非整数阶微积分理论的研究可以追溯到Leibniz,Liouville,Grunwald, Letnikov和Riemann. 在过去的几十年中,许多学者指出, 因为分数阶导数可以有效的描述物理中的非线性现象, 所以分数阶微积分在动力学理论^{[1, 2]}, 统计力学^{[3, 4]}以及其他领域^{[5, 6, 7]}中有广泛地应用.

在孤子理论中,寻找孤子方程新的可积系统是一个重要的课题. 屠格式^[8]是获得可积哈密顿系统的 一个有效方法. 应用屠格式, 研究人员已经获得了大量的可积系统及其它们的哈密顿结构. 吴国成和张盛^[9]提出了广义 屠格式, 可以构造分数阶可积孤子族的哈密顿结构. 岳超和夏铁成给出了分数阶二次型恒等式^[10], 并得到分数阶AKNS族的一个可积耦合的哈密顿结构. 王惠等将超迹恒等式推广为分数阶的超级恒等式,本文中 我们基于李超代数, 应用分数阶超级恒等式构造出分数阶超Yang族以及它的分数阶超哈密顿结构.

首先,考虑一组分数阶矩阵等谱问题 \begin{equation} D_{x}^{\alpha}\psi=U\psi=U(u,\lambda)\psi,~~D_{t}^{\beta}\psi=V\psi,(2.1) \end{equation}, 其中 $D_{x}^{\alpha}$是修正导数或者经典分数阶导数意义上的分数阶导数, $\psi$ 是 $n$ 维函数向量. 利用相容条件(2.1),可以得到广义零曲率方程 \begin{equation} D_{t}^{\beta}U-D_{x}^{\alpha}V+[U,V]=0,(2.2) \end{equation} 其中,$[U,V]=UV-VU$,当 $\alpha=\beta=1$, (2.2)式可以化简为经典的零曲率方程.

现在,给定圈李超代数$G$,我们引进超迹,使其满足

(1)~ 对称性 \begin{equation} {\rm str}(ad_{a}ad_{b})={\rm str}(ad_{b}ad_{a}),~~~~~a,b\in G;(2.3) \end{equation}

(2)~ 李积不变性 \begin{equation} {\rm str}(ad_{a}ad_{[b,c]})={\rm str}(ad_{[a,b]}ad_{c}),~~~~~a,b,c\in G,(2.4) \end{equation} $ad_{a}$是$a\in G$在$G$上的伴随表示, 其定义为 \begin{equation} ad_{a}b=[a,b],~~~~~~b\in G,(2.5) \end{equation} 此处 $[\cdot,\cdot]$是$G$的李超括号.

定义泛函 \begin{equation} W=\int[{\rm str}(ad_{V}ad_{U_{\lambda}})+{\rm str}(ad_{\Lambda }ad_{D_{x}^{\alpha}-[U,V]})]{\rm d}x,(2.6) \end{equation} 其中,$U_{\lambda}=\frac{\partial U}{\partial \lambda}$,$U$, $V$满足(2.1)式,$\Lambda\in \tilde{G}$是待定乘子.

从分数导数的意义上,泛函$R$关于$a$的梯度$\nabla_{a}R\in G$定义为 \begin{equation} \int {\rm str}(ad_{\nabla_{a}R}ad_{b}){\rm d}x=\frac{\partial}{\partial \epsilon}R(a+\epsilon b)|_{\epsilon=0},~~~~~a,b\in G.(2.7) \end{equation} 显然,由Killing型的非退化性,可得 \begin{equation} \nabla_{b}\int {\rm str}(ad_{a}ad_{b}){\rm d}x=a,~~~\nabla_{b}\int {\rm str}(ad_{a}ad_{D_{x}^{\alpha}b}){\rm d}x=-D_{x}^{\alpha}a.(2.8) \end{equation} 所以,我们有关于泛函$W$的变分运算的约束条件,如下 \begin{equation} \nabla_{V}W=U_{\lambda}-D_{x}^{\alpha}\Lambda+[U,\Lambda],(2.9) \end{equation} \begin{equation} \nabla_{\Lambda}W=D_{x}^{\alpha}V-[U,V],(2.10) \end{equation} 方程(2.9),(2.10)决定$V$和$\Lambda$,这意味着$V$和 $\Lambda$与$U$都与势函数$u$有关. 而且满足 \begin{equation} \frac{\delta}{\delta u}\int {\rm str}(ad_{V}ad_{U_{\lambda}}){\rm d}x=\frac{\delta W}{\delta u},(2.11) \end{equation} 其中,$\frac{\delta}{\delta u}$是关于$u$的变分导数. 显然, 计算$\frac{\delta W}{\delta u}$仅需考虑独立变量$u$, 而且如果$\nabla_{a}R(a)=0$,则 $$ \Big(\frac{\delta}{\delta u}\Big)R(a(u))=0. $$ 因此由(2.4)式不变性,我们有 \begin{equation} \frac{\delta}{\delta u}\int {\rm str}(ad_{V}ad_{U_{\lambda}}){\rm d}x=\frac{\delta W}{\delta u}={\rm str}(ad_{V}ad_{\frac{\partial u_{\lambda}}{\partial u}})+{\rm str}(ad_{[\Lambda,V]}ad_{\frac{\partial U}{\partial u}}).(2.12) \end{equation} 利用方程(2.9),(2.10)和 Jacobi恒等式,可得 \begin{eqnarray} D_{x}^{\alpha}[\Lambda,V] &=&[D_{x}^{\alpha}\Lambda,V]+[\Lambda,D_{x}^{\alpha}V]=[U_{\lambda}+[U,\Lambda],V]+[\Lambda,[U,V]] \nonumber\\ &=&[U_{\lambda},V]+[V,[\Lambda,U]]+[\Lambda,[U,V]]=[U_{\lambda},V]+[U,[\Lambda,V]],(2.13) \end{eqnarray} 同时,从(2.10)式有, \begin{equation} D_{x}^{\alpha}V=[U_{\lambda},V]+[U,V_{\lambda}].(2.14) \end{equation} 因此,$Z=[\Lambda,V]-V_{\lambda}$满足 \begin{equation} D_{x}^{\alpha}Z=[U,Z].(2.15) \end{equation} 根据唯一性以及${\rm rank}(Z)={\rm rank}(V_{\lambda})={\rm rank}(\frac{V}{\lambda})$, 则存在一个常数$\gamma$满足 \begin{equation} [\Lambda,V]-V_{\lambda}=\frac{\gamma}{\lambda}V, (2.16) \end{equation} 因为$\frac{V}{\lambda}$是$D_{x}^{\alpha}=[U,V]$的根. 最后,(2.12)式可以表示为 \begin{eqnarray} \frac{\delta}{\delta u}\int {\rm str}(ad_{V}ad_{U_{\lambda}}){\rm d}x &=&{\rm str}(ad_{V}ad_{\partial U_{\lambda}/\partial u})+{\rm str}(ad_{V_{\lambda}}ad_{\partial U/\partial u}) +\frac{\gamma}{\lambda}{\rm str}(ad_{V}ad_{\partial U/\partial u})\nonumber\\ &=&\frac{\partial}{\partial \lambda}{\rm str}(ad_{V}ad_{\partial U/\partial u}) +(\lambda^{-\gamma}\frac{\partial}{\partial \lambda}\lambda^{\gamma})({\rm str}(ad_{V}ad_{\partial U/\partial u})) \nonumber\\ &=&\lambda^{-\gamma}\frac{\partial}{\partial \lambda}\lambda^{\gamma}({\rm str}(ad_{V}ad_{\partial U/\partial u})).(2.17) \end{eqnarray} 这样我们得到了李超代数上的分数阶的超迹恒等式^[11].

定理2.1^[11]} 设$U=U(u,\lambda)\in G$是齐次秩的$n$阶矩阵, 且定态零曲率方程$D_{x}^{\alpha}=[U,V]$除常数因子外有唯一解$V\in G$, 则有如下分数阶超迹恒等式 \begin{equation} \frac{\delta}{\delta u}\int {\rm str}(ad_{V}ad_{U_{\lambda}}){\rm d}x=\lambda^{-\gamma}\frac{\partial}{\partial \lambda}\lambda^{\gamma}({\rm str}(ad_{V}ad_{\partial U/\partial u})),~~~~\gamma = \mbox{常数.}(2.18) \end{equation} (2.18)式中的变分导数是修正Riemman-Liouville意义下的分数阶导数.

定理2.^[11]} 设$V$是定态零曲率方程$D_{x}^{\alpha}=[U,V]$的解,则分数阶超级恒等式中的常数$\gamma$由下列式子给出 \begin{equation} \gamma=-\frac{\lambda}{2}\frac{\rm d}{{\rm d}\lambda}\ln|{\rm str}(ad_{V}ad_{V})|,~~~{\rm str}(ad_{V}ad_{V})\neq0.(2.19) \end{equation}

我们考虑下面的李超代数 $G_{2}= \Big\{\sum\limits_{i=1}^{5}\lambda_{i}e_{i}\mid i=1,2,3,4,5\Big\}$ (参见文献^[12]). $$e_{1}=\left(\begin{array}{ccc}0&1&0\\1&~~0~~&0\\0&0&0\\ \end{array}\right),\quad e_{2}=\left(\begin{array}{ccc}0&~~1~~&0\\-1&0&0\\0&0&0\\ \end{array}\right),\quad e_{3}=\left(\begin{array}{ccc}1&~~0~~&0\\0&-1&0\\0&0&0\\ \end{array}\right),$$ $$e_{4}=\left(\begin{array}{ccc}0&~~0~~&1\\0&0&0\\0&-1&0\\ \end{array}\right),\quad e_{5}=\left(\begin{array}{ccc}0&~~0~~&0\\0&0&1\\1&0&0\\ \end{array}\right).$$ 满足 $$ [e_{1},e_{2}]=-2e_{3},[e_{1},e_{3}]=-2e_{2},[e_{2},e_{3}]=-2e_{1},[e_{1},e_{4}]=[e_{5},e_{3}]=[e_{4},e_{2}]=e_{5}, $$ $$ [e_{1},e_{5}]=[e_{2},e_{5}]=[e_{3},e_{4}]=e_{4},[e_{4},e_{4}]_{+}=-(e_{1}+e_{2}),[e_{5},e_{5}]_{+}=e_{1}-e_{2}, $$ $$ [e_{4},e_{5}]_{+}=[e_{5},e_{4}]_{+}=e_{3}, $$ 其中,$e_{1},e_{2},e_{3}$是偶元,$e_{4},e_{5}$是奇元, $[\cdot,\cdot]$和$[\cdot,\cdot]_{+}$表示交换子和反交换子. 对应的圈李超代数 $\tilde{G_{2}}$定义为 $$e_{i}(n)=\lambda^{n}e_{i},~~ [e_{i}(m),e_{j}(n)]=[e_{i},e_{j}]\lambda^{m+n}, ~~ i,j=1,2,3,4,5,~ m,n\in {\bf Z}. $$

我们考虑分数阶超Yang等谱问题 $$D_{x}^{\alpha}\psi=U\psi,$$ \begin{equation} U=\left(\begin{array}{ccc}s&~~\lambda+q+r~~&u_{1}\\-\lambda-q+r&-s&u_{2}\\u_{2}&-u_{1}&0\\ \end{array}\right),\quad u=\left(\begin{array}{c}q\\r\\s\\u_{1}\\u_{2}\\ \end{array}\right),\quad \lambda_{t}=0.(3.1) \end{equation}

设$V=ae_{3}+be_{2}+ce_{1}+de_{4}+fe_{5}$. 则解定态零曲率方程$D_{x}^{\alpha}V=[U,V]$可得 $$ D_{x}^{\alpha}a_{m}=2c_{m+1}+2qc_{m}-2rb_{m}+u_{1}f_{m}+u_{2}d_{m}, $$ $$ D_{x}^{\alpha}b_{m}=-2ra_{m}+2sc_{m}-u_{1}d_{m}-u_{2}f_{m}, $$ $$ D_{x}^{\alpha}c_{m}=-2a_{m+1}-2qa_{m}+2sb_{m}-u_{1}d_{m}+u_{2}f_{m}, $$ $$ D_{x}^{\alpha}d_{m}=f_{m+1}+qf_{m}+rf_{m}+sd_{m}-u_{1}a_{m}-u_{2}b_{m}-u_{2}c_{m}, $$ $$ D_{x}^{\alpha}f_{m}=-d_{m+1}-qd_{m}+rd_{m}-sf_{m}+u_{2}a_{m}+u_{1}b_{m}-u_{1}c_{m}, $$ $$ a_{0}=c_{0}=d_{0}=e_{0}=0,b_{0}=1,a_{1}=s,c_{1}=r,d_{1}=u_{1},e_{1}=u_{2},b_{1}=0. $$

记 \begin{eqnarray*} V_{+}^{(n)} &=&\sum_{m=0}^{m}[a_{m}e_{3}(n-m)+b_{m}e_{2}(n-m) +c_{m}e_{1}(n-m)+d_{m}e_{4}(n-m)+f_{m}e_{5}(n-m)] \\ &=&\lambda^{n}V-V_{-}^{(n)}, \end{eqnarray*} 通过计算,我们有如下的结果 $$-D_{x}^{\alpha}V_{+}^{(n)}+[U,V_{+}^{(n)}]=-2c_{n+1}e_{3}(0)+2a_{n+1}e_{1}(0)-f_{n+1}e_{4}(0)+d_{n+1}e_{5}(0).$$ 取$V^{(n)}=V_{+}^{(n)}+b_{n+1}e_{2}(0)$,则有 \begin{eqnarray} -D_{x}^{\alpha}V^{(n)}+[U,V^{(n)}]&=&(-2c_{n+1}-2rb_{n+1})e_{3}(0)+(2a_{n+1}+2sb_{n+1})e_{1}(0)-b_{n+1x}e_{2}(0) \nonumber\\ &&+(-f_{n+1}-u_{2}b_{n+1})e_{4}(0)+(d_{n+1}+u_{1}b_{n+1})e_{5}(0), (3.2) \end{eqnarray} 因此由分数阶零曲率方程 \begin{equation} D_{t}^{\beta}-D_{x}^{\alpha}V^{(n)}+[U,V^{(n)}]=0 , (3.3) \end{equation} 决定了如下分数阶方程族 \begin{eqnarray} D_{t}^{\beta}u&=&\left(\begin{array}{c}D_{t}^{\beta}q\\D_{t}^{\beta}r\\D_{t}^{\beta}s\\D_{t}^{\beta}u_{1}\\D_{t}^{\beta}u_{2}\\\end{array}\right) =\left(\begin{array}{c}D_{x}^{\alpha}b_{n+1}\\-2a_{n+1}-2sb_{n+1}\\2c_{n+1}+2rb_{n+1}\\f_{n+1}+u_{2}b_{n+1}\\-d_{n+1}-u_{1}b_{n+1}\\\end{array}\right) \nonumber\\ &=&\left(\begin{array}{ccccc}-D_{x}^{\alpha}& ~~-s~~&r&~~ -\frac{1}{2}u_{2}~~& \frac{1}{2}u_{1}\\[2mm] s&0&-1&0&0\\-r&1&0&0&0\\[2mm] -\frac{1}{2}u_{2}&0&0& -\frac{1}{2}&0\\[3mm] \frac{1}{2}u_{1}&0&0&0& -\frac{1}{2}\end{array}\right)\left(\begin{array}{c}-2b_{n+1}\\2c_{n+1}\\2a_{n+1}\\-2f_{n+1}\\2d_{n+1}\\\end{array}\right) \nonumber\\ &=&JP_{n+1}=JLP_{n},(3.4) \end{eqnarray} 其中 $$L=\left(\begin{array}{ccccc}A&B&C&D&E\\[2mm] -r&-q& \frac{1}{2}D_{x}^{\alpha}& \frac{1}{2}u_{1}& -\frac{1}{2}u_{2}\\[3mm] -s&~~ -\frac{1}{2}D_{x}^{\alpha}~~&-q& -\frac{1}{2}u_{2}& -\frac{1}{2}u_{1}\\[2mm] u_{2}&-u_{2}&-u_{1}&~~-(q+r)~~&-D_{x}^{\alpha}+s\\ -u_{1}&-u_{1}&u_{2}&D_{x}^{\alpha}+s&-q+r \end{array}\right),$$ $A=-D_{x}^{-\alpha}(u_{1}^{2}+u_{2}^{2})$, $B=D_{x}^{-\alpha}(2qs-u_{1}^{2}+u_{2}^{2})-D_{x}^{-\alpha}rD_{x}^{\alpha}$, $C=2D_{x}^{-\alpha}(u_{1}u_{2}-qr)-D_{x}^{-\alpha}sD_{x}^{\alpha}$, $D=D_{x}^{-\alpha}u_{1}D_{x}^{\alpha}+D_{x}^{-\alpha}qu_{2}$, $E=D_{x}^{-\alpha}u_{2}D_{x}^{\alpha}-D_{x}^{-\alpha}qu_{1}$. 如果在系统(3.4)式中 取$u_{1}=u_{2}=0$, 则化简为参考文献^[13]中的超Yang族, 所以称系统(3.4)式为分数阶超Yang族.

下面将应用分数阶超级恒等式构造分数阶超Yang族的哈密顿结构. 对于$\forall a=a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{4}+a_{5}e_{5}\in G$, 通过计算即有 $$ad_{a}=\left(\begin{array}{ccccc}0&~~-a_{3}~~&a_{2}&a_{5}&a_{4}\\ -2a_{2}&2a_{1}&0&~~-2a_{4}~~&0\\2a_{3}&0&-2a_{1}&0&2a_{5}\\-a_{4}&-a_{5}&0&a_{1}&a_{2}\\a_{5}&0&-a_{4}&a_{3}&-a_{1}\\\end{array}\right), $$ 这里的$ad_{a}$是由(2.5)式决定的. 如果定义超迹 \begin{equation} {\rm str}(c)=c_{11}+c_{22}-c_{33},~~~~c=ab,~~~~~a,b\in G,(4.1) \end{equation} 其中$c=(c_{ij})_{3\times3}$且$ab$是 $a$和$b$的矩阵积, 则Killing型可按如下公式计算 \begin{equation} {\rm str}(ad_{a}ad_{b})=3{\rm str}(ab),~~~~~a,b\in G.(4.2) \end{equation} 利用(4.2)式可推得 $$ {\rm str}(ad_{V}ad_{u_{\lambda}})=-6b,{\rm str}(ad_{V}ad_{u_{q}})=-6b,{\rm str}(ad_{V}ad_{u_{r}})=6c, $$ $$ {\rm str}(ad_{V}ad_{u_{s}})=6a,{\rm str}(ad_{V}ad_{u_{u_{1}}})=-6e,{\rm str}(ad_{V}ad_{u_{u_{2}}})=6d, $$ 利用分数阶超级恒等式(2.18),即得 \begin{equation} \frac{\delta}{\delta u}\int(-6b){\rm d}x=\lambda^{-\gamma}\frac{\partial}{\partial \lambda}\lambda^{\gamma}\left(\begin{array}{c}-6b\\6c\\6a\\-6e\\6d\\ \end{array}\right).(4.3) \end{equation} 比较$\lambda^{-n-1}$的系数,等式满足 \begin{equation} \frac{\delta}{\delta u}\int(-6b_{n+1}){\rm d}x=(-n+\gamma)\left(\begin{array}{c}-6b_{n}\\6c_{n}\\6a_{n}\\-6e_{n}\\6d_{n}\\ \end{array}\right),~~~n\geq0.(4.4) \end{equation} 为了确定$\gamma$,我们首先设$n=0$,由此可得$\gamma=0$. 所以,我们有 \begin{equation} \frac{\delta}{\delta u}\int\Big(\frac{6b_{n+2}}{n+1}\Big){\rm d}x=\left(\begin{array}{c}-6b_{n+1}\\6c_{n+1}\\6a_{n+1}\\-6e_{n+1}\\6d_{n+1}\\ \end{array}\right),~~~n\geq0.(4.5) \end{equation} 因此,分数阶超Yang族具有如下超哈密顿结构 \begin{equation} D_{t}^{\beta}u=J\frac{\delta H_{n}}{\delta u},~~~~n\geq0,(4.6) \end{equation} 其中分数阶超哈密顿算子$J$和哈密顿函数$H_{n}$分别为 \begin{equation} J=\left(\begin{array}{ccccc}-D_{x}^{\alpha}&~~-s~~&r& ~~ -\frac{1}{2}u_{2}~~& \frac{1}{2}u_{1}\\[2mm] s&0&-1&0&0\\ -r&1&0&0&0\\[2mm] -\frac{1}{2}u_{2}&0&0& -\frac{1}{2}&0\\[3mm] \frac{1}{2}u_{1}&0&0&0& -\frac{1}{2}\end{array}\right) ,~~~~H_{n}=\int\frac{6b_{n+2}}{n+1},~~~~~n\geq0.(4.7) \end{equation} 显然,该分数阶超Yang族具有超双哈密顿结构 \begin{equation} D_{t}^{\beta}u=J\frac{\delta H_{n}}{\delta u}=M\frac{\delta H_{n-1}}{\delta u},~~~~n\geq0,(4.8) \end{equation} 其中,$M=LJ$^[12]. 当$n=1$,我们可得到分数阶非线性超Yang系统 $$D_{t}^{\beta}q=rD_{x}^{\alpha}r+sD_{x}^{\alpha}s+D_{x}^{\alpha}(u_{1}u_{2}),$$ $$D_{t}^{\beta}r=D_{x}^{\alpha}r+2qs-sr^{2}-s^{3}-2su_{1}u_{2},$$ $$D_{t}^{\beta}s=D_{x}^{\alpha}s-2qr+r^{3}+rs^{2}+2ru_{1}u_{2}$$ $$D_{t}^{\beta}u_{1}=D_{x}^{\alpha}u_{1}-qu_{2}+\frac{1}{2}u_{2}r^{2}+\frac{1}{2}u_{2}s^{2},$$ \begin{equation}D_{t}^{\beta}u_{2}=D_{x}^{\alpha}u_{2}+qu_{1}-\frac{1}{2}u_{1}r^{2}-\frac{1}{2}u_{1}s^{2}.(4.9) \end{equation}

显然,当$\alpha=\beta=1$,系统(4.9)式化简为经典非线性超Yang系统 $$q_{t}=rr_{x}+ss_{x}+u_{1x}u_{2}+u_{1}u_{2x},$$ $$r_{t}=r_{x}+2qs-sr^{2}-s^{3}-2su_{1}u_{2},$$ $$s_{t}=s_{x}-2qr+r^{3}+rs^{2}+2ru_{1}u_{2},$$ $$u_{1t}=u_{1x}-qu_{2}+\frac{1}{2}u_{2}r^{2}+\frac{1}{2}u_{2}s^{2},$$ \begin{equation}u_{2t}=u_{2x}+qu_{1}-\frac{1}{2}u_{1}r^{2}-\frac{1}{2}u_{1}s^{2}.(4.10) \end{equation}

本文不但从等谱问题(3.1)获得分数阶超Yang族, 而且应用分数阶超迹恒等式构造了分数阶超 Yang族的分数阶超哈密顿结构. 这种构造分数阶孤子方程族的方法可以被推广到其他方程族上.