Bendixson-Dulac theorem

If $D$ is simply connected and a $C^1$ function $B$ makes $\nabla \cdot (BF)$ have one sign and not vanish identically on any open subset of $D$, then the planar system $\dot{x}=F(x)$ has no periodic orbit lying in $D$.

A multiplier $B(x,y)$ whose rescaled planar vector field has divergence of one strict sign can exclude periodic orbits in a simply connected region. For a system $\dot{x}=xF$, $\dot{y}=yG$ in the positive quadrant, $B=1/(xy)$ is often effective.

For $\dot{x}=2x-x^2-y$, $\dot{y}=y-kx$ with $k>2$, take the positive Dulac function in the title. More generally, $\psi =e^{a(x+y)}$ givesThe equal coefficients in the exponential cancel the $y$ term. Setting $a=-2/(k-2)$ also cancels the linear $x$ term, leaving $\psi (3+2x^2/(k-2))>0$. The Bendixson-Dulac theorem excludes periodic orbits in the open positive quadrant. The axes are not invariant: the vector field points out of the quadrant at every nonzero axis point. A closed orbit staying in the nonnegative quadrant therefore cannot touch an axis, and cannot pass through the equilibrium at the origin by uniqueness. This extends the exclusion to the closed quadrant. Choosing an exponential multiplier turns a differential search for a useful Dulac function into cancellation of polynomial coefficients.