Periodic differential equation has periodic solutions

Question

The Exercise

Consider the equation $x'(t)=f(t,x)$, where $f$ is continuously differentiable on $x,t$ and periodic in $t$. If there are constants $p,q$ such that $f(t,p)>0,f(t,q)<0$ for all $t$, then there is a periodic solution $x(t)$ such that $p<x(0)<q$.

The Attempt

From the middle value theorem, there exists a function $y(t)$ such that $p<y(t)<q$ and $f(t,y(t))=0$ for all $t$. Since $f(t,y(t))=0$ for all $t$, $y$ is an equilibrium solution to the equation $x'(t)=f(t,x)$. Therefore, $y(t)=k$, a constant.

Note that, since $p<q$, $\partial_xf(t,x)|_{x=k}<0$ for all $t$. Then, $k$ is a sink. Now, since $f$ is periodic, the equation allows for some periodic solution $x(t)$ orbiting around $k$. We may set $x(0)=k$, so that $p<x(0)<q$.

The Problem

The second paragraph seems sketchy to me, rather than rigorous, but I can't see what I might be missing here.

Answer 1

I think your proof does not work. In general there are no constant solutions. I outline a proof I know.

Let $f$ be $T$-periodic in $t$, that is $f(t,x)=f(t+T,x)$ $((t,x) \in \mathbb{R}^2)$. Consider the IVP $$ x'(t)=f(t,x(t)), \quad x(0)=s \in [p,q], $$ to the right, with nonextendable solution $x(\cdot,s):[0, \omega_+(s)) \to \mathbb{R}$. First note that there is some $t_0 =t_0(s) > 0$ such that $p< x(t,s) < q$ for $0<t < t_0$: If $s \in (p,q)$ this follows from continuity of $t \mapsto x(t,s)$, and if $s \in \{p,q\}$ this follows from $f(0,p)>0$ and $f(0,q)< 0$, respectively. Now we can prove that $x(t,s) \in (p,q)$ for all $t \in (0, \omega_+(s))$: Assume that $x$ is somewhere $> q$. Then there is a first time $t_1 \ge t_0$ such that $x(t_1,s)=q$ and $x(t,s)<q$ $(t\in(0,t_1))$. Hence $$ 0 \le \lim_{t \to t_1-} \frac{x(t,s)-x(t_1,s)}{t-t_1}=x'(t_1,s) =f(t_1,q) < 0, $$ a contradiction. The conclusion for $p$ follows the same line. Thus $t \mapsto x(t,s)$ stays in $(p,q)$ for $t> 0$, in particular $\omega_+(s)=\infty$ for each $s \in [p,q]$. Since the solutions of the IVP depend continuously on $s$ we get a continuous shooting function $g:[p,q] \to (p,q)$, $s \mapsto x(T,s)$, which has a fixed point $s_0 \in (p,q)$ by the intermediate value theorem, that is $s_0=g(s_0)=x(T,s_0)$. Now $x(\cdot,s_0)$ is a $T$-periodic solution of $x'=f(t,x)$. Note that we need continuous dependence on the initial value, which holds if $f$ is continuously differentiable in $x$ (even if $f$ is locally Lipschitz in $x$).