 Solution

Assume the vector field and multiplier are continuously differentiable, as required for Green's theorem and uniqueness of trajectories. A nonconstant periodic trajectory of a planar autonomous system is a simple closed orbit: uniqueness prevents crossings, and one traversal can be taken over its least period. Let $\Omega$ be its bounded interior. Along the orbit $dx=fdt$, $dy=gdt$, soGreen's theorem identifies this with ${\iint }_{\Omega }[{\partial }_x(\varphi f)+{\partial }_y(\varphi g)]dxdy$, which is strictly positive by hypothesis, a contradiction. Thus there is no nonconstant periodic orbit. This standard periodic-orbit criterion does not by itself exclude stationary solutions.

For the first concrete system choose $\varphi =-1$. ThenThere are also no equilibria: $f=0$ requires $y=1$, but then $g=-e^{-x}-x^2<0$. Consequently it has no periodic solution, including constant ones.

For the second system the origin is outside the domain of the printed vector field. For $r>0$,Thus for an admissible initial point with $\rho =\sqrt{a^2+b^2}>0$,where $a=\rho \cos \alpha$, $b=\rho \sin \alpha$. These formulas are valid while $r>0$, in particular for every $t\ge 0$. Trajectories rotate counterclockwise, spiral outward when $0<r<1$, and inward when $r>1$, approaching the unit circle. On $r=1$ they give $x(t)=\cos (\alpha +t)$, $y(t)=\sin (\alpha +t)$, a periodic orbit with least period $2\pi$. It is an attracting limit cycle. There is no contradiction with the preceding criterion: this vector field is singular at the origin and has the exhibited closed orbit.