# Circular equation of motion

I'm reading an article which describes a generation of artificial AKG signal. The article states that they use the following equation of motion:

$\left \lbrace \begin{array} {ccc} \dot x &=& \alpha x - \omega y \\ \dot y &=& \alpha y + \omega x\end{array}\right .$

where $\alpha = 1- \sqrt{x^2 + y^2}$ and $\omega$ is the angular velocity.

I have figured out the easy part where I use the $\theta$ and $r(=1)$ coordinates thus:

$\left \lbrace \begin{array} {ccc} \dot x &=& \frac{d}{dt} \cos (\theta) = -\sin (\theta) \frac{d}{dt} = -\omega y \\ \dot y &=& \frac{d}{dt} sin (\theta) = \cos (\theta) \frac{d}{dt} = \omega x\end {array}\right .$

Which has made me assume that $r = r(t)$ and $\langle r\rangle = 1$. I have tried to solve this simple deferential equation:

$\frac{dr}{dt} = 1-r$

but the solution I have found was:

$r(t) = 1-c\text{e}^{t}$

Which has many problems, units to say the least.

It is clear to me that the units problem is originated from the definition of $\alpha$ but there is no reference to that in the article.

I was hoping somebody of you has encountered such a problem or is handling such approach.

Edit I have forgotten to mentioned the very important fact that they are using a unit circle. Also, my main issue is with the $\alpha$ part which, on the unit circle should be 0...

Edit A link to the article

-
I don't understand a few things: what do you mean when you say "$r(=1)$"? How about $\langle r\rangle=1$? Where do you get the differential equation ${dr\over dt}=1-r$? – Ted Bunn Jul 7 '11 at 17:28
After reading your added comment (Edit), I'm more confused than before! If you're assuming $r=1$, then what do you mean by saying that you assume $r=r(t)$? With $r=1$, we have $\alpha=0$, and $(x,y)=(\cos\omega t,\sin\omega t)$. What's the problem? – Ted Bunn Jul 7 '11 at 17:46
Can you provide a link to the article? That might help. – Ted Bunn Jul 7 '11 at 17:46
Thanks for the link. Now a followup question: Why do you say that "they are using a unit circle." Is it because of the comment in the paper that the unit circle is a limit cycle of this system? That just means that the solution (sometimes) tends to the unit circle as $t\to\infty$, not that it's always there. Fig. 2, for instance, shows a solution that's not always on the unit circle. – Ted Bunn Jul 8 '11 at 15:26

As far as units are concerned, the initial equations only make sense if everything's dimensionless. $x,y$ must be dimensionless for $\alpha=1-\sqrt{x^2+y^2}$ to make sense. Then $\alpha x$ is dimensionless. The first equation therefore implies that $t$ and $\omega$ are dimensionless.
Anyway, we can go ahead and solve this system of equations. Start with your initial equations, and make the substitutions $x=r\cos\theta$ and $y=r\sin\theta$. All of $x,y,r,\theta$ are functions of $t$, so that, e.g., $\dot x=\dot r\cos\theta-r\dot\theta\sin\theta$. If you express everything in terms of $r,\theta$ and simplify a bit, the two equations reduce to $$\dot r=r-r^2,$$ $$\dot\theta=\omega.$$ The solution for $\theta$ is just $\theta=\theta_0+\omega t$, of course. The solution for $r$ is $$r(t)={1\over 1-Ce^{-t}},$$ where $C$ is chosen to match your initial conditions.
I have forgotten to mention that everything is on the unit circle which force me to assume that $\alpha = 0$... – Yotam Jul 7 '11 at 17:38
@Yotam: The case where you start on the unit circle is given by $C = 0$ in Ted's equation above. In this case $r=1$ for all time. I agree that in this case the $\alpha$ terms vanish. I can't really say why they included them if they care about $r = 1$. – BebopButUnsteady Jul 7 '11 at 18:06