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No. Consider two uncoupled oscillators with nonzero amplitudes:The orbit is bounded. If it had a period $T>0$, then ${\omega }_{1}T$ and ${\omega }_{2}T$ would both be integer multiples of $2\pi$, forcing ${\omega }_1/{\omega }_2\in \mathbb{Q}$. Choosing an irrational frequency ratio gives an incommensurate harmonic-oscillator orbit, which is quasiperiodic on an invariant torus and not periodic.

For $a=b=0$, take the Logarithmic first integral of the Lotka-Volterra equationsAlong a trajectory,Thus $V$ is a first integral. The scalar function $h(s)=s-\log s-1$ is nonnegative for $s>0$, vanishes only at $s=1$, and tends to infinity as $s\downarrow 0$ or $s\to \infty$. Moreover,is positive definite, so $V$ is strictly convex. It follows that $(1,1)$ is its unique critical point and every level set $V=c>0$ is a compact smooth simple closed curve in the positive quadrant. The vector field is nonzero on such a level curve, so uniqueness of solutions makes each connected level curve one periodic orbit. The zero level consists of the fixed point $(1,1)$. Hence every trajectory in $\Lambda$ is either periodic or the fixed point $(1,1)$.

The Bendixson-Dulac criterion says the following. Let $D\subseteq {\mathbb{R}}^2$ be a simply connected domain, let $F=(f,g)$ be continuously differentiable, and suppose there is a continuously differentiable function $B$ such thathas one sign throughout $D$ and does not vanish identically on any open subset. Then $\dot{z}=F(z)$ has no periodic orbit contained in $D$.

Indeed, if a periodic orbit $\Gamma$ existed, simple connectivity would put its interior $U$ inside $D$. The vector field $F$, and hence $BF$, is tangent to $\Gamma$, so its outward flux is zero. The divergence theorem would givecontradicting the sign hypothesis.

The Poincare-Bendixson theorem states that a nonempty compact omega-limit set of a continuously differentiable planar flow that contains no equilibrium point is a periodic orbit. More generally, when the omega-limit set contains only finitely many equilibria, it is either an equilibrium, a periodic orbit, or a union of equilibria and connecting trajectories.

On the two vertical sides of the square, the $x$ component of the vector field satisfiesOn the horizontal sides,Thus the vector field never points outward through the boundary, so the closed square is a positively invariant set and trajectories cannot leave $S$.

At a fixed point, $\dot{y}=0$ gives $x=-y^3$. Substitution into $\dot{x}=0$ givesFor $0<y<1$, both $y^3(2-y^6)$ and $\sin (2y)$ are positive; for $-1<y<0$, both are negative. Hence $y=0$, and then $x=0$, is the unique equilibrium point in $S$.

This alone does not prove that $S$ lies in the domain of stability. A compact trapping region containing a unique equilibrium can also contain a periodic orbit or another nontrivial omega-limit set.

Dulac theorem states the following. Let $D$ be a simply connected region for the planar system $\dot{z}=f(z)$, and suppose there is a continuously differentiable function $B:D\to \mathbb{R}$ for whichhas one strict sign throughout $D$. Then there is no periodic orbit contained in $D$.

To prove it, suppose a periodic orbit $\Gamma$ existed and let $A$ be its interior. Simple connectedness ensures $A\subset D$. By the planar divergence theorem,The vector field $f$ is tangent to its trajectory $\Gamma$, so its scalar product with the outward normal $n$ vanishes. The right-hand side is therefore zero. The left-hand side cannot be zero because its integrand has one strict sign, a contradiction.

The Poincare-Bendixson theorem states that if a forward trajectory of a smooth planar system remains in a compact set and its nonempty omega-limit set contains no equilibrium point, then that omega-limit set is a periodic orbit.

At an equilibrium point of a dynamical system, the second equation givesThe choice $x=0$ is impossible because then $\dot{x}=r>0$. Hence $y=s/x$, and substitution into the first equation givesThus the unique fixed point is

Choosewhich satisfies $0<\alpha <r$. LetWe check the direction of the vector field on its four edges.

On $x=\alpha$,On $y=0$,On $y=s/\alpha$, where $x\ge \alpha$,Finally, throughout the system,On the sloping edge $x+y=r+s/\alpha$, the bound $y\le s/\alpha$ implies $x\ge r$, and hence $\dot{x}+\dot{y}\le 0$. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact trapping region and then remain there. This is the Brusselator trapping region.

The Jacobian matrix at the unique fixed point isIts determinant and trace areWhen $s-1>r^2$, the trace is positive, so linear stability of a planar equilibrium shows that the fixed point is a repeller. It lies in the interior of $R$.

Choose a trajectory in $R$ other than the equilibrium. Compactness of $R$ gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the Poincare-Bendixson theorem gives a periodic orbit. HenceThis is the Brusselator periodic-orbit criterion.

A simply connected region containing no equilibrium, or containing equilibria whose total Poincare index is not $+1$, cannot contain a periodic orbit that encloses exactly those equilibria.