Let $I\subset\mathbb{R}$ and $U\subset\mathbb{R}^n$ be open, and let $D\subset I\times U$ be a simply connected [open set](/page/Open%20Set). Let $\Omega\subset I\times U\times\mathbb{R}^n$ be open, and let $L:\Omega\to\mathbb{R}$ be $C^2$. Assume that the Legendre transform is defined on the velocities used below: for a $C^1$ velocity field $q:D\to\mathbb{R}^n$, suppose $(x,y,q(x,y))\in\Omega$ for every $(x,y)\in D$, define the momentum field

\begin{align*} P(x,y):=\partial_vL(x,y,q(x,y)), \end{align*}

and assume that the Hamiltonian $H$ is defined on every $(x,y,P(x,y))$ by

\begin{align*} H(x,y,P(x,y)):=P(x,y)\cdot q(x,y)-L(x,y,q(x,y)). \end{align*}

Define the Hilbert one-form on $D$ by

\begin{align*} \omega:=L(x,y,q(x,y))\,dx+P(x,y)\cdot(dy-q(x,y)\,dx). \end{align*}

Equivalently,

\begin{align*} \omega=-H(x,y,P(x,y))\,dx+P(x,y)\cdot dy. \end{align*}

Then $d\omega=0$ if and only if there exists a $C^2$ function $S:D\to\mathbb{R}$ such that

\begin{align*} \nabla_yS(x,y)=P(x,y) \end{align*}

and

\begin{align*} \partial_xS(x,y)+H(x,y,\nabla_yS(x,y))=0 \end{align*}

for every $(x,y)\in D$.