# Trapping region of Van der Pol system

Van der Pol's system of differential equations is given by $$\dot{x}=y \\ \dot{y}=-(\mu+x)y$$ where $\mu\geq0$ is a constant. I want to show that any disk centred at $(0,0)$ is a trapping region, i.e. a simply-connected compact subset $R$ of $\mathbb{R}^2$ such that for all $(x_0,y_0)\in R$ the corresponding solution $(x(t), y(t))\in R$ for all $t\geq0$.

A point on the boundary of the trapping region obeys $r^2=x^2+y^2=H(x,y)$, so to show that it is a trapping region, it is enough to show that $$\nabla H\cdot(\dot{x}, \dot{y})\leq0$$ That is, I need to show that $$2x\dot{x}+2y\dot{y}\leq0$$ I have $$2x\dot{x}+2y\dot{y}=2xy+2y(-(\mu+x)y)=2y(x(1-y)-\mu y)$$ But I am not sure if I am doing this right, and if I am, how to proceed.

The Van der Pol system actually has the form \begin{align} \dot x&=y\\ \dot y&=\mu(1-x^2)y-x \end{align}

Try your approach for this system.

Edit: after @Did's comment. As Did show in comments your system also has a trapping region that includes $(0,0)$. For exact formulas see the comments.

• The region $2y\geqslant-(x+\mu)^2$ if $x\geqslant-\mu$, $y\geqslant0$ if $x\leqslant-\mu$, is a trapping region. If $\mu\ne0$, this contains a neighborhood of $(0,0)$ hence OP's system does have some trapping regions around $(0,0)$.
– Did
Apr 12, 2013 at 7:49
• @Did A trapping region is by definition bounded. Your example is not. Apr 12, 2013 at 12:37
• And it is quite easy to deduce a bounded trapping region from it, say $0\lt(x+\mu)^2+2y\lt2\mu^2$, $x^2\lt\mu^2$, hence the sentence "Yours does not have a trapping region" in your answer is wrong.
– Did
Apr 12, 2013 at 13:19
• @Did. I edited the post. Apr 12, 2013 at 13:38