Question about the period of specific Hamiltonian flows Consider the Hamiltonian of the form $H(q,p)=p^2/2 + |q|^{\beta}/\beta$ for $\beta\in (1,2)$.
In the case of $\beta=2$, this is simply the harmonic system and we know that all contours, i.e. those of the form $\{(q,p):H(q,p)=E\}$ for fixed $E>0$ are circles and has a circumference of radius $\sqrt{q^2+p^2}=\sqrt{2H(q,p)}$ and the speed of the Hamiltonian flow is $\sqrt{J\nabla H(q,p)}=\sqrt{q^2+p^2}$ for $J=\begin{pmatrix}0 && 1 \\ -1 && 0 \end{pmatrix}$.
So the relation time=distance/speed tells us that the period length is $2\pi$ independent of the starting position.
My question is what happens to the period length of the contours as $|q|$ gets large, or the energy level increases? The speed will be $\sqrt{p^2 + |q|^{2\beta - 2}}$ where $2\beta-2 \in (0,2)$.
So the distance between $q$ for each level contour will increase as $|q|$ increases. I think the period length increases as a result of this but I am not sure about this. I would greatly appreciate any insight into this.
 A: A nice problem. Being in two dimensions, you may use action-angle variables to calculate the period of the motion as a function of energy using a simple area formula (see e.g. V.I.Arnold, Math methods of classical mechanics, either chap 50 or the Problem on page 20, albeit without any solution given).
More precisely, given $E=H_\beta(p,q)>0$ and $|p|\leq \sqrt{2E}$ we have in the present case that $q=\pm f_\beta(p,E)$ with
$$ f_\beta(p,E) = (\beta (E-p^2/2))^{1/\beta} .$$
The area of the set  $\{(p,q): H_\beta(p,q)\leq E\}$ is then given by:
$$ A_\beta(E) =\int_{H_\beta\leq E} dq\wedge dp = \oint_{H_\beta=E} q \; dp = 4\int_0^{\sqrt{2E}} f_\beta(p,E)dp$$
where I used Green's formula and
in the last formula the symmetry of the orbit. Making the substitution $p=\sqrt{2E} \; u$ we obtain:
$$A_\beta(E) = E^{1/2+1/\beta} C_\beta.$$
Here $C_\beta= 4\sqrt{2} \beta^{1/\beta} \int_0^1 \left(1-u^2\right)^{1/\beta}$ only depends upon $\beta$.
The formula given in Arnold's book now states that
the period of the orbit is simply given by the derivative of this area with respect to energy, i.e.:
$$ T_\beta(E) = \frac{\partial A_\beta}{\partial E} =
  E^{1/\beta-1/2} \left( \frac12 + \frac{1}{\beta}\right)
 4\sqrt{2} \beta^{1/\beta} \int_0^1 \left(1-u^2\right)^{1/\beta}.$$
Even if we don't know the explicit value of proportionality,  we may still deduce the behavior of the period when changing the energy.
We see  that $T_2(E)$ is independent of $E$ (as we knew). For $\beta>2$, $T_\beta$ is decreasing with $E$ while for $0<\beta<2$, $T_\beta$ is increasing with $E$. In some cases we may calculate an explicit formulae (possibly needing interpretation as the dynamical system is not regular for some of the $\beta$-values):
$$ T_{1/2}(E) = \frac{2}{3} \sqrt{2E}^3, \ \ \ T_1(E) = 4 \sqrt{2E}, \ \ \ T_2(E)=2\pi, \ \ \ T_\infty(E)= \frac{4}{\sqrt{2E}}.$$
I believe this should answer your question.
