2
$\begingroup$

A physicist colleague asks me the following question. I have no idea to answer him. Your answer is very appreciated.

Is there a global first integral on $\mathbb{R}^3$ for the following vector field?

$$\begin{cases}x'=sin(y) \\ y'=sin(z)\\ z'=sin(x) \end{cases}$$

Improve this question
$\endgroup$
5
  • $\begingroup$ Is this to be read as $x$, $y$, $z$ depending on one parameter, say, "$t$" and the prime denotes the derivative w.r.t. $t$? And what is sought is the solution of this set of first-order differential equations? $\endgroup$
    – Michael Engelhardt
    Commented Sep 7, 2019 at 23:24
  • 1
    $\begingroup$ @MichaelEngelhardt yes $x'=dx/dt$. $\endgroup$
    – Ali Taghavi
    Commented Sep 7, 2019 at 23:41
  • 1
    $\begingroup$ @MichaelEngelhardt by the second part of your comment do you mean "what is the motivation for consideration of this system"? I have no idea of such a motovation. I can ask him. $\endgroup$
    – Ali Taghavi
    Commented Sep 7, 2019 at 23:46
  • 1
    $\begingroup$ No no, the second part of my comment was simply to make sure I understood what form of answer was expected - I suppose that was a bit redundant once the first part was clarified. The second part was simply reinforcing the first part. Still, it would be interesting where this arises - but don't go to any length investigating, it's not essential of course. $\endgroup$
    – Michael Engelhardt
    Commented Sep 8, 2019 at 1:42
  • 2
    $\begingroup$ Now, for small $x$, $y$, $z$, the solutions of course are of the form $x=\exp (t)$, $\exp (\exp (i 2\pi /3) t)$ or $\exp (\exp (i 4\pi /3) t)$, with $y$, $z$ obtained by differentiation, but this is probably obvious to your colleague, since he explicitly asks for a global answer ... $\endgroup$
    – Michael Engelhardt
    Commented Sep 8, 2019 at 1:53

1 Answer 1

Reset to default
3
$\begingroup$

One can find a solution of the form $x=y=z$, namely, $x=2$arccot$(\exp (-t-a))$ with the free parameter $a$. Of course, there should be more. Note also the symmetries of the problem: For any solution $(x,y,z)$, also $(-x,-y,-z)$ is a solution, as is $(x+2k\pi,y+2l\pi,z+2m\pi)$, with arbitrary integers $k,l,m$.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ The latter symmetries presumably provide useful guardrails if one ultimately wants to resort to a numerical solution - one only has to interpolate between two adjacent analytic solutions bounding the direction of the vector $(x,y,z)$. $\endgroup$
    – Michael Engelhardt
    Commented Sep 8, 2019 at 14:11
  • 1
    $\begingroup$ In particular, for large $x$, $y$, $z$, one is always very close to one of the analytic solutions. In combination with the form of solutions for small $x$, $y$, $z$ in comments above, there isn't that much wiggle room left. $\endgroup$
    – Michael Engelhardt
    Commented Sep 8, 2019 at 14:20
  • $\begingroup$ Thank you very much for your answer. I will forward your answer to him. by global first integral I mean a function $f:\mathbb{R}63 \to \mathbb{R}$ with $f_x sin(y)+f_y sin(z)+f_z sin(x0=0$. may be your answer can help to find this globaly defined first integral. Thanks again for consideration of the question. $\endgroup$
    – Ali Taghavi
    Commented Sep 9, 2019 at 21:17
  • 1
    $\begingroup$ Oh - looks like we talked past each other then ... this is what I was trying to clarify with my initial question, but it seems we still didn't quite match minds. Still, maybe this helps anyway. I'll update if I think of anything else. $\endgroup$
    – Michael Engelhardt
    Commented Sep 9, 2019 at 23:47
  • 1
    $\begingroup$ Indeed, the trajectories $(x(t),y(t),z(t))$ I describe are by construction equipotential lines, i.e., lines of constant $f$, to the extent that $f$ exists. So they at least provide a scaffolding for a more complete construction of $f$. $\endgroup$
    – Michael Engelhardt
    Commented Sep 10, 2019 at 19:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .