# Why is the motion described by $F = -kx^{2n+1}$ periodic?

I understand the motion when:

$$F=-kx$$

which leads to the an ODE that then has solutions expressed as:

$$x(t)=A\cos(\omega t)+B\sin(\omega t)$$

This is periodic since a cosine function is periodic. However when:

$$F = -k x^{2n+1}$$

the differential equation is no longer a simple homogeneous linear differential equation. So how would one prove this is periodic? Any insight would be helpful.

• The answer to your question completely depends on the value of $n$; clearly when $n=0$ the motion described by this force is periodic – nluigi Jan 25 at 7:52
• Hint: Try solving the differential equation $m\ddot x=−kx^{2n+1}$ numerically and see what you get (i.e., if the motion is actually periodic or not for various $n \in \mathbb Z$). Another interesting route might be to solve Hamilton's equations of motion for the Lagrangian $L = \frac{1}{2}m{\dot x}^2 - \frac{1}{2(n + 1)}kx^{2(n + 1)}$ – GodotMisogi Jan 25 at 10:52

First, the force does not depend explicitly on time ($$\partial F/\partial t=0$$), so the energy is conserved, i.e., the potential energy transforms into the kinetic energy and vice versa. Second, the force "returns" the particle, or attracts it, and its potential is some sort of a symmetric potential well: $$V(x)\propto x^{2n+2}$$. Thus, the particle will move from one "wall" to the "opposite one", keeping its energy constant. Thist is a periodic, although not a harmonic, motion.

• Would it be correct to then generalize as when d/dt(Force)=0 that energy is conserved. Also if its periodic, would this mean a period T is solvable and if so how would one approach such a problem? Thanks for the help. – Rick Jan 25 at 7:44
• In fact, if the potential $V$ does not depend on time explicitly, the total energy is conserved; this is the exact formulation. Your $V(x)$ depends on time, but via $x(t)$, i.e., implicitly. It is OK. Next, if your particle has the total energy $E$, you can find the "turning points" $x_1$ and $x_2$, where $E=V(x)$ and particle "reflects" from the "wall". As the energy $E=m\dot{x}^2/2+V(x)$ is constant, you resolve this relationship to find $\dot{x}$ via $E-V(x)$. From this you find $dt\propto dx/\sqrt{E-V(x)}$. Integrating it over x between the turning points gives the half period $T/2$. – Vladimir Kalitvianski Jan 25 at 8:00

Since the energy is conserved, the kinetic energy have an upper bound. Also, for this reason and for the fact that the potential energy increases with the distance in this case (central potential), the potential energy has an upper bound. Therefore, energy conservation and the fact that the potential is a central potential, mandates that the motion of the system in the phase space is limited to a finite region. In this case only two things may happen:

1) The motion is periodic with a finite period T;

2) The motion of the system is chaotic (period T=$$\infty$$);

Since the system is 1-dimensional, the 2nd possibility is excluded, because a chaotic behavior can emerge only in 3 or more dimension (he Poincaré–Bendixson theorem).

In other words, the system is not complex enough to have a nonperiodic behavior.

• It is an interesting question whether one can have a noqperiodic behavior in the case of a central nonlinear potential in 3 dimensions... – sintetico Jan 25 at 8:06
• Woud you have the proof of this theorem : periodic or chaotic. If it is correct, the result apply to any potential well in one dimension : always periodic. It is rather intuitive but not evident ? – Vincent Fraticelli Jan 25 at 12:47
• Well, not to any potential, but only to central potential such that the potential energy increases with the distance from a fixed point – sintetico Jan 28 at 3:26
• It's what I wanted to say. – Vincent Fraticelli Jan 28 at 6:27

@VladimirKalitvianski's answer is correct. To build intuition, it is helpful to look at the problem using phase space. For this problem, phase space consists of $$x$$ and $$v=\dot{x}$$. In general, phase space is the minimum number of variables at a given time needed to predict its future behavior.

Now, as @VladimirKalitvianski said, energy is conserved here. To see why, just apply the work-energy theorem (multiply both sides of $$F=ma$$ by $$v$$, then integrate with respect to $$t$$). You'll find that the energy is given by $$E = \frac{1}{2}mv^2 + \frac{k}{2n+2}x^{2n+2}.$$

Because there are only two variables, we can represent phase space as a simple graph, and the state of the system is a point on that graph. As a system evolves with time it will move along a continuous curve through this plane.

Now, we can draw all of the points on the graph that have the same energy by just graphing $$v= \pm\sqrt{\frac{2E}{m} - \frac{2k}{4n+4}x^{2n+2}}$$ for different values of $$E$$. Notice that these curves are all closed curves. Because energy is conserved, any point on one of these curves must stay on that curve, so any point with a given energy will be stuck on a path that leads back to its current position.