bounded derivate

Question

Given a dynamical system $\dot{x}=f(x)$ with bounded solutions, i.e. $\Vert x\Vert \le c$, with $c<\infty$. Then, $\dot{x}$ is bounded if $f(x)$ is bounded. I wonder if there are less restrictive condition on the function $f(x)$ (conditions that do not rely on the boundedness of $f(\cdot)$) allowing to conclude that $\dot{x}$ is bounded.

Answer 1

First, it's quite easy to give an example where $f$ is unbounded, but all orbits are bounded; the most straightforward one is \begin{align} \dot{x} &= y,\\ \dot{y} &= -x, \end{align} i.e. the harmonic oscillator with unit frequency.

In this case, the boundedness of the orbits is a direct consequence of the fact that the above system is conservative.

For non-conservative (i.e. dissipative) systems, the question 'which orbits are bounded' is tantamount to fully understanding the state space of the system, in particular its attractors. This is in general a highly nontrivial exercise. I would recommend to read the Scholarpedia article on 'Basin of attraction' and the Wikipedia article on 'Attractor'; the lemma at MathWorld also gives some useful information. For a more general take on conservative systems, see the Scholarpedia article on Hamiltonian systems.