Im trying to solve the following problem.
Consider $\mathbb{R}^{2}$ with coordinates $(x,y)$. Let $H$ be a smooth function on $\mathbb{R}^{2}$. Also, consider the Hamilton equations: $$\dfrac{dx}{dt}=\dfrac{\partial H}{\partial y}, \qquad \dfrac{dy}{dt}=-\dfrac{\partial H}{\partial x}.$$ The Liouville $1-$form is $\Theta=y\,dx$. If $(x,y)$ lies on an integral curve $\Gamma^{+}$, we call the action to be the number $$I(x,y)=\int_{\Gamma^{+}}\Theta.$$
$1.$ What is the relation between the action and the area of the region inside $\Gamma$?
$2.$ Is possible to express the period of $\Gamma$ as a line integral over $\Gamma$? Such integral is always improper?
For the first part I used Stokes' theorem as follows: If $\Sigma$ is the region on the phase plane, whose boundary is $\Gamma$, and $\Sigma$ is oriented in such way that the induced orientation on $\Gamma$ is positive, we have $$I(x,y)=\int_{\Gamma^{+}}\Theta=\int_{\partial \Sigma^{+}}y\,dx=\int_{\Sigma^{+}}d(y\,dx)=\int_{\Sigma^{+}}dy\wedge dx=-\int_{\Sigma^{+}}dx\wedge dy=-\text{area}(\Sigma).$$
For the second part, we know from Hamilton equations that $dt=\left(\dfrac{\partial H}{\partial y}\right)^{-1}dx$. Now, let E be a value of Energy and $\Gamma$ a curve of level $E$ on the phase plane. Then the period $T$ of $\Gamma$ is: $$T=\int_{\Gamma^{+}}dt=\int_{\Gamma^{+}}\left(\dfrac{\partial H}{\partial y}\right)^{-1}dx.$$ Of course, the last integral is always improper, because that $1-$form is not defined on the whole plane, just on an open subset not containing any periodic orbit.
I would like to know if there is another $1-$ form defined on the whole plane, whose integral is the period.
I have the next idea but I'm not sure. For a curve $\Gamma$ of level $E$, define $\tau=\dfrac{dI}{E}$. So, $$T=\int_{\Gamma^{+}}\tau.$$ This is suggested by a problem in Arnold's book, Mathtematical Methods of Classical Mechanics, where a problem is to show that $T=\dfrac{dI}{dE}$.
Thank's!
I think it's easier to see the nice solution when you forget about your coordinates for a while and just look at it from an abstract perspective. Let $X = H_y e_x - H_x e_y$ so that the system is the flow of $X$, i.e. $$\gamma'(t) = \frac{dx}{dt}e_x + \frac{dy}{dt}e_y = X(\gamma(t)).$$ Then we are just looking for a one-form $\vartheta$ such that $\vartheta(X) = 1$. This is clearly never possible when $X=0$, but where $X \ne 0$ (i.e. away from critical points of $H$) we can choose $\vartheta = |X|^{-2} X^\flat$ so that $$\vartheta(X) = |X|^{-2} \langle X,X \rangle = 1.$$ For our $X$ this is
$$ \vartheta = \frac1{H_x^2 + H_y^2}\left(H_y dx - H_x dy\right).$$