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Suppose $a \in \mathbb{R}$ and $g\in C(\mathbb{R})$ is a nonnegative periodic function $g(t+1)=g(t)$. Find conditions on $a,g$ such that the linear inhomogeneous equation $$\dot{x}=ax+g(t)$$ has a periodic solution. When is this solution unique?

I can also use Poincare map

I should use the fact that the solution of the inhomogeneous equation $$\dot{x} = a(t)x +g(t)$$ is given by $$x(t) = x_{0}A(t,t_{0})+\int_{t_{0}}^{t} A(t,s)g(s)ds,$$ where $$A(t,s) = e^{\int_{s}^{t}a(s)ds}$$.+

So my idea was to show that $A(t,s) \neq 0$ but I am not sure if that´s right path at all. I would appreciate if someone could give me a suggestion or help to formulate the solution mathematically correct?

I´ve seen this solution already Find periodic solution of differential equation but there is no mention of $a$ being a constant and $g$ a periodic function or how does that relate to uniqness.

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  • $\begingroup$ How about this:$\frac{d x}{dt}-ax=g(t)$ It's integrating factor is $e^{-at}$, so $x=e^{at} \int e^{-at} g(t) dt+ Ce^{at}$ $\endgroup$
    – Z Ahmed
    Apr 2, 2020 at 17:33
  • $\begingroup$ And how from there folows that the solution is period and which are the conditions for a and g in that case? $\endgroup$ Apr 2, 2020 at 17:53

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Insert the periodicity condition into the solution formula $$ x(t)=e^{at} \int_{t_0}^t e^{-as} g(s) ds+ Ce^{at} $$ that is, $$ x(t)=x(t+1)=e^{at+a} \int_{t_0}^{t+1} e^{-as} g(s) ds+ Ce^{at+a} $$ so that comparing both sides \begin{align} C(e^a-1)&=\int_{t_0}^t e^{-as} g(s) ds - e^a\int_{t_0}^{t+1} e^{-as} g(s) ds \\ &=\int_{t_0}^t e^{-as} g(s) ds-\int_{t_0-1}^{t} e^{-as} g(s+1) ds \\ &=-\int_{t_0-1}^{t_0} e^{-as} g(s+1) ds \end{align} Now the question reduces to the computability of $C$ and its uniqueness.

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  • $\begingroup$ Thank you! Can you help me further by explaining how should $a$ and $g$ be explicit conditioned at the end to get the periodic solution? $\endgroup$ Apr 2, 2020 at 18:38
  • $\begingroup$ Obviously you want $a\ne 0$, and if $a=0$, you have to consider the case that the right side is zero. $\endgroup$ Apr 2, 2020 at 18:44
  • $\begingroup$ And for function $g$ there are no special conditions but what we already assumed? $\endgroup$ Apr 2, 2020 at 18:49
  • $\begingroup$ Yes. $g$ gives the numerator, so that there is nothing happening except in the case where you solve $C\cdot 0=0$. $\endgroup$ Apr 2, 2020 at 19:29

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