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.



mechanics: convergence to an equilibrium point
Consider the total energy
\begin{equation}
E = x'^2/2 + V(x)
\end{equation}
and assume that $V$ is bounded below and $V(x) \rightarrow \infty$ as $||x||\rightarrow \infty$ 
(i.e., V is radially unbounded). Since
\begin{equation}
E' = -\gamma x'^2 < 0, \quad \forall x' \neq 0,
\end{equation}
it follows from LaSalle's invariance principle that all solutions
tend to the largest invariant set in {$\;(x,x') \;|\; x' = 0\;$}, namely
\begin{equation}
M = \left[\;(x,x') \;|\; x'=0, \nabla V(x) = 0 \; \right].
\end{equation}
If every point in this set is isolated you will have convergence to an 
equilibrium point (which need not be stable). Otherwise, you may have quasi-convergence, meaning
that while every solution approaches $M$, $\lim\_{t\rightarrow\infty} (x'(t),x(t))$ 
may not exist.


(Parts of this answer has of course already been provided above.)


(Also, if $V$ is radially unbounded (and nice), the level sets {$\; (x,x') \; | \; E(x,x') \leq E(x\_{0},x'\_{0})\;$} are compact so the assumption on boundeness below can be replaced by saying, e.g., that $V$ should be continuously differentiable.)


