We have been studying a Hamiltonian system that possesses a one-parameter family of periodic orbits, depending on the energy level $h$. We "know" via various non-rigorous means that these periodic orbits are stable for $h<\frac{1}{8}$ and unstable for $h>\frac{1}{8}$ but haven't been able to prove it.


We know that if the linearization possesses a periodic orbit at the critical value $h=\frac{1}{8}$, then this value of $h$ lies on the boundary between stability and instability. We have numerically calculated this periodic orbit using ultra-high-precision arithmetic and a 30th order ODE solver. We have also used Hill's method of harmonic balance to high order in computer algebra to show that $h\_{\rm critical}$ agrees with $1/8$ to double precision, using no ODE solves.


If we could simply show that the following ODE has a $2\pi$-periodic orbit, we would have a proof. Does anybody know how find such a solution?


**Corrected since original posting**


$$\frac{d}{dt} \vec{x} = A(t) \vec{x},$$


where
$$ A(t) = 
\left(
\begin{array}{cc}
 -\frac{4 \sin (2 t)}{\sqrt{8 \cos (2 t)+17}} & \frac{8 \cos ^2(2 t)-12 \cos (2 t)+3 \sqrt{8 \cos (2 t)+17}-11}{2 (1-\cos
 (2 t)) \sqrt{8 \cos (2 t)+17}} \\
 \frac{-8 \cos ^2(2 t)-4 \cos (2 t)-\sqrt{8 \cos (2 t)+17}+7}{2 (\cos (2 t)+1) \sqrt{8 \cos (2 t)+17}} & \frac{4 \sin (2
 t)}{\sqrt{8 \cos (2 t)+17}} \\
\end{array}
\right)
$$


This would answer the one big question we left unanswered in [this paper](https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.4.124703) also available on [my website](https://web.njit.edu/%7Egoodman/publication/leapfrogprf/).



Exact solution to a periodic linear ODE sought
Rather incredibly, your (corrected) system does have a closed-form solution, which I found with Maple's help.
$$ x(t) = 1+4\,\cos \left( 2\,t \right) +3\,\sqrt {8\,\cos \left( 2\,t \right) +
17}$$
$y(t)$ is a rather complicated beast that I'll just write in text rather than LaTeX: when pretty-printed, it still doesn't look pretty.



```
-16384/(5-(16*cos(t)^2+9)^(1/2))^(1/4)/(4*cos(2*t)-5+(8*cos(2*t)+17)^(1/2))^(1/
2)/(8*cos(2*t)+17)^(3/4)/((16*cos(t)^2+9)^(1/2)-3)^(1/4)*(cos(t)+1)*((8*cos(2*t
)+17)^(1/2)+5)^(3/4)/((16*cos(t)^2+9)^(1/2)+5)^(3/4)/(16*cos(t)^2+9)^(1/4)/((8*
cos(2*t)+17)^(1/2)+3)^(1/4)*(-(8*cos(2*t)+17)^(1/2)+5)^(1/4)*((16*cos(t)^2+9)^(
1/2)+3)^(1/4)*(8*cos(t)^2-9+(16*cos(t)^2+9)^(1/2))^(1/2)*(-1/4*sin(2*t)*((5/8*
cos(2*t)^3-1/2*cos(2*t)^2-11/32*cos(2*t)-31/64)*(8*cos(2*t)+17)^(1/2)+(cos(2*t)
^3-2*cos(2*t)^2+5/4*cos(2*t)+7/8)*(cos(2*t)+17/8))*(16*cos(t)^2+9)^(1/2)+cos(t)
*((cos(2*t)^3-2*cos(2*t)^2+1/8*cos(2*t)+73/32)*(8*cos(2*t)+17)^(1/2)+cos(2*t)^4
-cos(2*t)^3+15/4*cos(2*t)^2-5/4*cos(2*t)-305/32)*sin(t)*(cos(2*t)+17/8))*((8*
cos(2*t)+17)^(1/2)-3)^(1/4)*(cos(t)-1)/(4*cos(2*t)^2*(8*cos(2*t)+17)^(1/2)+16*
cos(2*t)^3-44*cos(2*t)^2-13*(8*cos(2*t)+17)^(1/2)+20*cos(2*t)+53)/(32*cos(t)^4-
56*cos(t)^2+9+3*(16*cos(t)^2+9)^(1/2))

```

The solution is manifestly $2\pi$-periodic, with $y(0) = y(2\pi) = 0$.


EDIT: Of course $y(t)$ can be obtained from $x(t)$ and $x'(t)$ by looking at the differential equation for $x'(t)$.


$$y \left( t \right) =-2\,{\frac { \left( 4\,\sin \left( 2\,t \right) x
 \left( t \right) + \left( {\frac {\rm d}{{\rm d}t}}x \left( t
 \right) \right) \sqrt {8\,\cos \left( 2\,t \right) +17} \right) 
 \left( \cos \left( 2\,t \right) -1 \right) }{8\, \left( \cos \left( 2
\,t \right) \right) ^{2}-12\,\cos \left( 2\,t \right) +3\,\sqrt {8\,
\cos \left( 2\,t \right) +17}-11}}$$


