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I am trying to plot a Floquet number $c(k,g)$ figure, numerically, of the equation $y^{\prime\prime} + k y + g^2 \sin^2(\eta+b) y = 0$, where the prime denote the derivative to $t$, $t = \ln \eta $ and $b$ is a non-zero constant which make $\sin(\eta+b)$ may not be the standard sine or cosine form.

The key point to use the Floquet theorem is that the last term of the equation should be periodic, but $\sin(e^{t})$ is not periodic.

So I want to introduce a new variable $f(t) \sin(e^{t})$ which makes $f^2(t) \sin^2(e^{t})$ periodic and get the floquet number. About the Floquet number plot I can just rescale the parameter $g$ to $g_{\star}= \frac{g}{f(t)}$ and shows the system evolve in $c(k,g_{\star})$ plot.

Is there a function $f(t)$ multiply to $\sin(e^{t})$ can make it periodic?

JieJiang
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  • What is a "function multiply"? Are you just multiplying $f(t)\sin(e^t)$, or are you composing $\sin(e^t)$ with $f$ getting one of $f(\sin(e^t))$ or $\sin(e^{f(t)})$? Also, if former, does the trivial function $f(t)=0$ count? –  Oct 01 '19 at 13:54
  • Oh sorry, I mean I want a nontrivial function $f(t)$ to make $f(t)\sin(e^{t})$ periodic. – JieJiang Oct 01 '19 at 14:01
  • Welcome to stackexchange. I suspect the answer to your question is that there is no (nice) function $f$ that does the job. If you [edit] the question to tell us why you are looking for such a function perhaps we can help. – Ethan Bolker Oct 01 '19 at 14:12

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You want a continuous function $f(t)$ such that $f(t)$ is not identically 0, and such that $f(t)\sin(e^t)$ is periodic.

Unfortunately such a function does not exist, which I will now prove.

First, it will be shown that the spaces between zeros of $\sin(e^t)$ can be arbitrarily small. Let $t_1=\ln(\pi k)$ and $t_2=\ln(\pi k+\pi))$ for some $k\in\mathbb{N}$. Then, $t_1$ and $t_2$ are both zeros of $\sin(e^t)$ and $$\begin{align}t_2-t_1&=\ln(\pi k+\pi)-\ln(\pi k)\\ &=\ln\left(\frac{\pi k+\pi}{\pi k}\right)\\ \end{align}$$ which approaches $0$ as $k\to\infty$.

So the set of zeros of $f(t)\sin(e^t)$, if the function is periodic, must be dense. A continuous real function is completely determined by its values on a dense set, so $f(t)\sin(e^t)$ must be identically $0$.

This means $f(t)$ must equal $0$ wherever $\sin(e^t)\neq0$. But $\sin(e^t)\neq0$ has countably many zeros, and so $f(t)$ must be $0$ almost everywhere; this, along with its continuity, imply $f(t)$ is identically $0$.

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You want a function $f(t)$ such that $f(t)$ is not identically 0, and such that $f(t)\sin(e^t)$ is periodic.

How about $$f(t)=\begin{cases}1,&\sin(e^t)=0\\ 0,&\text{otherwise}\end{cases}$$

Then $f(t)\sin(e^t)=0$ for all $t$ and is therefore periodic.

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