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Suppose $f$ is a convex function with the property that level sets of $f$ are compact. I know that any solution of the differential equation $$\dot{x}=-\nabla f(x)$$ is bounded because $(d/dt) f(x) = - ||\nabla f(x)||^2$ so that $x_t$ always stays within the level set $\{ u ~| f(u) \leq f(x_0)\}$.

My question is: suppose I have instead $$m \ddot{x} + c \dot{x} = - \nabla f(x),$$ where $m > 0$ and $c \geq 0$. It seems natural to guess that this will stay bounded as well. Is that true?

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Define the Hamiltonian function: $$ H(x, \dot{x}) = \frac{1}{2}\,m\, {|\dot{x}|}^2 \;+\; f(x)\, . $$

Then: \begin{align} \frac{d}{dt}H(x, \dot{x}) &= m\, \dot{x}\cdot\ddot{x} \;+\; \dot{x}\cdot\nabla f(x)\\[0.05in] &= \dot{x}\cdot\left[m\, \ddot{x} \;+\; \nabla f(x)\right]\\[0.05in] &=\dot{x}\cdot\left[-c\,\dot{x} \;-\; \nabla f(x)\;+\; \nabla f(x)\right]\\[0.05in] &= -c\, {|\dot{x}|}^2 \end{align} Thus $x(t)$ is bounded in the same sense as in your original example, except with $H$ in place of $f$.

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