Talk:Bendixson–Dulac theorem

I have previously seen this stated as requiring it have the same sign across the region, rather than not being equal to zero. Which is correct?
 * Well, the proof linked to in the article only works if it has the same sign almost everywhere. As stated here it is absolutely false. Counterexample: take f = -y, g = x, R = x+y then (Rf)_x + (Rg)_y = - y + x, which is \neq 0 except for a set of measure 0 ( x = y), but of course x = cos(t), y = sin(t) is a nice periodic orbit. Micheal Hardy is essentially correct, it would hold if it is the same (non-zero) sign everywhere, except for a set of measure 0 (it could even be of a different sign, but only on a measure zero set. Continuity would prohibit this of course...) Mathjj (talk) 14:00, 10 May 2010 (UTC)

It would also be nice to include a section on the Bendixson-Dulac theorem in multiple dimensions —Preceding unsigned comment added by Quantum7 (talk • contribs)


 * I suspect a continuity assumption should be there. A continuous real-valued function on a connected region would have the same sign everywhere in the region if it is never 0.  Then there's the question of having the same sign is satisfied if it's &ge;0 everywhere, as opposed to being satisfied only if the inequality is strict. Michael Hardy (talk) 00:10, 29 November 2009 (UTC)