Doubt: Arnold's "Mathematical Methods of Classical Mechanics", pg. 18

Question

I am having trouble understanding the following diagram from pg. 18 of Arnold's book:

I am unable to see why local maxima of the potential energy correspond to unstable equilibria (and, reciprocally, why local minima correspond to stable equilibria).

My reasoning leads me to the opposite conclusion - for a local maxima $x_M$, $f'(x) > 0$ for $x < x_M$, so that things should increase to $x_M$, whereas $f'(x) < 0$ for $x > x_M$, so that things should decrease to $x_M$. However, I am being imprecise when I say "things", since I am not sure whether what I refer to is the same "thing" which is considered in the phase portrait.

Answer 1

Force is equal to minus the potential gradient, ie $F=-\dfrac {dU}{dx}$.

Local minimum at $x_{\rm min}$

For $x<x_{\rm min}$, $\dfrac {dU}{dx}<0$, ie $F>0$ ie in positive x-direction

For $x>x_{\rm min}$, $\dfrac {dU}{dx}>0$, ie $F<0$ ie in negative x-direction

Thus a small displacement from $x_{\rm min}$ will result in a restoring force back to $x_{\rm min}$, ie stable equilibrium.

Local maximum at $x_{\rm max}$

For $x<x_{\rm max}$, $\dfrac {dU}{dx}>0$, ie $F<0$ ie in negative x-direction

For $x>x_{\rm max}$, $\dfrac {dU}{dx}<0$, ie $F>0$ ie in positive x-direction

Thus a small displacement from $x_{\rm max}$ will result in a restoring force away from $x_{\rm max}$, ie unstable equilibrium.

In terms of the phase diagram just consider what happens if a little more energy is added to the system.
Inside the thick black line "contour" the trajectory will still be around the position of the local minimum.
At the thick black line "contour" the trajectory will no longer be just around one of the local minima.