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I have been looking at this question for a while without any progress.

Question. Maximize $$ I[\eta] = \int_0^\infty e^{-s} \Big[\sin\big(\eta(s)\big) + \sin\big(\sqrt{2}\eta(s)\big)\Big]\;ds$$ subjected to $\eta:[0,\infty) \rightarrow \mathbb{R}$ is absolutely continuous with $\eta(0) = 0$ and $\dot{\eta}(s) \in [-1,1]$.

What are the tools (if any) available to study this kind of question (both analytically and numerically)?

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    $\begingroup$ As written, the maximizer is, obviously, $\eta\equiv 0$. Check the statement of the question; I strongly suspect that you meant something else. $\endgroup$
    – fedja
    Commented Jan 6, 2020 at 17:52
  • $\begingroup$ You are right, I meant $\sin$ instead of $\cos$. $\endgroup$
    – Sean
    Commented Jan 9, 2020 at 16:09

1 Answer 1

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The first maximum of $f(s)=\sin s+\sin\sqrt 2 s$ is attained at some $s_0\in [1.2,1.3]$ and exceeds $1.9$. The derivative of $f$ is at most $1+\sqrt 2\le\frac 52$. One obvious strategy is to go at the highest speed $1$ to $s_0$ (i.e., to put $\eta(s)=s$ on $[0,s_0]$) and then stay there forever. Any attempt to go to another maximum will force you to go through $0$ first. Up to that point, say, $S\ge s_0$, you can only lose compared to the obvious strategy, but beyond that point, even if you increase $f$ at the maximal available rate $\frac 52$ and reach the maximal possible value $2$, you'll get only $\int_0^{4/5}\frac 52se^{-s}ds+\int_{4/5}^{\infty}2e^{-s}ds=2\frac{1-e^{-4/5}}{4/5}<1.4$ times $e^{-S}$ compared to at least $1.9e^{-S}$ the obvious strategy yields. So the obvious strategy is the best one here. Of course, this is rather ad hoc and heavily based on the properties of the exponential weight, but it solves the problem as posed, so I'll stop here.

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