I have a differential equation which solution is periodic. What can I tell about right-hand side of such equation?

Question

I have equation of form $$ \frac{dx}{dt} = f(x), $$ and know and for some initial value $x_0$ its solution is periodic with unknown period. What can I tell about $f(x)$ apart from non-linearity (or $f(x) \equiv 0$)?

Also, may be I'm wrong but I think that period is somehow related to "non-linearity" of $f(x)$, being longer for "more non-linear" functions. Is there something behind this intuition?

Answer 1

I'm going to assume you mean there exists a periodic function $x(t)$ such that its derivative exists everywhere and equals $f(x(t))$ at time $t$, i.e. $\frac{d}{dt}x(t) = f(x(t))$.

If this is the case, you can conclude that $f(x_0) = 0$. Differentiability implies continuity, and the intermediate value theorem and the fundamental theorem of calculus can give you a quick contradiction.

EDIT: Let $T$ denote the [unknown] first time $x(T) = x_0$ again. Suppose there exists a $t$ such that $x(t) \neq x_0$ and $0 < t < T$. By the intermediate value theorem, there exists times $t_1, t_2$ such that $x(t_1) = x(t_2) = (x(t) + x_0)/2$ and $0 < t_1 < t < t_2 < T$. However, since your dynamics don't depend on time at all, the functions $y(t) = x(t - t_1)$ (A solution to your differential equation starting with initial condition $x(t_1)$) must equal $y'(t) = x(t - t_2)$. This argument, when repeated $T/(t_2 - t_1)$ times, contradicts the periodicity. (There may be a quicker way to prove/state this, but this is what came to mind first.)

Answer 2

If uniqueness of solution holds for the equation $y'=f(y)$, then the only periodic solutions are equilibria.

Suppose that $x(t)$ is a periodic solution of period $T>0$: $x(0)=x(T)$. By Rolle's theorem the is a $\tau\in(0,T)$ such that $x'(\tau)=0$. Then $f(x(\tau))=0$. This implies that $\bar x(t)=x(\tau)$ is a solution and $\bar x(\tau)=x(\tau)$. By uniqueness $x$ and $\bar x$ are the same solution.

I just realized that uniqueness in not needed. It is enough that $f$ be continuos. Suppose that $f(a)=0$, $f(b)=0$ and $f$ has constant sign on $(a,b)$, say $f>0$. Then any solution $x(t)$ with $a<x(0)<b$ is strictly increasing, unless $x(\tau)=b$ for some $\tau>0$. In order for $x$ to be periodic, at some time $x$ must decrease to the value $x(0)$. This is impossible because $x'=f(x)>0$ if $a<x<b$.