Circular equation of motion

Question

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

Answer 1

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.