{{({S^{\rm{T}}}{s^A})}_{33}}({\lambda _j})}\limits^. {m_{23}}(s)({\lambda _j})}}{{\rm{e}}^{{\theta _{31}}({\lambda _j})}}{[H({\lambda _j})]_1}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + \frac{{{s_{11}}({\lambda _j}){S_{32}}({\lambda _j}) - {s_{31}}({\lambda _j}){S_{13}}({\lambda _j})}}{{\mathop {{{({S^{\rm{T}}}{s^A})}_{33}}({\lambda _j})}\limits^. {m_{23}}(s)({\lambda _j})}}{{\rm{e}}^{{\theta _{31}}({\lambda _j})}}{[H({\lambda _j})]_2}, \end{array}\\ {{n_0} + 1 \le j \le {n_1}, \quad {\lambda _j} \in {D_2}.} \end{array}} \right.$ | (33) |
$ \left\{ {\begin{array}{*{20}{c}} {\rm Res}_{\lambda = \lambda_j}[H_3]_1 = \frac{m_{12}(s)(\lambda_j) m_{33}(S)(\lambda_j)-m_{32}(s)(\lambda_j)m_{13}(S)(\lambda_j)}{\dot{(s^{\rm T}S^A)_{33}(\lambda_j)}s_{23}(\lambda_j)} {\rm e}^{\theta_{13}(\lambda_j)}[H(\lambda_j)]_3 , \\ n_1+1\leqslant j\leqslant n_2, \quad\lambda_j\in D_3. \end{array}} \right. $ | (34) |
$ \left\{ {\begin{array}{*{20}{c}} {\rm Res}_{\lambda = \lambda_j}[H_3]_2 = \frac{m_{21}(s) (\lambda_j)m_{33}(S)(\lambda_j)-m_{31}(s)(\lambda_j)m_{23}(S) (\lambda_j)}{\dot{(s^{\rm T}S^A)_{33}(\lambda_j)}s_{13}(\lambda_j)} {\rm e}^{\theta_{13}(\lambda_j)}[H(\lambda_j)]_3 , \\ n_1+1\leqslant j\leqslant n_2, \quad\lambda_j\in D_3. \end{array}} \right. $ | (35) |
$ \begin{align} {\rm Res}_{\lambda = \lambda_j}[H_4]_1 = \frac{m_{12}(s)(\lambda_j)}{\dot{s_{33}(\lambda_j)}s_{23}(\lambda_j)}{\rm e}^{\theta_{13}(\lambda_j)}[H(\lambda_j)]_3 , \quad n_2+1\leqslant j\leqslant N, \quad\lambda_j\in D_4. \end{align} $ | (36) |
$ \begin{align} {\rm Res}_{\lambda = \lambda_j}[H_4]_2 = \frac{m_{12}(s)(\lambda_j)}{\dot{s_{33}(\lambda_j)}s_{13}(\lambda_j)}{\rm e}^{\theta_{13}(\lambda_j)}[H(\lambda_j)]_3 , \quad n_2+1\leqslant j\leqslant N, \quad\lambda_j\in D_4. \end{align} $ | (37) |
其中:
$ \begin{align} \theta_{ij}(x, t, \lambda) = (a_i-a_j)x-(b_i-b_j)t, \quad i, j = 1, 2, 3. \end{align} $ | (38) |
证 明 这里我们只证明式(32), 而式(33)-(37)可以类似证明.由式(30)知
$ \begin{align} H_1(x, t, \lambda) = \Phi_2(x, t, \lambda){\rm e}^{({\rm i}\lambda x+4{\rm i}\lambda^3t)\hat\sigma} S_1(\lambda), \quad \lambda\in D_1. \end{align} $ | (39) |
再结合式(31)中给出的
$ \begin{align} {[H_1]}_{1} = {[\Phi_2]}_{1}s_{11}+{[\Phi_2]}_{2}s_{21}{\rm e}^{\theta_{21}}+{[\Phi_2]}_{3}s_{31}{\rm e}^{\theta_{31}}, \end{align} $ | (40) |
$ \begin{align} {[H_1]}_{2} = {[\Phi_2]}_{1}s_{12}{\rm e}^{\theta_{12}}+{[\Phi_2]}_{2}s_{22}+{[\Phi_2]}_{3}s_{32}{\rm e}^{\theta_{32}}, \end{align} $ | (41) |
$ \begin{align} {[H_1]}_{3} = {[\Phi_2]}_{3}\frac{1}{m_{33}(s)}. \end{align} $ | (42) |
从式(40)和式(41)中解出
$ \begin{align} {[H_1]}_{3} = \frac{s_{12}}{m_{33}(s)m_{23}(s)}{[H_1]}_{1}{\rm e}^{\theta_{31}} -\frac{s_{11}}{m_{33}(s)m_{23}(s)}{[H_1]}_{2}{\rm e}^{\theta_{31}} -\frac{1}{m_{23}(s)}{[\Phi_2]}_{2}{\rm e}^{\theta_{23}}. \end{align} $ | (43) |
在
由方程(27)定义的谱函数
$ \begin{align} \Phi_3(x, t, \lambda) = \Phi_1(x, t, \lambda){\rm e}^{({\rm i}\lambda x+4{\rm i}\lambda^3t)\hat\sigma} S^{-1}(\lambda)s(\lambda), \quad \lambda\in(C_{+}, C_{+}, C_{-}). \end{align} $ | (44) |
因为
$ \begin{align} S^{-1}(\lambda)s(\lambda) = {\rm e}^{-4{\rm i}\lambda^3T\hat\sigma}c(T, \lambda), \quad \lambda\in(C_{+}, C_{+}, C_{-}), \end{align} $ | (45) |
其中
在上节中定义的分片连续函数
定理 2.1 假设
$ \begin{align} u(x, t) = 2{\rm i}\lim\limits_{\lambda\rightarrow\infty}[\lambda H(x, t, \lambda)]_{13}, \end{align} $ | (46) |
其中
$ \begin{align} H_n(x, t, \lambda) = H_m\, (x, t, \lambda)J_{m, n}(x, t, \lambda), \quad \lambda\in \bar D_n\cup \bar D_m, \quad n\neq m. \end{align} $ | (47) |
则
证 明 这里只需要证明式(46).该式可以由特征函数的大
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