# 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 ProofOfBendixsonsNegativeCriterion 2013-03-22 13:31:07 2013-03-22 13:31:07 Daume (40) Daume (40) 5 Daume (40) Proof msc 34C25