Discussion
If the Hamiltonian of the dynamical system is given by \(H = pq\), then as \(t \to \infty\).
Answer: A
As \(t \to \infty\), \(q \to \infty\) and \(p \to 0\).
Hamilton's equations for \(H = pq\):
\[ \begin{aligned} \dot{q} &= \frac{\partial H}{\partial p} = q \\ \dot{p} &= -\frac{\partial H}{\partial q} = -p \end{aligned} \]
So \(q = q_{0}e^{t}\) and \(p = p_{0}e^{-t}\). For a non-zero starting \(q_{0}\) (positive, as the option assumes), \(q\) grows without bound, while \(p\) decays to 0. The product \(pq = H\) stays constant, as energy conservation requires.
Both tending to 0 or both to \(\infty\) would change \(pq\), and the last option reverses the two equations.
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