Question 8.3: Proving existence of a limit cycle Prove that the autonomous...
Proving existence of a limit cycle
Prove that the autonomous system of ODEs
\begin{aligned}&\dot{x}=x-y-\left(x^{2}+y^{2}\right) x, \\&\dot{y}=x+y-\left(x^{2}+y^{2}\right) y,\end{aligned}
has a limit cycle.
This system clearly has an equilibrium point at the origin x = y = 0, and a little algebra shows that there are no others. Although we have not proved this result, it is true that any periodic solution (simple closed loop) in the phase plane must have an equilibrium point lying inside it. In the present case, it follows that, if a periodic solution exists, then it must enclose the origin. This suggests taking the domain D to be the annular region between two circles centred on the origin.
It is convenient to express the system of equations in polar coordinates r, θ. The transformed equations are (see Problem 8.5)
\dot{r}=\frac{x_{1} \dot{x}_{1}+x_{2} \dot{x}_{2}}{r}, \quad \dot{\theta}=\frac{x_{1} \dot{x}_{2}-x_{2} \dot{x}_{1}}{r^{2}},
where x_{1}=r \cos \theta \text { and } x_{2}=r \sin \theta. In the present case, the polar equations take the simple form
\dot{r}=r\left(1-r^{2}\right), \quad \dot{\theta}=1 .
These equations can actually be solved explicitly, but, in order to illustrate the method, we will make no use of this fact. Let D be the annular domain a < r < b, where 0 < a < 1 and b > 1. On the circle r = b, \dot{r}=b\left(1-b^{2}\right)<0 . Thus a phase point that starts anywhere on the outer boundary r = b enters the domain D. Similarly, on the circle r = a, \dot{r}=a\left(1-a^{2}\right)>0 and so a phase point that starts anywhere on the inner boundary r = a also enters the domain D. It follows that any phase path that starts in the annular domain D 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, the system must have a periodic solution lying in the annulus a < r < b.
We can say more. Phase paths that begin on either boundary of D 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 the system must have (at least one) limit cycle lying in the domain D. [The explicit solution shows that the circle r = 1 is a limit cycle and that there are no other periodic solutions.]