Problem Setup
-------------


Suppose we have the following scalar, linear time-varying (LTV) system with parameter $\mu \in [0,\pi[$:


\begin{cases}
\dot{x\_1}(t,\mu) = a(t,\mu)x\_1(t,\mu) + b(t,\mu) \\
x\_1(0,\mu) = 0
\end{cases}


where


* $a(t + \pi,\mu) = a(t,\mu)$ and $a(-t,\mu) = -a(t,\mu)$ $\forall (t,\mu)$
* $b(t + \pi,\mu) = b(t,\mu)$ and $b(-t,\mu) = b(t,\mu)$ $\forall (t,\mu)$
* $a(t,0) = b(t,0) = 0$ $\forall t$


Define $A(t,\mu) = \int \limits\_0^t a(\sigma,\mu) \, d\sigma$. Then $A(t,\mu)$ is also odd and $\pi$-periodic in $t$, which means that $x\_1(t,\mu)$ can be written as:


$$ x\_1(t,\mu) = \exp\left\{ A(t,\mu)\right\} \int \limits\_0^t \exp\left\{-A(\tau,\mu)\right\}b(\tau,\mu) \, d\tau$$


Question
--------


My conjecture is that $x\_1(t,\mu)$ will NOT be $\pi$-periodic regardless of your choice of $\mu$. I have not been able to show this nor find any sources which support or reject this claim. I would appreciate any help in proving or disproving this conjecture.




---


Context
-------


I am trying to show that a particular system has no periodic solutions, but cannot solve for $a(t,\mu)$ nor $b(t,\mu)$ analytically - they depend on $\mu$ through the (unknown) solution to a nonlinear system $\dot{x}\_0 = f(t,x\_0)$ where $x\_0(0) = \mu$.


I have tried doing a sensitivity analysis to show $\frac{\partial x\_1}{\partial \mu} \neq 0$ $\forall \mu$, but this requires solving for $x\_1(t)$ analytically (which cannot be done). I have also tried extending this to the cylindrical system $(\dot{x\_1}, \dot{t} = 1)$ and finding a Dulac-Cherkas function to prove there are no closed orbits, with no success. 


Finally, I can show numerically that $b(t,\mu)$ has zero-average for some $\mu$ and non-zero average for other $\mu$, so I cannot put any further assumptions on the average of $b$ in my conjecture.




---



Does a scalar LTV system with odd-periodic coefficients and even-periodic inputs have no periodic solutions?
In fact, $x\_1$ can be $\pi$-periodic. Indeed, let $m:=\mu$, 
$$a(t,m):=m\sin2t,\quad b(t,m):=m\cos2t-c\_m,$$
where $c\_m$ is the unique solution to the equation
$$\int\_0^\pi e^{-A(s,m)}(m\cos2s-c\_m)\,ds=0,$$
with 
$$A(t,m)=\int\_0^t a(s,m)\,ds=\frac m2\,\sin^2t. $$
Then all your conditions on $a$ and $b$ hold, and 
$$x\_1(t,m)=e^{A(t,m)}\int\_0^t e^{-A(s,m)}(m\cos2s-c\_m)\,ds$$
will be $\pi$-periodic in $t$.




---


Here one can similarly use any other (say) bounded measurable odd and even $2\pi$-periodic functions instead of $\sin$ and $\cos$, respectively. 


