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Section8.4Limit cycles

less than 1 lecture, discussed in §6.1 and §6.4 in [EP] , §9.7 in [BD]

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.

Example8.4.1

The Van der Pol oscillator 1 Named for the Dutch physicist Balthasar van der Pol (1889–1959). is the following equation

\begin{equation*} x''-\mu(1-x^2) x' + x = 0, \end{equation*}

where \(\mu\) is some positive constant. The Van der Pol oscillator originated with electrical circuits, but finds applications in diverse fields such as biology, seismology, and other physical sciences.

For simplicity, let us use \(\mu = 1\text{.}\) A phase diagram is given in the left hand plot in Figure 9.4.2. 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 \(x'(0) = 0.1\text{.}\) Notice how the solution quickly tends to a periodic solution.

Figure9.4.2The phase portrait (left) and a graph of a sample solution 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 \(\mu = 0\text{,}\) we get \(x''+x=0\text{,}\) which is a linear system with a center and all trajectories become circles.

A trajectory in the phase portrait that is a closed curve (a curve that is a loop) is called a closed trajectory. A limit cycle is a closed trajectory such that at least one other trajectory spirals into it (or spirals out of it). For example, the closed curve in the phase portrait for the Van der Pol equation is a limit cycle. 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 closed trajectory on an autonomous system, any solution that starts on it is periodic. 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 \(y(t) = y(t+P)\text{.}\)

Consider the system

\begin{equation} x' = f(x,y), \qquad y' = g(x,y) ,\label{nlin_gensys}\tag{2} \end{equation}

where the functions \(f\) and \(g\) have continuous derivatives in some region \(R\) in the plane.

The main point of the theorem is that if you find one solution that exists for all \(t\) large enough (that is, as \(t\) goes 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 or tends to a critical point. That is, in the long term, the behavior is very close to a periodic function. Note that a constant solution at a critical point is periodic (with any period). The theorem is more a qualitative statement rather than something to help us in computations. In practice it is hard to find analytic solutions and so hard to show rigorously that they exist for all time. But if we think the solution exists we numerically solve for a large time to approximate the limit cycle. Another caveat 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 (2) is defined on a simply connected region, that is, a region with no holes we can 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.

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 curves 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.

Example8.4.2

Let us look at \(x'=y+y^2e^x\text{,}\) \(y'=x\) in the entire plane (see Example 8.2.2). The entire plane is simply connected and so we can apply the theorem. We compute \(\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = y^2e^x+ 0\text{.}\) The function \(y^2e^x\) is always positive except on the line \(y=0\text{.}\) 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 \(R\text{.}\)

Example8.4.3

Let us look at a somewhat more complicated example. Take the system \(x'=-y-x^2\text{,}\) \(y'=-x+y^2\) (see Example 8.2.1). We compute \(\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = -2x + 2y=2(-x+y)\text{.}\) 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 \(-x+y \geq 0\text{.}\) Via the theorem, there is no closed trajectory in that set. Similarly, there is no closed trajectory in the set \(-x+y \leq 0\text{.}\) We cannot conclude (yet) that there is no closed trajectory in the entire plane. Perhaps half of it is in the set where \(-x+y \geq 0\) and the other half is in the set where \(-x+y \leq 0\text{.}\)

The key is to look at the line where \(-x+y=0\text{,}\) or \(x=y\text{.}\) On this line \(x' = -y-x^2 = -x-x^2\) and \(y' = -x+y^2 = -x+x^2\text{.}\) In particular, when \(x=y\) then \(x' \leq y'\text{.}\) That means that the arrows, the vectors \((x',y')\text{,}\) always point into the set where \(-x+y \geq 0\text{.}\) There is no way we can start in the set where \(-x+y \geq 0\) and go into the set where \(-x+y \leq 0\text{.}\) Once we are in the set where \(-x+y \geq 0\text{,}\) we stay there. So no closed trajectory can have points in both sets.

Subsection8.4.1Exercises

Exercise8.4.1

Show that the following systems have no closed trajectories.
a) \(x'=x^3+y,\quad y'=y^3+x^2\text{,}\)
b) \(x'=e^{x-y},\quad y'=e^{x+y}\text{,}\)
c) \(x'=x+3y^2-y^3,\quad y'=y^3+x^2\text{.}\)

Exercise8.4.2

Formulate a condition for a 2-by-2 linear system \({\vec{x}\,}' = A \vec{x}\) to not be a center using the Bendixson-Dulac theorem. That is, the theorem says something about certain elements of \(A\text{.}\)

Exercise8.4.3

Explain why the Bendixson-Dulac Theorem does not apply for any conservative system \(x''+h(x) = 0\text{.}\)

Exercise8.4.4

A system such as \(x'=x, y'=y\) has solutions that exist for all time \(t\text{,}\) yet there are no closed trajectories. Explain why the Poincarè-Bendixson Theorem does not apply.

Exercise8.4.5

Differential equations can also be given in different coordinate systems. Suppose we have the system \(r' = 1-r^2\text{,}\) \(\theta' = 1\) given in polar coordinates. Find all the closed trajectories and check if they are limit cycles and if so, if they are asymptotically stable or not.

Exercise8.4.101

Show that the following systems have no closed trajectories.
a) \(x'=x+y^2,\quad y'=y+x^2\text{,}\)        b) \(x'=-x\sin^2(y),\quad y'=e^x\text{,}\)        c) \(x'=xy,\quad y'=x+x^2\text{.}\)

Answer

Use Bendixson-Dulac Theorem. a) \(f_x+g_y = 1+1 > 0\text{,}\) so no closed trajectories. b) \(f_x+g_y = -\sin^2(y)+0 < 0\) for all \(x,y\) except the lines given by \(y=k\pi\) (where we get zero), so no closed trajectories. c) \(f_x+g_y = y + 0 > 0\) for all \(x,y\) except the line given by \(y=0\) (where we get zero), so no closed trajectories.

Exercise8.4.102

Suppose an autonomous system in the plane has a solution \(x=\cos(t)+e^{-t}\text{,}\) \(y=\sin(t)+e^{-t}\text{.}\) What can you say about the system (in particular about limit cycles and periodic solutions)?

Answer

Using Poincarè-Bendixson Theorem, the system has a limit cycle, which is the unit circle centered at the origin as \(x=\cos(t)+e^{-t}\text{,}\) \(y=\sin(t)+e^{-t}\) gets closer and closer to the unit circle. Thus we also have that \(x=\cos(t)\text{,}\) \(y=\sin(t)\) is the periodic solution.

Exercise8.4.103

Show that the limit cycle of the Van der Pol oscillator (for \(\mu > 0\)) must not lie completely in the set where \(-1 < x < 1\text{.}\) Compare with Figure 9.4.2.

Answer

\(f(x,y) = y\text{,}\) \(g(x,y) = \mu(1-x^2)y-x\text{.}\) So \(f_x+g_y = \mu(1-x^2)\text{.}\) The Bendixson-Dulac Theorem says there is no closed trajectory lying entirely in the set \(x^2 < 1\text{.}\)

Exercise8.4.104

Suppose we have the system \(r' = \sin(r)\text{,}\) \(\theta' = 1\) given in polar coordinates. Find all the closed trajectories.

Answer

The closed trajectories are those where \(\sin(r) = 0\text{,}\) therefore, all the circles with radius a multiple of \(\pi\) are closed trajectories.

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