Stability of periodic solution I am stuck on this question:

$$x'(t) = x(x-p(t))(x-1),$$ where $p$ is $1$-periodic and $0< p(t) <1$ for all $t$. We want to show the existence of a STABLE periodic solution $\mu$ such that $0< \mu(t) <1$ for all $t$.

If we require $0< x_0 <1$, where we impose the initial condition that $x(0) = x_0$, then it is clear that there exists a solution $\mu$ that satisfies the ODE with this initial condition. Standard result also shows the existence of such periodic solutions. My question is how to show the stability of such solutions by the definition of stable periodic solutions.

Okay, so here is a proof that $f$ is continuous (which takes care of a detail in Evgeny’s proof of existence of a fixed point of $f$).
Setup:
Recall the ODE is:
$$ x’ = x(x-p)(x-1) $$
Assume $p$ is integrable and there are values $p_{min}, p_{max}$ such that for all $t$ we have
$$ 0<p_{min}\leq p(t) \leq p_{max} < 1 $$
Further impose any additional mild assumptions so that solutions with a given initial condition exist, are unique, and do not cross (perhaps we need $p$ continuous for that?)
Let $a,b$ be values such that $0<a<p_{min}<p_{max}<b<1$.
Define Evgeny’s function $f:[a,b]\rightarrow\mathbb{R}$ such that for each $z \in [a,b]$, the value $f(z)$ is defined as  $x(1)$ for a solution $x(t)$ to the ODE with initial condition $x(0)=z$.  We want to show $f$ is continuous over $[a,b]$.
Proof:
Fix $z \in [a,b]$.  Fix a small value $\delta \neq 0$ such that $z+\delta \in [a,b]$.  We want to show that $|f(z+\delta)-f(z)| \leq |\delta|D$ for some positive constant $D$.   Define $x(t)$ as the ODE solution with $x(0)=z$.  Define $y(t)$ as the ODE solution with $y(0)=z+\delta$.  Define $g(t) = y(t) - x(t)$, so that $g(0) = \delta$.  Since $y=x+g$ we get:
$$ y’ = x’ + g’  = x(x-p)(x-1) + g’ $$
On the other hand, since $y$ satisfies the ODE we get:
$$ y’ = y(y-p)(y-1) = (x+g)(x+g-p)(x+g-1) $$
Equating the above two expressions gives:
$$ x(x-p)(x-1) + g’ = (x+g)(x+g-p)(x+g-1) $$
and so:
$$ g’ = y’ - x’ = g\underbrace{[g^2 - gp + 3gx - g - 2px + p + 3x^2 -2x]}_{R(t)} $$
That is, for all $t$ we have
$ g’(t) = g(t)R(t) $.
Now, we already know (from above comments) that any solution to the ODE with initial condition in $[a,b]$ must stay in $[a,b]$.  Thus, $x,y, g$ are bounded  for all time, and there is some positive constant $C$ such that $|R(t)| \leq C$ for all $t$.
For all $t$ we have  $g(t) \neq 0$  (since $\delta\neq 0$ and solutions cannot cross). Thus:
$$ \frac{g’(t)}{g(t)} = R(t)$$
So for all $t \geq 0$:
$$ \log[|g(t)|/|g(0)|] = \int_0^t R(\tau)d\tau $$
So
$$ |g(t)| = |g(0)|e^{\int_0^t R(\tau)d\tau} = |\delta|  e^{\int_0^tR(\tau)d\tau} \leq |\delta| e^{Ct} $$
Thus:
$$ |f(z+\delta) - f(z)| = |y(1) - x(1)| = |g(1)| \leq |\delta| e^{C} $$
$\Box$
