Non-autonomous ODE $x'=q(t)\cos^2x-\sin^2x$, where $q(t)$ has period $1$.

Question

Consider the first-order non-autonomous ODE: \begin{align} \dfrac{\mathrm{d}x}{\mathrm{d}t}=q(t)\cos^2 x-\sin ^2 x, \qquad \tag{$\star$} \end{align} where $q(t)$ is a continuous function with period $1$. Assume that this equation has a bounded solution.

(a) Show that any solution of $(\star)$ is bounded.

(b) For any $x\in \mathbb{R}$, denote $\phi(t,x)$ be the solution of $(\star)$ with initial value $\phi(0,x)=x$. Define the function $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=\phi(1,x)$. Show that $f$ has exactly two fixed points $x_{1},x_{2}$ in $[0,\pi]$ with $f'(x_{i})\neq 1, i=1,2$.

For question (a), let $\phi_{0}$ be a bounded solution, then for any $n\in\mathbb{Z}$, $\phi_{0}+n\pi$ is also a bounded solution. By the uniqueness of solutions, for any two solutions of $(\star)$, their curves cannot intersect. Therefore, for any given solution of $(\star)$, its curve must be between $x=\phi_{0}(t)+n\pi$ and $x=\phi_{0}(t)+(n+1)\pi$ for some $n$, hence bounded.

However, I have no idea about (b). I would appreciate any answers and comments

Answer 1

I have an idea of finding one fixed point of $f$.

\begin{align} \dfrac{\mathrm{d}x}{\mathrm{d}t}=q(t)\cos^2x-\sin^2x. \tag{$\star$} \end{align}

Fact 1: Due to the periodicity of $q$ and $\sin,\cos$, we can easily verify if $\phi(t)$ is a solution of $(\star)$, then so are $\phi(t+1)$ and $\phi(t)+\pi$.

For any $x\in\mathbb{R}$, let $\phi(t,x)$ be the solution of $(\star)$ with initial value $\phi(0,x)=x$.

Claim: $f(x)=\phi(1,x)$ satisfies:
(1) $f^{n}(x):=f(f(\cdots f(x)))=\phi(n,x)$;
(2) $f(x+\pi)=f(x)+\pi$;
(3) $f$ is strictly increasing.

Proof: For (1), $\phi(t+1,x)$ is the solution of $(\star)$ with initial value $\phi(1,x)=f(x)$, thus $\phi(t+1,x)=\phi(t,f(x))$. In particular, take $t=1$, we have $\phi(2,x)=f(f(x)):=f^2(x)$. Inductively, we can get $\phi(n,x)=f(f(\cdots f(x)))=f^{n}(x)$.

For (2), $\phi(t,x)+\pi$ is the solution of $(\star)$ with initial value $x+\pi$, thus $\phi(t,x)+\pi=\phi(t,x+\pi)$. In particular, take $t=1$, we have $f(x)+\pi=f(x+\pi)$.

To see (3), we compute the equation that $\partial_{x}\phi(t,x)$ satisfies. \begin{align} \dfrac{\partial }{\partial t}\left(\dfrac{\partial}{\partial x}\phi(t,x)\right)&=\dfrac{\partial }{\partial x}\left(\dfrac{\partial}{\partial t}\phi(t,x)\right)=\dfrac{\partial }{\partial x}\left(q(t)\cos^2\phi(t,x)-\sin^2\phi(t,x)\right)\\ &=-2(q(t)+1)\sin\phi(t,x)\cos\phi(t,x) \dfrac{\partial }{\partial x}\phi(t,x). \end{align} Therefore, \begin{align*} \dfrac{\partial }{\partial x}\phi(t,x)=\exp\left\{\int_{0}^{t}-2(q(s)+1)\sin\phi(s,x)\cos\phi(s,x)\mathrm{d}s\right\}>0. \end{align*} In particular, $f'(x)=\dfrac{\partial }{\partial x}\phi(1,x)>0$, hence $f$ is strictly increasing.

Now, by the property of rotation number, the limit \begin{align} \rho(f):=\lim_{n\to\infty}\dfrac{f^{n}(x)-x}{n} \end{align} exists, and it is independent on $x$. In particular, take $x=0$. By $(2)$, $f^{n}(0)=\phi(n,0)$. Since $\phi(t,0)$ is bounded, then we must have \begin{align} \rho(f)=\lim_{n\to\infty}\dfrac{f^{n}(0)}{n}=0. \end{align} In fact, this implies $f$ has a fixed point in $[0,\pi]$. If not, then $g(x):=f(x)-x$ has no zero in $[0,\pi]$. WLOG, assume $g>0$. Note that $g(x+\pi)=f(x+\pi)-x-\pi=f(x)-x=g(x)$, then $g$ has period $\pi$. Now, we can assume that $\min_{\mathbb{R}}g=\min_{[0,\pi]}g=\delta>0$. Then \begin{align} f(x)-x&=g(x)\ge \delta\\ f(f(x))-f(x)&=g(f(x))\ge \delta\\ &\vdots\\ f^{n}(x)-f^{n-1}(x)&=g(f^{n-1}(x))\ge\delta \end{align} Therefore, $f^{n}(x)-x\ge n\delta$, hence $0=\lim_{n\to\infty}\dfrac{f^{n}(x)-x}{n}\ge\delta>0$, a contradiction.