KP可积系列^[1-2]是一类重要的经典可积系列,在数学和物理领域有很广泛的应用.约束KP(cKP)可积系列^[3-9]可以看做是KP可积系列的约化.另一个重要的可积系列是mKP可积系列^[10-14],它通过Miura变换和反- Miura变换^{[12, 15-16]}与KP可积系列相联系.存在许多版本的mKP可积系列,但在本文只考虑Kupershmidt-Kiso版本^[13-14]的mKP可积系列.近年来,有很多关于Kupershmidt-Kiso版本的mKP可积系列的结果,例如:规范变换^[17], tau函数^[18],平方本征对称^[19-20]和附加对称^[18]等.

平方本征(SE)对称^[19-22]又叫作"ghost"对称^[22],通过本征函数和共轭本征函数来定义,在可积系列中是一种重要的对称. SE对称有两个很重要的应用: 1) SE对称可以看作是附加对称^[22-24]的生成算子,附加对称是依赖于时间和空间变量的对称; 2) SE对称可以用来定义对称约束^[20]和扩展可积系列^[25-26].近期,已经研究了BKP可积系列的SE对称^[27], Toda晶格可积系列以及B和C类型的子可积系列^[28-29].本文研究KP和mKP可积系列及其约束的平方本征对称, Miura变换和反- Miura变换.

Miura变换^{[12, 15-16, 30-31]}在经典可积系列中扮演着重要的角色,它展示了不同可积性质^{[15, 30, 32-33]}之间的联系.本文中, Miura变换是指从mKP可积系列到KP可积系列,反- Miura变换是从KP可积系列到mKP可积系列.存在两种类型的反- Miura变换,分别由本征函数和共轭本征函数生成.相应的也有两种类型的Miura变换.研究Miura变换和反- Miura变换下SE对称的变化是很有趣的.尽管在文献[19-20]中已经研究了由本征函数生成的反- Miura变换,但它缺乏对其它Miura变换和反- Miura变换的研究.本文将考虑这个问题.

为了叙述方便,首先介绍一些符号.考虑拟微分算子^[2]

$ g = \bigg\{\sum\limits_{ i \ll\infty}u_i \partial^{i}\bigg\}, $

这里$ \partial = \partial_{x} $且系数为$ u_i = u_i(t_1 \equiv x, t_2, \cdots) $.任给函数$ f $, $ \partial^{i} $与$ f $的乘积满足Leibnitz规则^[2]

(2.1) $ \begin{equation} \partial^{i} f = \sum\limits_{j\geq 0}\bigg(\begin{array}{cc}i\\ {j}\end{array}\bigg)f^{(j)} \partial^{i-j} , \quad i\in {\Bbb Z}, \end{equation} $

这里$ f^{(j)} = f_{\underbrace{x\cdots x}_{j}} = \frac{ \partial^{j}f}{{ \partial x}^{j}} $.假设$ A, B\in g $且$ f $是一个函数.共轭$ * $定义如下: $ (AB)^* = B^*A^* $, $ \partial^* = - \partial $, $ f^* = f $. $ A f $表示$ A $乘$ f $, $ A(f) $表示$ A $作用于$ f $.若$ A = \sum\limits_{i}a_{i} \partial^{i}\in g $,记$ res_ \partial A = a_{-1} $, $ A_{\geq k} = \sum\limits_{i\geq k}a_{i} \partial^{i} $且$ A_{< k} = \sum\limits_{i< k}a_{i} \partial^{i} $.

通过Lax方程^{[2, 12]}介绍KP和mKP可积系列

(2.2) $ \begin{equation} L_{t_{m}} = [(L^{m})_{\geq k}, L], \quad k = 0, 1. \end{equation} $

Lax算子$ L $为

(2.3) $ \begin{eqnarray} L = \left\{ \begin{array}{ll} \partial+u_1 \partial^{-1}+u_2 \partial^{-2}+u_3 \partial^{-3}+\cdots, & \quad k = 0\quad (KP) , \\ \partial+v_0+v_1 \partial^{-1}+v_2 \partial^{-2}+v_3 \partial^{-3}+\cdots, &\quad k = 1\quad (mKP). \end{array} \right. \end{eqnarray} $

KP和mKP可积系列的Lax算子$ L $可以分别通过dressing算子$ S $和$ Z $^{[2, 17]}给出

(2.4) $ \begin{eqnarray} L = \left\{ \begin{array}{ll} S \partial S^{-1}, \quad S = 1+s_1 \partial^{-1}+s_2 \partial^{-2}+s_3 \partial^{-3}+\cdots, &\quad k = 0, \\ Z \partial Z^{-1}, \quad Z = z_0+z_1 \partial^{-1}+z_2 \partial^{-2}+z_3 \partial^{-3}+\cdots, &\quad k = 1. \end{array} \right. \end{eqnarray} $

系数$ s_i $和$ z_j $都是具有无穷多变量$ t = (t_1 = x, t_2, t_3, \cdots) $的函数.

定义本征函数$ \phi $和共轭本征函数$ \psi $如下^[19-20]

(2.5) $ \begin{eqnarray} \phi_{t_m} = (L^{m})_{\geq k}(\phi), \quad \psi_{t_m} = -( \partial^{-k}(L^{m})_{\geq k}^{*} \partial^{k})(\psi), \quad k = 0, 1. \end{eqnarray} $

约束KP和约束mKP可积系列通过在Lax算子上施加以下约束来定义

(2.6) $ \begin{equation} (L^{n})_{<k} = \sum\limits_{j = 1}^{l}q_j \partial^{-1}r_j \partial^{k}, \quad k = 0, 1. \end{equation} $

平方特征函数势$ \Omega $和$ \hat{\Omega} $定义如下^[19-20]

(2.7) $ \begin{eqnarray} &&\Omega(\psi^{(k)}, \phi)_x = \psi^{(k)}\phi, \quad\Omega(\psi^{(k)}, \phi)_{t_n} = res_{ \partial}( \partial^{-1}\psi^{(k)}(L^{n})_{\geq k}\phi \partial^{-1}), \end{eqnarray} $

(2.8) $ \begin{eqnarray} &&\hat{\Omega}(\psi, \phi^{(k)})_x = \psi\phi^{(k)}, \quad\hat{\Omega}(\psi, \phi^{(k)})_{t_n} = res_{ \partial}( \partial^{-1}\psi(L^{n})_{\geq k}\phi^{(k)} \partial^{-1}), \end{eqnarray} $

这里$ \psi^{(k)} $为$ \psi $, $ \psi_x $当$ k = 0, 1 $时($ \phi^{(k)} $类似).它们关系为

(2.9) $ \begin{eqnarray} \hat{\Omega}(\psi, \phi^{(k)}) = \left\{ \begin{array}{ll} \Omega(\psi, \phi), &\quad k = 0, \\ \Omega(\psi_x, \phi)-\psi\phi, &\quad k = 1. \end{array} \right. \end{eqnarray} $

引理2.1  对于任意的拟微分算子$ A\in g $和任意函数$ f $,有

(2.10) $ \begin{eqnarray} (f^{-1}Af)_{\geq1}& = &f^{-1}A_{\geq0}f-f^{-1}A_{\geq0}(f), \end{eqnarray} $

(2.11) $ \begin{eqnarray} ( \partial^{-1}fAf^{-1} \partial)_{\geq1}& = & \partial^{-1}fA_{\geq0}f^{-1} \partial- \partial^{-1}A_{\geq0}^{*}(f)f^{-1} \partial. \end{eqnarray} $

首先介绍Miura变换和反- Miura变换的定义.

命题3.1^{[12, 15-16]}  令$ L $, $ \phi $, $ \bar{\phi} $分别是KP可积系列的Lax算子,本征函数和共轭本征函数.从KP可积系列到mKP可积系列有两种类型的反- Miura变换

(3.1) $ \begin{eqnarray} L\rightarrow \tilde{L} = \left\{ \begin{array}{ll} T_mLT_m^{-1}, & \quad T_m = \phi^{-1}, \\ T_nLT_n^{-1}, &\quad T_{n} = \partial^{-1}\bar{\phi}. \end{array} \right. \end{eqnarray} $

这里$ \tilde{L} $是mKP可积系列的Lax算子.

命题3.2^[15-16]   令$ L $是mKP可积系列的Lax算子,且$ z_0 $是dressing算子$ Z $中$ \partial^{0} $前面的系数.从mKP可积系列到KP可积系列的Miura变换有以下形式

(3.2) $ \begin{eqnarray} L\rightarrow \tilde{L} = \left\{ \begin{array}{ll} T_{\mu}LT_{\mu}^{-1}, &\quad T_\mu = z_0^{-1}, \\ T_\nu LT_\nu^{-1}, &\quad T_\nu = z_0^{-1} \partial, \end{array} \right. \end{eqnarray} $

这里$ \tilde{L} $是KP可积系列的Lax算子.

现在考虑KP和mKP可积系列的SE对称和反- Miura变换的关系.对于反- Miura变换的$ m $情形,在文献[19]中已经研究过了,由下面的命题给出.

命题3.3  ^[19]假设$ L $是KP可积系列的Lax算子,满足Lax方程$ L_{t_n} = [(L^{n})_{\geq0}, L] $和SE对称

(3.3) $ \begin{equation} L_ \alpha = \bigg[\sum\limits_{i = 1}^{l}\phi_i \partial^{-1}\psi_i, L\bigg], \end{equation} $

这里$ \phi_i $和$ \psi_i $分别是本征函数和共轭本征函数.令$ \phi $是本征函数并且满足$ \phi_{ \alpha} = \sum\limits_{i = 1}^{l}\phi_i\Omega(\psi_i, \phi) $.则$ \tilde{L} = \phi^{-1}L\phi $满足mKP可积系列的SE对称流

(3.4) $ \begin{equation} \tilde{L}_ \alpha = \bigg[\sum\limits_{i = 1}^{l}\tilde{\phi}_i \partial^{-1}\tilde{\psi}_i \partial, \tilde{L}\bigg], \end{equation} $

这里$ \tilde{\phi}_i = \phi^{-1}\phi_i $和$ \tilde{\psi}_i = -\Omega(\psi_i, \phi) $分别是本征函数和共轭本征函数.

现在讨论反- Miura变换的$ n $情形.

命题3.4   KP可积系列中Lax算子$ L $的SE对称流定义为$ L_ \alpha = [\sum\limits_{i = 1}^{l}\phi_i \partial^{-1}\psi_i, L] $,这里$ \phi_i $和$ \psi_i $分别是本征函数和共轭本征函数.则$ \tilde{L} = \partial^{-1}\bar{\phi}L\bar{\phi}^{-1} \partial $ ($ \bar{\phi} $为共轭本征函数)满足mKP可积系列的SE对称流

(3.5) $ \begin{eqnarray} \tilde{L}_ \alpha = \bigg[\sum\limits_{i = 1}^{l}\tilde{\phi}_i \partial^{-1}\tilde{\psi}_i \partial, \tilde{L}\bigg], \end{eqnarray} $

这里$ \tilde{\phi}_i = \partial^{-1}(\bar{\phi}\phi_i) $和$ \tilde{\psi}_i = \bar{\phi}^{-1}\psi_i $分别是$ \tilde{L} $的本征函数和共轭本征函数.

证  首先,需要证明$ \tilde{\phi}_{i} $和$ \tilde{\psi}_{i} $分别是mKP可积系列的本征函数和共轭本征函数.根据(2.11)式,有

(3.6) $ \begin{eqnarray} (\tilde{L}^{m})_{\geq1}(\tilde{\phi}_i) & = &\Big( \partial^{-1}\bar{\phi}(L^{m})_{\geq0}\bar{\phi}^{-1} \partial- \partial^{-1}(L^{m})_{\geq0}^{*}(\bar{\phi})\bar{\phi}^{-1} \partial\Big)\big( \partial^{-1}(\bar{\phi}\phi_i)\big)\\ & = & \partial^{-1}(\bar{\phi}\phi_{it_m})+ \partial^{-1}(\bar{\phi}_{t_m}\phi_i) = \tilde{\phi}_{it_m} \end{eqnarray} $

和

(3.7) $ \begin{eqnarray} \partial^{-1}(\tilde{L}^{m})_{\geq1}^{*} \partial(\tilde{\psi}_i) & = &(\bar{\phi}^{-1}(L^{m})_{\geq0}^{*}\bar{\phi}-(L^{m})_{\geq0}^{*}(\bar{\phi})\bar{\phi}^{-1})(\bar{\phi}^{-1}\psi_i)\\ & = &-\bar{\phi}^{-1}\psi_{it_m}-(\bar{\phi}^{-1})_{t_m}\psi_i = -\tilde{\psi}_{it_m}. \end{eqnarray} $

现在讨论$ \tilde{L} $的SE对称

(3.8) $ \begin{eqnarray} \tilde{L}_ \alpha& = & \partial^{-1}\bar{\phi}_{ \alpha}L\bar{\phi}^{-1} \partial+ \partial^{-1}\bar{\phi}L_{ \alpha}\bar{\phi}^{-1} \partial- \partial^{-1}\bar{\phi}L\bar{\phi}^{-1}\bar{\phi}_{ \alpha}\bar{\phi}^{-1} \partial\\ & = &\bigg[ \partial^{-1}\bar{\phi}_{ \alpha}\bar{\phi}^{-1} \partial+\sum\limits_{i = 1}^{l} \partial^{-1}\bar{\phi}\phi_i \partial^{-1}\psi_i \bar{\phi}^{-1} \partial, \tilde{L}\bigg] = \bigg[\sum\limits_{i = 1}^{l}\tilde{\phi}_i \partial^{-1}\tilde{\psi}_i \partial, \tilde{L}\bigg]. \end{eqnarray} $

证毕.

接下来考虑从mKP可积系列到KP可积系列在Miura变换下的SE对称的变化.

命题3.5  令$ L $是mKP可积系列的Lax算子,且满足Lax方程$ L_{t_n} = [(L^{n})_{\geq1}, L] $和SE对称$ L_ \alpha = [\sum\limits_{i = 1}^{l}\phi_i \partial^{-1}\psi_i \partial, L] $. Miura变换$ T = T_{\mu} $或$ T = T_{\nu} $ (见命题3.2), $ \tilde{L} = TLT^{-1} $满足KP可积系列的SE对称

(3.9) $ \begin{eqnarray} \tilde{L}_ \alpha = \bigg[\sum\limits_{i = 1}^{l}\tilde{\phi}_i \partial^{-1}\tilde{\psi}_i, \tilde{L}\bigg], \end{eqnarray} $

这里$ \tilde{\phi}_i = T(\phi_i) $和$ \tilde{\psi}_i = -( \partial T^{-1})^{*}(\psi_i) $分别是KP可积系列的本征函数和共轭本征函数.

证  只需证$ \mu $情形, $ \nu $情形的证明与之类似.首先需要证$ \tilde{\phi}_{i} $和$ \tilde{\psi}_{i} $分别是KP可积系列的本征函数和共轭本征函数.考虑$ (z_0^{-1})_{t_m} = (\tilde{L}^{m})_{\geq0}(z_0^{-1}) $,可得

(3.10) $ \begin{eqnarray} \tilde{\phi}_{it_m}& = &(z_0^{-1})_{t_m}\phi_i+z_0^{-1}\phi_{it_m} = (\tilde{L}^{m})_{\geq0}(z_0^{-1})\phi_i+z_0^{-1}(L^{m})_{\geq1}(\phi_i)\\ & = &(\tilde{L}^{m})_{\geq0}(z_0^{-1})\phi_i+(\tilde{L}^{m})_{\geq0}(z_0^{-1}\phi_i)-(\tilde{L}^{m})_{\geq0}(z_0^{-1})\phi_i\\ & = &(\tilde{L}^{m})_{\geq0}(z_0^{-1})\phi_i+z_0^{-1}(\tilde{L}^{m})_{\geq0}(\phi_i) = (\tilde{L}^{m})_{\geq0}(\tilde{\phi}_i). \end{eqnarray} $

这里用到(2.11)式.类似的有

(3.11) $ \begin{eqnarray} \tilde{\psi}_{it_m}& = &(-z_{0}(\psi_i)_{x})_{t_m} = -z_{0t_m}(\psi_i)_{x}-z_{0} \partial\big(- \partial^{-1}(L^{m})_{\geq1}^{*} \partial(\psi_i) \big)\\ & = &z_0^{2}(\tilde{L}^{m})_{\geq0}(z_0^{-1})(\psi_{ix})+(\tilde{L}^{m}_{\geq0})^{*}(z_0)\psi_{ix}+z_0(\tilde{L}^{m}_{\geq0})^{*}(\psi_{ix})- z_0^{2}(\tilde{L}^{m})_{\geq0}(z_0^{-1})(\psi_{ix})\\ & = &-(\tilde{L}^{m})_{\geq0}^{*}(\tilde{\psi}_i). \end{eqnarray} $

最后, $ \tilde{L} $的SE对称可以通过以下方式计算

(3.12) $ \begin{eqnarray} \tilde{L}_{ \alpha}& = &-z_0^{-1}z_{0 \alpha}z_0^{-1}Lz_0+z_0^{-1}L_{ \alpha}z_0+z_0^{-1}Lz_0 z_0^{-1}z_{0 \alpha}\\ & = &\bigg[-z_0^{-1}z_{0 \alpha}+\sum\limits_{i = 1}^{l}z_0^{-1}\phi_i \partial^{-1}\psi_i \partial z_0, \tilde{L} \bigg ] = \bigg[\sum\limits_{i = 1}^{l}\tilde{\phi}_i \partial^{-1}\tilde{\psi}_i, \tilde{L}\bigg]. \end{eqnarray} $

证毕.

首先,在下个命题中回顾约束KP和约束mKP可积系列与SE对称的关系.

命题4.1^[20]   在SE流下约束(2.6)保持不变

(4.1) $ \begin{eqnarray} L_{ \alpha} = \bigg[\sum\limits_{i = 1}^{l}\phi_i \partial^{-1}\psi_i \partial^{k}, L\bigg], \quad q_{j \alpha} = \sum\limits_{i = 1}^{l}\phi_i\Omega(\psi_i, q_j^{(k)}), \quad r_{j \alpha} = (-1)^{k}\sum\limits_{i = 1}^{l}\Omega(r_j^{(k)}, \phi_i)\psi_i, \end{eqnarray} $

如果$ \phi_i $和$ \psi_i $ ($ i = 1, 2, \cdots, l $)满足

(4.2) $ \begin{equation} (L^{n})_{\geq k}(\phi_i)+\sum\limits_{j = 1}^{l}q_j\Omega(r_j, \phi_i^{(k)}) = \lambda_{i}\phi_i, \end{equation} $

(4.3) $ \begin{equation} \partial^{-k}(L^{n})_{\geq k}^{*} \partial^{k}(\psi_i)-(-1)^{k}\Omega(\psi_i^{(k)}, q_j)r_j = \lambda_i\psi_i, \end{equation} $

对于任意的光谱参数$ \lambda_i \in C $.

文献[15, 20]已研究了约束KP和约束mKP可积系列的反- Miura变换.

命题4.2^{[15, 20]}   Case I  令$ L $满足约束KP可积系列$ (L^{n})_{<0} = \sum\limits_{j = 1}^{l}q_j \partial^{-1}r_j $的Lax方程$ L_{t_m} = [(L^{m})_{\geq0}, L] $.对于反- Miura变换$ T = T_{m} $或$ T = T_{n} $ (见命题3.1), $ \tilde{L} = TLT^{-1} $满足约束mKP可积系列$ (\tilde{L}^{n})_{<1} = \sum\limits_{j = 1}^{l+1}\tilde{q}_j \partial^{-1}\tilde{r}_j \partial $有

(4.4) $ \begin{eqnarray} \left\{ \begin{array}{ll} \tilde{q}_{j} = \phi^{-1}q_j, \quad \tilde{r}_{j} = -\Omega(r_j, \phi), \quad j = 1, 2, \cdots, l, \\ { } \tilde{q}_{l+1} = \phi^{-1}\Big((L^{n})_{\geq 0}(\phi)+\sum\limits_{j = 1}^{l}q_j\Omega(r_j, \phi) \Big), \quad \tilde{r}_{l+1} = 1, \end{array} \right. \quad (T = T_m), \end{eqnarray} $

或

(4.5) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } \tilde{q}_{j} = \Omega(\bar{\phi}, q_j), \quad \tilde{r}_{j} = r_j\bar{\phi}^{-1}, \quad j = 1, 2, \cdots, l, \\ { } \tilde{q}_{l+1} = 1, \quad \tilde{r}_{l+1} = (L^{n})_{\geq0}^{*}(\bar{\phi})\bar{\phi}^{-1}-\sum\limits_{j = 1}^{l}\Omega(\bar{\phi}, q_j)r_j\bar{\phi}^{-1}, \end{array} \right. \quad (T = T_n). \end{eqnarray} $

Case II考虑约束mKP可积系列的Lax算子$ (L^{n})_{<1} = \sum\limits_{j = 1}^{l}q_j \partial^{-1}r_j \partial $,对于Miura变换$ T = T_{\mu} $或$ T = T_{\nu} $ (见命题3.2), $ \tilde{L} = TLT^{-1} $满足约束KP可积系列$ (\tilde{L}^{n})_{<0} = \sum\limits_{j = 1}^{l}\tilde{q}_j \partial^{-1}\tilde{r}_j $.有

(4.6) $ \begin{eqnarray} \tilde{q}_{j} = T(q_j), \quad \tilde{r}_{j} = -( \partial T^{-1})^{*}(r_{j}). \end{eqnarray} $

接下来讨论约束KP和约束mKP可积系列的SE对称与反- Backlund变换的关系.

命题4.3  令$ L $满足约束KP可积系列$ (L^{n})_{<0} = \sum\limits_{j = 1}^{l}q_j \partial^{-1}r_j $和$ L_{t_m} = [(L^{m})_{\geq0}, L] $,则对于反- Miura变换$ T = T_{m} $或$ T = T_{n} $ (见命题3.1), $ \tilde{L} = TLT^{-1} $满足SE对称

(4.7) $ \begin{eqnarray} \tilde{L}_ \alpha = \bigg[\sum\limits_{i = 1}^{l}\tilde{\phi}_i \partial^{-1}\tilde{\psi}_i \partial, \tilde{L}\bigg], \end{eqnarray} $

这里$ \tilde{\phi}_i = T(\phi_i) $, $ \tilde{\psi}_i = (T^{-1} \partial^{-1})^{*}(\psi_i) $.另外, $ \tilde{q}_{j} $和$ \tilde{r}_{j} $ ($ j = 1, 2, \cdots, l+1 $) (见命题4.2 Case I)满足

(4.8) $ \begin{eqnarray} \tilde{q}_{j \alpha} = \sum\limits_{i = 1}^{l}\tilde{\phi}_i\Omega(\tilde{\psi}_i, \tilde{q}_{jx}), \quad \tilde{r}_{j \alpha} = -\sum\limits_{i = 1}^{l}\Omega(\tilde{r}_{jx}, \tilde{\phi}_i)\tilde{\psi}_i \end{eqnarray} $

和

(4.9) $ \begin{equation} (\tilde{L}^{n})_{\geq 1}(\tilde{\phi}_i)+\sum\limits_{j = 1}^{l+1}\tilde{q}_j\Omega(\tilde{r}_j, \tilde{\phi}_{ix}) = \lambda_{i}\tilde{\phi}_i, \end{equation} $

(4.10) $ \begin{equation} \partial^{-1}(\tilde{L}^{n})_{\geq 1}^{*} \partial(\tilde{\psi}_i)+\sum\limits_{j = 1}^{l+1}\Omega(\tilde{\psi}_{ix}, \tilde{q}_j)\tilde{r}_j = \lambda_i\tilde{\psi}_i. \end{equation} $

证  只需证$ m $情形.首先

(4.11) $ \begin{eqnarray} \sum\limits_{i = 1}^{l}\phi^{-1}\phi_i \partial^{-1}\big( \partial^{-1}(\psi_i\phi)(\phi^{-1}q_j)_x \big) = \sum\limits_{i = 1}^{l}\phi^{-1}\phi_i\Big( \partial^{-1}(\psi_i\phi)\phi^{-1}q_j- \partial^{-1}(q_j\psi_i) \Big), \end{eqnarray} $

因此得到

(4.12) $ \begin{eqnarray} \tilde{q}_{j \alpha}& = &-\phi^{-1}\phi_{ \alpha}\phi^{-1}q_j+\phi^{-1}q_{j \alpha}\\ & = &\sum\limits_{i = 1}^{l}\phi^{-1}\phi_i \partial^{-1}\big( \partial^{-1}(\psi_i\phi)(\phi^{-1}q_j)_x \big) = \sum\limits_{i = 1}^{l}\tilde{\phi}_i\Omega(\tilde{\psi}_i, \tilde{q}_{jx}) \end{eqnarray} $

和

(4.13) $ \begin{eqnarray} \tilde{r}_{j \alpha}& = &- \partial^{-1}(r_{j \alpha}\phi)- \partial^{-1}(r_j\phi_{ \alpha})\\ & = &-\sum\limits_{i = 1}^{l} \partial^{-1}(r_j\phi_i) \partial^{-1}(\psi_i\phi) = -\sum\limits_{i = 1}^{l}\Omega(\tilde{r}_{jx}, \tilde{\phi}_i)\tilde{\psi}_i. \end{eqnarray} $

现证明(4.9)式.根据(2.10)式,可得

(4.14) $ \begin{eqnarray} (\tilde{L}^{n})_{\geq 1}(\tilde{\phi}_i)& = &(\phi^{-1}L^{n}\phi)_{\geq1}(\phi^{-1}\phi_i) = \Big(\phi^{-1}(L^{n})_{\geq0}\phi-\phi^{-1}(L^{n})_{\geq0}(\phi) \Big)(\phi^{-1}\phi_i)\\ & = &\phi^{-1}(L^{n})_{\geq0}(\phi_i)-\phi^{-1}(L^{n})_{\geq0}(\phi)\phi^{-1}\phi_i \end{eqnarray} $

和

(4.15) $ \begin{eqnarray} \sum\limits_{j = 1}^{l+1}\tilde{q}_j\Omega(\tilde{r}_j, \tilde{\phi}_{ix}) & = &-\sum\limits_{j = 1}^{l}\phi^{-1}q_j \partial^{-1}(r_j\phi)\phi^{-1}\phi_i+\sum\limits_{j = 1}^{l}\phi^{-1}q_j \partial^{-1}(r_j\phi\phi^{-1}\phi_i)\\ &&+\phi^{-1}(L^{n})_{\geq0}(\phi)\phi^{-1}\phi_i+\sum\limits_{j = 1}^{l}\phi^{-1}q_j \partial^{-1}(r_j\phi)\phi^{-1}\phi_i, \end{eqnarray} $

因此根据(4.14), (4.15)和(4.2)式($ k = 0 $),可得

(4.16) $ \begin{eqnarray} (\tilde{L}^{n})_{\geq 1}(\tilde{\phi}_i)+\sum\limits_{j = 1}^{l+1}\tilde{q}_j\Omega(\tilde{r}_j, \tilde{\phi}_{ix}) = \phi^{-1}(L^{n})_{\geq0}(\phi_i)+\sum\limits_{j = 1}^{l}\phi^{-1}q_j \partial^{-1}(r_j\phi_i) = \lambda_i\tilde{\phi}_i. \end{eqnarray} $

接下来关键是证明(4.10)式.根据(2.10)式,有

(4.17) $ \begin{eqnarray} \partial^{-1}(\tilde{L}^{n})_{\geq 1}^{*} \partial(\tilde{\psi}_i) & = & \partial^{-1}(\phi^{-1}L^{n}\phi)_{\geq 1}^{*} \partial\big(- \partial^{-1}(\psi_i\phi)\big)\\ & = &- \partial^{-1}\big(\phi(L^{n})_{\geq0}^{*}(\psi_i) \big)+ \partial^{-1}\big((L^{n})_{\geq0}(\phi)\psi_i \big) \end{eqnarray} $

和

(4.18) $ \begin{eqnarray} \sum\limits_{j = 1}^{l+1}\Omega(\tilde{\psi}_{ix}, \tilde{q}_j)\tilde{r}_j & = &\sum\limits_{j = 1}^{l} \partial^{-1}(\psi_iq_j) \partial^{-1}(r_j\phi)- \partial^{-1}\big(\psi_i(L^{n})_{\geq0}(\phi) \big)- \partial^{-1}\Big(\psi_i\sum\limits_{j = i}^{l}q_j \partial^{-1}(r_j\phi)\Big)\\ & = &\sum\limits_{j = 1}^{l} \partial^{-1}(\psi_iq_j) \partial^{-1}(r_j\phi)- \partial^{-1}\big(\psi_i(L^{n})_{\geq0}(\phi) \big)\\ &&-\sum\limits_{j = 1}^{l} \partial^{-1}(\psi_iq_j) \partial^{-1}(r_j\phi)+\sum\limits_{j = i}^{l} \partial^{-1}\Big( \partial^{-1}(\psi_iq_j)r_j\phi \Big). \end{eqnarray} $

利用(4.3)式($ k = 0 $),有

(4.19) $ \begin{equation} \partial^{-1}(\tilde{L}^{n})_{\geq 1}^{*} \partial(\tilde{\psi}_i)+\sum\limits_{j = 1}^{l+1}\Omega(\tilde{\psi}_{ix}, \tilde{q}_j)\tilde{r}_j = - \partial^{-1}\Big((L^{n})_{\geq0}^{*}(\psi_i)-\sum\limits_{j = 1}^{l} \partial^{-1}(\psi_iq_j)r_j \Big)\phi = \lambda_i\tilde{\psi}_i. \end{equation} $

证毕.

最后,考虑从约束mKP到约束KP可积系列的SE对称与Miura变换的关系.

命题4.4  约束mKP可积系列的Lax算子为$ (L^{n})_{<1} = \sum\limits_{j = 1}^{l}q_j \partial^{-1}r_j \partial $,考虑Miura变换$ T = T_{\mu} $或$ T = T_{\nu} $ (见命题3.2), $ \tilde{L} = TLT^{-1} $满足约束KP可积系列的SE对称

(4.20) $ \begin{eqnarray} \tilde{L}_ \alpha = \bigg[\sum\limits_{i = 1}^{l}\tilde{\phi}_i \partial^{-1}\tilde{\psi}_i, \tilde{L}\bigg], \end{eqnarray} $

这里$ \tilde{\phi}_i = T(\phi_i) $和$ \tilde{\psi}_i = -( \partial T^{-1})^{*}(\psi_i) $.且$ \tilde{q}_{j} $和$ \tilde{r}_{j} $ (见命题4.2 Case II)满足

(4.21) $ \begin{eqnarray} \tilde{q}_{j \alpha} = \sum\limits_{i = 1}^{l}\tilde{\phi}_i\Omega(\tilde{\psi}_i, \tilde{q}_{j}), \quad \tilde{r}_{j \alpha} = \sum\limits_{i = 1}^{l}\Omega(\tilde{r}_{j}, \tilde{\phi}_i)\tilde{\psi}_i \end{eqnarray} $

和

(4.22) $ \begin{equation} (\tilde{L}^{n})_{\geq 0}(\tilde{\phi}_i)+\sum\limits_{j = 1}^{l}\tilde{q}_j\Omega(\tilde{r}_j, \tilde{\phi}_{i}) = \lambda_{i}\tilde{\phi}_i, \end{equation} $

(4.23) $ \begin{equation} (\tilde{L}^{n})_{\geq 0}^{*}(\tilde{\psi}_i)-\sum\limits_{j = 1}^{l}\Omega(\tilde{\psi}_{i}, \tilde{q}_j)\tilde{r}_j = \lambda_i\tilde{\psi}_i. \end{equation} $

证  只需证Miura变换的$ \mu $情形.首先,根据$ z_{0 \alpha} = \sum\limits_{i = 1}^{l}\phi_i\psi_iz_0 $,有

(4.24) $ \begin{eqnarray} \tilde{q}_{j \alpha}& = &-z_0^{-1}\sum\limits_{i = 1}^{l}\phi_i\psi_iz_0z_0^{-1}q_j+z_0^{-1}\sum\limits_{i = 1}^{l}\phi_i \partial^{-1}(\psi_iq_{jx})\\ & = &-\sum\limits_{i = 1}^{l}z_0^{-1}\phi_i \partial^{-1}(\psi_{ix}z_0z_0^{-1}q_j) = \sum\limits_{i = 1}^{l}\tilde{\phi}_i\Omega(\tilde{\psi}_i, \tilde{q}_{j}) \end{eqnarray} $

和

(4.25) $ \begin{eqnarray} \tilde{r}_{j \alpha}& = &\sum\limits_{i = 1}^{l}\Big( \partial^{-1}(r_{jx}\phi_i)\psi_i \Big)_xz_0-\sum\limits_{i = 1}^{l}r_{jx}\phi_i\psi_iz_0\\ & = &\sum\limits_{i = 1}^{l} \partial^{-1}(r_{jx}\phi_i)\psi_{ix}z_0 = \sum\limits_{i = 1}^{l}\Omega(\tilde{r}_{j}, \tilde{\phi}_i)\tilde{\psi}_i. \end{eqnarray} $

至于(4.22)式,因为

(4.26) $ \begin{eqnarray} (\tilde{L}^{n})_{\geq0}& = &(z_0^{-1}L^{n}z_0)_{\geq0} = \Big(z_0^{-1}(L^{n})_{\geq1}z_0 +z_0^{-1}(L^{n})_{<1}z_0 \Big)_{\geq0}\\ & = &z_0^{-1}(L^{n})_{\geq1}z_0+\sum\limits_{j = 1}^{l}q_jr_j, \end{eqnarray} $

有

(4.27) $ \begin{equation} (\tilde{L}^{n})_{\geq0}(\tilde{\phi}_i) = \big(z_0^{-1}(L^{n})_{\geq1}z_0+\sum\limits_{j = 1}^{l}q_jr_j \big)(z_0^{-1}\phi_i) = z_0^{-1}(L^{n})_{\geq1}(\phi_i)+\sum\limits_{j = 1}^{l}q_jr_jz_0^{-1}\phi_i. \end{equation} $

并且

(4.28) $ \begin{equation} \sum\limits_{j = 1}^{l}\tilde{q}_j\Omega(\tilde{r}_j, \tilde{\phi}_{i}) = -\sum\limits_{j = 1}^{l}z_0^{-1}q_jr_j\phi_i+\sum\limits_{j = 1}^{l}z_0^{-1}q_j \partial^{-1}(r_j\phi_{ix}). \end{equation} $

因此

(4.29) $ \begin{equation} (\tilde{L}^{n})_{\geq0}(\tilde{\phi}_i)+\sum\limits_{j = 1}^{l}\tilde{q}_j\Omega(\tilde{r}_j, \tilde{\phi}_{i}) = z_0^{-1}(L^{n})_{\geq1}(\phi_i)+\sum\limits_{j = 1}^{l}\tilde{q}_j\Omega(\tilde{r}_j, \tilde{\phi}_{i}) = z_0^{-1} \lambda_i\phi_i = \lambda_i\tilde{\phi}_i. \end{equation} $

最后考虑(4.23)式.通过直接计算

(4.30) $ \begin{eqnarray} &&(\tilde{L}^{n})_{\geq 0}^{*}(\tilde{\psi}_i)-\sum\limits_{j = 1}^{l}\Omega(\tilde{\psi}_{i}, \tilde{q}_j)\tilde{r}_j \\ & = &\Big( z_0(L^{n})_{\geq1}^{*}z_{0}^{-1}+\sum\limits_{j = 1}^{l}q_jr_j \Big)(z_0\psi_{ix}) +\sum\limits_{j = 1}^{l} \partial^{-1}(z_0\psi_{ix}z_0^{-1}q_j)r_{jx}z_0\\ & = &z_0(L^{n})_{\geq1}^{*}(\psi_{ix})+\sum\limits_{j = 1}^{l}q_jr_jz_0\psi_{ix}+\sum\limits_{j = 1}^{l} \partial^{-1}(\psi_{ix}q_j)r_{jx}z_0. \end{eqnarray} $

如果用$ z_0 \partial $作用于(4.3)式($ k = 1 $)左右两端,则

(4.31) $ \begin{equation} z_0(L^{n})_{\geq1}^{*}(\psi_{ix})+\sum\limits_{j = 1}^{l} \partial^{-1}(\psi_{ix}q_j)r_{jx}z_0+\sum\limits_{j = 1}^{l}\psi_{ix}q_jr_jz_0 = \lambda_iz_0\psi_{ix}. \end{equation} $

可得

(4.32) $ \begin{equation} (\tilde{L}^{n})_{\geq 0}^{*}(\tilde{\psi}_i)-\sum\limits_{j = 1}^{l}\Omega(\tilde{\psi}_{i}, \tilde{q}_j)\tilde{r}_j = \lambda_i\tilde{\psi}_i. \end{equation} $

证毕.