Let $H(p,q) = T(q) + U(p)$ be a Hamiltonian function that defines a Hamiltonian system, i.e., \begin{align} &\frac{dp}{dt} = \frac{\partial H}{\partial q}(p,q) = \frac{dT}{dq},\\ &\frac{dq}{dt} = -\frac{\partial H}{\partial p}(p,q) = -\frac{dU}{dp}. \end{align} If I apply the symplectic Euler method to this system, I will get a discretized system (with step size $\eta$): \begin{align} &p_{n+1} =p_n + \eta \frac{dT}{dq}(q_n),\\ &q_{n+1} =q_n - \eta \frac{d U}{dp}(p_{n+1}). \end{align} It is well known that the symplectic Euler method is a special kind of "Geometric numerical method", and the map $(p_n,q_n) \to (p_{n+1},q_{n+1})$ is a symplectic map. This makes symplectic Euler method behaves better than naive Euler method in several aspects, such as nearly preserving energy in exponentially long time.
My question is, if the original Hamiltonian system defined by $H(p,q)$ has only bounded orbits, i.e., $(p(t),q(t))$ are bounded function of $t$, will the discrete system resulting from symplectic Euler method also has only bounded orbits ?
