For one-dimensional conservative motion, the potential-energy function $U(x)$ contains the same force information as the force law.

For a small displacement $dx$, the work done by a conservative force is

$$dW=F(x),dx,$$

while the associated potential-energy change satisfies

$$dU=-dW.$$

Therefore

$$dU=-F(x),dx,$$

so

$$\boxed{F(x)=-\frac{dU}{dx}}.$$

The force points toward decreasing potential energy. A steep potential corresponds to a large force magnitude, while a locally flat potential corresponds to a small force.

Equilibrium points

An equilibrium position $x_0$ has zero net force, so

$$F(x_0)=0.$$

For a conservative one-dimensional system this means

$$\boxed{U'(x_0)=0}.$$

The shape of $U$ near the stationary point determines the local stability.

• At a stable equilibrium, a small displacement produces a restoring force back toward equilibrium. A nondegenerate local minimum satisfies $$U'(x_0)=0,\qquad U''(x_0)>0.$$
• At an unstable equilibrium, a small displacement produces a force farther away. A nondegenerate local maximum satisfies $$U'(x_0)=0,\qquad U''(x_0)<0.$$
• If the potential is locally flat over a range, the system can have neutral equilibrium: small displacements do not generate a restoring or destabilizing force.

The second-derivative test is useful for nondegenerate stationary points, but if $U''(x_0)=0$ the higher-order shape must be examined.

Worked example

Consider

$$U(x)=ax^4-bx^2,$$

with $a>0$ and $b>0$. The force is

$$F(x)=-\frac{dU}{dx}=-(4ax^3-2bx)=2bx-4ax^3.$$

Equilibria satisfy

$$U'(x)=4ax^3-2bx=0,$$

so

$$x=0$$

or

$$x=\pm\sqrt{\frac{b}{2a}}.$$

The curvature is

$$U''(x)=12ax^2-2b.$$

At $x=0$,

$$U''(0)=-2b<0,$$

so the origin is unstable. At

$$x=\pm\sqrt{\frac{b}{2a}},$$

$$U''=4b>0,$$

so those two equilibria are stable.

A potential-energy function therefore does more than store energy values: its slope gives the force, and its local shape reveals equilibrium and stability.