Conservative differential equations "in the wild" - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:19:18Z http://mathoverflow.net/feeds/question/93239 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93239/conservative-differential-equations-in-the-wild Conservative differential equations "in the wild" Anthony Quas 2012-04-05T18:59:18Z 2012-04-05T19:35:31Z <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&amp;=y-x+1\cr \dot y&amp;=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#93241 Answer by Will Sawin for Conservative differential equations "in the wild" Will Sawin 2012-04-05T19:14:13Z 2012-04-05T19:35:31Z <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#93244 Answer by Michael Renardy for Conservative differential equations "in the wild" Michael Renardy 2012-04-05T19:30:53Z 2012-04-05T19:30:53Z <p>Divergence free vector fields in the plane are associated with Hamiltonian systems. Your vector field is easily seen to be divergence free.</p>