Based on this question:Poincaré-Bendixon show periodic solutions.

Show that the system $x^{'}=x-y-x^{3}$,$y^{'}=x+y-y^{3}$ has a unique periodic orbit on annulus $A:=\{(x,y): 1\le x^2+y^2\le 2\}$ and this periodic solution is asymptotic stable.

I consider function $V(x,y)=(x^2+y^2)/2$ and show that $\dot V(x,y)>0$ on the circle $x^2+y^2=1$ and $\dot V\le 0$ on $x^2+y^2=2$. So $A$ is positively invariant. By Poincare-Bendixson theorem, there is at least one periodic orbit in $A$.

Question: But how to apply one theorem to show that is unique asymptotic stable?

I have the following theorem:

Let $p(t)$ be a $T$-periodic orbit of our system. If $\int_0^T div(f(p(t))dt<0$, then $p(t)$ is orbitally asymptotic stable.

I am confused how to find our $p(t)$?

Use the polar coordinate transform $$ % \begin{align} % x &= r \cos \theta \\ % y &= r \sin \theta \\ % \end{align} % $$ which implies $$ r^{2} = x^{2} + y^{2} \tag{2} $$ Compute the derivative with respect to time for $(2)$ and use the definitions in $(1)$. This leads to $$ \dot{r} = r - r^{3} \left( \cos^{4} \theta + \sin^{4} \theta \right) = r \left( 1 - \frac{1}{4} \left( 3 + \cos 4 \theta \right) r^{2} \right) \tag{3} $$


2 Answers 2


Consider a periodic orbit $p(t) = (x(t),y(t))$. By the Jordan curve theorem, $p$ splits the plane in two connected components, one of which (call it $D$) is bounded. Either $D$ is a subset of the annulus $A$ or $D$ contains the ball $B$ of radius 1 around the origin (think of a 'winding number' of the orbit being zero or nonzero that determines which case we are in). We will show that the first case cannot happen.

Define a vector field $F$ by $$ F(x,y) = \begin{pmatrix} -(x+y-y^3) \\ x-y-x^3 \end{pmatrix} = \begin{pmatrix} F_1(x,y) \\ F_2(x,y) \end{pmatrix}. $$ Note that $F(p(t)) \cdot \dot{p}(t) = -y'x'+x'y' = 0$. Stokes' theorem tells us $$ 0 = \int_0^T F(p(t)) \cdot \dot{p}(t) dt = \iint_D \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) dxdy = \iint_D ( 2 - 3 (x^2+y^2) ) dxdy. $$ The integrand $2 - 3 (x^2+y^2) =: g(x,y)$ is strictly negative in the annulus $A$, so this integral being zero implies that $D$ cannot be a subset of $A$.

Now suppose there were two distinct periodic orbits $p_1$ and $p_2$, which bound $D_1$ and $D_2$ respectively. Since both $D_1$ and $D_2$ contain the ball $B$, they are not disjoint. At the same time, the boundaries of $D_1$ and $D_2$ cannot intersect because distinct orbits cannot intersect. Thus, $D_1$ is a subset of $D_2$ or vice versa. But $$ \iint_{D_2 \backslash D_1} g(x,y) dxdy = \iint_{D_2} g(x,y) dxdy - \iint_{D_1} g(x,y) dxdy = 0-0 = 0 $$ and $D_2 \backslash D_1$ is a subset of $A$ on which $g$ is strictly negative, so $D_2 \backslash D_1$ has Lebesgue measure zero. But two periodic orbits cannot bound a set of measure zero between them (more precisely, $D_2 \backslash \bar{D}_1$ is nonempty and open, hence, has positive Lebesgue measure). This derives a contradiction to the existence of two distinct periodic orbits.

I do not recognise the theorem you quote about asymptotic stability of the orbit, but it adds up nicely as follows. I assume the $f$ in that theorem is the vector field generating the system, that is $f = (F_2,-F_1)$. Then $\mathrm{div}(f) = g$ is the integrand from before, which is strictly negative in the entire annulus $A$, so we can deduce that $\int_0^T \mathrm{div}(f)(p(t)) dt$ is strictly negative without explicitly knowing the orbit $p$.


You need to follow the procedure detailed in the first answer to this question:

First order system in polar coordinates : finding the trap region and showing it has a limit cycle

In particular, you can say that

$$0.5 < \cos^4(x) + \sin^4(x) <1$$

And from that you see that

  • $\dot{r}<0$ for all $r>2$;
  • $\dot{r}>0$ for all $r<1$;
  • $\begingroup$ This is not an answer. What you state was in the question from the start. The question is about the uniqueness of the limit cycle. There could be multiple concentric ones in the trapping region. $\endgroup$ Dec 4, 2023 at 21:25
  • 1
    $\begingroup$ Uh you are right, I think that the other answer above provides a good explanation. I thought that the problem was the existence. $\endgroup$
    – Ouden
    Dec 5, 2023 at 23:39

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