Consider the coupled system $$\dot{x}=-x+y, \,\,\, \dot{y}=-x-y.$$ Working with polar coordinates $r$ and $\theta$ such that $$x=r\cos\theta, \,\,\, y=r\sin\theta,$$ Obtain uncoupled differential equations for $r$ and $\theta$.

I did some research online and found out how to convert a differential equation to polar coordinates. I used the relationship $$r^2 = x^2+y^2$$ Then I differentiated this equation, and plugged in our $\dot{x}$ and $\dot{y}$ values given.

I got $\dot{r}=-r.$ Which I'm nearly certain is correct. Then to find $\dot{\theta}$ I used $\tan \theta = \dfrac{r\sin\theta}{r\cos\theta} =\dfrac{y}{x},$ differentiate it and I found $\dot{\theta} = -1$.

Solve the two differential equations from (i) to obtain $r(t)$ and $\theta(t)$

This seems ok, I just integrated my $\dot{r}$ and my $\dot{\theta}$ equations. I got $$r(t) = -\dfrac{r^2}{2}\,\,\, \text{and} \,\,\, \theta(t)= -\theta$$

Using the results from (ii) sketch the trajectory and identify the type and stability of the equilibrium point $q=(0,0)$.

This is the part that I'm writing the question about. I just want make sure I can understand this right. Our $\theta$ is negative does that mean it will be spiralling anti-clockwise. Also $r$ is negative so does that mean it will be decaying towards the equilibrium? If that's the case wouldn't we have a stable attractive node around the equilibrium?

Thanks very much.

  • 1
    $\begingroup$ Integration of $\dot r$, $\dot\theta$ should be with respect to $t$: $\dot r=-r$ $\Rightarrow$ $r(t)=e^{-t}r_0$ etc $\endgroup$ – A.Γ. May 7 '17 at 10:36
  • $\begingroup$ Ah thanks very much. Yeah I solved both separable differential equations. $\theta= -t$. Was my idea about the phase plane correct. I haven't actually solved any trajectories for polar coordinates. I know you can solve other systems with eigenvalues and eigenvectors and plot their trajectories. Not sure how it works in polar coordinates. $r(t)$ is also decaying. $\endgroup$ – Patrick Moloney May 7 '17 at 10:56
  • 1
    $\begingroup$ $\dot\theta<0$, hence, rotating clockwise (the negative angle direction), $\dot r<0$, hence, $r$ decreasing. The origin is the stable equilibrium, correct. $\endgroup$ – A.Γ. May 7 '17 at 11:03
  • $\begingroup$ Of course $\theta>0$ is taken as clockwise. I understand. I appreciate the support. $\endgroup$ – Patrick Moloney May 7 '17 at 11:06
  • $\begingroup$ Just to summarize what's been done here, the solution is $r=r_0e^{i\theta_0}e^{-(1+i)\theta}$. The first term is a scaling parameter, the second term is a rotation parameter, and the last term shows that this is logarithmic spiral with a flair coefficient of $-1$ (shrinking) and rotating clockwise. $\endgroup$ – Cye Waldman May 9 '17 at 20:48

Your solutions of the differential equations are both wrong. The independent variable is $t$. $\dot r=-r$ has the well-known solution $r(t)=r_0·e^{-t}$ and $\dot θ=-1$ trivially $θ(t)=θ_0-t$.

Yes, you are right that the spirals are clock-wise. As the radius gets smaller in time, the image of the curve is an anti-clockwise spiral. This more intuitive interpretation can belongs to the time-inversed problem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.