Let $a,b: \mathbb{R} \to \mathbb{R}$ be periodic continuous functions with period $T$. Consider the following differential equation:
$y' = a(x)y + b(x)$
Let $u: \mathbb{R} \to \mathbb{R}$ be a solution to the equation. Prove that $u$ is $T$ periodic if and only if $u(0) = u(T)$. Indication: for the second implication, define the function $\phi(x) = u(x+T) - u(x)$, find the ODE it satisfies and apply the Theorem of Existence and Uniqueness properly to conclude.
The first implication is pretty straightforward, but I'm having a bit of trouble when it comes to the second. Here's what I've got so far:
$\Rightarrow$| Using the indication, we can identify the following Cauchy problem:
$\begin{equation} \phi' = a(x)\phi\end{equation}$
$\phi(0) = 0$
We know that $a(x)$ is continuous, so, to properly apply the TEU, $\phi$ must be Lipschitz. In this case, my thought was that it must be globally Lipschitz to assure the periodicity of $u$, but I can't find a way to proof the global condition.
If we asume that $\phi$ is, in fact, globally Lipschitz, then there's a unique solution to the Cauchy Problem given by $\phi(x) = 0$, where we can conclude that $u(x) = u(x+T)$.
My trouble lies in proving the global Lipschitz condition. Is the local condition enough in this case?
Can somebody help me with that?
You have an equation of the form $\phi^\prime = F(x,\phi)$. In your case, $F$ is Lipschitz in $\phi$. Since $a$ is continuous and periodic, we know that it is bounded. Pick $M > 0$ such that $a \leq M$. Then, for any fixed $x$, $$|F(x,\phi_1) - F(x,\phi_2)| = |a(x)\phi_1 - a(x)\phi_2| \leq M |\phi_1 - \phi_2|.$$ We already know that $F$ is continuous in $x$ by hypothesis. This gives you global existence of a unique solution. Since $0$ is a solution, this implies that $\phi \equiv 0$.
The idea is that since $F$ is globally Lipschitz with the same constant, we can glue together local solutions.