Given information: Let $f$ be a continuous function defined for all $(x,y) \in \mathbb{R}^2$. Let $f$ also satisfy a Lipschitz condition with respect to $y$. Let $f$ be periodic with respect to $x$ of period $w$, and let $y_1, y_2$ be such values that $f(x,y_1)f(x,y_2) \lt 0$ for all $x$.
Question: Show that the equation $y'=f(x,y)$ has at least one periodic solution of period $w$. Then apply this result to the equation $y'+p(x)y = q(x)$ where $p(x) \not = 0$ and $q(x)$ are continuous period functions of period $w$.
I'm given a hint that is:
Consider the map
$$p: \mathbb{R} \to \mathbb{R}, y_0 \mapsto p(y_0):=y_w$$
where $y_{w}=y(w)$ with $y$ being the solution of the initial value problem
$$y'=f(x,y), y(0)=y_0$$
The map $p$ is a Poincaré map, which maps the $y$-value $y_0$ of the solution curve through the point $(x,y)=(0,y_0)$ to the $y$-value of this solution curve $x=w$.
A periodic solution of $y'=f(x,y), $ corresponds to a fixed Poincare map, i.e. the existence of a $y^* \in \mathbb{R}$ with $p(y^{*})=y^*$
Random thoughts: I have to prove the existence of a fixed point by noting that the Poincare map is continuous as follows from a theorem.
Please help !
Best Answer
Assume $y_1<y_2$ and $f(x,y_1)>0$, $\ f(x,y_2)<0$ for all $x$. By general principles about ODE's any solution $x\mapsto \phi_\eta(x)$ starting at a point $(0,\eta)$, $\ y_1\leq\eta\leq y_2$, will finally leave the rectangle $R:=[0,w]\times[y_1,y_2]$. But it cannot do so along the horizontal edges of this rectangle. It follows that the solution $\phi_\eta$ will pass through a point $(w,\eta')$, $\ y_1\leq\eta'\leq y_2$, on the right edge of $R$. In this way a so-called Poincaré map $$\Phi:\quad [y_1,y_2]\to[y_1,y_2],\qquad \eta\mapsto \eta'=:\Phi(\eta)$$ is defined. Again by general principles this $\Phi$ is continuous. By Brouwer's fixed point theorem (or using the intermediate value theorem) it follows that $\Phi$ has a fixed point $\eta_*\in[y_1,y_2]$. The solution $\phi_{\eta_*}$ starting at $(0,\eta_*)$ is then periodic.
If instead of the initial assumption on $f$ we have $f(x,y_1)<0$, $\ f(x,y_2)>0$ we start the argument at $x=w$ and proceed to the left.
In the example $y'+p(x) y=q(x)$ we have $$f(x,y)=p(x)\left({q(x)\over p(x)} -y\right)\ .$$ As $p$ and $q$ are periodic and continuous, and $p(x)\ne0$ for all $x$ there is an $M>0$ such that $$-M<{q(x)\over p(x)}<M\qquad\forall x\ .$$ It follows that by choosing $y_1:=-M$, $\ y_2:=M$ we can fulfill the assumptions of the "theorem".