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

Question

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.

Thanks in advance!

Answer 1

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 $T = \frac{2\pi}{3} n$.

Answer 2

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 $A=0$.

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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 $T=2\pi/3$.

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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 $t=0$.