\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 17, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/17\hfil Reduction of infinite dimensional equations] {Reduction of infinite dimensional equations} \author[Z. Li, T. Xu\hfil EJDE-2006/17\hfilneg] {Zhongding Li, Taixi Xu} % in alphabetical order \address{Zhongding Li\hfill\break Department of Mathematics, Shijiazhuang Railway Institute, Hebei, China} \address{Taixi Xu \hfill\break Department of Mathematics \\ Southern Polytechnic State University\\ 1100 South Marietta Parkway\\ Marietta, GA 30060, USA} \email{txu@spsu.edu} \date{} \thanks{Submitted February 11, 2005. Published February 2, 2006.} \subjclass[2000]{37K15, 37K40} \keywords{Soliton equations; Hamiltonian equation; Euler-Lagrange equation; \hfill\break\indent integrable systems; Legendre transformation; involutive system; symmetries of equations; \hfill\break\indent invariant manifold; Poisson bracket; symplectic space} \begin{abstract} In this paper, we use the general Legendre transformation to show the infinite dimensional integrable equations can be reduced to a finite dimensional integrable Hamiltonian system on an invariant set under the flow of the integrable equations. Then we obtain the periodic or quasi-periodic solution of the equation. This generalizes the results of Lax and Novikov regarding the periodic or quasi-periodic solution of the KdV equation to the general case of isospectral Hamiltonian integrable equation. And finally, we discuss the AKNS hierarchy as a special example. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} \label{section6.1} Soliton equations emerged about 40 years ago \cite{Abl,Theoryofsolitons}. C.W. Cao discovered the nonlinearization method \cite{Cao-1}-\cite{Guchaohao} to obtain the finite dimensional integrable systems \cite{Arnold,Moser} associated with soliton equations. This way works well for many soliton equations \cite{Qiao-1}-\cite{Xu-7}. Its main drawback, however, is that there is no single approach for finding the Lax pair \cite{Lax-1}-\cite{Lax-2} of a soliton equation. More precisely, different soliton equations require very different ways of finding their Lax pairs. Furthermore, this method does not work for every soliton equation, and for some equations, we have both the Bargmann and Neumann systems, for some others, we only have the Bargmann systems. So, it is natural to seek how to explain this drawback and to ask whether there is a single method that works for every infinite dimensional system (i.e., soliton equation). We have been trying to answer these questions for the last few years. Even through we have been unable to characterize the conditions that ensure the existence of both the Bargmann and Neumann systems, we have found, however, a new method which works for every soliton equation. More specifically, for every existing infinite dimensional integrable Hamiltonian system, we can obtain the associated finite dimensional integrable Hamiltonian system without knowing its Lax pairs for the corresponding higher order soliton equations. Let $J$ be a Hamiltonian operator, $u_{t} = J \frac{\delta H_{1}} {\delta u}$ be an infinite dimensional integrable Hamiltonian equation $(u = (u_{1}, \dots, u_{N})^{T})$, and $\{ H_{m} \}_{m=0}^{\infty}$ be the first integrals of $u_{t} = J \frac{\delta H_{1}}{\delta u}$. Its higher order equations are $u_{t_{m}} = J \frac{\delta I} {\delta u}$ (here $I = \sum_{l=0}^{m}C_{m-l}H_{l}$, $C_{0}=1$, $C_{m-l}$ are constants, $m=0,1,2,\dots)$. We use the general Legendre transformation to show that the infinite dimensional integrable equation $u_{t_{m}} = J \frac{\delta H_{m}}{\delta u}$ can be reduced to a finite dimensional integrable Hamiltonian system on an invariant set $S$. Then we obtain the periodic or quasi-periodic solution of equations $u_{t_{m}} = J \frac{\delta I}{\delta u}$ $(m=0,1,2,\dots)$. This generalizes the results of Lax (\cite{Lax-1,Lax-2}) and Novikov \cite{Novikov} regarding the periodic or quasi-periodic solution of the KdV \cite{KdV1895,GGKM1967} equation to the general case of isospectral Hamiltonian integrable equations. As a special example, we will discuss the AKNS \cite{Abl} hierarchy. Generally, looking for the periodic or quasi-periodic solution of infinite dimensional integrable equations is very difficult. In \cite{Cao-1}, the nonlinear Schr\"{o}dinger equation is investigated and its periodic solution is obtained. Flaschka \cite{Flaschka} and Lax \cite{Lax-1} discussed the algebraic structure of the KdV equation and obtained its periodic or quasi-periodic solution. Novikov \cite{Novikov} studied in details the relationship between the KdV equation and its stationary equation and obtained its periodic solution. Cao \cite{Cao-1} used the nonlinearization of Lax pairs and obtained the periodic or quasi-periodic solutions (involutive solutions) of the AKNS, the KdV, and the Harry Dym \cite{Xu-2,Fordy} equations. Now we synthesize their results and generalize them to the (general) isospectral integrable equations, and obtain a (general) method to solve general infinite dimensional integrable equations $u_{t_{m}} = J \frac{\delta H_{m}}{\delta u}$ for periodic or quasi-periodic solutions. We also discuss the algebraic and geometric properties of the vector field of the Hamiltonian integrable equations $u_{t_{m}} = J \frac{\delta H_{m}}{\delta u}$ $(m\geq 0)$, and prove that they can be reduced on an invariant subset $S$ to a finite dimensional integrable Hamiltonian system $\dot{q}_{i} = \frac{\partial T_{m}}{\partial p_{i}}, \quad \dot{p}_{i} = - \frac{\partial T_{m}}{\partial q_{i}}, \quad i = 1, 2, \dots, n$ where $\dot{q}_{i} = \frac{\partial q_{i}}{\partial t_{m}}$, $\dot{p}_{i} = \frac{\partial p_{i}}{\partial t_{m}}$, $T_{m}$ is determined by $\frac{dT_{m}}{dx} = - \frac{\delta I}{\delta u} J \frac{\delta H_{m}}{\delta u}$ and is a function of $(q,p)$ $(m = 0, 1, 2, \dots)$. \section{The general Legendre transformation} \label{section6.2} Let $L = L(u, u', u'', \dots, u^{(n)})$ be the Lagrangian, which depends only on $u = (u_{1}, \dots, u_{N})^{T}$ and its derivatives with respect to x : $u^{(j)} = \frac{d^{j}u}{dx^{j}}, j = 1, 2, \dots, n.$ Let $I = \int _{\Omega} Ldx$. The Euler-Lagrange equation is $$\frac{\delta I}{\delta u} = \sum_{l=0}^{n}(-1)^{l}\frac{d^{l}}{dx^{l}} \frac{\partial L}{\partial u^{(l)}} = 0 \label{eq:per2.1}$$ where, when $\Omega = ( -\infty, +\infty)$, $u$ and $u^{(j)}(j = 1, 2, \dots)$ decrease rapidly as $x \to \infty$ and when $\Omega = [\alpha, \alpha+T]$, $u(x+T) = u(x)$ for $T > 0$ and $\alpha$ a constant. We introduce the following canonical coordinates $q_{i}$, $p_{i}$ ($i = 1, 2, \dots, n$). \label{eq:per2.2} \begin{aligned} q_{i} & = (q_{1i}, q_{2i}, \dots, q_{Ni})^{T} = u^{(i-1)}, \\ p_{i} & = (p_{1i}, p_{2i}, \dots, p_{Ni})^{T} = \sum_{l=0}^{n-i}(-1)^{l}\frac{d^{l}}{dx^{l}} \frac{\partial L}{\partial u^{(i+l)}} \\ & = \sum_{l=0}^{n-i}(-1)^{l}\frac{d^{l}}{dx^{l}} \Big( \frac{\partial L}{\partial u_{1}^{(i+l)}}, \frac{\partial L}{\partial u_{2}^{(i+l)}}, \dots, \frac{\partial L}{\partial u_{N}^{(i+l)}} \Big)^{T}, \end{aligned} where $i = 1, 2, \dots, n$. Let $$H(q, p) = \sum_{i=1}^{n}q'_{i}p_{i} - L = \sum_{i=1}^{n}\sum_{l=1}^{N}q'_{li}p_{li} - L \label{eq:per2.3}$$ where $q = (q_{1}, q_{2}, \dots, q_{n})^{T}$, $p = (p_{1}, p_{2}, \dots, p_{n})^{T}$. Equation (\ref{eq:per2.3}) is called the general Legendre transformation. \begin{definition} \rm A Lagrangian $L$ is said to be non-singular if equation (\ref{eq:per2.2}) can be uniquely solved in the form $$u^{(i)} = u^{(i)}(q, p), \quad i = 0, 1, \dots, 2n-1. \label{eq:per2.4}$$ \end{definition} \begin{lemma} \label{lem2.2} If the Lagrange function $L$ satisfies the condition $$\det Q = \det \Big( \frac{\partial^{2}L} {\partial u_{\alpha}^{(n)}\partial u_{\beta}^{(n)}} \Big) \neq 0 \label{eq:per2.5}$$ where $\alpha = 1,2,\dots,N$, $\beta = 1,2,\dots,N$, then $L$ is non-singular. \label{Lemma6.2.2} \end{lemma} \begin{proof} Since $L = L(u, u', \dots, u^{(n)}) = L (q, u^{(n)})$, $p_{n} = \frac{\partial L}{\partial u^{(n)}} = f(q, u^{(n)}),$ and the Jacobi determinant $J(u^{(n)}) = \left| \frac{\partial p_{n}}{\partial u^{(n)}} \right| = \left| \frac{\partial^{2}L} {\partial u^{(n)}_{\alpha} \partial u^{(n)}_{\beta}} \right| = \det Q \neq 0,$ we obtain $$u^{(n)} = f_{n}(q, p_{n}). \label{eq:per2.6}$$ Next, \begin{align*} p_{n-1} & = \frac{\partial L}{\partial u^{(n-1)}} - \frac{d}{dx}\frac{\partial L}{\partial u^{(n)}} \\ & = \frac{\partial L}{\partial u^{(n-1)}} - Q\cdot u^{(n+1)} - \sum_{j=0}^{n-1}\frac{\partial^{2}L} {\partial u^{(n)}\partial u^{(j)}}\cdot u^{(j+1)}. \end{align*} So $\det Q \neq 0$ yields $$u^{(n+1)} = f_{n-1}(q, p_{n-1}, p_{n}). \hspace{2.5in} \label{eq:per2.7}$$ Similarly, we obtain $$u^{(n+k)} = f_{n-k}(q, p_{n-k}, p_{n-k+1}, \dots, p_{n}), \ \ k = 0, 1, \dots, n-1. \label{eq:per2.8}$$ Thus, the Lagrangian is non-singular. \end{proof} \begin{lemma} \label{lrm2.3} If the Lagrangian $L$ has the form $L = a(u^{(n)})^{2} + L_{0}(u, u', \dots, u^{(n-1)})$ where $a \neq 0$ is a constant, then $L$ is non-singular. \end{lemma} For the proof of the above lemma, use $\det Q = 2aN \neq 0$ and Lemma \ref{Lemma6.2.2}. \begin{remark} \rm If the Lagrangian $L$ is non-singular, then the general Legendre transformation $(u, u', \dots, u^{(2n-1)})^{T} \longrightarrow (q, p)^{T}$ satisfies the relations \begin{gather} q'_{i} = \frac{\partial H}{\partial p_{i}}, \quad i = 1, 2, \dots, n, \label{eq:per2.9} \\ p'_{1} = - \frac{\partial H}{\partial q_{1}} - \frac{\delta I}{\delta u}, \quad p'_{i} = - \frac{\partial H}{\partial q_{i}}, \quad i = 2, 3, \dots, n. \label{eq:per2.10} \end{gather} \end{remark} For calculating the above expressions vote that $q_{i}' = q_{i+1}$ ($i = 1,2,\dots, n-1$). we have $H = \sum_{i=1}^{n}q_{i}'p_{i} - L = q_{2}p_{1} + q_{3}p_{2} + \dots + q_{n}p_{n-1} + q_{n}'p_{n} - L,$ which implies (\ref{eq:per2.9}). For (\ref{eq:per2.10}), we have \begin{align*} p_{1}' & = \frac{d}{dx}p_{1} = \frac{d}{dx}\Big( \sum_{l=0}^{n-1} (-1)^{l} \frac{d^{l}}{dx^{l}} \frac{\partial L}{\partial u^{(1+l)}} \Big) \\ & = \sum_{l=0}^{n-1} (-1)^{l}\frac{d^{l+1}}{dx^{l+1}} \frac{\partial L}{\partial u^{(1+l)}} = - \sum_{k=1}^{n}(-1)^{k}\frac{d^{k}}{dx^{k}} \frac{\partial L}{\partial u^{(k)}} \\ & = \frac{\partial L}{\partial u} - \sum_{k=0}^{n}(-1)^{k}\frac{d^{k}}{dx^{k}} \frac{\partial L}{\partial u^{(k)}}\\ & = - \frac{\partial H}{\partial q_{1}} - \frac{\delta I}{\delta u}\,, \end{align*} which is the first formula of (\ref{eq:per2.10}). Next, \begin{align*} p_{i}' & = \frac{d}{dx}p_{i} = \frac{d}{dx} \left( \sum_{l=0}^{n-i} (-1)^{l}\frac{d^{l}}{dx^{l}} \frac{\partial L}{\partial u^{(i+l)}} \right) = \sum_{l=0}^{n-i} (-1)^{l}\frac{d^{l+1}}{dx^{l+1}} \frac{\partial L}{\partial u^{(i+l)}} \\ & = - \sum_{k=1}^{n-i}(-1)^{k}\frac{d^{k}}{dx^{k}} \frac{\partial L}{\partial u^{(i+k-1)}} \\ & = - \sum_{k=0}^{n-(i-1)}(-1)^{k}\frac{d^{k}}{dx^{k}} \frac{\partial L}{\partial u^{(i-1+k)}} + \frac{\partial L}{\partial u^{(i-1)}} \\ & = - p_{i-1} + \frac{\partial L}{\partial u^{(i-1)}} \\ & = - p_{i-1} + \frac{\partial L}{\partial q_{i}} \\ & = - \frac{\partial H}{\partial q_{i}}, \end{align*} which is the second formula of (\ref{eq:per2.10}). \begin{theorem} \label{thm2.5} For non-singular Lagrangian $L$, the Euler-Lagrange equation $(\ref{eq:per2.1})$ is equivalent to the Hamiltonian system $$q'_{i} = \frac{\partial H}{\partial p_{i}}, \quad p'_{i} = - \frac{\partial H}{\partial q_{i}} \quad (i = 1, 2, \dots, n) \label{eq:per2.11}$$ where $H$ is given by $(\ref{eq:per2.3})$, and $\frac{dH}{dx} = - u' \frac{\delta I}{\delta u} = - \sum_{l=1}^{N} u'_{l} \frac{\delta I}{\delta u_{l}},$ where the symbol \,' indicates $\frac{\partial}{\partial x}$. \end{theorem} \begin{proof} Since (\ref{eq:per2.1}) is $\frac{\delta I}{\delta u} = 0$, (\ref{eq:per2.9}) and (\ref{eq:per2.10}) are equivalent to (\ref{eq:per2.11}). By direct calculation, we obtain $\frac{dH}{dx} = - u' \frac{\delta I}{\delta u}$. \end{proof} \begin{remark} \rm The non-singular general Legendre transformation is invertible, i.e., we can determine $u^{(i)}$ from (\ref{eq:per2.2}): $$u^{(k)} = h_{k} (q, p),\quad k = 0, 1, 2, \dots, 2n-1. \label{eq:per2.12}$$ \label{remark:6.2.6} \end{remark} \section{The reduction} \label{section6.3} Suppose $$u_{t} = K(u) = J \frac{\delta H}{\delta u} \label{eq:per3.1}$$ is an infinite dimensional integrable Hamiltonian equation, $H_{m} = \int_{\Omega} L_{m} dx \quad (m = -1, 0, 1, 2, \dots)$ are its infinitely many involutive first integrals in pair, where $H = H_{1}$. Its $m$th-order equations are defined by $$u_{t_{m}} = J \frac{\delta H_{m}}{\delta u} \label{eq:per3.2}$$ where $J$ is a differential operator for the corresponding soliton equations. For examples, for the KdV equation, $J = \frac{\partial}{\partial x}$, for the AKNS equation, $J = \begin{pmatrix}0 & -1 \\ 1& 0 \end{pmatrix}$. The general higher-order stationary equations are defined by $$\sum_{l=0}^{n} C_{n-l} J \frac{\delta H_{l}}{\delta u} = 0 \label{eq:3.3.1}$$ or $$\frac{\delta I}{\delta u} = 0 \label{eq:per3.3}$$ where $I = \sum_{l=-1}^{n} C_{n-l}H_{l} = \int_{\Omega} L( u, \dots, u^{(n)})dx, \quad J \frac{\delta H_{-1}}{\delta u} = 0, \quad n \geq 0,$ $C_{0} = 1$, $C_{i}$ ($i = -1, 1, 2, \dots, n+1$) are constants. Suppose the functional space is $E = \big\{ F : F = \int_{\Omega} P(u, u', \dots, u^{(m)})dx \big\}, \quad ( m \geq 0 ).$ The Poisson bracket on space $E$ is defined as $$\{ H, F \}(x) = \frac{d}{dt}\big|_{t=0} H( g_{F}^{t}u(x)) = \int_{\Omega} \frac{\delta H}{\delta u} \cdot J \frac{\delta F}{\delta u} dx =\big( \frac{\delta H}{\delta u}, J \frac{\delta F}{\delta u} \big) \label{eq:per3.4}$$ where $H, F \in E$, $g_{F}^{t}$ is the solution operator of equation $u_{t} = J \frac{\delta F}{\delta u}$, and $( \cdot,\cdot )$ is the standard inner product in $L_{2}(\Omega)$. Hence, $$\{ H_{m}, H_{n} \} = 0, \quad m,n = -1, 0, 1, 2, \dots. \label{eq:per3.5}$$ \begin{definition} \rm In the space $E$, the Hamiltonian vector field $\vec{F}$ of a functional $F$ is defined by $$L_{\vec{F}}(H) = \frac{d}{dt}\big|_{t=0} H( g_{F}^{t}u(x)) = \{ H, F \}, \label{eq:per3.6}$$ and for any $H \in E$, $L_{\lambda \vec{F}}(H) = \lambda \{ H, F \} = \lambda L_{\vec{F}}(H) \quad (\lambda \mbox{ is constant}).$ \end{definition} \begin{lemma} \label{lem3.2} All the Hamiltonian vector fields form a Lie algebra. Its Lie bracket $[\vec{H}, \vec{F}]$ is defined by $$L_{[\vec{H}, \vec{F}]} = L_{\vec{F}}L_{\vec{H}} - L_{\vec{H}}L_{\vec{F}}. \label{eq:per3.7}$$ \end{lemma} \begin{proof} The bilinearity and anti-symmetry are obvious. From the definition of Lie bracket we have $L_{ [ [\vec{H}, \vec{F}], \vec{A}] } = L_{\vec{A}}L_{\vec{F}}L_{\vec{H}} - L_{\vec{A}}L_{\vec{H}}L_{\vec{F}} + L_{\vec{H}}L_{\vec{F}}L_{\vec{A}} - L_{\vec{F}}L_{\vec{H}}L_{\vec{A}}$ and $L_{ [ [\vec{H}, \vec{F}], \vec{A}] } + L_{ [ [\vec{F}, \vec{A}], \vec{H}] } + L_{ [ [\vec{A}, \vec{H}], \vec{F}] }$ has $12$ terms in total, and every term appears twice with opposite signs. So the Jacobi identity holds. \end{proof} \begin{lemma} \label{lma5.3.3} The vector field $\vec{F}$ of $F = \{ F_{1}, F_{2} \}$ can be represented by $$\vec{F} = [ \vec{F}_{1}, \vec{F}_{2} ]. \label{eq:per3.8}$$ \end{lemma} \begin{proof} By the Jacobi identity of the Poisson bracket of functionals on the space $E$ we have $L_{\vec{F}}(H) = \{H, F\} = \{H, \{F_{1}, F_{2} \} \} = -\{F_{1}, \{F_{2}, H\}\} - \{F_{2}, \{ H, F_{1}\}\}.$ On the other hand, \begin{align*} L_{ [ \vec{F}_{1}, \vec{F}_{2} ] } (H) &= ( L_{\vec{F}_{2}}L_{\vec{F}_{1}} - L_{\vec{F}_{1}}L_{\vec{F}_{2}})(H) \\ &= L_{\vec{F}_{2}}L_{\vec{F}_{1}}(H) - L_{\vec{F}_{1}}L_{\vec{F}_{2}}(H) \\ &= L_{\vec{F}_{2}}(\{H,F_{1}\}) - L_{\vec{F}_{1}}(\{H, F_{2}\}) \\ &= \{ \{H,F_{1}\}, F_{2} \} - \{ \{H, F_{2}\}, F_{1} \} \\ &= - \{ F_{2}, \{H,F_{1}\} \} + \{ F_{1}, \{H, F_{2}\} \} \\ &= - \{ F_{1}, \{ F_{2}, H\} \} - \{ F_{2}, \{H,F_{1}\} \}. \end{align*} Comparing the above two equations we obtain $L_{\vec{F}}(H) = L_{ [ \vec{F}_{1}, \vec{F}_{2} ] } (H)$ for arbitrary $H \in E$. \end{proof} \begin{corollary} \label{coro3.4} The map of the Lie algebra of functionals on $E$ onto the Lie algebra of Hamiltonian vector fields is an algebra homomorphism. \end{corollary} \begin{lemma} \label{lma5.3.4} $$J \frac{\delta \{H, F\}}{\delta u} = \left[ J \frac{\delta H}{\delta u}, J \frac{\delta F}{\delta u} \right] \label{eq:per3.9}$$ where $[a, b] = a'[b] - b'[a],\ \ a'[b] = \frac{d}{d\varepsilon}\big|_{\varepsilon =0} a(u + \varepsilon b).$ \end{lemma} \begin{proof} For any $A \in E$, by the symmetry of $( \frac{\delta A}{\delta u})'$, we can show that $\big( \frac{\delta A}{\delta u}, J \frac{\delta \{H, F\}}{\delta u} \big) = \Big( \frac{\delta A}{\delta u}, \big( J \frac{\delta H}{\delta u} \big)' \big[ J \frac{\delta F}{\delta u} \big] - \big( J \frac{\delta F}{\delta u}\big)' \big[ J \frac{\delta H}{\delta u}\big] \Big).$ Hence (\ref{eq:per3.9}) holds. \end{proof} \begin{corollary} \label{coro3.6} $\{J\frac{\delta H_{m}}{\delta u}\}$ are the symmetries of $u_{t_{i}} = J \frac{\delta H_{i}}{\delta u}$ ($i = 0, 1, 2, \dots$). \end{corollary} \begin{proof} Since $\{H_{m}, H_{i}\} = 0$, Lemma \ref{lma5.3.4} implies $\big[ J \frac{\delta H_{m}}{\delta u}, \frac{\delta H_{i}}{\delta u} \big] = J \frac {\delta \{ H_{m}, H_{i} \} }{\delta u} = 0.$ So, by a property of symmetry: $\sigma$ is a symmetry of $u_{t} = K(u)$ if and only if $[K, \sigma ] = 0$, this corollary is proved. \end{proof} \begin{theorem} \label{thm3.7} The flows defined by \eqref{eq:per3.2} commute with each other. \label{thm5.3.6} \end{theorem} The above result follows from $[\vec{H}_{m}, \vec{H}_{m}] = 0$ and Lemma~\ref{lma5.3.3} or Lemma~\ref{lma5.3.4}. \begin{theorem} \label{Theorem6.3.8} The solutions of the stationary equation \eqref{eq:per3.3} form an invariant manifold $S$ of the flows defined by equation \eqref{eq:per3.2}. \end{theorem} \begin{proof} First we prove that $I = \sum_{l=-1}^{n} C_{n-l}H_{l}$ is a conserved functional of equation \eqref{eq:per3.2}. It suffices to show $\{ I, H_{m} \} = 0$. In fact, \begin{align*} \{I, H_{m}\} &= \big( \frac{\delta I}{\delta u}, J \frac{\delta H_{m}}{\delta u} \big) = \Big( \sum_{l=-1}^{n} \frac{\delta H_{l}}{\delta u}, J \frac{\delta H_{m}}{\delta u} \Big) \\ &= \sum_{l=0}^{n} C_{n-l} \big( \frac{\delta H_{l}}{\delta u}, J \frac{\delta H_{m}}{\delta u} \big) \\ &= \sum_{l=0}^{n} C_{n-l} \{ H_{l}, H_{m} \} = 0. \end{align*} By a theorem given by Lax (see \cite{Lax-1,Lax-2}), if $I$ is a conserved functional of \eqref{eq:per3.2}, then the set of stationary points of $I$, i.e., the solution set of \eqref{eq:per3.3}, forms an invariant set for the flow \eqref{eq:per3.2}. \end{proof} Using Remark~\ref{remark:6.2.6}, we can reduce an arbitrary function $P = P(u, u', \dots, u^{(n)})$ to a function $P_{1} = P_{1}(q, p)$ where $u^{(k)} = h_{k}(q, p)$ is on the manifold $S$. We call $P_{1}$ the reduction of $P$ through equations \eqref{eq:per2.12} and \eqref{eq:per2.2}, but we still use $P$ to denote $P_{1}$. Let $J\frac{\delta H_{i}}{\delta u}$ be the reduction of $J\frac{\delta H_{i}}{\delta u}$, and define $T_{i}$ by: $$\frac{dT_{i}}{dx} = - \frac{\delta I}{\delta u} J\frac{\delta H_{i}}{\delta u}, \quad i = 0, 1, 2, \dots. \label{eq:per3.10}$$ \begin{theorem} \label{maintheorem} If the Lagrangian of equation $\eqref{eq:per3.3}$ is non-singular, then via the Legendre transformation \eqref{eq:per2.2}, the stationary equation $\eqref{eq:per3.3}$ is transformed into a classical integrable Hamiltonian system: $$q'_{i} = \frac{\partial H}{\partial p_{i}}, \quad p'_{i} = - \frac{\partial H}{\partial q_{i}} \quad (i = 1, 2, \dots, n) \label{eq:per3.11}$$ where the Hamiltonian function \begin{gather} H(q,p) = \sum_{i=1}^{n}q'_{i}p_{i} - L(q_{1}, \dots, q_{n}, q'_{n}), \label{eq:per3.12} \\ \frac{dH}{dx} = - u_{x} \frac{\delta I}{\delta u}, \label{eq:per3.13} \end{gather} and the involutive first integrals are $T_{i}$ ($i = 0, 1, 2, \dots$). \end{theorem} To prove this theorem, we define the Poisson bracket $\{\cdot,\cdot\}$ in the symplectic space $({\bf{R}}^{2n}, \omega^{2})$ and prove that $\{ T_{i} \}$ is an involutive system. \begin{definition} \rm In the symplectic space $({\bf{R}}^{2n},\ \omega^{2})$, we define the Poisson bracket as $$\{A, B\} = \frac{d}{dt}\big|_{t=0} A(g_{B}^{t}(q(x), p(x))) = \sum_{i=1}^{n}\frac{\partial A}{\partial q_{i}} \frac{\partial B}{\partial p_{i}} - \frac{\partial A}{\partial p_{i}} \frac{\partial B}{\partial q_{i}} \label{eq:per3.14}$$ where the symplectic structure $\omega^{2} = \sum_{i=1}^{n} dp_{i}\wedge dq_{i} = \sum_{i=1}^{n}\sum_{l=1}^{N}dp_{li}\wedge dq_{li}$. \end{definition} To prove $\{T_{i}, T_{j}\} = 0$, $\{T_{i}, H\} = 0$ ($i,j = 1, 2, \dots$), we first give the following theorem. \begin{theorem} Under the reduction through equations \eqref{eq:per2.12} and \eqref{eq:per2.2}, the infinite dimensional Hamiltonian integrable system $$\frac{\partial u}{\partial t_{m}} = J \frac{\delta H_{m}}{\delta u} \label{eq:per3.15}$$ is transformed into the finite dimensional Hamiltonian system on $S$, $$\dot{q}_{j} = \frac{\partial T_{m}}{\partial p_{j}}, \quad \dot{p}_{j} = - \frac{\partial T_{m}}{\partial q_{j}} \label{eq:per3.16}$$ where $\dot{q}_{j} = \frac{\partial q_{j}}{\partial t_{m}}, \quad \dot{p}_{j} = \frac{\partial p_{j}}{\partial t_{m}}, \quad j = 1, 2, \dots, n\,.$ \end{theorem} \begin{proof} By \eqref{eq:per3.12}, we have \begin{gather} \frac{\partial^{2}H}{\partial q_{j}\partial p_{j-1}} = 1 \quad (j = 1, 2, \dots, n-1), \label{eq:per3.17} \\ \frac{\partial^{2}H}{\partial q_{l}\partial p_{s}} = 0 \quad (l \leq s < n \mbox{ or } 1 \leq s < l-1), \label{eq:per3.18} \\ \frac{\partial^{2}H}{\partial p_{n}\partial p_{j}} = 0 \quad (j < n). \label{eq:per3.19} \end{gather} From Remark \ref{eq:per2.4} we see that $\frac{\delta I}{\delta u} = 0$ is only one of the Hamiltonian equations \eqref{eq:per3.11}. Same as \cite{Novikov}, the other equations of \eqref{eq:per3.11} can be considered as a relation between $(q_{1}, \dots, q_{n}, p_{1}, \dots, p_{n})$ and $(u, u', \dots, u^{(2n-1)})$. Hence, from the identity formulas $(\ref{eq:per2.9})-(\ref{eq:per2.11})$ we obtain $$\frac{dT_{m}}{dx} = - \frac{\delta I}{\delta u} J \frac{\delta H_{m}}{\delta u} = \big( \frac{\partial H}{\partial q_{1}} + p'_{1} \big) J \frac{\delta H_{m}}{\delta u}. \label{eq:per3.20}$$ On the other hand, \begin{align*} \frac{dT_{m}}{dx} &= \frac{\partial T_{m}}{\partial p_{1}} p'_{1} + \sum_{l=2}^{n} \frac{\partial T_{m}}{\partial p_{l}} p_{l}' + \sum_{l=1}^{n} \frac{\partial T_{m}}{\partial q_{l}} q_{l}' \\ &= \frac{\partial T_{m}}{\partial p_{1}} p'_{1} - \sum_{l=2}^{n} \frac{\partial T_{m}}{\partial p_{l}} \frac{\partial H}{\partial q_{l}} + \sum_{l=1}^{n} \frac{\partial T_{m}}{\partial q_{l}} \frac{\partial H}{\partial p_{l}}. \end{align*} The identity \eqref{eq:per2.10} is true for arbitrary $u(x)$. Hence in \eqref{eq:per3.20} we can consider that $p'_{1}$ is arbitrary. Thus, comparing the above two expressions, we obtain $$J \frac{\delta H_{m}}{\partial u} = \frac{\partial T_{m}}{\partial p_{1}}. \label{eq:per3.21}$$ By \eqref{eq:per3.15} and $q_{1} = u$, on $S$ we have $\dot{q}_{1} = \frac{dq_{1}}{dt_{m}} = \frac{du}{dt_{m}} = J \frac{\delta H_{m}}{\delta u} = \frac{\partial T_{m}}{\partial p_{1}},$ i.e., $\dot{q}_{1} = \frac{\partial T_{m}}{\partial p_{1}}.$ We prove this theorem by mathematical induction. Let us assume $\dot{q}_{j} = \frac{\partial T_{m}}{\partial p_{j}}$ ($j = 1,2,\dots,k$), then as $j = k + 1$, we have $$\dot{q}_{k+1} = (\dot{q}_{k})' = \frac{d}{dx}\big( \frac{\partial T_{m}}{\partial p_{k}}\big). \label{eq:per3.22}$$ Let $A = (A_{1}, \dots, A_{N})^{T}, \ B$ and $A_{i}$ are the functionals with respect to $(q, p) = (q_{1}, q_{2},\dots, q_{n}, p_{1},p_{2}, \dots, p_{n}) \in {\bf R}^{2n}.$ We define $\{A, B\} \equiv ( \{A_{1}, B\}, \dots, \{A_{N}, B\} )^{T}.$ Using of the Jacobi identity of the Poisson bracket $(\ref{eq:per3.14})$ and $\{T_{m}, H\} = 0$ on $S$, we have \begin{aligned} \dot{q}_{k+1} &= \frac{dq_{k+1}}{dt} = \frac{dq_{k}'}{dt} = \frac{d}{dx}(\dot{q}_{k}) \\ &= \frac{d}{dx}\big( \frac{\partial T_{m}}{\partial p_{k}}\big) = \frac{d}{dx}\{q_{k}, T_{m}\} = \{ \{q_{k}, T_{m}\}, H \} \\ &= - \{ \{ T_{m}, H \}, q_{k} \} - \{ \{ H, q_{k} \}, T_{m} \}\\ &= \{ \{ q_{k}, H \}, T_{m} \} \\ &= \{ q'_{k}, T_{m} \}\\ &= \frac{\partial T_{m}}{\partial p_{k+1}}. \end{aligned} \label{eq:per3.23} Hence %www $\dot{q}_{j} = \frac{\partial T_{m}}{\partial p_{j}},\quad j = 1, 2, \dots, n\,.$ Next we prove the formulas $\dot{p}_{j} = - \frac{\partial T_{m}}{\partial q_{j}}, \quad j = 1, 2, \dots, n.$ Since $(\dot{q}_{n})' = (q'_{n})^{\cdot}$, we have $$\big( \frac{\partial T_{m}}{\partial p_{n}} \big)' = \big( \frac{\partial H}{\partial p_{n}} \big)^{\cdot}. \label{eq:per3.24}$$ By $(\ref{eq:per3.17})-(\ref{eq:per3.19})$ and the Jacobi identity of the the Poisson bracket, we have on $S$ $$\big( \frac{\partial H}{\partial p_{n}} \big)^{\cdot} = \sum_{j=1}^{n} \frac{\partial^{2} H}{\partial p_{n} \partial q_{j}} \frac{\partial T_{m}}{\partial p_{j}} + \frac{\partial^{2} H}{\partial p_{n}^{2}} \dot{p}_{n}. \label{eq:per3.25}$$ On the other hand, \begin{align*} \big( \frac{\partial T_{m}}{\partial p_{n}} \big)' & = (\dot{q}_{n})' = \{ \dot{q}_{n}, H \} \\ & = \big\{\frac{\partial T_{m}}{\partial p_{n}}, H \big\} = \{ \{ q_{n}, T_{m} \}, H \} \\ & = - \{ \{ T_{m}, H \}, q_{n} \} - \{ \{ H, q_{n} \}, T_{m} \} \\ & = \{ \{ q_{n}, H \}, T_{m} \} = \big\{\frac{\partial H}{\partial p_{n}}, T_{m} \big\} \end{align*} That is $$\big( \frac{\partial T_{m}}{\partial p_{n}} \big)' = \sum_{j=1}^{n} \frac{\partial^{2} H}{\partial p_{n} \partial q_{j}} \frac{\partial T_{m}}{\partial p_{j}} - \frac{\partial^{2} H}{\partial p_{n}^{2}} \frac{\partial T_{m}}{\partial q_{n}}. \label{eq:per3.25-2}$$ Comparing $(\ref{eq:per3.25})$ with $(\ref{eq:per3.25-2})$, we have $\dot{p}_{n} = - \frac{\partial T_{m}}{\partial q_{n}}.$ Now we use mathematical induction again. Let us assume $\dot{p}_{j} = - \frac{\partial T_{m}}{\partial q_{j}}$ $(j = n, n-1, \dots, k)$. Then when $j = k - 1$, similarly we obtain \begin{gather} (\dot{p}_{k})' = - \sum_{j=1}^{n} \frac{\partial^{2} H}{\partial q_{k} \partial q_{j}} \frac{\partial T_{m}}{\partial q_{j}} + \frac{\partial^{2} H}{\partial q_{k} \partial p_{n}} \frac{\partial T_{m}}{\partial q_{n}} + \frac{\partial T_{m}}{\partial q_{k-1}}, \label{eq:per3.26} \\ (p'_{k})^{\cdot} = - \sum_{j=1}^{n} \frac{\partial^{2} H}{\partial q_{k} \partial q_{j}} \frac{\partial T_{m}}{\partial q_{j}} + \frac{\partial^{2} H}{\partial q_{k} \partial p_{n}} \frac{\partial T_{m}}{\partial q_{n}} - \dot{p}_{k-1}. \label{eq:per3.27} \end{gather} Using \eqref{eq:per3.17} and $(p'_{k})^{\cdot} = (\dot{p}_{k})'$ we obtain $$\dot{p}_{k-1} = - \frac{\partial T_{m}}{\partial q_{k-1}} \quad (k = n, n-1, \dots, 3, 2) \label{eq:per3.28}$$ which completes the proof. \end{proof} \begin{corollary} \label{coro5.3.10} The flows defined by equation \eqref{eq:per3.16} commute with each other on $S$. \end{corollary} \begin{proof} From Theorem~\ref{thm5.3.6}, the solution operators of $u_{t_{m}} = J \frac{\delta H_{m}}{\delta u}$ commute, and when $n=1$, $H_{1} = H$. Denoting $t_{1}$ by $x$, the solution operators of $u_{t_{n}} = J \frac{\delta H_{n}}{\delta u}$ and $u_{x}= J \frac{\delta H}{\delta u}$ commute. By the invertibility of the general Legendre transformation $( u, u', \dots, u^{(2n-1}) \longrightarrow (q,p),$ we obtain that the solution operators determined by \eqref{eq:per3.16} commute. \end{proof} \begin{theorem} \label{Theorem6.3.13} System $\{T_{i}\}$ defined by $(\ref{eq:per3.10})$ is an involutive system. \end{theorem} \begin{proof} By Corollary \ref{coro5.3.10}, the flows defined by equation \eqref{eq:per3.16} commute. By a theorem given by V.I. Arnold (see \cite[page 211]{Arnold}), two flows commute if and only if the Poisson bracket of their corresponding vector fields is equal to zero. Thus, we obtain $\{ T_{i}, T_{j}\} = 0$. \end{proof} The involutivity of $\{T_{i}\}$ implies the following theorem. \begin{theorem} The Hamiltonian system defined by \eqref{eq:per3.16} is a FIH system in the symplectic space $({\bf R}^{2n}, \omega^{2})$ on $S$. \end{theorem} \begin{proof}[Proof of Theorem \ref{maintheorem}] Since we have $\{T_{i}, H \} = 0$ ($i=0,1,2,\dots$) and $\{ T_{i}, T_{j}\} = 0$ ($i,j=0,1,2, \dots$), the finite dimensional Hamiltonian system defined by \eqref{eq:per3.11} is a FIH system in the symplectic space $({\bf R}^{2n}, \omega^{2})$ on $S$. \end{proof} \begin{remark} \rm The first component $q_{1} = u$ of system \eqref{eq:per3.16} is the solution of the higher order equation $u_{t_{m}} = J \frac{\delta H_{m}}{\delta u}$. When $\Omega = [ \alpha, \alpha + T]$, we can obtain its periodic solution and when $\Omega = ( -\infty, +\infty)$, we can obtain the rapid decreasing solution at infinity. \end{remark} \section{A special example: The AKNS hierarchy} \label{section6.4} By \cite[page 54]{Abl} the $m$th order AKNS equation can be written in the Hamiltonian form $$u_{t_{n}} = J \frac{\delta H_{n}}{\delta u}\,, \label{eq:per4.1}$$ where \begin{align*} u & = ( v, w )^{T}, \\ J & = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},\\ H_{n} & = H_{n}(v,w) = \int_{\Omega}\mu_{n}dx, \\ \mu_{0} & = -vw, \\ \mu_{1} & = -wv_{x}, \\ \mu_{2} & = -wv_{xx} + (wv)^{2}, \\ \mu_{3} & = -wv_{xxx} + 4vw^{2}v_{x} + wv^{2}w_{x}, \\ \mu_{4} & = -wv_{xxxx} + 6vwv_{x}w_{x} + 5w^{2}v_{x}^{2} + 6vw^{2}v_{xx} -2(wv)^{3} + wv^{2}w_{xx}, \\ & \vdots \\ \mu_{n+1} & = w\left( \frac{\mu_{n}}{w}\right)_{x} + \sum_{k=0}^{n-1}\mu_{k}\mu_{n-1-k}. \end{align*} Using integration by parts, we rewrite $H_{n}$ as follows. \begin{align*} H_{0} & = -\int_{\Omega} wvdx, \\ H_{1} & = -\int_{\Omega} wv_{x}dx = \int_{\Omega} L_{1} dx, \\ H_{2} & = \int_{\Omega} [w_{x}v_{x} + (wv)^{2}] dx = \int_{\Omega}L_{2}dx, \\ H_{3} & = \int_{\Omega} [w_{x}v_{xx} + 4w^{2}vv_{x} + wv^{2}w_{x}]dx = \int_{\Omega}L_{3}dx, \\ H_{4} & = \int_{\Omega} [-w_{xx}v_{xx} - w^{2}v_{x}^{2} - v^{2}w_{x}^{2} - 8ww_{x}vv_{x} - 2(vw)^{3}] dx = \int_{\Omega}L_{4}dx, \\ & \vdots \\ H_{n} & = \int_{\Omega} L_{n}(u, u', \dots, u^{(m)})dx, \quad n = 0, 1, 2, \dots \end{align*} where $m = \begin{cases} k & \mbox{if } n = 2k, \\ k + 1 & \mbox{if } n = 2k + 1, \end{cases} \quad (k = 0, 1, 2, \dots).$ \begin{lemma} \label{lem4.1} The Lagrangian $L_{2n} = L_{2n}(u, u', \dots, u^{(n)})$ is non-singular. \end{lemma} \begin{proof} The result of this lemma follows from the following formulas $L_{2n} = (-1)^{n+1}v^{(n)}w^{(n)} + (\mbox{terms with order less than n}),$ and $Q = (-1)^{(n+1)}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$ Thus, $\det Q \neq 0$, and $L_{2n}$ is non-singular. \end{proof} Consider the $2n$th order stationary equation $\sum_{l=0}^{2n} C_{2n-l} J\frac{\delta H_{1}}{\delta u} = J\frac{\delta I}{\delta u} = 0$ or $$\frac{\delta I}{\delta u} = 0 \label{eq:per4.2}$$ where $$I = \sum_{l=0}^{2n}C_{2n-l}H_{l} = \int_{\Omega}\sum_{l=0}^{2n}C_{2n-l}L_{l}dx = \int_{\Omega}L(u, u', \dots, u^{(n)})dx, \label{eq:I}$$ $C_{0} = 1$, and $C_{i}$ are constants. \begin{theorem} \label{thm4.2} Under the reduction of \eqref{eq:per2.2} and \eqref{eq:per2.12}, where $N = 2$, the infinite dimensional integrable AKNS hierarchy $(\ref{eq:per4.1})$ can be transformed into the finite dimensional integrable Hamiltonian system \eqref{eq:per3.16} on $S$. Where $S$ is the solution set of equation \eqref{eq:per4.2}. \end{theorem} For example, let $n = 2$, $C_{i} = 0$ ($i \neq 0$), $C_{0} = 1,$ the equation \eqref{eq:per4.2} has the form \label{eq:per4.3} \begin{aligned} \frac{\delta I}{\delta u} &= \begin{bmatrix} -w_{xxxx} + 6vw_{x}^{2} + 8wvw_{xx} + 4ww_{x}v_{x} + 2w^{2}v_{xx} - 6w^{3}v^{2} \\ -v_{xxxx} + 6wv_{x}^{2} + 8wvv_{xx} + 4vv_{x}w_{x} + 2v^{2}w_{xx} - 6v^{3}w^{2} \end{bmatrix}\\ &= \begin{bmatrix}{c} 0 \\ 0 \end{bmatrix}. \end{aligned} The corresponding Legendre transformation is \begin{align*} q_{1} & = (v,w)^{T} = (q_{11}, q_{21})^{T}, \\ q_{2} & = (v,w)_{x}^{T} = (q_{12}, q_{22})^{T}, \\ p_{1} & = \frac{\partial L_{4}}{\partial u'} - \big( \frac{\partial L_{4}}{\partial u''} \big)' = \begin{bmatrix} -2v_{x}w^{2} - 8vww_{x} + w_{xxx} \\ -2w_{x}v^{2} - 8vwv_{x} + v_{xxx} \end{bmatrix} = \begin{bmatrix} p_{11} \\ p_{21} \end{bmatrix}, \\ p_{2} & = \frac{\partial L_{4}}{\partial u''} = \begin{bmatrix} -w_{xx} \\ -v_{xx} \end{bmatrix} = \begin{bmatrix} p_{12} \\ p_{22} \end{bmatrix}. \end{align*} It can be solved for $u^{(i)}$: \begin{align*} (v,w) & = (q_{11}, q_{21}), \\ (v,w)_{x} & = (q_{12}, q_{22}), \\ (v,w)_{xx} & = -(p_{22}, p_{12}), \\ (v,w)_{xxx} & = (p_{21} + 2q_{11}^{2}q_{22} + 8q_{11}q_{12}q_{21},\ p_{11} + 2q_{21}^{2}q_{12} + 8q_{11}q_{21}q_{22}). \end{align*} The Hamiltonian $H$ corresponding to the Lagrangian $L$ is $H = (q_{12}p_{11} + q_{22}p_{21} - p_{12}p_{22}) + 2(q_{11}q_{21})^{3} + q_{12}^{2}q_{21}^{2} + 8q_{11}q_{12}q_{21}q_{22} + q_{11}^{2}q_{22}^{2}.$ Thus, the Euler-Lagrange equation (\ref{eq:per4.3}) is equivalent to the following classical integrable Hamiltonian equation: $$q'_{j} = \frac{\partial H}{\partial p_{j}}, \quad p'_{j} = - \frac{\partial H}{\partial q_{j}}, \quad j = 1, 2. \label{eq:H}$$ And the involutive first integrals in pair are $T_{i} = - \int \frac{\delta I}{\delta u} J \frac{\delta H_{i}}{\delta u} dx, \quad i = 0, 1, 2, \dots.$ Here the integral constants are zeros. By direct calculations we have: \begin{align*} T_{0} & = q_{11}p_{11} + q_{12}p_{12} - (q_{21}p_{21} + q_{22}p_{22}), \\ T_{1} & = H = q_{12}p_{11} + q_{22}p_{21} - p_{12}p_{22} + 2(q_{11}q_{21})^{3} \\ &\quad + q_{12}^{2}q_{21}^{2} + 8q_{11}q_{12}q_{21}q_{22} + q_{11}^{2}q_{22}^{2}, \\ T_{2} & = - p_{11}p_{22} + p_{21}p_{12} \\ &\quad +2(q_{21}q_{12}^{2}q_{22} - q_{11}q_{22}^{2}q_{12} + q_{11}q_{21}^{2}p_{21} - q_{21}q_{11}^{2}p_{11}) \\ &\quad + 4(-q_{11}q_{12}q_{21}p_{22} + q_{11}q_{22}q_{21}p_{12} ) + 6(q_{11}^{3}q_{21}^{2}q_{22} - q_{21}^{3}q_{11}^{2}q_{12}), \\ T_{3} & = (2q_{21}^{2}q_{12} + 2q_{11}q_{21}q_{22} + p_{11}) (2q_{11}^{2}q_{22} + 2q_{11}q_{21}q_{12} + p_{21}) \\ &\quad - (q_{21}p_{22} + q_{11}p_{12} + q_{12}q_{22} + 3q_{11}^{2}q_{21}^{2})^{3}. \end{align*} Moreover, $$\frac{\partial}{\partial t_{i}} \begin{pmatrix} q_{j} \\ p_{j} \end{pmatrix} = \begin{pmatrix} \frac{\partial T_{i}}{\partial p_{j}} \\ - \frac{\partial T_{i}}{\partial q_{j}} \end{pmatrix} \label{eq:6.4.4}$$ are the constraining AKNS equations on $S$. Let $i=2$ and $i=3$ in \eqref{eq:6.4.4}, we obtain the following systems \begin{align} \label{eq:H2} \frac{\partial}{\partial t_{2}} \begin{bmatrix} q_{11} \\ q_{21} \end{bmatrix} & = \begin{bmatrix} -p_{22} - 2q_{21}q_{11}^{2} \\ p_{12} + 2q_{11}q_{21}^{2} \end{bmatrix}, \\ \frac{\partial}{\partial t_{3}} \begin{bmatrix} q_{11} \\ q_{21} \end{bmatrix} & = \begin{bmatrix} p_{21} + 2q_{22}q_{11}^{2} + 2q_{11}q_{21}q_{12} \\ p_{11} + 2q_{12}q_{21}^{2} + 2q_{11}q_{21}q_{22} \end{bmatrix}. \label{eq:H3} \end{align} Or \begin{gather*} \frac{\partial u}{\partial t_{2}} = \begin{bmatrix} v \\ w \end{bmatrix}_{t_{2}} = \begin{bmatrix} v_{xx} - 2wv^{2} \\ -w_{xx} + 2vw^{2} \end{bmatrix} = J \frac{\delta H_{2}}{\delta u}, \\ \frac{\partial u}{\partial t_{3}} = \begin{bmatrix} v \\ w \end{bmatrix}_{t_{3}} = \begin{bmatrix} v_{xxx} - 6vwv_{x} \\ w_{xxx} - 6vww_{x} \end{bmatrix} = J \frac{\delta H_{3}}{\delta u}. \end{gather*} \begin{remark} \rm Note that systems \eqref{eq:H}, \eqref{eq:H2} and \eqref{eq:H3} are new finite dimensional completely integrable Hamiltonian systems derived from the infinite dimensional integrable AKNS system using our new method. 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