If the differential equation $\frac{d^2x}{dt^2} = f\left(x, \frac{dx}{dt}\right)$ has no constant solutions, it has no periodic solutions either

Question

I want to prove the following statement:

If the differential equation $\frac{d^2x}{dt^2} = f\left(x, \frac{dx}{dt}\right)$ has no constant solutions, it has no periodic solutions either.

Having no constant solutions to me means that we don't have $f(x, 0) \equiv 0$ for any $x$, but I don't know how to then turn this into insight on periodic functions.

Answer 1

A periodic solution $x(t)$ results in a closed curve $(x(t),\dot x(t))$ in the phase space. The index of the vector field along that curve, its winding number, is $\pm1$. This is an invariant under homotopic deformations of the curve, as long as the homotopy remains in the region where the vector field is non-zero. Now contract the periodic solution to a point inside its enclosed region. The winding number along a point is always zero, so the contraction has to cross a zero of the vector field.

---

As another possibility, consider the vector field obtained by rotation by $90^\circ$ so that it points inward along the periodic cycle. Then the solution for the rotated vector field starting at the periodic solution stay always inside, so either converge to another limit cycle or a zero of the vector field. In the first case, we have the same situation as at the start, only for a smaller region.