Editing Bendixson–Dulac theorem (section)

==Proof==
Without loss of generality, let there exist a function <math> \varphi(x, y)</math> such that

:<math>\frac { \partial (\varphi f) }{ \partial x } +\frac { \partial (\varphi g) }{ \partial y } >0</math>

in simply connected region <math>R</math>. Let <math>C</math> be a closed trajectory of the plane autonomous system in <math>R</math>. Let <math>D</math> be the interior of <math>C</math>. Then by [[Green's theorem]],

: <math>
\begin{align}
& \iint_D \left( \frac { \partial (\varphi f) }{ \partial x } +\frac { \partial (\varphi g) }{ \partial y }  \right) \,dx\,dy 
=\iint_D \left( \frac { \partial (\varphi \dot { x }) }{ \partial x } +\frac { \partial (\varphi \dot { y }) }{ \partial y }  \right) \,dx\,dy \\[6pt]
= {} & \oint_C \varphi \left( -\dot { y } \,dx+\dot { x } \,dy\right)
=\oint_C \varphi \left( -\dot { y }\dot { x }+\dot { x }\dot { y }\right)\,dt=0
\end{align}
</math>

Because of the constant sign, the left-hand integral in the previous line must evaluate to a positive number. But on <math>C</math>, <math>dx=\dot { x } \,dt</math> and <math>dy=\dot { y } \,dt</math>, so the bottom integrand is in fact 0 everywhere and for this reason the right-hand integral evaluates to&nbsp;0. This is a contradiction, so there can be no such closed trajectory <math>C</math>.