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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.

A trajectory's limit set consists of its accumulation points along sequences of times tending to positive or negative infinity. The forward-time set is its omega-limit set and the backward-time set its alpha-limit set.

For Hopf thresholds in a two-pool calcium model, with the unique equilibrium parametrized by $u=\mu$, locate changes of stability by solving the two equationsIf there are two simple roots ${\mu }_{-}<{\mu }_{+}$ with positive trace between them, the equilibrium is a repeller throughout that interval, since its determinant stays positive. At a nondegenerate crossing a Hopf bifurcation is possible, with linear frequency $[-f_v/ϵ{]}^{1/2}$. A complete classification of criticality requires a nonlinear coefficient, not merely the trace calculation. At leading singular order the thresholds approach the folds determined by $f=0$ and $f_v-\gamma f_u=0$, equivalently $V^{\prime }=-1/\gamma$.

Periodic solutions are also supported globally in the positive quadrant. The axes point inward for positive input, $v^{\prime }\le (\beta -\delta v)/ϵ$ bounds $v$, and the $u$ equation then bounds $u$ by a linear restoring term plus a bounded source. A large rectangle can thus be chosen forward invariant. For an initial point distinct from the repelling unique equilibrium, its omega-limit set cannot contain that equilibrium or a boundary equilibrium. The Poincaré-Bendixson theorem supplies a periodic orbit. In the strong separation regime it follows the two attracting branches and jumps between them, a relaxation oscillation.

The interval requires the maximum of $f_v-\gamma f_u$ along the critical manifold to exceed $ϵ$; the listed scale hierarchy alone is not a substitute for that test for arbitrary remaining constants. As an explicit example, $\beta =0.1$, $\delta =10^{-5}$, $ϵ=5\times 10^{-4}$ and $\gamma =0.1$ give approximately ${\mu }_{-}=0.22618$ and ${\mu }_{+}=0.37350$. At $\mu =0.3$ the trace is positive, whereas at $0.03$ and $1$ it is negative. The figures use this example and label their singular-limit portions accordingly.

The fixed points are $(0,0)$ and $(\pm 1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, so it is a saddle with stable direction the y-axis. At either other fixed point, the characteristic polynomial is ${\lambda }^2+\lambda +2$, giving eigenvalues $(-1\pm i\sqrt{7})/2$. Thus both are stable foci. The nullclines are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.

A global argument determines the requested omega-limit sets, rather than guessing from the local phase portrait. For $x\ne 0$, use the quadratic-logarithmic Lyapunov function for a feedback systemIts sublevel sets are compact and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the omega-limit set. Every point of that set therefore has $y=1$. Invariance then requires $\dot{y}=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp ({\int }_0^t(1-y(s))ds)$ preserves its sign.

ConsequentlyOn $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The phase portrait below shows this stable axis, the nullclines and the spiral attraction in each half-plane.

Introduce the vector $x=(t,y,y^{\prime },\dots ,y^{(k-1)})\in {\mathbb{R}}^{k+1}$. Its autonomous ordinary differential equation iswhere the coordinate indices start at zero. Appending the clock removes the explicit time dependence.

For a vector field with local existence and uniqueness, its flow ${\varphi }_t(x)$ is the solution at elapsed time $t$ starting at $x$. The uniqueness theorem for ordinary differential equations gives ${\varphi }_s({\varphi }_t(x))={\varphi }_{s+t}(x)$ whenever both sides are defined. The orbit is $O(x)=\{{\varphi }_t(x):t\text{ lies in its maximal interval}\}$. For a forward-complete solution its omega-limit set isequivalently the set of limits along sequences $t_n\to \infty$. A homoclinic orbit is a nonconstant complete orbit approaching the same equilibrium as $t\to -\infty$ and $t\to \infty$. This last definition concerns a general autonomous system; the appended clock itself cannot approach an equilibrium.

A form of the Poincaré-Bendixson theorem states that a nonempty omega-limit set of a forward orbit of a $C^1$ planar autonomous vector field, contained in a compact region and containing no equilibrium, is a periodic orbit.

Set $\rho =x^2+y^2$. Direct calculation givesbecause $0\le x^2{y}^2\le {\rho }^2/4$. At the inner boundary $\rho =2/(2+k^2)$ the flow points into or along the annulus, and at the outer boundary $\rho =1$ it does likewise. Smoothness and these inequalities make the compact annulus forward invariant.

For nonzero $(x,y)$, write $x=\sqrt{\rho }\cos \theta$, $y=\sqrt{\rho }\sin \theta$. Thenthroughout the annulus. Hence it contains no equilibrium. Any orbit starting there remains compactly trapped, has a nonempty omega-limit set, and the Poincaré-Bendixson theorem supplies the required periodic orbit in the stated annulus. For $k=0$ the annulus reduces to $\rho =1$ itself, and there $\dot{\theta }=1$, giving the periodic orbit directly.

For a nonconstant invasion travelling wave, the relevant end states are $w(-\infty )=k$ and $w(+\infty )=0$. These fronts do not exist at every nonnegative speed: the phase plane calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv 0$ and $w\equiv k$ are, of course, nonnegative travelling waves for every speed, including zero.

Put $z=x-ct$ and $p=w^{\prime }$. The travelling-wave reduction of a reaction-diffusion system givesThe equilibrium points in the phase plane are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative eigenvalue. Thus it is a saddle equilibrium. Its descending unstable manifold has $w<k$, $p<0$ and locallyAt $(0,0)$ the eigenvalues areFor $c>2\sqrt{k}$ it is a stable node; for $0<c<2\sqrt{k}$ it is a stable spiral. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.

Suppose $c\ge 2\sqrt{k}$. Choose $a>0$ with $a(c-a)\ge k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular regionOn the horizontal side $p=0$, the vector field has $p^{\prime }=-w(k-w)\le 0$, pointing into $\mathcal{T}$. On the vertical side $w=k$, it has $w^{\prime }=p\le 0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an equilibrium point and cannot be crossed in finite time by a nonconstant trajectory, by uniqueness theorem for ordinary differential equations. This proves forward invariance of $\mathcal{T}$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.

The descending unstable manifold of $(k,0)$ initially enters $\mathcal{T}$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p^{\prime }<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell \in [0,k)$. To identify the limit, its omega-limit set lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w^{\prime }=0$; its second equation then forces $\ell (k-\ell )=0$. Since $\ell <k$, it follows that $\ell =0$, and the only possible omega-limit set is $(0,0)$. Backward in $z$, the unstable manifold tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing heteroclinic orbit and hence a nonnegative travelling wave for every $c\ge 2\sqrt{k}$.

For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu =\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero equilibrium point the system becomesFor a nonzero trajectory its polar angle satisfiesA trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integralis conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no heteroclinic orbit connects them. ConsequentlyThis is the phase-plane construction of a Fisher travelling front, including the critical speed. The statement for all $c\ge 0$ must distinguish these nonconstant fronts from the two constant travelling waves.

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 Poincaré-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 Poincaré-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 Poincaré-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.