# mechanics: convergence to an equilibrium point

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.

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.

• 1) "Bounded from below" isn't enough. You rather need "is greater than the initial value of $E$ near $\infty$. 2) The limit point doesn't need to be a minimum in general. 3) Contrary to your belief, you can oscillate forever. Imagine an infinite road carved into a gentle mountain slope like a trough that spirals into a flat disk-shaped valley. – fedja Dec 29 '11 at 12:30
• I agree with point 1, in order to avoid to a trajectory to escape to infinity: I edit my post. For point 2), that's why I wrote "extremum" and not "minimum". My problem is with your point 3. Once you are in the flat valley, the friction is still operating and your velocity decreases exponentially and you should stop somewhere ! Can you exhibit a concrete example of the behaviour you mention. – Damien S. Dec 29 '11 at 13:18
• 2) A saddle is perfectly possible as well. The right word in English is "a critical point". 3) a) You never reach the valley: the road makes infinitely many loops on the slope. b) OK, I'll post something a bit later. – fedja Dec 29 '11 at 13:39
• Thanks for the saddle point ! in general, we require only $\nabla V=0$. I was making my drawings in 1D and thus I skipped it... For your 3), I am very interested in your example that I still do not understand. What happened if we restrict to $d=1$ ? – Damien S. Dec 29 '11 at 14:05
• OK, I posted the 2D example as an answer. When $d=1$, such effect is impossible, so the statement is true. – fedja Dec 29 '11 at 20:02

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.

(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.)
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.
• $\int |x'|^2dt<+\infty$, $x''$ is bounded. Hence $x'\to 0$. Hence $V$ tends to some limit. That much is always true. On the line, if the limit of $x$ fails to exist, there exists a point from which you depart and go fixed distance in both directions and to which you return infinitely many times with arbitrarily low velocity. Moreover, this point is a (non-strict) local minimum (if not, the potential nearby is less and once the velocity drops low enough, the return is impossible). But you cannot go far from a local minimum if you do not have much kinetic energy. – fedja Dec 29 '11 at 21:00
• In general, you get attracted to some connected closed set where $V$ is constant and $\nabla V=0$ but that's all. – fedja Dec 29 '11 at 21:03