proof of Bendixson’s negative criterion

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

$\displaystyle\int\!\!\!\int_{D}\nabla\cdot\textbf{f}\>\mathrm{d}x\,\mathrm{d}y$	$\displaystyle=$	$\displaystyle\int\!\!\!\int_{D}\frac{\partial\textbf{X}}{\partial x}+\frac{% \partial\textbf{Y}}{\partial y}\>\mathrm{d}x\,\mathrm{d}y$
	$\displaystyle=$	$\displaystyle\oint_{\Gamma}(\textbf{X}\>\mathrm{d}y-\textbf{Y}\>\mathrm{d}x)$
	$\displaystyle=$	$\displaystyle\int_{0}^{T}(\textbf{X}\dot{y}-\textbf{Y}\dot{x})\>\mathrm{d}t$
	$\displaystyle=$	$\displaystyle\int_{0}^{T}(\textbf{XY}-\textbf{YX})\>\mathrm{d}t$
	$\displaystyle=$	$\displaystyle 0$

Since $\nabla\cdot\textbf{f}$ is not identically zero by hypothesis and is of one sign, the double integral on the left must be non zero and of that sign. This leads to a contradiction 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