proof of Bendixson’s negative criterion

Suppose that there exists a periodic solution called Γ which has a period of T and lies in E. Let the interior of Γ be denoted by D. Then by Green’s Theorem we can observe that

D𝐟dxdy = D𝐗x+𝐘ydxdy
= Γ(𝐗dy-𝐘dx)
= 0T(𝐗y˙-𝐘x˙)dt
= 0T(𝐗𝐘-𝐘𝐗)dt
= 0

Since 𝐟 is not identically zero by hypothesisMathworldPlanetmath and is of one sign, the double integral on the left must be non zero and of that sign. This leads to a contradictionMathworldPlanetmathPlanetmath since the right hand side is equal to zero. Therefore there does not exists a periodic solution in the simply connected region E.

Title proof of Bendixson’s negative criterion
Canonical name ProofOfBendixsonsNegativeCriterion
Date of creation 2013-03-22 13:31:07
Last modified on 2013-03-22 13:31:07
Owner Daume (40)
Last modified by Daume (40)
Numerical id 5
Author Daume (40)
Entry type Proof
Classification msc 34C25