# proof of Bendixson’s negative criterion

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

$\int {\int}_{D}\nabla \cdot \text{\mathbf{f}}dxdy$ | $=$ | $\int {\int}_{D}\frac{\partial \text{\mathbf{X}}}{\partial x}}+{\displaystyle \frac{\partial \text{\mathbf{Y}}}{\partial y}}\mathrm{d}x\mathrm{d}y$ | ||

$=$ | ${\oint}_{\mathrm{\Gamma}}}(\text{\mathbf{X}}\mathrm{d}y-\text{\mathbf{Y}}\mathrm{d}x)$ | |||

$=$ | ${\int}_{0}^{T}}(\text{\mathbf{X}}\dot{y}-\text{\mathbf{Y}}\dot{x})dt$ | |||

$=$ | ${\int}_{0}^{T}}(\text{\mathbf{X}\mathbf{Y}}-\text{\mathbf{Y}\mathbf{X}})dt$ | |||

$=$ | $0$ |

Since $\nabla \cdot \text{\mathbf{f}}$ 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 |