Periodic Solution of inhomogeneous ODE

Question

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.

Answer 1

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.