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The alpha-limit set of a trajectory consists of the points approached along sequences of times tending to negative infinity. It is the omega-limit set for the time-reversed flow.

Let ${\varphi }_t(x)$ denote the flow through $x$. Its alpha-limit set and omega-limit set areEquivalently, they contain the limits of ${\varphi }_{t_j}(x)$ along sequences $t_j\to -\infty$ and $t_j\to +\infty$, respectively.

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.

The Jacobian matrix at the origin isIts eigenvalues areThus, when $k>3$, the origin is a hyperbolic repelling spiral. By part (ii), every forward trajectory is eventually trapped in a compact disc. For $k<10$ the hypothesis says that the origin is the only equilibrium; choose any nonstationary trajectory sufficiently close to it. Its omega-limit set is nonempty and compact, and it cannot contain the repelling origin. The Poincare-Bendixson theorem then says that this omega-limit set is a periodic orbit. Combining the restrictions gives

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.

Put $y=\dot{x}$. The first-order system isAt an equilibrium point of a dynamical system, $y=0$ and $x(1-x^2{)}^2=0$, so the three fixed points are

DefineThen $U^{\prime }(x)=2x(1-x^2{)}^2=-F(x)$ andMoreover,so $E$ is positive definite at the origin. It is therefore a damped mechanical energy as a Lyapunov function, and the First Lyapunov theorem proves that the origin is Lyapunov stable.

Since $U(\pm 1)=1/3$, choose a compact energy sublevel $\{E\le c\}$ with $0<c<1/3$. In this set, $\dot{E}=0$ means $y=0$. A trajectory can remain in $y=0$ only when $F(x)=0$, and the chosen sublevel excludes $x=\pm 1$. Thus the largest invariant subset of $\{\dot{E}=0\}$ is the origin. The LaSalle invariance principle proves that the origin is asymptotically stable.

Finally, $E$ is radially unbounded because its leading terms are $x^6/3+y^2/2$. Every forward trajectory is therefore bounded. LaSalle's principle puts its omega-limit set insideAn omega-limit set of a bounded continuous trajectory is nonempty and connected. A connected subset of this three-point set is a singleton, so every trajectory has precisely one of the three fixed points as its omega-limit set.