Conservative differential equations "in the wild" - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T23:19:18Zhttp://mathoverflow.net/feeds/question/93239http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/93239/conservative-differential-equations-in-the-wildConservative differential equations "in the wild"Anthony Quas2012-04-05T18:59:18Z2012-04-05T19:35:31Z
<p>Dear MO world,</p>
<p>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!</p>
<p>The system was
$$
\eqalign{\dot x&=y-x+1\cr \dot y&=y-rx^2.}
$$</p>
<p>The phase portrait (with $r=0.15$) looks like this: <img src="http://www.math.uvic.ca/faculty/aquas/q2portrait.jpg"></p>
<p>Some computation with mathematica reveals that $2rx^3+3y^2+6y-6xy$ is a first integral. The question is <i>why?</i> 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 $\ddot x=f(x)$).</p>
<p>So: should I go and buy a lottery ticket, or is there some reason that this I shouldn't be surprised by this?</p>
http://mathoverflow.net/questions/93239/conservative-differential-equations-in-the-wild/93241#93241Answer by Will Sawin for Conservative differential equations "in the wild"Will Sawin2012-04-05T19:14:13Z2012-04-05T19:35:31Z<p>Let $a=y-x+1$, then $\dot{x}=a$ and $\dot{a}=-rx^2+x-1$, so $\ddot{x}=-rx^2+x-1$.</p>
<p>So perhaps it is possible to accidentally bump into a system of the form $\ddot{x}=f(x)$.</p>
<p>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.</p>
<p>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 $y$.</p>
http://mathoverflow.net/questions/93239/conservative-differential-equations-in-the-wild/93244#93244Answer by Michael Renardy for Conservative differential equations "in the wild"Michael Renardy2012-04-05T19:30:53Z2012-04-05T19:30:53Z<p>Divergence free vector fields in the plane are associated with Hamiltonian systems. Your vector field is easily seen to be divergence free.</p>