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}


  • $a(t + \pi,\mu) = a(t,\mu)$ 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:

$$ x_1(t,\mu) = \exp\left\{ A(t,\mu)\right\} \int \limits_0^t \exp\left\{-A(\tau,\mu)\right\}b(\tau,\mu) \, d\tau$$


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.


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.


1 Answer 1


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.

  • $\begingroup$ 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 $\mu$. $\endgroup$
    – fibonatic
    May 23, 2020 at 6:51
  • $\begingroup$ @fibonatic : Your counterexample will not satisfy the conditions that $b(t,0)=0$ and that $x_1$ be periodic. $\endgroup$ May 24, 2020 at 2:03
  • $\begingroup$ 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. $\endgroup$
    – fibonatic
    May 24, 2020 at 4:58
  • 1
    $\begingroup$ @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 $t$. $\endgroup$ May 24, 2020 at 11:55

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