mechanics: convergence to an equilibrium point

Question

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.

Answer 1

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.)

Answer 2

OK, here is an explicit construction. Let $\gamma=1$. Consider $V(r,\theta)=[1.1+\sin(\frac 1{r-1}+\theta)]f(r)$ in polar coordinates where, $f(r)=0$ on $[0,1]$, and $f(r)=\exp(-(r-1)^{-1/2})$ for $r\ge 1$. Then $\nabla V\ne 0$ for $r>1$. If you start with velocity $-\nabla V$ in the trough where $r$ is slightly greater than $1$ and $\theta$ is chosen so that $\sin=-1$, you won't ever be able to go over the ridges where $\sin=1$.

The reason is that we can control the quantity $u=x'+\nabla V(x)$ pretty well. Indeed, $|u'+u|<0.00001|x'|$ because the second differential of $V$ is very small for $r$ close to $1$. Let $G(r)=2\frac{f(r)}{(r-1)^2}$. Note that $G$ dominates $|\nabla V|$ and is comparable to it up to a factor of $4$ when $\sin=0$. Hence, $|u'+u|\le 0.1|u|$ whenever $|u|>0.01 G$. Note also that $G$ doesn't change noticeably within a single turn of the trough and it takes at least constant time to accomplish one revolution staying in the trough. Thus, $|u|\le 0.02G$ as long as we follow the trough at all, but as long as $|u|<0.03G$, we cannot even cross the middle of the trough wall $\sin=0$ because $-\nabla V$ looks almost directly towards the bottom of the trough there.