Question 8.4: Rayleigh’s equation has a limit cycle Show that Rayleigh’s e...

Rayleigh’s equation arose in his theory of the bowing of a violin string. In the context of particle oscillations however, it corresponds to a simple harmonic oscillator with a strange damping term. When |\dot{x}|>1, we have ordinary (positive) damping and the motion decays. However, when |\dot{x}|<1 , we have negative damping and the motion grows. The possibility arises then of a periodic motion which is positively damped on some parts of its cycle and negatively damped on others. Somewhat surprisingly, this actually exists.
Rayleigh’s equation is equivalent to the autonomous system of ODEs

\begin{aligned}&\dot{x}=v \\&\dot{v}=-x-\epsilon v\left(v^{2}-1\right)\end{aligned}                    (8.20)

for which the only equilibrium position is at x = v = 0. It follows that, if there is a periodic solution, then it must enclose the origin. At first, we proceed as in the first example. In polar form, the equations (8.20) become

\begin{aligned}&\dot{r}=-\epsilon r \sin ^{2} \theta\left(r^{2} \sin ^{2} \theta-1\right) \\&\dot{\theta}=-1-\epsilon \sin ^{2} \theta\left(r^{2} \sin ^{2} \theta-1\right)\end{aligned}                      (8.21)

Let r = c be a circle with centre at the origin and radius less than unity. Then \dot{r}>0 everywhere on r = c except at the two points x = ±c, v = 0, where it is zero. Hence, except for these two points, we can deduce that a phase point that starts on the circle r = c enters the domain r > c. Fortunately, these exceptional points can be disregarded. It does not matter if there are a finite number of points on r = c where the phase paths go the ‘wrong’ way, since this provides only a finite number of escape routes! The circle r = c thus provides a suitable inner boundary C _{1} of the domain D.

Sadly, one cannot simply take a large circle to be the outer boundary of D since \dot{r} has the wrong sign on those segments of the circle that lie in the strip −1 <v< 1.
This allows any number of phase paths to escape and so invalidates our argument.
However, this does not prevent us from choosing a boundary of a different shape. A suitable outer boundary for D is the contour C _{2} shown in Figure 8.6. This contour is made up from four segments. The first segment AB is part of an actual phase path of the system which starts at A(−a, 1) and continues as far as B(b, 1). The form of this phase path can be deduced from equations (8.21). When v(= r sin θ) > 1, \dot{r}<0 \text { and } \dot{\theta}<-1 , so that the phase point moves clockwise around the origin with r decreasing. In particular, B must be closer to the origin than A so that b < a, as shown. Similarly, the segment A′ B′ is part of a second actual phase path that begins at A′ (a, −1). Because of the symmetry of the equations (8.20) under the transformation x → −x, v → −v, this segment is just the reflection of the segment AB in the origin; the point B′ is therefore (−b, −1). The contour is closed by inserting the straight line segments B A′ and B′ A.

We will now show that, when C _{2} is made sufficiently large, it is a suitable outer boundary for our domain D. Consider first the segment AB. Since this is a phase path, no other phase path may cross it (in either direction); the same applies to the segment A′ B′ . Now consider the straight segment BA′ . Because a > b, the outward unit normal n shown in Figure 8.6 makes a positive acute angle α with the axis Ox.
Now the ‘phase plane velocity’ of a phase point is

\dot{x} i +\dot{v} j =v i -\left(\epsilon v\left(v^{2}-1\right)+x\right) j

and the component of this ‘velocity’ in the n-direction is therefore

for (x, v) on B A′ . We wish to say that this expression is negative so that phase points that begin on B A′ enter the domain D. This is true if the contour C _{2} is made large enough. If we let a tend to infinity, then b also tends to infinity and α tends to π/2.
It follows that, whatever the value of the parameter ε, we can make b sin α>(1 + ε) by taking a large enough. A similar argument applies to the segment B′ A. Thus the contour C _{2} is a suitable outer boundary for the domain D. It follows that any phase path that starts in the domain D enclosed by C _{1} and C _{2} can never leave. Since D is a bounded domain with no equilibrium points within it or on its boundaries, it follows from Poincaré–Bendixson that any such path must either be a simple closed loop or tend to a limit cycle. In either case, Rayleigh’s equation must have a periodic solution lying in D.

We can say more. Phase paths that begin on either of the straight segments of the outer boundary C _{2} enter D and can never leave. These phase paths cannot close themselves (that would mean leaving D) and so can only tend to a limit cycle. It follows that Rayleigh’s equation must have (at least one) limit cycle lying in the domain D. [There is in fact only one.]