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Suppose that $\phi$ is a solution to the ODE \begin{align} x' = f(x) + \sin(t) \end{align} What condition can we put on $f$ to ensure that $\phi$ is periodic of period $2\pi$? That is, what do we do to $f$ so that $\phi(0) = \phi(2\pi).

Okay, now for my approach. Please tell me if I've made an error I'm pretty new to this stuff. So, I want to make $f$ such that... (and there might be something much, much more specific and better)

Denote by $x(t,\alpha)$ the solution of the ODE, $x(t)$ such that $x(0) = \alpha$. Now, consider the function \begin{align} g(\alpha) = x(2\pi,\alpha). \end{align} where $\alpha \in [a,b] \subset [-1,1].$

Now, here's my "condition" on $f$: Choose $f$ so that $f$ is defined on the interval $[a,b]$ such that $[-1,1] \subset [a,b]$. Let $f$ have the condition that if $x > 1, x'<0$ (This means that for $x > 1$, then $f(x) < -1$. Similarly if $x < -1, f(x) > 1$, and thus $x' > 0$.

Therefore, solutions that start in $[a,b]$ remain in $[a,b]$. Now, this means that \begin{align} g: [a,b] \rightarrow [a,b], \end{align} and we can use Brouwer's fixed point theorem to guarantee that there exists $\alpha^{*}$ such that $g(\alpha^{*}) = \alpha^{*}$, and thus the solution $x(t,\alpha^{*})$ is $2\pi$ periodic.

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I think you need to revise the statement. You first say that $\phi$ is a solution with period $2\pi$ and then you say what conditions can you put on $f$ so that $\phi$ has period $2\pi$. – Spock Jan 15 '14 at 0:08
I mean, I think that what it's saying is that, "$f$ has been chosen so that this is the case, what did we do to f?" Should I still revise? – DRich Jan 15 '14 at 0:10
It'd be better plus there's no harm in doing that. :-) – Spock Jan 15 '14 at 0:12
@DRich I think you mean to stipulate that $f(x) > 1$ if $x < -1$; is that correct? If so, you should fix the typo. – A Blumenthal Jan 15 '14 at 5:33
Yes, that is what I mean. I'll revise. Thanks! – DRich Jan 15 '14 at 17:45

Your idea is correct. This is somewhat simpler and more general. Choose any interval $[a,b]$ and $f\colon[a,b]\to\mathbb{R}$ (smooth enough, for instance Lipschitz) such that $f(a)\ge1$ and $f(b)\le-1$. Let $x_1(t)=a$, $x_2(t)=b$. Then $$ x_1'(t)\le f(x_1(t))+\sin t\text{ and }x_2'(t)\ge f(x_2(t))+\sin t, $$ that is, $x_1$ is a sub solution and $x_2$ a super solution. Then for any $\alpha\in[a,b]$ $$ a\le x(t,\alpha)\le b,\quad 0\le t\le2\,\pi. $$ Then $g([a,b])\subset[a,b]$. Also $g$ is continuous, and $g$ has a fixed poit.

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