About the curvature of solutions of Hamilton's equations

I am a math major and have recently stumbled on the Hamilton's system of equations in the context of Hamiltonian Monte Carlo Markov chains on a continuous state space, say $$\mathbb{R}^d$$. I am trying to understand two things.

1. Are the Hamilton's system of equations analyzed in a compact subset of $$\mathbb{R}^d$$ rather than $$\mathbb{R}^d$$?
2. How do the solutions to the system of equations look like at least qualitatively? (Quantitative analysis would be great). In particular, are they curves for most starting position and momentum variables?

The reason for asking these questions is that in the context of Markov chains, firstly if the system is analyzed for a compact subset of $$\mathbb{R}^d$$, then one likes to know if the solutions are curves that avoid the boundary. This yields faster mixing of Markov chains.

All of this depends on the definition of $$H(\textbf{p},\textbf{q})$$ of course. $$\textbf{p},\textbf{q}$$ are momentum and position respectively. Let $$H(\textbf{p},\textbf{q})=\left(\frac{1}{2\pi}\right)^{d/2}e^{-\frac{\|\textbf{q}\|^2}{2}}+\frac{\|\textbf{p}\|^2}{2}$$. If it is easier explaining with another Hamiltonian, that works too. I know that the Hamiltonian is preserved by the dynamics and the system of equations is

$$\frac{d\textbf{q}}{dt}=\frac{\partial H}{\partial \textbf{p}}\,,\quad \frac{d\textbf{p}}{dt}=-\frac{\partial H}{\partial \textbf{q}}$$

• Apr 14 at 9:23

1. In Hamiltonian mechanics the system of equations is analyzed in phase space which is an even dimensional space since it is the direct product of the two spaces the canonical coordinates $$\boldsymbol{q}$$ and $$\boldsymbol{p}$$ live in. The phase space is not a compact subset of $$\mathbb R^d\,.$$
2. If $$\boldsymbol{q}$$ and $$\boldsymbol{p}$$ are one dimensional then you can nicely visualize the Phase portrait $$\tag{1} t\mapsto \left(\begin{matrix}\boldsymbol{q}(t)\\\boldsymbol{p}(t)\end{matrix}\right)$$ because you just need to plot the vector field $$\left(\begin{matrix}\displaystyle\frac{\partial H}{\partial \boldsymbol{p}}\\\displaystyle-\frac{\partial H}{\partial \boldsymbol{q}}\end{matrix}\right)$$ in a region of $$\mathbb R^2$$. This won't give you closed form solution for (1) but quite often that's enough or allows to guess what the solutions could be.
• Note, however, that if the Hamiltonian is not explicitly a function of time, then the quantity $H$ is automatically a first integral of the equations of motion; and if the level sets of $H$ are closed and bounded (as they often are, including for the OP's example Hamiltonian) then the trajectories are confined to a compact region of phase space. Mar 10 at 18:05
• @MichaelSeifert Thanks, may I know why does the level sets of $H$ being closed and bounded imply that the trajectories are confined to a compact region of phase space? Mar 10 at 20:02
• Heine-Borel theorem. Any closed & bounded set of $\mathbb{R}^n$ is compact. However, I should emphasize that for a given value of $H$, the orbits are confined to a compact region of phase space (namely the level set for that particular value.) One can typically find orbits that are outside this submanifold of phase space simply by changing $H$. Mar 10 at 20:25
• @MichaelSeifert How are the level sets of the Hamiltonian suggested by OP closed and bounded? The potential is repulsive, so the level sets extend to spatial infinity. In 1D this is seen in an example implicit plot of $1 = x^2 + \exp(-y^2)$. Or do I make a mistake here? Mar 10 at 23:10