Hello,

this is a math forum, I know, but my question is about classical mechanics. I am looking for a general (but simple proof) of the very intuitive idea physicists have about the following problem.

We consider a particle in $\mathbb{R}^d$ evolving in a potential $V$ and with a friction coefficient $\gamma$. The differential equation is thus $$ x''= -\nabla V(x) - \gamma x' $$ I assume that the potential is as smooth as we want and is bounded from below. Edit: I also assume that V is "large enough" at $\pm\infty$: there exists $R$ such that there exists $x_{-} < R$ and $x_+>R$ such that $V(x_\pm)>E_0$ where $E_0=x'(0)^2/2+V(x_0)$ is the initial energy. In this case, the particle cannot go beyond these points.

The intuition says that the particle will stop in an extremum of V (that depends on the initial condition). How do we actually prove it ?

It is easy to see that, if it stops, it is necessarily an extremum of V. My question is more about the fact that it stops...

I would like a proof that does not require any abstract ideas as lagrangians, so that it can presented to first or second year students. There are probably multiple references but I do not know any.

Thank you in advance. Damien.

EDIT: of course, it is easy to prove that the mechanical energy $E=x'^2/2+V(x)$ is decreasing and bounded from below, and thus converges; but coming back to x and x' doesn't look so easy.