TITLE: Is this periodic solution unique? (ODE)
QUESTION [5 upvotes]: So, for the ODE
\begin{align}
x' = -x^{3} + \sin t,
\end{align}
we can show that there exists a $2\pi$ periodic solution.  To do this, we denote by
\begin{align}
x(t,\alpha)
\end{align}
The solution $x(t)$ of the ODE such that $x(0) = \alpha$.  Then, let $\alpha \in [-2,2]$.  Consider the function $f(\alpha) = x(2\pi, \alpha)$
\begin{align}
x > 1 &\Rightarrow x' < 0 \\
x < 1 &\Rightarrow x' > 0.
\end{align}
Thus, solutions beginning in $[-2,2]$ stay there and we can use Brouwer's fixed point theorem to show that there exists a fixed point of $f$.  Therefore there is some $\alpha^{*}$ such that $f(\alpha^{*}) = \alpha^{*}$, which represents the periodic solution $x(t,\alpha)$.
Now, my question: Is this $2\pi$- periodic solution unique? Are there other $2\pi$ periodic solutions of this ODE?  I have started by trying to subtract two periodic solutions from one another but this hasn't taken me anywhere productive.

REPLY [2 votes]: Let $v(t,α)=\frac{\partial x}{\partial α}(t,α)$, then

$v(0,α)=1$ as $x(0,α)=α$ and
the $α$ derivative of the ODE gives
$$
v'(t,α)=-3x(t,α)^2v(t,α)
$$
which means that $v$ is falling towards zero, moreover
$$
v(2π,α)=\exp\left(-3\int_0^{2π}x(t,α)^2\,dt\right)
$$
and as there are no solutions that are constant zero, $0<v(2π,α)<1$.

By the usual contraction argument this excludes the existence of multiple fixed points of $f$, thus proving the uniqueness of the periodic solution.
A recent duplicate to this question with the same ODE asked also after the uniqueness after a not so recent duplicate only asked after the existence. I felt that this more specific question was the more appropriate place to answer.