Equilibrium — Question 5

Question 5

The force slope distinguishes restoring and destabilizing equilibria

A particle constrained to the $x$ axis has potential energy $U(x)=a(x^2-b^2)^2$, with $a,b>0$. No other force acts along the axis.

1. Find all equilibrium positions.

2. Classify their response to a small displacement using the local force.

Question 5 - Solution

The force slope distinguishes restoring and destabilizing equilibria

(a) Derivation and calculation

Differentiate the potential: $$F(x)=-U'(x)=-4ax(x^2-b^2).$$ Equilibrium requires $F=0$, hence $$\boxed{x_*=0,\quad x_*=+b,\quad x_*=-b}.$$

(b) Result and interpretation

For a small displacement $\xi=x-x_*$, $$F(x_*+\xi)\simeq-U''(x_*)\xi,\qquad U''(x)=12ax^2-4ab^2.$$ At the origin, $$U''(0)=-4ab^2,\quad F\simeq+4ab^2\xi\quad\Rightarrow\quad\boxed{\text{unstable}}.$$ At either outer point, $$U''(\pm b)=8ab^2,\quad F\simeq-8ab^2\xi\quad\Rightarrow\quad\boxed{\text{stable}}.$$

Physical check

The outer equilibria are potential minima; the origin is a local maximum.