Notes on Diffy Qs: Differential Equations for Engineers

Section8.4Limit cycles

Note: 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 8.10. Notice how the trajectories seem to very quickly settle on a closed curve. On the right-hand side is the plot of a single solution for $$t=0$$ to $$t=30$$ with initial conditions $$x(0) = 0.1$$ and $$x'(0) = 0.1\text{.}$$ The solution quickly tends to a periodic solution.

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
$$x' = f(x,y), \qquad y' = g(x,y) ,\tag{8.2}$$
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.
The theorem applies to all solutions in the Van der Pol oscillator. Solutions that start at any point except the origin $$(0,0)$$ tend to the periodic solution around the limit cycle, and the initial condition of $$(0,0)$$ gives the constant solution $$x=0\text{,}$$ $$y=0\text{.}$$

Example8.4.2.

Consider
\begin{equation*} x' = y + {(x^2+y^2-1)}^2 x, \qquad y' = -x + {(x^2+y^2-1)}^2 y. \end{equation*}
A vector field along with solutions with initial conditions $$(1.02,0)\text{,}$$ $$(0.9,0)\text{,}$$ and $$(0.1,0)$$ are drawn in Figure 8.11.

Notice that points on the unit circle (distance one from the origin) satisfy $$x^2+y^2-1=0\text{.}$$ And $$x(t) = \sin(t)\text{,}$$ $$y = \cos(t)$$ is a solution of the system. Therefore we have a closed trajectory. For points off the unit circle, the second term in $$x'$$ pushes the solution further away from the $$y$$-axis than the system $$x' = y\text{,}$$ $$y' = -x\text{,}$$ and $$y'$$ pushes the solution further away from the $$x$$-axis than the linear system $$x'=y\text{,}$$ $$y' = -x\text{.}$$ In other words for all other initial conditions the trajectory will spiral out.
This means that for initial conditions inside the unit circle, the solution spirals out towards the periodic solution on the unit circle, and for initial conditions outside the unit circle the solutions spiral off towards infinity. Therefore the unit circle is a limit cycle, but not an asymptotically stable one. The Poincaré–Bendixson Theorem applies to the initial points inside the unit circle, as those solutions stay bounded, but not to those outside, as those solutions go off to infinity.
A very similar analysis applies to the system
\begin{equation*} x' = y + {(x^2+y^2-1)} x, \qquad y' = -x + {(x^2+y^2-1)} y. \end{equation*}
We still obtain a closed trajectory on the unit circle, and points outside the unit circle spiral out to infinity, but now points inside the unit circle spiral towards the critical point at the origin. So this system does not have a limit cycle, even though it has a closed trajectory.
Due to the Picard theorem (Theorem 3.1.1) we find that no matter where we are in the plane we can always find a solution a little bit further in time, as long as $$f$$ and $$g$$ have continuous derivatives. So if we find a closed trajectory in an autonomous system, then for every initial point inside the closed trajectory, the solution will exist for all time and it will stay bounded (it will stay inside the closed trajectory). So the moment we found the solution above going around the unit circle, we knew that for every initial point inside the circle, the solution exists for all time and the Poincaré–Bendixson theorem applies.
Let us next look for conditions when limit cycles (or periodic orbits) do not exist. We assume the equation (8.2) 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 region 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.

Example8.4.3.

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 example above 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 $$R\text{.}$$

Example8.4.4.

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. For all we know, 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.

Example8.4.5.

Consider $$x' = y+(x^2+y^2-1)x\text{,}$$ $$y' = -x +(x^2+y^2-1)y\text{,}$$ and consider the region $$R$$ given by $$x^2+y^2 > \frac{1}{2}\text{.}$$ That is, $$R$$ is the region outside a circle of radius $$\frac{1}{\sqrt{2}}$$ centered at the origin. Then there is a closed trajectory in $$R\text{,}$$ namely $$x=\cos(t)\text{,}$$ $$y=\sin(t)\text{.}$$ Furthermore,
\begin{equation*} \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x} = 4x^2+4y^2-2 , \end{equation*}
which is always positive on $$R\text{.}$$ So what is going on? The Bendixson–Dulac theorem does not apply since the region $$R$$ is not simply connected—it has a hole, the circle we cut out!

ExercisesExercises

8.4.1.

Show that the following systems have no closed trajectories.
1. $$x'=x^3+y,\quad y'=y^3+x^2\text{,}$$
2. $$x'=e^{x-y},\quad y'=e^{x+y}\text{,}$$
3. $$x'=x+3y^2-y^3,\quad y'=y^3+x^2\text{.}$$

8.4.2.

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 $$A\text{.}$$

8.4.3.

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

8.4.4.

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 Poincaré–Bendixson Theorem does not apply.

8.4.5.

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.

8.4.101.

Show that the following systems have no closed trajectories.
1. $$x'=x+y^2,\quad y'=y+x^2\text{,}$$
2. $$x'=-x\sin^2(y),\quad y'=e^x\text{,}$$
3. $$x'=xy^2,\quad y'=x+x^2\text{.}$$
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) $$f_x+g_y = y^2 + 0 > 0$$ for all $$x,y$$ except the line given by $$y=0$$ (where we get zero), so no closed trajectories.

8.4.102.

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)?
Using Poincaré–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 $$x=\cos(t)\text{,}$$ $$y=\sin(t)$$ is the periodic solution.

8.4.103.

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 8.10.
$$f(x,y) = y\text{,}$$ $$g(x,y) = \mu(1-x^2)y-x\text{.}$$ 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 $$x^2 < 1\text{.}$$
Suppose we have the system $$r' = \sin(r)\text{,}$$ $$\theta' = 1$$ given in polar coordinates. Find all the closed trajectories.
The closed trajectories are those where $$\sin(r) = 0\text{,}$$ therefore, all the circles centered at the origin with radius that is a multiple of $$\pi$$ are closed trajectories.