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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 also available on my website.

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  • $\begingroup$ Are you sure you have written the system correctly? I tried the system numerically in Maple, and it's nowhere near to $2\pi$-periodic. $\endgroup$
    – Robert Israel
    Commented Jun 12, 2020 at 1:30
  • $\begingroup$ Periodic with period $2\pi$? With initial condition $(1,0)$, Maple gives me $[0.822717619142367, 0.627509890225599]$. It looks to me like (nearly) a period of $21\pi$. $\endgroup$
    – Robert Israel
    Commented Jun 12, 2020 at 21:00
  • $\begingroup$ Then there's a problem with the formula that I posted above. My student found that the solution returned to within $10^{-120}$ of it's initial condition at $t=2\pi$. We'll look into it. $\endgroup$
    – Roy Goodman
    Commented Jun 13, 2020 at 3:51
  • $\begingroup$ Okay. Here is the corrected matrix $$A(t) = \begin{pmatrix} -\frac{4 \sin (2 t)}{\sqrt{8 \cos (2 t)+17}} & \frac{8 \cos ^2(2 t)-12 \cos (2 t)+3 \sqrt{8 \cos (2 t)+17}-11}{2 (1-\cos (2 t)) \sqrt{8 \cos (2 t)+17}} \\ \frac{-8 \cos ^2(2 t)-4 \cos (2 t)-\sqrt{8 \cos (2 t)+17}+7}{2 (\cos (2 t)+1) \sqrt{8 \cos (2 t)+17}} & \frac{4 \sin (2 t)}{\sqrt{8 \cos (2 t)+17}} \end{pmatrix}$$ $\endgroup$
    – Roy Goodman
    Commented Jun 17, 2020 at 2:14
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    $\begingroup$ @manroygood: You really should correct the formula in the question, not just put the correction in the comments. It's trivial to do, since you have already have the formula in LaTeX. Unfortunately, only you can copy the formula you have in your comment and use it to correct the question. $\endgroup$
    – Robert Bryant
    Commented Jun 19, 2020 at 18:40

1 Answer 1

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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 $x'(t)$.

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

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  • $\begingroup$ Thanks. That does it. Mathematica's DSolve fails to integrate it. Is there any way to get an insight into how Maple fountain answer? $\endgroup$
    – Roy Goodman
    Commented Jun 21, 2020 at 14:05
  • $\begingroup$ Solving for $y(t)$ from the first component of the differential equation, and doing a bunch of trigonometric gymnastics yields in the end $$y(t)=-\frac{ \left(1+ 4 \cos {2t}+\sqrt{8 \cos {2t}+17}\right)\sin {2t}}{1 +\cos {2t}}$$ $\endgroup$
    – Roy Goodman
    Commented Jun 22, 2020 at 14:26
  • $\begingroup$ We would like to submit this as an addendum to a published paper. What is the best way to acknowledge your contribution? $\endgroup$
    – Roy Goodman
    Commented Jun 24, 2020 at 16:39
  • $\begingroup$ Just write something like "We thank Robert Israel for ..." $\endgroup$
    – Robert Israel
    Commented Jun 25, 2020 at 14:37

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