Limit Cycles and Planar Limit-Set Theorems, differential equations lesson

A frictionless spring repeats its motion at whatever amplitude it is given, so its phase plane is filled with closed orbits. Other oscillating systems settle onto one preferred oscillation from a wide range of starts. What singles out that one orbit, and how can its existence be proved when the equations cannot be solved? This lesson describes such isolated closed orbits, the limit cycles, and two planar tools with their hypotheses, the Poincaré-Bendixson theorem, which proves that a closed orbit exists, and the Bendixson-Dulac criterion, which rules closed orbits out. Throughout, $t$ is time and a dot means a derivative with respect to $t$ .

Key terms

Periodic orbit

Definition Consider a planar autonomous system

\begin{gathered} \dot x = P(x, y), \\[0.4em] \dot y = Q(x, y), \end{gathered}

with state $X = (x, y)^{T}$ , written $\dot X = F(X)$ with $F = (P, Q)^{T}$ continuously differentiable on an open set $D \subseteq \mathbb{R}^2$ . A periodic solution is a nonconstant solution with $X(t + T) = X(t)$ for all real $t$ and some $T > 0$ , and the smallest such $T$ is its period. Its orbit, the set of points the solution visits, is a periodic orbit, also called a closed orbit. It is a simple closed curve in $D$ that contains no equilibrium, because a solution through an equilibrium is constant. By uniqueness, a nonconstant solution that returns to a point it has visited before is periodic.

In plain words A periodic orbit is a loop in the phase plane that the state goes around forever, taking the same time for every lap.

Example The main example of this lesson is the system

\begin{aligned} \dot x &= x(1 - x^2 - y^2) - y, \\[0.4em] \dot y &= y(1 - x^2 - y^2) + x. \end{aligned}

For $x = \cos t$ and $y = \sin t$ the factor $1 - x^2 - y^2$ is $0$ , so the right sides reduce to $-\sin t$ and $\cos t$ , which are exactly the derivatives of $\cos t$ and $\sin t$ . This is a periodic solution of period $2\pi$ , and its orbit is the unit circle. Without that factor, the system $\dot x = -y$ , $\dot y = x$ has the solutions $(c\cos(t + \phi), c\sin(t + \phi))$ with $c > 0$ , so every circle centered at the origin is a periodic orbit.

Limit cycle

Definition A limit cycle is a periodic orbit $\Gamma$ that is isolated, meaning that some open set containing $\Gamma$ , such as a thin ring around it, contains no other periodic orbit. It is attracting if every solution starting close enough to $\Gamma$ has a distance to $\Gamma$ that tends to $0$ as $t \to \infty$ , repelling if the same happens as $t \to -\infty$ , and semistable if it attracts on one side and repels on the other. Attraction is measured by the distance to the curve as a set. Different starts can approach the curve at different phases, so attraction does not mean that all nearby solutions fall into step with one chosen periodic solution.

In plain words A limit cycle is a loop with no other loops right next to it. An attracting one pulls nearby states onto itself, so the oscillation ends up with the same size and shape from many different starts.

Example For the main system, use polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ . The rotation terms cancel in $x\dot x + y\dot y$ , and the remaining terms give

\begin{aligned} x\dot x + y\dot y &= r^2(1 - r^2), \\[0.4em] x\dot y - y\dot x &= r^2. \end{aligned}

Since $r\dot r = x\dot x + y\dot y$ and $r^2\dot\theta = x\dot y - y\dot x$ , for $r > 0$ this gives $\dot r = r(1 - r^2)$ and $\dot\theta = 1$ , and the origin is the only equilibrium. The radius increases for $0 < r < 1$ and decreases for $r > 1$ , and the derivative of $r - r^3$ at $r = 1$ is $1 - 3 = -2 < 0$ . Every start with $r_0 > 0$ has, for $t \ge 0$ ,

r(t) = \frac{r_0}{\sqrt{r_0^2 + (1 - r_0^2)e^{-2t}}},

which tends to $1$ . For $r_0 \ne 1$ this radius is strictly monotone, so there is no other periodic orbit, and the unit circle is an attracting limit cycle. For $r_0 = \frac12$ the formula reads $r(t) = (1 + 3e^{-2t})^{-1/2}$ . The circles of $\dot x = -y$ , $\dot y = x$ are periodic orbits but not limit cycles, since none of them is isolated.

Trapping region

Definition A trapping region for $\dot X = F(X)$ is a closed and bounded set $K \subseteq D$ such that every solution starting in $K$ stays in $K$ for all $t \ge 0$ . Such a solution exists for all $t \ge 0$ , because a solution that stays in a closed and bounded part of $D$ can neither become unbounded nor reach the edge of $D$ . A common way to find one is to bound $K$ by curves on which the vector field points strictly into $K$ .

In plain words A trapping region is a fenced part of the phase plane where every gate swings inward. Whatever starts inside stays inside forever.

Example For the main system take the closed ring $\frac12 \le r \le \frac32$ . Since $\dot r = r(1 - r^2)$ ,

\begin{gathered} \dot r\,\big\vert_{r = 1/2} = \tfrac38 > 0, \\[0.4em] \dot r\,\big\vert_{r = 3/2} = -\tfrac{15}{8} < 0, \end{gathered}

so the flow crosses the inner circle outward and the outer circle inward, into the ring in both cases. Because $\dot r$ depends on $r$ alone, a radius that starts in $[\frac12, \frac32]$ stays there, and the ring is a trapping region. It contains no equilibrium, since the only equilibrium is the origin. The closed disk $r \le \frac32$ is a trapping region too, but it contains the origin.

Omega-limit set

Definition Let $X(t)$ be a solution that exists for all $t \ge 0$ , with $X(0) = X_0$ . Its omega-limit set $\omega(X_0)$ is the set of points $p$ for which there are times $t_n \to \infty$ with $X(t_n) \to p$ . If the forward orbit stays in a closed and bounded set $K \subseteq D$ , then $\omega(X_0)$ is nonempty, closed, bounded, connected and contained in $K$ . It consists of whole orbits, and the distance from $X(t)$ to $\omega(X_0)$ tends to $0$ as $t \to \infty$ .

In plain words The omega-limit set is where a solution ends up in the long run, the set of points it keeps coming back close to as time grows. The solution need not ever reach them.

Example A solution that tends to an equilibrium has that single point as its omega-limit set, and a periodic solution has its own orbit. For the main system with $r_0 > 0$ the radius tends to $1$ while $\theta(t) = \theta_0 + t$ keeps turning. For any angle $\phi$ , the times $t_n = \phi - \theta_0 + 2\pi n$ give $X(t_n) \to (\cos\phi, \sin\phi)$ as $n \to \infty$ , so the omega-limit set is the whole unit circle, although a start with $r_0 \ne 1$ never lies on it. The start at the origin has the omega-limit set $\{(0, 0)\}$ .

Poincaré-Bendixson theorem

Definition Let $F$ be continuously differentiable on an open set $D \subseteq \mathbb{R}^2$ , and let $X(t)$ be a solution of $\dot X = F(X)$ that stays in a closed and bounded set $K \subseteq D$ for all $t \ge 0$ . If $\omega(X_0)$ contains no equilibrium, then $\omega(X_0)$ is a periodic orbit. Either $X(t)$ is itself periodic, or it approaches that orbit as $t \to \infty$ . In particular, a trapping region that contains no equilibrium contains at least one periodic orbit. The theorem does not say that the orbit is isolated, attracting or unique. The plane is essential. In three dimensions a bounded solution can have an omega-limit set that is neither an equilibrium nor a periodic orbit, as happens for the Lorenz system.

In plain words In the plane, a solution trapped in a bounded region, with no rest point among its limit points, has only one long-run behavior available, which is to cycle. The reason is that solutions cannot cross each other, and in the plane that leaves no room for anything more complicated.

Example The ring $\frac12 \le r \le \frac32$ is a trapping region of the main system with no equilibrium, so the theorem guarantees a periodic orbit inside it, and the unit circle is that orbit. That it is isolated and attracting came from the radial equation, not from the theorem. The same ring is a trapping region for $\dot x = -y$ , $\dot y = x$ , again without equilibria, and there the theorem is satisfied by a whole family of circles, none of which is a limit cycle.

Bendixson-Dulac criterion

Definition Let $U \subseteq D$ be a simply connected open set, so that $U$ has no holes, and let $B$ be a continuously differentiable real function on $U$ . If

\nabla \cdot (BF) = \frac{\partial (BP)}{\partial x} + \frac{\partial (BQ)}{\partial y}

is positive at every point of $U$ , or negative at every point of $U$ , then no periodic orbit of $\dot X = F(X)$ lies entirely in $U$ . With $B = 1$ this is Bendixson’s criterion. The reason is Green’s theorem. A periodic orbit $\Gamma$ in $U$ would enclose a region $R$ , which lies in $U$ because $U$ has no holes, and

\begin{gathered} \iint_R \nabla \cdot (BF)\,dA \\[0.4em] = \oint_\Gamma B\,(P\,dy - Q\,dx) = 0, \end{gathered}

since along $\Gamma$ we have $dx = P\,dt$ and $dy = Q\,dt$ . A continuous function of one strict sign cannot have integral $0$ over $R$ .

In plain words A closed orbit is tangent to the field everywhere, so nothing flows across it, and the divergence then has to add up to zero over the region inside. A divergence of one sign everywhere makes that impossible. A suitable multiplier can produce one sign when the field itself does not, and a failed test proves nothing either way.

Example For the damped oscillator $\dot x = y$ , $\dot y = -x - y$ with $B = 1$ , the divergence is $0 + (-1) = -1$ throughout the plane, so there is no periodic orbit at all, and its spirals never close up. For the main system

\begin{aligned} \nabla \cdot F &= (1 - 3x^2 - y^2) \\[0.4em] &\quad + (1 - x^2 - 3y^2) \\[0.4em] &= 2 - 4r^2, \end{aligned}

which changes sign at $r = 1/\sqrt{2}$ , so the test with $B = 1$ is inconclusive in the whole plane. That is not evidence for a cycle, which came from the radial equation. On the open disk $r < 1/\sqrt{2}$ , which has no holes, the divergence is positive, so no periodic orbit lies in that disk, in agreement with the unit circle being the only one. The criterion cannot be applied on the ring $\frac12 < r < \frac32$ , which has a hole.

Check yourself

The system $\dot x = xh - y$ , $\dot y = yh + x$ with $h = (1 - x^2 - y^2)(4 - x^2 - y^2)$ has the polar form $\dot r = r(1 - r^2)(4 - r^2)$ , $\dot\theta = 1$ . Find its periodic orbits and decide for each whether it is an attracting, repelling or semistable limit cycle.

Show the answer

The radius solves the scalar equation $\dot r = g(r)$ with $g(r) = r(1 - r^2)(4 - r^2)$ , and a nonconstant solution of such an equation is strictly monotone. So a periodic orbit must have a constant radius $c > 0$ with $g(c) = 0$ , which leaves the circles $r = 1$ and $r = 2$ . Each is traversed in time $2\pi$ , since $\dot\theta = 1$ . The sign of $g$ is positive for $0 < r < 1$ , negative for $1 < r < 2$ and positive for $r > 2$ . So the circle $r = 1$ attracts from both sides and is an attracting limit cycle, and the circle $r = 2$ repels on both sides and is a repelling limit cycle. Both are isolated, because nearby radii move monotonically. The derivative test agrees. Expanding gives $g(r) = 4r - 5r^3 + r^5$ , so $g'(r) = 4 - 15r^2 + 5r^4$ , with $g'(1) = -6 < 0$ and $g'(2) = 24 > 0$ .

A classmate notes that the closed disk $r \le \frac32$ is a trapping region for the main system and concludes from the Poincaré-Bendixson theorem that every solution in it approaches a periodic orbit. What is wrong, and what can a trapping region that contains an equilibrium prove about cycles on its own?

Show the answer

The disk contains the equilibrium at the origin. The solution that starts there stays there, and its omega-limit set is the equilibrium itself, so the hypothesis that the omega-limit set contains no equilibrium fails and the theorem says nothing about that solution. For the other starts in the disk the conclusion happens to be true, but it needs another argument, such as the radial equation or the theorem applied on a ring that leaves out the origin. On its own such a trapping region proves nothing about cycles. For $\dot x = -x$ , $\dot y = -y$ the unit disk is a trapping region, every solution tends to the origin, and there is no periodic orbit at all.

For the competition model $\dot x = x(3 - x - 2y)$ , $\dot y = y(2 - x - y)$ , show that the test with $B = 1$ is inconclusive on the open first quadrant $x > 0$ , $y > 0$ . Then use $B = 1/(xy)$ to show that no periodic orbit passes through a point of the open quadrant.

Show the answer

With $B = 1$ the divergence is $(3 - 2x - 2y) + (2 - x - 2y) = 5 - 3x - 4y$ , which is positive near the origin and negative far from it, so the test gives no conclusion. With $B = 1/(xy)$ ,

\begin{gathered} BP = \frac{3 - x - 2y}{y}, \\[0.4em] BQ = \frac{2 - x - y}{x}, \\[0.4em] \nabla \cdot (BF) = -\frac{1}{y} - \frac{1}{x} < 0. \end{gathered}

The open quadrant is convex, so it has no holes, and $B$ is continuously differentiable there, so no periodic orbit lies entirely in it. The axes are made of whole orbits, because $\dot x = 0$ when $x = 0$ and $\dot y = 0$ when $y = 0$ . An orbit through a point of the open quadrant therefore never reaches an axis and stays in the open quadrant, so by the criterion it cannot be periodic.

Limit Cycles and Planar Limit-Set Theorems

Key terms

Periodic orbit

Limit cycle

Trapping region

Omega-limit set

Poincaré-Bendixson theorem

Bendixson-Dulac criterion

Check yourself

How this lesson connects

Builds on

Used later in