# 8.4: Limit cycles

For nonlinear systems, trajectories do not simply need to approach or leave a single point. They may in fact approach a larger set, such as a circle or another closed curve.

Example \(\PageIndex{1}\): The Van der Pol oscillator

The Van der Pol oscillator is the following equation

\[x''-\mu(1-x^2) x' + x = 0,\]

where \(\mu\) is some positive constant. The Van der Pol oscillator comes up often in applications, for example in electrical circuits.

For simplicity, let us use \(\mu = 1\). A phase diagram is given in the left hand plot in Figure 8.10. Notice how the trajectories seem to very quickly settle on a closed curve. On the right hand plot we have the plot of a single solution for \(t=0\) to \(t=30\) with initial conditions \(x(0) = 0.1\) and \(x'(0) = 0.1\). Notice how the solution quickly tends to a periodic solution.

**Figure 8.10**: The phase portrait (left) and graphs of sample solutions of the Van der Pol oscillator.

The Van der Pol oscillator is an example of so-called *relaxation oscillation*. The word relaxation comes from the sudden jump (the very steep part of the solution). For larger \(\mu\) the steep part becomes even more pronounced, for small \(\mu\) the limit cycle looks more like a circle. In fact setting \(\mu = 0\), we get \(x''+x=0\), which is a linear system with a center and all trajectories become circles.

The closed curve in the phase portrait above is called a* limit cycle*. A limit cycle is a closed trajectory such that at least one other trajectory spirals into it (or spirals out of it). If all trajectories that start near the limit cycle spiral into it, the limit cycle is called *asymptotically stable*. The limit cycle in the Van der Pol oscillator is asymptotically stable.

Given a limit cycle on an autonomous system, any solution that starts on it is periodic. In fact, this is true for any trajectory that is a closed curve (a so-called *closed trajectory*). Such a curve is called a *periodic orbit. *More precisely, if \(\bigl(x(t),y(t)\bigr)\) is a solution such that for some \(t_0\) the point \(\bigl(x(t_0),y(t_0)\bigr)\) lies on a periodic orbit, then both \(x(t)\) and \(y(t)\) are periodic functions (with the same period). That is, there is some number \(P\) such that \(x(t) = x(t+P)\) and \(y(t) = y(t+P)\).

Consider the system

\[ x' = f(x,y), ~~~~~ y' = g(x,y) ,\]

where the functions \(f\) and \(g\) have continuous derivatives.

Theorem 8.4.1 (Poincarè-Bendixson). Suppose \(R\) is a closed bounded region (a region in the plane that includes its boundary and does not have points arbitrarily far from the origin). Suppose \(\bigl(x(t), y(t)\bigr)\) is a solution of (8.4.2) in \(R\) that exists for all \(t \geq t_0\). Then either the solution is a periodic function, or the solution spirals towards a periodic solution in \(R\).

The main point of the theorem is that if you find one solution that exists for all \(t\) large enough (that is, we can let \(t\) go to infinity) and stays within a bounded region, then you have found either a periodic orbit, or a solution that spirals towards a limit cycle. That is, in the long term, the behavior will be very close to a periodic function. We should take the theorem more as a qualitative statement rather than something to help us in computations. In practice it is hard to find solutions and therefore hard to show rigorously that they exist for all time. Another caveat to consider is that the theorem only works in two dimensions. In three dimensions and higher, there is simply too much room.

Let us next look when limit cycles (or periodic orbits) do not exist. We will assume the equation (8.4.2) is defined on a *simply connected region*, that is, a region with no holes we could go around. For example the entire plane is a simply connected region, and so is the inside of the unit disc. However, the entire plane minus a point is not a simply connected domain as it has a "hole" at the origin.

Theorem 8.4.2 (Bendixson-Dulac). Suppose \(f\) and \(g\) are defined in a simply connected region \(R\). If the expression

\[ \frac{\partial f}{\partial x} + \frac{\partial g}{\partial y}\]

is either always positive or always negative on \(R\) (except perhaps a small set such as on isolated points or curves) then the system (8.4.2) has no closed trajectory inside \(R\).

The theorem gives us a way of ruling out the existence of a closed trajectory, and hence a way of ruling out limit cycles. The exception about points or lines really means that we can allow the expression to be zero at a few points, or perhaps on a curve, but not on any larger set.

Example \(\PageIndex{2}\):

Let us look at \(x'=y+y^2e^x\), \(y'=x\) in the entire plane (see Example 8.2.2.) The entire plane is simply connected and so we can apply the theorem. We compute \(\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = y^2e^x+ 0\). The function \(y^2e^x\) is always positive except on the line \(y=0\). Therefore, via the theorem, the system has no closed trajectories.

In some books (or the internet) the theorem is not stated carefully and it concludes there are no periodic solutions. That is not quite right. The above example has two critical points and hence it has constant solutions, and constant functions are periodic. The conclusion of the theorem should be that there exist no trajectories that form closed curves. Another way to state the conclusion of the theorem would be to say that there exist no nonconstant periodic solutions that stay in \(R\).

Example \(\PageIndex{3}\):

Let us look at a somewhat more complicated example. Take the system \(x'=-y-x^2\), \(y'=-x+y^2\) (see Example 8.2.1). We compute \(\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = 2x + 2y\). This expression takes on both signs, so if we are talking about the whole plane we cannot simply apply the theorem. However, we could apply it on the set where \(x+y > 0\). Via the theorem, there is no closed trajectory in that set. Similarly, there is no closed trajectory in the set \(x+y < 0\). We cannot conclude (yet) that there is no closed trajectory in the entire plane. Perhaps half of it is in the set where \(x+y >0\) and the other half is in the set where \(x+y < 0\).

The key is to look at the set \(x+y=0\), or \(x=-y\). Let us make a substitution \(x=z\) and \(y=-z\) (so that \(x=-y\)). Both equations become \(z'=z-z^2\). So any solution of \(z'=z-z^2\), gives us a solution \(x(t)=z(t)\), \(y(t)=-z(t)\). In particular, any solution that starts out on the line \(x+y=0\), stays on the line \(x+y = 0\). In other words, there cannot be a closed trajectory that starts on the set where \(x+y > 0\) and goes through the set where \(x+y < 0\), as it would have to pass through \(x+y = 0\).

### Contributors

- Jiří Lebl (Oklahoma State University).These pages were supported by NSF grants DMS-0900885 and DMS-1362337.