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Dear MO world,

I'm teaching an undergraduate course on "fun with chaos". As part of a test (on bifurcations in differential equations), I asked students to sketch phase portraits for a family of (2d) differential equations. While preparing my solutions, I fed the differential equations (with one value of the parameter) into a phase portrait plotter on the web and found that there were families of closed curves as solutions, so that the system seemed to have a first integral. Changing the parameter, this persisted. I'm hoping someone can tell me why this should have been obvious to me!

The system was $$ \eqalign{\dot x&=y-x+1\cr \dot y&=y-rx^2.} $$

The phase portrait (with $r=0.15$) looks like this:

Some computation with mathematica reveals that $2rx^3+3y^2+6y-6xy$ is a first integral. The question is why? Is this an outrageous coincidence? (I thought that your chance of bumping into a conservative differential equation by accident were nil unless the system was of the form $\ddot x=f(x)$).

So: should I go and buy a lottery ticket, or is there some reason that this I shouldn't be surprised by this?

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Divergence free vector fields in the plane are associated with Hamiltonian systems. Your vector field is easily seen to be divergence free.

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    $\begingroup$ In fact, a vector field $Z = f(x,y)\partial_x + g(x,y)\partial_y$, is divergence-free if and only if the $1$-form $\omega = - g(x,y)\ dx + f(x,y)\ dy$ is closed. In this case, there exists a function $H$ on the plane such that $dH = \omega$, and this $H$ will be a conservation law for the flow of $Z$. $\endgroup$
    – Robert Bryant
    Commented Apr 6, 2012 at 15:39
  • $\begingroup$ Thank you. A bit embarrassed not to have figured this out for myself... $\endgroup$
    – Anthony Quas
    Commented Apr 6, 2012 at 20:14
  • $\begingroup$ @Robert Bryant: to make explicit the transition from $Z$ to $\omega,$ I would trivially observe that $Z$ is the symplectic gradient of $\omega$ through the canonical symplectic form $\mathrm{d}x\wedge\mathrm{d}y$ on $\mathbb{R}^2,$ so the exactness of $\omega$ is the same as the hamiltonicity of $Z.$ $\endgroup$
    – agt
    Commented Apr 7, 2012 at 10:01
  • $\begingroup$ In fluid mechanics, Robert Bryant's $H$ is called a stream function for the velocity field $Z$. $\endgroup$
    – Bob Terrell
    Commented Apr 15, 2012 at 0:53
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Let $a=y-x+1$, then $\dot{x}=a$ and $\dot{a}=-rx^2+x-1$, so $\ddot{x}=-rx^2+x-1$.

So perhaps it is possible to accidentally bump into a system of the form $\ddot{x}=f(x)$.

How to tell this from your integral: naively trying to put it into a simpler form by completing the square seemed to work pretty well.

How to tell this from your differential equation: Compute $\ddot{x}$ and $\ddot{y}$ and see if you can write them in terms of only $x$ or only $y$.

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  • $\begingroup$ Thanks for this answer. It still looks a bit surprising to me, but at least it gives some mechanism... $\endgroup$
    – Anthony Quas
    Commented Apr 5, 2012 at 19:26
  • $\begingroup$ A bit more generally, the system $\dot{x} = a x + b y + c$, $\dot{y} = f(x)+d y$ will be of this form if and only if $d = -a$. $\endgroup$
    – Robert Israel
    Commented Apr 5, 2012 at 19:46

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