1

I'd like to show that there is a periodic orbit for the system $$ x' =20-x-\frac{4xy}{1+x^{2}}, y'=3x\left(1-\frac{y}{1+x^{2}}\right)$$ in the upper right quadrant. To do this, I would like to use the Poincare-Bendixson theorem.

The only fixed point occurs at $(4,17)$ and it is repelling. I was able to find an outer rectangle which is positively invariant and would contain the inner periodic orbit. It contains the fixed point, and so as usual we need to find an inner contour where trajectories that start on it always leave in order to use PB.

My question is, how I can find an inner contour to form a positively invariant set that does not contain the fixed point and has this property?

The usual Lyapunov function $(x-4)^2 + (y-17)^2$ does not work, since the derivative is not always positive.

It was suggested to me that I use the linearized system to obtain an elliptical contour which will work, but I am not sure how to do this at all.

Any help with this question would be appreciated.

QuantumINF0
  • 41
  • You could try a Lyaponuov function with elliptical level sets, like $V(x,y) = a(x-4)^2 + (y-17)^2$ and find a value of $a$ that works. You could add more parameters to account for ellipses rotated at different angles like here – whpowell96 Nov 25 '24 at 04:20

1 Answers1

1

In principle, the fact that the equilibrium is repelling (as shown by linearization) is already enough to tell you that there exists such a curve. But if you really want to find the curve explicitly, you can use the Lyapunov equation to compute a $V$ that works.

The Jacobian matrix of the system at the equilibrium point $(4,17)$ is $\tfrac{1}{17} A$, where $$ A = \begin{pmatrix} 43 & -16 \\ 96 & -12 \end{pmatrix} . $$ Now pick some positive definite symmetric $2 \times 2$-matrix $Q$, and solve the Lyapunov equation $$ A^T P + P A = Q $$ for the unknown symmetric $2 \times 2$-matrix $P$. The numbers may become slightly horrible, depending on your choice of $Q$, but I played around with Mathematica a bit and found that $$ Q = \begin{pmatrix} 4 & 1 \\ 1 & 5 \end{pmatrix} $$ happens to give the reasonably simple solution $$ P = \frac{1}{744} \begin{pmatrix} 624 & -264 \\ -264 & 197 \end{pmatrix} , $$ which is positive definite. Shift the coordinates by letting $(h,k)=(x,y)-(4,17)$, so that the equilibrium is at $(h,k)=(0,0)$, and take $$ V(h,k) = 624 h^2 - 528 hk + 197 k^2 , $$ as the matrix $P$ tells you. Then the Maclaurin expansion of $\dot V$ will be a positive constant times $4 h^2 + 2hk + 5 k^2$ (the positive definite quadratic form corresponding to $Q$) plus higher-order terms, and hence $\dot V > 0$ in some punctured neighbourhood of the equilibrium. So an ellipse $V(h,k)=C$ with $C>0$ small enough will be the curve that you seek. (Of course, if you want to know how small $C$ needs to be, then you must bound the higher-order terms somehow.)

Hans Lundmark
  • 55,038