# What are the possible periods of the solutions to this ODE?

Consider the non-autonomous equation $$\ddot x + x = \epsilon f(x,\dot x) + \epsilon \sin(3t)$$ Assume that this equation has (a) periodic solution(s). What are the possible periods?

I need to assume that this equation has (a) period solution(s), e.g., solutions $$x = \phi(t)$$ for which $$\phi(t + T) = \phi(t)$$ for some $$T$$ and I need to find for which $$T$$ this is possible. As a start I entered $$\ddot x + x = \epsilon\sin(3t)$$ ($$f(x,\dot x) = 0$$) in Wolfram for different $$\epsilon$$ and it seems as if $$\epsilon$$ doesn't have an impact on the period of the solution for this equation. I therefore think that the possible periods depend on the function $$f(x,\dot x)$$, but I'm not really sure which options for $$f(x,\dot x)$$ I should consider. Furthermore, I think that I should do something with $$\sin(3t)$$ since I think this term will definitely have an impact on the possible periods.

My Question: I really have no clue how I should solve this problem, and I'm not necessarily looking for the direct solution. I would like to know where to start and which questions I should ask so that I can find the solution by myself.

• Try to start with this reasoning. Suppose that $x = \phi(t)$ is a periodic solution with period $T$. If you plug it in the original equation, you'll get $\phi''(t) + \phi(t) \equiv \epsilon f(\phi(t), \phi'(t)) + \epsilon \sin 3t$. Obviously, if $\phi(t)$ is periodic, its derivatives are periodic too. Why not check this expression for $t = 0$ and $t = T$? Apr 13, 2019 at 11:28

I'll follow the hint that I've left in the comments. Suppose that $$x = \phi(t)$$ is a periodic solution of an equation $$\ddot{x} + x = \epsilon f(x, \dot{x}) + \epsilon \sin{3t}.$$ Let $$\phi(t)$$ be $$T$$-periodic, $$\phi(t) \equiv \phi(t + T)$$. Since $$\phi(t)$$ is a periodic function, its derivatives and compositions are also periodic; i.e., $$\dot{\phi}(t)$$, $$\ddot{\phi}(t)$$, $$f(\phi(t), \dot{\phi}(t))$$ are all periodic functions. So, the following $$\ddot{\phi}(t) + \phi(t) \equiv \epsilon f(\phi(t), \dot{\phi}(t)) + \epsilon \sin{3t}$$ holds for all $$t$$, thus this $$\ddot{\phi}(t+T) + \phi(t+T) \equiv \epsilon f(\phi(t+T), \dot{\phi}(t+T)) + \epsilon \sin{3(t+T)}$$ also holds for any $$t$$. Subtracting the first from the second and taking into account the periodicity of $$\phi(t)$$, we get $$\sin{3(t+T)} \equiv \sin{3t},$$ thus $$T = \frac{2\pi}{3} k,\; k\in \mathbb{N}$$. This essentially means that the only possible periods are multiples of the forcing period.

Of course, the periods that can be really observed in such equation strongly depend on the $$f(x, \dot{x})$$. If $$f(x, \dot{x}) = x$$, you can have $$T$$-periodic solution for any $$T = \frac{2\pi}{3} n$$.

For $$f=0$$ you should get $$x(t)=A\cos(t+\phi_0)-\fracϵ8\sin(3t).$$ This has period $$T=2\pi$$ for $$A\ne 0$$ and period $$T=2\pi/3$$ for $$A=0$$.

Next consider $$f(x)=x$$. Then the solution changes to $$x(t)=A\cos(\sqrt{1-ϵ} \,t+\phi_0)-\fracϵ{8+ϵ}\sin(3t).$$ Now the frequencies $$3$$ and $$\sqrt{1-ϵ}$$ are in general not commensurable, there is no minimal frequency and thus no period except for $$A=0$$ where again the frequency is $$3$$, the period $$T=2\pi/3$$.

As a non-linear example take the Van der Pol oscillator with the added forcing term per the task, that is, set $$f(x,\dot x)=-(1-x^2)\dot x$$. Then it is known that the unforced equation $$\ddot x+ϵ(1-x^2)\dot x+x=0$$ has a source at the origin and limit cycle close to the circle with radius $$r=2$$ and period of about $$T=2\pi(1+\frac{ϵ^2}{16})$$. Again the frequencies for the "large" solution are not commensurable, which blurs the trajectories around the unforced limit circle, and can further lead to temporary near-resonance moving the trajectory dramatically away from that limit cycle. Again there is an $$O(ϵ)$$ size solution around $$x(t)\approx-\fracϵ8\sin(3t)$$ with period $$T=2\pi/3$$ from the forcing blurring the center.

This is demonstrated in the following picture for $$ϵ=0.5$$ (that large for visual effect), left the small limit cycle, the dot indicating the $$t=0$$ position. Right the almost periodic solutions close to $$r=2$$ are traced. A BVP solver was used to find a nearly periodic solution close to some point on the circle of radius $$2$$, which was then continued as IVP. The dots on the graphs indicate the gap to from the position at $$t=2\pi$$ to $$t=0$$.

• What is your conclusion then? This question can be treated analytically, without using numerics, so what exactly do you want to say? Apr 14, 2019 at 17:57
• The first two cases should answer that. You always get a periodic orbit $x(t)=-\fracϵ8\sin(3t)+O(ϵ^2)$ around the origin, and only in exceptional cases will you get periodic orbits of period $2\pi$ inherited from the unperturbed problem. All of this under the assumption that $f(0,0)=0$, else there might be an additional $O(ϵ)$ constant term involved. Apr 14, 2019 at 18:29
• Not true. Even if you consider $f(x, \dot{x}) = x$, you have a much wider variety of choice for period. Pick any $\epsilon$ such that $\sqrt{1-\epsilon}$ is $\frac{m}{n}$: forcing has $\frac{2\pi}{3}$ as a period, and a cosine term in your solution is $\frac{2\pi n}{m}$-periodic. A resulting period won't be a simply $2\pi$ or $\frac{2\pi}{3}$. Apr 14, 2019 at 18:37
• $ϵ$ is some small undefined perturbation parameter. Usually in these questions it is only interesting what holds asymptotically. There is no single finite period from that pattern that is present asymptotically, a countable amount of pairs $(m,n)$ is still smaller than the continuum. // The usual interpretation of "in general" allows to exclude sets of measure zero. Apr 14, 2019 at 20:16
• I read the question as it is written. Nothing has been said about how small $\epsilon$ is. Nothing has been said about the nature of non-linearity. And also nothing has been said about asymptotics. The only question is: suppose that an equation of such form has a periodic solution, what its period could possibly be? Apr 14, 2019 at 20:39