$y'+a(x)y=0$ is an homogeneous linear differential equation, where $a$ is continuous in $-\infty<x<\infty$ with periodicity $\xi>0$, i.e. $a(x+\xi)=a(x)~\forall x$. I have to show three things:
- If $\phi$ is a non-trivial solution and $\psi(x)=\phi(x+\xi)$, show that $\psi$ is also a solution.
- Show that there exists a constante $c$ such that $\phi(x+\xi)=c\phi(x)~\forall x$. Also prove that $$c=e^{-\int\limits_0^\xi a(t)dt}$$.
- Which condition needs $a$ so that there exists a non-trivial solution, with periodicity $\xi$.
I could manage to prove item 1, we know
$$\phi'(x+\xi)+a(x+\xi)\phi(x+\xi)=0$$ $$\phi'(x+\xi)+a(x)\phi(x+\xi)=0$$ $$\psi'(x)+a(x)\psi(x)=0$$ Hence $\psi$ is a solution.
I'm struggling with item 2. Not sure where to start, the only thing I tried is the following: $$y'+a(x)y=0$$ $$\dfrac{1}{y}\dfrac{dy}{dx}=-a(x)$$ $$|y|=e^{-\int^x a(t)dt}e^c$$ which leads me to nowhere, but it has a similar form of the $c$ given.
I also thought about using Mean Value Theorem but not sure how to apply it here because there is a $c$ multiplying $\phi(x)$. Tried working with $\dfrac{\phi(x+\xi)}{\phi(x)}$ but still failed.
Let $\phi$ be a nontrivial solution of $y'+a(x)y=0$ on $\mathbb{R}$. Then $\phi$ is a fundamental system of this first order homogeneous linear differential equation, that is each solution on $\mathbb{R}$ is of the form $c \phi$ for some $c \in \mathbb{R}$. You already know that $x \mapsto \phi(x+ \xi)$ is a solution on $\mathbb{R}$. Thus $\psi(x):=\phi(x+ \xi)=c\phi(x)$ for some $c\in \mathbb{R}$. Moreover, $\phi(x)=\exp(\int_0^x -a(t) dt)\phi(0)$. Thus $$ c=\psi(0)/\phi(0)=\phi(\xi)/\phi(0)=\exp(-\int_0^\xi a(t) dt). $$ Now, you see that if $\int_0^\xi a(t) dt=0$, then $\psi(x)=\phi(x)$, that is $\phi$ is periodic with period $\xi$.