# Does a scalar LTV system with odd-periodic coefficients and even-periodic inputs have no periodic solutions?

## 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.

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.
• A simpler example would be $a(t,\mu)=0$ and $b(t,\mu)=\mu$, which just means that $x_1(t,\mu)$ is just the integral of $\mu$. May 23 '20 at 6:51
• @fibonatic : Your counterexample will not satisfy the conditions that $b(t,0)=0$ and that $x_1$ be periodic. May 24 '20 at 2:03
• How does $b(t,\mu)=\mu$ not satisfy $b(t,0)=0$? And indeed I should have specified that $\mu$ should be periodic and the average value of $\mu$ over such period should be zero. May 24 '20 at 4:58
• @fibonatic : You are right about the case $\mu=0$. However, $\mu$ is a parameter, not depending on $t$, and thus cannot be made periodic in $t$. May 24 '20 at 11:55